Research ArticleQuasi-Jordan Banach Algebras
Reem K Alhefthi Akhlaq A Siddiqui and Fatmah B Jamjoom
Department of Mathematics College of Science King Saud University PO Box 2455-5 Riyadh 11451 Saudi Arabia
Correspondence should be addressed to Fatmah B Jamjoom fatmahjyahoocom
Received 16 November 2013 Accepted 14 January 2014 Published 25 March 2014
Academic Editor Marco Sabatini
Copyright copy 2014 Reem K Alhefthi et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
We initiate a study of quasi-Jordan normed algebras It is demonstrated that any quasi-Jordan Banach algebra with a norm 1 unitcan be given an equivalent norm making the algebra isometrically isomorphic to a closed right ideal of a unital split quasi-JordanBanach algebra the set of invertible elements may not be open the spectrum of any element is nonempty but it may be neitherbounded nor closed and hence not compact Some characterizations of the unbounded spectrum of an element in a split quasi-Jordan Banach algebra with certain examples are given in the end
1 Introduction
Looking back at the development ofmodernmathematics wesee that early formal study of algebrawasmostly commutativeand associative With an abstract study of functions andmatrices it became noncommutative but still associativethen introduction of nonassociative structures such as Liestructures due to Sophus Lie [1] and Jordan structures dueto Jordan [2] has led us to the mathematics which at presentis noncommutative as well as nonassociative
There is a strong relationship between Lie algebras andJordan algebras [3] Jordan structures have been extensivelystudied by a large number of mathematicians P Jordanvon Neumann E Wigner N Jacobson K McCrimmon RBraun M Koecher E Neher and O Loos to name but afew A vast literature containing many important results onJordan algebras has been developed (cf [3 4]) After themid-1960s people began studying Jordan structures fromthe point of view of functional analysis Interesting theoriesof Jordan Banach algebras 119869119861-algebras 119869119861lowast-algebras and119869119861
lowast-triples have been developed which closely resemblethat of 119862lowast-algebras and have found surprisingly importantapplications in a wide range of mathematical disciplinesincluding analysis physics and biology (cf [5ndash7])
A Jordan algebra is a nonassociative algebra J with theproduct 119909 ∘ 119910 satisfying 119909 ∘ 119910 = 119910 ∘ 119909 and the Jordan identity(119909
2
∘ 119910) ∘ 119909 = 1199092
∘ (119910 ∘ 119909) where 1199092
= 119909 ∘ 119909 Any associativealgebra 119860 becomes a Jordan algebra 119860
+ with the same linear
space structure and the Jordan product119909∘119910 = (12)(119909119910+119910119909)it becomes a Lie algebra under the skew-symmetric product[119909 119910] = 119909119910 minus 119910119909 so called the Lie bracket (cf [4]) For anyJordan algebra J there is a Lie algebra L(J) such that Jis a linear subspace of L(J) and the product of J can beexpressed in terms of the Lie bracket inL(J) Moreover theuniversal enveloping algebra of a Lie algebra has the structureof an associative algebra see the original work due to KantorKoecher and Tits appearing in [8ndash10]
Loday introduced a generalization of Lie algebras calledthe Leibniz algebras [11 12] and successfully demonstratedthat the relationship between Lie algebras and associativealgebras can be translated into an analogous relationshipbetween Leibniz algebras and the so-called dialgebras (cf[13]) a dialgebra over a field 119870 is a 119870-module 119863 equippedwith two bilinear associative maps ⊣ ⊢ 119863 times 119863 rarr 119863
satisfying 119909 ⊣ (119910 ⊢ 119911) = 119909 ⊣ (119910 ⊣ 119911) (119909 ⊢ 119910) ⊣ 119911 =
119909 ⊢ (119910 ⊣ 119911) and (119909 ⊣ 119910) ⊢119911 = (119909 ⊢119910) ⊢119911 The maps ⊣ and ⊢
are called the left product and the right product respectivelyAny dialgebra (119863 ⊣ ⊢) becomes a Leibniz algebra under theLeibniz bracket [119909 119910] = 119909 ⊣ 119910 minus 119910⊢119909 and the universalenveloping algebra of a Leibniz algebra has the structure of adialgebra for details see [12 13]
Recently in [14] Velasquez and Felipe introduced thenotion of quasi-Jordan algebras which have relation with theLeibniz algebras similar to the existing relationship betweenthe Jordan algebras and the Lie algebras [15] The quasi-Jordan algebras are a generalization of Jordan algebras where
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 690806 11 pageshttpdxdoiorg1011552014690806
2 Abstract and Applied Analysis
the commutative law is replaced by a quasicommutative iden-tity and a special form of the Jordan identity is retainedThesefacts indicate the significance of studying the quasi-Jordanalgebras within a few years time many mathematiciansincluding M K Kinyon M R Bremner L A Peresi J San-chez-Ortega and V Voronin have got their interests in thisnew area In [16] Felipe made an attempt to study dialge-bras from the functional analytic point of view
A Jordan Banach algebra is a real or complex Jordanalgebra with a complete norm sdot satisfying 119909119910 le 119909119910basics of Jordan Banach algebras may be seen in [7] In thispaper we initiate a study of the quasi-Jordan normed alge-bras The class of complete quasi-Jordan normed algebrascalled quasi-Jordan Banach algebras properly includes allJordan Banach algebras and hence all 119862lowast-algebras (cf [7])This study may provide a better mathematical foundationfor some important areas such as quantum mechanics Weare interested specially in extending as much as possiblethe theory of Jordan Banach algebras to the general settingof quasi-Jordan Banach algebras We investigate the notionsof invertibility and spectrum of elements in the setting ofunital quasi-JordanBanach algebras Among other results wedemonstrate that any quasi-Jordan Banach algebra I with anorm 1 unit can be given an equivalent norm that makes thealgebra I isometrically isomorphic to a norm closed rightideal of a unital split quasi-Jordan Banach algebra We showthat the set of invertible elements in a unital quasi-JordanBanach algebra generally is not open and that the spectrumofany element is nonempty but it may be neither bounded norclosed hence not compact Moreover if the spectrum of anelement in a complex unital split quasi-JordanBanach algebra(see below) is unbounded then it coincides with the wholecomplex plane and vice versa Some examples are also givenin the end
2 Quasi-Jordan Banach Algebras
Webegin by recalling some basics of the quasi-Jordan algebratheory from [14 17] A quasi-Jordan algebra is a vector spaceI over a field119870 of characteristic = 2 equipped with a bilinearmap ⊲ I times I rarr I called the quasi-Jordan productsatisfying 119909 ⊲ (119910 ⊲ 119911) = 119909 ⊲ (119911 ⊲ 119910) (right commutativity)and (119910 ⊲ 119909) ⊲ 119909
2
= (119910 ⊲ 1199092
) ⊲ 119909 (right Jordan identity)for all 119909 119910 119911 isin I Here 1199092 = 119909 ⊲ 119909 and 119909
119899
= 119909119899minus1
⊲ 119909
for 119899 ge 2 in the sequel we will see that 1199092 ⊲ 119909 may notcoincide with 119909 ⊲ 119909
2 Every quasi-Jordan algebra I includestwo important sets Iann
= the linear span of the elements119909 ⊲ 119910 minus 119910 ⊲ 119909 119908119894119905ℎ 119909 119910 isin I and 119885(I) = 119911 isin I 119909 ⊲
119911 = 0 forall119909 isin I respectively called the annihilator and thezero part of I It follows from the right commutativity thatIann
sube 119885(I) and that the quasi-Jordan algebra I is a Jordanalgebra if and only ifIann
= 0
Example 1 Let (119863 ⊣ ⊢) be a dialgebra over a field119870 of char-acteristic = 2 One can define another product⊲ 119863times119863 rarr 119863
by 119909 ⊲ 119910 = (12)(119909 ⊣ 119910 + 119910 ⊢ 119909) for all 119909 119910 isin 119863 whichsatisfies the identities 119909 ⊲ (119910 ⊲ 119911) = 119909 ⊲ (119911 ⊲ 119910) (119910 ⊲
119909) ⊲ 1199092
= (119910 ⊲ 1199092
) ⊲ 119909 and 1199092
⊲ (119909 ⊲ 119910) = 119909 ⊲
(1199092
⊲ 119910) however this quasi-Jordan product generally isnot commutativeTherefore (119863 ⊲) is a quasi-Jordan algebrawhich may not be a Jordan algebra (cf [14]) This quasi-Jordan algebra is denoted by 119863
+ called a plus quasi-Jordanalgebra Any quasi-Jordan algebra which is a homomorphicimage of a plus quasi-Jordan algebra is called special
We define a quasi-Jordan normed algebra as a quasi-Jor-dan algebra (I ⊲) over the field C of complex numbersendowed with a norm sdot satisfying 119909 ⊲ 119910 le 119909119910 for all119909 isin IThus the quasi-Jordan product ldquo⊲rdquo in a normed quasi-Jordan algebraI is continuous A quasi-Jordan normed alge-bra is called a quasi-Jordan Banach algebra if it is completeas a normed space
If there is an element 119890 in a quasi-Jordan algebra Isatisfying 119890 ⊲ 119909 = 119909 for all 119909 isin I called the left unit thenI becomes commutative and hence a Jordan algebra Due tothis fact we will consider only the right unit elements hence-forth unit in a quasi-Jordan algebra would mean a right unitAn element 119890 in a quasi-Jordan algebraI is called a unit if119909 ⊲
119890 = 119909 for all 119909 isin I A quasi-Jordan algebra may have many(right) units (cf [15 17]) If the dialgebra 119863 has a bar-unit 119890(ie 119909 ⊣ 119890 = 119909 = 119890⊢119909) then 119909 ⊲ 119890 = (12)(119909 ⊣ 119890 + 119890 ⊢ 119909) = 119909for all 119909 isin 119863 and hence 119890 is a unit of the plus quasi-Jordanalgebra119863
+For any quasi-Jordan algebraIwith unit 119890 we know from
[17] thatIann and119885(I) are two-sided ideals ofIIann= 119909 isin
I 119890 ⊲ 119909 = 0 = 119885(I) and 119880(I) = 119909 + 119890 119909 isin 119885(I)where 119880(I) denotes the set of all (right) units inI
One can always attach a (two-sided) unit to any Jordanalgebra by following the standard unitization process Thisunitization process no longer works for quasi-Jordan algebras(cf [17 pages 210-211]) Adding a unit to a quasi-Jordan alge-bra is yet an open problem In giving a partial solution to thisunitization problemVelasquez andFelipe [17] introduced thefollowing special class of quasi-Jordan algebras called splitquasi-Jordan algebras letI be a quasi-Jordan algebra and let119868 be an ideal inI such thatIann
sube 119868 sube 119885(I) We say thatI isa split quasi-Jordan algebra (more precisely I is split over 119868)if there exists a subalgebra 119869 ofI such thatI = 119869oplus119868 the directsum of 119869 and 119868 It is easily seen that such a subalgebra 119869 is aJordan algebra One can attach a unit to any split quasi-Jordanalgebra details of such a unitization process are given in [17]If the algebra I has a unit then Iann
= 119885(I) Thus a quasi-Jordan algebra I with unit is a split quasi-Jordan algebra ifand only if I = 119869 oplus 119885(I) for some subalgebra 119869 of I thealgebra 119869 is called the Jordan part of I In such a case eachelement 119909 isin I has a unique representation 119909 = 119909
119869+ 119909
119885with
119909119869isin 119869 and 119909
119885isin 119885(I) respectively called the Jordan part and
the zero part of 119909 Moreover ifI is split quasi-Jordan algebrawith unit 119890 then there exists a unique element 119890
119869isin 119869 which
acts as a unit of the quasi-Jordan algebra I and at the sametime a unit of the Jordan algebra 119869
Proposition 2 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan Banach algebra with unit 119890 isin 119869 Then the Jordan part119869 is a norm closed subalgebra of I and hence a unital JordanBanach algebra
Abstract and Applied Analysis 3
Proof From the above discussion it is clear that the Jordanpart 119869 is a unital Jordan normed algebra Next let 119909
119899 be any
fixed Cauchy sequence in 119869 Then the same 119909119899 is Cauchy
sequence also in the quasi-Jordan Banach algebraI since 119869 isa normed subspace ofI Hence 119909
119899rarr 119909 for some 119909 isin I
Now since each119909119899isin 119869 and since 119890 is the unit of the Jordan
algebra 119869 under the (restricted) product ldquo⊲rdquo we get 119890 ⊲ 119909119899=
119909119899for all 119899 So by the continuity of the product 119909
119899= 119890 ⊲
119909119899
rarr 119890 ⊲ 119909 Hence by the uniqueness of the limit of anyconvergent sequence in a normed space 119890 ⊲ 119909 = 119909 However119909 = 119890 ⊲ (119909
119869+ 119909
119885) = 119890 ⊲ 119909
119869+ 119890 ⊲ 119909
119885= 119890 ⊲ 119909
119869+ 0 = 119909
119869isin 119869
Thus the required result follows
We know from [17] that for any quasi-Jordan algebra II
1= (119909 119877
119910) 119909 119910 isin I equipped with the sum (119909 119877
119910) +
(119886 119877119887) = (119909 + 119886 119877
119910+119887) scalar multiplication 120582(119909 119877
119910) =
(120582119909 119877120582119910) and product (119909 119877
119910) ⊲ (119886 119877
119887) = (119909 ⊲ 119887 119877
119910⊲119887)
is a split quasi-Jordan algebra and the map 119909 997891rarr 120593(119909) =
(119909 119877119909) is an embedding of I into I
1 where 119877
119911stands for
the usual right multiplication operator on I It is easily seenthat 0 times 119877(I) is the Jordan part of I
1and the zero part
119885(I1) = I times 0 Moreover the embedding 120593 preserves the
units clearly (119886 119877119887) ⊲ (119890 119877
119890) = (119886 ⊲ 119890 119877
119887⊲119890) = (119886 119877
119887) for
all (119886 119877119887) isin I
1so that (119890 119877
119890) is a unit in I
1whenever 119890 is a
unit in I In fact (119909 119877119890) is a unit in I
1 for all 119909 isin I it may
be noted here that (0 119877119890) is the only unit in the Jordan part of
I1From [17] we also know that 119877(I) = 119877
119909 119909 isin I with
product ldquo∙rdquo defined by 119877119909∙ 119877
119910= 119877
119909⊲119910 for all 119909 119910 isin I is
a quasi-Jordan algebra MoreoverI1is the direct product of
the quasi-Jordan algebrasI and 119877(I)Indeed the split quasi-Jordan algebraI
1is a quasi-Jordan
Banach algebra with unit (119890 119877119890)wheneverI is a quasi-Jordan
Banach algebra with a norm 1 unit 119890 and that 120593(I) =
(119909 119877119909) 119909 isin I is a closed unital quasi-Jordan normed
subalgebra ofI1 To justify this claim we need the following
result
Proposition 3 Suppose I is a quasi-Jordan Banach algebrawith a norm 1 unit 119890Then the algebra119877(I) as above is a quasi-Jordan Banach algebra with unit 119877
119890of norm 1
Proof Clearly each 119877119909is a bounded linear operator with
119877119909 le 119909 Hence the usual operator norm is a norm on
the quasi-Jordan algebra 119877(I)with the quasi-Jordan product119877119909∙ 119877
119910= 119877
119909⊲119910 Further we observe that
10038171003817100381710038171003817119877119909∙ 119877
119910
10038171003817100381710038171003817=
10038171003817100381710038171003817119877119909⊲119910
10038171003817100381710038171003817
= sup119911isinI119911=1
10038171003817100381710038171003817119877119909⊲119910
(119911)10038171003817100381710038171003817
= sup119911isinI119911=1
1003817100381710038171003817119911 ⊲ (119909 ⊲ 119910)1003817100381710038171003817
= sup119911isinI119911=1
1003817100381710038171003817119911 ⊲ ((119890 ⊲ 119909) ⊲ 119910)1003817100381710038171003817
= sup119911isinI119911=1
10038171003817100381710038171003817119911 ⊲ (119877
119910(119877
119909(119890)))
10038171003817100381710038171003817
le sup119911isinI119911=1
11991110038171003817100381710038171003817119877119910(119877
119909(119890))
10038171003817100381710038171003817
=10038171003817100381710038171003817119877119910(119877
119909119890)10038171003817100381710038171003817le
1003817100381710038171003817119877119909
1003817100381710038171003817
10038171003817100381710038171003817119877119910
10038171003817100381710038171003817119890
=1003817100381710038171003817119877119909
1003817100381710038171003817
10038171003817100381710038171003817119877119910
10038171003817100381710038171003817
(1)
Alternately by exploiting the right commutativity of thequasi-Jordan product inI we get
10038171003817100381710038171003817119877119909∙ 119877
119910
10038171003817100381710038171003817= sup
119911isinI119911=1
1003817100381710038171003817119911 ⊲ (119909 ⊲ 119910)1003817100381710038171003817
= sup119911isinI119911=1
1003817100381710038171003817119911 ⊲ ((119909 ⊲ 119890) ⊲ (119910 ⊲ 119890))1003817100381710038171003817
= sup119911isinI119911=1
1003817100381710038171003817119911 ⊲ ((119909 ⊲ 119890) ⊲ (119890 ⊲ 119910))1003817100381710038171003817
= sup119911isinI119911=1
1003817100381710038171003817119911 ⊲ ((119890 ⊲ 119910) ⊲ (119909 ⊲ 119890))1003817100381710038171003817
= sup119911isinI119911=1
1003817100381710038171003817119911 ⊲ ((119890 ⊲ 119910) ⊲ (119890 ⊲ 119909))1003817100381710038171003817
= sup119911isinI119911=1
10038171003817100381710038171003817119911 ⊲ (119877
119910(119890) ⊲ 119877
119909(119890))
10038171003817100381710038171003817
le sup119911isinI119911=1
11991110038171003817100381710038171003817119877119910
10038171003817100381710038171003817
1003817100381710038171003817119877119909
1003817100381710038171003817 1198902
=1003817100381710038171003817119877119909
1003817100381710038171003817
10038171003817100381710038171003817119877119910
10038171003817100381710038171003817
(2)
Thus 119877(I) together with the operator norm is a quasi-Jordan normed algebra Moreover for any 119909 isin 119877(I) wehave 119877
119909∙ 119877
119890= 119877
119909⊲119890= 119877
119909and 1 = 119890 ge 119877
119890 =
sup0 = 119911isinI(119877119890
(119911)119911) ge 119877119890(119890)119890 = 1 that is 119877
119890is a
norm 1 (right) unit in 119877(I)Suppose 119877
119909119899
is any fixed Cauchy sequence in 119877(I)Then for any fixed 119886 isin I 119886 ⊲ 119909
119898minus 119886 ⊲ 119909
119899 = 119877
119909119898
(119886) minus
119877119909119899
(119886) = (119877119909119898
minus 119877119909119899
)(119886) le 119877119909119898
minus 119877119909119899
119886 rarr 0 as119898 119899 rarr infin and so 119886 ⊲ 119909
119899 is a Cauchy sequence in I
But I is complete Hence the sequence 119886 ⊲ 119909119899 for any
119886 isin I is convergent inI In particular the sequence 119890 ⊲ 119909119899
converges to some 119910 isin I Moreover the Cauchy sequence119877
119909119899
converges to 119877119910isin 119877(I) in the operator norm because
(119877119909119899
minus 119877119910)(119911) = 119911 ⊲ (119909
119899minus 119910) = 119911 ⊲ 119909
119899minus 119911 ⊲ 119910 = 119911 ⊲
(119909119899⊲ 119890) minus 119911 ⊲ 119910 = 119911 ⊲ (119890 ⊲ 119909
119899) minus 119911 ⊲ 119910 = 119911 ⊲ (119890 ⊲
119909119899minus 119910) le 119911119890 ⊲ 119909
119899minus 119910 rarr 0 as 119899 rarr infin for all 119911 isin I
Thus the quasi-Jordan normed algebra 119877(I) is complete
Now we show that the corresponding algebra I1is a
unital split quasi-Jordan Banach algebra
Proposition 4 Let I be a quasi-Jordan Banach algebra witha norm 1 unit 119890 and let I
1be as above Then I
1is the direct
product of the quasi-Jordan algebras I and 119877(I) MoreoverI
1equipped with the product norm (119909 119877
119910) = 119909 + 119877
119910
is a split quasi-Jordan Banach algebra with unit (119890 119877119890) and
120593(I) = (119909 119877119909) 119909 isin I is a closed right ideal in I
1in the
norm topology
4 Abstract and Applied Analysis
Proof For the first part see [17] Clearly (119909 119877119910) = 119909 +
119877119910 is a norm and it satisfies
10038171003817100381710038171003817(119909 119877
119910) ⊲ (119911 119877
119908)10038171003817100381710038171003817=
10038171003817100381710038171003817(119909 ⊲ 119908 119877
119910⊲119908)10038171003817100381710038171003817
=10038171003817100381710038171003817(119909 ⊲ 119908 119877
119910∙ 119877
119908)10038171003817100381710038171003817
=1003817100381710038171003817119877119908
(119909)1003817100381710038171003817 +
10038171003817100381710038171003817119877119910∙ 119877
119908
10038171003817100381710038171003817
le (119909 +10038171003817100381710038171003817119877119910
10038171003817100381710038171003817)1003817100381710038171003817119877119908
1003817100381710038171003817
le (119909 +10038171003817100381710038171003817119877119910
10038171003817100381710038171003817) (119911 +
1003817100381710038171003817119877119908
1003817100381710038171003817)
=10038171003817100381710038171003817(119909 119877
119910)10038171003817100381710038171003817
1003817100381710038171003817(119911 119877119908)1003817100381710038171003817
(3)
for all 119908 119909 119910 119911 isin I Keeping in view Proposition 3 wededuce that I
1being the product of complete spaces I and
119877(I) is complete in the product norm Thus I1is a split
quasi-Jordan Banach algebra with (right) unit (119890 119877119890)
Clearly 120593(I) is a subspace ofI1with (119909 119877
119909) ⊲ (119910 119877
119911) =
(119909 ⊲ 119911 119877119909⊲119911
) isin 120593(I) for all (119909 119877119909) isin 120593(I) (119910 119877
119911) isin I
1 so
that 120593(I) is a quasi-Jordan normed subalgebra with the unit(119890 119877
119890) included Further let (119909
119899 119877
119909119899
) be any fixed Cauchysequence in 120593(I) Then 119909
119898minus 119909
119899 le 119909
119898minus 119909
119899 + 119877
119909119898
minus
119877119909119899
= (119909119898
minus 119909119899 119877
119909119898
minus 119877119909119899
) = (119909119898 119877
119909119898
) minus (119909119899 119877
119909119899
) rarr
0 as 119898 119899 rarr infin so that 119909119899 is a Cauchy sequence in (the
complete space) I and hence it converges to some 119909 isin INow by using the fact 119877
119911le 119911 for all 119911 isin I (since
119890 is a norm 1 unit in I see above) we get the convergenceof arbitrarily fixed Cauchy sequence (119909
119899 119877
119909119899
) to (119909 119877119909) isin
120593(I) since (119909119899 119877
119909119899
)minus(119909 119877119909) = (119909
119899minus119909 119877
119909119899minus119909
) = 119909119899minus119909+
119877119909119899minus119909
le 2119909119899minus119909 rarr 0 as 119899 rarr infinThus 120593(I) being com-
plete is a closed right ideal inI1in the norm topology
Next we observe the isometry between I and 120593(I)
Proposition 5 Let I be a quasi-Jordan Banach algebra withnorm 1 unit Then there exists an equivalent norm that makesI isometrically isomorphic to a quasi-Jordan closed subalgebra120593(I) of the split quasi-Jordan Banach algebra I
1
Proof Clearly 119909∘
= 119909 + 119877119909 defines a norm on the
quasi-Jordan algebra I It follows that (I sdot ∘) is a quasi-
Jordan Banach algebraMoreoverI is isomorphic to the sub-algebra (119909 119877
119909) 119909 isin I of I
1under the isomorphism
120593 119909 997891rarr (119909 119877119909) as seen above Further we observe that 120593
is an isometry since 119909∘= 119909 + 119877
119909 = (119909 119877
119909) = 120593(119909)
Finally we note that the two norms sdot and sdot ∘on I are
equivalent since 119909 le 119909 + 119877119909 = 119909
∘= 119909 + 119877
119909 le
2119909
3 Invertible Elements
As in [17] an element 119909 in a quasi-Jordan algebra I is calledinvertible with respect to a unit 119890 isin I if there exists 119910 isin I
such that 119910 ⊲ 119909 = 119890 + (119890 ⊲ 119909 minus 119909) and 119910 ⊲ 1199092
= 119909 + (119890 ⊲
119909 minus 119909) + (119890 ⊲ 1199092
minus 1199092
) such an element 119910 is called an inverseof119909 with respect to 119890 Let 119890
⊲(119909) denote the element 119890 ⊲ 119909 minus 119909
Then119909 has an inverse119910with respect to 119890 hArr 119910 ⊲ 119909 = 119890+119890⊲(119909)
and 119910 ⊲ 1199092
= 119909 + 119890⊲(119909) + 119890
⊲(119909
2
)We know from the above discussion that the embedding
119909 997891rarr 120593(119909) = (119909 119877119909) of a quasi-Jordan algebraI into the split
quasi-Jordan algebra I1preserves the units The embedding
120593 also preserves the corresponding invertible elements if 119910 isan inverse of 119909 with respect to a unit 119890 in I then 119910 ⊲ 119909 =
119890 + (119890 ⊲ 119909 minus 119909) and 119910 ⊲ 1199092
= 119909 + (119890 ⊲ 119909 minus 119909) + (119890 ⊲ 1199092
minus 1199092
)Hence
(119910 119877119910) ⊲ (119909 119877
119909) = (119910 ⊲ 119909 119877
119910⊲119909)
= (119890 + (119890 ⊲ 119909 minus 119909) 119877119890+(119890⊲119909minus119909)
)
= (119890 + (119890 ⊲ 119909 minus 119909) 119877119890+ 119877
119890⊲119909minus 119877
119909)
= (119890 119877119890) + (119890 ⊲ 119909 119877
119890⊲119909) minus (119909 119877
119909)
= (119890 119877119890) + (119890 119877
119890) ⊲ (119909 119877
119909) minus (119909 119877
119909)
(119910 119877119910) ⊲ (119909 119877
119909)2
= (119910 ⊲ 1199092
119877119910⊲1199092)
= (119909 119877119909) + ((119890 ⊲ 119909 119877
119890⊲119909) minus (119909 119877
119909))
+ ((119890 ⊲ 1199092
119877119890⊲1199092) minus (119909
2
1198771199092))
= (119909 119877119909) + ((119890 119877
119890) ⊲ (119909 119877
119909) minus (119909 119877
119909))
+ ((119890 119877119890) ⊲ (119909 119877
119909)2
minus (119909 119877119909)2
)
(4)
Thus (119910 119877119910) is an inverse of (119909 119877
119909) with respect to the unit
(119890 119877119890) inI
1
Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebraThen 119877
119909(119910) = 119910 ⊲ 119909 = 119910 ⊲ (119909
119869+ 119909
119885) = 119910 ⊲ 119909
119869+ 119910 ⊲ 119909
119885=
119910 ⊲ 119909119869+ 0 = 119877
119909119869
(119910) for all 119909 119910 isin I Thus 119877119909= 119877
119909119869
for all119909 isin I
In this section we demonstrate that the set of invertibleelements with respect to a fixed unit in a quasi-JordanBanach algebra may not be open For this we proceed asfollows
Proposition 6 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebra with unit 119890 isin 119869 If 119909 isin I is invertible withrespect to 119890 then so is 120582119909 for all 120582 = 0
Proof Let119910 be an inverse of 119909 inIwith respect to 119890We showthat 1199101015840
= (1120582)119910119869+ 119910
119885is an inverse of 120582119909 with respect to 119890
Observe that
119910119869⊲ 119909 + 119910
119885⊲ 119909 = (119910
119869+ 119910
119885) ⊲ 119909
= 119910 ⊲ 119909 = 119890 + 119890⊲(119909)
= 119890 + 119890⊲(119909
119869+ 119909
119885) = 119890 + (minus119909
119885)
Abstract and Applied Analysis 5
119910119869⊲ 119909
2
+ 119910119885⊲ 119909
2
= 119910 ⊲ 1199092
= 119909 + 119890⊲(119909
119869+ 119909
119885) + 119890
⊲((119909
119869+ 119909
119885)2
)
= 119909119869+ (minus119909
119885⊲ 119909)
(5)
Hence by the uniqueness of the representation as sum of Jor-dan and zero parts we get 119910
119869⊲ 119909 = 119890 119910
119885⊲ 119909 = minus119909
119885 119910
119869⊲
1199092
= 119909119869 and 119910
119885⊲ 119909
2
= minus119909119885
⊲ 119909 Therefore 1199101015840
⊲ 120582119909 =
120582(((1120582)119910119869+ 119910
119885) ⊲ 119909) = 119910
119869⊲ 119909 + 120582119910
119885⊲ 119909 = 119890 minus 120582119909
119885=
119890 + 119890⊲(120582119909) and 119910
1015840
⊲ (120582119909)2
= 1205822
((1120582)119910119869⊲ 119909
2
+ 119910119885
⊲ 1199092
) =
120582119909119869minus 120582
2
119909119885⊲ 119909 = 120582119909 + 119890
⊲(120582119909) + 119890
⊲((120582119909)
2
) because 119890⊲(120582119909) =
119890 ⊲ (120582119909) minus 120582119909 = minus120582119909119885and 119890
⊲((120582119909)
2
) = 119890 ⊲ 1205822
1199092
minus 1205822
1199092
=
1205822
(119890⊲(119909
2
)) = minus1205822
119909119885⊲ 119909
Proposition 7 LetI be a quasi-Jordan normed algebrawith aunit 119890 Let119866
119890(I) = 119909 isin I 119909 is invertible with respect to 119890
be an open set and 119909 isin 119866119890(I) Then 119909 + 119911 isin 119866
119890(I) for all
119911 isin 119885(I)
Proof Suppose 119909 isin 119866119890(I) and 119911 isin 119885(I) If 119911 = 0 then 119909+119911 =
119909 isin 119866119890(I) Next suppose 119911 = 0 Since 119866
119890(I) is an open set
there exists 120598 gt 0 such that 119886 isin 119866119890(I) whenever 119909 minus 119886 lt 120598
Hence 119909 + 119911∘isin 119866
119890(I) with 119911
∘= (1205982119911)119911
Let 119910 and 119910∘be inverses of 119909 and 119909 + 119911
∘with respect to 119890
respectively Then
119910∘⊲ (119909 + 119911
∘) = 119890 + 119890
⊲(119909 + 119911
∘)
= 119890 + 119890⊲(119909) minus 119911
∘
= 119910 ⊲ 119909 minus 119911∘
119910∘⊲ (119909 + 119911
∘)2
= 119909 + 119911∘+ 119890
⊲(119909 + 119911
∘) + 119890
⊲((119909 + 119911
∘)2
)
= 119909 + 119890⊲(119909) + 119890
⊲(119909
2
) minus 119911∘⊲ 119909
= 119910 ⊲ 1199092
minus 119911∘⊲ 119909
(6)
Hence by setting 1199101= 119910 + 120572(119910
∘minus 119910) with 120572 = 2119911120598 we see
that1199101⊲ (119909 + 119911) = (119910 + 120572 (119910
∘minus 119910)) ⊲ 119909
= 119910 ⊲ 119909 + 120572 (119910∘⊲ (119909 + 119911
∘) minus 119910 ⊲ 119909)
= 119890 + 119890⊲(119909) + 120572 (minus119911
∘)
= 119890 + 119890⊲(119909) minus 119911
= 119890 + 119890⊲(119909 + 119911)
1199101⊲ (119909 + 119911)
2
= (119910 + 120572 (119910∘minus 119910)) ⊲ 119909
2
= 119910 ⊲ 1199092
+ 120572 (119910∘⊲ (119909 + 119911
∘)2
minus 119910 ⊲ 1199092
)
= 119909 + 119890⊲(119909) + 119890
⊲(119909
2
) minus 120572 (119911∘⊲ 119909)
= 119909 + 119890⊲(119909) + 119890
⊲(119909
2
) minus 119911 ⊲ 119909
= (119909 + 119911) + 119890⊲(119909 + 119911) + 119890
⊲((119909 + 119911)
2
)
(7)
since 119911 isin 119885(I) Thus 1199101is an inverse of 119909 + 119911 with respect to
the unit 119890
Corollary 8 Under the hypothesis of Proposition 7 120582119890 minus 119909 isin
119866119890(I) implies 120582119890 minus (119909 + 119911) isin 119866
119890(I) for all 119911 isin 119885(I)
Proposition 9 LetI = 119869 oplus 119885(I) be a split quasi-Jordan nor-med algebra and let 119890 isin 119869 be a unit inI such that the set119866
119890(I)
is open Then 119909 ⊲ 1199092
= 1199092
⊲ 119909 for all 119909 isin I
Proof Of course the element 119890 is the unit of the Jordanalgebra 119869 Then for any fixed element 119886 isin 119869 there exists120582 isin C such that 120582119890 minus 119886 is invertible in 119869 otherwise wewould get the negation of the well-known fact that spectrumof an element of a unital Jordan algebra is bounded Thatis there exists 119910 isin 119869 such that 119910 ⊲ (120582119890 minus 119886) = 119890 and119910 ⊲ (120582119890minus119886)
2
= (120582119890minus119886) However 119890⊲(119886) = 0 = 119890
⊲(119886
2
) Hence119910 is an inverse of 119886 in the quasi-Jordan algebraIwith respectto the unit 119890 By Corollary 8 120582119890 minus (119886 + 119911) is also invertiblefor any 119911 isin 119885(I) This in turn gives the existence of 119887 isin I
satisfying 119887 ⊲ (120582119890 minus (119886 + 119911)) = 119890 + 119911 and 119887 ⊲ (120582119890 minus (119886 + 119911))2
=
(120582119890minus119886)+119911 ⊲ (120582119890minus119886) Multiplying the first equation from theright by (120582119890 minus (119886 + 119911))
2 the second equation by (120582119890minus (119886+119911))and using the right Jordan identity we get
(119890 + 119911) ⊲ (120582119890 minus (119886 + 119911))2
= ((120582119890 minus 119886) + 119911 ⊲ (120582119890 minus 119886)) ⊲ (120582119890 minus (119886 + 119911))
(8)
Hence
(119890 + 119911) ⊲ (120582119890 minus 119886)2
= ((120582119890 minus 119886) + 119911 ⊲ (120582119890 minus 119886)) ⊲ (120582119890 minus 119886)
(9)
so that
119890 ⊲ (120582119890 minus 119886)2
+ 1205822
119911 minus 2120582 (119911 ⊲ 119886) + 119911 ⊲ 1198862
= (120582119890 minus 119886)2
+ 1205822
119911 minus 2120582119911 ⊲ 119886 + (119911 ⊲ 119886) ⊲ 119886
(10)
This last equation reduces to 119890 ⊲ (120582119890 minus 119886)2
= (120582119890 minus 119886)2 since
(120582119890 minus 119886)2
isin 119869 and 119890 is the unit of 119869 Hence 119911 ⊲ 1198862
= (119911 ⊲
119886) ⊲ 119886 for all 119886 isin 119869 and 119911 isin 119885(I) Now for any 119909 isin I thelast equation with 119886 = 119909
119869and 119911 = 119909
119885gives 119909
119885⊲ 119909
2
119869= (119909
119885⊲
119909119869) ⊲ 119909
119869Thus 119909 ⊲ 119909
2
= 119909119869⊲ 119909
2
119869+119909
119885⊲ 119909
2
119869= 119909
2
119869⊲ 119909
119869+(119909
119885⊲
119909119869) ⊲ 119909
119869= 119909
2
⊲ 119909 for all 119909 isin I
Corollary 10 If a unital split quasi-Jordan algebra has anelement 119909 with 119909 ⊲ 119909
2
= 1199092
⊲ 119909 then the set of invertibleelements with respect to the unit of the Jordan part is not open
In the sequel we will show the existence of a unital splitquasi-Jordan Banach algebra with elements 119909 such that 119909 ⊲
1199092
= 1199092
⊲ 119909 Thus the above result establishes that the set ofinvertible elements with respect to a fixed unit in a quasi-Jordan Banach algebra may not be open
6 Abstract and Applied Analysis
4 The Spectrum of Elements in a UnitalQuasi-Jordan Algebra
As usual we define the spectrum of an element 119909 in aunital quasi-Jordan algebra (I 119890) denoted by 120590
(I119890)(119909) tobe the collection of all complex numbers 120582 for which 120582119890 minus
119909 is not invertible Thus 120590(I119890)(119909) = 120582 isin C 120582119890 minus
119909 is not invertible Here the subscript 119890 indicates that theinvariability depends on the choice of unit 119890 which generallyis not unique
Proposition 11 Let I be a unital quasi-Jordan algebra Then120590(I119890)(119890
1015840
) = 1 for all 119890 1198901015840 isin 119880(I)
Proof Let 119890 1198901015840 isin 119880(I) Then for any 120582 = 1 119910 = (1(120582minus1))1198901015840
is an inverse of 120582119890 minus 1198901015840 with respect to the unit 119890 because 119910 ⊲
(120582119890 minus 1198901015840
) = (1(120582 minus 1))1198901015840
⊲ (120582119890 minus 1198901015840
) = 119890 + 119890⊲(120582119890 minus 119890
1015840
) and119910 ⊲ (120582119890 minus 119890
1015840
)2
= (120582119890 minus 1198901015840
) + 119890⊲(120582119890 minus 119890
1015840
) + 119890⊲((120582119890 minus 119890
1015840
)2
)
Proposition 12 Let I be a quasi-Jordan algebra with unit 119890Then 120590
(I119890)(119911) = 0 for all 119911 isin 119885(I)
Proof For any fixed 119911 isin 119885(I) and nonzero scalar 120582 thevector 119910 = (1120582)(119890 + 119911) satisfies 119910 ⊲ (120582119890 minus 119911) = 120582119910 = 119890 + 119911 =
119890+ 119890⊲(120582119890minus 119911) and 119910 ⊲ (120582119890 minus 119911)
2
= 1205822
119910 = 120582119890+120582119911 = (120582119890minus119911) +
119890⊲(120582119890 minus 119911) + 119890
⊲((120582119890 minus 119911)
2
) So (1120582)(119890 + 119911) is an inverse of120582119890 minus 119911 with respect to 119890 This means 120582 notin 120590
(I119890)(119911) for all 120582 = 0However the zero vector is not invertible Thus 120590
(I119890)(119911) =
0
Proposition 13 Let I = 119869 oplus 119885(I) be a split quasi-Jordanalgebra with unit 119890 isin 119869 Then 120590
(I119890)(119901) sube 0 1 for all idempo-tents 119901 isin I (ie 1199012
= 119901)
Proof Let 119901 be any fixed idempotent inI Since 119901 isin I 119901 hasa unique representation 119901 = 119901
119869+ 119901
119885with 119901
119869isin 119869 and 119901
119885isin
119885(I) Clearly 1199012
119869+ 119901
119885⊲ 119901
119869= 119901
2
= 119901 = 119901119869+ 119901
119885 Then
by uniqueness of the representation in the split quasi-Jordanalgebra I 119901
119885⊲ 119901
119869= 119901
119885and 119901
2
119869= 119901
119869 this means 119901
119869is an
idempotent in the Jordan algebra 119869 Hence 120590(119869119890)
(119901119869) sube 0 1
Thus 119901120582= 120582119890 minus 119901
119869is invertible in 119869 with the unique inverse
119901minus1
120582 for all 120582 notin 0 1We show that 119910 = 119901
minus1
120582+ (1(120582 minus 1))119901
119885is an inverse of
120582119890 minus 119901 in I with respect to the unit 119890 for this we note that120582119890 minus 119901 = 119901
120582minus 119901
119885 (119901
120582minus 119901
119885)2
= 1199012
120582minus 119901
119885⊲ 119901
120582 119890
⊲(120582119890 minus
119901) = 119890⊲(119901
120582minus 119901
119885) = 119890 ⊲ (119901
120582minus 119901
119885) minus (119901
120582minus 119901
119885) = 119901
119885 and
119890⊲((120582119890 minus 119901)
2
) = 119890Δ((119901
120582minus 119901
119885)2
) = 119890⊲(119901
2
120582minus 119901
119885⊲ 119901
120582) = 119901
119885⊲
119901120582= 119901
119885⊲ (120582119890minus119901
119869) = 120582119901
119885minus119901
119885⊲ 119901
119869= 120582119901
119885minus119901
119885= (120582minus1)119901
119885
Hence 119910 ⊲ (120582119890 minus 119901) = 119910 ⊲ (119901120582minus 119901
119885) = 119910 ⊲ 119901
120582= (119901
minus1
120582+
(1(120582 minus 1))119901119885) ⊲ 119901
120582= 119901
minus1
120582⊲ 119901
120582+ (1(120582 minus 1))119901
119885⊲ 119901
120582= 119890 +
(1(120582 minus 1))119901119885
⊲ (120582119890 minus 119901119869) = 119890 + (1(120582 minus 1))(120582119901
119885minus 119901
119885⊲
119901119869) = 119890 + (1(120582 minus 1))(120582119901
119885minus 119901
119885) = 119890 + 119901
119885= 119890 + 119890
⊲(120582119890 minus 119901)
and 119910 ⊲ (120582119890 minus 119901)2
= 119910 ⊲ (1199012
120582minus 119901
119885⊲ 119901
120582) = 119910 ⊲ 119901
2
120582=
(119901minus1
120582+(1(120582minus1))119901
119885) ⊲ 119901
2
120582= 119901
120582+(1(120582minus1))119901
119885⊲ (120582119890minus119901
119869)2
=
119901120582+(1(120582minus1))119901
119885⊲ (120582
2
119890minus2120582119901119869+119901
2
119869) = 119901
120582+(1(120582minus1))119901
119885⊲
(1205822
119890 minus 2120582119901119869+ 119901
119869) = 119901
120582+ (1(120582 minus 1))(120582
2
119901119885
⊲ 119890 minus 2120582119901119885
⊲
119901119869+ 119901
119885⊲ 119901
119869) = 119901
120582+ (1(120582 minus 1))(120582
2
119901119885
minus 2120582119901119885
+ 119901119885) =
119901120582+ (1(120582 minus 1))(120582 minus 1)
2
119901119885= 120582119890 minus119901
119869+ (120582 minus 1)119901
119885= (120582119890 minus119901) +
119890⊲(120582119890 minus 119901) + 119890
⊲((120582119890 minus 119901)
2
)
As mentioned in Section 2 if I is a quasi-Jordan algebrawith a unit 119890 then the set 119890 + 119911 119911 isin 119885(I) coincides withthe set 119880(I) of all units inI
Proposition 14 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebrawith unit 119890 isin 119869 and119909 isin I invertiblewith respectto some 1198901015840 isin 119880(I)Then119909
119869is invertible with respect to the unit
119890 in the Jordan algebra 119869
Proof Clearly 119890 ⊲ 119909 minus 119909 = minus119909119885and 119890 ⊲ 119909
2
minus 1199092
= minus119909119885⊲ 119909
119869
Since 1198901015840
isin 119880(I) we have 1198901015840
= 119890 + 119911 for some 119911 isin 119885(I)Hence the invertibility of 119909 in I with respect to the unit 1198901015840gives the existence of 119910 isin I such that 119910
119869⊲ 119909
119869+ 119910
119885⊲ 119909
119869=
119910 ⊲ 119909119869= 119910 ⊲ 119909 = 119890 + 119911 + (119890 + 119911) ⊲ 119909 minus 119909 = 119890 + 119911 + 119911 ⊲ 119909 minus 119909
119885
and 119910119869⊲ 119909
2
119869+ 119910
119885⊲ 119909
2
119869= 119910 ⊲ 119909
2
119869= 119910 ⊲ 119909
2
= (119890 + 119911) ⊲
119909 + (119890 + 119911) ⊲ 1199092
minus 1199092
= 119909119869+ 119911 ⊲ 119909 + 119911 ⊲ 119909
2
minus 119909119885⊲ 119909
119869 So by
the uniqueness of the representations in the split algebra Iwe get 119910
119869⊲ 119909
119869= 119890 and 119910
119869⊲ 119909
2
119869= 119909
119869 Thus 119910
119869is the inverse
of 119909119869in the Jordan algebra 119869
Next we observe that the spectrum of 119909 in a unital splitquasi-Jordan algebra with respect to any unit includes thespectrum of 119909
119869in the Jordan part 119869
Corollary 15 LetI = 119869 oplus119885(I) be a unital split quasi-Jordanalgebra with unit 119890 isin 119869 and let 119909 isin I Then 120590
(119869119890)(119909
119869) sube
120590(I1198901015840)(119909) for all 1198901015840 isin 119880(I)
Proof Let 1198901015840 = 119890 + 119911 with 119911 isin 119885(I) and let 120582 notin 120590(I1198901015840)(119909)
Then 1205821198901015840 minus 119909 is invertible in I with respect to the unit 1198901015840Hence its Jordanpart120582119890minus119909
119869is invertible in the Jordan algebra
119869 by Proposition 14 Thus 120582 notin 120590(119869119890)
(119909119869)
It is well known that the spectrum of any element ina unital Jordan Banach algebra is nonempty (cf [7]) Thistogether with Proposition 2 and Corollary 15 gives thefollowing result
Corollary 16 The spectrum of any element in a unital splitquasi-Jordan Banach algebra is nonempty
The next result extends Corollary 16 to any quasi-JordanBanach algebra with a norm 1 unit
Proposition 17 The spectrum of any element in a quasi-Jordan Banach algebra with a norm 1 unit is nonempty
Proof Let I be a unital quasi-Jordan Banach algebra witha norm 1 unit 119890 From Section 2 we know that the map120593 119909 997891rarr 120593(119909) = (119909 119877
119909) embeds I into the unital split
quasi-Jordan Banach algebra I1
= (119909 119877119910) 119909 119910 isin I
equipped with the sum (119909 119877119910) + (119886 119877
119887) = (119909 + 119886 119877
119910+119887) sca-
lar multiplication 120582(119909 119877119910) = (120582119909 119877
120582119910) and product (119909
119877119910) ⊲ (119886 119877
119887) = (119909 ⊲ 119887 119877
119910⊲119887) and the image 120593(I) is a
norm closed right ideal isomorphic to I with norm 1 unit(119890 119877
119890) Moreover it is seen in Section 3 that the embedding
Abstract and Applied Analysis 7
120593 also preserves the corresponding invertible elements thatis (119910 119877
119910) is an inverse of (119909 119877
119909) with respect to the unit
(119890 119877119890) in I
1whenever 119910 is an inverse of 119909 with respect
to a unit 119890 in I Hence by Corollary 16 it follows that120601 = 120590
(I1(119890119877119890))(119909 119877
119909) sube 120590
(120593(I)(119890119877119890))(119909 119877
119909) = 120590
(I119890)(119909) for all119909 isin I
Proposition 18 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebra with unit 119890 isin 119869 and let 119909 isin 119869 Then 120590
(119869119890)(119909) =
120590(I119890)(119909)
Proof By Corollary 15 120590(119869119890)
(119909) sube 120590(I119890)(119909) For the reverse
inclusion let 120582 notin 120590(119869119890)
(119909) then 120582119890 minus 119909 is invertible in 119869 thatis there exists 119910 isin 119869 such that 119910 ⊲ (120582119890 minus 119909) = 119890 and 119910 ⊲
(120582119890 minus 119909)2
= 120582119890 minus 119909 However 119890⊲(120582119890 minus 119909) = 0 = 119890
⊲((120582119890 minus 119909)
2
)Hence 119910 is an inverse of 120582119890 minus 119909 in I with respect to the unit119890 that is 120582 notin 120590
(I119890)(119909) Thus 120590(I119890)(119909) sube 120590
(119869119890)(119909)
Proposition 19 Let I be a unital quasi-Jordan normedalgebra and let 119890 be a unit inI for which 119866
119890(I) is open Then
120590(I119890)(119909) = 120590
(I119890)(119909 + 119911) for all 119911 isin 119885(I)
Proof By Corollary 8 120582119890 minus 119909 is invertible if and only if 120582119890 minus
(119909 + 119911) is invertible for all 119911 isin 119885(I) Thus 120582 notin 120590(I119890)(119909) if
and only if 120582 notin 120590(I119890)(119909 + 119911) for all 119911 isin 119885(I)
Corollary 20 LetI = 119869oplus119885(I) be a unital split quasi-Jordannormed algebra and let 119890 be a unit in I such that 119866
119890(I) is
open Then 120590(I119890)(119909) = 120590
(I119890)(119909119869) for all 119909 = 119909119869+ 119909
119885isin I
Further if the unit 119890 isin 119869 then120590(I119890)(119909) = 120590
(I119890)(119909119869) = 120590(119869119890)
(119909119869)
Lemma 21 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordanalgebra with a unit 119890 isin 119869 and let 119909 isin I be invertible withrespect to 119890 Then 119909
119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
Proof As 119909 is invertible there exists 119910 isin I such that 119910 ⊲ 119909 =
119890 + (119890 ⊲ 119909 minus 119909) = 119890 minus 119909119885and 119910 ⊲ 119909
2
= 119909 + (119890 ⊲ 119909 minus 119909) + (119890 ⊲
1199092
minus 1199092
) = 119909119869minus 119909
119885⊲ 119909 Hence by the uniqueness of the
representation in a split quasi-Jordan algebra we obtain
119910119885⊲ 119909 = minus119909
119885 (11)
119910119885⊲ 119909
2
= minus119909119885⊲ 119909 (12)
Thus
119909119885⊲ 119909
2
= minus (119910119885⊲ 119909) ⊲ 119909
2
(by (11))
= minus (119910119885⊲ 119909
2
) ⊲ 119909 (by the right Jordan identity)
= (119909119885⊲ 119909) ⊲ 119909 (by (12))
(13)
Proposition 22 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebra with a unit 119890 isin 119869 119909 = 119909
119869+ 119909
119885isin I satisfies
120590(I119890)(119909) =C Then
(1) 119909119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
(2) 1199092 ⊲ 119909 = 119909 ⊲ 1199092
Proof (1) Let 120582 notin 120590(I119890)(119909) Then 120582119890 minus 119909 = (120582119890 minus 119909
119869) + (minus119909
119885)
is invertible with respect to 119890 By Lemma 21 we have
119909119885⊲ (120582119890 minus 119909)
2
= (119909119885⊲ (120582119890 minus 119909)) ⊲ (120582119890 minus 119909) (14)
since the zero part of 120582119890 minus 119909 is minus119909119885 However
119909119885⊲ (120582119890 minus 119909)
2
= 119909119885⊲ (120582
2
119890 minus 120582119890 ⊲ 119909 minus 120582119909 ⊲ 119890 + 1199092
)
= 1205822
(119909119885⊲ 119890) minus 120582 (119909
119885⊲ (119890 ⊲ 119909))
minus 120582 (119909119885⊲ (119909 ⊲ 119890)) + 119909
119885⊲ 119909
2
= 1205822
119909119885minus 2120582119909
119885⊲ 119909
+ 119909119885⊲ 119909
2
(by the right commutativity of ⊲)
(119909119885⊲ (120582119890 minus 119909)) ⊲ (120582119890 minus 119909)
= (120582 (119909119885⊲ 119890) minus 119909
119885⊲ 119909) ⊲ (120582119890 minus 119909)
= (120582119909119885minus 119909
119885⊲ 119909) ⊲ (120582119890 minus 119909)
= 1205822
(119909119885⊲ 119890) minus 120582 (119909
119885⊲ 119909) ⊲ 119890
minus 120582119909119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
= 1205822
119909119885minus 2120582119909
119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
(15)
Therefore (14) becomes
1205822
119909119885minus 2120582119909
119885⊲ 119909 + 119909
119885⊲ 119909
2
= 1205822
119909119885minus 2120582119909
119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
(16)
which after simplification reduces to the required equation119909119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
(2) Since 1199092 = (119909119869+ 119909
119885)2
= 1199092
119869+ 119909
119885⊲ 119909
119869 we have
1199092
⊲ 119909 = (1199092
119869+ 119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909
119869+ (119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909
119869+ 119909
119885⊲ 119909
2
(17)
by the part (1) But 1199092119869⊲ 119909
119869= 119909
119869⊲ 119909
2
119869since 119909
119869is in the Jor-
dan algebra 119869 Therefore 1199092 ⊲ 119909 = 1199092
119869⊲ 119909
119869+ 119909
119885⊲ 119909
2
= 119909119869⊲
1199092
119869+ 119909
119885⊲ 119909
2
= 119909 ⊲ 1199092
Remark 23 In any quasi-Jordan algebra if an element 119909
satisfies 1199093
= 1199092
⊲ 119909 = 119909 ⊲ 1199092 then 119909
119899
⊲ 1199092
= 119909119899+2 for
all positive integers 119899 For this suppose 119909 satisfies 1199092 ⊲ 119909 =
