Notes on Multi-Linear Algebra and Tensor Calculus.pdf

8/11/2019 Notes on Multi-Linear Algebra and Tensor Calculus.pdf

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Notes on Multi-Linear Algebra and Tensor Calculus.

by Valter Moretti

Department of Mathematics,Faculty of Science,

University of Trento

2002-2003

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1 Multi-linear Mappings and Tensors.

Within this section we introduce basic concepts concerning multi-linear algebra and tensors.The theory of vector spaces and linear mappings is assumed to be well known.

1.1 Dual space and conjugate space.

As a first step we introduce the dual space and the conjugate space of a given vector space.

Def.1.1. (Dual Space, Conjugate Dual Space and Conjugate space.) LetV be a vectorspace on the field eitherK = R orC.(a) The dual space of V, V, is the vector space of linear functionals on V, i.e., the linearmappingsf :V K.(b)IfK = C, theconjugate dual spaceofV,V, is the vector space of anti-linear functionalson V, i.e., the antilinear mappings g : V C. Finally theconjugate space of V, V is thespace(V)

Comments.(1) IfV and V are vector spaces on C, a mapping f : VV is said anti linearor conjugatelinearif it satisfies

f(u + v) =f(u) + f(v)

for all u, v V and , C, denoting the complex conjugate of C. IfV = C the givendefinition reduces to the definition ofanti-linear functional.(2) V, V and V turn out to be vector spaces on the field K when the composition rule ofvectors and the product with elements of the field are defined in the usual way. For instace, iff, g

V or V, and

K then f+ g and fare functions such that:

(f+ g)(u) :=f(u) + g(u)

and(f)(u) :=f(u)

for all ofuV.

Def.1.2. (Dual Basis, Conjugate Dual Basis and Conjugate Basis.) Let V be a finite-dimensional vector space on either K = R or C. Let{ei}iI be a vector basis of V. The set{ej}jI V whose elements are defined by

ej (ei) :=ji

for alli, jI is called thedual basis of{ei}iI.Similarly, ifK = C, the set of elements{ej}jI V defined by:

ej (ei) :=ji

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for alli, jI is called theconjugate dual basis of{ei}iI.Finally the set of elements{ej}jI V defined by:

ep(eq

) :=qq

for allp, qI is called theconjugate basisof{ei}iI.

Comments.(1)We have explicitly assumed that V is finite dimensional. Anyway, each vector space admitsa vector basis (i.e. a possibly infinite set of vectors such that each vector ofV can be obtainedas a finite linear combination of those vectors), therefore one could discuss on the validity of thegiven definitions in the general case of a non-finite dimensional vector space. We shall not followthat way because we are interested on algebraic features only and the infinite-dimensional caseshould be approached by convenient topological tools which are quite far form the goals of theseintroductory notes.(2) If the vector space V with field Kadmits the vector basis

{ei

}i

I, each linear or anti-linear

mapping f : V K is completely defined by giving the values f(ei) for all of i I. This isbecause, if v V then v =iIv ciei for some numbers ci K, Iv I being finite. Thenthe linearity off yelds f(v) =

iIv c

if(ei). (This fact holds true no matter if dimV


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dimV


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Let us prove that vF(v) with (F(v)) (u) :=u(v) is linear, injective and surjective.(Linearity.) We have to show that, for all , K, v , vV,

F(v+ v ) =F(v) + F(v) .

This is equivalent to, by the definition ofF given above,

u(v+ v ) =u(v) + u(v) ,

for all uV. This is obvious because, u is a linear functional on V.(Injectivity.) We have to show that F(v) = F(v) implies v = v . (F(v)) (u) = (F(v)) (u) canbe re-written as u(v) =u(v) or u(v v) = 0. In our hypotheses, this holds true for all uV.Then, definee1:= v v and notice that, ifv v= 0, one can complete e1 with other vectors toget a vector basis ofV,{ei}iI. Since u is arbitrary, choosing u = e1 we should have u(e1) = 1by definition of dual basis but this contradicts u(e1) = 0, i.e., u(v v) = 0. By consequencev v = 0 and v= v .(Surjectivity.) We have to show that iff(V), there is vf V such that F(vf) =f.Fix a basis {ei}iI inV and the dual one inV. vf :=f(ei)eifulfills the requirement F(vf) =f.

Def.1.3. (Pairing.) LetV be a finite-dimensional vector space onK = C orR with dual spaceV. The bi-linear map ., .: V V K such that

u, v:= v(u)

for alluV, vV, is calledpairing.

Comment. Because of the theorem proven above, we may indifferently thinku, vas represent-ing either the action ofu(V) on vV or the action ofvV on uV.

Notation 1.2. From now onVW

indicates that the vector spaces V and W are isomorphic under some natural isomorphism. Ifthe field ofV and W is C,

V Windicates that there is a natural anti-isomorphism, i.e. there is an injective, surjective, anti-linearmapping G: V

Wbuilt up using the definition ofV and V only.

Exercises 1.1.1.1.1. Show that ifvV thenv =v, ejej , where {ej}jIis any basis of the finite dimensionalvector space V.(Hint. Decompose v = ciei, apply e

k and use the uniqueness of the decomposition of a vector

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along a vector basis.)1.1.2. Show that V V if dimV


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for all(v1, . . . , vn)V1 . . . Vn andwW.