119909 ⊲ 1199092 and 119909
119898
⊲ 1199092
= 119909119898+2 for any fixed119898 ge 1 Then 119909
119898+1
⊲
1199092
= (119909119898
⊲ 119909) ⊲ 1199092
= (119909119898
⊲ 1199092
) ⊲ 119909 (by the right Jor-dan identity) = 119909
119898+2
⊲ 119909 = 119909119898+3
Proposition 24 Let I be a unital quasi-Jordan Banachalgebra with unit 119890 and let 119909 isin I satisfy (119890 minus 119909) ⊲ (119890 minus 119909)
2
=
(119890 minus 119909)2
⊲ (119890 minus 119909) Then 119909 isin 119866119890(I) whenever 119890 minus 119909 lt 1
8 Abstract and Applied Analysis
Proof First note that 119890minus119909 lt 1 gives (119890 minus 119909)119899
le 119890 minus 119909119899
lt
1 for all 119899 = 1 2 3 Hence the infinite geometric series119890 +sum
infin
119899=1(119890 minus119909)
119899 converges absolutely to some element 119910 isin IWe show that the geometric series sum 119910 is an inverse of 119909with respect to the unit 119890 For any fixed positive integer 119899 let119910119899= 119890+sum
119899
119896=1(119890minus119909)
119896Then the sequence 119910119899 of partial sums
converges to 119910 By setting 119908 = 119890 minus 119909 we get
119910119899⊲ 119909 = (119890 +
119899
sum
119896=1
119908119896
) ⊲ (119890 minus 119908)
= 119890 +
119899
sum
119896=1
119908119896
minus (119890 ⊲ 119908 +
119899+1
sum
119896=2
119908119896
)
= 119890 + 119908 minus 119890 ⊲ 119908 minus 119908119899+1
= 119890 + (119890 minus 119909) minus 119890 ⊲ (119890 minus 119909) minus (119890 minus 119909)119899+1
= 119890 + (119890 minus 119909 minus 119890 + 119890 ⊲ 119909) minus (119890 minus 119909)119899+1
= 119890 + (119890 ⊲ 119909 minus 119909) minus (119890 minus 119909)119899+1
(18)
Thus by allowing 119899 rarr infin we obtain 119909 ⊲ 119910 = 119890 + (119890 ⊲
119909 minus 119909) = 119890 + 119890⊲(119909) since 119890 minus 119909 lt 1
Next by Remark 23 we have
119910119899⊲ 119909
2
= 119910119899⊲ (119890 minus 119908)
2
= 119910119899⊲ (119890 minus 119890 ⊲ 119908 minus 119908 ⊲ 119890 + 119908
2
)
= 119910119899⊲ (119890 minus 2119908 + 119908
2
)
= 119910119899⊲ 119890 minus 2119910
119899⊲ 119908 + 119910
119899⊲ 119908
2
= 119910119899minus 2119910
119899⊲ 119908 + 119910
119899⊲ 119908
2
= (119890 +
119899
sum
119896=1
119908119896
) minus 2(119890 +
119899
sum
119896=1
119908119896
) ⊲ 119908
+ (119890 +
119899
sum
119896=1
119908119896
) ⊲ 1199082
= (119890 +
119899
sum
119896=1
119908119896
) minus 2(119890 ⊲ 119908 +
119899+1
sum
119896=2
119908119896
)
+ (119890 ⊲ 1199082
+
119899+2
sum
119896=3
119908119896
)
= 119890 + 119908 minus 2119890 ⊲ 119908 minus 1199082
+ 119890 ⊲ 1199082
minus 119908119899+1
+ 119908119899+2
(19)
Taking the limit as 119899 rarr infin we get
119910 ⊲ 1199092
= 119890 + 119908 minus 2119890 ⊲ 119908 minus 1199082
+ 119890 ⊲ 1199082
= 119890 + (119890 minus 119909) minus 2119890 ⊲ (119890 minus 119909) minus (119890 minus 119909)2
+ 119890 ⊲ (119890 minus 119909)2
= 2119890 minus 119909 minus 2119890 + 2119890 ⊲ 119909 minus 119890 + 119890 ⊲ 119909 + 119909 minus 1199092
+ 119890 minus 2119890 ⊲ 119909 + 119890 ⊲ 1199092
= 119909 + (119890 ⊲ 119909 minus 119909) + (119890 ⊲ 1199092
minus 1199092
)
(20)
Proposition 25 Let I be a unital split quasi-Jordan Banachalgebra with unit 119890 If 119909 isin I with (119890 minus 119909) ⊲ (119890 minus 119909)
2
=
(119890 minus 119909)2
⊲ (119890 minus 119909) then |120582| le 119909 for all 120582 isin 120590(I119890)(119909)
Proof If 120582 isin 120590(I119890)(119909) with 120582 = 0 then the noninvertibility of
120582119890 minus 119909 means the noninvertibility of 119890 minus (1120582)119909 with respectto the unit 119890 However by Proposition 24 119890 minus (1120582)119909must beinvertible with respect to the unit 119890 whenever (1|120582|)119909 lt 1It follows that |120582| le 119909 for all 120582 isin 120590
(I119890)(119909)
5 Unbounded and Nonclosed Spectrum
In this section we show that the spectrum of an element ina split quasi-Jordan Banach algebra may be neither boundednor closed and hence not compactThe following result givesa couple of characterizations of the unbounded spectrum ofan element in a split quasi-Jordan Banach algebra
Proposition 26 Let 119909 be an element of a unital split quasi-Jordan Banach algebra I = 119869 oplus 119885(I) with a unit 119890 isin 119869 Thenthe following statements are equivalent
(1) 120590(I119890)(119909) =C
(2) 1199092 ⊲ 119909 = 119909 ⊲ 1199092
(3) |120582| le 119909 for all 120582 isin 120590(I119890)(119909)
Proof (1rArr2) See Proposition 22(2rArr3) Suppose 1199092 ⊲ 119909 = 119909 ⊲ 119909
2 Then
(119890 minus 119909)2
⊲ (119890 minus 119909) = (119890 minus 119890 ⊲ 119909 minus 119909 + 1199092
) ⊲ (119890 minus 119909)
= 119890 minus 2119890 ⊲ 119909 minus 119909 + 21199092
+ (119890 ⊲ 119909) ⊲ 119909 minus 1199092
⊲ 119909
= 119890 minus 2119909119869minus 119909 + 2119909
2
+ 119890 ⊲ 1199092
minus 1199093
(119890 minus 119909) ⊲ (119890 minus 119909)2
= (119890 minus 119909) ⊲ (119890 minus 2119909 + 1199092
)
= 119890 minus 2 (119890 ⊲ 119909) + 119890 ⊲ 1199092
minus 119909
+ 21199092
minus 119909 ⊲ 1199092
= 119890 minus 2119909119869+ 119890 ⊲ 119909
2
minus 119909 + 21199092
minus 1199093
(21)
Abstract and Applied Analysis 9
From (21) we get
(119890 minus 119909)2
⊲ (119890 minus 119909) = (119890 minus 119909) ⊲ (119890 minus 119909)2
(22)
Hence |120582| le 119909 for all 120582 isin 120590(I119890)(119909) by Proposition 25
(3rArr1) Immediate
Remark 27 There do exist unital split quasi-Jordan algebrascontaining elements that have the spectrum with respect tothe unit of the Jordan part equal to the whole ofC and henceunbounded To justify this claim we proceed as follows
Let 119860 be a unital associative algebra and let 119872 be an 119860-bimodule Let 119891 119872 rarr 119860 be an 119860-bimodule map (ie anadditive map satisfying 119891(119886119909) = 119886119891(119909) and 119891(119909119886) = 119891(119909)119886for all 119886 isin 119860 119909 isin 119872) Then one can put a dialgebra structureon 119872 as follows 119909 ⊣ 119910 = 119909119891(119910) and 119909⊢119910 = 119891(119909)119910 (cf [13Example 22(d)]) Hence119872+ is a quasi-Jordan algebra underthe quasi-Jordan product ldquo⊲rdquo given by119909 ⊲ 119910 = (12)(119909119891(119910)+
119891(119910)119909) Further for any 119909 isin 119872 we observe that
1199092
⊲ 119909 =1
2(119909119891 (119909) + 119891 (119909) 119909) ⊲ 119909
=1
4((119909119891 (119909) + 119891 (119909) 119909) 119891 (119909)
+119891 (119909) (119909119891 (119909) + 119891 (119909) 119909))
=1
4(119909119891 (119909) 119891 (119909) + 2119891 (119909) 119909119891 (119909) + 119891 (119909) 119891 (119909) 119909)
119909 ⊲ 1199092
=1
2119909 ⊲ (119909119891 (119909) + 119891 (119909) 119909)
=1
4(119909119891 (119909119891 (119909) + 119891 (119909) 119909) + 119891 (119909119891 (119909) + 119891 (119909) 119909) 119909)
=1
4(2119909119891 (119909) 119891 (119909) + 2119891 (119909) 119891 (119909) 119909)
(23)
However the right hand sides of the above equations (23)may not be equal see the following example (Example 28)For such elements 119909 we have 119909
2
⊲ 119909 = 119909 ⊲ 1199092 Hence by
Proposition 26 the spectrum of 119909 is unbounded whenever119872
+ is a unital split quasi-Jordan Banach algebra
Example 28 Let 119872 be the collection of 2 times 2 matrices withentries from the fieldC and let119860 be the algebra of allmatricesof the form [
120572 0
0 120573] with 120572 120573 isin C Then it is easily seen
that 119872 is an 119860-bimodule Next we define 119891 119872 rarr 119860
by 119891 [11988611
11988612
11988621
11988622
] = [11988611
0
0 11988622
] Of course 119891 is an additive mapsatisfying
119891([120572 0
0 120573] [
11988611
11988612
11988621
11988622
]) = [120572 0
0 120573]119891([
11988611
11988612
11988621
11988622
])
119891([11988611
11988612
11988621
11988622
] [120572 0
0 120573]) = 119891([
11988611
11988612
11988621
11988622
]) [120572 0
0 120573]
(24)
Hence119891 is an119860-bimodule mapThus by Remark 27119872+ is aquasi-Jordan algebra with the quasi-Jordan product as below
[11988611
11988612
11988621
11988622
] ⊲ [11988711
11988712
11988721
11988722
] =1
2([
11988611
11988612
11988621
11988622
]119891([11988711
11988712
11988721
11988722
])
+119891([11988711
11988712
11988721
11988722
]) [11988611
11988612
11988621
11988622
])
=[[
[
1198861111988711
11988612
11988711
+ 11988722
2
11988621
11988711
+ 11988722
21198862211988722
]]
]
(25)
Indeed 119872+
= 119869 oplus 119885(119872+
) where 119869 = [119886 0
0 119887] 119886 119887 isin C
is a subalgebra of 119872+ and 119885(119872+
) = [0 119886
119887 0] 119886 119887 isin C Any
matrix of the form [1 119886
119887 1] with 119886 119887 isin C is a (right) unit in119872
+Thus119872+ is a unital split quasi-Jordan algebrawith thematrix119868 = [
1 0
0 1] as the unit of its Jordan part 119869
Further a natural norm is defined on119872+ as follows
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
=100381610038161003816100381611988611
1003816100381610038161003816 +100381610038161003816100381611988612
1003816100381610038161003816 +100381610038161003816100381611988621
1003816100381610038161003816 +100381610038161003816100381611988622
1003816100381610038161003816 (26)
This norm also satisfies10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
] ⊲ [11988711
11988712
11988721
11988722
]
10038171003817100381710038171003817100381710038171003817
=1
2(
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
] [11988711
0
0 11988722
]
+ [11988711
0
0 11988722
] [11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
)
le
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
10038171003817100381710038171003817100381710038171003817
[11988711
0
0 11988722
]
10038171003817100381710038171003817100381710038171003817
le
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
10038171003817100381710038171003817100381710038171003817
[11988711
11988712
11988721
11988722
]
10038171003817100381710038171003817100381710038171003817
(27)
Next for any 119909 = [11988611
11988612
11988621
11988622
] isin 119872+ we observe that
1199092
= [11988611
11988612
11988621
11988622
] ⊲ [11988611
11988612
11988621
11988622
]
=[[
[
1198862
1111988612
11988611
+ 11988622
2
11988621
11988611
+ 11988622
21198862
22
]]
]
(28)
so that
1199092
⊲ 119909 = [
[
1198863
11
11988612
4(119886
11+ 119886
22)2
11988621
4(119886
11+ 119886
22)2
1198863
22
]
]
119909 ⊲ 1199092
= [
[
1198863
11
11988612
2(119886
2
11+ 119886
2
22)
11988621
2(119886
2
11+ 119886
2
22) 119886
3
22
]
]
(29)
So for any 119909 = [11988611
11988612
11988621
11988622
] isin 119872+ 1199092 ⊲ 119909 = 119909 ⊲ 119909
2
hArr
11988612
= 11988621
= 0 or 11988611
= 11988622 Thus by Proposition 26
10 Abstract and Applied Analysis
120590(119872+119868)([
11988611
11988612
11988621
11988622
]) = C whenever 11988611
= 11988622
and 11988612 119886
21 = 0
In particular for 119909 = [1 1
0 2] isin 119872
+ we have
1199092
⊲ 119909 =[[
[
13
1
4(1 + 2)
2
0
4(1 + 2)
2
23
]]
]
= [1
9
4
0 8
]
119909 ⊲ 1199092
=[[
[
13
1
2(1
2
+ 22
)
0
2(1
2
+ 22
) 23
]]
]
= [1
5
2
0 8
]
(30)
Concerning the inequality between the right hand sidesof (23) in Remark 27 we observe for 119909 as above that
1
4(119909119891 (119909) 119891 (119909) + 2119891 (119909) 119909119891 (119909) + 119891 (119909) 119891 (119909) 119909)
=1
4([
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 1
0 2] [
1 0
0 2]
+ [1 0
0 2] [
1 0
0 2] [
1 1
0 2]) = [
19
4
0 8
]
(31)
but
1
4(2119909119891 (119909) 119891 (119909) + 2119891 (119909) 119891 (119909) 119909)
=1
4(2 [
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 0
0 2] [
1 1
0 2])
= [1
5
2
0 8
]
(32)
Hence 1199092 ⊲ 119909 = 119909 ⊲ 1199092 Thus 120590
(119872+119868)([
1 1
0 2]) = C by
Proposition 26Further suppose the matrix [
11988611198862
11988631198864
] isin 119872+ is invertible
with respect to the unit 119868 that is [ 1198861 119886211988631198864
] isin 119866119868(119872
+
) Thenthere exists [ 119887
11198872
11988731198874
] isin 119872+ such that
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] = 119868 + 119868⊲([
1198861
1198862
1198863
1198864
]) = [1 minus119886
2
minus1198863
1]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
= [1198861
1198862
1198863
1198864
] + 119868⊲([
1198861
1198862
1198863
1198864
])
+ 119868⊲([
1198861
1198862
1198863
1198864
]
2
)
=[[
[
1198861
minus1198862
1198861+ 119886
4
2
minus1198863
1198861+ 119886
4
21198864
]]
]
(33)
However
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] =[[
[
11988711198861
1198872
1198861+ 119886
4
2
1198873
1198861+ 119886
4
211988741198864
]]
]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
=
[[[
[
11988711198861
1198872
1198862
1+ 119886
2
4
2
1198873
1198862
1+ 119886
2
4
211988741198862
4
]]]
]
(34)
It follows that 1198861
= 0 1198864
= 0 1198871= 1119886
1 119887
4= 1119886
4 119887
2((119886
1+
1198864)2) = minus119886
2 119887
3((119886
1+119886
4)2) = minus119886
3 119887
2((119886
2
1+119886
2
4)2) = minus119886
2((119886
1+
1198864)2) and 119887
3((119886
2
1+ 119886
2
4)2) = minus119886
3((119886
1+ 119886
4)2) From these
equations we get 1198872(119886
1minus 119886
4)2
= 0 and 1198873(119886
1minus 119886
4)2
= 0 so that(119887
2minus119887
3)(119886
1minus 119886
4)2
= 0Then for 1198861
= 1198864 we obtain 119887
2= 119887
3= 0
and hence 1198862= 119886
3= 0 Therefore
119866119868(119872
+
) = [119886 119887
119888 119886] 119909 = [
120572 0
0 120573] isin 119872
+
119886 = 0
120572 = 0 120573 = 0 120572 = 120573
(35)
The set 119866119868(119872
+
) is not open clearly [1 1
0 1] isin 119866
119868(119872
+
) forany 120598 gt 0
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[1 1
0 1] minus [
[
1 minus120598
41
0 1 +120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[
[
120598
40
0 minus120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=120598
2lt 120598
(36)
but [ 1minus1205984 1
0 1+1205984] notin 119866
119868(119872
+
)Now if 119860 = [
minus1 1
0 1] then 120582119868 minus 119860 = [
120582+1 minus1
0 120582minus1] notin 119866
119868(119872
+
)and so 120582 isin 120590
(119872+119868)(119860) for all 120582 isin C Thus 120590
(119872+119868)(119860) = C an
unbounded spectrum
Next we observe that the spectrum of an element withrespect to a unit is closed whenever the corresponding set ofinvertibles is open
Proposition 29 LetI be a quasi-Jordan normed algebra witha unit 119890 such that119866
119890(I) is openThen 120590
(I119890)(119909) is closed for all119909 isin I
Proof Define 119891 C rarr I by 119891(120582) = 120582119890 minus 119909 Since 119891 iscontinuous the inverse image of the open set 119866
119890(I) is open
in C and so its complement 120590(I119890)(119909) is closed
We conclude this paper with the following example of anonclosed spectrum
Example 30 Let 119872 and 119872+ be as in Example 28 Let 119864 =
[1 1
0 1] and 119860 = [
119886 0
0 119887] with 119886 = 119887 both different from 1 Then
119864 is a unit in 119872+ We show that 120590
(119872+119864)
(119860) = C 1 Forthis let us first investigate when can an element of the form119861 = [
120572 120573
0 120574] be invertible Assuming that 119861 is invertible we get
the existence of an element 119862 isin 119863+ such that
119862 ⊲ 119861 = 119864 + 119864⊲(119861)
119862 ⊲ 1198612
= 119861 + 119864⊲(119861) + 119864
⊲(119861
2
)
(37)
Abstract and Applied Analysis 11
From these equations we get 120572 = 0 120574 = 0 and
119862 =[[
[
1
120572120573∘
01
120574
]]
]
(38)
where 120573∘satisfies the following two equations
120573∘(120572 + 120574
2) = 1 minus 120573 +
120572 + 120574
2
120573∘(1205722
+ 1205742
2) =
120572 + 120574
2+
1205722
+ 1205742
2minus 120573
120572 + 120574
2
(39)
Multiplying the last equation by 2(120572 + 120574)(1205722
+ 1205742
) and thenusing the other equation we get
2 (1 minus 120573) + (120572 + 120574) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
+ (120572 + 120574) (40)
or equivalently
2 (1 minus 120573) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
(41)
this equation is satisfied for 120573 = 1 or 120572 = 120574 Hence thematrix119861 is invertible if and only if 119861 = [
120572 1
0 120574] or 119861 = [
120572 120573
0 120572
] for 120572 120574 isin
C 0 and 120573 isin CWe conclude that 119860
120582= 120582119864 minus 119860 = [
120582minus119886 120582
0 120582minus119887] is invertible
with respect to 119864 if and only if 120582 minus 119887 = 0 120582 minus 119886 = 0 and 120582 = 1that is119860
120582is invertible if 119886 = 120582 and 119887 = 120582 and 120582 = 1 Hence119860
120582
is invertible only if 120582 = 1 as we assumed that 119886 = 119887 and bothare not 1 So for all 120582 = 1 119860
120582notin 119866
119864(119872
+
) Thus 120590(119872+119864)
(119860) =
C 1 which is neither bounded nor closed
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This project was supported by King Saud University Dean-ship of Scientific Research College of Science ResearchCenter
References
[1] B Fritzsche ldquoSophus Lie a sketch of his life and workrdquo Journalof Lie Theory vol 9 no 1 pp 1ndash38 1999
[2] P Jordan ldquoUber eine klasse nichtassoziativer hyperkomplexeralgebrenrdquo Nachrichten von der Gesellschaft der Wissenschaftenzu Gottingen pp 569ndash575 1932
[3] K McCrimmon A Taste of Jordan Algebras UniversitextSpringer New York NY USA 2004
[4] N Jacobson Structure and Representations of Jordan Algebrasvol 39 of American Mathematical Society Colloquium Publica-tions American Mathematical Society Providence RI USA1968
[5] F Gursey and C-H Tze On the Role of Division Jordan andRelatedAlgebras in Particle PhysicsWorld Scientific River EdgeNJ USA 1996
[6] R Iordanescu Jordan Structures in Analysis Geometry andPhysics EdituraAcademiei Romane Bucharest Romania 2009
[7] H Upmeier Symmetric Banach Manifolds and Jordan Clowast-Alge-bras vol 104 of North-Holland Mathematics Studies North-Holland Amsterdam The Netherlands 1985
[8] I L Kantor ldquoClassification of irreducible transitive differentialgroupsrdquo Doklady Akademii Nauk SSSR vol 158 pp 1271ndash12741964
[9] M Koecher ldquoUber eine Gruppe von rationalen AbbildungenrdquoInventiones Mathematicae vol 3 pp 136ndash171 1967
[10] J Tits ldquoUne classe drsquoalgebres de Lie en relation avec les algebresde Jordanrdquo Indagationes Mathematicae vol 24 pp 530ndash5351962
[11] J-L Loday ldquoUne version non commutative des algebres de Lieles algebres de Leibnizrdquo LrsquoEnseignement Mathematique vol 39no 3-4 pp 269ndash293 1993
[12] J-L Loday and T Pirashvili ldquoUniversal enveloping algebras ofLeibniz algebras and (co)homologyrdquo Mathematische Annalenvol 296 no 1 pp 139ndash158 1993
[13] J-L Loday ldquoDialgebrasrdquo in Dialgebras and Related Operadsvol 1763 of Lecture Notes in Math pp 7ndash66 Springer BerlinGermany 2001
[14] R Velasquez and R Felipe ldquoQuasi-Jordan algebrasrdquo Communi-cations in Algebra vol 36 no 4 pp 1580ndash1602 2008
[15] R Velasquez and R Felipe ldquoOn K-B quasi-Jordan algebrasand their relation with Liebniz algebrasrdquo Comunicaciones delCIMAT No 1-10-1015-12-2010
[16] R Felipe ldquoAn analogue to functional analysis in dialgebrasrdquoInternational Mathematical Forum vol 2 no 21ndash24 pp 1069ndash1091 2007
[17] R Velasquez and R Felipe ldquoSplit dialgebras split quasi-Jordanalgebras and regular elementsrdquo Journal of Algebra and ItsApplications vol 8 no 2 pp 191ndash218 2009
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Complex AnalysisJournal of
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OptimizationJournal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Discrete Dynamics in Nature and Society
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Stochastic AnalysisInternational Journal of
2 Abstract and Applied Analysis
the commutative law is replaced by a quasicommutative iden-tity and a special form of the Jordan identity is retainedThesefacts indicate the significance of studying the quasi-Jordanalgebras within a few years time many mathematiciansincluding M K Kinyon M R Bremner L A Peresi J San-chez-Ortega and V Voronin have got their interests in thisnew area In [16] Felipe made an attempt to study dialge-bras from the functional analytic point of view
A Jordan Banach algebra is a real or complex Jordanalgebra with a complete norm sdot satisfying 119909119910 le 119909119910basics of Jordan Banach algebras may be seen in [7] In thispaper we initiate a study of the quasi-Jordan normed alge-bras The class of complete quasi-Jordan normed algebrascalled quasi-Jordan Banach algebras properly includes allJordan Banach algebras and hence all 119862lowast-algebras (cf [7])This study may provide a better mathematical foundationfor some important areas such as quantum mechanics Weare interested specially in extending as much as possiblethe theory of Jordan Banach algebras to the general settingof quasi-Jordan Banach algebras We investigate the notionsof invertibility and spectrum of elements in the setting ofunital quasi-JordanBanach algebras Among other results wedemonstrate that any quasi-Jordan Banach algebra I with anorm 1 unit can be given an equivalent norm that makes thealgebra I isometrically isomorphic to a norm closed rightideal of a unital split quasi-Jordan Banach algebra We showthat the set of invertible elements in a unital quasi-JordanBanach algebra generally is not open and that the spectrumofany element is nonempty but it may be neither bounded norclosed hence not compact Moreover if the spectrum of anelement in a complex unital split quasi-JordanBanach algebra(see below) is unbounded then it coincides with the wholecomplex plane and vice versa Some examples are also givenin the end
2 Quasi-Jordan Banach Algebras
Webegin by recalling some basics of the quasi-Jordan algebratheory from [14 17] A quasi-Jordan algebra is a vector spaceI over a field119870 of characteristic = 2 equipped with a bilinearmap ⊲ I times I rarr I called the quasi-Jordan productsatisfying 119909 ⊲ (119910 ⊲ 119911) = 119909 ⊲ (119911 ⊲ 119910) (right commutativity)and (119910 ⊲ 119909) ⊲ 119909
2
= (119910 ⊲ 1199092
) ⊲ 119909 (right Jordan identity)for all 119909 119910 119911 isin I Here 1199092 = 119909 ⊲ 119909 and 119909
119899
= 119909119899minus1
⊲ 119909
for 119899 ge 2 in the sequel we will see that 1199092 ⊲ 119909 may notcoincide with 119909 ⊲ 119909
2 Every quasi-Jordan algebra I includestwo important sets Iann
= the linear span of the elements119909 ⊲ 119910 minus 119910 ⊲ 119909 119908119894119905ℎ 119909 119910 isin I and 119885(I) = 119911 isin I 119909 ⊲
119911 = 0 forall119909 isin I respectively called the annihilator and thezero part of I It follows from the right commutativity thatIann
sube 119885(I) and that the quasi-Jordan algebra I is a Jordanalgebra if and only ifIann
= 0
Example 1 Let (119863 ⊣ ⊢) be a dialgebra over a field119870 of char-acteristic = 2 One can define another product⊲ 119863times119863 rarr 119863
by 119909 ⊲ 119910 = (12)(119909 ⊣ 119910 + 119910 ⊢ 119909) for all 119909 119910 isin 119863 whichsatisfies the identities 119909 ⊲ (119910 ⊲ 119911) = 119909 ⊲ (119911 ⊲ 119910) (119910 ⊲
119909) ⊲ 1199092
= (119910 ⊲ 1199092
) ⊲ 119909 and 1199092
⊲ (119909 ⊲ 119910) = 119909 ⊲
(1199092
⊲ 119910) however this quasi-Jordan product generally isnot commutativeTherefore (119863 ⊲) is a quasi-Jordan algebrawhich may not be a Jordan algebra (cf [14]) This quasi-Jordan algebra is denoted by 119863
+ called a plus quasi-Jordanalgebra Any quasi-Jordan algebra which is a homomorphicimage of a plus quasi-Jordan algebra is called special
We define a quasi-Jordan normed algebra as a quasi-Jor-dan algebra (I ⊲) over the field C of complex numbersendowed with a norm sdot satisfying 119909 ⊲ 119910 le 119909119910 for all119909 isin IThus the quasi-Jordan product ldquo⊲rdquo in a normed quasi-Jordan algebraI is continuous A quasi-Jordan normed alge-bra is called a quasi-Jordan Banach algebra if it is completeas a normed space
If there is an element 119890 in a quasi-Jordan algebra Isatisfying 119890 ⊲ 119909 = 119909 for all 119909 isin I called the left unit thenI becomes commutative and hence a Jordan algebra Due tothis fact we will consider only the right unit elements hence-forth unit in a quasi-Jordan algebra would mean a right unitAn element 119890 in a quasi-Jordan algebraI is called a unit if119909 ⊲
119890 = 119909 for all 119909 isin I A quasi-Jordan algebra may have many(right) units (cf [15 17]) If the dialgebra 119863 has a bar-unit 119890(ie 119909 ⊣ 119890 = 119909 = 119890⊢119909) then 119909 ⊲ 119890 = (12)(119909 ⊣ 119890 + 119890 ⊢ 119909) = 119909for all 119909 isin 119863 and hence 119890 is a unit of the plus quasi-Jordanalgebra119863
+For any quasi-Jordan algebraIwith unit 119890 we know from
[17] thatIann and119885(I) are two-sided ideals ofIIann= 119909 isin
I 119890 ⊲ 119909 = 0 = 119885(I) and 119880(I) = 119909 + 119890 119909 isin 119885(I)where 119880(I) denotes the set of all (right) units inI
One can always attach a (two-sided) unit to any Jordanalgebra by following the standard unitization process Thisunitization process no longer works for quasi-Jordan algebras(cf [17 pages 210-211]) Adding a unit to a quasi-Jordan alge-bra is yet an open problem In giving a partial solution to thisunitization problemVelasquez andFelipe [17] introduced thefollowing special class of quasi-Jordan algebras called splitquasi-Jordan algebras letI be a quasi-Jordan algebra and let119868 be an ideal inI such thatIann
sube 119868 sube 119885(I) We say thatI isa split quasi-Jordan algebra (more precisely I is split over 119868)if there exists a subalgebra 119869 ofI such thatI = 119869oplus119868 the directsum of 119869 and 119868 It is easily seen that such a subalgebra 119869 is aJordan algebra One can attach a unit to any split quasi-Jordanalgebra details of such a unitization process are given in [17]If the algebra I has a unit then Iann
= 119885(I) Thus a quasi-Jordan algebra I with unit is a split quasi-Jordan algebra ifand only if I = 119869 oplus 119885(I) for some subalgebra 119869 of I thealgebra 119869 is called the Jordan part of I In such a case eachelement 119909 isin I has a unique representation 119909 = 119909
119869+ 119909
119885with
119909119869isin 119869 and 119909
119885isin 119885(I) respectively called the Jordan part and
the zero part of 119909 Moreover ifI is split quasi-Jordan algebrawith unit 119890 then there exists a unique element 119890
119869isin 119869 which
acts as a unit of the quasi-Jordan algebra I and at the sametime a unit of the Jordan algebra 119869
Proposition 2 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan Banach algebra with unit 119890 isin 119869 Then the Jordan part119869 is a norm closed subalgebra of I and hence a unital JordanBanach algebra
Abstract and Applied Analysis 3
Proof From the above discussion it is clear that the Jordanpart 119869 is a unital Jordan normed algebra Next let 119909
119899 be any
fixed Cauchy sequence in 119869 Then the same 119909119899 is Cauchy
sequence also in the quasi-Jordan Banach algebraI since 119869 isa normed subspace ofI Hence 119909
119899rarr 119909 for some 119909 isin I
Now since each119909119899isin 119869 and since 119890 is the unit of the Jordan
algebra 119869 under the (restricted) product ldquo⊲rdquo we get 119890 ⊲ 119909119899=
119909119899for all 119899 So by the continuity of the product 119909
119899= 119890 ⊲
119909119899
rarr 119890 ⊲ 119909 Hence by the uniqueness of the limit of anyconvergent sequence in a normed space 119890 ⊲ 119909 = 119909 However119909 = 119890 ⊲ (119909
119869+ 119909
119885) = 119890 ⊲ 119909
119869+ 119890 ⊲ 119909
119885= 119890 ⊲ 119909
119869+ 0 = 119909
119869isin 119869
Thus the required result follows
We know from [17] that for any quasi-Jordan algebra II
1= (119909 119877
119910) 119909 119910 isin I equipped with the sum (119909 119877
119910) +
(119886 119877119887) = (119909 + 119886 119877
119910+119887) scalar multiplication 120582(119909 119877
119910) =
(120582119909 119877120582119910) and product (119909 119877
119910) ⊲ (119886 119877
119887) = (119909 ⊲ 119887 119877
119910⊲119887)
is a split quasi-Jordan algebra and the map 119909 997891rarr 120593(119909) =
(119909 119877119909) is an embedding of I into I
1 where 119877
119911stands for
the usual right multiplication operator on I It is easily seenthat 0 times 119877(I) is the Jordan part of I
1and the zero part
119885(I1) = I times 0 Moreover the embedding 120593 preserves the
units clearly (119886 119877119887) ⊲ (119890 119877
119890) = (119886 ⊲ 119890 119877
119887⊲119890) = (119886 119877
119887) for
all (119886 119877119887) isin I
1so that (119890 119877
119890) is a unit in I
1whenever 119890 is a
unit in I In fact (119909 119877119890) is a unit in I
1 for all 119909 isin I it may
be noted here that (0 119877119890) is the only unit in the Jordan part of
I1From [17] we also know that 119877(I) = 119877
119909 119909 isin I with
product ldquo∙rdquo defined by 119877119909∙ 119877
119910= 119877
119909⊲119910 for all 119909 119910 isin I is
a quasi-Jordan algebra MoreoverI1is the direct product of
the quasi-Jordan algebrasI and 119877(I)Indeed the split quasi-Jordan algebraI
1is a quasi-Jordan
Banach algebra with unit (119890 119877119890)wheneverI is a quasi-Jordan
Banach algebra with a norm 1 unit 119890 and that 120593(I) =
(119909 119877119909) 119909 isin I is a closed unital quasi-Jordan normed
subalgebra ofI1 To justify this claim we need the following
result
Proposition 3 Suppose I is a quasi-Jordan Banach algebrawith a norm 1 unit 119890Then the algebra119877(I) as above is a quasi-Jordan Banach algebra with unit 119877
119890of norm 1
Proof Clearly each 119877119909is a bounded linear operator with
119877119909 le 119909 Hence the usual operator norm is a norm on
the quasi-Jordan algebra 119877(I)with the quasi-Jordan product119877119909∙ 119877
119910= 119877
119909⊲119910 Further we observe that
10038171003817100381710038171003817119877119909∙ 119877
119910
10038171003817100381710038171003817=
10038171003817100381710038171003817119877119909⊲119910
10038171003817100381710038171003817
= sup119911isinI119911=1
10038171003817100381710038171003817119877119909⊲119910
(119911)10038171003817100381710038171003817
= sup119911isinI119911=1
1003817100381710038171003817119911 ⊲ (119909 ⊲ 119910)1003817100381710038171003817
= sup119911isinI119911=1
1003817100381710038171003817119911 ⊲ ((119890 ⊲ 119909) ⊲ 119910)1003817100381710038171003817
= sup119911isinI119911=1
10038171003817100381710038171003817119911 ⊲ (119877
119910(119877
119909(119890)))
10038171003817100381710038171003817
le sup119911isinI119911=1
11991110038171003817100381710038171003817119877119910(119877
119909(119890))
10038171003817100381710038171003817
=10038171003817100381710038171003817119877119910(119877
119909119890)10038171003817100381710038171003817le
1003817100381710038171003817119877119909
1003817100381710038171003817
10038171003817100381710038171003817119877119910
10038171003817100381710038171003817119890
=1003817100381710038171003817119877119909
1003817100381710038171003817
10038171003817100381710038171003817119877119910
10038171003817100381710038171003817
(1)
Alternately by exploiting the right commutativity of thequasi-Jordan product inI we get
10038171003817100381710038171003817119877119909∙ 119877
119910
10038171003817100381710038171003817= sup
119911isinI119911=1
1003817100381710038171003817119911 ⊲ (119909 ⊲ 119910)1003817100381710038171003817
= sup119911isinI119911=1
1003817100381710038171003817119911 ⊲ ((119909 ⊲ 119890) ⊲ (119910 ⊲ 119890))1003817100381710038171003817
= sup119911isinI119911=1
1003817100381710038171003817119911 ⊲ ((119909 ⊲ 119890) ⊲ (119890 ⊲ 119910))1003817100381710038171003817
= sup119911isinI119911=1
1003817100381710038171003817119911 ⊲ ((119890 ⊲ 119910) ⊲ (119909 ⊲ 119890))1003817100381710038171003817
= sup119911isinI119911=1
1003817100381710038171003817119911 ⊲ ((119890 ⊲ 119910) ⊲ (119890 ⊲ 119909))1003817100381710038171003817
= sup119911isinI119911=1
10038171003817100381710038171003817119911 ⊲ (119877
119910(119890) ⊲ 119877
119909(119890))
10038171003817100381710038171003817
le sup119911isinI119911=1
11991110038171003817100381710038171003817119877119910
10038171003817100381710038171003817
1003817100381710038171003817119877119909
1003817100381710038171003817 1198902
=1003817100381710038171003817119877119909
1003817100381710038171003817
10038171003817100381710038171003817119877119910
10038171003817100381710038171003817
(2)
Thus 119877(I) together with the operator norm is a quasi-Jordan normed algebra Moreover for any 119909 isin 119877(I) wehave 119877
119909∙ 119877
119890= 119877
119909⊲119890= 119877
119909and 1 = 119890 ge 119877
119890 =
sup0 = 119911isinI(119877119890
(119911)119911) ge 119877119890(119890)119890 = 1 that is 119877
119890is a
norm 1 (right) unit in 119877(I)Suppose 119877
119909119899
is any fixed Cauchy sequence in 119877(I)Then for any fixed 119886 isin I 119886 ⊲ 119909
119898minus 119886 ⊲ 119909
119899 = 119877
119909119898
(119886) minus
119877119909119899
(119886) = (119877119909119898
minus 119877119909119899
)(119886) le 119877119909119898
minus 119877119909119899
119886 rarr 0 as119898 119899 rarr infin and so 119886 ⊲ 119909
119899 is a Cauchy sequence in I
But I is complete Hence the sequence 119886 ⊲ 119909119899 for any
119886 isin I is convergent inI In particular the sequence 119890 ⊲ 119909119899
converges to some 119910 isin I Moreover the Cauchy sequence119877
119909119899
converges to 119877119910isin 119877(I) in the operator norm because
(119877119909119899
minus 119877119910)(119911) = 119911 ⊲ (119909
119899minus 119910) = 119911 ⊲ 119909
119899minus 119911 ⊲ 119910 = 119911 ⊲
(119909119899⊲ 119890) minus 119911 ⊲ 119910 = 119911 ⊲ (119890 ⊲ 119909
119899) minus 119911 ⊲ 119910 = 119911 ⊲ (119890 ⊲
119909119899minus 119910) le 119911119890 ⊲ 119909
119899minus 119910 rarr 0 as 119899 rarr infin for all 119911 isin I
Thus the quasi-Jordan normed algebra 119877(I) is complete
Now we show that the corresponding algebra I1is a
unital split quasi-Jordan Banach algebra
Proposition 4 Let I be a quasi-Jordan Banach algebra witha norm 1 unit 119890 and let I
1be as above Then I
1is the direct
product of the quasi-Jordan algebras I and 119877(I) MoreoverI
1equipped with the product norm (119909 119877
119910) = 119909 + 119877
119910
is a split quasi-Jordan Banach algebra with unit (119890 119877119890) and
120593(I) = (119909 119877119909) 119909 isin I is a closed right ideal in I
1in the
norm topology
4 Abstract and Applied Analysis
Proof For the first part see [17] Clearly (119909 119877119910) = 119909 +
119877119910 is a norm and it satisfies
10038171003817100381710038171003817(119909 119877
119910) ⊲ (119911 119877
119908)10038171003817100381710038171003817=
10038171003817100381710038171003817(119909 ⊲ 119908 119877
119910⊲119908)10038171003817100381710038171003817
=10038171003817100381710038171003817(119909 ⊲ 119908 119877
119910∙ 119877
119908)10038171003817100381710038171003817
=1003817100381710038171003817119877119908
(119909)1003817100381710038171003817 +
10038171003817100381710038171003817119877119910∙ 119877
119908
10038171003817100381710038171003817
le (119909 +10038171003817100381710038171003817119877119910
10038171003817100381710038171003817)1003817100381710038171003817119877119908
1003817100381710038171003817
le (119909 +10038171003817100381710038171003817119877119910
10038171003817100381710038171003817) (119911 +
1003817100381710038171003817119877119908
1003817100381710038171003817)
=10038171003817100381710038171003817(119909 119877
119910)10038171003817100381710038171003817
1003817100381710038171003817(119911 119877119908)1003817100381710038171003817
(3)
for all 119908 119909 119910 119911 isin I Keeping in view Proposition 3 wededuce that I
1being the product of complete spaces I and
119877(I) is complete in the product norm Thus I1is a split
quasi-Jordan Banach algebra with (right) unit (119890 119877119890)
Clearly 120593(I) is a subspace ofI1with (119909 119877
119909) ⊲ (119910 119877
119911) =
(119909 ⊲ 119911 119877119909⊲119911
) isin 120593(I) for all (119909 119877119909) isin 120593(I) (119910 119877
119911) isin I
1 so
that 120593(I) is a quasi-Jordan normed subalgebra with the unit(119890 119877
119890) included Further let (119909
119899 119877
119909119899
) be any fixed Cauchysequence in 120593(I) Then 119909
119898minus 119909
119899 le 119909
119898minus 119909
119899 + 119877
119909119898
minus
119877119909119899
= (119909119898
minus 119909119899 119877
119909119898
minus 119877119909119899
) = (119909119898 119877
119909119898
) minus (119909119899 119877
119909119899
) rarr
0 as 119898 119899 rarr infin so that 119909119899 is a Cauchy sequence in (the
complete space) I and hence it converges to some 119909 isin INow by using the fact 119877
119911le 119911 for all 119911 isin I (since
119890 is a norm 1 unit in I see above) we get the convergenceof arbitrarily fixed Cauchy sequence (119909
119899 119877
119909119899
) to (119909 119877119909) isin
120593(I) since (119909119899 119877
119909119899
)minus(119909 119877119909) = (119909
119899minus119909 119877
119909119899minus119909
) = 119909119899minus119909+
119877119909119899minus119909
le 2119909119899minus119909 rarr 0 as 119899 rarr infinThus 120593(I) being com-
plete is a closed right ideal inI1in the norm topology
Next we observe the isometry between I and 120593(I)
Proposition 5 Let I be a quasi-Jordan Banach algebra withnorm 1 unit Then there exists an equivalent norm that makesI isometrically isomorphic to a quasi-Jordan closed subalgebra120593(I) of the split quasi-Jordan Banach algebra I
1
Proof Clearly 119909∘
= 119909 + 119877119909 defines a norm on the
quasi-Jordan algebra I It follows that (I sdot ∘) is a quasi-
Jordan Banach algebraMoreoverI is isomorphic to the sub-algebra (119909 119877
119909) 119909 isin I of I
1under the isomorphism
120593 119909 997891rarr (119909 119877119909) as seen above Further we observe that 120593
is an isometry since 119909∘= 119909 + 119877
119909 = (119909 119877
119909) = 120593(119909)
Finally we note that the two norms sdot and sdot ∘on I are
equivalent since 119909 le 119909 + 119877119909 = 119909
∘= 119909 + 119877
119909 le
2119909
3 Invertible Elements
As in [17] an element 119909 in a quasi-Jordan algebra I is calledinvertible with respect to a unit 119890 isin I if there exists 119910 isin I
such that 119910 ⊲ 119909 = 119890 + (119890 ⊲ 119909 minus 119909) and 119910 ⊲ 1199092
= 119909 + (119890 ⊲
119909 minus 119909) + (119890 ⊲ 1199092
minus 1199092
) such an element 119910 is called an inverseof119909 with respect to 119890 Let 119890
⊲(119909) denote the element 119890 ⊲ 119909 minus 119909
Then119909 has an inverse119910with respect to 119890 hArr 119910 ⊲ 119909 = 119890+119890⊲(119909)
and 119910 ⊲ 1199092
= 119909 + 119890⊲(119909) + 119890
⊲(119909
2
)We know from the above discussion that the embedding
119909 997891rarr 120593(119909) = (119909 119877119909) of a quasi-Jordan algebraI into the split
quasi-Jordan algebra I1preserves the units The embedding
120593 also preserves the corresponding invertible elements if 119910 isan inverse of 119909 with respect to a unit 119890 in I then 119910 ⊲ 119909 =
119890 + (119890 ⊲ 119909 minus 119909) and 119910 ⊲ 1199092
= 119909 + (119890 ⊲ 119909 minus 119909) + (119890 ⊲ 1199092
minus 1199092
)Hence
(119910 119877119910) ⊲ (119909 119877
119909) = (119910 ⊲ 119909 119877
119910⊲119909)
= (119890 + (119890 ⊲ 119909 minus 119909) 119877119890+(119890⊲119909minus119909)
)
= (119890 + (119890 ⊲ 119909 minus 119909) 119877119890+ 119877
119890⊲119909minus 119877
119909)
= (119890 119877119890) + (119890 ⊲ 119909 119877
119890⊲119909) minus (119909 119877
119909)
= (119890 119877119890) + (119890 119877
119890) ⊲ (119909 119877
119909) minus (119909 119877
119909)
(119910 119877119910) ⊲ (119909 119877
119909)2
= (119910 ⊲ 1199092
119877119910⊲1199092)
= (119909 119877119909) + ((119890 ⊲ 119909 119877
119890⊲119909) minus (119909 119877
119909))
+ ((119890 ⊲ 1199092
119877119890⊲1199092) minus (119909
2
1198771199092))
= (119909 119877119909) + ((119890 119877
119890) ⊲ (119909 119877
119909) minus (119909 119877
119909))
+ ((119890 119877119890) ⊲ (119909 119877
119909)2
minus (119909 119877119909)2
)
(4)
Thus (119910 119877119910) is an inverse of (119909 119877
119909) with respect to the unit
(119890 119877119890) inI
1
Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebraThen 119877
119909(119910) = 119910 ⊲ 119909 = 119910 ⊲ (119909
119869+ 119909
119885) = 119910 ⊲ 119909
119869+ 119910 ⊲ 119909
119885=
119910 ⊲ 119909119869+ 0 = 119877
119909119869
(119910) for all 119909 119910 isin I Thus 119877119909= 119877
119909119869
for all119909 isin I
In this section we demonstrate that the set of invertibleelements with respect to a fixed unit in a quasi-JordanBanach algebra may not be open For this we proceed asfollows
Proposition 6 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebra with unit 119890 isin 119869 If 119909 isin I is invertible withrespect to 119890 then so is 120582119909 for all 120582 = 0
Proof Let119910 be an inverse of 119909 inIwith respect to 119890We showthat 1199101015840
= (1120582)119910119869+ 119910
119885is an inverse of 120582119909 with respect to 119890
Observe that
119910119869⊲ 119909 + 119910
119885⊲ 119909 = (119910
119869+ 119910
119885) ⊲ 119909
= 119910 ⊲ 119909 = 119890 + 119890⊲(119909)
= 119890 + 119890⊲(119909
119869+ 119909
119885) = 119890 + (minus119909
119885)
Abstract and Applied Analysis 5
119910119869⊲ 119909
2
+ 119910119885⊲ 119909
2
= 119910 ⊲ 1199092
= 119909 + 119890⊲(119909
119869+ 119909
119885) + 119890
⊲((119909
119869+ 119909
119885)2
)
= 119909119869+ (minus119909
119885⊲ 119909)
(5)
Hence by the uniqueness of the representation as sum of Jor-dan and zero parts we get 119910
119869⊲ 119909 = 119890 119910
119885⊲ 119909 = minus119909
119885 119910
119869⊲
1199092
= 119909119869 and 119910
119885⊲ 119909
2
= minus119909119885
⊲ 119909 Therefore 1199101015840
⊲ 120582119909 =
120582(((1120582)119910119869+ 119910
119885) ⊲ 119909) = 119910
119869⊲ 119909 + 120582119910
119885⊲ 119909 = 119890 minus 120582119909
119885=
119890 + 119890⊲(120582119909) and 119910
1015840
⊲ (120582119909)2
= 1205822
((1120582)119910119869⊲ 119909
2
+ 119910119885
⊲ 1199092
) =
120582119909119869minus 120582
2
119909119885⊲ 119909 = 120582119909 + 119890
⊲(120582119909) + 119890
⊲((120582119909)
2
) because 119890⊲(120582119909) =
119890 ⊲ (120582119909) minus 120582119909 = minus120582119909119885and 119890
⊲((120582119909)
2
) = 119890 ⊲ 1205822
1199092
minus 1205822
1199092
=
1205822
(119890⊲(119909
2
)) = minus1205822
119909119885⊲ 119909
Proposition 7 LetI be a quasi-Jordan normed algebrawith aunit 119890 Let119866
119890(I) = 119909 isin I 119909 is invertible with respect to 119890
be an open set and 119909 isin 119866119890(I) Then 119909 + 119911 isin 119866
119890(I) for all
119911 isin 119885(I)
Proof Suppose 119909 isin 119866119890(I) and 119911 isin 119885(I) If 119911 = 0 then 119909+119911 =
119909 isin 119866119890(I) Next suppose 119911 = 0 Since 119866
119890(I) is an open set
there exists 120598 gt 0 such that 119886 isin 119866119890(I) whenever 119909 minus 119886 lt 120598
Hence 119909 + 119911∘isin 119866
119890(I) with 119911
∘= (1205982119911)119911
Let 119910 and 119910∘be inverses of 119909 and 119909 + 119911
∘with respect to 119890
respectively Then
119910∘⊲ (119909 + 119911
∘) = 119890 + 119890
⊲(119909 + 119911
∘)
= 119890 + 119890⊲(119909) minus 119911
∘
= 119910 ⊲ 119909 minus 119911∘
119910∘⊲ (119909 + 119911
∘)2
= 119909 + 119911∘+ 119890
⊲(119909 + 119911
∘) + 119890
⊲((119909 + 119911
∘)2
)
= 119909 + 119890⊲(119909) + 119890
⊲(119909
2
) minus 119911∘⊲ 119909
= 119910 ⊲ 1199092
minus 119911∘⊲ 119909
(6)
Hence by setting 1199101= 119910 + 120572(119910
∘minus 119910) with 120572 = 2119911120598 we see
that1199101⊲ (119909 + 119911) = (119910 + 120572 (119910
∘minus 119910)) ⊲ 119909
= 119910 ⊲ 119909 + 120572 (119910∘⊲ (119909 + 119911
∘) minus 119910 ⊲ 119909)
= 119890 + 119890⊲(119909) + 120572 (minus119911
∘)
= 119890 + 119890⊲(119909) minus 119911
= 119890 + 119890⊲(119909 + 119911)
1199101⊲ (119909 + 119911)
2
= (119910 + 120572 (119910∘minus 119910)) ⊲ 119909
2
= 119910 ⊲ 1199092
+ 120572 (119910∘⊲ (119909 + 119911
∘)2
minus 119910 ⊲ 1199092
)
= 119909 + 119890⊲(119909) + 119890
⊲(119909
2
) minus 120572 (119911∘⊲ 119909)
= 119909 + 119890⊲(119909) + 119890
⊲(119909
2
) minus 119911 ⊲ 119909
= (119909 + 119911) + 119890⊲(119909 + 119911) + 119890
⊲((119909 + 119911)
2
)
(7)
since 119911 isin 119885(I) Thus 1199101is an inverse of 119909 + 119911 with respect to
the unit 119890
Corollary 8 Under the hypothesis of Proposition 7 120582119890 minus 119909 isin
119866119890(I) implies 120582119890 minus (119909 + 119911) isin 119866
119890(I) for all 119911 isin 119885(I)
Proposition 9 LetI = 119869 oplus 119885(I) be a split quasi-Jordan nor-med algebra and let 119890 isin 119869 be a unit inI such that the set119866
119890(I)
is open Then 119909 ⊲ 1199092
= 1199092
⊲ 119909 for all 119909 isin I
Proof Of course the element 119890 is the unit of the Jordanalgebra 119869 Then for any fixed element 119886 isin 119869 there exists120582 isin C such that 120582119890 minus 119886 is invertible in 119869 otherwise wewould get the negation of the well-known fact that spectrumof an element of a unital Jordan algebra is bounded Thatis there exists 119910 isin 119869 such that 119910 ⊲ (120582119890 minus 119886) = 119890 and119910 ⊲ (120582119890minus119886)
2
= (120582119890minus119886) However 119890⊲(119886) = 0 = 119890
⊲(119886
2
) Hence119910 is an inverse of 119886 in the quasi-Jordan algebraIwith respectto the unit 119890 By Corollary 8 120582119890 minus (119886 + 119911) is also invertiblefor any 119911 isin 119885(I) This in turn gives the existence of 119887 isin I
satisfying 119887 ⊲ (120582119890 minus (119886 + 119911)) = 119890 + 119911 and 119887 ⊲ (120582119890 minus (119886 + 119911))2
=
(120582119890minus119886)+119911 ⊲ (120582119890minus119886) Multiplying the first equation from theright by (120582119890 minus (119886 + 119911))
2 the second equation by (120582119890minus (119886+119911))and using the right Jordan identity we get
(119890 + 119911) ⊲ (120582119890 minus (119886 + 119911))2
= ((120582119890 minus 119886) + 119911 ⊲ (120582119890 minus 119886)) ⊲ (120582119890 minus (119886 + 119911))
(8)
Hence
(119890 + 119911) ⊲ (120582119890 minus 119886)2
= ((120582119890 minus 119886) + 119911 ⊲ (120582119890 minus 119886)) ⊲ (120582119890 minus 119886)
(9)
so that
119890 ⊲ (120582119890 minus 119886)2
+ 1205822
119911 minus 2120582 (119911 ⊲ 119886) + 119911 ⊲ 1198862
= (120582119890 minus 119886)2
+ 1205822
119911 minus 2120582119911 ⊲ 119886 + (119911 ⊲ 119886) ⊲ 119886
(10)
This last equation reduces to 119890 ⊲ (120582119890 minus 119886)2
= (120582119890 minus 119886)2 since
(120582119890 minus 119886)2
isin 119869 and 119890 is the unit of 119869 Hence 119911 ⊲ 1198862
= (119911 ⊲
119886) ⊲ 119886 for all 119886 isin 119869 and 119911 isin 119885(I) Now for any 119909 isin I thelast equation with 119886 = 119909
119869and 119911 = 119909
119885gives 119909
119885⊲ 119909
2
119869= (119909
119885⊲
119909119869) ⊲ 119909
119869Thus 119909 ⊲ 119909
2
= 119909119869⊲ 119909
2
119869+119909
119885⊲ 119909
2
119869= 119909
2
119869⊲ 119909
119869+(119909
119885⊲
119909119869) ⊲ 119909
119869= 119909
2
⊲ 119909 for all 119909 isin I
Corollary 10 If a unital split quasi-Jordan algebra has anelement 119909 with 119909 ⊲ 119909
2
= 1199092
⊲ 119909 then the set of invertibleelements with respect to the unit of the Jordan part is not open
In the sequel we will show the existence of a unital splitquasi-Jordan Banach algebra with elements 119909 such that 119909 ⊲
1199092
= 1199092
⊲ 119909 Thus the above result establishes that the set ofinvertible elements with respect to a fixed unit in a quasi-Jordan Banach algebra may not be open
6 Abstract and Applied Analysis
4 The Spectrum of Elements in a UnitalQuasi-Jordan Algebra
As usual we define the spectrum of an element 119909 in aunital quasi-Jordan algebra (I 119890) denoted by 120590
(I119890)(119909) tobe the collection of all complex numbers 120582 for which 120582119890 minus
119909 is not invertible Thus 120590(I119890)(119909) = 120582 isin C 120582119890 minus
119909 is not invertible Here the subscript 119890 indicates that theinvariability depends on the choice of unit 119890 which generallyis not unique
Proposition 11 Let I be a unital quasi-Jordan algebra Then120590(I119890)(119890
1015840
) = 1 for all 119890 1198901015840 isin 119880(I)
Proof Let 119890 1198901015840 isin 119880(I) Then for any 120582 = 1 119910 = (1(120582minus1))1198901015840
is an inverse of 120582119890 minus 1198901015840 with respect to the unit 119890 because 119910 ⊲
(120582119890 minus 1198901015840
) = (1(120582 minus 1))1198901015840
⊲ (120582119890 minus 1198901015840
) = 119890 + 119890⊲(120582119890 minus 119890
1015840
) and119910 ⊲ (120582119890 minus 119890
1015840
)2
= (120582119890 minus 1198901015840
) + 119890⊲(120582119890 minus 119890
1015840
) + 119890⊲((120582119890 minus 119890
1015840
)2
)
Proposition 12 Let I be a quasi-Jordan algebra with unit 119890Then 120590
(I119890)(119911) = 0 for all 119911 isin 119885(I)
Proof For any fixed 119911 isin 119885(I) and nonzero scalar 120582 thevector 119910 = (1120582)(119890 + 119911) satisfies 119910 ⊲ (120582119890 minus 119911) = 120582119910 = 119890 + 119911 =
119890+ 119890⊲(120582119890minus 119911) and 119910 ⊲ (120582119890 minus 119911)
2
= 1205822
119910 = 120582119890+120582119911 = (120582119890minus119911) +
119890⊲(120582119890 minus 119911) + 119890
⊲((120582119890 minus 119911)
2
) So (1120582)(119890 + 119911) is an inverse of120582119890 minus 119911 with respect to 119890 This means 120582 notin 120590
(I119890)(119911) for all 120582 = 0However the zero vector is not invertible Thus 120590
(I119890)(119911) =
0
Proposition 13 Let I = 119869 oplus 119885(I) be a split quasi-Jordanalgebra with unit 119890 isin 119869 Then 120590
(I119890)(119901) sube 0 1 for all idempo-tents 119901 isin I (ie 1199012
= 119901)
Proof Let 119901 be any fixed idempotent inI Since 119901 isin I 119901 hasa unique representation 119901 = 119901
119869+ 119901
119885with 119901
119869isin 119869 and 119901
119885isin
119885(I) Clearly 1199012
119869+ 119901
119885⊲ 119901
119869= 119901
2
= 119901 = 119901119869+ 119901
119885 Then
by uniqueness of the representation in the split quasi-Jordanalgebra I 119901
119885⊲ 119901
119869= 119901
119885and 119901
2
119869= 119901
119869 this means 119901
119869is an
idempotent in the Jordan algebra 119869 Hence 120590(119869119890)
(119901119869) sube 0 1
Thus 119901120582= 120582119890 minus 119901
119869is invertible in 119869 with the unique inverse
119901minus1
120582 for all 120582 notin 0 1We show that 119910 = 119901
minus1
120582+ (1(120582 minus 1))119901
119885is an inverse of
120582119890 minus 119901 in I with respect to the unit 119890 for this we note that120582119890 minus 119901 = 119901
120582minus 119901
119885 (119901
120582minus 119901
119885)2
= 1199012
120582minus 119901
119885⊲ 119901
120582 119890
⊲(120582119890 minus
119901) = 119890⊲(119901
120582minus 119901
119885) = 119890 ⊲ (119901
120582minus 119901
119885) minus (119901
120582minus 119901
119885) = 119901
119885 and
119890⊲((120582119890 minus 119901)
2
) = 119890Δ((119901
120582minus 119901
119885)2
) = 119890⊲(119901
2
120582minus 119901
119885⊲ 119901
120582) = 119901
119885⊲
119901120582= 119901
119885⊲ (120582119890minus119901
119869) = 120582119901
119885minus119901
119885⊲ 119901
119869= 120582119901
119885minus119901
119885= (120582minus1)119901
119885
Hence 119910 ⊲ (120582119890 minus 119901) = 119910 ⊲ (119901120582minus 119901
119885) = 119910 ⊲ 119901
120582= (119901
minus1
120582+
(1(120582 minus 1))119901119885) ⊲ 119901
120582= 119901
minus1
120582⊲ 119901
120582+ (1(120582 minus 1))119901
119885⊲ 119901
120582= 119890 +
(1(120582 minus 1))119901119885
⊲ (120582119890 minus 119901119869) = 119890 + (1(120582 minus 1))(120582119901
119885minus 119901
119885⊲
119901119869) = 119890 + (1(120582 minus 1))(120582119901
119885minus 119901
119885) = 119890 + 119901
119885= 119890 + 119890
⊲(120582119890 minus 119901)
and 119910 ⊲ (120582119890 minus 119901)2
= 119910 ⊲ (1199012
120582minus 119901
119885⊲ 119901
120582) = 119910 ⊲ 119901
2
120582=
(119901minus1
120582+(1(120582minus1))119901
119885) ⊲ 119901
2
120582= 119901
120582+(1(120582minus1))119901
119885⊲ (120582119890minus119901
119869)2
=
119901120582+(1(120582minus1))119901
119885⊲ (120582
2
119890minus2120582119901119869+119901
2
119869) = 119901
120582+(1(120582minus1))119901
119885⊲
(1205822
119890 minus 2120582119901119869+ 119901
119869) = 119901
120582+ (1(120582 minus 1))(120582
2
119901119885
⊲ 119890 minus 2120582119901119885
⊲
119901119869+ 119901
119885⊲ 119901
119869) = 119901
120582+ (1(120582 minus 1))(120582
2
119901119885
minus 2120582119901119885
+ 119901119885) =
119901120582+ (1(120582 minus 1))(120582 minus 1)
2
119901119885= 120582119890 minus119901
119869+ (120582 minus 1)119901
119885= (120582119890 minus119901) +
119890⊲(120582119890 minus 119901) + 119890
⊲((120582119890 minus 119901)
2
)
As mentioned in Section 2 if I is a quasi-Jordan algebrawith a unit 119890 then the set 119890 + 119911 119911 isin 119885(I) coincides withthe set 119880(I) of all units inI
Proposition 14 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebrawith unit 119890 isin 119869 and119909 isin I invertiblewith respectto some 1198901015840 isin 119880(I)Then119909
119869is invertible with respect to the unit
119890 in the Jordan algebra 119869
Proof Clearly 119890 ⊲ 119909 minus 119909 = minus119909119885and 119890 ⊲ 119909
2
minus 1199092
= minus119909119885⊲ 119909
119869
Since 1198901015840
isin 119880(I) we have 1198901015840
= 119890 + 119911 for some 119911 isin 119885(I)Hence the invertibility of 119909 in I with respect to the unit 1198901015840gives the existence of 119910 isin I such that 119910
119869⊲ 119909
119869+ 119910
119885⊲ 119909
119869=
119910 ⊲ 119909119869= 119910 ⊲ 119909 = 119890 + 119911 + (119890 + 119911) ⊲ 119909 minus 119909 = 119890 + 119911 + 119911 ⊲ 119909 minus 119909
119885
and 119910119869⊲ 119909
2
119869+ 119910
119885⊲ 119909
2
119869= 119910 ⊲ 119909
2
119869= 119910 ⊲ 119909
2
= (119890 + 119911) ⊲
119909 + (119890 + 119911) ⊲ 1199092
minus 1199092
= 119909119869+ 119911 ⊲ 119909 + 119911 ⊲ 119909
2
minus 119909119885⊲ 119909
119869 So by
the uniqueness of the representations in the split algebra Iwe get 119910
119869⊲ 119909
119869= 119890 and 119910
119869⊲ 119909
2
119869= 119909
119869 Thus 119910
119869is the inverse
of 119909119869in the Jordan algebra 119869
Next we observe that the spectrum of 119909 in a unital splitquasi-Jordan algebra with respect to any unit includes thespectrum of 119909
119869in the Jordan part 119869
Corollary 15 LetI = 119869 oplus119885(I) be a unital split quasi-Jordanalgebra with unit 119890 isin 119869 and let 119909 isin I Then 120590
(119869119890)(119909
119869) sube
120590(I1198901015840)(119909) for all 1198901015840 isin 119880(I)
Proof Let 1198901015840 = 119890 + 119911 with 119911 isin 119885(I) and let 120582 notin 120590(I1198901015840)(119909)
Then 1205821198901015840 minus 119909 is invertible in I with respect to the unit 1198901015840Hence its Jordanpart120582119890minus119909
119869is invertible in the Jordan algebra
119869 by Proposition 14 Thus 120582 notin 120590(119869119890)
(119909119869)
It is well known that the spectrum of any element ina unital Jordan Banach algebra is nonempty (cf [7]) Thistogether with Proposition 2 and Corollary 15 gives thefollowing result
Corollary 16 The spectrum of any element in a unital splitquasi-Jordan Banach algebra is nonempty
The next result extends Corollary 16 to any quasi-JordanBanach algebra with a norm 1 unit
Proposition 17 The spectrum of any element in a quasi-Jordan Banach algebra with a norm 1 unit is nonempty
Proof Let I be a unital quasi-Jordan Banach algebra witha norm 1 unit 119890 From Section 2 we know that the map120593 119909 997891rarr 120593(119909) = (119909 119877
119909) embeds I into the unital split
quasi-Jordan Banach algebra I1
= (119909 119877119910) 119909 119910 isin I
equipped with the sum (119909 119877119910) + (119886 119877
119887) = (119909 + 119886 119877
119910+119887) sca-
lar multiplication 120582(119909 119877119910) = (120582119909 119877
120582119910) and product (119909
119877119910) ⊲ (119886 119877
119887) = (119909 ⊲ 119887 119877
119910⊲119887) and the image 120593(I) is a
norm closed right ideal isomorphic to I with norm 1 unit(119890 119877
119890) Moreover it is seen in Section 3 that the embedding
Abstract and Applied Analysis 7
120593 also preserves the corresponding invertible elements thatis (119910 119877
119910) is an inverse of (119909 119877
119909) with respect to the unit
(119890 119877119890) in I
1whenever 119910 is an inverse of 119909 with respect
to a unit 119890 in I Hence by Corollary 16 it follows that120601 = 120590
(I1(119890119877119890))(119909 119877
119909) sube 120590
(120593(I)(119890119877119890))(119909 119877
119909) = 120590
(I119890)(119909) for all119909 isin I
Proposition 18 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebra with unit 119890 isin 119869 and let 119909 isin 119869 Then 120590
(119869119890)(119909) =
120590(I119890)(119909)
Proof By Corollary 15 120590(119869119890)
(119909) sube 120590(I119890)(119909) For the reverse
inclusion let 120582 notin 120590(119869119890)
(119909) then 120582119890 minus 119909 is invertible in 119869 thatis there exists 119910 isin 119869 such that 119910 ⊲ (120582119890 minus 119909) = 119890 and 119910 ⊲
(120582119890 minus 119909)2
= 120582119890 minus 119909 However 119890⊲(120582119890 minus 119909) = 0 = 119890
⊲((120582119890 minus 119909)
2
)Hence 119910 is an inverse of 120582119890 minus 119909 in I with respect to the unit119890 that is 120582 notin 120590
(I119890)(119909) Thus 120590(I119890)(119909) sube 120590
(119869119890)(119909)
Proposition 19 Let I be a unital quasi-Jordan normedalgebra and let 119890 be a unit inI for which 119866
119890(I) is open Then
120590(I119890)(119909) = 120590
(I119890)(119909 + 119911) for all 119911 isin 119885(I)
Proof By Corollary 8 120582119890 minus 119909 is invertible if and only if 120582119890 minus
(119909 + 119911) is invertible for all 119911 isin 119885(I) Thus 120582 notin 120590(I119890)(119909) if
and only if 120582 notin 120590(I119890)(119909 + 119911) for all 119911 isin 119885(I)
Corollary 20 LetI = 119869oplus119885(I) be a unital split quasi-Jordannormed algebra and let 119890 be a unit in I such that 119866
119890(I) is
open Then 120590(I119890)(119909) = 120590
(I119890)(119909119869) for all 119909 = 119909119869+ 119909
119885isin I
Further if the unit 119890 isin 119869 then120590(I119890)(119909) = 120590
(I119890)(119909119869) = 120590(119869119890)
(119909119869)
Lemma 21 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordanalgebra with a unit 119890 isin 119869 and let 119909 isin I be invertible withrespect to 119890 Then 119909
119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
Proof As 119909 is invertible there exists 119910 isin I such that 119910 ⊲ 119909 =
119890 + (119890 ⊲ 119909 minus 119909) = 119890 minus 119909119885and 119910 ⊲ 119909
2
= 119909 + (119890 ⊲ 119909 minus 119909) + (119890 ⊲
1199092
minus 1199092
) = 119909119869minus 119909
119885⊲ 119909 Hence by the uniqueness of the
representation in a split quasi-Jordan algebra we obtain
119910119885⊲ 119909 = minus119909
119885 (11)
119910119885⊲ 119909
2
= minus119909119885⊲ 119909 (12)
Thus
119909119885⊲ 119909
2
= minus (119910119885⊲ 119909) ⊲ 119909
2
(by (11))
= minus (119910119885⊲ 119909
2
) ⊲ 119909 (by the right Jordan identity)
= (119909119885⊲ 119909) ⊲ 119909 (by (12))
(13)
Proposition 22 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebra with a unit 119890 isin 119869 119909 = 119909
119869+ 119909
119885isin I satisfies
120590(I119890)(119909) =C Then
(1) 119909119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
(2) 1199092 ⊲ 119909 = 119909 ⊲ 1199092
Proof (1) Let 120582 notin 120590(I119890)(119909) Then 120582119890 minus 119909 = (120582119890 minus 119909
119869) + (minus119909
119885)
is invertible with respect to 119890 By Lemma 21 we have
119909119885⊲ (120582119890 minus 119909)
2
= (119909119885⊲ (120582119890 minus 119909)) ⊲ (120582119890 minus 119909) (14)
since the zero part of 120582119890 minus 119909 is minus119909119885 However
119909119885⊲ (120582119890 minus 119909)
2
= 119909119885⊲ (120582
2
119890 minus 120582119890 ⊲ 119909 minus 120582119909 ⊲ 119890 + 1199092
)
= 1205822
(119909119885⊲ 119890) minus 120582 (119909
119885⊲ (119890 ⊲ 119909))
minus 120582 (119909119885⊲ (119909 ⊲ 119890)) + 119909
119885⊲ 119909
2
= 1205822
119909119885minus 2120582119909
119885⊲ 119909
+ 119909119885⊲ 119909
2
(by the right commutativity of ⊲)
(119909119885⊲ (120582119890 minus 119909)) ⊲ (120582119890 minus 119909)
= (120582 (119909119885⊲ 119890) minus 119909
119885⊲ 119909) ⊲ (120582119890 minus 119909)
= (120582119909119885minus 119909
119885⊲ 119909) ⊲ (120582119890 minus 119909)
= 1205822
(119909119885⊲ 119890) minus 120582 (119909
119885⊲ 119909) ⊲ 119890
minus 120582119909119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
= 1205822
119909119885minus 2120582119909
119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
(15)
Therefore (14) becomes
1205822
119909119885minus 2120582119909
119885⊲ 119909 + 119909
119885⊲ 119909
2
= 1205822
119909119885minus 2120582119909
119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
(16)
which after simplification reduces to the required equation119909119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
(2) Since 1199092 = (119909119869+ 119909
119885)2
= 1199092
119869+ 119909
119885⊲ 119909
119869 we have
1199092
⊲ 119909 = (1199092
119869+ 119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909
119869+ (119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909
119869+ 119909
119885⊲ 119909
2
(17)
by the part (1) But 1199092119869⊲ 119909
119869= 119909
119869⊲ 119909
2
119869since 119909
119869is in the Jor-
dan algebra 119869 Therefore 1199092 ⊲ 119909 = 1199092
119869⊲ 119909
119869+ 119909
119885⊲ 119909
2
= 119909119869⊲
1199092
119869+ 119909
119885⊲ 119909
2
= 119909 ⊲ 1199092
Remark 23 In any quasi-Jordan algebra if an element 119909
satisfies 1199093
= 1199092
⊲ 119909 = 119909 ⊲ 1199092 then 119909
119899
⊲ 1199092
= 119909119899+2 for
all positive integers 119899 For this suppose 119909 satisfies 1199092 ⊲ 119909 =
119909 ⊲ 1199092 and 119909
119898
⊲ 1199092
= 119909119898+2 for any fixed119898 ge 1 Then 119909
119898+1
⊲
1199092
= (119909119898
⊲ 119909) ⊲ 1199092
= (119909119898
⊲ 1199092
) ⊲ 119909 (by the right Jor-dan identity) = 119909
119898+2
⊲ 119909 = 119909119898+3
Proposition 24 Let I be a unital quasi-Jordan Banachalgebra with unit 119890 and let 119909 isin I satisfy (119890 minus 119909) ⊲ (119890 minus 119909)
2
=
(119890 minus 119909)2
⊲ (119890 minus 119909) Then 119909 isin 119866119890(I) whenever 119890 minus 119909 lt 1
8 Abstract and Applied Analysis
Proof First note that 119890minus119909 lt 1 gives (119890 minus 119909)119899
le 119890 minus 119909119899
lt
1 for all 119899 = 1 2 3 Hence the infinite geometric series119890 +sum
infin
119899=1(119890 minus119909)
119899 converges absolutely to some element 119910 isin IWe show that the geometric series sum 119910 is an inverse of 119909with respect to the unit 119890 For any fixed positive integer 119899 let119910119899= 119890+sum
119899
119896=1(119890minus119909)
119896Then the sequence 119910119899 of partial sums
converges to 119910 By setting 119908 = 119890 minus 119909 we get
119910119899⊲ 119909 = (119890 +
119899
sum
119896=1
119908119896
) ⊲ (119890 minus 119908)
= 119890 +
119899
sum
119896=1
119908119896
minus (119890 ⊲ 119908 +
119899+1
sum
119896=2
119908119896
)
= 119890 + 119908 minus 119890 ⊲ 119908 minus 119908119899+1
= 119890 + (119890 minus 119909) minus 119890 ⊲ (119890 minus 119909) minus (119890 minus 119909)119899+1
= 119890 + (119890 minus 119909 minus 119890 + 119890 ⊲ 119909) minus (119890 minus 119909)119899+1
= 119890 + (119890 ⊲ 119909 minus 119909) minus (119890 minus 119909)119899+1
(18)
Thus by allowing 119899 rarr infin we obtain 119909 ⊲ 119910 = 119890 + (119890 ⊲
119909 minus 119909) = 119890 + 119890⊲(119909) since 119890 minus 119909 lt 1
Next by Remark 23 we have
119910119899⊲ 119909
2
= 119910119899⊲ (119890 minus 119908)
2
= 119910119899⊲ (119890 minus 119890 ⊲ 119908 minus 119908 ⊲ 119890 + 119908
2
)
= 119910119899⊲ (119890 minus 2119908 + 119908
2
)
= 119910119899⊲ 119890 minus 2119910
119899⊲ 119908 + 119910
119899⊲ 119908
2
= 119910119899minus 2119910
119899⊲ 119908 + 119910
119899⊲ 119908
2
= (119890 +
119899
sum
119896=1
119908119896
) minus 2(119890 +
119899
sum
119896=1
119908119896
) ⊲ 119908
+ (119890 +
119899
sum
119896=1
119908119896
) ⊲ 1199082
= (119890 +
119899
sum
119896=1
119908119896
) minus 2(119890 ⊲ 119908 +
119899+1
sum
119896=2
119908119896
)
+ (119890 ⊲ 1199082
+
119899+2
sum
119896=3
119908119896
)
= 119890 + 119908 minus 2119890 ⊲ 119908 minus 1199082
+ 119890 ⊲ 1199082
minus 119908119899+1
+ 119908119899+2
(19)
Taking the limit as 119899 rarr infin we get
119910 ⊲ 1199092
= 119890 + 119908 minus 2119890 ⊲ 119908 minus 1199082
+ 119890 ⊲ 1199082
= 119890 + (119890 minus 119909) minus 2119890 ⊲ (119890 minus 119909) minus (119890 minus 119909)2
+ 119890 ⊲ (119890 minus 119909)2
= 2119890 minus 119909 minus 2119890 + 2119890 ⊲ 119909 minus 119890 + 119890 ⊲ 119909 + 119909 minus 1199092
+ 119890 minus 2119890 ⊲ 119909 + 119890 ⊲ 1199092
= 119909 + (119890 ⊲ 119909 minus 119909) + (119890 ⊲ 1199092
minus 1199092
)
(20)
Proposition 25 Let I be a unital split quasi-Jordan Banachalgebra with unit 119890 If 119909 isin I with (119890 minus 119909) ⊲ (119890 minus 119909)
2
=
(119890 minus 119909)2
⊲ (119890 minus 119909) then |120582| le 119909 for all 120582 isin 120590(I119890)(119909)
Proof If 120582 isin 120590(I119890)(119909) with 120582 = 0 then the noninvertibility of
120582119890 minus 119909 means the noninvertibility of 119890 minus (1120582)119909 with respectto the unit 119890 However by Proposition 24 119890 minus (1120582)119909must beinvertible with respect to the unit 119890 whenever (1|120582|)119909 lt 1It follows that |120582| le 119909 for all 120582 isin 120590
(I119890)(119909)
5 Unbounded and Nonclosed Spectrum
In this section we show that the spectrum of an element ina split quasi-Jordan Banach algebra may be neither boundednor closed and hence not compactThe following result givesa couple of characterizations of the unbounded spectrum ofan element in a split quasi-Jordan Banach algebra
Proposition 26 Let 119909 be an element of a unital split quasi-Jordan Banach algebra I = 119869 oplus 119885(I) with a unit 119890 isin 119869 Thenthe following statements are equivalent
(1) 120590(I119890)(119909) =C
(2) 1199092 ⊲ 119909 = 119909 ⊲ 1199092
(3) |120582| le 119909 for all 120582 isin 120590(I119890)(119909)
Proof (1rArr2) See Proposition 22(2rArr3) Suppose 1199092 ⊲ 119909 = 119909 ⊲ 119909
2 Then
(119890 minus 119909)2
⊲ (119890 minus 119909) = (119890 minus 119890 ⊲ 119909 minus 119909 + 1199092
) ⊲ (119890 minus 119909)
= 119890 minus 2119890 ⊲ 119909 minus 119909 + 21199092
+ (119890 ⊲ 119909) ⊲ 119909 minus 1199092
⊲ 119909
= 119890 minus 2119909119869minus 119909 + 2119909
2
+ 119890 ⊲ 1199092
minus 1199093
(119890 minus 119909) ⊲ (119890 minus 119909)2
= (119890 minus 119909) ⊲ (119890 minus 2119909 + 1199092
)
= 119890 minus 2 (119890 ⊲ 119909) + 119890 ⊲ 1199092
minus 119909
+ 21199092
minus 119909 ⊲ 1199092
= 119890 minus 2119909119869+ 119890 ⊲ 119909
2
minus 119909 + 21199092
minus 1199093
(21)
Abstract and Applied Analysis 9
From (21) we get
(119890 minus 119909)2
⊲ (119890 minus 119909) = (119890 minus 119909) ⊲ (119890 minus 119909)2
(22)
Hence |120582| le 119909 for all 120582 isin 120590(I119890)(119909) by Proposition 25
(3rArr1) Immediate
Remark 27 There do exist unital split quasi-Jordan algebrascontaining elements that have the spectrum with respect tothe unit of the Jordan part equal to the whole ofC and henceunbounded To justify this claim we proceed as follows
Let 119860 be a unital associative algebra and let 119872 be an 119860-bimodule Let 119891 119872 rarr 119860 be an 119860-bimodule map (ie anadditive map satisfying 119891(119886119909) = 119886119891(119909) and 119891(119909119886) = 119891(119909)119886for all 119886 isin 119860 119909 isin 119872) Then one can put a dialgebra structureon 119872 as follows 119909 ⊣ 119910 = 119909119891(119910) and 119909⊢119910 = 119891(119909)119910 (cf [13Example 22(d)]) Hence119872+ is a quasi-Jordan algebra underthe quasi-Jordan product ldquo⊲rdquo given by119909 ⊲ 119910 = (12)(119909119891(119910)+
119891(119910)119909) Further for any 119909 isin 119872 we observe that
1199092
⊲ 119909 =1
2(119909119891 (119909) + 119891 (119909) 119909) ⊲ 119909
=1
4((119909119891 (119909) + 119891 (119909) 119909) 119891 (119909)
+119891 (119909) (119909119891 (119909) + 119891 (119909) 119909))
=1
4(119909119891 (119909) 119891 (119909) + 2119891 (119909) 119909119891 (119909) + 119891 (119909) 119891 (119909) 119909)
119909 ⊲ 1199092
=1
2119909 ⊲ (119909119891 (119909) + 119891 (119909) 119909)
=1
4(119909119891 (119909119891 (119909) + 119891 (119909) 119909) + 119891 (119909119891 (119909) + 119891 (119909) 119909) 119909)
=1
4(2119909119891 (119909) 119891 (119909) + 2119891 (119909) 119891 (119909) 119909)
(23)
However the right hand sides of the above equations (23)may not be equal see the following example (Example 28)For such elements 119909 we have 119909
2
⊲ 119909 = 119909 ⊲ 1199092 Hence by
Proposition 26 the spectrum of 119909 is unbounded whenever119872
+ is a unital split quasi-Jordan Banach algebra
Example 28 Let 119872 be the collection of 2 times 2 matrices withentries from the fieldC and let119860 be the algebra of allmatricesof the form [
120572 0
0 120573] with 120572 120573 isin C Then it is easily seen
that 119872 is an 119860-bimodule Next we define 119891 119872 rarr 119860
by 119891 [11988611
11988612
11988621
11988622
] = [11988611
0
0 11988622
] Of course 119891 is an additive mapsatisfying
119891([120572 0
0 120573] [
11988611
11988612
11988621
11988622
]) = [120572 0
0 120573]119891([
11988611
11988612
11988621
11988622
])
119891([11988611
11988612
11988621
11988622
] [120572 0
0 120573]) = 119891([
11988611
11988612
11988621
11988622
]) [120572 0
0 120573]
(24)
Hence119891 is an119860-bimodule mapThus by Remark 27119872+ is aquasi-Jordan algebra with the quasi-Jordan product as below
[11988611
11988612
11988621
11988622
] ⊲ [11988711
11988712
11988721
11988722
] =1
2([
11988611
11988612
11988621
11988622
]119891([11988711
11988712
11988721
11988722
])
+119891([11988711
11988712
11988721
11988722
]) [11988611
11988612
11988621
11988622
])
=[[
[
1198861111988711
11988612
11988711
+ 11988722
2
11988621
11988711
+ 11988722
21198862211988722
]]
]
(25)
Indeed 119872+
= 119869 oplus 119885(119872+
) where 119869 = [119886 0
0 119887] 119886 119887 isin C
is a subalgebra of 119872+ and 119885(119872+
) = [0 119886
119887 0] 119886 119887 isin C Any
matrix of the form [1 119886
119887 1] with 119886 119887 isin C is a (right) unit in119872
+Thus119872+ is a unital split quasi-Jordan algebrawith thematrix119868 = [
1 0
0 1] as the unit of its Jordan part 119869
Further a natural norm is defined on119872+ as follows
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
=100381610038161003816100381611988611
1003816100381610038161003816 +100381610038161003816100381611988612
1003816100381610038161003816 +100381610038161003816100381611988621
1003816100381610038161003816 +100381610038161003816100381611988622
1003816100381610038161003816 (26)
This norm also satisfies10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
] ⊲ [11988711
11988712
11988721
11988722
]
10038171003817100381710038171003817100381710038171003817
=1
2(
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
] [11988711
0
0 11988722
]
+ [11988711
0
0 11988722
] [11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
)
le
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
10038171003817100381710038171003817100381710038171003817
[11988711
0
0 11988722
]
10038171003817100381710038171003817100381710038171003817
le
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
10038171003817100381710038171003817100381710038171003817
[11988711
11988712
11988721
11988722
]
10038171003817100381710038171003817100381710038171003817
(27)
Next for any 119909 = [11988611
11988612
11988621
11988622
] isin 119872+ we observe that
1199092
= [11988611
11988612
11988621
11988622
] ⊲ [11988611
11988612
11988621
11988622
]
=[[
[
1198862
1111988612
11988611
+ 11988622
2
11988621
11988611
+ 11988622
21198862
22
]]
]
(28)
so that
1199092
⊲ 119909 = [
[
1198863
11
11988612
4(119886
11+ 119886
22)2
11988621
4(119886
11+ 119886
22)2
1198863
22
]
]
119909 ⊲ 1199092
= [
[
1198863
11
11988612
2(119886
2
11+ 119886
2
22)
11988621
2(119886
2
11+ 119886
2
22) 119886
3
22
]
]
(29)
So for any 119909 = [11988611
11988612
11988621
11988622
] isin 119872+ 1199092 ⊲ 119909 = 119909 ⊲ 119909
2
hArr
11988612
= 11988621
= 0 or 11988611
= 11988622 Thus by Proposition 26
10 Abstract and Applied Analysis
120590(119872+119868)([
11988611
11988612
11988621
11988622
]) = C whenever 11988611
= 11988622
and 11988612 119886
21 = 0
In particular for 119909 = [1 1
0 2] isin 119872
+ we have
1199092
⊲ 119909 =[[
[
13
1
4(1 + 2)
2
0
4(1 + 2)
2
23
]]
]
= [1
9
4
0 8
]
119909 ⊲ 1199092
=[[
[
13
1
2(1
2
+ 22
)
0
2(1
2
+ 22
) 23
]]
]
= [1
5
2
0 8
]
(30)
Concerning the inequality between the right hand sidesof (23) in Remark 27 we observe for 119909 as above that
1
4(119909119891 (119909) 119891 (119909) + 2119891 (119909) 119909119891 (119909) + 119891 (119909) 119891 (119909) 119909)
=1
4([
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 1
0 2] [
1 0
0 2]
+ [1 0
0 2] [
1 0
0 2] [
1 1
0 2]) = [
19
4
0 8
]
(31)
but
1
4(2119909119891 (119909) 119891 (119909) + 2119891 (119909) 119891 (119909) 119909)
=1
4(2 [
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 0
0 2] [
1 1
0 2])
= [1
5
2
0 8
]
(32)
Hence 1199092 ⊲ 119909 = 119909 ⊲ 1199092 Thus 120590
(119872+119868)([
1 1
0 2]) = C by
Proposition 26Further suppose the matrix [
11988611198862
11988631198864
] isin 119872+ is invertible
with respect to the unit 119868 that is [ 1198861 119886211988631198864
] isin 119866119868(119872
+
) Thenthere exists [ 119887
11198872
11988731198874
] isin 119872+ such that
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] = 119868 + 119868⊲([
1198861
1198862
1198863
1198864
]) = [1 minus119886
2
minus1198863
1]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
= [1198861
1198862
1198863
1198864
] + 119868⊲([
1198861
1198862
1198863
1198864
])
+ 119868⊲([
1198861
1198862
1198863
1198864
]
2
)
=[[
[
1198861
minus1198862
1198861+ 119886
4
2
minus1198863
1198861+ 119886
4
21198864
]]
]
(33)
However
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] =[[
[
11988711198861
1198872
1198861+ 119886
4
2
1198873
1198861+ 119886
4
211988741198864
]]
]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
=
[[[
[
11988711198861
1198872
1198862
1+ 119886
2
4
2
1198873
1198862
1+ 119886
2
4
211988741198862
4
]]]
]
(34)
It follows that 1198861
= 0 1198864
= 0 1198871= 1119886
1 119887
4= 1119886
4 119887
2((119886
1+
1198864)2) = minus119886
2 119887
3((119886
1+119886
4)2) = minus119886
3 119887
2((119886
2
1+119886
2
4)2) = minus119886
2((119886
1+
1198864)2) and 119887
3((119886
2
1+ 119886
2
4)2) = minus119886
3((119886
1+ 119886
4)2) From these
equations we get 1198872(119886
1minus 119886
4)2
= 0 and 1198873(119886
1minus 119886
4)2
= 0 so that(119887
2minus119887
3)(119886
1minus 119886
4)2
= 0Then for 1198861
= 1198864 we obtain 119887
2= 119887
3= 0
and hence 1198862= 119886
3= 0 Therefore
119866119868(119872
+
) = [119886 119887
119888 119886] 119909 = [
120572 0
0 120573] isin 119872
+
119886 = 0
120572 = 0 120573 = 0 120572 = 120573
(35)
The set 119866119868(119872
+
) is not open clearly [1 1
0 1] isin 119866
119868(119872
+
) forany 120598 gt 0
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[1 1
0 1] minus [
[
1 minus120598
41
0 1 +120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[
[
120598
40
0 minus120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=120598
2lt 120598
(36)
but [ 1minus1205984 1
0 1+1205984] notin 119866
119868(119872
+
)Now if 119860 = [
minus1 1
0 1] then 120582119868 minus 119860 = [
120582+1 minus1
0 120582minus1] notin 119866
119868(119872
+
)and so 120582 isin 120590
(119872+119868)(119860) for all 120582 isin C Thus 120590
(119872+119868)(119860) = C an
unbounded spectrum
Next we observe that the spectrum of an element withrespect to a unit is closed whenever the corresponding set ofinvertibles is open
Proposition 29 LetI be a quasi-Jordan normed algebra witha unit 119890 such that119866
119890(I) is openThen 120590
(I119890)(119909) is closed for all119909 isin I
Proof Define 119891 C rarr I by 119891(120582) = 120582119890 minus 119909 Since 119891 iscontinuous the inverse image of the open set 119866
119890(I) is open
in C and so its complement 120590(I119890)(119909) is closed
We conclude this paper with the following example of anonclosed spectrum
Example 30 Let 119872 and 119872+ be as in Example 28 Let 119864 =
[1 1
0 1] and 119860 = [
119886 0
0 119887] with 119886 = 119887 both different from 1 Then
119864 is a unit in 119872+ We show that 120590
(119872+119864)
(119860) = C 1 Forthis let us first investigate when can an element of the form119861 = [
120572 120573
0 120574] be invertible Assuming that 119861 is invertible we get
the existence of an element 119862 isin 119863+ such that
119862 ⊲ 119861 = 119864 + 119864⊲(119861)
119862 ⊲ 1198612
= 119861 + 119864⊲(119861) + 119864
⊲(119861
2
)
(37)
Abstract and Applied Analysis 11
From these equations we get 120572 = 0 120574 = 0 and
119862 =[[
[
1
120572120573∘
01
120574
]]
]
(38)
where 120573∘satisfies the following two equations
120573∘(120572 + 120574
2) = 1 minus 120573 +
120572 + 120574
2
120573∘(1205722
+ 1205742
2) =
120572 + 120574
2+
1205722
+ 1205742
2minus 120573
120572 + 120574
2
(39)
Multiplying the last equation by 2(120572 + 120574)(1205722
+ 1205742
) and thenusing the other equation we get
2 (1 minus 120573) + (120572 + 120574) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
+ (120572 + 120574) (40)
or equivalently
2 (1 minus 120573) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
(41)
this equation is satisfied for 120573 = 1 or 120572 = 120574 Hence thematrix119861 is invertible if and only if 119861 = [
120572 1
0 120574] or 119861 = [
120572 120573
0 120572
] for 120572 120574 isin
C 0 and 120573 isin CWe conclude that 119860
120582= 120582119864 minus 119860 = [
120582minus119886 120582
0 120582minus119887] is invertible
with respect to 119864 if and only if 120582 minus 119887 = 0 120582 minus 119886 = 0 and 120582 = 1that is119860
120582is invertible if 119886 = 120582 and 119887 = 120582 and 120582 = 1 Hence119860
120582
is invertible only if 120582 = 1 as we assumed that 119886 = 119887 and bothare not 1 So for all 120582 = 1 119860
120582notin 119866
119864(119872
+
) Thus 120590(119872+119864)
(119860) =
C 1 which is neither bounded nor closed
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This project was supported by King Saud University Dean-ship of Scientific Research College of Science ResearchCenter
References
[1] B Fritzsche ldquoSophus Lie a sketch of his life and workrdquo Journalof Lie Theory vol 9 no 1 pp 1ndash38 1999
[2] P Jordan ldquoUber eine klasse nichtassoziativer hyperkomplexeralgebrenrdquo Nachrichten von der Gesellschaft der Wissenschaftenzu Gottingen pp 569ndash575 1932
[3] K McCrimmon A Taste of Jordan Algebras UniversitextSpringer New York NY USA 2004
[4] N Jacobson Structure and Representations of Jordan Algebrasvol 39 of American Mathematical Society Colloquium Publica-tions American Mathematical Society Providence RI USA1968
[5] F Gursey and C-H Tze On the Role of Division Jordan andRelatedAlgebras in Particle PhysicsWorld Scientific River EdgeNJ USA 1996
[6] R Iordanescu Jordan Structures in Analysis Geometry andPhysics EdituraAcademiei Romane Bucharest Romania 2009
[7] H Upmeier Symmetric Banach Manifolds and Jordan Clowast-Alge-bras vol 104 of North-Holland Mathematics Studies North-Holland Amsterdam The Netherlands 1985
[8] I L Kantor ldquoClassification of irreducible transitive differentialgroupsrdquo Doklady Akademii Nauk SSSR vol 158 pp 1271ndash12741964
[9] M Koecher ldquoUber eine Gruppe von rationalen AbbildungenrdquoInventiones Mathematicae vol 3 pp 136ndash171 1967
[10] J Tits ldquoUne classe drsquoalgebres de Lie en relation avec les algebresde Jordanrdquo Indagationes Mathematicae vol 24 pp 530ndash5351962
[11] J-L Loday ldquoUne version non commutative des algebres de Lieles algebres de Leibnizrdquo LrsquoEnseignement Mathematique vol 39no 3-4 pp 269ndash293 1993
[12] J-L Loday and T Pirashvili ldquoUniversal enveloping algebras ofLeibniz algebras and (co)homologyrdquo Mathematische Annalenvol 296 no 1 pp 139ndash158 1993
[13] J-L Loday ldquoDialgebrasrdquo in Dialgebras and Related Operadsvol 1763 of Lecture Notes in Math pp 7ndash66 Springer BerlinGermany 2001
[14] R Velasquez and R Felipe ldquoQuasi-Jordan algebrasrdquo Communi-cations in Algebra vol 36 no 4 pp 1580ndash1602 2008
[15] R Velasquez and R Felipe ldquoOn K-B quasi-Jordan algebrasand their relation with Liebniz algebrasrdquo Comunicaciones delCIMAT No 1-10-1015-12-2010
[16] R Felipe ldquoAn analogue to functional analysis in dialgebrasrdquoInternational Mathematical Forum vol 2 no 21ndash24 pp 1069ndash1091 2007
[17] R Velasquez and R Felipe ldquoSplit dialgebras split quasi-Jordanalgebras and regular elementsrdquo Journal of Algebra and ItsApplications vol 8 no 2 pp 191ndash218 2009
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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International Journal of
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Operations ResearchAdvances in
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
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Volume 2014
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Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 3
Proof From the above discussion it is clear that the Jordanpart 119869 is a unital Jordan normed algebra Next let 119909
119899 be any
fixed Cauchy sequence in 119869 Then the same 119909119899 is Cauchy
sequence also in the quasi-Jordan Banach algebraI since 119869 isa normed subspace ofI Hence 119909
119899rarr 119909 for some 119909 isin I
Now since each119909119899isin 119869 and since 119890 is the unit of the Jordan
algebra 119869 under the (restricted) product ldquo⊲rdquo we get 119890 ⊲ 119909119899=
119909119899for all 119899 So by the continuity of the product 119909
119899= 119890 ⊲
119909119899
rarr 119890 ⊲ 119909 Hence by the uniqueness of the limit of anyconvergent sequence in a normed space 119890 ⊲ 119909 = 119909 However119909 = 119890 ⊲ (119909
119869+ 119909
119885) = 119890 ⊲ 119909
119869+ 119890 ⊲ 119909
119885= 119890 ⊲ 119909
119869+ 0 = 119909
119869isin 119869
Thus the required result follows
We know from [17] that for any quasi-Jordan algebra II
1= (119909 119877
119910) 119909 119910 isin I equipped with the sum (119909 119877
119910) +
(119886 119877119887) = (119909 + 119886 119877
119910+119887) scalar multiplication 120582(119909 119877
119910) =
(120582119909 119877120582119910) and product (119909 119877
119910) ⊲ (119886 119877
119887) = (119909 ⊲ 119887 119877
119910⊲119887)
is a split quasi-Jordan algebra and the map 119909 997891rarr 120593(119909) =
(119909 119877119909) is an embedding of I into I
1 where 119877
119911stands for
the usual right multiplication operator on I It is easily seenthat 0 times 119877(I) is the Jordan part of I
1and the zero part
119885(I1) = I times 0 Moreover the embedding 120593 preserves the
units clearly (119886 119877119887) ⊲ (119890 119877
119890) = (119886 ⊲ 119890 119877
119887⊲119890) = (119886 119877
119887) for
all (119886 119877119887) isin I
1so that (119890 119877
119890) is a unit in I
1whenever 119890 is a
unit in I In fact (119909 119877119890) is a unit in I
1 for all 119909 isin I it may
be noted here that (0 119877119890) is the only unit in the Jordan part of
I1From [17] we also know that 119877(I) = 119877
119909 119909 isin I with
product ldquo∙rdquo defined by 119877119909∙ 119877
119910= 119877
119909⊲119910 for all 119909 119910 isin I is
a quasi-Jordan algebra MoreoverI1is the direct product of
the quasi-Jordan algebrasI and 119877(I)Indeed the split quasi-Jordan algebraI
1is a quasi-Jordan
Banach algebra with unit (119890 119877119890)wheneverI is a quasi-Jordan
Banach algebra with a norm 1 unit 119890 and that 120593(I) =
(119909 119877119909) 119909 isin I is a closed unital quasi-Jordan normed
subalgebra ofI1 To justify this claim we need the following
result
Proposition 3 Suppose I is a quasi-Jordan Banach algebrawith a norm 1 unit 119890Then the algebra119877(I) as above is a quasi-Jordan Banach algebra with unit 119877
119890of norm 1
Proof Clearly each 119877119909is a bounded linear operator with
119877119909 le 119909 Hence the usual operator norm is a norm on
the quasi-Jordan algebra 119877(I)with the quasi-Jordan product119877119909∙ 119877
119910= 119877
119909⊲119910 Further we observe that
10038171003817100381710038171003817119877119909∙ 119877
119910
10038171003817100381710038171003817=
10038171003817100381710038171003817119877119909⊲119910
10038171003817100381710038171003817
= sup119911isinI119911=1
10038171003817100381710038171003817119877119909⊲119910
(119911)10038171003817100381710038171003817
= sup119911isinI119911=1
1003817100381710038171003817119911 ⊲ (119909 ⊲ 119910)1003817100381710038171003817
= sup119911isinI119911=1
1003817100381710038171003817119911 ⊲ ((119890 ⊲ 119909) ⊲ 119910)1003817100381710038171003817
= sup119911isinI119911=1
10038171003817100381710038171003817119911 ⊲ (119877
119910(119877
119909(119890)))
10038171003817100381710038171003817
le sup119911isinI119911=1
11991110038171003817100381710038171003817119877119910(119877
119909(119890))
10038171003817100381710038171003817
=10038171003817100381710038171003817119877119910(119877
119909119890)10038171003817100381710038171003817le
1003817100381710038171003817119877119909
1003817100381710038171003817
10038171003817100381710038171003817119877119910
10038171003817100381710038171003817119890
=1003817100381710038171003817119877119909
1003817100381710038171003817
10038171003817100381710038171003817119877119910
10038171003817100381710038171003817
(1)
Alternately by exploiting the right commutativity of thequasi-Jordan product inI we get
10038171003817100381710038171003817119877119909∙ 119877
119910
10038171003817100381710038171003817= sup
119911isinI119911=1
1003817100381710038171003817119911 ⊲ (119909 ⊲ 119910)1003817100381710038171003817
= sup119911isinI119911=1
1003817100381710038171003817119911 ⊲ ((119909 ⊲ 119890) ⊲ (119910 ⊲ 119890))1003817100381710038171003817
= sup119911isinI119911=1
1003817100381710038171003817119911 ⊲ ((119909 ⊲ 119890) ⊲ (119890 ⊲ 119910))1003817100381710038171003817
= sup119911isinI119911=1
1003817100381710038171003817119911 ⊲ ((119890 ⊲ 119910) ⊲ (119909 ⊲ 119890))1003817100381710038171003817
= sup119911isinI119911=1
1003817100381710038171003817119911 ⊲ ((119890 ⊲ 119910) ⊲ (119890 ⊲ 119909))1003817100381710038171003817
= sup119911isinI119911=1
10038171003817100381710038171003817119911 ⊲ (119877
119910(119890) ⊲ 119877
119909(119890))
10038171003817100381710038171003817
le sup119911isinI119911=1
11991110038171003817100381710038171003817119877119910
10038171003817100381710038171003817
1003817100381710038171003817119877119909
1003817100381710038171003817 1198902
=1003817100381710038171003817119877119909
1003817100381710038171003817
10038171003817100381710038171003817119877119910
10038171003817100381710038171003817
(2)
Thus 119877(I) together with the operator norm is a quasi-Jordan normed algebra Moreover for any 119909 isin 119877(I) wehave 119877
119909∙ 119877
119890= 119877
119909⊲119890= 119877
119909and 1 = 119890 ge 119877
119890 =
sup0 = 119911isinI(119877119890
(119911)119911) ge 119877119890(119890)119890 = 1 that is 119877
119890is a
norm 1 (right) unit in 119877(I)Suppose 119877
119909119899
is any fixed Cauchy sequence in 119877(I)Then for any fixed 119886 isin I 119886 ⊲ 119909
119898minus 119886 ⊲ 119909
119899 = 119877
119909119898
(119886) minus
119877119909119899
(119886) = (119877119909119898
minus 119877119909119899
)(119886) le 119877119909119898
minus 119877119909119899
119886 rarr 0 as119898 119899 rarr infin and so 119886 ⊲ 119909
119899 is a Cauchy sequence in I
But I is complete Hence the sequence 119886 ⊲ 119909119899 for any
119886 isin I is convergent inI In particular the sequence 119890 ⊲ 119909119899
converges to some 119910 isin I Moreover the Cauchy sequence119877
119909119899
converges to 119877119910isin 119877(I) in the operator norm because
(119877119909119899
minus 119877119910)(119911) = 119911 ⊲ (119909
119899minus 119910) = 119911 ⊲ 119909
119899minus 119911 ⊲ 119910 = 119911 ⊲
(119909119899⊲ 119890) minus 119911 ⊲ 119910 = 119911 ⊲ (119890 ⊲ 119909
119899) minus 119911 ⊲ 119910 = 119911 ⊲ (119890 ⊲
119909119899minus 119910) le 119911119890 ⊲ 119909
119899minus 119910 rarr 0 as 119899 rarr infin for all 119911 isin I
Thus the quasi-Jordan normed algebra 119877(I) is complete
Now we show that the corresponding algebra I1is a
unital split quasi-Jordan Banach algebra
Proposition 4 Let I be a quasi-Jordan Banach algebra witha norm 1 unit 119890 and let I
1be as above Then I
1is the direct
product of the quasi-Jordan algebras I and 119877(I) MoreoverI
1equipped with the product norm (119909 119877
119910) = 119909 + 119877
119910
is a split quasi-Jordan Banach algebra with unit (119890 119877119890) and
120593(I) = (119909 119877119909) 119909 isin I is a closed right ideal in I
1in the
norm topology
4 Abstract and Applied Analysis
Proof For the first part see [17] Clearly (119909 119877119910) = 119909 +
119877119910 is a norm and it satisfies
10038171003817100381710038171003817(119909 119877
119910) ⊲ (119911 119877
119908)10038171003817100381710038171003817=
10038171003817100381710038171003817(119909 ⊲ 119908 119877
119910⊲119908)10038171003817100381710038171003817
=10038171003817100381710038171003817(119909 ⊲ 119908 119877
119910∙ 119877
119908)10038171003817100381710038171003817
=1003817100381710038171003817119877119908
(119909)1003817100381710038171003817 +
10038171003817100381710038171003817119877119910∙ 119877
119908
10038171003817100381710038171003817
le (119909 +10038171003817100381710038171003817119877119910
10038171003817100381710038171003817)1003817100381710038171003817119877119908
1003817100381710038171003817
le (119909 +10038171003817100381710038171003817119877119910
10038171003817100381710038171003817) (119911 +
1003817100381710038171003817119877119908
1003817100381710038171003817)
=10038171003817100381710038171003817(119909 119877
119910)10038171003817100381710038171003817
1003817100381710038171003817(119911 119877119908)1003817100381710038171003817
(3)
for all 119908 119909 119910 119911 isin I Keeping in view Proposition 3 wededuce that I
1being the product of complete spaces I and
119877(I) is complete in the product norm Thus I1is a split
quasi-Jordan Banach algebra with (right) unit (119890 119877119890)
Clearly 120593(I) is a subspace ofI1with (119909 119877
119909) ⊲ (119910 119877
119911) =
(119909 ⊲ 119911 119877119909⊲119911
) isin 120593(I) for all (119909 119877119909) isin 120593(I) (119910 119877
119911) isin I
1 so
that 120593(I) is a quasi-Jordan normed subalgebra with the unit(119890 119877
119890) included Further let (119909
119899 119877
119909119899
) be any fixed Cauchysequence in 120593(I) Then 119909
119898minus 119909
119899 le 119909
119898minus 119909
119899 + 119877
119909119898
minus
119877119909119899
= (119909119898
minus 119909119899 119877
119909119898
minus 119877119909119899
) = (119909119898 119877
119909119898
) minus (119909119899 119877
119909119899
) rarr
0 as 119898 119899 rarr infin so that 119909119899 is a Cauchy sequence in (the
complete space) I and hence it converges to some 119909 isin INow by using the fact 119877
119911le 119911 for all 119911 isin I (since
119890 is a norm 1 unit in I see above) we get the convergenceof arbitrarily fixed Cauchy sequence (119909
119899 119877
119909119899
) to (119909 119877119909) isin
120593(I) since (119909119899 119877
119909119899
)minus(119909 119877119909) = (119909
119899minus119909 119877
119909119899minus119909
) = 119909119899minus119909+
119877119909119899minus119909
le 2119909119899minus119909 rarr 0 as 119899 rarr infinThus 120593(I) being com-
plete is a closed right ideal inI1in the norm topology
Next we observe the isometry between I and 120593(I)
Proposition 5 Let I be a quasi-Jordan Banach algebra withnorm 1 unit Then there exists an equivalent norm that makesI isometrically isomorphic to a quasi-Jordan closed subalgebra120593(I) of the split quasi-Jordan Banach algebra I
1
Proof Clearly 119909∘
= 119909 + 119877119909 defines a norm on the
quasi-Jordan algebra I It follows that (I sdot ∘) is a quasi-
Jordan Banach algebraMoreoverI is isomorphic to the sub-algebra (119909 119877
119909) 119909 isin I of I
1under the isomorphism
120593 119909 997891rarr (119909 119877119909) as seen above Further we observe that 120593
is an isometry since 119909∘= 119909 + 119877
119909 = (119909 119877
119909) = 120593(119909)
Finally we note that the two norms sdot and sdot ∘on I are
equivalent since 119909 le 119909 + 119877119909 = 119909
∘= 119909 + 119877
119909 le
2119909
3 Invertible Elements
As in [17] an element 119909 in a quasi-Jordan algebra I is calledinvertible with respect to a unit 119890 isin I if there exists 119910 isin I
such that 119910 ⊲ 119909 = 119890 + (119890 ⊲ 119909 minus 119909) and 119910 ⊲ 1199092
= 119909 + (119890 ⊲
119909 minus 119909) + (119890 ⊲ 1199092
minus 1199092
) such an element 119910 is called an inverseof119909 with respect to 119890 Let 119890
⊲(119909) denote the element 119890 ⊲ 119909 minus 119909
Then119909 has an inverse119910with respect to 119890 hArr 119910 ⊲ 119909 = 119890+119890⊲(119909)
and 119910 ⊲ 1199092
= 119909 + 119890⊲(119909) + 119890
⊲(119909
2
)We know from the above discussion that the embedding
119909 997891rarr 120593(119909) = (119909 119877119909) of a quasi-Jordan algebraI into the split
quasi-Jordan algebra I1preserves the units The embedding
120593 also preserves the corresponding invertible elements if 119910 isan inverse of 119909 with respect to a unit 119890 in I then 119910 ⊲ 119909 =