Proof. Let us consider the mappingFdefined above. It is obvious that f L(V1, . . . , V n, W)for allf

L(V1, . . . , V n

|W), we leave the simple proof (based on multi linearity offand linearity

of the pairing) to the reader.We want to show that F is linear, injective and surjective.(Linearity.) We have to prove that f+g =f+gfor all , K and f, g L(V1, . . . , V n|W).In fact, making use of the left-hand linearity of the pairing, one has

f+g (v1, . . . , vn, w) =(f+g)(v1, . . . , vn), w= f(v1, . . . , vn), w+g(v1, . . . , vn), w

and this is nothing but:

f+g (v1, . . . , vn, w) = (f+ g)(v1, . . . , vn, w) .

Since (v1

, . . . , vn, w

) is arbitrary, the thesis is proven.(Injectivity.) We have to show that if f = f then f=f

.In fact, if f(v1, . . . , vn, w

) = f(v1, . . . , vn, w) for all (v1, . . . , vn, w), using the definition ofg we havef(v1, . . . , vn), w=f(v1, . . . , vn), w, or, equivalently

f(v1, . . . , vn) f(v1, . . . , vn), w= 0 ,

for all (v1, . . . , vn) and w. Then define e1 := f(v1, . . . , vn)f(v1, . . . , vn), if e1= 0 we can

complete it to a basis ofW. Fixing w = e1 we should have

f(v1, . . . , vn) f(v1, . . . , vn), w= 1

which contradicts the statement above. Therefore f(v1

, . . . , vn

)

f(v1

, . . . , vn

) = 0 for all(v1, . . . , vn)V1 . . . Vn, in other words f =f.(Surjectivity.) We have to show that for each L(V1, . . . , V n, W) there is a f L(V1, . . . , V n|W)with f = .To this end, fix vector bases{er,ir}irIr Vr and a (dual) basis{ek}kI W. Then, takeL(V1, . . . , V n, W) and define the mapping f L(V1, . . . , V n|W) given by

f(v1, . . . , vn) :=vi11 vinn (e1,i1 , . . . , en,in , ek)ek.

By construction that mapping is multilinear and, using multilinearity and Exercise 1.2.1, wefind

f(v1, . . . , vn, w) =vi11 v

inn (e1,i1 , . . . , en,in , e

k

)ek, wheh

= (v1, . . . , vn, w) ,for all (v1, . . . , vn, w

) and this is equivalent to f = .

Theorem 1.3 allow us to restrict our study to the spaces of multi-linear functionalsL(V1, . . . , V k),since the spaces of multi-linear mapsare completely encompassed. Let us introduce the concept

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oftensor productof vector spaces. The following definitions can be extended to encompass thecase of non-finite dimensional vector spaces by introducing suitable topological notions (e.g.,Hilbert spaces). To introduce the tensor product we need a trivial but relevant preliminarydefinition.

Def.1.4. (Linear action of a vector space.) Let U and V two finite-dimensional vectorspace on the same fieldK= R orC. We say thatU linearly acts on V by means of F, ifF :U V is a natural isomorphism.

Examples.(1)IfV is a finite-dimensional vector space on K = Ror C, V linearly acts on V byF =idV .(2) IfV is that as in (1), V linearly acts on V by means of the natural isomorphism betweenV and (V).

In the following ifUlinearly acts on Vby means ofF, and u

U, v

V we write u(v) insteadof (F(u))(v) whenever F is unambiguously determined. For instance, in the second exampleabove,v(u) meansv, uwhen vV and uV.

Def.1.5. (Tensor product.) LetU1, . . . , U n be n1 finite-dimensional vector spaces on thecommon field K = R or C, suppose each Ui linearly acts on Vi (by means of some Fi), fori= 1, . . . , n.(1)if(u1, . . . , un)U1. . .Un,u1. . .un denotes the multi linear mapping inL(V1, . . . , V n)defined by

(u1 . . . un)(v1, . . . , vn) :=u1(v1) un(vn) ,for all(v1, . . . , vn)

V1

. . .

Vn. u1

. . .

un is calledtensor product of vectorsu1, . . . , un.

(2)The mapping: U1 . . . Un L(V1, . . . , V n) given by: : (u1, . . . , un)u1 . . . un,is saidtensor product map.(3) The vector subspace ofL(V1, . . . , V n) generated by all ofu1 . . . un for all(u1, . . . , un)U1 . . . Un is calledtensor product of spaces U1, . . . , U n and is indicated byU1 . . . Un.The vectors inU1 . . . Un are saidtensors.

Remarks.(1) U1 . . . Un is made of all the linear combinations of the form

Nj=1 j u1,j . . . un,j ,

wherej K, uk,j Uk and N= 1, 2, . . ..(2) It is trivially proven that the tensor product map:

(u1, . . . , un)u1 . . . un,is multi linear.(3) From the given definition, if V1, . . . , V n are given finite-dimensional vector spaces on thesame field K = R or C, it is clear what we mean by V1 . . . Vn or V1 . . . Vn but also, forinstance, V1 V2 V3 or V1 V2 V3 . One simply has to take the natural linear action ofV

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on V and V on V into account.