119890 + (119890 ⊲ 119909 minus 119909) and 119910 ⊲ 1199092
= 119909 + (119890 ⊲ 119909 minus 119909) + (119890 ⊲ 1199092
minus 1199092
)Hence
(119910 119877119910) ⊲ (119909 119877
119909) = (119910 ⊲ 119909 119877
119910⊲119909)
= (119890 + (119890 ⊲ 119909 minus 119909) 119877119890+(119890⊲119909minus119909)
)
= (119890 + (119890 ⊲ 119909 minus 119909) 119877119890+ 119877
119890⊲119909minus 119877
119909)
= (119890 119877119890) + (119890 ⊲ 119909 119877
119890⊲119909) minus (119909 119877
119909)
= (119890 119877119890) + (119890 119877
119890) ⊲ (119909 119877
119909) minus (119909 119877
119909)
(119910 119877119910) ⊲ (119909 119877
119909)2
= (119910 ⊲ 1199092
119877119910⊲1199092)
= (119909 119877119909) + ((119890 ⊲ 119909 119877
119890⊲119909) minus (119909 119877
119909))
+ ((119890 ⊲ 1199092
119877119890⊲1199092) minus (119909
2
1198771199092))
= (119909 119877119909) + ((119890 119877
119890) ⊲ (119909 119877
119909) minus (119909 119877
119909))
+ ((119890 119877119890) ⊲ (119909 119877
119909)2
minus (119909 119877119909)2
)
(4)
Thus (119910 119877119910) is an inverse of (119909 119877
119909) with respect to the unit
(119890 119877119890) inI
1
Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebraThen 119877
119909(119910) = 119910 ⊲ 119909 = 119910 ⊲ (119909
119869+ 119909
119885) = 119910 ⊲ 119909
119869+ 119910 ⊲ 119909
119885=
119910 ⊲ 119909119869+ 0 = 119877
119909119869
(119910) for all 119909 119910 isin I Thus 119877119909= 119877
119909119869
for all119909 isin I
In this section we demonstrate that the set of invertibleelements with respect to a fixed unit in a quasi-JordanBanach algebra may not be open For this we proceed asfollows
Proposition 6 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebra with unit 119890 isin 119869 If 119909 isin I is invertible withrespect to 119890 then so is 120582119909 for all 120582 = 0
Proof Let119910 be an inverse of 119909 inIwith respect to 119890We showthat 1199101015840
= (1120582)119910119869+ 119910
119885is an inverse of 120582119909 with respect to 119890
Observe that
119910119869⊲ 119909 + 119910
119885⊲ 119909 = (119910
119869+ 119910
119885) ⊲ 119909
= 119910 ⊲ 119909 = 119890 + 119890⊲(119909)
= 119890 + 119890⊲(119909
119869+ 119909
119885) = 119890 + (minus119909
119885)
Abstract and Applied Analysis 5
119910119869⊲ 119909
2
+ 119910119885⊲ 119909
2
= 119910 ⊲ 1199092
= 119909 + 119890⊲(119909
119869+ 119909
119885) + 119890
⊲((119909
119869+ 119909
119885)2
)
= 119909119869+ (minus119909
119885⊲ 119909)
(5)
Hence by the uniqueness of the representation as sum of Jor-dan and zero parts we get 119910
119869⊲ 119909 = 119890 119910
119885⊲ 119909 = minus119909
119885 119910
119869⊲
1199092
= 119909119869 and 119910
119885⊲ 119909
2
= minus119909119885
⊲ 119909 Therefore 1199101015840
⊲ 120582119909 =
120582(((1120582)119910119869+ 119910
119885) ⊲ 119909) = 119910
119869⊲ 119909 + 120582119910
119885⊲ 119909 = 119890 minus 120582119909
119885=
119890 + 119890⊲(120582119909) and 119910
1015840
⊲ (120582119909)2
= 1205822
((1120582)119910119869⊲ 119909
2
+ 119910119885
⊲ 1199092
) =
120582119909119869minus 120582
2
119909119885⊲ 119909 = 120582119909 + 119890
⊲(120582119909) + 119890
⊲((120582119909)
2
) because 119890⊲(120582119909) =
119890 ⊲ (120582119909) minus 120582119909 = minus120582119909119885and 119890
⊲((120582119909)
2
) = 119890 ⊲ 1205822
1199092
minus 1205822
1199092
=
1205822
(119890⊲(119909
2
)) = minus1205822
119909119885⊲ 119909
Proposition 7 LetI be a quasi-Jordan normed algebrawith aunit 119890 Let119866
119890(I) = 119909 isin I 119909 is invertible with respect to 119890
be an open set and 119909 isin 119866119890(I) Then 119909 + 119911 isin 119866
119890(I) for all
119911 isin 119885(I)
Proof Suppose 119909 isin 119866119890(I) and 119911 isin 119885(I) If 119911 = 0 then 119909+119911 =
119909 isin 119866119890(I) Next suppose 119911 = 0 Since 119866
119890(I) is an open set
there exists 120598 gt 0 such that 119886 isin 119866119890(I) whenever 119909 minus 119886 lt 120598
Hence 119909 + 119911∘isin 119866
119890(I) with 119911
∘= (1205982119911)119911
Let 119910 and 119910∘be inverses of 119909 and 119909 + 119911
∘with respect to 119890
respectively Then
119910∘⊲ (119909 + 119911
∘) = 119890 + 119890
⊲(119909 + 119911
∘)
= 119890 + 119890⊲(119909) minus 119911
∘
= 119910 ⊲ 119909 minus 119911∘
119910∘⊲ (119909 + 119911
∘)2
= 119909 + 119911∘+ 119890
⊲(119909 + 119911
∘) + 119890
⊲((119909 + 119911
∘)2
)
= 119909 + 119890⊲(119909) + 119890
⊲(119909
2
) minus 119911∘⊲ 119909
= 119910 ⊲ 1199092
minus 119911∘⊲ 119909
(6)
Hence by setting 1199101= 119910 + 120572(119910
∘minus 119910) with 120572 = 2119911120598 we see
that1199101⊲ (119909 + 119911) = (119910 + 120572 (119910
∘minus 119910)) ⊲ 119909
= 119910 ⊲ 119909 + 120572 (119910∘⊲ (119909 + 119911
∘) minus 119910 ⊲ 119909)
= 119890 + 119890⊲(119909) + 120572 (minus119911
∘)
= 119890 + 119890⊲(119909) minus 119911
= 119890 + 119890⊲(119909 + 119911)
1199101⊲ (119909 + 119911)
2
= (119910 + 120572 (119910∘minus 119910)) ⊲ 119909
2
= 119910 ⊲ 1199092
+ 120572 (119910∘⊲ (119909 + 119911
∘)2
minus 119910 ⊲ 1199092
)
= 119909 + 119890⊲(119909) + 119890
⊲(119909
2
) minus 120572 (119911∘⊲ 119909)
= 119909 + 119890⊲(119909) + 119890
⊲(119909
2
) minus 119911 ⊲ 119909
= (119909 + 119911) + 119890⊲(119909 + 119911) + 119890
⊲((119909 + 119911)
2
)
(7)
since 119911 isin 119885(I) Thus 1199101is an inverse of 119909 + 119911 with respect to
the unit 119890
Corollary 8 Under the hypothesis of Proposition 7 120582119890 minus 119909 isin
119866119890(I) implies 120582119890 minus (119909 + 119911) isin 119866
119890(I) for all 119911 isin 119885(I)
Proposition 9 LetI = 119869 oplus 119885(I) be a split quasi-Jordan nor-med algebra and let 119890 isin 119869 be a unit inI such that the set119866
119890(I)
is open Then 119909 ⊲ 1199092
= 1199092
⊲ 119909 for all 119909 isin I
Proof Of course the element 119890 is the unit of the Jordanalgebra 119869 Then for any fixed element 119886 isin 119869 there exists120582 isin C such that 120582119890 minus 119886 is invertible in 119869 otherwise wewould get the negation of the well-known fact that spectrumof an element of a unital Jordan algebra is bounded Thatis there exists 119910 isin 119869 such that 119910 ⊲ (120582119890 minus 119886) = 119890 and119910 ⊲ (120582119890minus119886)
2
= (120582119890minus119886) However 119890⊲(119886) = 0 = 119890
⊲(119886
2
) Hence119910 is an inverse of 119886 in the quasi-Jordan algebraIwith respectto the unit 119890 By Corollary 8 120582119890 minus (119886 + 119911) is also invertiblefor any 119911 isin 119885(I) This in turn gives the existence of 119887 isin I
satisfying 119887 ⊲ (120582119890 minus (119886 + 119911)) = 119890 + 119911 and 119887 ⊲ (120582119890 minus (119886 + 119911))2
=
(120582119890minus119886)+119911 ⊲ (120582119890minus119886) Multiplying the first equation from theright by (120582119890 minus (119886 + 119911))
2 the second equation by (120582119890minus (119886+119911))and using the right Jordan identity we get
(119890 + 119911) ⊲ (120582119890 minus (119886 + 119911))2
= ((120582119890 minus 119886) + 119911 ⊲ (120582119890 minus 119886)) ⊲ (120582119890 minus (119886 + 119911))
(8)
Hence
(119890 + 119911) ⊲ (120582119890 minus 119886)2
= ((120582119890 minus 119886) + 119911 ⊲ (120582119890 minus 119886)) ⊲ (120582119890 minus 119886)
(9)
so that
119890 ⊲ (120582119890 minus 119886)2
+ 1205822
119911 minus 2120582 (119911 ⊲ 119886) + 119911 ⊲ 1198862
= (120582119890 minus 119886)2
+ 1205822
119911 minus 2120582119911 ⊲ 119886 + (119911 ⊲ 119886) ⊲ 119886
(10)
This last equation reduces to 119890 ⊲ (120582119890 minus 119886)2
= (120582119890 minus 119886)2 since
(120582119890 minus 119886)2
isin 119869 and 119890 is the unit of 119869 Hence 119911 ⊲ 1198862
= (119911 ⊲
119886) ⊲ 119886 for all 119886 isin 119869 and 119911 isin 119885(I) Now for any 119909 isin I thelast equation with 119886 = 119909
119869and 119911 = 119909
119885gives 119909
119885⊲ 119909
2
119869= (119909
119885⊲
119909119869) ⊲ 119909
119869Thus 119909 ⊲ 119909
2
= 119909119869⊲ 119909
2
119869+119909
119885⊲ 119909
2
119869= 119909
2
119869⊲ 119909
119869+(119909
119885⊲
119909119869) ⊲ 119909
119869= 119909
2
⊲ 119909 for all 119909 isin I
Corollary 10 If a unital split quasi-Jordan algebra has anelement 119909 with 119909 ⊲ 119909
2
= 1199092
⊲ 119909 then the set of invertibleelements with respect to the unit of the Jordan part is not open
In the sequel we will show the existence of a unital splitquasi-Jordan Banach algebra with elements 119909 such that 119909 ⊲
1199092
= 1199092
⊲ 119909 Thus the above result establishes that the set ofinvertible elements with respect to a fixed unit in a quasi-Jordan Banach algebra may not be open
6 Abstract and Applied Analysis
4 The Spectrum of Elements in a UnitalQuasi-Jordan Algebra
As usual we define the spectrum of an element 119909 in aunital quasi-Jordan algebra (I 119890) denoted by 120590
(I119890)(119909) tobe the collection of all complex numbers 120582 for which 120582119890 minus
119909 is not invertible Thus 120590(I119890)(119909) = 120582 isin C 120582119890 minus
119909 is not invertible Here the subscript 119890 indicates that theinvariability depends on the choice of unit 119890 which generallyis not unique
Proposition 11 Let I be a unital quasi-Jordan algebra Then120590(I119890)(119890
1015840
) = 1 for all 119890 1198901015840 isin 119880(I)
Proof Let 119890 1198901015840 isin 119880(I) Then for any 120582 = 1 119910 = (1(120582minus1))1198901015840
is an inverse of 120582119890 minus 1198901015840 with respect to the unit 119890 because 119910 ⊲
(120582119890 minus 1198901015840
) = (1(120582 minus 1))1198901015840
⊲ (120582119890 minus 1198901015840
) = 119890 + 119890⊲(120582119890 minus 119890
1015840
) and119910 ⊲ (120582119890 minus 119890
1015840
)2
= (120582119890 minus 1198901015840
) + 119890⊲(120582119890 minus 119890
1015840
) + 119890⊲((120582119890 minus 119890
1015840
)2
)
Proposition 12 Let I be a quasi-Jordan algebra with unit 119890Then 120590
(I119890)(119911) = 0 for all 119911 isin 119885(I)
Proof For any fixed 119911 isin 119885(I) and nonzero scalar 120582 thevector 119910 = (1120582)(119890 + 119911) satisfies 119910 ⊲ (120582119890 minus 119911) = 120582119910 = 119890 + 119911 =
119890+ 119890⊲(120582119890minus 119911) and 119910 ⊲ (120582119890 minus 119911)
2
= 1205822
119910 = 120582119890+120582119911 = (120582119890minus119911) +
119890⊲(120582119890 minus 119911) + 119890
⊲((120582119890 minus 119911)
2
) So (1120582)(119890 + 119911) is an inverse of120582119890 minus 119911 with respect to 119890 This means 120582 notin 120590
(I119890)(119911) for all 120582 = 0However the zero vector is not invertible Thus 120590
(I119890)(119911) =
0
Proposition 13 Let I = 119869 oplus 119885(I) be a split quasi-Jordanalgebra with unit 119890 isin 119869 Then 120590
(I119890)(119901) sube 0 1 for all idempo-tents 119901 isin I (ie 1199012
= 119901)
Proof Let 119901 be any fixed idempotent inI Since 119901 isin I 119901 hasa unique representation 119901 = 119901
119869+ 119901
119885with 119901
119869isin 119869 and 119901
119885isin
119885(I) Clearly 1199012
119869+ 119901
119885⊲ 119901
119869= 119901
2
= 119901 = 119901119869+ 119901
119885 Then
by uniqueness of the representation in the split quasi-Jordanalgebra I 119901
119885⊲ 119901
119869= 119901
119885and 119901
2
119869= 119901
119869 this means 119901
119869is an
idempotent in the Jordan algebra 119869 Hence 120590(119869119890)
(119901119869) sube 0 1
Thus 119901120582= 120582119890 minus 119901
119869is invertible in 119869 with the unique inverse
119901minus1
120582 for all 120582 notin 0 1We show that 119910 = 119901
minus1
120582+ (1(120582 minus 1))119901
119885is an inverse of
120582119890 minus 119901 in I with respect to the unit 119890 for this we note that120582119890 minus 119901 = 119901
120582minus 119901
119885 (119901
120582minus 119901
119885)2
= 1199012
120582minus 119901
119885⊲ 119901
120582 119890
⊲(120582119890 minus
119901) = 119890⊲(119901
120582minus 119901
119885) = 119890 ⊲ (119901
120582minus 119901
119885) minus (119901
120582minus 119901
119885) = 119901
119885 and
119890⊲((120582119890 minus 119901)
2
) = 119890Δ((119901
120582minus 119901
119885)2
) = 119890⊲(119901
2
120582minus 119901
119885⊲ 119901
120582) = 119901
119885⊲
119901120582= 119901
119885⊲ (120582119890minus119901
119869) = 120582119901
119885minus119901
119885⊲ 119901
119869= 120582119901
119885minus119901
119885= (120582minus1)119901
119885
Hence 119910 ⊲ (120582119890 minus 119901) = 119910 ⊲ (119901120582minus 119901
119885) = 119910 ⊲ 119901
120582= (119901
minus1
120582+
(1(120582 minus 1))119901119885) ⊲ 119901
120582= 119901
minus1
120582⊲ 119901
120582+ (1(120582 minus 1))119901
119885⊲ 119901
120582= 119890 +
(1(120582 minus 1))119901119885
⊲ (120582119890 minus 119901119869) = 119890 + (1(120582 minus 1))(120582119901
119885minus 119901
119885⊲
119901119869) = 119890 + (1(120582 minus 1))(120582119901
119885minus 119901
119885) = 119890 + 119901
119885= 119890 + 119890
⊲(120582119890 minus 119901)
and 119910 ⊲ (120582119890 minus 119901)2
= 119910 ⊲ (1199012
120582minus 119901
119885⊲ 119901
120582) = 119910 ⊲ 119901
2
120582=
(119901minus1
120582+(1(120582minus1))119901
119885) ⊲ 119901
2
120582= 119901
120582+(1(120582minus1))119901
119885⊲ (120582119890minus119901
119869)2
=
119901120582+(1(120582minus1))119901
119885⊲ (120582
2
119890minus2120582119901119869+119901
2
119869) = 119901
120582+(1(120582minus1))119901
119885⊲
(1205822
119890 minus 2120582119901119869+ 119901
119869) = 119901
120582+ (1(120582 minus 1))(120582
2
119901119885
⊲ 119890 minus 2120582119901119885
⊲
119901119869+ 119901
119885⊲ 119901
119869) = 119901
120582+ (1(120582 minus 1))(120582
2
119901119885
minus 2120582119901119885
+ 119901119885) =
119901120582+ (1(120582 minus 1))(120582 minus 1)
2
119901119885= 120582119890 minus119901
119869+ (120582 minus 1)119901
119885= (120582119890 minus119901) +
119890⊲(120582119890 minus 119901) + 119890
⊲((120582119890 minus 119901)
2
)
As mentioned in Section 2 if I is a quasi-Jordan algebrawith a unit 119890 then the set 119890 + 119911 119911 isin 119885(I) coincides withthe set 119880(I) of all units inI
Proposition 14 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebrawith unit 119890 isin 119869 and119909 isin I invertiblewith respectto some 1198901015840 isin 119880(I)Then119909
119869is invertible with respect to the unit
119890 in the Jordan algebra 119869
Proof Clearly 119890 ⊲ 119909 minus 119909 = minus119909119885and 119890 ⊲ 119909
2
minus 1199092
= minus119909119885⊲ 119909
119869
Since 1198901015840
isin 119880(I) we have 1198901015840
= 119890 + 119911 for some 119911 isin 119885(I)Hence the invertibility of 119909 in I with respect to the unit 1198901015840gives the existence of 119910 isin I such that 119910
119869⊲ 119909
119869+ 119910
119885⊲ 119909
119869=
119910 ⊲ 119909119869= 119910 ⊲ 119909 = 119890 + 119911 + (119890 + 119911) ⊲ 119909 minus 119909 = 119890 + 119911 + 119911 ⊲ 119909 minus 119909
119885
and 119910119869⊲ 119909
2
119869+ 119910
119885⊲ 119909
2
119869= 119910 ⊲ 119909
2
119869= 119910 ⊲ 119909
2
= (119890 + 119911) ⊲
119909 + (119890 + 119911) ⊲ 1199092
minus 1199092
= 119909119869+ 119911 ⊲ 119909 + 119911 ⊲ 119909
2
minus 119909119885⊲ 119909
119869 So by
the uniqueness of the representations in the split algebra Iwe get 119910
119869⊲ 119909
119869= 119890 and 119910
119869⊲ 119909
2
119869= 119909
119869 Thus 119910
119869is the inverse
of 119909119869in the Jordan algebra 119869
Next we observe that the spectrum of 119909 in a unital splitquasi-Jordan algebra with respect to any unit includes thespectrum of 119909
119869in the Jordan part 119869
Corollary 15 LetI = 119869 oplus119885(I) be a unital split quasi-Jordanalgebra with unit 119890 isin 119869 and let 119909 isin I Then 120590
(119869119890)(119909
119869) sube
120590(I1198901015840)(119909) for all 1198901015840 isin 119880(I)
Proof Let 1198901015840 = 119890 + 119911 with 119911 isin 119885(I) and let 120582 notin 120590(I1198901015840)(119909)
Then 1205821198901015840 minus 119909 is invertible in I with respect to the unit 1198901015840Hence its Jordanpart120582119890minus119909
119869is invertible in the Jordan algebra
119869 by Proposition 14 Thus 120582 notin 120590(119869119890)
(119909119869)
It is well known that the spectrum of any element ina unital Jordan Banach algebra is nonempty (cf [7]) Thistogether with Proposition 2 and Corollary 15 gives thefollowing result
Corollary 16 The spectrum of any element in a unital splitquasi-Jordan Banach algebra is nonempty
The next result extends Corollary 16 to any quasi-JordanBanach algebra with a norm 1 unit
Proposition 17 The spectrum of any element in a quasi-Jordan Banach algebra with a norm 1 unit is nonempty
Proof Let I be a unital quasi-Jordan Banach algebra witha norm 1 unit 119890 From Section 2 we know that the map120593 119909 997891rarr 120593(119909) = (119909 119877
119909) embeds I into the unital split
quasi-Jordan Banach algebra I1
= (119909 119877119910) 119909 119910 isin I
equipped with the sum (119909 119877119910) + (119886 119877
119887) = (119909 + 119886 119877
119910+119887) sca-
lar multiplication 120582(119909 119877119910) = (120582119909 119877
120582119910) and product (119909
119877119910) ⊲ (119886 119877
119887) = (119909 ⊲ 119887 119877
119910⊲119887) and the image 120593(I) is a
norm closed right ideal isomorphic to I with norm 1 unit(119890 119877
119890) Moreover it is seen in Section 3 that the embedding
Abstract and Applied Analysis 7
120593 also preserves the corresponding invertible elements thatis (119910 119877
119910) is an inverse of (119909 119877
119909) with respect to the unit
(119890 119877119890) in I
1whenever 119910 is an inverse of 119909 with respect
to a unit 119890 in I Hence by Corollary 16 it follows that120601 = 120590
(I1(119890119877119890))(119909 119877
119909) sube 120590
(120593(I)(119890119877119890))(119909 119877
119909) = 120590
(I119890)(119909) for all119909 isin I
Proposition 18 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebra with unit 119890 isin 119869 and let 119909 isin 119869 Then 120590
(119869119890)(119909) =
120590(I119890)(119909)
Proof By Corollary 15 120590(119869119890)
(119909) sube 120590(I119890)(119909) For the reverse
inclusion let 120582 notin 120590(119869119890)
(119909) then 120582119890 minus 119909 is invertible in 119869 thatis there exists 119910 isin 119869 such that 119910 ⊲ (120582119890 minus 119909) = 119890 and 119910 ⊲
(120582119890 minus 119909)2
= 120582119890 minus 119909 However 119890⊲(120582119890 minus 119909) = 0 = 119890
⊲((120582119890 minus 119909)
2
)Hence 119910 is an inverse of 120582119890 minus 119909 in I with respect to the unit119890 that is 120582 notin 120590
(I119890)(119909) Thus 120590(I119890)(119909) sube 120590
(119869119890)(119909)
Proposition 19 Let I be a unital quasi-Jordan normedalgebra and let 119890 be a unit inI for which 119866
119890(I) is open Then
120590(I119890)(119909) = 120590
(I119890)(119909 + 119911) for all 119911 isin 119885(I)
Proof By Corollary 8 120582119890 minus 119909 is invertible if and only if 120582119890 minus
(119909 + 119911) is invertible for all 119911 isin 119885(I) Thus 120582 notin 120590(I119890)(119909) if
and only if 120582 notin 120590(I119890)(119909 + 119911) for all 119911 isin 119885(I)
Corollary 20 LetI = 119869oplus119885(I) be a unital split quasi-Jordannormed algebra and let 119890 be a unit in I such that 119866
119890(I) is
open Then 120590(I119890)(119909) = 120590
(I119890)(119909119869) for all 119909 = 119909119869+ 119909
119885isin I
Further if the unit 119890 isin 119869 then120590(I119890)(119909) = 120590
(I119890)(119909119869) = 120590(119869119890)
(119909119869)
Lemma 21 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordanalgebra with a unit 119890 isin 119869 and let 119909 isin I be invertible withrespect to 119890 Then 119909
119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
Proof As 119909 is invertible there exists 119910 isin I such that 119910 ⊲ 119909 =
119890 + (119890 ⊲ 119909 minus 119909) = 119890 minus 119909119885and 119910 ⊲ 119909
2
= 119909 + (119890 ⊲ 119909 minus 119909) + (119890 ⊲
1199092
minus 1199092
) = 119909119869minus 119909
119885⊲ 119909 Hence by the uniqueness of the
representation in a split quasi-Jordan algebra we obtain
119910119885⊲ 119909 = minus119909
119885 (11)
119910119885⊲ 119909
2
= minus119909119885⊲ 119909 (12)
Thus
119909119885⊲ 119909
2
= minus (119910119885⊲ 119909) ⊲ 119909
2
(by (11))
= minus (119910119885⊲ 119909
2
) ⊲ 119909 (by the right Jordan identity)
= (119909119885⊲ 119909) ⊲ 119909 (by (12))
(13)
Proposition 22 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebra with a unit 119890 isin 119869 119909 = 119909
119869+ 119909
119885isin I satisfies
120590(I119890)(119909) =C Then
(1) 119909119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
(2) 1199092 ⊲ 119909 = 119909 ⊲ 1199092
Proof (1) Let 120582 notin 120590(I119890)(119909) Then 120582119890 minus 119909 = (120582119890 minus 119909
119869) + (minus119909
119885)
is invertible with respect to 119890 By Lemma 21 we have
119909119885⊲ (120582119890 minus 119909)
2
= (119909119885⊲ (120582119890 minus 119909)) ⊲ (120582119890 minus 119909) (14)
since the zero part of 120582119890 minus 119909 is minus119909119885 However
119909119885⊲ (120582119890 minus 119909)
2
= 119909119885⊲ (120582
2
119890 minus 120582119890 ⊲ 119909 minus 120582119909 ⊲ 119890 + 1199092
)
= 1205822
(119909119885⊲ 119890) minus 120582 (119909
119885⊲ (119890 ⊲ 119909))
minus 120582 (119909119885⊲ (119909 ⊲ 119890)) + 119909
119885⊲ 119909
2
= 1205822
119909119885minus 2120582119909
119885⊲ 119909
+ 119909119885⊲ 119909
2
(by the right commutativity of ⊲)
(119909119885⊲ (120582119890 minus 119909)) ⊲ (120582119890 minus 119909)
= (120582 (119909119885⊲ 119890) minus 119909
119885⊲ 119909) ⊲ (120582119890 minus 119909)
= (120582119909119885minus 119909
119885⊲ 119909) ⊲ (120582119890 minus 119909)
= 1205822
(119909119885⊲ 119890) minus 120582 (119909
119885⊲ 119909) ⊲ 119890
minus 120582119909119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
= 1205822
119909119885minus 2120582119909
119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
(15)
Therefore (14) becomes
1205822
119909119885minus 2120582119909
119885⊲ 119909 + 119909
119885⊲ 119909
2
= 1205822
119909119885minus 2120582119909
119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
(16)
which after simplification reduces to the required equation119909119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
(2) Since 1199092 = (119909119869+ 119909
119885)2
= 1199092
119869+ 119909
119885⊲ 119909
119869 we have
1199092
⊲ 119909 = (1199092
119869+ 119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909
119869+ (119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909
119869+ 119909
119885⊲ 119909
2
(17)
by the part (1) But 1199092119869⊲ 119909
119869= 119909
119869⊲ 119909
2
119869since 119909
119869is in the Jor-
dan algebra 119869 Therefore 1199092 ⊲ 119909 = 1199092
119869⊲ 119909
119869+ 119909
119885⊲ 119909
2
= 119909119869⊲
1199092
119869+ 119909
119885⊲ 119909
2
= 119909 ⊲ 1199092
Remark 23 In any quasi-Jordan algebra if an element 119909
satisfies 1199093
= 1199092
⊲ 119909 = 119909 ⊲ 1199092 then 119909
119899
⊲ 1199092
= 119909119899+2 for
all positive integers 119899 For this suppose 119909 satisfies 1199092 ⊲ 119909 =
119909 ⊲ 1199092 and 119909
119898
⊲ 1199092
= 119909119898+2 for any fixed119898 ge 1 Then 119909
119898+1
⊲
1199092
= (119909119898
⊲ 119909) ⊲ 1199092
= (119909119898
⊲ 1199092
) ⊲ 119909 (by the right Jor-dan identity) = 119909
119898+2
⊲ 119909 = 119909119898+3
Proposition 24 Let I be a unital quasi-Jordan Banachalgebra with unit 119890 and let 119909 isin I satisfy (119890 minus 119909) ⊲ (119890 minus 119909)
2
=
(119890 minus 119909)2
⊲ (119890 minus 119909) Then 119909 isin 119866119890(I) whenever 119890 minus 119909 lt 1
8 Abstract and Applied Analysis
Proof First note that 119890minus119909 lt 1 gives (119890 minus 119909)119899
le 119890 minus 119909119899
lt
1 for all 119899 = 1 2 3 Hence the infinite geometric series119890 +sum
infin
119899=1(119890 minus119909)
119899 converges absolutely to some element 119910 isin IWe show that the geometric series sum 119910 is an inverse of 119909with respect to the unit 119890 For any fixed positive integer 119899 let119910119899= 119890+sum
119899
119896=1(119890minus119909)
119896Then the sequence 119910119899 of partial sums
converges to 119910 By setting 119908 = 119890 minus 119909 we get
119910119899⊲ 119909 = (119890 +
119899
sum
119896=1
119908119896
) ⊲ (119890 minus 119908)
= 119890 +
119899
sum
119896=1
119908119896
minus (119890 ⊲ 119908 +
119899+1
sum
119896=2
119908119896
)
= 119890 + 119908 minus 119890 ⊲ 119908 minus 119908119899+1
= 119890 + (119890 minus 119909) minus 119890 ⊲ (119890 minus 119909) minus (119890 minus 119909)119899+1
= 119890 + (119890 minus 119909 minus 119890 + 119890 ⊲ 119909) minus (119890 minus 119909)119899+1
= 119890 + (119890 ⊲ 119909 minus 119909) minus (119890 minus 119909)119899+1
(18)
Thus by allowing 119899 rarr infin we obtain 119909 ⊲ 119910 = 119890 + (119890 ⊲
119909 minus 119909) = 119890 + 119890⊲(119909) since 119890 minus 119909 lt 1
Next by Remark 23 we have
119910119899⊲ 119909
2
= 119910119899⊲ (119890 minus 119908)
2
= 119910119899⊲ (119890 minus 119890 ⊲ 119908 minus 119908 ⊲ 119890 + 119908
2
)
= 119910119899⊲ (119890 minus 2119908 + 119908
2
)
= 119910119899⊲ 119890 minus 2119910
119899⊲ 119908 + 119910
119899⊲ 119908
2
= 119910119899minus 2119910
119899⊲ 119908 + 119910
119899⊲ 119908
2
= (119890 +
119899
sum
119896=1
119908119896
) minus 2(119890 +
119899
sum
119896=1
119908119896
) ⊲ 119908
+ (119890 +
119899
sum
119896=1
119908119896
) ⊲ 1199082
= (119890 +
119899
sum
119896=1
119908119896
) minus 2(119890 ⊲ 119908 +
119899+1
sum
119896=2
119908119896
)
+ (119890 ⊲ 1199082
+
119899+2
sum
119896=3
119908119896
)
= 119890 + 119908 minus 2119890 ⊲ 119908 minus 1199082
+ 119890 ⊲ 1199082
minus 119908119899+1
+ 119908119899+2
(19)
Taking the limit as 119899 rarr infin we get
119910 ⊲ 1199092
= 119890 + 119908 minus 2119890 ⊲ 119908 minus 1199082
+ 119890 ⊲ 1199082
= 119890 + (119890 minus 119909) minus 2119890 ⊲ (119890 minus 119909) minus (119890 minus 119909)2
+ 119890 ⊲ (119890 minus 119909)2
= 2119890 minus 119909 minus 2119890 + 2119890 ⊲ 119909 minus 119890 + 119890 ⊲ 119909 + 119909 minus 1199092
+ 119890 minus 2119890 ⊲ 119909 + 119890 ⊲ 1199092
= 119909 + (119890 ⊲ 119909 minus 119909) + (119890 ⊲ 1199092
minus 1199092
)
(20)
Proposition 25 Let I be a unital split quasi-Jordan Banachalgebra with unit 119890 If 119909 isin I with (119890 minus 119909) ⊲ (119890 minus 119909)
2
=
(119890 minus 119909)2
⊲ (119890 minus 119909) then |120582| le 119909 for all 120582 isin 120590(I119890)(119909)
Proof If 120582 isin 120590(I119890)(119909) with 120582 = 0 then the noninvertibility of
120582119890 minus 119909 means the noninvertibility of 119890 minus (1120582)119909 with respectto the unit 119890 However by Proposition 24 119890 minus (1120582)119909must beinvertible with respect to the unit 119890 whenever (1|120582|)119909 lt 1It follows that |120582| le 119909 for all 120582 isin 120590
(I119890)(119909)
5 Unbounded and Nonclosed Spectrum
In this section we show that the spectrum of an element ina split quasi-Jordan Banach algebra may be neither boundednor closed and hence not compactThe following result givesa couple of characterizations of the unbounded spectrum ofan element in a split quasi-Jordan Banach algebra
Proposition 26 Let 119909 be an element of a unital split quasi-Jordan Banach algebra I = 119869 oplus 119885(I) with a unit 119890 isin 119869 Thenthe following statements are equivalent
(1) 120590(I119890)(119909) =C
(2) 1199092 ⊲ 119909 = 119909 ⊲ 1199092
(3) |120582| le 119909 for all 120582 isin 120590(I119890)(119909)
Proof (1rArr2) See Proposition 22(2rArr3) Suppose 1199092 ⊲ 119909 = 119909 ⊲ 119909
2 Then
(119890 minus 119909)2
⊲ (119890 minus 119909) = (119890 minus 119890 ⊲ 119909 minus 119909 + 1199092
) ⊲ (119890 minus 119909)
= 119890 minus 2119890 ⊲ 119909 minus 119909 + 21199092
+ (119890 ⊲ 119909) ⊲ 119909 minus 1199092
⊲ 119909
= 119890 minus 2119909119869minus 119909 + 2119909
2
+ 119890 ⊲ 1199092
minus 1199093
(119890 minus 119909) ⊲ (119890 minus 119909)2
= (119890 minus 119909) ⊲ (119890 minus 2119909 + 1199092
)
= 119890 minus 2 (119890 ⊲ 119909) + 119890 ⊲ 1199092
minus 119909
+ 21199092
minus 119909 ⊲ 1199092
= 119890 minus 2119909119869+ 119890 ⊲ 119909
2
minus 119909 + 21199092
minus 1199093
(21)
Abstract and Applied Analysis 9
From (21) we get
(119890 minus 119909)2
⊲ (119890 minus 119909) = (119890 minus 119909) ⊲ (119890 minus 119909)2
(22)
Hence |120582| le 119909 for all 120582 isin 120590(I119890)(119909) by Proposition 25
(3rArr1) Immediate
Remark 27 There do exist unital split quasi-Jordan algebrascontaining elements that have the spectrum with respect tothe unit of the Jordan part equal to the whole ofC and henceunbounded To justify this claim we proceed as follows
Let 119860 be a unital associative algebra and let 119872 be an 119860-bimodule Let 119891 119872 rarr 119860 be an 119860-bimodule map (ie anadditive map satisfying 119891(119886119909) = 119886119891(119909) and 119891(119909119886) = 119891(119909)119886for all 119886 isin 119860 119909 isin 119872) Then one can put a dialgebra structureon 119872 as follows 119909 ⊣ 119910 = 119909119891(119910) and 119909⊢119910 = 119891(119909)119910 (cf [13Example 22(d)]) Hence119872+ is a quasi-Jordan algebra underthe quasi-Jordan product ldquo⊲rdquo given by119909 ⊲ 119910 = (12)(119909119891(119910)+
119891(119910)119909) Further for any 119909 isin 119872 we observe that
1199092
⊲ 119909 =1
2(119909119891 (119909) + 119891 (119909) 119909) ⊲ 119909
=1
4((119909119891 (119909) + 119891 (119909) 119909) 119891 (119909)
+119891 (119909) (119909119891 (119909) + 119891 (119909) 119909))
=1
4(119909119891 (119909) 119891 (119909) + 2119891 (119909) 119909119891 (119909) + 119891 (119909) 119891 (119909) 119909)
119909 ⊲ 1199092
=1
2119909 ⊲ (119909119891 (119909) + 119891 (119909) 119909)
=1
4(119909119891 (119909119891 (119909) + 119891 (119909) 119909) + 119891 (119909119891 (119909) + 119891 (119909) 119909) 119909)
=1
4(2119909119891 (119909) 119891 (119909) + 2119891 (119909) 119891 (119909) 119909)
(23)
However the right hand sides of the above equations (23)may not be equal see the following example (Example 28)For such elements 119909 we have 119909
2
⊲ 119909 = 119909 ⊲ 1199092 Hence by
Proposition 26 the spectrum of 119909 is unbounded whenever119872
+ is a unital split quasi-Jordan Banach algebra
Example 28 Let 119872 be the collection of 2 times 2 matrices withentries from the fieldC and let119860 be the algebra of allmatricesof the form [
120572 0
0 120573] with 120572 120573 isin C Then it is easily seen
that 119872 is an 119860-bimodule Next we define 119891 119872 rarr 119860
by 119891 [11988611
11988612
11988621
11988622
] = [11988611
0
0 11988622
] Of course 119891 is an additive mapsatisfying
119891([120572 0
0 120573] [
11988611
11988612
11988621
11988622
]) = [120572 0
0 120573]119891([
11988611
11988612
11988621
11988622
])
119891([11988611
11988612
11988621
11988622
] [120572 0
0 120573]) = 119891([
11988611
11988612
11988621
11988622
]) [120572 0
0 120573]
(24)
Hence119891 is an119860-bimodule mapThus by Remark 27119872+ is aquasi-Jordan algebra with the quasi-Jordan product as below
[11988611
11988612
11988621
11988622
] ⊲ [11988711
11988712
11988721
11988722
] =1
2([
11988611
11988612
11988621
11988622
]119891([11988711
11988712
11988721
11988722
])
+119891([11988711
11988712
11988721
11988722
]) [11988611
11988612
11988621
11988622
])
=[[
[
1198861111988711
11988612
11988711
+ 11988722
2
11988621
11988711
+ 11988722
21198862211988722
]]
]
(25)
Indeed 119872+
= 119869 oplus 119885(119872+
) where 119869 = [119886 0
0 119887] 119886 119887 isin C
is a subalgebra of 119872+ and 119885(119872+
) = [0 119886
119887 0] 119886 119887 isin C Any
matrix of the form [1 119886
119887 1] with 119886 119887 isin C is a (right) unit in119872
+Thus119872+ is a unital split quasi-Jordan algebrawith thematrix119868 = [
1 0
0 1] as the unit of its Jordan part 119869
Further a natural norm is defined on119872+ as follows
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
=100381610038161003816100381611988611
1003816100381610038161003816 +100381610038161003816100381611988612
1003816100381610038161003816 +100381610038161003816100381611988621
1003816100381610038161003816 +100381610038161003816100381611988622
1003816100381610038161003816 (26)
This norm also satisfies10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
] ⊲ [11988711
11988712
11988721
11988722
]
10038171003817100381710038171003817100381710038171003817
=1
2(
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
] [11988711
0
0 11988722
]
+ [11988711
0
0 11988722
] [11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
)
le
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
10038171003817100381710038171003817100381710038171003817
[11988711
0
0 11988722
]
10038171003817100381710038171003817100381710038171003817
le
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
10038171003817100381710038171003817100381710038171003817
[11988711
11988712
11988721
11988722
]
10038171003817100381710038171003817100381710038171003817
(27)
Next for any 119909 = [11988611
11988612
11988621
11988622
] isin 119872+ we observe that
1199092
= [11988611
11988612
11988621
11988622
] ⊲ [11988611
11988612
11988621
11988622
]
=[[
[
1198862
1111988612
11988611
+ 11988622
2
11988621
11988611
+ 11988622
21198862
22
]]
]
(28)
so that
1199092
⊲ 119909 = [
[
1198863
11
11988612
4(119886
11+ 119886
22)2
11988621
4(119886
11+ 119886
22)2
1198863
22
]
]
119909 ⊲ 1199092
= [
[
1198863
11
11988612
2(119886
2
11+ 119886
2
22)
11988621
2(119886
2
11+ 119886
2
22) 119886
3
22
]
]
(29)
So for any 119909 = [11988611
11988612
11988621
11988622
] isin 119872+ 1199092 ⊲ 119909 = 119909 ⊲ 119909
2
hArr
11988612
= 11988621
= 0 or 11988611
= 11988622 Thus by Proposition 26
10 Abstract and Applied Analysis
120590(119872+119868)([
11988611
11988612
11988621
11988622
]) = C whenever 11988611
= 11988622
and 11988612 119886
21 = 0
In particular for 119909 = [1 1
0 2] isin 119872
+ we have
1199092
⊲ 119909 =[[
[
13
1
4(1 + 2)
2
0
4(1 + 2)
2
23
]]
]
= [1
9
4
0 8
]
119909 ⊲ 1199092
=[[
[
13
1
2(1
2
+ 22
)
0
2(1
2
+ 22
) 23
]]
]
= [1
5
2
0 8
]
(30)
Concerning the inequality between the right hand sidesof (23) in Remark 27 we observe for 119909 as above that
1
4(119909119891 (119909) 119891 (119909) + 2119891 (119909) 119909119891 (119909) + 119891 (119909) 119891 (119909) 119909)
=1
4([
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 1
0 2] [
1 0
0 2]
+ [1 0
0 2] [
1 0
0 2] [
1 1
0 2]) = [
19
4
0 8
]
(31)
but
1
4(2119909119891 (119909) 119891 (119909) + 2119891 (119909) 119891 (119909) 119909)
=1
4(2 [
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 0
0 2] [
1 1
0 2])
= [1
5
2
0 8
]
(32)
Hence 1199092 ⊲ 119909 = 119909 ⊲ 1199092 Thus 120590
(119872+119868)([
1 1
0 2]) = C by
Proposition 26Further suppose the matrix [
11988611198862
11988631198864
] isin 119872+ is invertible
with respect to the unit 119868 that is [ 1198861 119886211988631198864
] isin 119866119868(119872
+
) Thenthere exists [ 119887
11198872
11988731198874
] isin 119872+ such that
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] = 119868 + 119868⊲([
1198861
1198862
1198863
1198864
]) = [1 minus119886
2
minus1198863
1]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
= [1198861
1198862
1198863
1198864
] + 119868⊲([
1198861
1198862
1198863
1198864
])
+ 119868⊲([
1198861
1198862
1198863
1198864
]
2
)
=[[
[
1198861
minus1198862
1198861+ 119886
4
2
minus1198863
1198861+ 119886
4
21198864
]]
]
(33)
However
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] =[[
[
11988711198861
1198872
1198861+ 119886
4
2
1198873
1198861+ 119886
4
211988741198864
]]
]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
=
[[[
[
11988711198861
1198872
1198862
1+ 119886
2
4
2
1198873
1198862
1+ 119886
2
4
211988741198862
4
]]]
]
(34)
It follows that 1198861
= 0 1198864
= 0 1198871= 1119886
1 119887
4= 1119886
4 119887
2((119886
1+
1198864)2) = minus119886
2 119887
3((119886
1+119886
4)2) = minus119886
3 119887
2((119886
2
1+119886
2
4)2) = minus119886
2((119886
1+
1198864)2) and 119887
3((119886
2
1+ 119886
2
4)2) = minus119886
3((119886
1+ 119886
4)2) From these
equations we get 1198872(119886
1minus 119886
4)2
= 0 and 1198873(119886
1minus 119886
4)2
= 0 so that(119887
2minus119887
3)(119886
1minus 119886
4)2
= 0Then for 1198861
= 1198864 we obtain 119887
2= 119887
3= 0
and hence 1198862= 119886
3= 0 Therefore
119866119868(119872
+
) = [119886 119887
119888 119886] 119909 = [
120572 0
0 120573] isin 119872
+
119886 = 0
120572 = 0 120573 = 0 120572 = 120573
(35)
The set 119866119868(119872
+
) is not open clearly [1 1
0 1] isin 119866
119868(119872
+
) forany 120598 gt 0
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[1 1
0 1] minus [
[
1 minus120598
41
0 1 +120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[
[
120598
40
0 minus120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=120598
2lt 120598
(36)
but [ 1minus1205984 1
0 1+1205984] notin 119866
119868(119872
+
)Now if 119860 = [
minus1 1
0 1] then 120582119868 minus 119860 = [
120582+1 minus1
0 120582minus1] notin 119866
119868(119872
+
)and so 120582 isin 120590
(119872+119868)(119860) for all 120582 isin C Thus 120590
(119872+119868)(119860) = C an
unbounded spectrum
Next we observe that the spectrum of an element withrespect to a unit is closed whenever the corresponding set ofinvertibles is open
Proposition 29 LetI be a quasi-Jordan normed algebra witha unit 119890 such that119866
119890(I) is openThen 120590
(I119890)(119909) is closed for all119909 isin I
Proof Define 119891 C rarr I by 119891(120582) = 120582119890 minus 119909 Since 119891 iscontinuous the inverse image of the open set 119866
119890(I) is open
in C and so its complement 120590(I119890)(119909) is closed
We conclude this paper with the following example of anonclosed spectrum
Example 30 Let 119872 and 119872+ be as in Example 28 Let 119864 =
[1 1
0 1] and 119860 = [
119886 0
0 119887] with 119886 = 119887 both different from 1 Then
119864 is a unit in 119872+ We show that 120590
(119872+119864)
(119860) = C 1 Forthis let us first investigate when can an element of the form119861 = [
120572 120573
0 120574] be invertible Assuming that 119861 is invertible we get
the existence of an element 119862 isin 119863+ such that
119862 ⊲ 119861 = 119864 + 119864⊲(119861)
119862 ⊲ 1198612
= 119861 + 119864⊲(119861) + 119864
⊲(119861
2
)
(37)
Abstract and Applied Analysis 11
From these equations we get 120572 = 0 120574 = 0 and
119862 =[[
[
1
120572120573∘
01
120574
]]
]
(38)
where 120573∘satisfies the following two equations
120573∘(120572 + 120574
2) = 1 minus 120573 +
120572 + 120574
2
120573∘(1205722
+ 1205742
2) =
120572 + 120574
2+
1205722
+ 1205742
2minus 120573
120572 + 120574
2
(39)
Multiplying the last equation by 2(120572 + 120574)(1205722
+ 1205742
) and thenusing the other equation we get
2 (1 minus 120573) + (120572 + 120574) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
+ (120572 + 120574) (40)
or equivalently
2 (1 minus 120573) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
(41)
this equation is satisfied for 120573 = 1 or 120572 = 120574 Hence thematrix119861 is invertible if and only if 119861 = [
120572 1
0 120574] or 119861 = [
120572 120573
0 120572
] for 120572 120574 isin
C 0 and 120573 isin CWe conclude that 119860
120582= 120582119864 minus 119860 = [
120582minus119886 120582
0 120582minus119887] is invertible
with respect to 119864 if and only if 120582 minus 119887 = 0 120582 minus 119886 = 0 and 120582 = 1that is119860
120582is invertible if 119886 = 120582 and 119887 = 120582 and 120582 = 1 Hence119860
120582
is invertible only if 120582 = 1 as we assumed that 119886 = 119887 and bothare not 1 So for all 120582 = 1 119860
120582notin 119866
119864(119872
+
) Thus 120590(119872+119864)
(119860) =
C 1 which is neither bounded nor closed
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This project was supported by King Saud University Dean-ship of Scientific Research College of Science ResearchCenter
References
[1] B Fritzsche ldquoSophus Lie a sketch of his life and workrdquo Journalof Lie Theory vol 9 no 1 pp 1ndash38 1999
[2] P Jordan ldquoUber eine klasse nichtassoziativer hyperkomplexeralgebrenrdquo Nachrichten von der Gesellschaft der Wissenschaftenzu Gottingen pp 569ndash575 1932
[3] K McCrimmon A Taste of Jordan Algebras UniversitextSpringer New York NY USA 2004
[4] N Jacobson Structure and Representations of Jordan Algebrasvol 39 of American Mathematical Society Colloquium Publica-tions American Mathematical Society Providence RI USA1968
[5] F Gursey and C-H Tze On the Role of Division Jordan andRelatedAlgebras in Particle PhysicsWorld Scientific River EdgeNJ USA 1996
[6] R Iordanescu Jordan Structures in Analysis Geometry andPhysics EdituraAcademiei Romane Bucharest Romania 2009
[7] H Upmeier Symmetric Banach Manifolds and Jordan Clowast-Alge-bras vol 104 of North-Holland Mathematics Studies North-Holland Amsterdam The Netherlands 1985
[8] I L Kantor ldquoClassification of irreducible transitive differentialgroupsrdquo Doklady Akademii Nauk SSSR vol 158 pp 1271ndash12741964
[9] M Koecher ldquoUber eine Gruppe von rationalen AbbildungenrdquoInventiones Mathematicae vol 3 pp 136ndash171 1967
[10] J Tits ldquoUne classe drsquoalgebres de Lie en relation avec les algebresde Jordanrdquo Indagationes Mathematicae vol 24 pp 530ndash5351962
[11] J-L Loday ldquoUne version non commutative des algebres de Lieles algebres de Leibnizrdquo LrsquoEnseignement Mathematique vol 39no 3-4 pp 269ndash293 1993
[12] J-L Loday and T Pirashvili ldquoUniversal enveloping algebras ofLeibniz algebras and (co)homologyrdquo Mathematische Annalenvol 296 no 1 pp 139ndash158 1993
[13] J-L Loday ldquoDialgebrasrdquo in Dialgebras and Related Operadsvol 1763 of Lecture Notes in Math pp 7ndash66 Springer BerlinGermany 2001
[14] R Velasquez and R Felipe ldquoQuasi-Jordan algebrasrdquo Communi-cations in Algebra vol 36 no 4 pp 1580ndash1602 2008
[15] R Velasquez and R Felipe ldquoOn K-B quasi-Jordan algebrasand their relation with Liebniz algebrasrdquo Comunicaciones delCIMAT No 1-10-1015-12-2010
[16] R Felipe ldquoAn analogue to functional analysis in dialgebrasrdquoInternational Mathematical Forum vol 2 no 21ndash24 pp 1069ndash1091 2007
[17] R Velasquez and R Felipe ldquoSplit dialgebras split quasi-Jordanalgebras and regular elementsrdquo Journal of Algebra and ItsApplications vol 8 no 2 pp 191ndash218 2009
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 Abstract and Applied Analysis
Proof For the first part see [17] Clearly (119909 119877119910) = 119909 +
119877119910 is a norm and it satisfies
10038171003817100381710038171003817(119909 119877
119910) ⊲ (119911 119877
119908)10038171003817100381710038171003817=
10038171003817100381710038171003817(119909 ⊲ 119908 119877
119910⊲119908)10038171003817100381710038171003817
=10038171003817100381710038171003817(119909 ⊲ 119908 119877
119910∙ 119877
119908)10038171003817100381710038171003817
=1003817100381710038171003817119877119908
(119909)1003817100381710038171003817 +
10038171003817100381710038171003817119877119910∙ 119877
119908
10038171003817100381710038171003817
le (119909 +10038171003817100381710038171003817119877119910
10038171003817100381710038171003817)1003817100381710038171003817119877119908
1003817100381710038171003817
le (119909 +10038171003817100381710038171003817119877119910
10038171003817100381710038171003817) (119911 +
1003817100381710038171003817119877119908
1003817100381710038171003817)
=10038171003817100381710038171003817(119909 119877
119910)10038171003817100381710038171003817
1003817100381710038171003817(119911 119877119908)1003817100381710038171003817
(3)
for all 119908 119909 119910 119911 isin I Keeping in view Proposition 3 wededuce that I
1being the product of complete spaces I and
119877(I) is complete in the product norm Thus I1is a split
quasi-Jordan Banach algebra with (right) unit (119890 119877119890)
Clearly 120593(I) is a subspace ofI1with (119909 119877
119909) ⊲ (119910 119877
119911) =
(119909 ⊲ 119911 119877119909⊲119911
) isin 120593(I) for all (119909 119877119909) isin 120593(I) (119910 119877
119911) isin I
1 so
that 120593(I) is a quasi-Jordan normed subalgebra with the unit(119890 119877
119890) included Further let (119909
119899 119877
119909119899
) be any fixed Cauchysequence in 120593(I) Then 119909
119898minus 119909
119899 le 119909
119898minus 119909
119899 + 119877
119909119898
minus
119877119909119899
= (119909119898
minus 119909119899 119877
119909119898
minus 119877119909119899
) = (119909119898 119877
119909119898
) minus (119909119899 119877
119909119899
) rarr
0 as 119898 119899 rarr infin so that 119909119899 is a Cauchy sequence in (the
complete space) I and hence it converges to some 119909 isin INow by using the fact 119877
119911le 119911 for all 119911 isin I (since
119890 is a norm 1 unit in I see above) we get the convergenceof arbitrarily fixed Cauchy sequence (119909
119899 119877
119909119899
) to (119909 119877119909) isin
120593(I) since (119909119899 119877
119909119899
)minus(119909 119877119909) = (119909
119899minus119909 119877
119909119899minus119909
) = 119909119899minus119909+
119877119909119899minus119909
le 2119909119899minus119909 rarr 0 as 119899 rarr infinThus 120593(I) being com-
plete is a closed right ideal inI1in the norm topology
Next we observe the isometry between I and 120593(I)
Proposition 5 Let I be a quasi-Jordan Banach algebra withnorm 1 unit Then there exists an equivalent norm that makesI isometrically isomorphic to a quasi-Jordan closed subalgebra120593(I) of the split quasi-Jordan Banach algebra I
1
Proof Clearly 119909∘
= 119909 + 119877119909 defines a norm on the
quasi-Jordan algebra I It follows that (I sdot ∘) is a quasi-
Jordan Banach algebraMoreoverI is isomorphic to the sub-algebra (119909 119877
119909) 119909 isin I of I
1under the isomorphism
120593 119909 997891rarr (119909 119877119909) as seen above Further we observe that 120593
is an isometry since 119909∘= 119909 + 119877
119909 = (119909 119877
119909) = 120593(119909)
Finally we note that the two norms sdot and sdot ∘on I are
equivalent since 119909 le 119909 + 119877119909 = 119909
∘= 119909 + 119877
119909 le
2119909
3 Invertible Elements
As in [17] an element 119909 in a quasi-Jordan algebra I is calledinvertible with respect to a unit 119890 isin I if there exists 119910 isin I
such that 119910 ⊲ 119909 = 119890 + (119890 ⊲ 119909 minus 119909) and 119910 ⊲ 1199092
= 119909 + (119890 ⊲
119909 minus 119909) + (119890 ⊲ 1199092
minus 1199092
) such an element 119910 is called an inverseof119909 with respect to 119890 Let 119890
⊲(119909) denote the element 119890 ⊲ 119909 minus 119909
Then119909 has an inverse119910with respect to 119890 hArr 119910 ⊲ 119909 = 119890+119890⊲(119909)
and 119910 ⊲ 1199092
= 119909 + 119890⊲(119909) + 119890
⊲(119909
2
)We know from the above discussion that the embedding
119909 997891rarr 120593(119909) = (119909 119877119909) of a quasi-Jordan algebraI into the split
quasi-Jordan algebra I1preserves the units The embedding
120593 also preserves the corresponding invertible elements if 119910 isan inverse of 119909 with respect to a unit 119890 in I then 119910 ⊲ 119909 =
119890 + (119890 ⊲ 119909 minus 119909) and 119910 ⊲ 1199092
= 119909 + (119890 ⊲ 119909 minus 119909) + (119890 ⊲ 1199092
minus 1199092
)Hence
(119910 119877119910) ⊲ (119909 119877
119909) = (119910 ⊲ 119909 119877
119910⊲119909)
= (119890 + (119890 ⊲ 119909 minus 119909) 119877119890+(119890⊲119909minus119909)
)
= (119890 + (119890 ⊲ 119909 minus 119909) 119877119890+ 119877
119890⊲119909minus 119877
119909)
= (119890 119877119890) + (119890 ⊲ 119909 119877
119890⊲119909) minus (119909 119877
119909)
= (119890 119877119890) + (119890 119877
119890) ⊲ (119909 119877
119909) minus (119909 119877
119909)
(119910 119877119910) ⊲ (119909 119877
119909)2
= (119910 ⊲ 1199092
119877119910⊲1199092)
= (119909 119877119909) + ((119890 ⊲ 119909 119877
119890⊲119909) minus (119909 119877
119909))
+ ((119890 ⊲ 1199092
119877119890⊲1199092) minus (119909
2
1198771199092))
= (119909 119877119909) + ((119890 119877
119890) ⊲ (119909 119877
119909) minus (119909 119877
119909))
+ ((119890 119877119890) ⊲ (119909 119877
119909)2
minus (119909 119877119909)2
)
(4)
Thus (119910 119877119910) is an inverse of (119909 119877
119909) with respect to the unit
(119890 119877119890) inI
1
Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebraThen 119877
119909(119910) = 119910 ⊲ 119909 = 119910 ⊲ (119909
119869+ 119909
119885) = 119910 ⊲ 119909
119869+ 119910 ⊲ 119909
119885=
119910 ⊲ 119909119869+ 0 = 119877
119909119869
(119910) for all 119909 119910 isin I Thus 119877119909= 119877
119909119869
for all119909 isin I
In this section we demonstrate that the set of invertibleelements with respect to a fixed unit in a quasi-JordanBanach algebra may not be open For this we proceed asfollows
Proposition 6 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebra with unit 119890 isin 119869 If 119909 isin I is invertible withrespect to 119890 then so is 120582119909 for all 120582 = 0
Proof Let119910 be an inverse of 119909 inIwith respect to 119890We showthat 1199101015840
= (1120582)119910119869+ 119910
119885is an inverse of 120582119909 with respect to 119890
Observe that
119910119869⊲ 119909 + 119910
119885⊲ 119909 = (119910
119869+ 119910
119885) ⊲ 119909
= 119910 ⊲ 119909 = 119890 + 119890⊲(119909)
= 119890 + 119890⊲(119909
119869+ 119909
119885) = 119890 + (minus119909
119885)
Abstract and Applied Analysis 5
119910119869⊲ 119909
2
+ 119910119885⊲ 119909
2
= 119910 ⊲ 1199092
= 119909 + 119890⊲(119909
119869+ 119909
119885) + 119890
⊲((119909
119869+ 119909
119885)2
)
= 119909119869+ (minus119909
119885⊲ 119909)
(5)
Hence by the uniqueness of the representation as sum of Jor-dan and zero parts we get 119910
119869⊲ 119909 = 119890 119910
119885⊲ 119909 = minus119909
119885 119910
119869⊲
1199092
= 119909119869 and 119910
119885⊲ 119909
2
= minus119909119885
⊲ 119909 Therefore 1199101015840
⊲ 120582119909 =
120582(((1120582)119910119869+ 119910
119885) ⊲ 119909) = 119910
119869⊲ 119909 + 120582119910
119885⊲ 119909 = 119890 minus 120582119909
119885=
119890 + 119890⊲(120582119909) and 119910
1015840
⊲ (120582119909)2
= 1205822
((1120582)119910119869⊲ 119909
2
+ 119910119885
⊲ 1199092
) =
120582119909119869minus 120582
2
119909119885⊲ 119909 = 120582119909 + 119890
⊲(120582119909) + 119890
⊲((120582119909)
2
) because 119890⊲(120582119909) =
119890 ⊲ (120582119909) minus 120582119909 = minus120582119909119885and 119890
⊲((120582119909)
2
) = 119890 ⊲ 1205822
1199092
minus 1205822
1199092
=
1205822
(119890⊲(119909
2
)) = minus1205822
119909119885⊲ 119909
Proposition 7 LetI be a quasi-Jordan normed algebrawith aunit 119890 Let119866
119890(I) = 119909 isin I 119909 is invertible with respect to 119890
be an open set and 119909 isin 119866119890(I) Then 119909 + 119911 isin 119866
119890(I) for all
119911 isin 119885(I)
Proof Suppose 119909 isin 119866119890(I) and 119911 isin 119885(I) If 119911 = 0 then 119909+119911 =
119909 isin 119866119890(I) Next suppose 119911 = 0 Since 119866
119890(I) is an open set
there exists 120598 gt 0 such that 119886 isin 119866119890(I) whenever 119909 minus 119886 lt 120598
Hence 119909 + 119911∘isin 119866
119890(I) with 119911
∘= (1205982119911)119911
Let 119910 and 119910∘be inverses of 119909 and 119909 + 119911
∘with respect to 119890
respectively Then
119910∘⊲ (119909 + 119911
∘) = 119890 + 119890
⊲(119909 + 119911
∘)
= 119890 + 119890⊲(119909) minus 119911
∘
= 119910 ⊲ 119909 minus 119911∘
119910∘⊲ (119909 + 119911
∘)2
= 119909 + 119911∘+ 119890
⊲(119909 + 119911
∘) + 119890
⊲((119909 + 119911
∘)2
)
= 119909 + 119890⊲(119909) + 119890
⊲(119909
2
) minus 119911∘⊲ 119909
= 119910 ⊲ 1199092
minus 119911∘⊲ 119909
(6)
Hence by setting 1199101= 119910 + 120572(119910
∘minus 119910) with 120572 = 2119911120598 we see
that1199101⊲ (119909 + 119911) = (119910 + 120572 (119910
∘minus 119910)) ⊲ 119909
= 119910 ⊲ 119909 + 120572 (119910∘⊲ (119909 + 119911
∘) minus 119910 ⊲ 119909)
= 119890 + 119890⊲(119909) + 120572 (minus119911
∘)
= 119890 + 119890⊲(119909) minus 119911
= 119890 + 119890⊲(119909 + 119911)
1199101⊲ (119909 + 119911)
2
= (119910 + 120572 (119910∘minus 119910)) ⊲ 119909
2
= 119910 ⊲ 1199092
+ 120572 (119910∘⊲ (119909 + 119911
∘)2
minus 119910 ⊲ 1199092
)
= 119909 + 119890⊲(119909) + 119890
⊲(119909
2
) minus 120572 (119911∘⊲ 119909)
= 119909 + 119890⊲(119909) + 119890
⊲(119909
2
) minus 119911 ⊲ 119909
= (119909 + 119911) + 119890⊲(119909 + 119911) + 119890
⊲((119909 + 119911)
2
)
(7)
since 119911 isin 119885(I) Thus 1199101is an inverse of 119909 + 119911 with respect to
the unit 119890
Corollary 8 Under the hypothesis of Proposition 7 120582119890 minus 119909 isin
119866119890(I) implies 120582119890 minus (119909 + 119911) isin 119866
119890(I) for all 119911 isin 119885(I)
Proposition 9 LetI = 119869 oplus 119885(I) be a split quasi-Jordan nor-med algebra and let 119890 isin 119869 be a unit inI such that the set119866
119890(I)
is open Then 119909 ⊲ 1199092
= 1199092
⊲ 119909 for all 119909 isin I
Proof Of course the element 119890 is the unit of the Jordanalgebra 119869 Then for any fixed element 119886 isin 119869 there exists120582 isin C such that 120582119890 minus 119886 is invertible in 119869 otherwise wewould get the negation of the well-known fact that spectrumof an element of a unital Jordan algebra is bounded Thatis there exists 119910 isin 119869 such that 119910 ⊲ (120582119890 minus 119886) = 119890 and119910 ⊲ (120582119890minus119886)
2
= (120582119890minus119886) However 119890⊲(119886) = 0 = 119890
⊲(119886
2
) Hence119910 is an inverse of 119886 in the quasi-Jordan algebraIwith respectto the unit 119890 By Corollary 8 120582119890 minus (119886 + 119911) is also invertiblefor any 119911 isin 119885(I) This in turn gives the existence of 119887 isin I
satisfying 119887 ⊲ (120582119890 minus (119886 + 119911)) = 119890 + 119911 and 119887 ⊲ (120582119890 minus (119886 + 119911))2
=
(120582119890minus119886)+119911 ⊲ (120582119890minus119886) Multiplying the first equation from theright by (120582119890 minus (119886 + 119911))
2 the second equation by (120582119890minus (119886+119911))and using the right Jordan identity we get
(119890 + 119911) ⊲ (120582119890 minus (119886 + 119911))2
= ((120582119890 minus 119886) + 119911 ⊲ (120582119890 minus 119886)) ⊲ (120582119890 minus (119886 + 119911))
(8)
Hence
(119890 + 119911) ⊲ (120582119890 minus 119886)2
= ((120582119890 minus 119886) + 119911 ⊲ (120582119890 minus 119886)) ⊲ (120582119890 minus 119886)
(9)
so that
119890 ⊲ (120582119890 minus 119886)2
+ 1205822
119911 minus 2120582 (119911 ⊲ 119886) + 119911 ⊲ 1198862
= (120582119890 minus 119886)2
+ 1205822
119911 minus 2120582119911 ⊲ 119886 + (119911 ⊲ 119886) ⊲ 119886
(10)
This last equation reduces to 119890 ⊲ (120582119890 minus 119886)2
= (120582119890 minus 119886)2 since
(120582119890 minus 119886)2
isin 119869 and 119890 is the unit of 119869 Hence 119911 ⊲ 1198862
= (119911 ⊲
119886) ⊲ 119886 for all 119886 isin 119869 and 119911 isin 119885(I) Now for any 119909 isin I thelast equation with 119886 = 119909
119869and 119911 = 119909
119885gives 119909
119885⊲ 119909
2
119869= (119909
119885⊲
119909119869) ⊲ 119909
119869Thus 119909 ⊲ 119909
2
= 119909119869⊲ 119909
2
119869+119909
119885⊲ 119909
2
119869= 119909
2
119869⊲ 119909
119869+(119909
119885⊲
119909119869) ⊲ 119909
119869= 119909
2
⊲ 119909 for all 119909 isin I
Corollary 10 If a unital split quasi-Jordan algebra has anelement 119909 with 119909 ⊲ 119909
2
= 1199092
⊲ 119909 then the set of invertibleelements with respect to the unit of the Jordan part is not open
In the sequel we will show the existence of a unital splitquasi-Jordan Banach algebra with elements 119909 such that 119909 ⊲
1199092
= 1199092
⊲ 119909 Thus the above result establishes that the set ofinvertible elements with respect to a fixed unit in a quasi-Jordan Banach algebra may not be open
6 Abstract and Applied Analysis
4 The Spectrum of Elements in a UnitalQuasi-Jordan Algebra
As usual we define the spectrum of an element 119909 in aunital quasi-Jordan algebra (I 119890) denoted by 120590
(I119890)(119909) tobe the collection of all complex numbers 120582 for which 120582119890 minus
119909 is not invertible Thus 120590(I119890)(119909) = 120582 isin C 120582119890 minus
119909 is not invertible Here the subscript 119890 indicates that theinvariability depends on the choice of unit 119890 which generallyis not unique
Proposition 11 Let I be a unital quasi-Jordan algebra Then120590(I119890)(119890
1015840
) = 1 for all 119890 1198901015840 isin 119880(I)
Proof Let 119890 1198901015840 isin 119880(I) Then for any 120582 = 1 119910 = (1(120582minus1))1198901015840
is an inverse of 120582119890 minus 1198901015840 with respect to the unit 119890 because 119910 ⊲
(120582119890 minus 1198901015840
) = (1(120582 minus 1))1198901015840
⊲ (120582119890 minus 1198901015840
) = 119890 + 119890⊲(120582119890 minus 119890
1015840
) and119910 ⊲ (120582119890 minus 119890
1015840
)2
= (120582119890 minus 1198901015840
) + 119890⊲(120582119890 minus 119890
1015840
) + 119890⊲((120582119890 minus 119890
1015840
)2
)
Proposition 12 Let I be a quasi-Jordan algebra with unit 119890Then 120590
(I119890)(119911) = 0 for all 119911 isin 119885(I)
Proof For any fixed 119911 isin 119885(I) and nonzero scalar 120582 thevector 119910 = (1120582)(119890 + 119911) satisfies 119910 ⊲ (120582119890 minus 119911) = 120582119910 = 119890 + 119911 =
119890+ 119890⊲(120582119890minus 119911) and 119910 ⊲ (120582119890 minus 119911)
2
= 1205822
119910 = 120582119890+120582119911 = (120582119890minus119911) +
119890⊲(120582119890 minus 119911) + 119890
⊲((120582119890 minus 119911)
2
) So (1120582)(119890 + 119911) is an inverse of120582119890 minus 119911 with respect to 119890 This means 120582 notin 120590
(I119890)(119911) for all 120582 = 0However the zero vector is not invertible Thus 120590
(I119890)(119911) =
0
Proposition 13 Let I = 119869 oplus 119885(I) be a split quasi-Jordanalgebra with unit 119890 isin 119869 Then 120590
(I119890)(119901) sube 0 1 for all idempo-tents 119901 isin I (ie 1199012
= 119901)
Proof Let 119901 be any fixed idempotent inI Since 119901 isin I 119901 hasa unique representation 119901 = 119901
119869+ 119901
119885with 119901
119869isin 119869 and 119901
119885isin
119885(I) Clearly 1199012
119869+ 119901
119885⊲ 119901
119869= 119901
2
= 119901 = 119901119869+ 119901
119885 Then
by uniqueness of the representation in the split quasi-Jordanalgebra I 119901
119885⊲ 119901
119869= 119901
119885and 119901
2
119869= 119901
119869 this means 119901
119869is an
idempotent in the Jordan algebra 119869 Hence 120590(119869119890)
(119901119869) sube 0 1
Thus 119901120582= 120582119890 minus 119901
119869is invertible in 119869 with the unique inverse
119901minus1
120582 for all 120582 notin 0 1We show that 119910 = 119901
minus1
120582+ (1(120582 minus 1))119901
119885is an inverse of
120582119890 minus 119901 in I with respect to the unit 119890 for this we note that120582119890 minus 119901 = 119901
120582minus 119901
119885 (119901
120582minus 119901
119885)2
= 1199012
120582minus 119901
119885⊲ 119901
120582 119890
⊲(120582119890 minus
119901) = 119890⊲(119901
120582minus 119901
119885) = 119890 ⊲ (119901
120582minus 119901
119885) minus (119901
120582minus 119901
119885) = 119901
119885 and
119890⊲((120582119890 minus 119901)
2
) = 119890Δ((119901
120582minus 119901
119885)2
) = 119890⊲(119901
2
120582minus 119901
119885⊲ 119901
120582) = 119901
119885⊲
119901120582= 119901
119885⊲ (120582119890minus119901
119869) = 120582119901
119885minus119901
119885⊲ 119901
119869= 120582119901
119885minus119901
119885= (120582minus1)119901
119885
Hence 119910 ⊲ (120582119890 minus 119901) = 119910 ⊲ (119901120582minus 119901
119885) = 119910 ⊲ 119901
120582= (119901
minus1
120582+
(1(120582 minus 1))119901119885) ⊲ 119901
120582= 119901
minus1
120582⊲ 119901
120582+ (1(120582 minus 1))119901
119885⊲ 119901
120582= 119890 +
(1(120582 minus 1))119901119885
⊲ (120582119890 minus 119901119869) = 119890 + (1(120582 minus 1))(120582119901
119885minus 119901
119885⊲
119901119869) = 119890 + (1(120582 minus 1))(120582119901
119885minus 119901
119885) = 119890 + 119901
119885= 119890 + 119890
⊲(120582119890 minus 119901)
and 119910 ⊲ (120582119890 minus 119901)2
= 119910 ⊲ (1199012
120582minus 119901
119885⊲ 119901
120582) = 119910 ⊲ 119901
2
120582=
(119901minus1
120582+(1(120582minus1))119901
119885) ⊲ 119901
2
120582= 119901
120582+(1(120582minus1))119901
119885⊲ (120582119890minus119901
119869)2
=
119901120582+(1(120582minus1))119901
119885⊲ (120582
2
119890minus2120582119901119869+119901
2
119869) = 119901
120582+(1(120582minus1))119901
119885⊲
(1205822
119890 minus 2120582119901119869+ 119901
119869) = 119901
120582+ (1(120582 minus 1))(120582
2
119901119885
⊲ 119890 minus 2120582119901119885
⊲
119901119869+ 119901
119885⊲ 119901
119869) = 119901
120582+ (1(120582 minus 1))(120582
2
119901119885
minus 2120582119901119885
+ 119901119885) =
119901120582+ (1(120582 minus 1))(120582 minus 1)
2
119901119885= 120582119890 minus119901
119869+ (120582 minus 1)119901
119885= (120582119890 minus119901) +
119890⊲(120582119890 minus 119901) + 119890
⊲((120582119890 minus 119901)
2
)
As mentioned in Section 2 if I is a quasi-Jordan algebrawith a unit 119890 then the set 119890 + 119911 119911 isin 119885(I) coincides withthe set 119880(I) of all units inI
Proposition 14 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebrawith unit 119890 isin 119869 and119909 isin I invertiblewith respectto some 1198901015840 isin 119880(I)Then119909
119869is invertible with respect to the unit
119890 in the Jordan algebra 119869
Proof Clearly 119890 ⊲ 119909 minus 119909 = minus119909119885and 119890 ⊲ 119909
2
minus 1199092
= minus119909119885⊲ 119909
119869
Since 1198901015840
isin 119880(I) we have 1198901015840
= 119890 + 119911 for some 119911 isin 119885(I)Hence the invertibility of 119909 in I with respect to the unit 1198901015840gives the existence of 119910 isin I such that 119910
119869⊲ 119909
119869+ 119910
119885⊲ 119909
119869=
119910 ⊲ 119909119869= 119910 ⊲ 119909 = 119890 + 119911 + (119890 + 119911) ⊲ 119909 minus 119909 = 119890 + 119911 + 119911 ⊲ 119909 minus 119909
119885
and 119910119869⊲ 119909
2
119869+ 119910
119885⊲ 119909
2
119869= 119910 ⊲ 119909
2
119869= 119910 ⊲ 119909
2
= (119890 + 119911) ⊲
119909 + (119890 + 119911) ⊲ 1199092
minus 1199092
= 119909119869+ 119911 ⊲ 119909 + 119911 ⊲ 119909
2
minus 119909119885⊲ 119909
119869 So by
the uniqueness of the representations in the split algebra Iwe get 119910
119869⊲ 119909
119869= 119890 and 119910
119869⊲ 119909
2
119869= 119909
119869 Thus 119910
119869is the inverse
of 119909119869in the Jordan algebra 119869
Next we observe that the spectrum of 119909 in a unital splitquasi-Jordan algebra with respect to any unit includes thespectrum of 119909
119869in the Jordan part 119869
Corollary 15 LetI = 119869 oplus119885(I) be a unital split quasi-Jordanalgebra with unit 119890 isin 119869 and let 119909 isin I Then 120590
(119869119890)(119909
119869) sube
120590(I1198901015840)(119909) for all 1198901015840 isin 119880(I)
Proof Let 1198901015840 = 119890 + 119911 with 119911 isin 119885(I) and let 120582 notin 120590(I1198901015840)(119909)
Then 1205821198901015840 minus 119909 is invertible in I with respect to the unit 1198901015840Hence its Jordanpart120582119890minus119909
119869is invertible in the Jordan algebra
119869 by Proposition 14 Thus 120582 notin 120590(119869119890)
(119909119869)
It is well known that the spectrum of any element ina unital Jordan Banach algebra is nonempty (cf [7]) Thistogether with Proposition 2 and Corollary 15 gives thefollowing result
Corollary 16 The spectrum of any element in a unital splitquasi-Jordan Banach algebra is nonempty
The next result extends Corollary 16 to any quasi-JordanBanach algebra with a norm 1 unit
Proposition 17 The spectrum of any element in a quasi-Jordan Banach algebra with a norm 1 unit is nonempty
Proof Let I be a unital quasi-Jordan Banach algebra witha norm 1 unit 119890 From Section 2 we know that the map120593 119909 997891rarr 120593(119909) = (119909 119877
119909) embeds I into the unital split
quasi-Jordan Banach algebra I1
= (119909 119877119910) 119909 119910 isin I
equipped with the sum (119909 119877119910) + (119886 119877
119887) = (119909 + 119886 119877
119910+119887) sca-
lar multiplication 120582(119909 119877119910) = (120582119909 119877
120582119910) and product (119909
119877119910) ⊲ (119886 119877
119887) = (119909 ⊲ 119887 119877
119910⊲119887) and the image 120593(I) is a
norm closed right ideal isomorphic to I with norm 1 unit(119890 119877
119890) Moreover it is seen in Section 3 that the embedding
Abstract and Applied Analysis 7
120593 also preserves the corresponding invertible elements thatis (119910 119877
119910) is an inverse of (119909 119877
119909) with respect to the unit
(119890 119877119890) in I
1whenever 119910 is an inverse of 119909 with respect
to a unit 119890 in I Hence by Corollary 16 it follows that120601 = 120590
(I1(119890119877119890))(119909 119877
119909) sube 120590
(120593(I)(119890119877119890))(119909 119877
119909) = 120590
(I119890)(119909) for all119909 isin I
Proposition 18 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebra with unit 119890 isin 119869 and let 119909 isin 119869 Then 120590
(119869119890)(119909) =
120590(I119890)(119909)
Proof By Corollary 15 120590(119869119890)
(119909) sube 120590(I119890)(119909) For the reverse
inclusion let 120582 notin 120590(119869119890)
(119909) then 120582119890 minus 119909 is invertible in 119869 thatis there exists 119910 isin 119869 such that 119910 ⊲ (120582119890 minus 119909) = 119890 and 119910 ⊲
(120582119890 minus 119909)2
= 120582119890 minus 119909 However 119890⊲(120582119890 minus 119909) = 0 = 119890
⊲((120582119890 minus 119909)
2
)Hence 119910 is an inverse of 120582119890 minus 119909 in I with respect to the unit119890 that is 120582 notin 120590
(I119890)(119909) Thus 120590(I119890)(119909) sube 120590
(119869119890)(119909)
Proposition 19 Let I be a unital quasi-Jordan normedalgebra and let 119890 be a unit inI for which 119866
119890(I) is open Then
120590(I119890)(119909) = 120590
(I119890)(119909 + 119911) for all 119911 isin 119885(I)
Proof By Corollary 8 120582119890 minus 119909 is invertible if and only if 120582119890 minus
(119909 + 119911) is invertible for all 119911 isin 119885(I) Thus 120582 notin 120590(I119890)(119909) if
and only if 120582 notin 120590(I119890)(119909 + 119911) for all 119911 isin 119885(I)
Corollary 20 LetI = 119869oplus119885(I) be a unital split quasi-Jordannormed algebra and let 119890 be a unit in I such that 119866
119890(I) is
open Then 120590(I119890)(119909) = 120590
(I119890)(119909119869) for all 119909 = 119909119869+ 119909
119885isin I
Further if the unit 119890 isin 119869 then120590(I119890)(119909) = 120590
(I119890)(119909119869) = 120590(119869119890)
(119909119869)
Lemma 21 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordanalgebra with a unit 119890 isin 119869 and let 119909 isin I be invertible withrespect to 119890 Then 119909
119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
Proof As 119909 is invertible there exists 119910 isin I such that 119910 ⊲ 119909 =
119890 + (119890 ⊲ 119909 minus 119909) = 119890 minus 119909119885and 119910 ⊲ 119909
2
= 119909 + (119890 ⊲ 119909 minus 119909) + (119890 ⊲
1199092
minus 1199092
) = 119909119869minus 119909
119885⊲ 119909 Hence by the uniqueness of the
representation in a split quasi-Jordan algebra we obtain
119910119885⊲ 119909 = minus119909
119885 (11)
119910119885⊲ 119909
2
= minus119909119885⊲ 119909 (12)
Thus
119909119885⊲ 119909
2
= minus (119910119885⊲ 119909) ⊲ 119909
2
(by (11))
= minus (119910119885⊲ 119909
2
) ⊲ 119909 (by the right Jordan identity)
= (119909119885⊲ 119909) ⊲ 119909 (by (12))
(13)
Proposition 22 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebra with a unit 119890 isin 119869 119909 = 119909
119869+ 119909
119885isin I satisfies
120590(I119890)(119909) =C Then
(1) 119909119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
(2) 1199092 ⊲ 119909 = 119909 ⊲ 1199092
Proof (1) Let 120582 notin 120590(I119890)(119909) Then 120582119890 minus 119909 = (120582119890 minus 119909
119869) + (minus119909
119885)
is invertible with respect to 119890 By Lemma 21 we have
119909119885⊲ (120582119890 minus 119909)
2
= (119909119885⊲ (120582119890 minus 119909)) ⊲ (120582119890 minus 119909) (14)
since the zero part of 120582119890 minus 119909 is minus119909119885 However
119909119885⊲ (120582119890 minus 119909)
2
= 119909119885⊲ (120582
2
119890 minus 120582119890 ⊲ 119909 minus 120582119909 ⊲ 119890 + 1199092
)
= 1205822
(119909119885⊲ 119890) minus 120582 (119909
119885⊲ (119890 ⊲ 119909))
minus 120582 (119909119885⊲ (119909 ⊲ 119890)) + 119909
119885⊲ 119909
2
= 1205822
119909119885minus 2120582119909
119885⊲ 119909
+ 119909119885⊲ 119909
2
(by the right commutativity of ⊲)
(119909119885⊲ (120582119890 minus 119909)) ⊲ (120582119890 minus 119909)
= (120582 (119909119885⊲ 119890) minus 119909
119885⊲ 119909) ⊲ (120582119890 minus 119909)
= (120582119909119885minus 119909
119885⊲ 119909) ⊲ (120582119890 minus 119909)
= 1205822
(119909119885⊲ 119890) minus 120582 (119909
119885⊲ 119909) ⊲ 119890
minus 120582119909119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
= 1205822
119909119885minus 2120582119909
119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
(15)
Therefore (14) becomes
1205822
119909119885minus 2120582119909
119885⊲ 119909 + 119909
119885⊲ 119909
2
= 1205822
119909119885minus 2120582119909
119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
(16)
which after simplification reduces to the required equation119909119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
(2) Since 1199092 = (119909119869+ 119909
119885)2
= 1199092
119869+ 119909
119885⊲ 119909
119869 we have
1199092
⊲ 119909 = (1199092
119869+ 119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909
119869+ (119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909
119869+ 119909
119885⊲ 119909
2
(17)
by the part (1) But 1199092119869⊲ 119909
119869= 119909
119869⊲ 119909
2
119869since 119909
119869is in the Jor-
dan algebra 119869 Therefore 1199092 ⊲ 119909 = 1199092
119869⊲ 119909
119869+ 119909
119885⊲ 119909
2
= 119909119869⊲
1199092
119869+ 119909
119885⊲ 119909
2
= 119909 ⊲ 1199092
Remark 23 In any quasi-Jordan algebra if an element 119909
satisfies 1199093
= 1199092
⊲ 119909 = 119909 ⊲ 1199092 then 119909
119899
⊲ 1199092
= 119909119899+2 for
all positive integers 119899 For this suppose 119909 satisfies 1199092 ⊲ 119909 =
119909 ⊲ 1199092 and 119909
119898
⊲ 1199092
= 119909119898+2 for any fixed119898 ge 1 Then 119909
119898+1
⊲
1199092
= (119909119898
⊲ 119909) ⊲ 1199092
= (119909119898
⊲ 1199092
) ⊲ 119909 (by the right Jor-dan identity) = 119909
119898+2
⊲ 119909 = 119909119898+3
Proposition 24 Let I be a unital quasi-Jordan Banachalgebra with unit 119890 and let 119909 isin I satisfy (119890 minus 119909) ⊲ (119890 minus 119909)
2
=
(119890 minus 119909)2
⊲ (119890 minus 119909) Then 119909 isin 119866119890(I) whenever 119890 minus 119909 lt 1
8 Abstract and Applied Analysis
Proof First note that 119890minus119909 lt 1 gives (119890 minus 119909)119899
le 119890 minus 119909119899
lt
1 for all 119899 = 1 2 3 Hence the infinite geometric series119890 +sum
infin
119899=1(119890 minus119909)
119899 converges absolutely to some element 119910 isin IWe show that the geometric series sum 119910 is an inverse of 119909with respect to the unit 119890 For any fixed positive integer 119899 let119910119899= 119890+sum
119899
119896=1(119890minus119909)
119896Then the sequence 119910119899 of partial sums
converges to 119910 By setting 119908 = 119890 minus 119909 we get
119910119899⊲ 119909 = (119890 +
119899
sum
119896=1
119908119896
) ⊲ (119890 minus 119908)
= 119890 +
119899
sum
119896=1
119908119896
minus (119890 ⊲ 119908 +
119899+1
sum
119896=2
119908119896
)
= 119890 + 119908 minus 119890 ⊲ 119908 minus 119908119899+1
= 119890 + (119890 minus 119909) minus 119890 ⊲ (119890 minus 119909) minus (119890 minus 119909)119899+1
= 119890 + (119890 minus 119909 minus 119890 + 119890 ⊲ 119909) minus (119890 minus 119909)119899+1
= 119890 + (119890 ⊲ 119909 minus 119909) minus (119890 minus 119909)119899+1
(18)
Thus by allowing 119899 rarr infin we obtain 119909 ⊲ 119910 = 119890 + (119890 ⊲
119909 minus 119909) = 119890 + 119890⊲(119909) since 119890 minus 119909 lt 1
Next by Remark 23 we have
119910119899⊲ 119909
2
= 119910119899⊲ (119890 minus 119908)
2
= 119910119899⊲ (119890 minus 119890 ⊲ 119908 minus 119908 ⊲ 119890 + 119908
2
)
= 119910119899⊲ (119890 minus 2119908 + 119908
2
)
= 119910119899⊲ 119890 minus 2119910
119899⊲ 119908 + 119910
119899⊲ 119908
2
= 119910119899minus 2119910
119899⊲ 119908 + 119910
119899⊲ 119908
2
= (119890 +
119899
sum
119896=1
119908119896
) minus 2(119890 +
119899
sum
119896=1
119908119896
) ⊲ 119908
+ (119890 +
119899
sum
119896=1
119908119896
) ⊲ 1199082
= (119890 +
119899
sum
119896=1
119908119896
) minus 2(119890 ⊲ 119908 +
119899+1
sum
119896=2
119908119896
)
+ (119890 ⊲ 1199082
+
119899+2
sum
119896=3
119908119896
)
= 119890 + 119908 minus 2119890 ⊲ 119908 minus 1199082
+ 119890 ⊲ 1199082
minus 119908119899+1
+ 119908119899+2
(19)
Taking the limit as 119899 rarr infin we get
119910 ⊲ 1199092
= 119890 + 119908 minus 2119890 ⊲ 119908 minus 1199082
+ 119890 ⊲ 1199082
= 119890 + (119890 minus 119909) minus 2119890 ⊲ (119890 minus 119909) minus (119890 minus 119909)2
+ 119890 ⊲ (119890 minus 119909)2
= 2119890 minus 119909 minus 2119890 + 2119890 ⊲ 119909 minus 119890 + 119890 ⊲ 119909 + 119909 minus 1199092
+ 119890 minus 2119890 ⊲ 119909 + 119890 ⊲ 1199092
= 119909 + (119890 ⊲ 119909 minus 119909) + (119890 ⊲ 1199092
minus 1199092
)
(20)
Proposition 25 Let I be a unital split quasi-Jordan Banachalgebra with unit 119890 If 119909 isin I with (119890 minus 119909) ⊲ (119890 minus 119909)
2
=
(119890 minus 119909)2
⊲ (119890 minus 119909) then |120582| le 119909 for all 120582 isin 120590(I119890)(119909)
Proof If 120582 isin 120590(I119890)(119909) with 120582 = 0 then the noninvertibility of
120582119890 minus 119909 means the noninvertibility of 119890 minus (1120582)119909 with respectto the unit 119890 However by Proposition 24 119890 minus (1120582)119909must beinvertible with respect to the unit 119890 whenever (1|120582|)119909 lt 1It follows that |120582| le 119909 for all 120582 isin 120590
(I119890)(119909)
5 Unbounded and Nonclosed Spectrum
In this section we show that the spectrum of an element ina split quasi-Jordan Banach algebra may be neither boundednor closed and hence not compactThe following result givesa couple of characterizations of the unbounded spectrum ofan element in a split quasi-Jordan Banach algebra
Proposition 26 Let 119909 be an element of a unital split quasi-Jordan Banach algebra I = 119869 oplus 119885(I) with a unit 119890 isin 119869 Thenthe following statements are equivalent
(1) 120590(I119890)(119909) =C
(2) 1199092 ⊲ 119909 = 119909 ⊲ 1199092
(3) |120582| le 119909 for all 120582 isin 120590(I119890)(119909)
Proof (1rArr2) See Proposition 22(2rArr3) Suppose 1199092 ⊲ 119909 = 119909 ⊲ 119909
2 Then
(119890 minus 119909)2
⊲ (119890 minus 119909) = (119890 minus 119890 ⊲ 119909 minus 119909 + 1199092
) ⊲ (119890 minus 119909)
= 119890 minus 2119890 ⊲ 119909 minus 119909 + 21199092
+ (119890 ⊲ 119909) ⊲ 119909 minus 1199092
⊲ 119909
= 119890 minus 2119909119869minus 119909 + 2119909
2
+ 119890 ⊲ 1199092
minus 1199093
(119890 minus 119909) ⊲ (119890 minus 119909)2
= (119890 minus 119909) ⊲ (119890 minus 2119909 + 1199092
)
= 119890 minus 2 (119890 ⊲ 119909) + 119890 ⊲ 1199092
minus 119909
+ 21199092
minus 119909 ⊲ 1199092
= 119890 minus 2119909119869+ 119890 ⊲ 119909
2
minus 119909 + 21199092
minus 1199093
(21)
Abstract and Applied Analysis 9
From (21) we get
(119890 minus 119909)2
⊲ (119890 minus 119909) = (119890 minus 119909) ⊲ (119890 minus 119909)2
(22)
Hence |120582| le 119909 for all 120582 isin 120590(I119890)(119909) by Proposition 25
(3rArr1) Immediate
Remark 27 There do exist unital split quasi-Jordan algebrascontaining elements that have the spectrum with respect tothe unit of the Jordan part equal to the whole ofC and henceunbounded To justify this claim we proceed as follows
Let 119860 be a unital associative algebra and let 119872 be an 119860-bimodule Let 119891 119872 rarr 119860 be an 119860-bimodule map (ie anadditive map satisfying 119891(119886119909) = 119886119891(119909) and 119891(119909119886) = 119891(119909)119886for all 119886 isin 119860 119909 isin 119872) Then one can put a dialgebra structureon 119872 as follows 119909 ⊣ 119910 = 119909119891(119910) and 119909⊢119910 = 119891(119909)119910 (cf [13Example 22(d)]) Hence119872+ is a quasi-Jordan algebra underthe quasi-Jordan product ldquo⊲rdquo given by119909 ⊲ 119910 = (12)(119909119891(119910)+
119891(119910)119909) Further for any 119909 isin 119872 we observe that
1199092
⊲ 119909 =1
2(119909119891 (119909) + 119891 (119909) 119909) ⊲ 119909
=1
4((119909119891 (119909) + 119891 (119909) 119909) 119891 (119909)
+119891 (119909) (119909119891 (119909) + 119891 (119909) 119909))
=1
4(119909119891 (119909) 119891 (119909) + 2119891 (119909) 119909119891 (119909) + 119891 (119909) 119891 (119909) 119909)
119909 ⊲ 1199092
=1
2119909 ⊲ (119909119891 (119909) + 119891 (119909) 119909)
=1
4(119909119891 (119909119891 (119909) + 119891 (119909) 119909) + 119891 (119909119891 (119909) + 119891 (119909) 119909) 119909)
=1
4(2119909119891 (119909) 119891 (119909) + 2119891 (119909) 119891 (119909) 119909)
(23)
However the right hand sides of the above equations (23)may not be equal see the following example (Example 28)For such elements 119909 we have 119909
2
⊲ 119909 = 119909 ⊲ 1199092 Hence by
Proposition 26 the spectrum of 119909 is unbounded whenever119872
+ is a unital split quasi-Jordan Banach algebra
Example 28 Let 119872 be the collection of 2 times 2 matrices withentries from the fieldC and let119860 be the algebra of allmatricesof the form [
120572 0
0 120573] with 120572 120573 isin C Then it is easily seen
that 119872 is an 119860-bimodule Next we define 119891 119872 rarr 119860
by 119891 [11988611
11988612
11988621
11988622
] = [11988611
0
0 11988622
] Of course 119891 is an additive mapsatisfying
119891([120572 0
0 120573] [
11988611
11988612
11988621
11988622
]) = [120572 0
0 120573]119891([
11988611
11988612
11988621
11988622
])
119891([11988611
11988612
11988621
11988622
] [120572 0
0 120573]) = 119891([
11988611
11988612
11988621
11988622
]) [120572 0
0 120573]
(24)
Hence119891 is an119860-bimodule mapThus by Remark 27119872+ is aquasi-Jordan algebra with the quasi-Jordan product as below
[11988611
11988612
11988621
11988622
] ⊲ [11988711
11988712
11988721
11988722
] =1
2([
11988611
11988612
11988621
11988622
]119891([11988711
11988712
11988721
11988722
])
+119891([11988711
11988712
11988721
11988722
]) [11988611
11988612
11988621
11988622
])
=[[
[
1198861111988711
11988612
11988711
+ 11988722
2
11988621
11988711
+ 11988722
21198862211988722
]]
]
(25)
Indeed 119872+
= 119869 oplus 119885(119872+
) where 119869 = [119886 0
0 119887] 119886 119887 isin C
is a subalgebra of 119872+ and 119885(119872+
) = [0 119886
119887 0] 119886 119887 isin C Any
matrix of the form [1 119886
119887 1] with 119886 119887 isin C is a (right) unit in119872
+Thus119872+ is a unital split quasi-Jordan algebrawith thematrix119868 = [
1 0
0 1] as the unit of its Jordan part 119869
Further a natural norm is defined on119872+ as follows
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
=100381610038161003816100381611988611
1003816100381610038161003816 +100381610038161003816100381611988612
1003816100381610038161003816 +100381610038161003816100381611988621
1003816100381610038161003816 +100381610038161003816100381611988622
1003816100381610038161003816 (26)
This norm also satisfies10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
] ⊲ [11988711
11988712
11988721
11988722
]
10038171003817100381710038171003817100381710038171003817
=1
2(
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
] [11988711
0
0 11988722
]
+ [11988711
0
0 11988722
] [11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
)
le
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
10038171003817100381710038171003817100381710038171003817
[11988711
0
0 11988722
]
10038171003817100381710038171003817100381710038171003817
le
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
10038171003817100381710038171003817100381710038171003817
[11988711
11988712
11988721
11988722
]
10038171003817100381710038171003817100381710038171003817
(27)
Next for any 119909 = [11988611
11988612
11988621
11988622
] isin 119872+ we observe that
1199092
= [11988611
11988612
11988621
11988622
] ⊲ [11988611
11988612
11988621
11988622
]
=[[
[
1198862
1111988612
11988611
+ 11988622
2
11988621
11988611
+ 11988622
21198862
22
]]
]
(28)
so that
1199092
⊲ 119909 = [
[
1198863
11
11988612
4(119886
11+ 119886
22)2
11988621
4(119886
11+ 119886
22)2
1198863
22
]
]
119909 ⊲ 1199092
= [
[
1198863
11
11988612
2(119886
2
11+ 119886
2
22)
11988621
2(119886
2
11+ 119886
2
22) 119886
3
22
]
]
(29)
So for any 119909 = [11988611
11988612
11988621
11988622
] isin 119872+ 1199092 ⊲ 119909 = 119909 ⊲ 119909
2
hArr
11988612
= 11988621
= 0 or 11988611
= 11988622 Thus by Proposition 26
10 Abstract and Applied Analysis
120590(119872+119868)([
11988611
11988612
11988621
11988622
]) = C whenever 11988611
= 11988622
and 11988612 119886
21 = 0
In particular for 119909 = [1 1
0 2] isin 119872
+ we have
1199092
⊲ 119909 =[[
[
13
1
4(1 + 2)
2
0
4(1 + 2)
2
23
]]
]
= [1
9
4
0 8
]
119909 ⊲ 1199092
=[[
[
13
1
2(1
2
+ 22
)
0
2(1
2
+ 22
) 23
]]
]
= [1
5
2
0 8
]
(30)
Concerning the inequality between the right hand sidesof (23) in Remark 27 we observe for 119909 as above that
1
4(119909119891 (119909) 119891 (119909) + 2119891 (119909) 119909119891 (119909) + 119891 (119909) 119891 (119909) 119909)
=1
4([
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 1
0 2] [
1 0
0 2]
+ [1 0
0 2] [
1 0
0 2] [
1 1
0 2]) = [
19
4
0 8
]
(31)
but
1
4(2119909119891 (119909) 119891 (119909) + 2119891 (119909) 119891 (119909) 119909)
=1
4(2 [
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 0
0 2] [
1 1
0 2])
= [1
5
2
0 8
]
(32)
Hence 1199092 ⊲ 119909 = 119909 ⊲ 1199092 Thus 120590
(119872+119868)([
1 1
0 2]) = C by
Proposition 26Further suppose the matrix [
11988611198862
11988631198864
] isin 119872+ is invertible
with respect to the unit 119868 that is [ 1198861 119886211988631198864
] isin 119866119868(119872
+
) Thenthere exists [ 119887
11198872
11988731198874
] isin 119872+ such that
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] = 119868 + 119868⊲([
1198861
1198862
1198863
1198864
]) = [1 minus119886
2
minus1198863
1]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
= [1198861
1198862
1198863
1198864
] + 119868⊲([
1198861
1198862
1198863
1198864
])
+ 119868⊲([
1198861
1198862
1198863
1198864
]
2
)
=[[
[
1198861
minus1198862
1198861+ 119886
4
2
minus1198863
1198861+ 119886
4
21198864
]]
]
(33)
However
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] =[[
[
11988711198861
1198872
1198861+ 119886
4
2
1198873
1198861+ 119886
4
211988741198864
]]
]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
=
[[[
[
11988711198861
1198872
1198862
1+ 119886
2
4
2
1198873
1198862
1+ 119886
2
4
211988741198862
4
]]]
]
(34)
It follows that 1198861
= 0 1198864
= 0 1198871= 1119886
1 119887
4= 1119886
4 119887
2((119886
1+
1198864)2) = minus119886
2 119887
3((119886
1+119886
4)2) = minus119886
3 119887
2((119886
2
1+119886
2
4)2) = minus119886
2((119886
1+
1198864)2) and 119887
3((119886
2
1+ 119886
2
4)2) = minus119886
3((119886
1+ 119886
4)2) From these
equations we get 1198872(119886
1minus 119886
4)2
= 0 and 1198873(119886
1minus 119886
4)2
= 0 so that(119887
2minus119887
3)(119886
1minus 119886
4)2
= 0Then for 1198861
= 1198864 we obtain 119887
2= 119887
3= 0
and hence 1198862= 119886
3= 0 Therefore
119866119868(119872
+
) = [119886 119887
119888 119886] 119909 = [
120572 0
0 120573] isin 119872
+
119886 = 0
120572 = 0 120573 = 0 120572 = 120573
(35)
The set 119866119868(119872
+
) is not open clearly [1 1
0 1] isin 119866
119868(119872
+
) forany 120598 gt 0
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[1 1
0 1] minus [
[
1 minus120598
41
0 1 +120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[
[
120598
40
0 minus120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=120598
2lt 120598
(36)
but [ 1minus1205984 1
0 1+1205984] notin 119866
119868(119872
+
)Now if 119860 = [
minus1 1
0 1] then 120582119868 minus 119860 = [
120582+1 minus1
0 120582minus1] notin 119866
119868(119872
+
)and so 120582 isin 120590
(119872+119868)(119860) for all 120582 isin C Thus 120590
(119872+119868)(119860) = C an
unbounded spectrum
Next we observe that the spectrum of an element withrespect to a unit is closed whenever the corresponding set ofinvertibles is open
Proposition 29 LetI be a quasi-Jordan normed algebra witha unit 119890 such that119866
119890(I) is openThen 120590
(I119890)(119909) is closed for all119909 isin I
Proof Define 119891 C rarr I by 119891(120582) = 120582119890 minus 119909 Since 119891 iscontinuous the inverse image of the open set 119866
119890(I) is open
in C and so its complement 120590(I119890)(119909) is closed
We conclude this paper with the following example of anonclosed spectrum
Example 30 Let 119872 and 119872+ be as in Example 28 Let 119864 =
[1 1
0 1] and 119860 = [
119886 0
0 119887] with 119886 = 119887 both different from 1 Then
119864 is a unit in 119872+ We show that 120590
(119872+119864)
(119860) = C 1 Forthis let us first investigate when can an element of the form119861 = [
120572 120573
0 120574] be invertible Assuming that 119861 is invertible we get
the existence of an element 119862 isin 119863+ such that
119862 ⊲ 119861 = 119864 + 119864⊲(119861)
119862 ⊲ 1198612
= 119861 + 119864⊲(119861) + 119864
⊲(119861
2
)
(37)
Abstract and Applied Analysis 11
From these equations we get 120572 = 0 120574 = 0 and
119862 =[[
[
1
120572120573∘
01
120574
]]
]
(38)
where 120573∘satisfies the following two equations
120573∘(120572 + 120574
2) = 1 minus 120573 +
120572 + 120574
2
120573∘(1205722
+ 1205742
2) =
120572 + 120574
2+
1205722
+ 1205742
2minus 120573
120572 + 120574
2
(39)
Multiplying the last equation by 2(120572 + 120574)(1205722
+ 1205742
) and thenusing the other equation we get
2 (1 minus 120573) + (120572 + 120574) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
+ (120572 + 120574) (40)
or equivalently
2 (1 minus 120573) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
(41)
this equation is satisfied for 120573 = 1 or 120572 = 120574 Hence thematrix119861 is invertible if and only if 119861 = [
120572 1
0 120574] or 119861 = [
120572 120573
0 120572
] for 120572 120574 isin
C 0 and 120573 isin CWe conclude that 119860
120582= 120582119864 minus 119860 = [
120582minus119886 120582
0 120582minus119887] is invertible
with respect to 119864 if and only if 120582 minus 119887 = 0 120582 minus 119886 = 0 and 120582 = 1that is119860
120582is invertible if 119886 = 120582 and 119887 = 120582 and 120582 = 1 Hence119860
120582
is invertible only if 120582 = 1 as we assumed that 119886 = 119887 and bothare not 1 So for all 120582 = 1 119860
120582notin 119866
119864(119872
+
) Thus 120590(119872+119864)
(119860) =
C 1 which is neither bounded nor closed
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This project was supported by King Saud University Dean-ship of Scientific Research College of Science ResearchCenter
References
[1] B Fritzsche ldquoSophus Lie a sketch of his life and workrdquo Journalof Lie Theory vol 9 no 1 pp 1ndash38 1999
[2] P Jordan ldquoUber eine klasse nichtassoziativer hyperkomplexeralgebrenrdquo Nachrichten von der Gesellschaft der Wissenschaftenzu Gottingen pp 569ndash575 1932
[3] K McCrimmon A Taste of Jordan Algebras UniversitextSpringer New York NY USA 2004
[4] N Jacobson Structure and Representations of Jordan Algebrasvol 39 of American Mathematical Society Colloquium Publica-tions American Mathematical Society Providence RI USA1968
[5] F Gursey and C-H Tze On the Role of Division Jordan andRelatedAlgebras in Particle PhysicsWorld Scientific River EdgeNJ USA 1996
[6] R Iordanescu Jordan Structures in Analysis Geometry andPhysics EdituraAcademiei Romane Bucharest Romania 2009
[7] H Upmeier Symmetric Banach Manifolds and Jordan Clowast-Alge-bras vol 104 of North-Holland Mathematics Studies North-Holland Amsterdam The Netherlands 1985
[8] I L Kantor ldquoClassification of irreducible transitive differentialgroupsrdquo Doklady Akademii Nauk SSSR vol 158 pp 1271ndash12741964
[9] M Koecher ldquoUber eine Gruppe von rationalen AbbildungenrdquoInventiones Mathematicae vol 3 pp 136ndash171 1967
[10] J Tits ldquoUne classe drsquoalgebres de Lie en relation avec les algebresde Jordanrdquo Indagationes Mathematicae vol 24 pp 530ndash5351962
[11] J-L Loday ldquoUne version non commutative des algebres de Lieles algebres de Leibnizrdquo LrsquoEnseignement Mathematique vol 39no 3-4 pp 269ndash293 1993
[12] J-L Loday and T Pirashvili ldquoUniversal enveloping algebras ofLeibniz algebras and (co)homologyrdquo Mathematische Annalenvol 296 no 1 pp 139ndash158 1993
[13] J-L Loday ldquoDialgebrasrdquo in Dialgebras and Related Operadsvol 1763 of Lecture Notes in Math pp 7ndash66 Springer BerlinGermany 2001
[14] R Velasquez and R Felipe ldquoQuasi-Jordan algebrasrdquo Communi-cations in Algebra vol 36 no 4 pp 1580ndash1602 2008
[15] R Velasquez and R Felipe ldquoOn K-B quasi-Jordan algebrasand their relation with Liebniz algebrasrdquo Comunicaciones delCIMAT No 1-10-1015-12-2010
[16] R Felipe ldquoAn analogue to functional analysis in dialgebrasrdquoInternational Mathematical Forum vol 2 no 21ndash24 pp 1069ndash1091 2007
[17] R Velasquez and R Felipe ldquoSplit dialgebras split quasi-Jordanalgebras and regular elementsrdquo Journal of Algebra and ItsApplications vol 8 no 2 pp 191ndash218 2009
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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International Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Discrete Dynamics in Nature and Society
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Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 5
119910119869⊲ 119909
2
+ 119910119885⊲ 119909
2
= 119910 ⊲ 1199092
= 119909 + 119890⊲(119909
119869+ 119909
119885) + 119890
⊲((119909
119869+ 119909
119885)2
)
= 119909119869+ (minus119909
119885⊲ 119909)
(5)
Hence by the uniqueness of the representation as sum of Jor-dan and zero parts we get 119910
119869⊲ 119909 = 119890 119910
119885⊲ 119909 = minus119909
119885 119910
119869⊲
1199092
= 119909119869 and 119910
119885⊲ 119909
2
= minus119909119885
⊲ 119909 Therefore 1199101015840
⊲ 120582119909 =
120582(((1120582)119910119869+ 119910
119885) ⊲ 119909) = 119910
119869⊲ 119909 + 120582119910
119885⊲ 119909 = 119890 minus 120582119909
119885=
119890 + 119890⊲(120582119909) and 119910
1015840
⊲ (120582119909)2
= 1205822
((1120582)119910119869⊲ 119909
2
+ 119910119885
⊲ 1199092
) =
120582119909119869minus 120582
2
119909119885⊲ 119909 = 120582119909 + 119890
⊲(120582119909) + 119890
⊲((120582119909)
2
) because 119890⊲(120582119909) =
119890 ⊲ (120582119909) minus 120582119909 = minus120582119909119885and 119890
⊲((120582119909)
2
) = 119890 ⊲ 1205822
1199092
minus 1205822
1199092
=
1205822
(119890⊲(119909
2
)) = minus1205822
119909119885⊲ 119909
Proposition 7 LetI be a quasi-Jordan normed algebrawith aunit 119890 Let119866
119890(I) = 119909 isin I 119909 is invertible with respect to 119890
be an open set and 119909 isin 119866119890(I) Then 119909 + 119911 isin 119866
119890(I) for all
119911 isin 119885(I)
Proof Suppose 119909 isin 119866119890(I) and 119911 isin 119885(I) If 119911 = 0 then 119909+119911 =
119909 isin 119866119890(I) Next suppose 119911 = 0 Since 119866
119890(I) is an open set
there exists 120598 gt 0 such that 119886 isin 119866119890(I) whenever 119909 minus 119886 lt 120598
Hence 119909 + 119911∘isin 119866
119890(I) with 119911
∘= (1205982119911)119911
Let 119910 and 119910∘be inverses of 119909 and 119909 + 119911
∘with respect to 119890
respectively Then
119910∘⊲ (119909 + 119911
∘) = 119890 + 119890
⊲(119909 + 119911
∘)
= 119890 + 119890⊲(119909) minus 119911
∘
= 119910 ⊲ 119909 minus 119911∘
119910∘⊲ (119909 + 119911
∘)2
= 119909 + 119911∘+ 119890
⊲(119909 + 119911
∘) + 119890
⊲((119909 + 119911
∘)2
)
= 119909 + 119890⊲(119909) + 119890
⊲(119909
2
) minus 119911∘⊲ 119909
= 119910 ⊲ 1199092
minus 119911∘⊲ 119909
(6)
Hence by setting 1199101= 119910 + 120572(119910
∘minus 119910) with 120572 = 2119911120598 we see
that1199101⊲ (119909 + 119911) = (119910 + 120572 (119910
∘minus 119910)) ⊲ 119909
= 119910 ⊲ 119909 + 120572 (119910∘⊲ (119909 + 119911
∘) minus 119910 ⊲ 119909)
= 119890 + 119890⊲(119909) + 120572 (minus119911
∘)
= 119890 + 119890⊲(119909) minus 119911
= 119890 + 119890⊲(119909 + 119911)
1199101⊲ (119909 + 119911)
2
= (119910 + 120572 (119910∘minus 119910)) ⊲ 119909
2
= 119910 ⊲ 1199092
+ 120572 (119910∘⊲ (119909 + 119911
∘)2
minus 119910 ⊲ 1199092
)
= 119909 + 119890⊲(119909) + 119890
⊲(119909
2
) minus 120572 (119911∘⊲ 119909)
= 119909 + 119890⊲(119909) + 119890
⊲(119909
2
) minus 119911 ⊲ 119909
= (119909 + 119911) + 119890⊲(119909 + 119911) + 119890
⊲((119909 + 119911)
2
)
(7)
since 119911 isin 119885(I) Thus 1199101is an inverse of 119909 + 119911 with respect to
the unit 119890
Corollary 8 Under the hypothesis of Proposition 7 120582119890 minus 119909 isin
119866119890(I) implies 120582119890 minus (119909 + 119911) isin 119866
119890(I) for all 119911 isin 119885(I)
Proposition 9 LetI = 119869 oplus 119885(I) be a split quasi-Jordan nor-med algebra and let 119890 isin 119869 be a unit inI such that the set119866
119890(I)
is open Then 119909 ⊲ 1199092
= 1199092
⊲ 119909 for all 119909 isin I
Proof Of course the element 119890 is the unit of the Jordanalgebra 119869 Then for any fixed element 119886 isin 119869 there exists120582 isin C such that 120582119890 minus 119886 is invertible in 119869 otherwise wewould get the negation of the well-known fact that spectrumof an element of a unital Jordan algebra is bounded Thatis there exists 119910 isin 119869 such that 119910 ⊲ (120582119890 minus 119886) = 119890 and119910 ⊲ (120582119890minus119886)
2
= (120582119890minus119886) However 119890⊲(119886) = 0 = 119890