The natural question which arises is ifU1 . . . Un is a propersubspace ofL(V1, . . . , V n) or,conversely, recovers the whole space L(V1, . . . , V n). The following theorem gives an answer tothat question.

Theorem 1.4. Referring to Def.1.4, it holds:

U1 . . . Un= L(V1, . . . , V n) .

Proof. It is sufficient to show that if f L(V1, . . . , V n) then f U1 . . . Un. To this endfix bases{ek,i}iIk Vk for k = 1, . . . , n. f above is completely determined by coefficientsfi1,...,in :=f(e1,i1, . . . , en,in). Then define the tensor tf

U1

. . .

Un by defining:

tf :=fi1...inei11 . . . einn ,

where we have identified each Uk with the corresponding space V

k which linearly acts on Vk.Then, by Def.1.5, one can directly prove that, by multi linearity

tf(v1, . . . , vn) =vi11 . . . v

inn fi1...in =f(v1, . . . , vn) ,

for all of (v1, . . . , vn)V1 . . . Vn. This is nothing but tf =f.

Another relevant result is stated by the theorem below.

Theorem 1.5. Referring to Def.1.4, the following statements hold.(a) The dimension ofU1 . . . Un is:

dim(U1 . . . Un) =n

k=1

dimUk =n

k=1

dimVk.

(b) If{ek,ik}ikIk is a basis ofUk, k= 1, . . . , n, then :{e1,i1 . . . en,in}(i1,...,in)I1...In is avector basis ofU1 . . . Un.(c) IftU1. . .Un, the components oftwith respect to a basis{e1,i1. . .en,in}(i1,...,in)I1...Inare given by (identifying eachVi with the correspondingU

i):

ti1...in =t(ei11 , . . . , einn )

and thus it holdst= ti1...ine1,i1 . . . en,in.

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Proof. Notice that (b) trivially implies (a) because the elements e1,i1 . . . en,in are exactlynk=1 dimUk. (Also

nk=1 dimUk =

nk=1 dimVk because each Uk is isomorphic to the corre-

sponding Vk and dimVk = dimV

k by definition of dual basis.) So it is sufficient to show thatthe second statement holds true. To this end, since elements e1,i

1. . .

en,in are generators of

U1 . . . Un, it is sufficient to show that they are linearly independent. Consider the genericvanishing linear combination

Ci1...ine1,i1 . . . en,in = 0 ,We want to show that all of the coefficients Ci1...in vanish. We may identify each Vk with thecorrespondingUk by means of the natural isomorphism ofUand (U

)and thus we may considereach dual-basis vector ejrr Uj as an element ofVj . Then the action of that linear combinationof multi-linear functionals on the generic element (ej11 , . . . , e

jnn ) V1 Vn produces the

resultCj1...j1 = 0 .

Since we can arbitrarily fix the indices j1, . . . jn this proves the thesis.Concerning (c), by the uniqueness of the components of a vector with respect to a basis, it issufficient to show that, defining

t := ti1...ine1,i1 . . . en,in,where

ti1...in :=t(ei11 , . . . , einn ) ,

it holdst(v1, . . . , vn) =t(v1, . . . , vn) ,

for all (v1, . . . , vn)V1 . . . Vn . By multi linearity,t(v1, . . . , vn) =v1i1 vn int(ei11 , . . . , einn ) ,

By multi linearity, the left-hand side is nothing but t(v1, . . . , vn). This concludes the proof.

There are two important theorems which, together with the identification of V and (V),imply that all of the spaces which one may build up coherently using the symbols ,Vk, and ()are naturally isomorphic to spaces which are of the form V

()i1

. . . V()in . The rule to producespaces naturally isomorphic to a given initial space is that one has to (1) ignore parentheses,(2) assume that is distributive with respect to and (3) assume that is involutive (i.e.(X)

X). For instance, one has:

((V1 V2) (V3 V4))V1 V2 V3 V4.Let us state the theorems corresponding to the rules (2) and (1) respectively (the rule (3) beingnothing but Theorem 1.2).

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Theorem 1.6. isdistributive with respect to by means ofnatural isomorphisms. Inother words, if V1 . . . V n are finite dimensional vector spaces on the same fieldK = R or C, itholds:

(V1

. . .

Vn)

V1

. . .

Vn

.

Theorem 1.7. The tensor product is associative by means ofnatural isomorphisms. Inother words considering a space which is made of tensor products of tensor products of finite-dimensional vector spaces on the same fieldR orC, one may omit parenthesis everywhere ob-taining a space which is naturally isomorphic to the initial one. So, for instance

(V1 V2) V3V1 V2 V3,

andV1

(V2

V3

)

V1

V2

V3

.

Sketch of proof. The natural isomorphisms required in the two theorems above can be built upas follows. In the former case, consider a linear mappingF :Vn . . . Vn(V1 . . . Vn)which satisfies

F :v1 . . . vn[v1 . . . vn](V1 . . . Vn),where, for all u1 . . . unV1 . . . Vn,

[v1 . . . vn](u1 . . . un) :=u1, v1 . . . un, vn .