⊲(119886
2
) Hence119910 is an inverse of 119886 in the quasi-Jordan algebraIwith respectto the unit 119890 By Corollary 8 120582119890 minus (119886 + 119911) is also invertiblefor any 119911 isin 119885(I) This in turn gives the existence of 119887 isin I
satisfying 119887 ⊲ (120582119890 minus (119886 + 119911)) = 119890 + 119911 and 119887 ⊲ (120582119890 minus (119886 + 119911))2
=
(120582119890minus119886)+119911 ⊲ (120582119890minus119886) Multiplying the first equation from theright by (120582119890 minus (119886 + 119911))
2 the second equation by (120582119890minus (119886+119911))and using the right Jordan identity we get
(119890 + 119911) ⊲ (120582119890 minus (119886 + 119911))2
= ((120582119890 minus 119886) + 119911 ⊲ (120582119890 minus 119886)) ⊲ (120582119890 minus (119886 + 119911))
(8)
Hence
(119890 + 119911) ⊲ (120582119890 minus 119886)2
= ((120582119890 minus 119886) + 119911 ⊲ (120582119890 minus 119886)) ⊲ (120582119890 minus 119886)
(9)
so that
119890 ⊲ (120582119890 minus 119886)2
+ 1205822
119911 minus 2120582 (119911 ⊲ 119886) + 119911 ⊲ 1198862
= (120582119890 minus 119886)2
+ 1205822
119911 minus 2120582119911 ⊲ 119886 + (119911 ⊲ 119886) ⊲ 119886
(10)
This last equation reduces to 119890 ⊲ (120582119890 minus 119886)2
= (120582119890 minus 119886)2 since
(120582119890 minus 119886)2
isin 119869 and 119890 is the unit of 119869 Hence 119911 ⊲ 1198862
= (119911 ⊲
119886) ⊲ 119886 for all 119886 isin 119869 and 119911 isin 119885(I) Now for any 119909 isin I thelast equation with 119886 = 119909
119869and 119911 = 119909
119885gives 119909
119885⊲ 119909
2
119869= (119909
119885⊲
119909119869) ⊲ 119909
119869Thus 119909 ⊲ 119909
2
= 119909119869⊲ 119909
2
119869+119909
119885⊲ 119909
2
119869= 119909
2
119869⊲ 119909
119869+(119909
119885⊲
119909119869) ⊲ 119909
119869= 119909
2
⊲ 119909 for all 119909 isin I
Corollary 10 If a unital split quasi-Jordan algebra has anelement 119909 with 119909 ⊲ 119909
2
= 1199092
⊲ 119909 then the set of invertibleelements with respect to the unit of the Jordan part is not open
In the sequel we will show the existence of a unital splitquasi-Jordan Banach algebra with elements 119909 such that 119909 ⊲
1199092
= 1199092
⊲ 119909 Thus the above result establishes that the set ofinvertible elements with respect to a fixed unit in a quasi-Jordan Banach algebra may not be open
6 Abstract and Applied Analysis
4 The Spectrum of Elements in a UnitalQuasi-Jordan Algebra
As usual we define the spectrum of an element 119909 in aunital quasi-Jordan algebra (I 119890) denoted by 120590
(I119890)(119909) tobe the collection of all complex numbers 120582 for which 120582119890 minus
119909 is not invertible Thus 120590(I119890)(119909) = 120582 isin C 120582119890 minus
119909 is not invertible Here the subscript 119890 indicates that theinvariability depends on the choice of unit 119890 which generallyis not unique
Proposition 11 Let I be a unital quasi-Jordan algebra Then120590(I119890)(119890
1015840
) = 1 for all 119890 1198901015840 isin 119880(I)
Proof Let 119890 1198901015840 isin 119880(I) Then for any 120582 = 1 119910 = (1(120582minus1))1198901015840
is an inverse of 120582119890 minus 1198901015840 with respect to the unit 119890 because 119910 ⊲
(120582119890 minus 1198901015840
) = (1(120582 minus 1))1198901015840
⊲ (120582119890 minus 1198901015840
) = 119890 + 119890⊲(120582119890 minus 119890
1015840
) and119910 ⊲ (120582119890 minus 119890
1015840
)2
= (120582119890 minus 1198901015840
) + 119890⊲(120582119890 minus 119890
1015840
) + 119890⊲((120582119890 minus 119890
1015840
)2
)
Proposition 12 Let I be a quasi-Jordan algebra with unit 119890Then 120590
(I119890)(119911) = 0 for all 119911 isin 119885(I)
Proof For any fixed 119911 isin 119885(I) and nonzero scalar 120582 thevector 119910 = (1120582)(119890 + 119911) satisfies 119910 ⊲ (120582119890 minus 119911) = 120582119910 = 119890 + 119911 =
119890+ 119890⊲(120582119890minus 119911) and 119910 ⊲ (120582119890 minus 119911)
2
= 1205822
119910 = 120582119890+120582119911 = (120582119890minus119911) +
119890⊲(120582119890 minus 119911) + 119890
⊲((120582119890 minus 119911)
2
) So (1120582)(119890 + 119911) is an inverse of120582119890 minus 119911 with respect to 119890 This means 120582 notin 120590
(I119890)(119911) for all 120582 = 0However the zero vector is not invertible Thus 120590
(I119890)(119911) =
0
Proposition 13 Let I = 119869 oplus 119885(I) be a split quasi-Jordanalgebra with unit 119890 isin 119869 Then 120590
(I119890)(119901) sube 0 1 for all idempo-tents 119901 isin I (ie 1199012
= 119901)
Proof Let 119901 be any fixed idempotent inI Since 119901 isin I 119901 hasa unique representation 119901 = 119901
119869+ 119901
119885with 119901
119869isin 119869 and 119901
119885isin
119885(I) Clearly 1199012
119869+ 119901
119885⊲ 119901
119869= 119901
2
= 119901 = 119901119869+ 119901
119885 Then
by uniqueness of the representation in the split quasi-Jordanalgebra I 119901
119885⊲ 119901
119869= 119901
119885and 119901
2
119869= 119901
119869 this means 119901
119869is an
idempotent in the Jordan algebra 119869 Hence 120590(119869119890)
(119901119869) sube 0 1
Thus 119901120582= 120582119890 minus 119901
119869is invertible in 119869 with the unique inverse
119901minus1
120582 for all 120582 notin 0 1We show that 119910 = 119901
minus1
120582+ (1(120582 minus 1))119901
119885is an inverse of
120582119890 minus 119901 in I with respect to the unit 119890 for this we note that120582119890 minus 119901 = 119901
120582minus 119901
119885 (119901
120582minus 119901
119885)2
= 1199012
120582minus 119901
119885⊲ 119901
120582 119890
⊲(120582119890 minus
119901) = 119890⊲(119901
120582minus 119901
119885) = 119890 ⊲ (119901
120582minus 119901
119885) minus (119901
120582minus 119901
119885) = 119901
119885 and
119890⊲((120582119890 minus 119901)
2
) = 119890Δ((119901
120582minus 119901
119885)2
) = 119890⊲(119901
2
120582minus 119901
119885⊲ 119901
120582) = 119901
119885⊲
119901120582= 119901
119885⊲ (120582119890minus119901
119869) = 120582119901
119885minus119901
119885⊲ 119901
119869= 120582119901
119885minus119901
119885= (120582minus1)119901
119885
Hence 119910 ⊲ (120582119890 minus 119901) = 119910 ⊲ (119901120582minus 119901
119885) = 119910 ⊲ 119901
120582= (119901
minus1
120582+
(1(120582 minus 1))119901119885) ⊲ 119901
120582= 119901
minus1
120582⊲ 119901
120582+ (1(120582 minus 1))119901
119885⊲ 119901
120582= 119890 +
(1(120582 minus 1))119901119885
⊲ (120582119890 minus 119901119869) = 119890 + (1(120582 minus 1))(120582119901
119885minus 119901
119885⊲
119901119869) = 119890 + (1(120582 minus 1))(120582119901
119885minus 119901
119885) = 119890 + 119901
119885= 119890 + 119890
⊲(120582119890 minus 119901)
and 119910 ⊲ (120582119890 minus 119901)2
= 119910 ⊲ (1199012
120582minus 119901
119885⊲ 119901
120582) = 119910 ⊲ 119901
2
120582=
(119901minus1
120582+(1(120582minus1))119901
119885) ⊲ 119901
2
120582= 119901
120582+(1(120582minus1))119901
119885⊲ (120582119890minus119901
119869)2
=
119901120582+(1(120582minus1))119901
119885⊲ (120582
2
119890minus2120582119901119869+119901
2
119869) = 119901
120582+(1(120582minus1))119901
119885⊲
(1205822
119890 minus 2120582119901119869+ 119901
119869) = 119901
120582+ (1(120582 minus 1))(120582
2
119901119885
⊲ 119890 minus 2120582119901119885
⊲
119901119869+ 119901
119885⊲ 119901
119869) = 119901
120582+ (1(120582 minus 1))(120582
2
119901119885
minus 2120582119901119885
+ 119901119885) =
119901120582+ (1(120582 minus 1))(120582 minus 1)
2
119901119885= 120582119890 minus119901
119869+ (120582 minus 1)119901
119885= (120582119890 minus119901) +
119890⊲(120582119890 minus 119901) + 119890
⊲((120582119890 minus 119901)
2
)
As mentioned in Section 2 if I is a quasi-Jordan algebrawith a unit 119890 then the set 119890 + 119911 119911 isin 119885(I) coincides withthe set 119880(I) of all units inI
Proposition 14 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebrawith unit 119890 isin 119869 and119909 isin I invertiblewith respectto some 1198901015840 isin 119880(I)Then119909
119869is invertible with respect to the unit
119890 in the Jordan algebra 119869
Proof Clearly 119890 ⊲ 119909 minus 119909 = minus119909119885and 119890 ⊲ 119909
2
minus 1199092
= minus119909119885⊲ 119909
119869
Since 1198901015840
isin 119880(I) we have 1198901015840
= 119890 + 119911 for some 119911 isin 119885(I)Hence the invertibility of 119909 in I with respect to the unit 1198901015840gives the existence of 119910 isin I such that 119910
119869⊲ 119909
119869+ 119910
119885⊲ 119909
119869=
119910 ⊲ 119909119869= 119910 ⊲ 119909 = 119890 + 119911 + (119890 + 119911) ⊲ 119909 minus 119909 = 119890 + 119911 + 119911 ⊲ 119909 minus 119909
119885
and 119910119869⊲ 119909
2
119869+ 119910
119885⊲ 119909
2
119869= 119910 ⊲ 119909
2
119869= 119910 ⊲ 119909
2
= (119890 + 119911) ⊲
119909 + (119890 + 119911) ⊲ 1199092
minus 1199092
= 119909119869+ 119911 ⊲ 119909 + 119911 ⊲ 119909
2
minus 119909119885⊲ 119909
119869 So by
the uniqueness of the representations in the split algebra Iwe get 119910
119869⊲ 119909
119869= 119890 and 119910
119869⊲ 119909
2
119869= 119909
119869 Thus 119910
119869is the inverse
of 119909119869in the Jordan algebra 119869
Next we observe that the spectrum of 119909 in a unital splitquasi-Jordan algebra with respect to any unit includes thespectrum of 119909
119869in the Jordan part 119869
Corollary 15 LetI = 119869 oplus119885(I) be a unital split quasi-Jordanalgebra with unit 119890 isin 119869 and let 119909 isin I Then 120590
(119869119890)(119909
119869) sube
120590(I1198901015840)(119909) for all 1198901015840 isin 119880(I)
Proof Let 1198901015840 = 119890 + 119911 with 119911 isin 119885(I) and let 120582 notin 120590(I1198901015840)(119909)
Then 1205821198901015840 minus 119909 is invertible in I with respect to the unit 1198901015840Hence its Jordanpart120582119890minus119909
119869is invertible in the Jordan algebra
119869 by Proposition 14 Thus 120582 notin 120590(119869119890)
(119909119869)
It is well known that the spectrum of any element ina unital Jordan Banach algebra is nonempty (cf [7]) Thistogether with Proposition 2 and Corollary 15 gives thefollowing result
Corollary 16 The spectrum of any element in a unital splitquasi-Jordan Banach algebra is nonempty
The next result extends Corollary 16 to any quasi-JordanBanach algebra with a norm 1 unit
Proposition 17 The spectrum of any element in a quasi-Jordan Banach algebra with a norm 1 unit is nonempty
Proof Let I be a unital quasi-Jordan Banach algebra witha norm 1 unit 119890 From Section 2 we know that the map120593 119909 997891rarr 120593(119909) = (119909 119877
119909) embeds I into the unital split
quasi-Jordan Banach algebra I1
= (119909 119877119910) 119909 119910 isin I
equipped with the sum (119909 119877119910) + (119886 119877
119887) = (119909 + 119886 119877
119910+119887) sca-
lar multiplication 120582(119909 119877119910) = (120582119909 119877
120582119910) and product (119909
119877119910) ⊲ (119886 119877
119887) = (119909 ⊲ 119887 119877
119910⊲119887) and the image 120593(I) is a
norm closed right ideal isomorphic to I with norm 1 unit(119890 119877
119890) Moreover it is seen in Section 3 that the embedding
Abstract and Applied Analysis 7
120593 also preserves the corresponding invertible elements thatis (119910 119877
119910) is an inverse of (119909 119877
119909) with respect to the unit
(119890 119877119890) in I
1whenever 119910 is an inverse of 119909 with respect
to a unit 119890 in I Hence by Corollary 16 it follows that120601 = 120590
(I1(119890119877119890))(119909 119877
119909) sube 120590
(120593(I)(119890119877119890))(119909 119877
119909) = 120590
(I119890)(119909) for all119909 isin I
Proposition 18 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebra with unit 119890 isin 119869 and let 119909 isin 119869 Then 120590
(119869119890)(119909) =
120590(I119890)(119909)
Proof By Corollary 15 120590(119869119890)
(119909) sube 120590(I119890)(119909) For the reverse
inclusion let 120582 notin 120590(119869119890)
(119909) then 120582119890 minus 119909 is invertible in 119869 thatis there exists 119910 isin 119869 such that 119910 ⊲ (120582119890 minus 119909) = 119890 and 119910 ⊲
(120582119890 minus 119909)2
= 120582119890 minus 119909 However 119890⊲(120582119890 minus 119909) = 0 = 119890
⊲((120582119890 minus 119909)
2
)Hence 119910 is an inverse of 120582119890 minus 119909 in I with respect to the unit119890 that is 120582 notin 120590
(I119890)(119909) Thus 120590(I119890)(119909) sube 120590
(119869119890)(119909)
Proposition 19 Let I be a unital quasi-Jordan normedalgebra and let 119890 be a unit inI for which 119866
119890(I) is open Then
120590(I119890)(119909) = 120590
(I119890)(119909 + 119911) for all 119911 isin 119885(I)
Proof By Corollary 8 120582119890 minus 119909 is invertible if and only if 120582119890 minus
(119909 + 119911) is invertible for all 119911 isin 119885(I) Thus 120582 notin 120590(I119890)(119909) if
and only if 120582 notin 120590(I119890)(119909 + 119911) for all 119911 isin 119885(I)
Corollary 20 LetI = 119869oplus119885(I) be a unital split quasi-Jordannormed algebra and let 119890 be a unit in I such that 119866
119890(I) is
open Then 120590(I119890)(119909) = 120590
(I119890)(119909119869) for all 119909 = 119909119869+ 119909
119885isin I
Further if the unit 119890 isin 119869 then120590(I119890)(119909) = 120590
(I119890)(119909119869) = 120590(119869119890)
(119909119869)
Lemma 21 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordanalgebra with a unit 119890 isin 119869 and let 119909 isin I be invertible withrespect to 119890 Then 119909
119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
Proof As 119909 is invertible there exists 119910 isin I such that 119910 ⊲ 119909 =
119890 + (119890 ⊲ 119909 minus 119909) = 119890 minus 119909119885and 119910 ⊲ 119909
2
= 119909 + (119890 ⊲ 119909 minus 119909) + (119890 ⊲
1199092
minus 1199092
) = 119909119869minus 119909
119885⊲ 119909 Hence by the uniqueness of the
representation in a split quasi-Jordan algebra we obtain
119910119885⊲ 119909 = minus119909
119885 (11)
119910119885⊲ 119909
2
= minus119909119885⊲ 119909 (12)
Thus
119909119885⊲ 119909
2
= minus (119910119885⊲ 119909) ⊲ 119909
2
(by (11))
= minus (119910119885⊲ 119909
2
) ⊲ 119909 (by the right Jordan identity)
= (119909119885⊲ 119909) ⊲ 119909 (by (12))
(13)
Proposition 22 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebra with a unit 119890 isin 119869 119909 = 119909
119869+ 119909
119885isin I satisfies
120590(I119890)(119909) =C Then
(1) 119909119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
(2) 1199092 ⊲ 119909 = 119909 ⊲ 1199092
Proof (1) Let 120582 notin 120590(I119890)(119909) Then 120582119890 minus 119909 = (120582119890 minus 119909
119869) + (minus119909
119885)
is invertible with respect to 119890 By Lemma 21 we have
119909119885⊲ (120582119890 minus 119909)
2
= (119909119885⊲ (120582119890 minus 119909)) ⊲ (120582119890 minus 119909) (14)
since the zero part of 120582119890 minus 119909 is minus119909119885 However
119909119885⊲ (120582119890 minus 119909)
2
= 119909119885⊲ (120582
2
119890 minus 120582119890 ⊲ 119909 minus 120582119909 ⊲ 119890 + 1199092
)
= 1205822
(119909119885⊲ 119890) minus 120582 (119909
119885⊲ (119890 ⊲ 119909))
minus 120582 (119909119885⊲ (119909 ⊲ 119890)) + 119909
119885⊲ 119909
2
= 1205822
119909119885minus 2120582119909
119885⊲ 119909
+ 119909119885⊲ 119909
2
(by the right commutativity of ⊲)
(119909119885⊲ (120582119890 minus 119909)) ⊲ (120582119890 minus 119909)
= (120582 (119909119885⊲ 119890) minus 119909
119885⊲ 119909) ⊲ (120582119890 minus 119909)
= (120582119909119885minus 119909
119885⊲ 119909) ⊲ (120582119890 minus 119909)
= 1205822
(119909119885⊲ 119890) minus 120582 (119909
119885⊲ 119909) ⊲ 119890
minus 120582119909119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
= 1205822
119909119885minus 2120582119909
119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
(15)
Therefore (14) becomes
1205822
119909119885minus 2120582119909
119885⊲ 119909 + 119909
119885⊲ 119909
2
= 1205822
119909119885minus 2120582119909
119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
(16)
which after simplification reduces to the required equation119909119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
(2) Since 1199092 = (119909119869+ 119909
119885)2
= 1199092
119869+ 119909
119885⊲ 119909
119869 we have
1199092
⊲ 119909 = (1199092
119869+ 119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909
119869+ (119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909
119869+ 119909
119885⊲ 119909
2
(17)
by the part (1) But 1199092119869⊲ 119909
119869= 119909
119869⊲ 119909
2
119869since 119909
119869is in the Jor-
dan algebra 119869 Therefore 1199092 ⊲ 119909 = 1199092
119869⊲ 119909
119869+ 119909
119885⊲ 119909
2
= 119909119869⊲
1199092
119869+ 119909
119885⊲ 119909
2
= 119909 ⊲ 1199092
Remark 23 In any quasi-Jordan algebra if an element 119909
satisfies 1199093
= 1199092
⊲ 119909 = 119909 ⊲ 1199092 then 119909
119899
⊲ 1199092
= 119909119899+2 for
all positive integers 119899 For this suppose 119909 satisfies 1199092 ⊲ 119909 =
119909 ⊲ 1199092 and 119909
119898
⊲ 1199092
= 119909119898+2 for any fixed119898 ge 1 Then 119909
119898+1
⊲
1199092
= (119909119898
⊲ 119909) ⊲ 1199092
= (119909119898
⊲ 1199092
) ⊲ 119909 (by the right Jor-dan identity) = 119909
119898+2
⊲ 119909 = 119909119898+3
Proposition 24 Let I be a unital quasi-Jordan Banachalgebra with unit 119890 and let 119909 isin I satisfy (119890 minus 119909) ⊲ (119890 minus 119909)
2
=
(119890 minus 119909)2
⊲ (119890 minus 119909) Then 119909 isin 119866119890(I) whenever 119890 minus 119909 lt 1
8 Abstract and Applied Analysis
Proof First note that 119890minus119909 lt 1 gives (119890 minus 119909)119899
le 119890 minus 119909119899
lt
1 for all 119899 = 1 2 3 Hence the infinite geometric series119890 +sum
infin
119899=1(119890 minus119909)
119899 converges absolutely to some element 119910 isin IWe show that the geometric series sum 119910 is an inverse of 119909with respect to the unit 119890 For any fixed positive integer 119899 let119910119899= 119890+sum
119899
119896=1(119890minus119909)
119896Then the sequence 119910119899 of partial sums
converges to 119910 By setting 119908 = 119890 minus 119909 we get
119910119899⊲ 119909 = (119890 +
119899
sum
119896=1
119908119896
) ⊲ (119890 minus 119908)
= 119890 +
119899
sum
119896=1
119908119896
minus (119890 ⊲ 119908 +
119899+1
sum
119896=2
119908119896
)
= 119890 + 119908 minus 119890 ⊲ 119908 minus 119908119899+1
= 119890 + (119890 minus 119909) minus 119890 ⊲ (119890 minus 119909) minus (119890 minus 119909)119899+1
= 119890 + (119890 minus 119909 minus 119890 + 119890 ⊲ 119909) minus (119890 minus 119909)119899+1
= 119890 + (119890 ⊲ 119909 minus 119909) minus (119890 minus 119909)119899+1
(18)
Thus by allowing 119899 rarr infin we obtain 119909 ⊲ 119910 = 119890 + (119890 ⊲
119909 minus 119909) = 119890 + 119890⊲(119909) since 119890 minus 119909 lt 1
Next by Remark 23 we have
119910119899⊲ 119909
2
= 119910119899⊲ (119890 minus 119908)
2
= 119910119899⊲ (119890 minus 119890 ⊲ 119908 minus 119908 ⊲ 119890 + 119908
2
)
= 119910119899⊲ (119890 minus 2119908 + 119908
2
)
= 119910119899⊲ 119890 minus 2119910
119899⊲ 119908 + 119910
119899⊲ 119908
2
= 119910119899minus 2119910
119899⊲ 119908 + 119910
119899⊲ 119908
2
= (119890 +
119899
sum
119896=1
119908119896
) minus 2(119890 +
119899
sum
119896=1
119908119896
) ⊲ 119908
+ (119890 +
119899
sum
119896=1
119908119896
) ⊲ 1199082
= (119890 +
119899
sum
119896=1
119908119896
) minus 2(119890 ⊲ 119908 +
119899+1
sum
119896=2
119908119896
)
+ (119890 ⊲ 1199082
+
119899+2
sum
119896=3
119908119896
)
= 119890 + 119908 minus 2119890 ⊲ 119908 minus 1199082
+ 119890 ⊲ 1199082
minus 119908119899+1
+ 119908119899+2
(19)
Taking the limit as 119899 rarr infin we get
119910 ⊲ 1199092
= 119890 + 119908 minus 2119890 ⊲ 119908 minus 1199082
+ 119890 ⊲ 1199082
= 119890 + (119890 minus 119909) minus 2119890 ⊲ (119890 minus 119909) minus (119890 minus 119909)2
+ 119890 ⊲ (119890 minus 119909)2
= 2119890 minus 119909 minus 2119890 + 2119890 ⊲ 119909 minus 119890 + 119890 ⊲ 119909 + 119909 minus 1199092
+ 119890 minus 2119890 ⊲ 119909 + 119890 ⊲ 1199092
= 119909 + (119890 ⊲ 119909 minus 119909) + (119890 ⊲ 1199092
minus 1199092
)
(20)
Proposition 25 Let I be a unital split quasi-Jordan Banachalgebra with unit 119890 If 119909 isin I with (119890 minus 119909) ⊲ (119890 minus 119909)
2
=
(119890 minus 119909)2
⊲ (119890 minus 119909) then |120582| le 119909 for all 120582 isin 120590(I119890)(119909)
Proof If 120582 isin 120590(I119890)(119909) with 120582 = 0 then the noninvertibility of
120582119890 minus 119909 means the noninvertibility of 119890 minus (1120582)119909 with respectto the unit 119890 However by Proposition 24 119890 minus (1120582)119909must beinvertible with respect to the unit 119890 whenever (1|120582|)119909 lt 1It follows that |120582| le 119909 for all 120582 isin 120590
(I119890)(119909)
5 Unbounded and Nonclosed Spectrum
In this section we show that the spectrum of an element ina split quasi-Jordan Banach algebra may be neither boundednor closed and hence not compactThe following result givesa couple of characterizations of the unbounded spectrum ofan element in a split quasi-Jordan Banach algebra
Proposition 26 Let 119909 be an element of a unital split quasi-Jordan Banach algebra I = 119869 oplus 119885(I) with a unit 119890 isin 119869 Thenthe following statements are equivalent
(1) 120590(I119890)(119909) =C
(2) 1199092 ⊲ 119909 = 119909 ⊲ 1199092
(3) |120582| le 119909 for all 120582 isin 120590(I119890)(119909)
Proof (1rArr2) See Proposition 22(2rArr3) Suppose 1199092 ⊲ 119909 = 119909 ⊲ 119909
2 Then
(119890 minus 119909)2
⊲ (119890 minus 119909) = (119890 minus 119890 ⊲ 119909 minus 119909 + 1199092
) ⊲ (119890 minus 119909)
= 119890 minus 2119890 ⊲ 119909 minus 119909 + 21199092
+ (119890 ⊲ 119909) ⊲ 119909 minus 1199092
⊲ 119909
= 119890 minus 2119909119869minus 119909 + 2119909
2
+ 119890 ⊲ 1199092
minus 1199093
(119890 minus 119909) ⊲ (119890 minus 119909)2
= (119890 minus 119909) ⊲ (119890 minus 2119909 + 1199092
)
= 119890 minus 2 (119890 ⊲ 119909) + 119890 ⊲ 1199092
minus 119909
+ 21199092
minus 119909 ⊲ 1199092
= 119890 minus 2119909119869+ 119890 ⊲ 119909
2
minus 119909 + 21199092
minus 1199093
(21)
Abstract and Applied Analysis 9
From (21) we get
(119890 minus 119909)2
⊲ (119890 minus 119909) = (119890 minus 119909) ⊲ (119890 minus 119909)2
(22)
Hence |120582| le 119909 for all 120582 isin 120590(I119890)(119909) by Proposition 25
(3rArr1) Immediate
Remark 27 There do exist unital split quasi-Jordan algebrascontaining elements that have the spectrum with respect tothe unit of the Jordan part equal to the whole ofC and henceunbounded To justify this claim we proceed as follows
Let 119860 be a unital associative algebra and let 119872 be an 119860-bimodule Let 119891 119872 rarr 119860 be an 119860-bimodule map (ie anadditive map satisfying 119891(119886119909) = 119886119891(119909) and 119891(119909119886) = 119891(119909)119886for all 119886 isin 119860 119909 isin 119872) Then one can put a dialgebra structureon 119872 as follows 119909 ⊣ 119910 = 119909119891(119910) and 119909⊢119910 = 119891(119909)119910 (cf [13Example 22(d)]) Hence119872+ is a quasi-Jordan algebra underthe quasi-Jordan product ldquo⊲rdquo given by119909 ⊲ 119910 = (12)(119909119891(119910)+
119891(119910)119909) Further for any 119909 isin 119872 we observe that
1199092
⊲ 119909 =1
2(119909119891 (119909) + 119891 (119909) 119909) ⊲ 119909
=1
4((119909119891 (119909) + 119891 (119909) 119909) 119891 (119909)
+119891 (119909) (119909119891 (119909) + 119891 (119909) 119909))
=1
4(119909119891 (119909) 119891 (119909) + 2119891 (119909) 119909119891 (119909) + 119891 (119909) 119891 (119909) 119909)
119909 ⊲ 1199092
=1
2119909 ⊲ (119909119891 (119909) + 119891 (119909) 119909)
=1
4(119909119891 (119909119891 (119909) + 119891 (119909) 119909) + 119891 (119909119891 (119909) + 119891 (119909) 119909) 119909)
=1
4(2119909119891 (119909) 119891 (119909) + 2119891 (119909) 119891 (119909) 119909)
(23)
However the right hand sides of the above equations (23)may not be equal see the following example (Example 28)For such elements 119909 we have 119909
2
⊲ 119909 = 119909 ⊲ 1199092 Hence by
Proposition 26 the spectrum of 119909 is unbounded whenever119872
+ is a unital split quasi-Jordan Banach algebra
Example 28 Let 119872 be the collection of 2 times 2 matrices withentries from the fieldC and let119860 be the algebra of allmatricesof the form [
120572 0
0 120573] with 120572 120573 isin C Then it is easily seen
that 119872 is an 119860-bimodule Next we define 119891 119872 rarr 119860
by 119891 [11988611
11988612
11988621
11988622
] = [11988611
0
0 11988622
] Of course 119891 is an additive mapsatisfying
119891([120572 0
0 120573] [
11988611
11988612
11988621
11988622
]) = [120572 0
0 120573]119891([
11988611
11988612
11988621
11988622
])
119891([11988611
11988612
11988621
11988622
] [120572 0
0 120573]) = 119891([
11988611
11988612
11988621
11988622
]) [120572 0
0 120573]
(24)
Hence119891 is an119860-bimodule mapThus by Remark 27119872+ is aquasi-Jordan algebra with the quasi-Jordan product as below
[11988611
11988612
11988621
11988622
] ⊲ [11988711
11988712
11988721
11988722
] =1
2([
11988611
11988612
11988621
11988622
]119891([11988711
11988712
11988721
11988722
])
+119891([11988711
11988712
11988721
11988722
]) [11988611
11988612
11988621
11988622
])
=[[
[
1198861111988711
11988612
11988711
+ 11988722
2
11988621
11988711
+ 11988722
21198862211988722
]]
]
(25)
Indeed 119872+
= 119869 oplus 119885(119872+
) where 119869 = [119886 0
0 119887] 119886 119887 isin C
is a subalgebra of 119872+ and 119885(119872+
) = [0 119886
119887 0] 119886 119887 isin C Any
matrix of the form [1 119886
119887 1] with 119886 119887 isin C is a (right) unit in119872
+Thus119872+ is a unital split quasi-Jordan algebrawith thematrix119868 = [
1 0
0 1] as the unit of its Jordan part 119869
Further a natural norm is defined on119872+ as follows
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
=100381610038161003816100381611988611
1003816100381610038161003816 +100381610038161003816100381611988612
1003816100381610038161003816 +100381610038161003816100381611988621
1003816100381610038161003816 +100381610038161003816100381611988622
1003816100381610038161003816 (26)
This norm also satisfies10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
] ⊲ [11988711
11988712
11988721
11988722
]
10038171003817100381710038171003817100381710038171003817
=1
2(
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
] [11988711
0
0 11988722
]
+ [11988711
0
0 11988722
] [11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
)
le
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
10038171003817100381710038171003817100381710038171003817
[11988711
0
0 11988722
]
10038171003817100381710038171003817100381710038171003817
le
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
10038171003817100381710038171003817100381710038171003817
[11988711
11988712
11988721
11988722
]
10038171003817100381710038171003817100381710038171003817
(27)
Next for any 119909 = [11988611
11988612
11988621
11988622
] isin 119872+ we observe that
1199092
= [11988611
11988612
11988621
11988622
] ⊲ [11988611
11988612
11988621
11988622
]
=[[
[
1198862
1111988612
11988611
+ 11988622
2
11988621
11988611
+ 11988622
21198862
22
]]
]
(28)
so that
1199092
⊲ 119909 = [
[
1198863
11
11988612
4(119886
11+ 119886
22)2
11988621
4(119886
11+ 119886
22)2
1198863
22
]
]
119909 ⊲ 1199092
= [
[
1198863
11
11988612
2(119886
2
11+ 119886
2
22)
11988621
2(119886
2
11+ 119886
2
22) 119886
3
22
]
]
(29)
So for any 119909 = [11988611
11988612
11988621
11988622
] isin 119872+ 1199092 ⊲ 119909 = 119909 ⊲ 119909
2
hArr
11988612
= 11988621
= 0 or 11988611
= 11988622 Thus by Proposition 26
10 Abstract and Applied Analysis
120590(119872+119868)([
11988611
11988612
11988621
11988622
]) = C whenever 11988611
= 11988622
and 11988612 119886
21 = 0
In particular for 119909 = [1 1
0 2] isin 119872
+ we have
1199092
⊲ 119909 =[[
[
13
1
4(1 + 2)
2
0
4(1 + 2)
2
23
]]
]
= [1
9
4
0 8
]
119909 ⊲ 1199092
=[[
[
13
1
2(1
2
+ 22
)
0
2(1
2
+ 22
) 23
]]
]
= [1
5
2
0 8
]
(30)
Concerning the inequality between the right hand sidesof (23) in Remark 27 we observe for 119909 as above that
1
4(119909119891 (119909) 119891 (119909) + 2119891 (119909) 119909119891 (119909) + 119891 (119909) 119891 (119909) 119909)
=1
4([
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 1
0 2] [
1 0
0 2]
+ [1 0
0 2] [
1 0
0 2] [
1 1
0 2]) = [
19
4
0 8
]
(31)
but
1
4(2119909119891 (119909) 119891 (119909) + 2119891 (119909) 119891 (119909) 119909)
=1
4(2 [
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 0
0 2] [
1 1
0 2])
= [1
5
2
0 8
]
(32)
Hence 1199092 ⊲ 119909 = 119909 ⊲ 1199092 Thus 120590
(119872+119868)([
1 1
0 2]) = C by
Proposition 26Further suppose the matrix [
11988611198862
11988631198864
] isin 119872+ is invertible
with respect to the unit 119868 that is [ 1198861 119886211988631198864
] isin 119866119868(119872
+
) Thenthere exists [ 119887
11198872
11988731198874
] isin 119872+ such that
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] = 119868 + 119868⊲([
1198861
1198862
1198863
1198864
]) = [1 minus119886
2
minus1198863
1]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
= [1198861
1198862
1198863
1198864
] + 119868⊲([
1198861
1198862
1198863
1198864
])
+ 119868⊲([
1198861
1198862
1198863
1198864
]
2
)
=[[
[
1198861
minus1198862
1198861+ 119886
4
2
minus1198863
1198861+ 119886
4
21198864
]]
]
(33)
However
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] =[[
[
11988711198861
1198872
1198861+ 119886
4
2
1198873
1198861+ 119886
4
211988741198864
]]
]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
=
[[[
[
11988711198861
1198872
1198862
1+ 119886
2
4
2
1198873
1198862
1+ 119886
2
4
211988741198862
4
]]]
]
(34)
It follows that 1198861
= 0 1198864
= 0 1198871= 1119886
1 119887
4= 1119886
4 119887
2((119886
1+
1198864)2) = minus119886
2 119887
3((119886
1+119886
4)2) = minus119886
3 119887
2((119886
2
1+119886
2
4)2) = minus119886
2((119886
1+
1198864)2) and 119887
3((119886
2
1+ 119886
2
4)2) = minus119886
3((119886
1+ 119886
4)2) From these
equations we get 1198872(119886
1minus 119886
4)2
= 0 and 1198873(119886
1minus 119886
4)2
= 0 so that(119887
2minus119887
3)(119886
1minus 119886
4)2
= 0Then for 1198861
= 1198864 we obtain 119887
2= 119887
3= 0
and hence 1198862= 119886
3= 0 Therefore
119866119868(119872
+
) = [119886 119887
119888 119886] 119909 = [
120572 0
0 120573] isin 119872
+
119886 = 0
120572 = 0 120573 = 0 120572 = 120573
(35)
The set 119866119868(119872
+
) is not open clearly [1 1
0 1] isin 119866
119868(119872
+
) forany 120598 gt 0
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[1 1
0 1] minus [
[
1 minus120598
41
0 1 +120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[
[
120598
40
0 minus120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=120598
2lt 120598
(36)
but [ 1minus1205984 1
0 1+1205984] notin 119866
119868(119872
+
)Now if 119860 = [
minus1 1
0 1] then 120582119868 minus 119860 = [
120582+1 minus1
0 120582minus1] notin 119866
119868(119872
+
)and so 120582 isin 120590
(119872+119868)(119860) for all 120582 isin C Thus 120590
(119872+119868)(119860) = C an
unbounded spectrum
Next we observe that the spectrum of an element withrespect to a unit is closed whenever the corresponding set ofinvertibles is open
Proposition 29 LetI be a quasi-Jordan normed algebra witha unit 119890 such that119866
119890(I) is openThen 120590
(I119890)(119909) is closed for all119909 isin I
Proof Define 119891 C rarr I by 119891(120582) = 120582119890 minus 119909 Since 119891 iscontinuous the inverse image of the open set 119866
119890(I) is open
in C and so its complement 120590(I119890)(119909) is closed
We conclude this paper with the following example of anonclosed spectrum
Example 30 Let 119872 and 119872+ be as in Example 28 Let 119864 =
[1 1
0 1] and 119860 = [
119886 0
0 119887] with 119886 = 119887 both different from 1 Then
119864 is a unit in 119872+ We show that 120590
(119872+119864)
(119860) = C 1 Forthis let us first investigate when can an element of the form119861 = [
120572 120573
0 120574] be invertible Assuming that 119861 is invertible we get
the existence of an element 119862 isin 119863+ such that
119862 ⊲ 119861 = 119864 + 119864⊲(119861)
119862 ⊲ 1198612
= 119861 + 119864⊲(119861) + 119864
⊲(119861
2
)
(37)
Abstract and Applied Analysis 11
From these equations we get 120572 = 0 120574 = 0 and
119862 =[[
[
1
120572120573∘
01
120574
]]
]
(38)
where 120573∘satisfies the following two equations
120573∘(120572 + 120574
2) = 1 minus 120573 +
120572 + 120574
2
120573∘(1205722
+ 1205742
2) =
120572 + 120574
2+
1205722
+ 1205742
2minus 120573
120572 + 120574
2
(39)
Multiplying the last equation by 2(120572 + 120574)(1205722
+ 1205742
) and thenusing the other equation we get
2 (1 minus 120573) + (120572 + 120574) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
+ (120572 + 120574) (40)
or equivalently
2 (1 minus 120573) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
(41)
this equation is satisfied for 120573 = 1 or 120572 = 120574 Hence thematrix119861 is invertible if and only if 119861 = [
120572 1
0 120574] or 119861 = [
120572 120573
0 120572
] for 120572 120574 isin
C 0 and 120573 isin CWe conclude that 119860
120582= 120582119864 minus 119860 = [
120582minus119886 120582
0 120582minus119887] is invertible
with respect to 119864 if and only if 120582 minus 119887 = 0 120582 minus 119886 = 0 and 120582 = 1that is119860
120582is invertible if 119886 = 120582 and 119887 = 120582 and 120582 = 1 Hence119860
120582
is invertible only if 120582 = 1 as we assumed that 119886 = 119887 and bothare not 1 So for all 120582 = 1 119860
120582notin 119866
119864(119872
+
) Thus 120590(119872+119864)
(119860) =
C 1 which is neither bounded nor closed
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This project was supported by King Saud University Dean-ship of Scientific Research College of Science ResearchCenter
References
[1] B Fritzsche ldquoSophus Lie a sketch of his life and workrdquo Journalof Lie Theory vol 9 no 1 pp 1ndash38 1999
[2] P Jordan ldquoUber eine klasse nichtassoziativer hyperkomplexeralgebrenrdquo Nachrichten von der Gesellschaft der Wissenschaftenzu Gottingen pp 569ndash575 1932
[3] K McCrimmon A Taste of Jordan Algebras UniversitextSpringer New York NY USA 2004
[4] N Jacobson Structure and Representations of Jordan Algebrasvol 39 of American Mathematical Society Colloquium Publica-tions American Mathematical Society Providence RI USA1968
[5] F Gursey and C-H Tze On the Role of Division Jordan andRelatedAlgebras in Particle PhysicsWorld Scientific River EdgeNJ USA 1996
[6] R Iordanescu Jordan Structures in Analysis Geometry andPhysics EdituraAcademiei Romane Bucharest Romania 2009
[7] H Upmeier Symmetric Banach Manifolds and Jordan Clowast-Alge-bras vol 104 of North-Holland Mathematics Studies North-Holland Amsterdam The Netherlands 1985
[8] I L Kantor ldquoClassification of irreducible transitive differentialgroupsrdquo Doklady Akademii Nauk SSSR vol 158 pp 1271ndash12741964
[9] M Koecher ldquoUber eine Gruppe von rationalen AbbildungenrdquoInventiones Mathematicae vol 3 pp 136ndash171 1967
[10] J Tits ldquoUne classe drsquoalgebres de Lie en relation avec les algebresde Jordanrdquo Indagationes Mathematicae vol 24 pp 530ndash5351962
[11] J-L Loday ldquoUne version non commutative des algebres de Lieles algebres de Leibnizrdquo LrsquoEnseignement Mathematique vol 39no 3-4 pp 269ndash293 1993
[12] J-L Loday and T Pirashvili ldquoUniversal enveloping algebras ofLeibniz algebras and (co)homologyrdquo Mathematische Annalenvol 296 no 1 pp 139ndash158 1993
[13] J-L Loday ldquoDialgebrasrdquo in Dialgebras and Related Operadsvol 1763 of Lecture Notes in Math pp 7ndash66 Springer BerlinGermany 2001
[14] R Velasquez and R Felipe ldquoQuasi-Jordan algebrasrdquo Communi-cations in Algebra vol 36 no 4 pp 1580ndash1602 2008
[15] R Velasquez and R Felipe ldquoOn K-B quasi-Jordan algebrasand their relation with Liebniz algebrasrdquo Comunicaciones delCIMAT No 1-10-1015-12-2010
[16] R Felipe ldquoAn analogue to functional analysis in dialgebrasrdquoInternational Mathematical Forum vol 2 no 21ndash24 pp 1069ndash1091 2007
[17] R Velasquez and R Felipe ldquoSplit dialgebras split quasi-Jordanalgebras and regular elementsrdquo Journal of Algebra and ItsApplications vol 8 no 2 pp 191ndash218 2009
Submit your manuscripts athttpwwwhindawicom
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Discrete MathematicsJournal of
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Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Abstract and Applied Analysis
4 The Spectrum of Elements in a UnitalQuasi-Jordan Algebra
As usual we define the spectrum of an element 119909 in aunital quasi-Jordan algebra (I 119890) denoted by 120590
(I119890)(119909) tobe the collection of all complex numbers 120582 for which 120582119890 minus
119909 is not invertible Thus 120590(I119890)(119909) = 120582 isin C 120582119890 minus
119909 is not invertible Here the subscript 119890 indicates that theinvariability depends on the choice of unit 119890 which generallyis not unique
Proposition 11 Let I be a unital quasi-Jordan algebra Then120590(I119890)(119890
1015840
) = 1 for all 119890 1198901015840 isin 119880(I)
Proof Let 119890 1198901015840 isin 119880(I) Then for any 120582 = 1 119910 = (1(120582minus1))1198901015840
is an inverse of 120582119890 minus 1198901015840 with respect to the unit 119890 because 119910 ⊲
(120582119890 minus 1198901015840
) = (1(120582 minus 1))1198901015840
⊲ (120582119890 minus 1198901015840
) = 119890 + 119890⊲(120582119890 minus 119890
1015840
) and119910 ⊲ (120582119890 minus 119890
1015840
)2
= (120582119890 minus 1198901015840
) + 119890⊲(120582119890 minus 119890
1015840
) + 119890⊲((120582119890 minus 119890
1015840
)2
)
Proposition 12 Let I be a quasi-Jordan algebra with unit 119890Then 120590
(I119890)(119911) = 0 for all 119911 isin 119885(I)
Proof For any fixed 119911 isin 119885(I) and nonzero scalar 120582 thevector 119910 = (1120582)(119890 + 119911) satisfies 119910 ⊲ (120582119890 minus 119911) = 120582119910 = 119890 + 119911 =
119890+ 119890⊲(120582119890minus 119911) and 119910 ⊲ (120582119890 minus 119911)
2
= 1205822
119910 = 120582119890+120582119911 = (120582119890minus119911) +
119890⊲(120582119890 minus 119911) + 119890
⊲((120582119890 minus 119911)
2
) So (1120582)(119890 + 119911) is an inverse of120582119890 minus 119911 with respect to 119890 This means 120582 notin 120590
(I119890)(119911) for all 120582 = 0However the zero vector is not invertible Thus 120590
(I119890)(119911) =
0
Proposition 13 Let I = 119869 oplus 119885(I) be a split quasi-Jordanalgebra with unit 119890 isin 119869 Then 120590
(I119890)(119901) sube 0 1 for all idempo-tents 119901 isin I (ie 1199012
= 119901)
Proof Let 119901 be any fixed idempotent inI Since 119901 isin I 119901 hasa unique representation 119901 = 119901
119869+ 119901
119885with 119901
119869isin 119869 and 119901
119885isin
119885(I) Clearly 1199012
119869+ 119901
119885⊲ 119901
119869= 119901
2
= 119901 = 119901119869+ 119901
119885 Then
by uniqueness of the representation in the split quasi-Jordanalgebra I 119901
119885⊲ 119901
119869= 119901
119885and 119901
2
119869= 119901
119869 this means 119901
119869is an
idempotent in the Jordan algebra 119869 Hence 120590(119869119890)
(119901119869) sube 0 1
Thus 119901120582= 120582119890 minus 119901
119869is invertible in 119869 with the unique inverse
119901minus1
120582 for all 120582 notin 0 1We show that 119910 = 119901
minus1
120582+ (1(120582 minus 1))119901
119885is an inverse of
120582119890 minus 119901 in I with respect to the unit 119890 for this we note that120582119890 minus 119901 = 119901
120582minus 119901
119885 (119901
120582minus 119901
119885)2
= 1199012
120582minus 119901
119885⊲ 119901
120582 119890
⊲(120582119890 minus
119901) = 119890⊲(119901
120582minus 119901
119885) = 119890 ⊲ (119901
120582minus 119901
119885) minus (119901
120582minus 119901
119885) = 119901
119885 and
119890⊲((120582119890 minus 119901)
2
) = 119890Δ((119901
120582minus 119901
119885)2
) = 119890⊲(119901
2
120582minus 119901
119885⊲ 119901
120582) = 119901
119885⊲
119901120582= 119901
119885⊲ (120582119890minus119901
119869) = 120582119901
119885minus119901
119885⊲ 119901
119869= 120582119901
119885minus119901
119885= (120582minus1)119901
119885
Hence 119910 ⊲ (120582119890 minus 119901) = 119910 ⊲ (119901120582minus 119901
119885) = 119910 ⊲ 119901
120582= (119901
minus1
120582+
(1(120582 minus 1))119901119885) ⊲ 119901
120582= 119901
minus1
120582⊲ 119901
120582+ (1(120582 minus 1))119901
119885⊲ 119901
120582= 119890 +
(1(120582 minus 1))119901119885
⊲ (120582119890 minus 119901119869) = 119890 + (1(120582 minus 1))(120582119901
119885minus 119901
119885⊲
119901119869) = 119890 + (1(120582 minus 1))(120582119901
119885minus 119901
119885) = 119890 + 119901
119885= 119890 + 119890
⊲(120582119890 minus 119901)
and 119910 ⊲ (120582119890 minus 119901)2
= 119910 ⊲ (1199012
120582minus 119901
119885⊲ 119901
120582) = 119910 ⊲ 119901
2
120582=
(119901minus1
120582+(1(120582minus1))119901
119885) ⊲ 119901
2
120582= 119901
120582+(1(120582minus1))119901
119885⊲ (120582119890minus119901
119869)2
=
119901120582+(1(120582minus1))119901
119885⊲ (120582
2
119890minus2120582119901119869+119901
2
119869) = 119901
120582+(1(120582minus1))119901
119885⊲
(1205822
119890 minus 2120582119901119869+ 119901
119869) = 119901
120582+ (1(120582 minus 1))(120582
2
119901119885
⊲ 119890 minus 2120582119901119885
⊲
119901119869+ 119901
119885⊲ 119901
119869) = 119901
120582+ (1(120582 minus 1))(120582
2
119901119885
minus 2120582119901119885
+ 119901119885) =
119901120582+ (1(120582 minus 1))(120582 minus 1)
2
119901119885= 120582119890 minus119901
119869+ (120582 minus 1)119901
119885= (120582119890 minus119901) +
119890⊲(120582119890 minus 119901) + 119890
⊲((120582119890 minus 119901)
2
)
As mentioned in Section 2 if I is a quasi-Jordan algebrawith a unit 119890 then the set 119890 + 119911 119911 isin 119885(I) coincides withthe set 119880(I) of all units inI
Proposition 14 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebrawith unit 119890 isin 119869 and119909 isin I invertiblewith respectto some 1198901015840 isin 119880(I)Then119909
119869is invertible with respect to the unit
119890 in the Jordan algebra 119869
Proof Clearly 119890 ⊲ 119909 minus 119909 = minus119909119885and 119890 ⊲ 119909
2
minus 1199092
= minus119909119885⊲ 119909
119869
Since 1198901015840
isin 119880(I) we have 1198901015840
= 119890 + 119911 for some 119911 isin 119885(I)Hence the invertibility of 119909 in I with respect to the unit 1198901015840gives the existence of 119910 isin I such that 119910
119869⊲ 119909
119869+ 119910
119885⊲ 119909
119869=
119910 ⊲ 119909119869= 119910 ⊲ 119909 = 119890 + 119911 + (119890 + 119911) ⊲ 119909 minus 119909 = 119890 + 119911 + 119911 ⊲ 119909 minus 119909
119885
and 119910119869⊲ 119909
2
119869+ 119910
119885⊲ 119909
2
119869= 119910 ⊲ 119909
2
119869= 119910 ⊲ 119909
2
= (119890 + 119911) ⊲
119909 + (119890 + 119911) ⊲ 1199092
minus 1199092
= 119909119869+ 119911 ⊲ 119909 + 119911 ⊲ 119909
2
minus 119909119885⊲ 119909
119869 So by
the uniqueness of the representations in the split algebra Iwe get 119910
119869⊲ 119909
119869= 119890 and 119910
119869⊲ 119909
2
119869= 119909
119869 Thus 119910
119869is the inverse
of 119909119869in the Jordan algebra 119869
Next we observe that the spectrum of 119909 in a unital splitquasi-Jordan algebra with respect to any unit includes thespectrum of 119909
119869in the Jordan part 119869
Corollary 15 LetI = 119869 oplus119885(I) be a unital split quasi-Jordanalgebra with unit 119890 isin 119869 and let 119909 isin I Then 120590
(119869119890)(119909
119869) sube
120590(I1198901015840)(119909) for all 1198901015840 isin 119880(I)
Proof Let 1198901015840 = 119890 + 119911 with 119911 isin 119885(I) and let 120582 notin 120590(I1198901015840)(119909)
Then 1205821198901015840 minus 119909 is invertible in I with respect to the unit 1198901015840Hence its Jordanpart120582119890minus119909
119869is invertible in the Jordan algebra
119869 by Proposition 14 Thus 120582 notin 120590(119869119890)
(119909119869)
It is well known that the spectrum of any element ina unital Jordan Banach algebra is nonempty (cf [7]) Thistogether with Proposition 2 and Corollary 15 gives thefollowing result
Corollary 16 The spectrum of any element in a unital splitquasi-Jordan Banach algebra is nonempty
The next result extends Corollary 16 to any quasi-JordanBanach algebra with a norm 1 unit
Proposition 17 The spectrum of any element in a quasi-Jordan Banach algebra with a norm 1 unit is nonempty
Proof Let I be a unital quasi-Jordan Banach algebra witha norm 1 unit 119890 From Section 2 we know that the map120593 119909 997891rarr 120593(119909) = (119909 119877
119909) embeds I into the unital split
quasi-Jordan Banach algebra I1
= (119909 119877119910) 119909 119910 isin I
equipped with the sum (119909 119877119910) + (119886 119877
119887) = (119909 + 119886 119877
119910+119887) sca-
lar multiplication 120582(119909 119877119910) = (120582119909 119877
120582119910) and product (119909
119877119910) ⊲ (119886 119877
119887) = (119909 ⊲ 119887 119877
119910⊲119887) and the image 120593(I) is a
norm closed right ideal isomorphic to I with norm 1 unit(119890 119877
119890) Moreover it is seen in Section 3 that the embedding
Abstract and Applied Analysis 7
120593 also preserves the corresponding invertible elements thatis (119910 119877
119910) is an inverse of (119909 119877
119909) with respect to the unit
(119890 119877119890) in I
1whenever 119910 is an inverse of 119909 with respect
to a unit 119890 in I Hence by Corollary 16 it follows that120601 = 120590
(I1(119890119877119890))(119909 119877
119909) sube 120590
(120593(I)(119890119877119890))(119909 119877
119909) = 120590
(I119890)(119909) for all119909 isin I
Proposition 18 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebra with unit 119890 isin 119869 and let 119909 isin 119869 Then 120590
(119869119890)(119909) =
120590(I119890)(119909)
Proof By Corollary 15 120590(119869119890)
(119909) sube 120590(I119890)(119909) For the reverse
inclusion let 120582 notin 120590(119869119890)
(119909) then 120582119890 minus 119909 is invertible in 119869 thatis there exists 119910 isin 119869 such that 119910 ⊲ (120582119890 minus 119909) = 119890 and 119910 ⊲
(120582119890 minus 119909)2
= 120582119890 minus 119909 However 119890⊲(120582119890 minus 119909) = 0 = 119890
⊲((120582119890 minus 119909)
2
)Hence 119910 is an inverse of 120582119890 minus 119909 in I with respect to the unit119890 that is 120582 notin 120590
(I119890)(119909) Thus 120590(I119890)(119909) sube 120590
(119869119890)(119909)
Proposition 19 Let I be a unital quasi-Jordan normedalgebra and let 119890 be a unit inI for which 119866
119890(I) is open Then
120590(I119890)(119909) = 120590
(I119890)(119909 + 119911) for all 119911 isin 119885(I)
Proof By Corollary 8 120582119890 minus 119909 is invertible if and only if 120582119890 minus
(119909 + 119911) is invertible for all 119911 isin 119885(I) Thus 120582 notin 120590(I119890)(119909) if
and only if 120582 notin 120590(I119890)(119909 + 119911) for all 119911 isin 119885(I)
Corollary 20 LetI = 119869oplus119885(I) be a unital split quasi-Jordannormed algebra and let 119890 be a unit in I such that 119866
119890(I) is
open Then 120590(I119890)(119909) = 120590
(I119890)(119909119869) for all 119909 = 119909119869+ 119909
119885isin I
Further if the unit 119890 isin 119869 then120590(I119890)(119909) = 120590
(I119890)(119909119869) = 120590(119869119890)
(119909119869)
Lemma 21 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordanalgebra with a unit 119890 isin 119869 and let 119909 isin I be invertible withrespect to 119890 Then 119909
119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
Proof As 119909 is invertible there exists 119910 isin I such that 119910 ⊲ 119909 =
119890 + (119890 ⊲ 119909 minus 119909) = 119890 minus 119909119885and 119910 ⊲ 119909
2
= 119909 + (119890 ⊲ 119909 minus 119909) + (119890 ⊲
1199092
minus 1199092
) = 119909119869minus 119909
119885⊲ 119909 Hence by the uniqueness of the
representation in a split quasi-Jordan algebra we obtain
119910119885⊲ 119909 = minus119909
119885 (11)
119910119885⊲ 119909
2
= minus119909119885⊲ 119909 (12)
Thus
119909119885⊲ 119909
2
= minus (119910119885⊲ 119909) ⊲ 119909
2
(by (11))
= minus (119910119885⊲ 119909
2
) ⊲ 119909 (by the right Jordan identity)
= (119909119885⊲ 119909) ⊲ 119909 (by (12))
(13)
Proposition 22 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebra with a unit 119890 isin 119869 119909 = 119909
119869+ 119909
119885isin I satisfies
120590(I119890)(119909) =C Then
(1) 119909119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
(2) 1199092 ⊲ 119909 = 119909 ⊲ 1199092
Proof (1) Let 120582 notin 120590(I119890)(119909) Then 120582119890 minus 119909 = (120582119890 minus 119909
119869) + (minus119909
119885)
is invertible with respect to 119890 By Lemma 21 we have
119909119885⊲ (120582119890 minus 119909)
2
= (119909119885⊲ (120582119890 minus 119909)) ⊲ (120582119890 minus 119909) (14)
since the zero part of 120582119890 minus 119909 is minus119909119885 However
119909119885⊲ (120582119890 minus 119909)
2
= 119909119885⊲ (120582
2
119890 minus 120582119890 ⊲ 119909 minus 120582119909 ⊲ 119890 + 1199092
)
= 1205822
(119909119885⊲ 119890) minus 120582 (119909
119885⊲ (119890 ⊲ 119909))
minus 120582 (119909119885⊲ (119909 ⊲ 119890)) + 119909
119885⊲ 119909
2
= 1205822
119909119885minus 2120582119909
119885⊲ 119909
+ 119909119885⊲ 119909
2
(by the right commutativity of ⊲)
(119909119885⊲ (120582119890 minus 119909)) ⊲ (120582119890 minus 119909)
= (120582 (119909119885⊲ 119890) minus 119909
119885⊲ 119909) ⊲ (120582119890 minus 119909)
= (120582119909119885minus 119909
119885⊲ 119909) ⊲ (120582119890 minus 119909)
= 1205822
(119909119885⊲ 119890) minus 120582 (119909
119885⊲ 119909) ⊲ 119890
minus 120582119909119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
= 1205822
119909119885minus 2120582119909
119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
(15)
Therefore (14) becomes
1205822
119909119885minus 2120582119909
119885⊲ 119909 + 119909
119885⊲ 119909
2
= 1205822
119909119885minus 2120582119909
119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
(16)
which after simplification reduces to the required equation119909119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
(2) Since 1199092 = (119909119869+ 119909
119885)2
= 1199092
119869+ 119909
119885⊲ 119909
119869 we have
1199092
⊲ 119909 = (1199092
119869+ 119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909
119869+ (119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909
119869+ 119909
119885⊲ 119909
2
(17)
by the part (1) But 1199092119869⊲ 119909
119869= 119909
119869⊲ 119909
2
119869since 119909
119869is in the Jor-
dan algebra 119869 Therefore 1199092 ⊲ 119909 = 1199092
119869⊲ 119909
119869+ 119909
119885⊲ 119909
2
= 119909119869⊲
1199092
119869+ 119909
119885⊲ 119909
2
= 119909 ⊲ 1199092
Remark 23 In any quasi-Jordan algebra if an element 119909
satisfies 1199093
= 1199092
⊲ 119909 = 119909 ⊲ 1199092 then 119909
119899
⊲ 1199092
= 119909119899+2 for
all positive integers 119899 For this suppose 119909 satisfies 1199092 ⊲ 119909 =
119909 ⊲ 1199092 and 119909
119898
⊲ 1199092
= 119909119898+2 for any fixed119898 ge 1 Then 119909
119898+1
⊲
1199092
= (119909119898
⊲ 119909) ⊲ 1199092
= (119909119898
⊲ 1199092
) ⊲ 119909 (by the right Jor-dan identity) = 119909
119898+2
⊲ 119909 = 119909119898+3
Proposition 24 Let I be a unital quasi-Jordan Banachalgebra with unit 119890 and let 119909 isin I satisfy (119890 minus 119909) ⊲ (119890 minus 119909)
2
=
(119890 minus 119909)2
⊲ (119890 minus 119909) Then 119909 isin 119866119890(I) whenever 119890 minus 119909 lt 1
8 Abstract and Applied Analysis
Proof First note that 119890minus119909 lt 1 gives (119890 minus 119909)119899
le 119890 minus 119909119899
lt
1 for all 119899 = 1 2 3 Hence the infinite geometric series119890 +sum
infin
119899=1(119890 minus119909)
119899 converges absolutely to some element 119910 isin IWe show that the geometric series sum 119910 is an inverse of 119909with respect to the unit 119890 For any fixed positive integer 119899 let119910119899= 119890+sum
119899
119896=1(119890minus119909)
119896Then the sequence 119910119899 of partial sums
converges to 119910 By setting 119908 = 119890 minus 119909 we get
119910119899⊲ 119909 = (119890 +
119899
sum
119896=1
119908119896
) ⊲ (119890 minus 119908)
= 119890 +
119899
sum
119896=1
119908119896
minus (119890 ⊲ 119908 +
119899+1
sum
119896=2
119908119896
)
= 119890 + 119908 minus 119890 ⊲ 119908 minus 119908119899+1
= 119890 + (119890 minus 119909) minus 119890 ⊲ (119890 minus 119909) minus (119890 minus 119909)119899+1
= 119890 + (119890 minus 119909 minus 119890 + 119890 ⊲ 119909) minus (119890 minus 119909)119899+1
= 119890 + (119890 ⊲ 119909 minus 119909) minus (119890 minus 119909)119899+1
(18)
Thus by allowing 119899 rarr infin we obtain 119909 ⊲ 119910 = 119890 + (119890 ⊲
119909 minus 119909) = 119890 + 119890⊲(119909) since 119890 minus 119909 lt 1
Next by Remark 23 we have
119910119899⊲ 119909
2
= 119910119899⊲ (119890 minus 119908)
2
= 119910119899⊲ (119890 minus 119890 ⊲ 119908 minus 119908 ⊲ 119890 + 119908
2
)
= 119910119899⊲ (119890 minus 2119908 + 119908
2
)
= 119910119899⊲ 119890 minus 2119910
119899⊲ 119908 + 119910
119899⊲ 119908
2
= 119910119899minus 2119910
119899⊲ 119908 + 119910
119899⊲ 119908
2
= (119890 +
119899
sum
119896=1
119908119896
) minus 2(119890 +
119899
sum
119896=1
119908119896
) ⊲ 119908
+ (119890 +
119899
sum
119896=1
119908119896
) ⊲ 1199082
= (119890 +
119899
sum
119896=1
119908119896
) minus 2(119890 ⊲ 119908 +
119899+1
sum
119896=2
119908119896
)
+ (119890 ⊲ 1199082
+
119899+2
sum
119896=3
119908119896
)
= 119890 + 119908 minus 2119890 ⊲ 119908 minus 1199082
+ 119890 ⊲ 1199082
minus 119908119899+1
+ 119908119899+2
(19)
Taking the limit as 119899 rarr infin we get
119910 ⊲ 1199092
= 119890 + 119908 minus 2119890 ⊲ 119908 minus 1199082
+ 119890 ⊲ 1199082
= 119890 + (119890 minus 119909) minus 2119890 ⊲ (119890 minus 119909) minus (119890 minus 119909)2
+ 119890 ⊲ (119890 minus 119909)2
= 2119890 minus 119909 minus 2119890 + 2119890 ⊲ 119909 minus 119890 + 119890 ⊲ 119909 + 119909 minus 1199092
+ 119890 minus 2119890 ⊲ 119909 + 119890 ⊲ 1199092
= 119909 + (119890 ⊲ 119909 minus 119909) + (119890 ⊲ 1199092
minus 1199092
)
(20)
Proposition 25 Let I be a unital split quasi-Jordan Banachalgebra with unit 119890 If 119909 isin I with (119890 minus 119909) ⊲ (119890 minus 119909)
2
=
(119890 minus 119909)2
⊲ (119890 minus 119909) then |120582| le 119909 for all 120582 isin 120590(I119890)(119909)
Proof If 120582 isin 120590(I119890)(119909) with 120582 = 0 then the noninvertibility of
120582119890 minus 119909 means the noninvertibility of 119890 minus (1120582)119909 with respectto the unit 119890 However by Proposition 24 119890 minus (1120582)119909must beinvertible with respect to the unit 119890 whenever (1|120582|)119909 lt 1It follows that |120582| le 119909 for all 120582 isin 120590
(I119890)(119909)
5 Unbounded and Nonclosed Spectrum
In this section we show that the spectrum of an element ina split quasi-Jordan Banach algebra may be neither boundednor closed and hence not compactThe following result givesa couple of characterizations of the unbounded spectrum ofan element in a split quasi-Jordan Banach algebra
Proposition 26 Let 119909 be an element of a unital split quasi-Jordan Banach algebra I = 119869 oplus 119885(I) with a unit 119890 isin 119869 Thenthe following statements are equivalent
(1) 120590(I119890)(119909) =C
(2) 1199092 ⊲ 119909 = 119909 ⊲ 1199092
(3) |120582| le 119909 for all 120582 isin 120590(I119890)(119909)
Proof (1rArr2) See Proposition 22(2rArr3) Suppose 1199092 ⊲ 119909 = 119909 ⊲ 119909
2 Then
(119890 minus 119909)2
⊲ (119890 minus 119909) = (119890 minus 119890 ⊲ 119909 minus 119909 + 1199092
) ⊲ (119890 minus 119909)
= 119890 minus 2119890 ⊲ 119909 minus 119909 + 21199092
+ (119890 ⊲ 119909) ⊲ 119909 minus 1199092
⊲ 119909
= 119890 minus 2119909119869minus 119909 + 2119909
2
+ 119890 ⊲ 1199092
minus 1199093
(119890 minus 119909) ⊲ (119890 minus 119909)2
= (119890 minus 119909) ⊲ (119890 minus 2119909 + 1199092
)
= 119890 minus 2 (119890 ⊲ 119909) + 119890 ⊲ 1199092
minus 119909
+ 21199092
minus 119909 ⊲ 1199092
= 119890 minus 2119909119869+ 119890 ⊲ 119909
2
minus 119909 + 21199092
minus 1199093
(21)
Abstract and Applied Analysis 9
From (21) we get
(119890 minus 119909)2
⊲ (119890 minus 119909) = (119890 minus 119909) ⊲ (119890 minus 119909)2
(22)
Hence |120582| le 119909 for all 120582 isin 120590(I119890)(119909) by Proposition 25
(3rArr1) Immediate
Remark 27 There do exist unital split quasi-Jordan algebrascontaining elements that have the spectrum with respect tothe unit of the Jordan part equal to the whole ofC and henceunbounded To justify this claim we proceed as follows
Let 119860 be a unital associative algebra and let 119872 be an 119860-bimodule Let 119891 119872 rarr 119860 be an 119860-bimodule map (ie anadditive map satisfying 119891(119886119909) = 119886119891(119909) and 119891(119909119886) = 119891(119909)119886for all 119886 isin 119860 119909 isin 119872) Then one can put a dialgebra structureon 119872 as follows 119909 ⊣ 119910 = 119909119891(119910) and 119909⊢119910 = 119891(119909)119910 (cf [13Example 22(d)]) Hence119872+ is a quasi-Jordan algebra underthe quasi-Jordan product ldquo⊲rdquo given by119909 ⊲ 119910 = (12)(119909119891(119910)+
119891(119910)119909) Further for any 119909 isin 119872 we observe that
1199092
⊲ 119909 =1
2(119909119891 (119909) + 119891 (119909) 119909) ⊲ 119909
=1
4((119909119891 (119909) + 119891 (119909) 119909) 119891 (119909)
+119891 (119909) (119909119891 (119909) + 119891 (119909) 119909))
=1
4(119909119891 (119909) 119891 (119909) + 2119891 (119909) 119909119891 (119909) + 119891 (119909) 119891 (119909) 119909)
119909 ⊲ 1199092
=1
2119909 ⊲ (119909119891 (119909) + 119891 (119909) 119909)
=1
4(119909119891 (119909119891 (119909) + 119891 (119909) 119909) + 119891 (119909119891 (119909) + 119891 (119909) 119909) 119909)
=1
4(2119909119891 (119909) 119891 (119909) + 2119891 (119909) 119891 (119909) 119909)
(23)
However the right hand sides of the above equations (23)may not be equal see the following example (Example 28)For such elements 119909 we have 119909
2
⊲ 119909 = 119909 ⊲ 1199092 Hence by
Proposition 26 the spectrum of 119909 is unbounded whenever119872
+ is a unital split quasi-Jordan Banach algebra
Example 28 Let 119872 be the collection of 2 times 2 matrices withentries from the fieldC and let119860 be the algebra of allmatricesof the form [
120572 0
0 120573] with 120572 120573 isin C Then it is easily seen
that 119872 is an 119860-bimodule Next we define 119891 119872 rarr 119860
by 119891 [11988611
11988612
11988621
11988622
] = [11988611
0
0 11988622
] Of course 119891 is an additive mapsatisfying
119891([120572 0
0 120573] [
11988611
11988612
11988621
11988622
]) = [120572 0
0 120573]119891([
11988611
11988612
11988621
11988622
])
119891([11988611
11988612
11988621
11988622
] [120572 0
0 120573]) = 119891([
11988611
11988612
11988621
11988622
]) [120572 0
0 120573]
(24)
Hence119891 is an119860-bimodule mapThus by Remark 27119872+ is aquasi-Jordan algebra with the quasi-Jordan product as below
[11988611
11988612
11988621
11988622
] ⊲ [11988711
11988712
11988721
11988722
] =1
2([
11988611
11988612
11988621
11988622
]119891([11988711
11988712
11988721
11988722
])
+119891([11988711
11988712
11988721
11988722
]) [11988611
11988612
11988621
11988622
])
=[[
[
1198861111988711
11988612
11988711
+ 11988722
2
11988621
11988711
+ 11988722
21198862211988722
]]
]
(25)
Indeed 119872+
= 119869 oplus 119885(119872+
) where 119869 = [119886 0
0 119887] 119886 119887 isin C
is a subalgebra of 119872+ and 119885(119872+
) = [0 119886
119887 0] 119886 119887 isin C Any
matrix of the form [1 119886
119887 1] with 119886 119887 isin C is a (right) unit in119872
+Thus119872+ is a unital split quasi-Jordan algebrawith thematrix119868 = [
1 0
0 1] as the unit of its Jordan part 119869
Further a natural norm is defined on119872+ as follows
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
=100381610038161003816100381611988611
1003816100381610038161003816 +100381610038161003816100381611988612
1003816100381610038161003816 +100381610038161003816100381611988621
1003816100381610038161003816 +100381610038161003816100381611988622
1003816100381610038161003816 (26)
This norm also satisfies10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
] ⊲ [11988711
11988712
11988721
11988722
]
10038171003817100381710038171003817100381710038171003817
=1
2(
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
] [11988711
0
0 11988722
]
+ [11988711
0
0 11988722
] [11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
)
le
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
10038171003817100381710038171003817100381710038171003817
[11988711
0
0 11988722
]
10038171003817100381710038171003817100381710038171003817
le
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
10038171003817100381710038171003817100381710038171003817
[11988711
11988712
11988721
11988722
]
10038171003817100381710038171003817100381710038171003817
(27)
Next for any 119909 = [11988611
11988612
11988621
11988622
] isin 119872+ we observe that
1199092
= [11988611
11988612
11988621
11988622
] ⊲ [11988611
11988612
11988621
11988622
]
=[[
[
1198862
1111988612
11988611
+ 11988622
2
11988621
11988611
+ 11988622
21198862
22
]]
]
(28)
so that
1199092
⊲ 119909 = [
[
1198863
11
11988612
4(119886
11+ 119886
22)2
11988621
4(119886
11+ 119886
22)2
1198863
22
]
]
119909 ⊲ 1199092
= [
[
1198863
11
11988612
2(119886
2
11+ 119886
2
22)
11988621
2(119886
2
11+ 119886
2
22) 119886
3
22
]
]
(29)
So for any 119909 = [11988611
11988612
11988621
11988622
] isin 119872+ 1199092 ⊲ 119909 = 119909 ⊲ 119909
2
hArr
11988612
= 11988621
= 0 or 11988611
= 11988622 Thus by Proposition 26
10 Abstract and Applied Analysis
120590(119872+119868)([
11988611
11988612
11988621
11988622
]) = C whenever 11988611
= 11988622
and 11988612 119886
21 = 0
In particular for 119909 = [1 1
0 2] isin 119872
+ we have
1199092
⊲ 119909 =[[
[
13
1
4(1 + 2)
2
0
4(1 + 2)
2
23
]]
]
= [1
9
4
0 8
]
119909 ⊲ 1199092
=[[
[
13
1
2(1
2
+ 22
)
0
2(1
2
+ 22
) 23
]]
]
= [1
5
2
0 8
]
(30)
Concerning the inequality between the right hand sidesof (23) in Remark 27 we observe for 119909 as above that
1
4(119909119891 (119909) 119891 (119909) + 2119891 (119909) 119909119891 (119909) + 119891 (119909) 119891 (119909) 119909)
=1
4([
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 1
0 2] [
1 0
0 2]
+ [1 0
0 2] [
1 0
0 2] [
1 1
0 2]) = [
19
4
0 8
]
(31)
but
1
4(2119909119891 (119909) 119891 (119909) + 2119891 (119909) 119891 (119909) 119909)
=1
4(2 [
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 0
0 2] [
1 1
0 2])
= [1
5
2
0 8
]
(32)
Hence 1199092 ⊲ 119909 = 119909 ⊲ 1199092 Thus 120590
(119872+119868)([
1 1
0 2]) = C by
Proposition 26Further suppose the matrix [
11988611198862
11988631198864
] isin 119872+ is invertible
with respect to the unit 119868 that is [ 1198861 119886211988631198864
] isin 119866119868(119872
+
) Thenthere exists [ 119887
11198872
11988731198874
] isin 119872+ such that
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] = 119868 + 119868⊲([
1198861
1198862
1198863
1198864
]) = [1 minus119886
2
minus1198863
1]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
= [1198861
1198862
1198863
1198864
] + 119868⊲([
1198861
1198862
1198863
1198864
])
+ 119868⊲([
1198861
1198862
1198863
1198864
]
2
)
=[[
[
1198861
minus1198862
1198861+ 119886
4
2
minus1198863
1198861+ 119886
4
21198864
]]
]
(33)
However
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] =[[
[
11988711198861
1198872
1198861+ 119886
4
2
1198873
1198861+ 119886
4
211988741198864
]]
]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
=
[[[
[
11988711198861
1198872
1198862
1+ 119886
2
4
2
1198873
1198862
1+ 119886
2
4
211988741198862
4
]]]
]
(34)
It follows that 1198861
= 0 1198864
= 0 1198871= 1119886
1 119887
4= 1119886
4 119887
2((119886
1+
1198864)2) = minus119886
2 119887
3((119886
1+119886
4)2) = minus119886
3 119887
2((119886
2
1+119886
2
4)2) = minus119886
2((119886
1+
1198864)2) and 119887
3((119886
2
1+ 119886
2
4)2) = minus119886
3((119886
1+ 119886
4)2) From these
equations we get 1198872(119886
1minus 119886
4)2
= 0 and 1198873(119886
1minus 119886
4)2
= 0 so that(119887
2minus119887
3)(119886
1minus 119886
4)2
= 0Then for 1198861
= 1198864 we obtain 119887
2= 119887
3= 0
and hence 1198862= 119886
3= 0 Therefore
119866119868(119872
+
) = [119886 119887
119888 119886] 119909 = [
120572 0
0 120573] isin 119872
+
119886 = 0
120572 = 0 120573 = 0 120572 = 120573
(35)
The set 119866119868(119872
+
) is not open clearly [1 1
0 1] isin 119866
119868(119872
+
) forany 120598 gt 0
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[1 1
0 1] minus [
[
1 minus120598
41
0 1 +120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[
[
120598
40
0 minus120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=120598
2lt 120598
(36)
but [ 1minus1205984 1
0 1+1205984] notin 119866
119868(119872
+
)Now if 119860 = [
minus1 1
0 1] then 120582119868 minus 119860 = [
120582+1 minus1
0 120582minus1] notin 119866
119868(119872
+
)and so 120582 isin 120590
(119872+119868)(119860) for all 120582 isin C Thus 120590
(119872+119868)(119860) = C an
unbounded spectrum
Next we observe that the spectrum of an element withrespect to a unit is closed whenever the corresponding set ofinvertibles is open
Proposition 29 LetI be a quasi-Jordan normed algebra witha unit 119890 such that119866
119890(I) is openThen 120590
(I119890)(119909) is closed for all119909 isin I
Proof Define 119891 C rarr I by 119891(120582) = 120582119890 minus 119909 Since 119891 iscontinuous the inverse image of the open set 119866
119890(I) is open
in C and so its complement 120590(I119890)(119909) is closed
We conclude this paper with the following example of anonclosed spectrum
Example 30 Let 119872 and 119872+ be as in Example 28 Let 119864 =
[1 1
0 1] and 119860 = [
119886 0
0 119887] with 119886 = 119887 both different from 1 Then
119864 is a unit in 119872+ We show that 120590
(119872+119864)
(119860) = C 1 Forthis let us first investigate when can an element of the form119861 = [
120572 120573
0 120574] be invertible Assuming that 119861 is invertible we get
the existence of an element 119862 isin 119863+ such that
119862 ⊲ 119861 = 119864 + 119864⊲(119861)
119862 ⊲ 1198612
= 119861 + 119864⊲(119861) + 119864
⊲(119861
2
)
(37)
Abstract and Applied Analysis 11
From these equations we get 120572 = 0 120574 = 0 and
119862 =[[
[
1
120572120573∘
01
120574
]]
]
(38)
where 120573∘satisfies the following two equations
120573∘(120572 + 120574
2) = 1 minus 120573 +
120572 + 120574
2
120573∘(1205722
+ 1205742
2) =
120572 + 120574
2+
1205722
+ 1205742
2minus 120573
120572 + 120574
2
(39)
Multiplying the last equation by 2(120572 + 120574)(1205722
+ 1205742
) and thenusing the other equation we get
2 (1 minus 120573) + (120572 + 120574) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
+ (120572 + 120574) (40)
or equivalently
2 (1 minus 120573) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
(41)
this equation is satisfied for 120573 = 1 or 120572 = 120574 Hence thematrix119861 is invertible if and only if 119861 = [
120572 1
0 120574] or 119861 = [
120572 120573
0 120572
] for 120572 120574 isin
C 0 and 120573 isin CWe conclude that 119860
120582= 120582119864 minus 119860 = [
120582minus119886 120582
0 120582minus119887] is invertible
with respect to 119864 if and only if 120582 minus 119887 = 0 120582 minus 119886 = 0 and 120582 = 1that is119860
120582is invertible if 119886 = 120582 and 119887 = 120582 and 120582 = 1 Hence119860
120582
is invertible only if 120582 = 1 as we assumed that 119886 = 119887 and bothare not 1 So for all 120582 = 1 119860
120582notin 119866
119864(119872
+
) Thus 120590(119872+119864)
(119860) =
C 1 which is neither bounded nor closed
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This project was supported by King Saud University Dean-ship of Scientific Research College of Science ResearchCenter
References
[1] B Fritzsche ldquoSophus Lie a sketch of his life and workrdquo Journalof Lie Theory vol 9 no 1 pp 1ndash38 1999
[2] P Jordan ldquoUber eine klasse nichtassoziativer hyperkomplexeralgebrenrdquo Nachrichten von der Gesellschaft der Wissenschaftenzu Gottingen pp 569ndash575 1932
[3] K McCrimmon A Taste of Jordan Algebras UniversitextSpringer New York NY USA 2004
[4] N Jacobson Structure and Representations of Jordan Algebrasvol 39 of American Mathematical Society Colloquium Publica-tions American Mathematical Society Providence RI USA1968
[5] F Gursey and C-H Tze On the Role of Division Jordan andRelatedAlgebras in Particle PhysicsWorld Scientific River EdgeNJ USA 1996
[6] R Iordanescu Jordan Structures in Analysis Geometry andPhysics EdituraAcademiei Romane Bucharest Romania 2009
[7] H Upmeier Symmetric Banach Manifolds and Jordan Clowast-Alge-bras vol 104 of North-Holland Mathematics Studies North-Holland Amsterdam The Netherlands 1985
[8] I L Kantor ldquoClassification of irreducible transitive differentialgroupsrdquo Doklady Akademii Nauk SSSR vol 158 pp 1271ndash12741964
[9] M Koecher ldquoUber eine Gruppe von rationalen AbbildungenrdquoInventiones Mathematicae vol 3 pp 136ndash171 1967
[10] J Tits ldquoUne classe drsquoalgebres de Lie en relation avec les algebresde Jordanrdquo Indagationes Mathematicae vol 24 pp 530ndash5351962
[11] J-L Loday ldquoUne version non commutative des algebres de Lieles algebres de Leibnizrdquo LrsquoEnseignement Mathematique vol 39no 3-4 pp 269ndash293 1993
[12] J-L Loday and T Pirashvili ldquoUniversal enveloping algebras ofLeibniz algebras and (co)homologyrdquo Mathematische Annalenvol 296 no 1 pp 139ndash158 1993
[13] J-L Loday ldquoDialgebrasrdquo in Dialgebras and Related Operadsvol 1763 of Lecture Notes in Math pp 7ndash66 Springer BerlinGermany 2001
[14] R Velasquez and R Felipe ldquoQuasi-Jordan algebrasrdquo Communi-cations in Algebra vol 36 no 4 pp 1580ndash1602 2008
[15] R Velasquez and R Felipe ldquoOn K-B quasi-Jordan algebrasand their relation with Liebniz algebrasrdquo Comunicaciones delCIMAT No 1-10-1015-12-2010
[16] R Felipe ldquoAn analogue to functional analysis in dialgebrasrdquoInternational Mathematical Forum vol 2 no 21ndash24 pp 1069ndash1091 2007
[17] R Velasquez and R Felipe ldquoSplit dialgebras split quasi-Jordanalgebras and regular elementsrdquo Journal of Algebra and ItsApplications vol 8 no 2 pp 191ndash218 2009
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
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Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 7
120593 also preserves the corresponding invertible elements thatis (119910 119877
119910) is an inverse of (119909 119877
119909) with respect to the unit
(119890 119877119890) in I
1whenever 119910 is an inverse of 119909 with respect
to a unit 119890 in I Hence by Corollary 16 it follows that120601 = 120590
(I1(119890119877119890))(119909 119877
119909) sube 120590
(120593(I)(119890119877119890))(119909 119877
119909) = 120590
(I119890)(119909) for all119909 isin I
Proposition 18 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebra with unit 119890 isin 119869 and let 119909 isin 119869 Then 120590
(119869119890)(119909) =
120590(I119890)(119909)
Proof By Corollary 15 120590(119869119890)
(119909) sube 120590(I119890)(119909) For the reverse
inclusion let 120582 notin 120590(119869119890)
(119909) then 120582119890 minus 119909 is invertible in 119869 thatis there exists 119910 isin 119869 such that 119910 ⊲ (120582119890 minus 119909) = 119890 and 119910 ⊲
(120582119890 minus 119909)2
= 120582119890 minus 119909 However 119890⊲(120582119890 minus 119909) = 0 = 119890
⊲((120582119890 minus 119909)
2
)Hence 119910 is an inverse of 120582119890 minus 119909 in I with respect to the unit119890 that is 120582 notin 120590
(I119890)(119909) Thus 120590(I119890)(119909) sube 120590
(119869119890)(119909)
Proposition 19 Let I be a unital quasi-Jordan normedalgebra and let 119890 be a unit inI for which 119866
119890(I) is open Then
120590(I119890)(119909) = 120590
(I119890)(119909 + 119911) for all 119911 isin 119885(I)
Proof By Corollary 8 120582119890 minus 119909 is invertible if and only if 120582119890 minus
(119909 + 119911) is invertible for all 119911 isin 119885(I) Thus 120582 notin 120590(I119890)(119909) if
and only if 120582 notin 120590(I119890)(119909 + 119911) for all 119911 isin 119885(I)
Corollary 20 LetI = 119869oplus119885(I) be a unital split quasi-Jordannormed algebra and let 119890 be a unit in I such that 119866
119890(I) is
open Then 120590(I119890)(119909) = 120590
(I119890)(119909119869) for all 119909 = 119909119869+ 119909
119885isin I
Further if the unit 119890 isin 119869 then120590(I119890)(119909) = 120590
(I119890)(119909119869) = 120590(119869119890)
(119909119869)
Lemma 21 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordanalgebra with a unit 119890 isin 119869 and let 119909 isin I be invertible withrespect to 119890 Then 119909
119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
Proof As 119909 is invertible there exists 119910 isin I such that 119910 ⊲ 119909 =
119890 + (119890 ⊲ 119909 minus 119909) = 119890 minus 119909119885and 119910 ⊲ 119909
2
= 119909 + (119890 ⊲ 119909 minus 119909) + (119890 ⊲
1199092
minus 1199092
) = 119909119869minus 119909
119885⊲ 119909 Hence by the uniqueness of the
representation in a split quasi-Jordan algebra we obtain
119910119885⊲ 119909 = minus119909
119885 (11)
119910119885⊲ 119909
2
= minus119909119885⊲ 119909 (12)
Thus
119909119885⊲ 119909
2
= minus (119910119885⊲ 119909) ⊲ 119909
2
(by (11))
= minus (119910119885⊲ 119909
2
) ⊲ 119909 (by the right Jordan identity)
= (119909119885⊲ 119909) ⊲ 119909 (by (12))
(13)
Proposition 22 Let I = 119869 oplus 119885(I) be a unital split quasi-Jordan algebra with a unit 119890 isin 119869 119909 = 119909
119869+ 119909
119885isin I satisfies
120590(I119890)(119909) =C Then
(1) 119909119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
(2) 1199092 ⊲ 119909 = 119909 ⊲ 1199092
Proof (1) Let 120582 notin 120590(I119890)(119909) Then 120582119890 minus 119909 = (120582119890 minus 119909
119869) + (minus119909
119885)
is invertible with respect to 119890 By Lemma 21 we have
119909119885⊲ (120582119890 minus 119909)
2
= (119909119885⊲ (120582119890 minus 119909)) ⊲ (120582119890 minus 119909) (14)
since the zero part of 120582119890 minus 119909 is minus119909119885 However
119909119885⊲ (120582119890 minus 119909)
2
= 119909119885⊲ (120582
2
119890 minus 120582119890 ⊲ 119909 minus 120582119909 ⊲ 119890 + 1199092
)
= 1205822
(119909119885⊲ 119890) minus 120582 (119909
119885⊲ (119890 ⊲ 119909))
minus 120582 (119909119885⊲ (119909 ⊲ 119890)) + 119909
119885⊲ 119909
2
= 1205822
119909119885minus 2120582119909
119885⊲ 119909
+ 119909119885⊲ 119909
2
(by the right commutativity of ⊲)
(119909119885⊲ (120582119890 minus 119909)) ⊲ (120582119890 minus 119909)
= (120582 (119909119885⊲ 119890) minus 119909
119885⊲ 119909) ⊲ (120582119890 minus 119909)
= (120582119909119885minus 119909
119885⊲ 119909) ⊲ (120582119890 minus 119909)
= 1205822
(119909119885⊲ 119890) minus 120582 (119909
119885⊲ 119909) ⊲ 119890
minus 120582119909119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
= 1205822
119909119885minus 2120582119909
119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
(15)
Therefore (14) becomes
1205822
119909119885minus 2120582119909
119885⊲ 119909 + 119909
119885⊲ 119909
2
= 1205822
119909119885minus 2120582119909
119885⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
(16)
which after simplification reduces to the required equation119909119885⊲ 119909
2
= (119909119885⊲ 119909) ⊲ 119909
(2) Since 1199092 = (119909119869+ 119909
119885)2
= 1199092
119869+ 119909
119885⊲ 119909
119869 we have
1199092
⊲ 119909 = (1199092
119869+ 119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909 + (119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909
119869+ (119909
119885⊲ 119909) ⊲ 119909
= 1199092
119869⊲ 119909
119869+ 119909
119885⊲ 119909
2
(17)
by the part (1) But 1199092119869⊲ 119909
119869= 119909
119869⊲ 119909
2
119869since 119909
119869is in the Jor-
dan algebra 119869 Therefore 1199092 ⊲ 119909 = 1199092
119869⊲ 119909
119869+ 119909
119885⊲ 119909
2
= 119909119869⊲
1199092
119869+ 119909
119885⊲ 119909
2
= 119909 ⊲ 1199092
Remark 23 In any quasi-Jordan algebra if an element 119909
satisfies 1199093
= 1199092
⊲ 119909 = 119909 ⊲ 1199092 then 119909
119899
⊲ 1199092
= 119909119899+2 for
all positive integers 119899 For this suppose 119909 satisfies 1199092 ⊲ 119909 =
119909 ⊲ 1199092 and 119909
119898
⊲ 1199092
= 119909119898+2 for any fixed119898 ge 1 Then 119909
119898+1
⊲
1199092
= (119909119898
⊲ 119909) ⊲ 1199092
= (119909119898
⊲ 1199092
) ⊲ 119909 (by the right Jor-dan identity) = 119909
119898+2
⊲ 119909 = 119909119898+3
Proposition 24 Let I be a unital quasi-Jordan Banachalgebra with unit 119890 and let 119909 isin I satisfy (119890 minus 119909) ⊲ (119890 minus 119909)
2
=
(119890 minus 119909)2
⊲ (119890 minus 119909) Then 119909 isin 119866119890(I) whenever 119890 minus 119909 lt 1
8 Abstract and Applied Analysis
Proof First note that 119890minus119909 lt 1 gives (119890 minus 119909)119899
le 119890 minus 119909119899
lt
1 for all 119899 = 1 2 3 Hence the infinite geometric series119890 +sum
infin
119899=1(119890 minus119909)
119899 converges absolutely to some element 119910 isin IWe show that the geometric series sum 119910 is an inverse of 119909with respect to the unit 119890 For any fixed positive integer 119899 let119910119899= 119890+sum
119899
119896=1(119890minus119909)
119896Then the sequence 119910119899 of partial sums
converges to 119910 By setting 119908 = 119890 minus 119909 we get
119910119899⊲ 119909 = (119890 +
119899
sum
119896=1
119908119896
) ⊲ (119890 minus 119908)
= 119890 +
119899
sum
119896=1
119908119896
minus (119890 ⊲ 119908 +
119899+1
sum
119896=2
119908119896
)
= 119890 + 119908 minus 119890 ⊲ 119908 minus 119908119899+1
= 119890 + (119890 minus 119909) minus 119890 ⊲ (119890 minus 119909) minus (119890 minus 119909)119899+1
= 119890 + (119890 minus 119909 minus 119890 + 119890 ⊲ 119909) minus (119890 minus 119909)119899+1
= 119890 + (119890 ⊲ 119909 minus 119909) minus (119890 minus 119909)119899+1
(18)
Thus by allowing 119899 rarr infin we obtain 119909 ⊲ 119910 = 119890 + (119890 ⊲
119909 minus 119909) = 119890 + 119890⊲(119909) since 119890 minus 119909 lt 1
Next by Remark 23 we have
119910119899⊲ 119909
2
= 119910119899⊲ (119890 minus 119908)
2
= 119910119899⊲ (119890 minus 119890 ⊲ 119908 minus 119908 ⊲ 119890 + 119908
2
)
= 119910119899⊲ (119890 minus 2119908 + 119908
2
)
= 119910119899⊲ 119890 minus 2119910
119899⊲ 119908 + 119910
119899⊲ 119908
2
= 119910119899minus 2119910
119899⊲ 119908 + 119910
119899⊲ 119908
2
= (119890 +
119899
sum
119896=1
119908119896
) minus 2(119890 +
119899
sum
119896=1
119908119896
) ⊲ 119908
+ (119890 +
119899
sum
119896=1
119908119896
) ⊲ 1199082
= (119890 +
119899
sum
119896=1
119908119896
) minus 2(119890 ⊲ 119908 +
119899+1
sum
119896=2
119908119896
)
+ (119890 ⊲ 1199082
+
119899+2
sum
119896=3
119908119896
)
= 119890 + 119908 minus 2119890 ⊲ 119908 minus 1199082
+ 119890 ⊲ 1199082
minus 119908119899+1
+ 119908119899+2
(19)
Taking the limit as 119899 rarr infin we get
119910 ⊲ 1199092
= 119890 + 119908 minus 2119890 ⊲ 119908 minus 1199082
+ 119890 ⊲ 1199082
= 119890 + (119890 minus 119909) minus 2119890 ⊲ (119890 minus 119909) minus (119890 minus 119909)2
+ 119890 ⊲ (119890 minus 119909)2
= 2119890 minus 119909 minus 2119890 + 2119890 ⊲ 119909 minus 119890 + 119890 ⊲ 119909 + 119909 minus 1199092
+ 119890 minus 2119890 ⊲ 119909 + 119890 ⊲ 1199092
= 119909 + (119890 ⊲ 119909 minus 119909) + (119890 ⊲ 1199092
minus 1199092
)
(20)
Proposition 25 Let I be a unital split quasi-Jordan Banachalgebra with unit 119890 If 119909 isin I with (119890 minus 119909) ⊲ (119890 minus 119909)
2
=
(119890 minus 119909)2
⊲ (119890 minus 119909) then |120582| le 119909 for all 120582 isin 120590(I119890)(119909)
Proof If 120582 isin 120590(I119890)(119909) with 120582 = 0 then the noninvertibility of
120582119890 minus 119909 means the noninvertibility of 119890 minus (1120582)119909 with respectto the unit 119890 However by Proposition 24 119890 minus (1120582)119909must beinvertible with respect to the unit 119890 whenever (1|120582|)119909 lt 1It follows that |120582| le 119909 for all 120582 isin 120590
(I119890)(119909)
5 Unbounded and Nonclosed Spectrum
In this section we show that the spectrum of an element ina split quasi-Jordan Banach algebra may be neither boundednor closed and hence not compactThe following result givesa couple of characterizations of the unbounded spectrum ofan element in a split quasi-Jordan Banach algebra
Proposition 26 Let 119909 be an element of a unital split quasi-Jordan Banach algebra I = 119869 oplus 119885(I) with a unit 119890 isin 119869 Thenthe following statements are equivalent
(1) 120590(I119890)(119909) =C
(2) 1199092 ⊲ 119909 = 119909 ⊲ 1199092
(3) |120582| le 119909 for all 120582 isin 120590(I119890)(119909)
Proof (1rArr2) See Proposition 22(2rArr3) Suppose 1199092 ⊲ 119909 = 119909 ⊲ 119909
2 Then
(119890 minus 119909)2
⊲ (119890 minus 119909) = (119890 minus 119890 ⊲ 119909 minus 119909 + 1199092
) ⊲ (119890 minus 119909)
= 119890 minus 2119890 ⊲ 119909 minus 119909 + 21199092
+ (119890 ⊲ 119909) ⊲ 119909 minus 1199092
⊲ 119909
= 119890 minus 2119909119869minus 119909 + 2119909
2
+ 119890 ⊲ 1199092
minus 1199093
(119890 minus 119909) ⊲ (119890 minus 119909)2
= (119890 minus 119909) ⊲ (119890 minus 2119909 + 1199092
)
= 119890 minus 2 (119890 ⊲ 119909) + 119890 ⊲ 1199092
minus 119909
+ 21199092
minus 119909 ⊲ 1199092
= 119890 minus 2119909119869+ 119890 ⊲ 119909
2
minus 119909 + 21199092
minus 1199093
(21)
Abstract and Applied Analysis 9
From (21) we get
(119890 minus 119909)2
⊲ (119890 minus 119909) = (119890 minus 119909) ⊲ (119890 minus 119909)2
(22)
Hence |120582| le 119909 for all 120582 isin 120590(I119890)(119909) by Proposition 25
(3rArr1) Immediate
Remark 27 There do exist unital split quasi-Jordan algebrascontaining elements that have the spectrum with respect tothe unit of the Jordan part equal to the whole ofC and henceunbounded To justify this claim we proceed as follows
Let 119860 be a unital associative algebra and let 119872 be an 119860-bimodule Let 119891 119872 rarr 119860 be an 119860-bimodule map (ie anadditive map satisfying 119891(119886119909) = 119886119891(119909) and 119891(119909119886) = 119891(119909)119886for all 119886 isin 119860 119909 isin 119872) Then one can put a dialgebra structureon 119872 as follows 119909 ⊣ 119910 = 119909119891(119910) and 119909⊢119910 = 119891(119909)119910 (cf [13Example 22(d)]) Hence119872+ is a quasi-Jordan algebra underthe quasi-Jordan product ldquo⊲rdquo given by119909 ⊲ 119910 = (12)(119909119891(119910)+
119891(119910)119909) Further for any 119909 isin 119872 we observe that
1199092
⊲ 119909 =1
2(119909119891 (119909) + 119891 (119909) 119909) ⊲ 119909
=1
4((119909119891 (119909) + 119891 (119909) 119909) 119891 (119909)
+119891 (119909) (119909119891 (119909) + 119891 (119909) 119909))
=1
4(119909119891 (119909) 119891 (119909) + 2119891 (119909) 119909119891 (119909) + 119891 (119909) 119891 (119909) 119909)
119909 ⊲ 1199092
=1
2119909 ⊲ (119909119891 (119909) + 119891 (119909) 119909)
=1
4(119909119891 (119909119891 (119909) + 119891 (119909) 119909) + 119891 (119909119891 (119909) + 119891 (119909) 119909) 119909)
=1
4(2119909119891 (119909) 119891 (119909) + 2119891 (119909) 119891 (119909) 119909)
(23)
However the right hand sides of the above equations (23)may not be equal see the following example (Example 28)For such elements 119909 we have 119909
2
⊲ 119909 = 119909 ⊲ 1199092 Hence by
Proposition 26 the spectrum of 119909 is unbounded whenever119872
+ is a unital split quasi-Jordan Banach algebra
Example 28 Let 119872 be the collection of 2 times 2 matrices withentries from the fieldC and let119860 be the algebra of allmatricesof the form [
120572 0
0 120573] with 120572 120573 isin C Then it is easily seen
that 119872 is an 119860-bimodule Next we define 119891 119872 rarr 119860
by 119891 [11988611
11988612
11988621
11988622
] = [11988611
0
0 11988622
] Of course 119891 is an additive mapsatisfying
119891([120572 0
0 120573] [
11988611
11988612
11988621
11988622
]) = [120572 0
0 120573]119891([
11988611
11988612
11988621
11988622
])
119891([11988611
11988612
11988621
11988622
] [120572 0
0 120573]) = 119891([
11988611
11988612
11988621
11988622
]) [120572 0
0 120573]
(24)
Hence119891 is an119860-bimodule mapThus by Remark 27119872+ is aquasi-Jordan algebra with the quasi-Jordan product as below
[11988611
11988612
11988621
11988622
] ⊲ [11988711
11988712
11988721
11988722
] =1
2([
11988611
11988612
11988621
11988622
]119891([11988711
11988712
11988721
11988722
])
+119891([11988711
11988712
11988721
11988722
]) [11988611
11988612
11988621
11988622
])
=[[
[
1198861111988711
11988612
11988711
+ 11988722
2
11988621
11988711
+ 11988722
21198862211988722
]]
]
(25)
Indeed 119872+
= 119869 oplus 119885(119872+
) where 119869 = [119886 0
0 119887] 119886 119887 isin C
is a subalgebra of 119872+ and 119885(119872+
) = [0 119886
119887 0] 119886 119887 isin C Any
matrix of the form [1 119886
119887 1] with 119886 119887 isin C is a (right) unit in119872
+Thus119872+ is a unital split quasi-Jordan algebrawith thematrix119868 = [
1 0
0 1] as the unit of its Jordan part 119869
Further a natural norm is defined on119872+ as follows
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
=100381610038161003816100381611988611
1003816100381610038161003816 +100381610038161003816100381611988612
1003816100381610038161003816 +100381610038161003816100381611988621
1003816100381610038161003816 +100381610038161003816100381611988622
1003816100381610038161003816 (26)
This norm also satisfies10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
] ⊲ [11988711
11988712
11988721
11988722
]
10038171003817100381710038171003817100381710038171003817
=1
2(
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
] [11988711
0
0 11988722
]
+ [11988711
0
0 11988722
] [11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
)
le
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
10038171003817100381710038171003817100381710038171003817
[11988711
0
0 11988722
]
10038171003817100381710038171003817100381710038171003817
le
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
10038171003817100381710038171003817100381710038171003817
[11988711
11988712
11988721
11988722
]
10038171003817100381710038171003817100381710038171003817
(27)
Next for any 119909 = [11988611
11988612
11988621
11988622
] isin 119872+ we observe that
1199092
= [11988611
11988612
11988621
11988622
] ⊲ [11988611
11988612
11988621
11988622
]
=[[
[
1198862
1111988612
11988611
+ 11988622
2
11988621
11988611
+ 11988622
21198862
22
]]
]
(28)
so that
1199092
⊲ 119909 = [
[
1198863
11
11988612
4(119886
11+ 119886
22)2
11988621
4(119886
11+ 119886
22)2
1198863
22
]
]
119909 ⊲ 1199092
= [
[
1198863
11
11988612
2(119886
2
11+ 119886
2
22)
11988621
2(119886
2
11+ 119886
2
22) 119886
3
22
]
]
(29)
So for any 119909 = [11988611
11988612
11988621
11988622
] isin 119872+ 1199092 ⊲ 119909 = 119909 ⊲ 119909
2
hArr
11988612
= 11988621
= 0 or 11988611
= 11988622 Thus by Proposition 26
10 Abstract and Applied Analysis
120590(119872+119868)([
11988611
11988612
11988621
11988622
]) = C whenever 11988611
= 11988622
and 11988612 119886
21 = 0
In particular for 119909 = [1 1
0 2] isin 119872
+ we have
1199092
⊲ 119909 =[[
[
13
1
4(1 + 2)
2
0
4(1 + 2)
2
23
]]
]
= [1
9
4
0 8
]
119909 ⊲ 1199092
=[[
[
13
1
2(1
2
+ 22
)
0
2(1
2
+ 22
) 23
]]
]
= [1
5
2
0 8
]
(30)
Concerning the inequality between the right hand sidesof (23) in Remark 27 we observe for 119909 as above that
1
4(119909119891 (119909) 119891 (119909) + 2119891 (119909) 119909119891 (119909) + 119891 (119909) 119891 (119909) 119909)
=1
4([
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 1
0 2] [
1 0
0 2]
+ [1 0
0 2] [
1 0
0 2] [
1 1
0 2]) = [
19
4
0 8
]
(31)
but
1
4(2119909119891 (119909) 119891 (119909) + 2119891 (119909) 119891 (119909) 119909)
=1
4(2 [
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 0
0 2] [
1 1
0 2])
= [1
5
2
0 8
]
(32)
Hence 1199092 ⊲ 119909 = 119909 ⊲ 1199092 Thus 120590
(119872+119868)([
1 1
0 2]) = C by
Proposition 26Further suppose the matrix [
11988611198862
11988631198864
] isin 119872+ is invertible
with respect to the unit 119868 that is [ 1198861 119886211988631198864
] isin 119866119868(119872
+
) Thenthere exists [ 119887
11198872
11988731198874
] isin 119872+ such that
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] = 119868 + 119868⊲([
1198861
1198862
1198863
1198864
]) = [1 minus119886
2
minus1198863
1]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
= [1198861
1198862
1198863
1198864
] + 119868⊲([
1198861
1198862
1198863
1198864
])
+ 119868⊲([
1198861
1198862
1198863
1198864
]
2
)
=[[
[
1198861
minus1198862
1198861+ 119886
4
2
minus1198863
1198861+ 119886
4
21198864
]]
]
(33)
However
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] =[[
[
11988711198861
1198872
1198861+ 119886
4
2
1198873
1198861+ 119886
4
211988741198864
]]
]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
=
[[[
[
11988711198861
1198872
1198862
1+ 119886
2
4
2
1198873
1198862
1+ 119886
2
4
211988741198862
4
]]]
]
(34)
It follows that 1198861
= 0 1198864
= 0 1198871= 1119886
1 119887
4= 1119886
4 119887
2((119886
1+
1198864)2) = minus119886
2 119887
3((119886
1+119886
4)2) = minus119886
3 119887
2((119886
2
1+119886
2
4)2) = minus119886
2((119886
1+
1198864)2) and 119887
3((119886
2
1+ 119886
2
4)2) = minus119886
3((119886
1+ 119886
4)2) From these
equations we get 1198872(119886
1minus 119886
4)2
= 0 and 1198873(119886
1minus 119886
4)2
= 0 so that(119887
2minus119887
3)(119886
1minus 119886
4)2
= 0Then for 1198861
= 1198864 we obtain 119887
2= 119887
3= 0
and hence 1198862= 119886
3= 0 Therefore
119866119868(119872
+
) = [119886 119887
119888 119886] 119909 = [
120572 0
0 120573] isin 119872
+
119886 = 0
120572 = 0 120573 = 0 120572 = 120573
(35)
The set 119866119868(119872
+
) is not open clearly [1 1
0 1] isin 119866
119868(119872
+
) forany 120598 gt 0
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[1 1
0 1] minus [
[
1 minus120598
41
0 1 +120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[
[
120598
40
0 minus120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=120598
2lt 120598
(36)
but [ 1minus1205984 1
0 1+1205984] notin 119866
119868(119872
+
)Now if 119860 = [
minus1 1
0 1] then 120582119868 minus 119860 = [
120582+1 minus1
0 120582minus1] notin 119866
119868(119872
+
)and so 120582 isin 120590
(119872+119868)(119860) for all 120582 isin C Thus 120590
(119872+119868)(119860) = C an
unbounded spectrum
Next we observe that the spectrum of an element withrespect to a unit is closed whenever the corresponding set ofinvertibles is open
Proposition 29 LetI be a quasi-Jordan normed algebra witha unit 119890 such that119866
119890(I) is openThen 120590
(I119890)(119909) is closed for all119909 isin I
Proof Define 119891 C rarr I by 119891(120582) = 120582119890 minus 119909 Since 119891 iscontinuous the inverse image of the open set 119866
119890(I) is open
in C and so its complement 120590(I119890)(119909) is closed
We conclude this paper with the following example of anonclosed spectrum
Example 30 Let 119872 and 119872+ be as in Example 28 Let 119864 =
[1 1
0 1] and 119860 = [
119886 0
0 119887] with 119886 = 119887 both different from 1 Then
119864 is a unit in 119872+ We show that 120590
(119872+119864)
(119860) = C 1 Forthis let us first investigate when can an element of the form119861 = [
120572 120573
0 120574] be invertible Assuming that 119861 is invertible we get
the existence of an element 119862 isin 119863+ such that
119862 ⊲ 119861 = 119864 + 119864⊲(119861)
119862 ⊲ 1198612
= 119861 + 119864⊲(119861) + 119864
⊲(119861
2
)
(37)
Abstract and Applied Analysis 11
From these equations we get 120572 = 0 120574 = 0 and
119862 =[[
[
1
120572120573∘
01
120574
]]
]
(38)
where 120573∘satisfies the following two equations
120573∘(120572 + 120574
2) = 1 minus 120573 +
120572 + 120574
2
120573∘(1205722
+ 1205742
2) =
120572 + 120574
2+
1205722
+ 1205742
2minus 120573
120572 + 120574
2
(39)
Multiplying the last equation by 2(120572 + 120574)(1205722
+ 1205742
) and thenusing the other equation we get
2 (1 minus 120573) + (120572 + 120574) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
+ (120572 + 120574) (40)
or equivalently
2 (1 minus 120573) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
(41)
this equation is satisfied for 120573 = 1 or 120572 = 120574 Hence thematrix119861 is invertible if and only if 119861 = [
120572 1
0 120574] or 119861 = [
120572 120573
0 120572
] for 120572 120574 isin
C 0 and 120573 isin CWe conclude that 119860
120582= 120582119864 minus 119860 = [
120582minus119886 120582
0 120582minus119887] is invertible
with respect to 119864 if and only if 120582 minus 119887 = 0 120582 minus 119886 = 0 and 120582 = 1that is119860
120582is invertible if 119886 = 120582 and 119887 = 120582 and 120582 = 1 Hence119860
120582
is invertible only if 120582 = 1 as we assumed that 119886 = 119887 and bothare not 1 So for all 120582 = 1 119860
120582notin 119866
119864(119872
+
) Thus 120590(119872+119864)
(119860) =
C 1 which is neither bounded nor closed
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This project was supported by King Saud University Dean-ship of Scientific Research College of Science ResearchCenter
References
[1] B Fritzsche ldquoSophus Lie a sketch of his life and workrdquo Journalof Lie Theory vol 9 no 1 pp 1ndash38 1999
[2] P Jordan ldquoUber eine klasse nichtassoziativer hyperkomplexeralgebrenrdquo Nachrichten von der Gesellschaft der Wissenschaftenzu Gottingen pp 569ndash575 1932
[3] K McCrimmon A Taste of Jordan Algebras UniversitextSpringer New York NY USA 2004
[4] N Jacobson Structure and Representations of Jordan Algebrasvol 39 of American Mathematical Society Colloquium Publica-tions American Mathematical Society Providence RI USA1968
[5] F Gursey and C-H Tze On the Role of Division Jordan andRelatedAlgebras in Particle PhysicsWorld Scientific River EdgeNJ USA 1996
[6] R Iordanescu Jordan Structures in Analysis Geometry andPhysics EdituraAcademiei Romane Bucharest Romania 2009
[7] H Upmeier Symmetric Banach Manifolds and Jordan Clowast-Alge-bras vol 104 of North-Holland Mathematics Studies North-Holland Amsterdam The Netherlands 1985