In the latter case, consider the first proposed example. The required natural isomorphism canbe build up as a linear mapping F1: V1 V2 V3(V1 V2) V3 such that

F1 : v1 v2 v3(v1 v2) v3,

for all v1 v2 v3V1 V2 V3.Using the linearity and the involved definitions the reader can simply prove that the mappingsF, F1 are surjective. Moreover, by the previously proven theorems

dim(V1 Vn)= dim(V1 Vn) =dim(V1 Vn )

and

dim((V1 V2) V3) =dim(V1 V2) dim(V3) =dim(V1 V2 V3) .As a consequence, F and F1 must be also injective and thus they are isomorphisms.

Comment. The important point is that the mappings F and F1 above have been given byspecifying their action on tensor products of elements (e.g, F(v1 . . . vn)) and not on linear

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combinationsof these tensor products of elements. Remind that, for instanceV1 . . . Vn isnot the set of products u1 . . . un but it is the set oflinear combinationsof those products.Hence, in order to completely define F and F1, one must require that F and F1 admit uniquelydetermined linear extensionson their initial domains in order to encompass the whole tensorspaces generated by linear combinations of simple tensor products. In other words one has tocomplete the given definition, in the former case, by adding the further requirement

F( u1 . . . un+ v1 . . . vn) =F(u1 . . . un) + F(v1 . . . vn) ,

and similarly in the latter case. Despite it could seem a trivial task it is not the case. Indeed it isworthwhile noticing that each product v1 . . . vn can be re-written using linear combinations,e.g., concerning F:

v1 . . . vn= [(v1+ u1) . . . vn] [u1 . . . vn] .

Now consider the identities, which has to hold as a consequence of the assumed linearity ofF,such as:F(v1 . . . vn) =F((v1+ u1) . . . vn) F(u1 . . . vn)

AboveF(v1 . . . vn), F(u1 . . . vn), F((v1+ u1) . . . vn) are independentlydefined aswe said at the beginning and there is no reason, in principle, for the validity of the constraint:

F(v1 . . . vn) =F((v1+ u1) . . . vn) F(u1 . . . vn) .

Similar problems arise concerning F1.The general problem which arises by the two considered cases can be stated as follows. Supposewe are given a tensor product of vector spaces V1 . . . Vn and we are interested in the possiblelinear extensionsof a mapping f on V1 . . . Vn, with values in some vector space W, when fis initially defined on simple products v1 . . . vnV1 . . . Vn only.Is there any general prescription on the specification of values f(v1. . .vn) which assuresthatf can be extended, uniquely, to a linear mapping fromV1 . . . Vn to W?An answer is given by the following very important universality theorem.

Theorem 1.8. (Universality Theorem.) Given n 1 finite-dimensional vector spacesU1, . . . , U n on the same fieldK = R orC, the following statements hold.(a) For each finite-dimensional vector spaceWand each multi-linear mappingf :U1. . .UnW, there is a unique linear mapping f : U1. . .Un W such that the diagram belowcommutes(in other words, f= f).

U1 . . . Un U1 . . . Un

W

f

f

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(b) ForU1, . . . , U n fixed as above, suppose there is another pair (T, UT) such that, for each Wand eachf withf :U1 . . . UnWmulti linear, there is a uniquefT :UT W such thatthe diagram below commute,

U1

. . .

Un

UT

W

T

f

fT

then there is a vector space natural isomorphism: U1 . . . UnUT withfT = f. Inother words, givenU1, . . . , U n as in the hypotheses, the pair(, U1 . . . Un) is determined upto vector spaces isomorphisms by the diagram property above.

Remark. The property of the pair (, U1 . . . Un) stated in the second part of the theoremis calleduniversality property.

Before proving the theorem, let us explain how it gives a precise answer to the stated question.The theorem says that a linear extensionof any function fwith values in W, initially definedon simple tensor products only, f(v1 . . . vn) with v1 . . . vnU1 . . . Un, doesexistonthe whole domain space U1 . . . Un and it is uniquelydetermined provided f(v1 . . . vn) =g(v1, . . . , vn) where g: U1 . . . UnW is some multi-linear function.Concerning the mappings F and F1 introduced above, we may profitably use the universalitytheorem to show that they are well-defined on the whole domain made of linear combinations ofsimple tensor products. In fact, considerF for example. We can define themulti-linearmappingG: V1 . . . Vn (V1 . . . Vn) such that

G(v1, . . . , vn) := [v1 . . . v1].(The reader should check on the multi linearity of that function.) Then the universality theoremwithUk = V

k and W = (V1 . . . Vn) assures the existence of a linear mappingF :V1 . . .

Vn (V1 . . . Vn) with the required properties because F :=G is such thatF(v1 . . . vn) = (G )(v1, . . . , vn) =G(v1, . . . , vn) = [v1 . . . v1].

A similar multi-linear mapping G1 can be found for F1:

G1 : (v1, v2, v3)(v1 v2) v3.F1:= G

1

can be used in the proof of Theorem 1.7.