[8] I L Kantor ldquoClassification of irreducible transitive differentialgroupsrdquo Doklady Akademii Nauk SSSR vol 158 pp 1271ndash12741964
[9] M Koecher ldquoUber eine Gruppe von rationalen AbbildungenrdquoInventiones Mathematicae vol 3 pp 136ndash171 1967
[10] J Tits ldquoUne classe drsquoalgebres de Lie en relation avec les algebresde Jordanrdquo Indagationes Mathematicae vol 24 pp 530ndash5351962
[11] J-L Loday ldquoUne version non commutative des algebres de Lieles algebres de Leibnizrdquo LrsquoEnseignement Mathematique vol 39no 3-4 pp 269ndash293 1993
[12] J-L Loday and T Pirashvili ldquoUniversal enveloping algebras ofLeibniz algebras and (co)homologyrdquo Mathematische Annalenvol 296 no 1 pp 139ndash158 1993
[13] J-L Loday ldquoDialgebrasrdquo in Dialgebras and Related Operadsvol 1763 of Lecture Notes in Math pp 7ndash66 Springer BerlinGermany 2001
[14] R Velasquez and R Felipe ldquoQuasi-Jordan algebrasrdquo Communi-cations in Algebra vol 36 no 4 pp 1580ndash1602 2008
[15] R Velasquez and R Felipe ldquoOn K-B quasi-Jordan algebrasand their relation with Liebniz algebrasrdquo Comunicaciones delCIMAT No 1-10-1015-12-2010
[16] R Felipe ldquoAn analogue to functional analysis in dialgebrasrdquoInternational Mathematical Forum vol 2 no 21ndash24 pp 1069ndash1091 2007
[17] R Velasquez and R Felipe ldquoSplit dialgebras split quasi-Jordanalgebras and regular elementsrdquo Journal of Algebra and ItsApplications vol 8 no 2 pp 191ndash218 2009
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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8 Abstract and Applied Analysis
Proof First note that 119890minus119909 lt 1 gives (119890 minus 119909)119899
le 119890 minus 119909119899
lt
1 for all 119899 = 1 2 3 Hence the infinite geometric series119890 +sum
infin
119899=1(119890 minus119909)
119899 converges absolutely to some element 119910 isin IWe show that the geometric series sum 119910 is an inverse of 119909with respect to the unit 119890 For any fixed positive integer 119899 let119910119899= 119890+sum
119899
119896=1(119890minus119909)
119896Then the sequence 119910119899 of partial sums
converges to 119910 By setting 119908 = 119890 minus 119909 we get
119910119899⊲ 119909 = (119890 +
119899
sum
119896=1
119908119896
) ⊲ (119890 minus 119908)
= 119890 +
119899
sum
119896=1
119908119896
minus (119890 ⊲ 119908 +
119899+1
sum
119896=2
119908119896
)
= 119890 + 119908 minus 119890 ⊲ 119908 minus 119908119899+1
= 119890 + (119890 minus 119909) minus 119890 ⊲ (119890 minus 119909) minus (119890 minus 119909)119899+1
= 119890 + (119890 minus 119909 minus 119890 + 119890 ⊲ 119909) minus (119890 minus 119909)119899+1
= 119890 + (119890 ⊲ 119909 minus 119909) minus (119890 minus 119909)119899+1
(18)
Thus by allowing 119899 rarr infin we obtain 119909 ⊲ 119910 = 119890 + (119890 ⊲
119909 minus 119909) = 119890 + 119890⊲(119909) since 119890 minus 119909 lt 1
Next by Remark 23 we have
119910119899⊲ 119909
2
= 119910119899⊲ (119890 minus 119908)
2
= 119910119899⊲ (119890 minus 119890 ⊲ 119908 minus 119908 ⊲ 119890 + 119908
2
)
= 119910119899⊲ (119890 minus 2119908 + 119908
2
)
= 119910119899⊲ 119890 minus 2119910
119899⊲ 119908 + 119910
119899⊲ 119908
2
= 119910119899minus 2119910
119899⊲ 119908 + 119910
119899⊲ 119908
2
= (119890 +
119899
sum
119896=1
119908119896
) minus 2(119890 +
119899
sum
119896=1
119908119896
) ⊲ 119908
+ (119890 +
119899
sum
119896=1
119908119896
) ⊲ 1199082
= (119890 +
119899
sum
119896=1
119908119896
) minus 2(119890 ⊲ 119908 +
119899+1
sum
119896=2
119908119896
)
+ (119890 ⊲ 1199082
+
119899+2
sum
119896=3
119908119896
)
= 119890 + 119908 minus 2119890 ⊲ 119908 minus 1199082
+ 119890 ⊲ 1199082
minus 119908119899+1
+ 119908119899+2
(19)
Taking the limit as 119899 rarr infin we get
119910 ⊲ 1199092
= 119890 + 119908 minus 2119890 ⊲ 119908 minus 1199082
+ 119890 ⊲ 1199082
= 119890 + (119890 minus 119909) minus 2119890 ⊲ (119890 minus 119909) minus (119890 minus 119909)2
+ 119890 ⊲ (119890 minus 119909)2
= 2119890 minus 119909 minus 2119890 + 2119890 ⊲ 119909 minus 119890 + 119890 ⊲ 119909 + 119909 minus 1199092
+ 119890 minus 2119890 ⊲ 119909 + 119890 ⊲ 1199092
= 119909 + (119890 ⊲ 119909 minus 119909) + (119890 ⊲ 1199092
minus 1199092
)
(20)
Proposition 25 Let I be a unital split quasi-Jordan Banachalgebra with unit 119890 If 119909 isin I with (119890 minus 119909) ⊲ (119890 minus 119909)
2
=
(119890 minus 119909)2
⊲ (119890 minus 119909) then |120582| le 119909 for all 120582 isin 120590(I119890)(119909)
Proof If 120582 isin 120590(I119890)(119909) with 120582 = 0 then the noninvertibility of
120582119890 minus 119909 means the noninvertibility of 119890 minus (1120582)119909 with respectto the unit 119890 However by Proposition 24 119890 minus (1120582)119909must beinvertible with respect to the unit 119890 whenever (1|120582|)119909 lt 1It follows that |120582| le 119909 for all 120582 isin 120590
(I119890)(119909)
5 Unbounded and Nonclosed Spectrum
In this section we show that the spectrum of an element ina split quasi-Jordan Banach algebra may be neither boundednor closed and hence not compactThe following result givesa couple of characterizations of the unbounded spectrum ofan element in a split quasi-Jordan Banach algebra
Proposition 26 Let 119909 be an element of a unital split quasi-Jordan Banach algebra I = 119869 oplus 119885(I) with a unit 119890 isin 119869 Thenthe following statements are equivalent
(1) 120590(I119890)(119909) =C
(2) 1199092 ⊲ 119909 = 119909 ⊲ 1199092
(3) |120582| le 119909 for all 120582 isin 120590(I119890)(119909)
Proof (1rArr2) See Proposition 22(2rArr3) Suppose 1199092 ⊲ 119909 = 119909 ⊲ 119909
2 Then
(119890 minus 119909)2
⊲ (119890 minus 119909) = (119890 minus 119890 ⊲ 119909 minus 119909 + 1199092
) ⊲ (119890 minus 119909)
= 119890 minus 2119890 ⊲ 119909 minus 119909 + 21199092
+ (119890 ⊲ 119909) ⊲ 119909 minus 1199092
⊲ 119909
= 119890 minus 2119909119869minus 119909 + 2119909
2
+ 119890 ⊲ 1199092
minus 1199093
(119890 minus 119909) ⊲ (119890 minus 119909)2
= (119890 minus 119909) ⊲ (119890 minus 2119909 + 1199092
)
= 119890 minus 2 (119890 ⊲ 119909) + 119890 ⊲ 1199092
minus 119909
+ 21199092
minus 119909 ⊲ 1199092
= 119890 minus 2119909119869+ 119890 ⊲ 119909
2
minus 119909 + 21199092
minus 1199093
(21)
Abstract and Applied Analysis 9
From (21) we get
(119890 minus 119909)2
⊲ (119890 minus 119909) = (119890 minus 119909) ⊲ (119890 minus 119909)2
(22)
Hence |120582| le 119909 for all 120582 isin 120590(I119890)(119909) by Proposition 25
(3rArr1) Immediate
Remark 27 There do exist unital split quasi-Jordan algebrascontaining elements that have the spectrum with respect tothe unit of the Jordan part equal to the whole ofC and henceunbounded To justify this claim we proceed as follows
Let 119860 be a unital associative algebra and let 119872 be an 119860-bimodule Let 119891 119872 rarr 119860 be an 119860-bimodule map (ie anadditive map satisfying 119891(119886119909) = 119886119891(119909) and 119891(119909119886) = 119891(119909)119886for all 119886 isin 119860 119909 isin 119872) Then one can put a dialgebra structureon 119872 as follows 119909 ⊣ 119910 = 119909119891(119910) and 119909⊢119910 = 119891(119909)119910 (cf [13Example 22(d)]) Hence119872+ is a quasi-Jordan algebra underthe quasi-Jordan product ldquo⊲rdquo given by119909 ⊲ 119910 = (12)(119909119891(119910)+
119891(119910)119909) Further for any 119909 isin 119872 we observe that
1199092
⊲ 119909 =1
2(119909119891 (119909) + 119891 (119909) 119909) ⊲ 119909
=1
4((119909119891 (119909) + 119891 (119909) 119909) 119891 (119909)
+119891 (119909) (119909119891 (119909) + 119891 (119909) 119909))
=1
4(119909119891 (119909) 119891 (119909) + 2119891 (119909) 119909119891 (119909) + 119891 (119909) 119891 (119909) 119909)
119909 ⊲ 1199092
=1
2119909 ⊲ (119909119891 (119909) + 119891 (119909) 119909)
=1
4(119909119891 (119909119891 (119909) + 119891 (119909) 119909) + 119891 (119909119891 (119909) + 119891 (119909) 119909) 119909)
=1
4(2119909119891 (119909) 119891 (119909) + 2119891 (119909) 119891 (119909) 119909)
(23)
However the right hand sides of the above equations (23)may not be equal see the following example (Example 28)For such elements 119909 we have 119909
2
⊲ 119909 = 119909 ⊲ 1199092 Hence by
Proposition 26 the spectrum of 119909 is unbounded whenever119872
+ is a unital split quasi-Jordan Banach algebra
Example 28 Let 119872 be the collection of 2 times 2 matrices withentries from the fieldC and let119860 be the algebra of allmatricesof the form [
120572 0
0 120573] with 120572 120573 isin C Then it is easily seen
that 119872 is an 119860-bimodule Next we define 119891 119872 rarr 119860
by 119891 [11988611
11988612
11988621
11988622
] = [11988611
0
0 11988622
] Of course 119891 is an additive mapsatisfying
119891([120572 0
0 120573] [
11988611
11988612
11988621
11988622
]) = [120572 0
0 120573]119891([
11988611
11988612
11988621
11988622
])
119891([11988611
11988612
11988621
11988622
] [120572 0
0 120573]) = 119891([
11988611
11988612
11988621
11988622
]) [120572 0
0 120573]
(24)
Hence119891 is an119860-bimodule mapThus by Remark 27119872+ is aquasi-Jordan algebra with the quasi-Jordan product as below
[11988611
11988612
11988621
11988622
] ⊲ [11988711
11988712
11988721
11988722
] =1
2([
11988611
11988612
11988621
11988622
]119891([11988711
11988712
11988721
11988722
])
+119891([11988711
11988712
11988721
11988722
]) [11988611
11988612
11988621
11988622
])
=[[
[
1198861111988711
11988612
11988711
+ 11988722
2
11988621
11988711
+ 11988722
21198862211988722
]]
]
(25)
Indeed 119872+
= 119869 oplus 119885(119872+
) where 119869 = [119886 0
0 119887] 119886 119887 isin C
is a subalgebra of 119872+ and 119885(119872+
) = [0 119886
119887 0] 119886 119887 isin C Any
matrix of the form [1 119886
119887 1] with 119886 119887 isin C is a (right) unit in119872
+Thus119872+ is a unital split quasi-Jordan algebrawith thematrix119868 = [
1 0
0 1] as the unit of its Jordan part 119869
Further a natural norm is defined on119872+ as follows
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
=100381610038161003816100381611988611
1003816100381610038161003816 +100381610038161003816100381611988612
1003816100381610038161003816 +100381610038161003816100381611988621
1003816100381610038161003816 +100381610038161003816100381611988622
1003816100381610038161003816 (26)
This norm also satisfies10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
] ⊲ [11988711
11988712
11988721
11988722
]
10038171003817100381710038171003817100381710038171003817
=1
2(
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
] [11988711
0
0 11988722
]
+ [11988711
0
0 11988722
] [11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
)
le
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
10038171003817100381710038171003817100381710038171003817
[11988711
0
0 11988722
]
10038171003817100381710038171003817100381710038171003817
le
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
10038171003817100381710038171003817100381710038171003817
[11988711
11988712
11988721
11988722
]
10038171003817100381710038171003817100381710038171003817
(27)
Next for any 119909 = [11988611
11988612
11988621
11988622
] isin 119872+ we observe that
1199092
= [11988611
11988612
11988621
11988622
] ⊲ [11988611
11988612
11988621
11988622
]
=[[
[
1198862
1111988612
11988611
+ 11988622
2
11988621
11988611
+ 11988622
21198862
22
]]
]
(28)
so that
1199092
⊲ 119909 = [
[
1198863
11
11988612
4(119886
11+ 119886
22)2
11988621
4(119886
11+ 119886
22)2
1198863
22
]
]
119909 ⊲ 1199092
= [
[
1198863
11
11988612
2(119886
2
11+ 119886
2
22)
11988621
2(119886
2
11+ 119886
2
22) 119886
3
22
]
]
(29)
So for any 119909 = [11988611
11988612
11988621
11988622
] isin 119872+ 1199092 ⊲ 119909 = 119909 ⊲ 119909
2
hArr
11988612
= 11988621
= 0 or 11988611
= 11988622 Thus by Proposition 26
10 Abstract and Applied Analysis
120590(119872+119868)([
11988611
11988612
11988621
11988622
]) = C whenever 11988611
= 11988622
and 11988612 119886
21 = 0
In particular for 119909 = [1 1
0 2] isin 119872
+ we have
1199092
⊲ 119909 =[[
[
13
1
4(1 + 2)
2
0
4(1 + 2)
2
23
]]
]
= [1
9
4
0 8
]
119909 ⊲ 1199092
=[[
[
13
1
2(1
2
+ 22
)
0
2(1
2
+ 22
) 23
]]
]
= [1
5
2
0 8
]
(30)
Concerning the inequality between the right hand sidesof (23) in Remark 27 we observe for 119909 as above that
1
4(119909119891 (119909) 119891 (119909) + 2119891 (119909) 119909119891 (119909) + 119891 (119909) 119891 (119909) 119909)
=1
4([
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 1
0 2] [
1 0
0 2]
+ [1 0
0 2] [
1 0
0 2] [
1 1
0 2]) = [
19
4
0 8
]
(31)
but
1
4(2119909119891 (119909) 119891 (119909) + 2119891 (119909) 119891 (119909) 119909)
=1
4(2 [
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 0
0 2] [
1 1
0 2])
= [1
5
2
0 8
]
(32)
Hence 1199092 ⊲ 119909 = 119909 ⊲ 1199092 Thus 120590
(119872+119868)([
1 1
0 2]) = C by
Proposition 26Further suppose the matrix [
11988611198862
11988631198864
] isin 119872+ is invertible
with respect to the unit 119868 that is [ 1198861 119886211988631198864
] isin 119866119868(119872
+
) Thenthere exists [ 119887
11198872
11988731198874
] isin 119872+ such that
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] = 119868 + 119868⊲([
1198861
1198862
1198863
1198864
]) = [1 minus119886
2
minus1198863
1]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
= [1198861
1198862
1198863
1198864
] + 119868⊲([
1198861
1198862
1198863
1198864
])
+ 119868⊲([
1198861
1198862
1198863
1198864
]
2
)
=[[
[
1198861
minus1198862
1198861+ 119886
4
2
minus1198863
1198861+ 119886
4
21198864
]]
]
(33)
However
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] =[[
[
11988711198861
1198872
1198861+ 119886
4
2
1198873
1198861+ 119886
4
211988741198864
]]
]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
=
[[[
[
11988711198861
1198872
1198862
1+ 119886
2
4
2
1198873
1198862
1+ 119886
2
4
211988741198862
4
]]]
]
(34)
It follows that 1198861
= 0 1198864
= 0 1198871= 1119886
1 119887
4= 1119886
4 119887
2((119886
1+
1198864)2) = minus119886
2 119887
3((119886
1+119886
4)2) = minus119886
3 119887
2((119886
2
1+119886
2
4)2) = minus119886
2((119886
1+
1198864)2) and 119887
3((119886
2
1+ 119886
2
4)2) = minus119886
3((119886
1+ 119886
4)2) From these
equations we get 1198872(119886
1minus 119886
4)2
= 0 and 1198873(119886
1minus 119886
4)2
= 0 so that(119887
2minus119887
3)(119886
1minus 119886
4)2
= 0Then for 1198861
= 1198864 we obtain 119887
2= 119887
3= 0
and hence 1198862= 119886
3= 0 Therefore
119866119868(119872
+
) = [119886 119887
119888 119886] 119909 = [
120572 0
0 120573] isin 119872
+
119886 = 0
120572 = 0 120573 = 0 120572 = 120573
(35)
The set 119866119868(119872
+
) is not open clearly [1 1
0 1] isin 119866
119868(119872
+
) forany 120598 gt 0
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[1 1
0 1] minus [
[
1 minus120598
41
0 1 +120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[
[
120598
40
0 minus120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=120598
2lt 120598
(36)
but [ 1minus1205984 1
0 1+1205984] notin 119866
119868(119872
+
)Now if 119860 = [
minus1 1
0 1] then 120582119868 minus 119860 = [
120582+1 minus1
0 120582minus1] notin 119866
119868(119872
+
)and so 120582 isin 120590
(119872+119868)(119860) for all 120582 isin C Thus 120590
(119872+119868)(119860) = C an
unbounded spectrum
Next we observe that the spectrum of an element withrespect to a unit is closed whenever the corresponding set ofinvertibles is open
Proposition 29 LetI be a quasi-Jordan normed algebra witha unit 119890 such that119866
119890(I) is openThen 120590
(I119890)(119909) is closed for all119909 isin I
Proof Define 119891 C rarr I by 119891(120582) = 120582119890 minus 119909 Since 119891 iscontinuous the inverse image of the open set 119866
119890(I) is open
in C and so its complement 120590(I119890)(119909) is closed
We conclude this paper with the following example of anonclosed spectrum
Example 30 Let 119872 and 119872+ be as in Example 28 Let 119864 =
[1 1
0 1] and 119860 = [
119886 0
0 119887] with 119886 = 119887 both different from 1 Then
119864 is a unit in 119872+ We show that 120590
(119872+119864)
(119860) = C 1 Forthis let us first investigate when can an element of the form119861 = [
120572 120573
0 120574] be invertible Assuming that 119861 is invertible we get
the existence of an element 119862 isin 119863+ such that
119862 ⊲ 119861 = 119864 + 119864⊲(119861)
119862 ⊲ 1198612
= 119861 + 119864⊲(119861) + 119864
⊲(119861
2
)
(37)
Abstract and Applied Analysis 11
From these equations we get 120572 = 0 120574 = 0 and
119862 =[[
[
1
120572120573∘
01
120574
]]
]
(38)
where 120573∘satisfies the following two equations
120573∘(120572 + 120574
2) = 1 minus 120573 +
120572 + 120574
2
120573∘(1205722
+ 1205742
2) =
120572 + 120574
2+
1205722
+ 1205742
2minus 120573
120572 + 120574
2
(39)
Multiplying the last equation by 2(120572 + 120574)(1205722
+ 1205742
) and thenusing the other equation we get
2 (1 minus 120573) + (120572 + 120574) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
+ (120572 + 120574) (40)
or equivalently
2 (1 minus 120573) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
(41)
this equation is satisfied for 120573 = 1 or 120572 = 120574 Hence thematrix119861 is invertible if and only if 119861 = [
120572 1
0 120574] or 119861 = [
120572 120573
0 120572
] for 120572 120574 isin
C 0 and 120573 isin CWe conclude that 119860
120582= 120582119864 minus 119860 = [
120582minus119886 120582
0 120582minus119887] is invertible
with respect to 119864 if and only if 120582 minus 119887 = 0 120582 minus 119886 = 0 and 120582 = 1that is119860
120582is invertible if 119886 = 120582 and 119887 = 120582 and 120582 = 1 Hence119860
120582
is invertible only if 120582 = 1 as we assumed that 119886 = 119887 and bothare not 1 So for all 120582 = 1 119860
120582notin 119866
119864(119872
+
) Thus 120590(119872+119864)
(119860) =
C 1 which is neither bounded nor closed
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This project was supported by King Saud University Dean-ship of Scientific Research College of Science ResearchCenter
References
[1] B Fritzsche ldquoSophus Lie a sketch of his life and workrdquo Journalof Lie Theory vol 9 no 1 pp 1ndash38 1999
[2] P Jordan ldquoUber eine klasse nichtassoziativer hyperkomplexeralgebrenrdquo Nachrichten von der Gesellschaft der Wissenschaftenzu Gottingen pp 569ndash575 1932
[3] K McCrimmon A Taste of Jordan Algebras UniversitextSpringer New York NY USA 2004
[4] N Jacobson Structure and Representations of Jordan Algebrasvol 39 of American Mathematical Society Colloquium Publica-tions American Mathematical Society Providence RI USA1968
[5] F Gursey and C-H Tze On the Role of Division Jordan andRelatedAlgebras in Particle PhysicsWorld Scientific River EdgeNJ USA 1996
[6] R Iordanescu Jordan Structures in Analysis Geometry andPhysics EdituraAcademiei Romane Bucharest Romania 2009
[7] H Upmeier Symmetric Banach Manifolds and Jordan Clowast-Alge-bras vol 104 of North-Holland Mathematics Studies North-Holland Amsterdam The Netherlands 1985
[8] I L Kantor ldquoClassification of irreducible transitive differentialgroupsrdquo Doklady Akademii Nauk SSSR vol 158 pp 1271ndash12741964
[9] M Koecher ldquoUber eine Gruppe von rationalen AbbildungenrdquoInventiones Mathematicae vol 3 pp 136ndash171 1967
[10] J Tits ldquoUne classe drsquoalgebres de Lie en relation avec les algebresde Jordanrdquo Indagationes Mathematicae vol 24 pp 530ndash5351962
[11] J-L Loday ldquoUne version non commutative des algebres de Lieles algebres de Leibnizrdquo LrsquoEnseignement Mathematique vol 39no 3-4 pp 269ndash293 1993
[12] J-L Loday and T Pirashvili ldquoUniversal enveloping algebras ofLeibniz algebras and (co)homologyrdquo Mathematische Annalenvol 296 no 1 pp 139ndash158 1993
[13] J-L Loday ldquoDialgebrasrdquo in Dialgebras and Related Operadsvol 1763 of Lecture Notes in Math pp 7ndash66 Springer BerlinGermany 2001
[14] R Velasquez and R Felipe ldquoQuasi-Jordan algebrasrdquo Communi-cations in Algebra vol 36 no 4 pp 1580ndash1602 2008
[15] R Velasquez and R Felipe ldquoOn K-B quasi-Jordan algebrasand their relation with Liebniz algebrasrdquo Comunicaciones delCIMAT No 1-10-1015-12-2010
[16] R Felipe ldquoAn analogue to functional analysis in dialgebrasrdquoInternational Mathematical Forum vol 2 no 21ndash24 pp 1069ndash1091 2007
[17] R Velasquez and R Felipe ldquoSplit dialgebras split quasi-Jordanalgebras and regular elementsrdquo Journal of Algebra and ItsApplications vol 8 no 2 pp 191ndash218 2009
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 9
From (21) we get
(119890 minus 119909)2
⊲ (119890 minus 119909) = (119890 minus 119909) ⊲ (119890 minus 119909)2
(22)
Hence |120582| le 119909 for all 120582 isin 120590(I119890)(119909) by Proposition 25
(3rArr1) Immediate
Remark 27 There do exist unital split quasi-Jordan algebrascontaining elements that have the spectrum with respect tothe unit of the Jordan part equal to the whole ofC and henceunbounded To justify this claim we proceed as follows
Let 119860 be a unital associative algebra and let 119872 be an 119860-bimodule Let 119891 119872 rarr 119860 be an 119860-bimodule map (ie anadditive map satisfying 119891(119886119909) = 119886119891(119909) and 119891(119909119886) = 119891(119909)119886for all 119886 isin 119860 119909 isin 119872) Then one can put a dialgebra structureon 119872 as follows 119909 ⊣ 119910 = 119909119891(119910) and 119909⊢119910 = 119891(119909)119910 (cf [13Example 22(d)]) Hence119872+ is a quasi-Jordan algebra underthe quasi-Jordan product ldquo⊲rdquo given by119909 ⊲ 119910 = (12)(119909119891(119910)+
119891(119910)119909) Further for any 119909 isin 119872 we observe that
1199092
⊲ 119909 =1
2(119909119891 (119909) + 119891 (119909) 119909) ⊲ 119909
=1
4((119909119891 (119909) + 119891 (119909) 119909) 119891 (119909)
+119891 (119909) (119909119891 (119909) + 119891 (119909) 119909))
=1
4(119909119891 (119909) 119891 (119909) + 2119891 (119909) 119909119891 (119909) + 119891 (119909) 119891 (119909) 119909)
119909 ⊲ 1199092
=1
2119909 ⊲ (119909119891 (119909) + 119891 (119909) 119909)
=1
4(119909119891 (119909119891 (119909) + 119891 (119909) 119909) + 119891 (119909119891 (119909) + 119891 (119909) 119909) 119909)
=1
4(2119909119891 (119909) 119891 (119909) + 2119891 (119909) 119891 (119909) 119909)
(23)
However the right hand sides of the above equations (23)may not be equal see the following example (Example 28)For such elements 119909 we have 119909
2
⊲ 119909 = 119909 ⊲ 1199092 Hence by
Proposition 26 the spectrum of 119909 is unbounded whenever119872
+ is a unital split quasi-Jordan Banach algebra
Example 28 Let 119872 be the collection of 2 times 2 matrices withentries from the fieldC and let119860 be the algebra of allmatricesof the form [
120572 0
0 120573] with 120572 120573 isin C Then it is easily seen
that 119872 is an 119860-bimodule Next we define 119891 119872 rarr 119860
by 119891 [11988611
11988612
11988621
11988622
] = [11988611
0
0 11988622
] Of course 119891 is an additive mapsatisfying
119891([120572 0
0 120573] [
11988611
11988612
11988621
11988622
]) = [120572 0
0 120573]119891([
11988611
11988612
11988621
11988622
])
119891([11988611
11988612
11988621
11988622
] [120572 0
0 120573]) = 119891([
11988611
11988612
11988621
11988622
]) [120572 0
0 120573]
(24)
Hence119891 is an119860-bimodule mapThus by Remark 27119872+ is aquasi-Jordan algebra with the quasi-Jordan product as below
[11988611
11988612
11988621
11988622
] ⊲ [11988711
11988712
11988721
11988722
] =1
2([
11988611
11988612
11988621
11988622
]119891([11988711
11988712
11988721
11988722
])
+119891([11988711
11988712
11988721
11988722
]) [11988611
11988612
11988621
11988622
])
=[[
[
1198861111988711
11988612
11988711
+ 11988722
2
11988621
11988711
+ 11988722
21198862211988722
]]
]
(25)
Indeed 119872+
= 119869 oplus 119885(119872+
) where 119869 = [119886 0
0 119887] 119886 119887 isin C
is a subalgebra of 119872+ and 119885(119872+
) = [0 119886
119887 0] 119886 119887 isin C Any
matrix of the form [1 119886
119887 1] with 119886 119887 isin C is a (right) unit in119872
+Thus119872+ is a unital split quasi-Jordan algebrawith thematrix119868 = [
1 0
0 1] as the unit of its Jordan part 119869
Further a natural norm is defined on119872+ as follows
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
=100381610038161003816100381611988611
1003816100381610038161003816 +100381610038161003816100381611988612
1003816100381610038161003816 +100381610038161003816100381611988621
1003816100381610038161003816 +100381610038161003816100381611988622
1003816100381610038161003816 (26)
This norm also satisfies10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
] ⊲ [11988711
11988712
11988721
11988722
]
10038171003817100381710038171003817100381710038171003817
=1
2(
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
] [11988711
0
0 11988722
]
+ [11988711
0
0 11988722
] [11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
)
le
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
10038171003817100381710038171003817100381710038171003817
[11988711
0
0 11988722
]
10038171003817100381710038171003817100381710038171003817
le
10038171003817100381710038171003817100381710038171003817
[11988611
11988612
11988621
11988622
]
10038171003817100381710038171003817100381710038171003817
10038171003817100381710038171003817100381710038171003817
[11988711
11988712
11988721
11988722
]
10038171003817100381710038171003817100381710038171003817
(27)
Next for any 119909 = [11988611
11988612
11988621
11988622
] isin 119872+ we observe that
1199092
= [11988611
11988612
11988621
11988622
] ⊲ [11988611
11988612
11988621
11988622
]
=[[
[
1198862
1111988612
11988611
+ 11988622
2
11988621
11988611
+ 11988622
21198862
22
]]
]
(28)
so that
1199092
⊲ 119909 = [
[
1198863
11
11988612
4(119886
11+ 119886
22)2
11988621
4(119886
11+ 119886
22)2
1198863
22
]
]
119909 ⊲ 1199092
= [
[
1198863
11
11988612
2(119886
2
11+ 119886
2
22)
11988621
2(119886
2
11+ 119886
2
22) 119886
3
22
]
]
(29)
So for any 119909 = [11988611
11988612
11988621
11988622
] isin 119872+ 1199092 ⊲ 119909 = 119909 ⊲ 119909
2
hArr
11988612
= 11988621
= 0 or 11988611
= 11988622 Thus by Proposition 26
10 Abstract and Applied Analysis
120590(119872+119868)([
11988611
11988612
11988621
11988622
]) = C whenever 11988611
= 11988622
and 11988612 119886
21 = 0
In particular for 119909 = [1 1
0 2] isin 119872
+ we have
1199092
⊲ 119909 =[[
[
13
1
4(1 + 2)
2
0
4(1 + 2)
2
23
]]
]
= [1
9
4
0 8
]
119909 ⊲ 1199092
=[[
[
13
1
2(1
2
+ 22
)
0
2(1
2
+ 22
) 23
]]
]
= [1
5
2
0 8
]
(30)
Concerning the inequality between the right hand sidesof (23) in Remark 27 we observe for 119909 as above that
1
4(119909119891 (119909) 119891 (119909) + 2119891 (119909) 119909119891 (119909) + 119891 (119909) 119891 (119909) 119909)
=1
4([
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 1
0 2] [
1 0
0 2]
+ [1 0
0 2] [
1 0
0 2] [
1 1
0 2]) = [
19
4
0 8
]
(31)
but
1
4(2119909119891 (119909) 119891 (119909) + 2119891 (119909) 119891 (119909) 119909)
=1
4(2 [
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 0
0 2] [
1 1
0 2])
= [1
5
2
0 8
]
(32)
Hence 1199092 ⊲ 119909 = 119909 ⊲ 1199092 Thus 120590
(119872+119868)([
1 1
0 2]) = C by
Proposition 26Further suppose the matrix [
11988611198862
11988631198864
] isin 119872+ is invertible
with respect to the unit 119868 that is [ 1198861 119886211988631198864
] isin 119866119868(119872
+
) Thenthere exists [ 119887
11198872
11988731198874
] isin 119872+ such that
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] = 119868 + 119868⊲([
1198861
1198862
1198863
1198864
]) = [1 minus119886
2
minus1198863
1]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
= [1198861
1198862
1198863
1198864
] + 119868⊲([
1198861
1198862
1198863
1198864
])
+ 119868⊲([
1198861
1198862
1198863
1198864
]
2
)
=[[
[
1198861
minus1198862
1198861+ 119886
4
2
minus1198863
1198861+ 119886
4
21198864
]]
]
(33)
However
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] =[[
[
11988711198861
1198872
1198861+ 119886
4
2
1198873
1198861+ 119886
4
211988741198864
]]
]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
=
[[[
[
11988711198861
1198872
1198862
1+ 119886
2
4
2
1198873
1198862
1+ 119886
2
4
211988741198862
4
]]]
]
(34)
It follows that 1198861
= 0 1198864
= 0 1198871= 1119886
1 119887
4= 1119886
4 119887
2((119886
1+
1198864)2) = minus119886
2 119887
3((119886
1+119886
4)2) = minus119886
3 119887
2((119886
2
1+119886
2
4)2) = minus119886
2((119886
1+
1198864)2) and 119887
3((119886
2
1+ 119886
2
4)2) = minus119886
3((119886
1+ 119886
4)2) From these
equations we get 1198872(119886
1minus 119886
4)2
= 0 and 1198873(119886
1minus 119886
4)2
= 0 so that(119887
2minus119887
3)(119886
1minus 119886
4)2
= 0Then for 1198861
= 1198864 we obtain 119887
2= 119887
3= 0
and hence 1198862= 119886
3= 0 Therefore
119866119868(119872
+
) = [119886 119887
119888 119886] 119909 = [
120572 0
0 120573] isin 119872
+
119886 = 0
120572 = 0 120573 = 0 120572 = 120573
(35)
The set 119866119868(119872
+
) is not open clearly [1 1
0 1] isin 119866
119868(119872
+
) forany 120598 gt 0
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[1 1
0 1] minus [
[
1 minus120598
41
0 1 +120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[
[
120598
40
0 minus120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=120598
2lt 120598
(36)
but [ 1minus1205984 1
0 1+1205984] notin 119866
119868(119872
+
)Now if 119860 = [
minus1 1
0 1] then 120582119868 minus 119860 = [
120582+1 minus1
0 120582minus1] notin 119866
119868(119872
+
)and so 120582 isin 120590
(119872+119868)(119860) for all 120582 isin C Thus 120590
(119872+119868)(119860) = C an
unbounded spectrum
Next we observe that the spectrum of an element withrespect to a unit is closed whenever the corresponding set ofinvertibles is open
Proposition 29 LetI be a quasi-Jordan normed algebra witha unit 119890 such that119866
119890(I) is openThen 120590
(I119890)(119909) is closed for all119909 isin I
Proof Define 119891 C rarr I by 119891(120582) = 120582119890 minus 119909 Since 119891 iscontinuous the inverse image of the open set 119866
119890(I) is open
in C and so its complement 120590(I119890)(119909) is closed
We conclude this paper with the following example of anonclosed spectrum
Example 30 Let 119872 and 119872+ be as in Example 28 Let 119864 =
[1 1
0 1] and 119860 = [
119886 0
0 119887] with 119886 = 119887 both different from 1 Then
119864 is a unit in 119872+ We show that 120590
(119872+119864)
(119860) = C 1 Forthis let us first investigate when can an element of the form119861 = [
120572 120573
0 120574] be invertible Assuming that 119861 is invertible we get
the existence of an element 119862 isin 119863+ such that
119862 ⊲ 119861 = 119864 + 119864⊲(119861)
119862 ⊲ 1198612
= 119861 + 119864⊲(119861) + 119864
⊲(119861
2
)
(37)
Abstract and Applied Analysis 11
From these equations we get 120572 = 0 120574 = 0 and
119862 =[[
[
1
120572120573∘
01
120574
]]
]
(38)
where 120573∘satisfies the following two equations
120573∘(120572 + 120574
2) = 1 minus 120573 +
120572 + 120574
2
120573∘(1205722
+ 1205742
2) =
120572 + 120574
2+
1205722
+ 1205742
2minus 120573
120572 + 120574
2
(39)
Multiplying the last equation by 2(120572 + 120574)(1205722
+ 1205742
) and thenusing the other equation we get
2 (1 minus 120573) + (120572 + 120574) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
+ (120572 + 120574) (40)
or equivalently
2 (1 minus 120573) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
(41)
this equation is satisfied for 120573 = 1 or 120572 = 120574 Hence thematrix119861 is invertible if and only if 119861 = [
120572 1
0 120574] or 119861 = [
120572 120573
0 120572
] for 120572 120574 isin
C 0 and 120573 isin CWe conclude that 119860
120582= 120582119864 minus 119860 = [
120582minus119886 120582
0 120582minus119887] is invertible
with respect to 119864 if and only if 120582 minus 119887 = 0 120582 minus 119886 = 0 and 120582 = 1that is119860
120582is invertible if 119886 = 120582 and 119887 = 120582 and 120582 = 1 Hence119860
120582
is invertible only if 120582 = 1 as we assumed that 119886 = 119887 and bothare not 1 So for all 120582 = 1 119860
120582notin 119866
119864(119872
+
) Thus 120590(119872+119864)
(119860) =
C 1 which is neither bounded nor closed
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This project was supported by King Saud University Dean-ship of Scientific Research College of Science ResearchCenter
References
[1] B Fritzsche ldquoSophus Lie a sketch of his life and workrdquo Journalof Lie Theory vol 9 no 1 pp 1ndash38 1999
[2] P Jordan ldquoUber eine klasse nichtassoziativer hyperkomplexeralgebrenrdquo Nachrichten von der Gesellschaft der Wissenschaftenzu Gottingen pp 569ndash575 1932
[3] K McCrimmon A Taste of Jordan Algebras UniversitextSpringer New York NY USA 2004
[4] N Jacobson Structure and Representations of Jordan Algebrasvol 39 of American Mathematical Society Colloquium Publica-tions American Mathematical Society Providence RI USA1968
[5] F Gursey and C-H Tze On the Role of Division Jordan andRelatedAlgebras in Particle PhysicsWorld Scientific River EdgeNJ USA 1996
[6] R Iordanescu Jordan Structures in Analysis Geometry andPhysics EdituraAcademiei Romane Bucharest Romania 2009
[7] H Upmeier Symmetric Banach Manifolds and Jordan Clowast-Alge-bras vol 104 of North-Holland Mathematics Studies North-Holland Amsterdam The Netherlands 1985
[8] I L Kantor ldquoClassification of irreducible transitive differentialgroupsrdquo Doklady Akademii Nauk SSSR vol 158 pp 1271ndash12741964
[9] M Koecher ldquoUber eine Gruppe von rationalen AbbildungenrdquoInventiones Mathematicae vol 3 pp 136ndash171 1967
[10] J Tits ldquoUne classe drsquoalgebres de Lie en relation avec les algebresde Jordanrdquo Indagationes Mathematicae vol 24 pp 530ndash5351962
[11] J-L Loday ldquoUne version non commutative des algebres de Lieles algebres de Leibnizrdquo LrsquoEnseignement Mathematique vol 39no 3-4 pp 269ndash293 1993
[12] J-L Loday and T Pirashvili ldquoUniversal enveloping algebras ofLeibniz algebras and (co)homologyrdquo Mathematische Annalenvol 296 no 1 pp 139ndash158 1993
[13] J-L Loday ldquoDialgebrasrdquo in Dialgebras and Related Operadsvol 1763 of Lecture Notes in Math pp 7ndash66 Springer BerlinGermany 2001
[14] R Velasquez and R Felipe ldquoQuasi-Jordan algebrasrdquo Communi-cations in Algebra vol 36 no 4 pp 1580ndash1602 2008
[15] R Velasquez and R Felipe ldquoOn K-B quasi-Jordan algebrasand their relation with Liebniz algebrasrdquo Comunicaciones delCIMAT No 1-10-1015-12-2010
[16] R Felipe ldquoAn analogue to functional analysis in dialgebrasrdquoInternational Mathematical Forum vol 2 no 21ndash24 pp 1069ndash1091 2007
[17] R Velasquez and R Felipe ldquoSplit dialgebras split quasi-Jordanalgebras and regular elementsrdquo Journal of Algebra and ItsApplications vol 8 no 2 pp 191ndash218 2009
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
10 Abstract and Applied Analysis
120590(119872+119868)([
11988611
11988612
11988621
11988622
]) = C whenever 11988611
= 11988622
and 11988612 119886
21 = 0
In particular for 119909 = [1 1
0 2] isin 119872
+ we have
1199092
⊲ 119909 =[[
[
13
1
4(1 + 2)
2
0
4(1 + 2)
2
23
]]
]
= [1
9
4
0 8
]
119909 ⊲ 1199092
=[[
[
13
1
2(1
2
+ 22
)
0
2(1
2
+ 22
) 23
]]
]
= [1
5
2
0 8
]
(30)
Concerning the inequality between the right hand sidesof (23) in Remark 27 we observe for 119909 as above that
1
4(119909119891 (119909) 119891 (119909) + 2119891 (119909) 119909119891 (119909) + 119891 (119909) 119891 (119909) 119909)
=1
4([
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 1
0 2] [
1 0
0 2]
+ [1 0
0 2] [
1 0
0 2] [
1 1
0 2]) = [
19
4
0 8
]
(31)
but
1
4(2119909119891 (119909) 119891 (119909) + 2119891 (119909) 119891 (119909) 119909)
=1
4(2 [
1 1
0 2] [
1 0
0 2] [
1 0
0 2] + 2 [
1 0
0 2] [
1 0
0 2] [
1 1
0 2])
= [1
5
2
0 8
]
(32)
Hence 1199092 ⊲ 119909 = 119909 ⊲ 1199092 Thus 120590
(119872+119868)([
1 1
0 2]) = C by
Proposition 26Further suppose the matrix [
11988611198862
11988631198864
] isin 119872+ is invertible
with respect to the unit 119868 that is [ 1198861 119886211988631198864
] isin 119866119868(119872
+
) Thenthere exists [ 119887
11198872
11988731198874
] isin 119872+ such that
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] = 119868 + 119868⊲([
1198861
1198862
1198863
1198864
]) = [1 minus119886
2
minus1198863
1]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
= [1198861
1198862
1198863
1198864
] + 119868⊲([
1198861
1198862
1198863
1198864
])
+ 119868⊲([
1198861
1198862
1198863
1198864
]
2
)
=[[
[
1198861
minus1198862
1198861+ 119886
4
2
minus1198863
1198861+ 119886
4
21198864
]]
]
(33)
However
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
] =[[
[
11988711198861
1198872
1198861+ 119886
4
2
1198873
1198861+ 119886
4
211988741198864
]]
]
[1198871
1198872
1198873
1198874
] ⊲ [1198861
1198862
1198863
1198864
]
2
=
[[[
[
11988711198861
1198872
1198862
1+ 119886
2
4
2
1198873
1198862
1+ 119886
2
4
211988741198862
4
]]]
]
(34)
It follows that 1198861
= 0 1198864
= 0 1198871= 1119886
1 119887
4= 1119886
4 119887
2((119886
1+
1198864)2) = minus119886
2 119887
3((119886
1+119886
4)2) = minus119886
3 119887
2((119886
2
1+119886
2
4)2) = minus119886
2((119886
1+
1198864)2) and 119887
3((119886
2
1+ 119886
2
4)2) = minus119886
3((119886
1+ 119886
4)2) From these
equations we get 1198872(119886
1minus 119886
4)2
= 0 and 1198873(119886
1minus 119886
4)2
= 0 so that(119887
2minus119887
3)(119886
1minus 119886
4)2
= 0Then for 1198861
= 1198864 we obtain 119887
2= 119887
3= 0
and hence 1198862= 119886
3= 0 Therefore
119866119868(119872
+
) = [119886 119887
119888 119886] 119909 = [
120572 0
0 120573] isin 119872
+
119886 = 0
120572 = 0 120573 = 0 120572 = 120573
(35)
The set 119866119868(119872
+
) is not open clearly [1 1
0 1] isin 119866
119868(119872
+
) forany 120598 gt 0
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[1 1
0 1] minus [
[
1 minus120598
41
0 1 +120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
[
[
120598
40
0 minus120598
4
]
]
100381710038171003817100381710038171003817100381710038171003817100381710038171003817
=120598
2lt 120598
(36)
but [ 1minus1205984 1
0 1+1205984] notin 119866
119868(119872
+
)Now if 119860 = [
minus1 1
0 1] then 120582119868 minus 119860 = [
120582+1 minus1
0 120582minus1] notin 119866
119868(119872
+
)and so 120582 isin 120590
(119872+119868)(119860) for all 120582 isin C Thus 120590
(119872+119868)(119860) = C an
unbounded spectrum
Next we observe that the spectrum of an element withrespect to a unit is closed whenever the corresponding set ofinvertibles is open
Proposition 29 LetI be a quasi-Jordan normed algebra witha unit 119890 such that119866
119890(I) is openThen 120590
(I119890)(119909) is closed for all119909 isin I
Proof Define 119891 C rarr I by 119891(120582) = 120582119890 minus 119909 Since 119891 iscontinuous the inverse image of the open set 119866
119890(I) is open
in C and so its complement 120590(I119890)(119909) is closed
We conclude this paper with the following example of anonclosed spectrum
Example 30 Let 119872 and 119872+ be as in Example 28 Let 119864 =
[1 1
0 1] and 119860 = [
119886 0
0 119887] with 119886 = 119887 both different from 1 Then
119864 is a unit in 119872+ We show that 120590
(119872+119864)
(119860) = C 1 Forthis let us first investigate when can an element of the form119861 = [
120572 120573
0 120574] be invertible Assuming that 119861 is invertible we get
the existence of an element 119862 isin 119863+ such that
119862 ⊲ 119861 = 119864 + 119864⊲(119861)
119862 ⊲ 1198612
= 119861 + 119864⊲(119861) + 119864
⊲(119861
2
)
(37)
Abstract and Applied Analysis 11
From these equations we get 120572 = 0 120574 = 0 and
119862 =[[
[
1
120572120573∘
01
120574
]]
]
(38)
where 120573∘satisfies the following two equations
120573∘(120572 + 120574
2) = 1 minus 120573 +
120572 + 120574
2
120573∘(1205722
+ 1205742
2) =
120572 + 120574
2+
1205722
+ 1205742
2minus 120573
120572 + 120574
2
(39)
Multiplying the last equation by 2(120572 + 120574)(1205722
+ 1205742
) and thenusing the other equation we get
2 (1 minus 120573) + (120572 + 120574) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
+ (120572 + 120574) (40)
or equivalently
2 (1 minus 120573) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
(41)
this equation is satisfied for 120573 = 1 or 120572 = 120574 Hence thematrix119861 is invertible if and only if 119861 = [
120572 1
0 120574] or 119861 = [
120572 120573
0 120572
] for 120572 120574 isin
C 0 and 120573 isin CWe conclude that 119860
120582= 120582119864 minus 119860 = [
120582minus119886 120582
0 120582minus119887] is invertible
with respect to 119864 if and only if 120582 minus 119887 = 0 120582 minus 119886 = 0 and 120582 = 1that is119860
120582is invertible if 119886 = 120582 and 119887 = 120582 and 120582 = 1 Hence119860
120582
is invertible only if 120582 = 1 as we assumed that 119886 = 119887 and bothare not 1 So for all 120582 = 1 119860
120582notin 119866
119864(119872
+
) Thus 120590(119872+119864)
(119860) =
C 1 which is neither bounded nor closed
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This project was supported by King Saud University Dean-ship of Scientific Research College of Science ResearchCenter
References
[1] B Fritzsche ldquoSophus Lie a sketch of his life and workrdquo Journalof Lie Theory vol 9 no 1 pp 1ndash38 1999
[2] P Jordan ldquoUber eine klasse nichtassoziativer hyperkomplexeralgebrenrdquo Nachrichten von der Gesellschaft der Wissenschaftenzu Gottingen pp 569ndash575 1932
[3] K McCrimmon A Taste of Jordan Algebras UniversitextSpringer New York NY USA 2004
[4] N Jacobson Structure and Representations of Jordan Algebrasvol 39 of American Mathematical Society Colloquium Publica-tions American Mathematical Society Providence RI USA1968
[5] F Gursey and C-H Tze On the Role of Division Jordan andRelatedAlgebras in Particle PhysicsWorld Scientific River EdgeNJ USA 1996
[6] R Iordanescu Jordan Structures in Analysis Geometry andPhysics EdituraAcademiei Romane Bucharest Romania 2009
[7] H Upmeier Symmetric Banach Manifolds and Jordan Clowast-Alge-bras vol 104 of North-Holland Mathematics Studies North-Holland Amsterdam The Netherlands 1985
[8] I L Kantor ldquoClassification of irreducible transitive differentialgroupsrdquo Doklady Akademii Nauk SSSR vol 158 pp 1271ndash12741964
[9] M Koecher ldquoUber eine Gruppe von rationalen AbbildungenrdquoInventiones Mathematicae vol 3 pp 136ndash171 1967
[10] J Tits ldquoUne classe drsquoalgebres de Lie en relation avec les algebresde Jordanrdquo Indagationes Mathematicae vol 24 pp 530ndash5351962
[11] J-L Loday ldquoUne version non commutative des algebres de Lieles algebres de Leibnizrdquo LrsquoEnseignement Mathematique vol 39no 3-4 pp 269ndash293 1993
[12] J-L Loday and T Pirashvili ldquoUniversal enveloping algebras ofLeibniz algebras and (co)homologyrdquo Mathematische Annalenvol 296 no 1 pp 139ndash158 1993
[13] J-L Loday ldquoDialgebrasrdquo in Dialgebras and Related Operadsvol 1763 of Lecture Notes in Math pp 7ndash66 Springer BerlinGermany 2001
[14] R Velasquez and R Felipe ldquoQuasi-Jordan algebrasrdquo Communi-cations in Algebra vol 36 no 4 pp 1580ndash1602 2008
[15] R Velasquez and R Felipe ldquoOn K-B quasi-Jordan algebrasand their relation with Liebniz algebrasrdquo Comunicaciones delCIMAT No 1-10-1015-12-2010
[16] R Felipe ldquoAn analogue to functional analysis in dialgebrasrdquoInternational Mathematical Forum vol 2 no 21ndash24 pp 1069ndash1091 2007
[17] R Velasquez and R Felipe ldquoSplit dialgebras split quasi-Jordanalgebras and regular elementsrdquo Journal of Algebra and ItsApplications vol 8 no 2 pp 191ndash218 2009
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 11
From these equations we get 120572 = 0 120574 = 0 and
119862 =[[
[
1
120572120573∘
01
120574
]]
]
(38)
where 120573∘satisfies the following two equations
120573∘(120572 + 120574
2) = 1 minus 120573 +
120572 + 120574
2
120573∘(1205722
+ 1205742
2) =
120572 + 120574
2+
1205722
+ 1205742
2minus 120573
120572 + 120574
2
(39)
Multiplying the last equation by 2(120572 + 120574)(1205722
+ 1205742
) and thenusing the other equation we get
2 (1 minus 120573) + (120572 + 120574) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
+ (120572 + 120574) (40)
or equivalently
2 (1 minus 120573) = (1 minus 120573)(120572 + 120574)
2
1205722+ 120574
2
(41)
this equation is satisfied for 120573 = 1 or 120572 = 120574 Hence thematrix119861 is invertible if and only if 119861 = [
120572 1
0 120574] or 119861 = [
120572 120573
0 120572
] for 120572 120574 isin
C 0 and 120573 isin CWe conclude that 119860
120582= 120582119864 minus 119860 = [
120582minus119886 120582
0 120582minus119887] is invertible
with respect to 119864 if and only if 120582 minus 119887 = 0 120582 minus 119886 = 0 and 120582 = 1that is119860
120582is invertible if 119886 = 120582 and 119887 = 120582 and 120582 = 1 Hence119860
120582
is invertible only if 120582 = 1 as we assumed that 119886 = 119887 and bothare not 1 So for all 120582 = 1 119860
120582notin 119866
119864(119872
+
) Thus 120590(119872+119864)
(119860) =
C 1 which is neither bounded nor closed
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This project was supported by King Saud University Dean-ship of Scientific Research College of Science ResearchCenter
References
[1] B Fritzsche ldquoSophus Lie a sketch of his life and workrdquo Journalof Lie Theory vol 9 no 1 pp 1ndash38 1999
[2] P Jordan ldquoUber eine klasse nichtassoziativer hyperkomplexeralgebrenrdquo Nachrichten von der Gesellschaft der Wissenschaftenzu Gottingen pp 569ndash575 1932
[3] K McCrimmon A Taste of Jordan Algebras UniversitextSpringer New York NY USA 2004
[4] N Jacobson Structure and Representations of Jordan Algebrasvol 39 of American Mathematical Society Colloquium Publica-tions American Mathematical Society Providence RI USA1968
[5] F Gursey and C-H Tze On the Role of Division Jordan andRelatedAlgebras in Particle PhysicsWorld Scientific River EdgeNJ USA 1996
[6] R Iordanescu Jordan Structures in Analysis Geometry andPhysics EdituraAcademiei Romane Bucharest Romania 2009
[7] H Upmeier Symmetric Banach Manifolds and Jordan Clowast-Alge-bras vol 104 of North-Holland Mathematics Studies North-Holland Amsterdam The Netherlands 1985
[8] I L Kantor ldquoClassification of irreducible transitive differentialgroupsrdquo Doklady Akademii Nauk SSSR vol 158 pp 1271ndash12741964
[9] M Koecher ldquoUber eine Gruppe von rationalen AbbildungenrdquoInventiones Mathematicae vol 3 pp 136ndash171 1967
[10] J Tits ldquoUne classe drsquoalgebres de Lie en relation avec les algebresde Jordanrdquo Indagationes Mathematicae vol 24 pp 530ndash5351962
[11] J-L Loday ldquoUne version non commutative des algebres de Lieles algebres de Leibnizrdquo LrsquoEnseignement Mathematique vol 39no 3-4 pp 269ndash293 1993
[12] J-L Loday and T Pirashvili ldquoUniversal enveloping algebras ofLeibniz algebras and (co)homologyrdquo Mathematische Annalenvol 296 no 1 pp 139ndash158 1993
[13] J-L Loday ldquoDialgebrasrdquo in Dialgebras and Related Operadsvol 1763 of Lecture Notes in Math pp 7ndash66 Springer BerlinGermany 2001
[14] R Velasquez and R Felipe ldquoQuasi-Jordan algebrasrdquo Communi-cations in Algebra vol 36 no 4 pp 1580ndash1602 2008
[15] R Velasquez and R Felipe ldquoOn K-B quasi-Jordan algebrasand their relation with Liebniz algebrasrdquo Comunicaciones delCIMAT No 1-10-1015-12-2010
[16] R Felipe ldquoAn analogue to functional analysis in dialgebrasrdquoInternational Mathematical Forum vol 2 no 21ndash24 pp 1069ndash1091 2007
[17] R Velasquez and R Felipe ldquoSplit dialgebras split quasi-Jordanalgebras and regular elementsrdquo Journal of Algebra and ItsApplications vol 8 no 2 pp 191ndash218 2009
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of