Proof of Theorem 1.8. (a) Fix bases{ek,ik}ikIk Uk, k = 1, . . . , n and{ej}jI W.Because of the multi-linearity property, the specification of coefficients fji1...in K,ik Ik, jI,k = 1, . . . , n uniquely determines a multi-linear mapping f : U1 . . . Un W by definingf

ji1...in

:=f(e1,i1 , . . . , en,in), ej. Conversely, a multi-linear mapping f : U1 . . . Un W

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uniquely determines coefficients fji1...in Kby the same rule. Similarly, because of the linearityproperty, a linear mapping g : U1 . . . Un W is completely and uniquely determined bycoefficientsgji1...in :=g(ei1 . . .ein), ej. Therefore, iff :U1 . . .UnWis given and it is

determined by coefficients f

j

i1...in as above, defining f : U1 . . . UnWby the coefficients(f)ji1...in :=f

ji1...in

, we have

f(v1 . . . vn), ej= vi1 . . . vinfji1...in =f(v1, . . . , vn), ej,

for all (v1, . . . , vn)U1 . . . Un and jI. This means that f(v1 . . . vn) =f(v1, . . . , vn)for all (v1, . . . , vn)U1 . . . Un. In other words f = f. The uniqueness of the mappingf is obvious: suppose there is another mappingg with g = f then (f g) = 0.This means in particular that

(f g)(ei1 . . . ein), ej= 0 ,

for all ikIk, k = 1, . . . , nand jI. Since the coefficients above completely determine a map,the considered mapping must be the null mapping and thus: f= g.

(b)Suppose that there is a pair (T, UT) as specified in the hypotheses. Then, using the part(a) with f = T and W = UTwe have the diagram below ,where we have a former diagramcommutation relation:

T = T .U1 . . . Un U1 . . . Un

UT

T

T

On the other hand, using the analogous property of the pair (T, UT) with f = and W =U1 . . . Un we also have the commutative diagram

U1 . . . Un UT

U1 . . . Un

T

T

which involves a second diagram commutation relation:

T T = .

The two obtained relations imply(T T) T =T

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and(T T) = ,

In other words, ifRanT UT is the range of the mapping T and Ran U1 . . . Un is theanalogue for:

(T T)|RanT =I dRanTand

(T T)|Ran= I dRanThen consider the latter. Notice that Ran is not a subspace, but the subspace spanned byRan coincides with U1 . . . Un by definition of tensorial product. SinceT T is linear(because composition of linear maps) and defined on the whole spaceU1 . . . Un, we concludethat

T T= I dU1...Un.Concerning the former identity we could conclude analogously, i.e.,

TT =I dUT ,

provided the subspace spanned by RanT, Span(RanT), coincides with the whole space UT.Anyway this holds true by the uniqueness property offT for a fixed f (see hypotheses aboutthe pair (T, UT)). (If Span(RanT)= UT, then UT = Span(RanT) S, S={0} being someproper subspace ofUT withSSpan(RanT) ={0}. Then, a mappingfT could not be uniquelydetermined by the requirement fT T = f because such a relation would be preserved undermodifications offT on the subspace S).We conclude that the linear mapping := T : U1 . . . Un UT is a natural isomorphismwith inverse

T.

As far as the property fT = f is concerned we may deal with as follows. Since :U1 . . .Un UT is an isomorphism, fT T = f implies (fT )(1 T) = f, but1 T =T T =, therefore (fT ) = f. By the uniqueness of the mapping f whichsatisfies the same identity, it has to hold fT = f.

Exercises 1.3.1.3.1. Consider a finite-dimensional vector space V and its dual V. Show by the universalitytheorem that there is a natural isomorphism such that

V VV V .

(Hint. Consider the bilinear mappingf : V V V V with f : (v1, v2) v2 v1. Showthat f is injective and thus surjective because dim(V V) = dim(V V).)

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2 Tensor algebra. Abstract Index Notation.

2.1 Tensor algebra generated by a vector space.

LetVbe a finite-dimensional vector space on the field K = R or C. Equipped with this algebraicstructure, in principle, we may build up several different tensor spaces. Notice that we have toconsider also K, Kand Vas admissible tensor factors. (In the following we shall not interestedin conjugate spaces.) Obviously, we are interested in tensor products which are not identifiableby some natural isomorphism.First consider the dual K when we consider K as a vector space on the field K itself. K ismade of linear functionals from Kto K. Each c K has the form c(k) :=c k, for all k K,where c K is a fixed field element which completely determines c. The mappingcc is a(natural) vector space isomorphism. Therefore

K K.

Then we pass to consider products K . . . K K . . . K = L(K, . . . ,K). Each multi-linear mapping f L(K, . . . ,K) is completely determined by the number f(1, . . . , 1), sincef(k1, . . . , kn) =k1 knf(1, . . . , 1). One can trivially show that the mapping ff(1, . . . , 1) isa (natural) vector space isomorphism between K . . . K and K itself. Therefore

K . . . K K .

Also notice that the found isomorphism trivially satisfies c1 . . . cnc1 cn, and thus thetensor product mapping reduces to the ordinary product of the field.We pass to consider the product K V = L(K, V)L(K, V). Each multi-linear functionalf in L(K, V) is completely determined by the element ofV, f(1,

) :V

K, which maps each

vV in f(1, v). Once again, it is a trivial task to show thatff(1, ) is a (natural) vectorspace isomorphism between K V and V itself.

K VV .

Notice that the found isomorphism satisfies k v kv and thus the tensor product mappingreduces to the ordinary product of a field element and a vector.Concluding, only the spaces K, V, V and whatever products ofV and V may be significantlydifferent. This result leads us to the following definition.

Def.2.1. (Tensor Algebra generated byV.) LetVbe a finite-dimensional vector space with

fieldK

.(1)Thetensor algebraAK(V)generated byV with fieldKis the class whose elements are thevector spaces: K, V, V and all of tensor products of factorsV andV in whatever order andnumber.(2) The tensors ofK are said scalars, the tensors of V are saidcontravariant vectors, thetensors ofV are saidcovariant vectors, the tensors of spacesVn:= V . . . V, whereV

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appearsn1 times, are said contravariant tensors of ordern ortensors of order(n, 0),the tensors of spacesVn:= V . . . V, whereV appearsn1 times, are saidcovarianttensors of order n or tensors of order (0, n), the remaining tensors which belong to spacescontainingn factorsV andm factorsV are saidtensors of order (n, m) not depending on theorder of factors.

Remark. Obviously, pairs of tensors spaces made of the same number of factors V and V indifferent order, are naturally isomorphic (see exercise 1.3.1). However, for practical reasons itis convenient to consider these spaces as different spaces and use the identifications when and ifnecessary.

2.2 The abstract index notation and rules to handle tensors.

Let us introduce the abstract index notation. Consider a finite-dimensional vector spaceV

with field K = R or C. After specification of a basis{ei}iI Vand the corresponding basis inV, each tensor is completely determined by giving its components with respect to the inducedbasis in the corresponding tensor space in AK(V). We are interested in the transformation ruleof these components under change of the base in V. Suppose to fix another basis{ej}jI Vwith ei = A

jiej . The coefficients A

ji determine a matrix [A] := [A

ji] in the matrix group

GL(dimV,K), i.e., the group (see the next section) ofdimV dimVmatrices with coefficientsin K and non-vanishing determinant. First consider a contravariant vector t = tiei, passingto the other basis, we have t = tiei = t

j ej and thus tj ej = t

iAj

iej . This is equivalent to

(tj Ajiti)ej = 0 which impliestj =Ajit

i ,

because of the linear independence of vectors ej . Similarly, if we specify a set of componentsin K,{ti}iI for each basis{ei}iI V and these components, changing the basis to{ej}jI,transform as

tj =Ajiti ,

where the coefficients Aji are defined by

ei = Aj

iej ,

then a contravariant tensor t is defined. It is determined by t := tiei in each basis. The proof isself evident.

Concerning covariant vectors, a similar result holds. Indeed a covariant vector u V is com-pletely determined by the specification of a set of components {ui}iIfor each basis {ei}iI V(dual basis of{ei}iIabove) when these components, changing the basis to{ej}jI(dual baseof{ej}jI above), transform as

uj =B i

j ui,

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where the coefficients B rl are defined by

ei =B ij ej .

What is the relation between the matrix [A] = [Aji] and the matrix [B] := [B hk ]? The answer

is clear: it must be

ij =ej , ei= Alj B ikel, ek= Alj B ikkl =Akj B ik .In other words it has to hold I=AtB , which is equivalent to

[B] = [A]1t .

(Notice that t and1 commute.)Proper tensors have components which transform similarly to the vector case. For instance,consider t

V

V, fix a basis

{ei

}i

I

V, the dual one

{ei

}i

I

V and consider that

induced inV V,{ej ej}(i,j)II. Thent = tij ei ej . By bi linearity of the tensor productmap, if we pass to consider another basis{ei}iI Vand those associated in the relevant spacesas above, concerning the components tkl oft in the new tensor space basis, one trivially gets

tkl = Ak

iB jl t

ij ,

where the matrices [A] = [Aji] and [B] := [B hk ] are those considered above. It is clear that

the specification of a tensor ofV V is completely equivalent to the specification of a set ofcomponents for each basis ofV V,{ej ej}(i,j)II, provided these components transformas specified above under change of basis.We can generalize the obtained results after a definition.

Def. 2.2 (Canonical bases.) Let AK(V) be the tensor algebra generated by the finite-dimensional vector space V on the field K (= C or R). If B ={ei}iI is a basis in V withdual basisB ={ei}iI V, thecanonical bases associated to the former are the bases inthe tensor spaces ofAK(V) obtained by tensor products of elements ofB andB

.

Remark. Notice that also{ei ej}i,jI is a basis ofV V if{ei}iI and{ej}jIare bases ofV.However,{ei ej}i,jI is notcanonical unless ei = ei for all iI.

Theorem 2.1. Consider the tensor algebraAK(V)generated by a finite-dimension vector spaceV with fieldK = R orC and take a tensor spaceVn Vm AK(V). The specification of atensortV

n

Vm

is completely equivalent to the specification of a set of components{ti1...inj1...jm}i1,...,in,j1,...,jmI

with respect to each canonical basis ofVn Vm,{ei1 . . . ein ej1 . . . ejn}i1,...in,j1,...jmI

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which, under change of basis:

{ei1 . . . ein ej1 . . . ejn}i1,...in,j1,...jmI

transform as:ti1...in

j1...jm=Ai1k1 . . . A

inkn

B l1j1 . . . B lm

jmtk1...knl1...lm ,

whereei = A

jiej ,

and the coefficientsB lj are those of the matrix:

[B] = [A]1t ,

with [A] := [Aji]. The associated tensor t is represented by

t= ti1...in

j1...jmei1 . . . ein ej1

. . . ejn

for each considered canonical basis. Analogous results hold for tensor spaces whose factors VandV take different positions.

Notation 2.1. In the abstract index notation a tensor is indicated by writing its genericcomponent in a non-specified basis. E.g. tV V is indicated byt ji .

Somewhere in the following we adopt a cumulative index notation, i.e., letters A, B , C , . . . denoteset of covariant, contravariant or mixed indices. For instance tijklm can be written as t

A withA =ijklm. SimilarlyeAdenotes the element of a canonical basiseiejekelem. Moreover, ifA

andB

are cumulative indices, the indices of the cumulative indexAB

are those ofA

immediatelyfollowed by those of B, e.g, if A =ijklm, and B =pq u n, AB =

ijklm

pqu

n.Let us specify the allowed mathematical rules to produces tensors from given tensors. To

this end we shall use both the synthetic and the index notation.

Linear combinations of tensors of a fixed tensor space. Take a tensor space S AK(V).This is a vector space by definition, and thus picking out s, t S, and , K, linear com-binations can be formed which still belong to S. in other words we may define the tensor ofS

u:= s + t or, in the abstract index notation uA =sA + tA .

The definition ofu above given by the abstract index notation means that the components ofu

are related to the components ofs and t by a linear combination which has the same form inwhatever canonical basis of the spaceSand the coefficients ,do not depend on the basis.

Products of tensors of generally different tensor spaces. Take two tensor spaces S, SAK(V) and pick out t S, t S. Then is well defined the tensor t t S S. Using the

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associativity of the tensor product by natural isomorphisms, we found a unique tensor spaceS AK(V) which is isomorphic to S S and thus we can identify t t with a tensor in Swhich we shall indicate by t t once again with a little misuse of notation. t t is called theproductof tensors t and t. Therefore, the product of tensors coincides with the usual tensorproduct of tensors up to a natural isomorphism. What about the abstract index notation? Weleave to the reader the trivial proof of the following fact.

(t t)AB =tAtB ,

so that, for instance, ifS=V V and S= V the tensors t and t are respectively indicatedbytij and sk and thus (t s)ijk =tij sk.

Contractions. Consider a tensor space ofAK(V) of the form

U1 . . . Uk V Uk+1 . . . Ul V Ul+1 . . . Unwhere Ui denotes either V or V

. Everything we are going to say can be re-stated for theanalogous space

U1 . . . Uk V Uk+1 . . . Ul V Ul+1 . . . Un.

Then consider the multi-linearmapping Cwith domain

U1 . . . Uk V Uk+1 . . . Ul V Ul+1 . . . Un,

and values inU1 . . . Uk Uk+1 . . . Ul Ul+1 . . . Un

defined by:

(u1, . . . , uk, v , uk+1, . . . , ul, v, vl+1, . . . , vn) v, vu1. . .uk uk+1. . .ulul+1. . .un .

By the universality theorem there is a linear mapping C, called contraction of V and V,defined on the whole tensor space

U1 . . . Uk V Uk+1 . . . Ul V Ul+1 . . . Untaking values in

U1 . . . Uk Uk+1 . . . Ul Ul+1 . . . Unsuch that, on simple products of vectors reduces to

u1. . .ukvuk+1. . .ulvul+1. . .un v, vu1. . .ukuk+1. . .ulul+1. . .un.

This linear mapping takes tensors in a tensor product space with n+ 2 factors and producestensors in a space withnfactors. The simplest case arises for n = 0. In that caseC :VV K

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is nothing but the bilinear pairing C : (v, v) v, vand C is the linear associated mappingby the universality theorem.Finally, let us represent the contraction mapping within the abstract index picture. It is quitesimple to show that,Ctakes a tensortAi Bj C whereA , B and C are arbitrary cumulative indices,and produces the tensor (Ct)ABC := tAk B kC where we remark the convention of summationof the double repeated indexk. To show that the abstract-index representation of contractionsis that above notice that the contractions are linear and thus

C(tAiBjCeA ei eB ej eC) =tAiBj CC(eA ei eB ej eC) =tAiBj Cji eA eB eC,

and thusC(tAiBj

CeA ei eB ej eC) =tAkBk CeA eB eC.This is nothing but:

(Ct)ABC :=tAkBkC .

To conclude we pass to consider a final theorem which shows that there is a one-to-one corre-spondence between linear mappings on tensors and tensors them-selves.

Theorem 2.2 (Linear mappings and tensors.) Let S, S be a pair of tensor spaces of atensor algebraAK(V). The vector space of linear mappings fromSto S

is naturally isomorphicto S S (which it is naturally isomorphic to the corresponding tensor space ofAK(V)). Theisomorphism F : L(S|S) S S is given by f t(f), where the tensor t(f) S S isthat (uniquely) determined by the requirement

t(f)(s, s) =s(f(s)) , for allsS, sS

.Moreover, fixing a basis{ei}iI in V, let{eA} denote the canonical basis induced in S,{eB}that induced inS and{eC} that induced inS. With those definitions

t(f)AC =f(eA), eC

and, ifs= sAEAS,f(s)C =sAt(f)A

C .

Proof. The linear mappings fromS

toS

define the vector spaceL

(S|S). FromTheorem 1.3,we know that L(S|S) L(S, S), the latter is S S. The mappingF : L(S|S)S S

defined above is exactly the natural isomorphism inTheorem 1.3specialized to the presentlyconsidered case. The tensorf(eA), eC satisfies t(f)(s, s) = s(f(s)) by construction andthus is t(f). The last identity can trivially be proven by direct inspection.

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Let us illustrate how one may use that theorem. For instance consider a linear mappingf : V V where V is finite-dimensional with field K = C or R. Then fdefines a tensor ofV V. In fact, fixing a basis{ei}iIin Vand considering those canonically associated, by thelinearity offand the pairing:

f(v) =f(v), ekek =vif(ei), ekek.We may define the tensor t(f)V V such that

t(f) :=f(ei), ekei ek.EmployingTheorem 2.1one can trivially show that the given definition is well posed. We leaveto the reader the proof that the obtained maps which associates linear functions with tensors isthe isomorphism of the theorem above.The action off on v can be represented in terms oft(f) and v using the rules presented above.Indeed, by the abstract index notation, one has

(f(v))k =v it(f) ki .

In other words the action off on v reduces to (1) a product of the involved tensors:

vit(f) kj ,

(2) followed by a convenient contraction:

(f(v))k =v it(f) ki .

More complicate cases can be treated similarly. For example, linear mappings f from V Vto V V V are determined by tensors t(f)

ijk

lm

ofV V V V Vand their action ontensors u qp ofV V is

f(u) lmk =u ji t(f)

ijk

lm

i.e., a product of tensors and two contractions. Obviously

t(f)ijklm =

f(ei ej)

(ek, e

l, em)

where we have used the multi-linear action of elements ofVVVon elements ofVVV.

2.3 Physical invariance of the form of laws and tensors.

A physically relevant result is that the rules given above to produce a new tensor from giventensors have the same form whatever basis one use to handle tensors. For that reason the ab-stract index notation makes sense. In physics the choice of a basis is associated with the choiceof a reference frame. As is well known, various relativity principles (Galileian Principle, SpecialRelativity Principleand General Relativity Principle) assume thatthe law of Physics can be written in such a way that they preserve their form whatever reference

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frame is used.The allowed reference frames range in a class singled out by the considered relativity principle,for instance in Special Relativity the relevant class is that of inertial reference frames.It is clear that the use of tensors and rules to compose and decompose tensors to represent phys-ical laws is very hepling in implementing relativity principles. In fact the theories of Relativitycan be profitably formulated in terms of tensors and operations of tensors just to assure theinvariance of the physical laws under change of the reference frame. When physical laws aregiven in terms of tensorial relations one says that those laws are covariant. It is worthwhilestressing that covariance is not the only way to state physical laws which preserve their formunder changes of reference frames. For instance the Hamiltonian formulation of mechanics, inrelativistic contexts, is invariant under change of reference frame but it is not formuled in termsof tensor relations (in spacetime).

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3 Some Applications.

In this section we present a few applications of the theory previously developed.

3.1 Tensor products of group representations.

As is known a groupis an algebraic structure, (G, ), where G is a set and : G GG is amapping called thecomposition ruleof the group. Moreover the following three conditions haveto be satisfied.(1)is associative, i.e.,

g1 (g2 g3) = (g1 g2) g3, for all g1, g2, g3G .

(2) There is a group unit, i.e., there is eG such that

e

g= g

e= g , for all g

G .

(3) Each element gG admits an inverse element, i.e.,

for each gG there is g1 G with g g1 =g1 g= e .

We remind the reader that the unit element turns out to be unique and so does the inverseelement for each element of the group (the reader might show those uniqueness properties as anexercise). A group (G, ) is saidcommutativeor Abelianifg g = g gfor each pair of elements,g, g G; otherwise it is said non-commutativeor non-Abelian. A subset G G of a group issaidsubgroupif it is a group with respect to the restriction to GGof the composition rule ofG.

If (G1, 1) and (G2, 2) are groups, a (group) homomorphism from G1 to G2 is a mappingh : G1 G2 which preserves the group structure, i.e., the following requirement has to befulfilled:

h(g 1g ) =h(g) 2h(g) for all g, gG1,As a consequence of the uniqueness of the unit element and inverse element, it also holds withobvious notations:

h(e1) =e2,

andh(g11) = (h(g))12 for each gG1.

A group isomorphismis a bijectivegroup homomorphism.

A simple example of group is GL(n,K) which is the set of the n n matrices A with compo-nents in the field K= C or R and detA= 0. The group composition rule is the usual productof matrices. If n > 0 the group is non-commutative. An important subgroup of GL(n,K) isSL(n,K), i.e., the set of matrices B in GL(n,K) with detB = 1. (The reader might show that

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SL(n,K) is a subgroup ofGL(n,K).) However there are groups which are not defined as groupof matrices, e.g., (Z, +). An non completely trivial example is given by the group of permuta-tions ofn elementswhich we shall consider in the next subsection.

Exercises 3.1.3.1.1. Prove the uniqueness of the unit element and the inverse element in any group.3.1.2. Show that in any group G the unique element e such that e2 =e is the unit element.3.1.3. Show that ifG is a subgroup ofG the unit element ofG must coincide with the unitelement ofG, and, ifgG, the inverse element g1 in G coincides with the inverse element inG.3.1.4. Show that if h : G1 G2 is a group homomorphism, then h(e1) = e2 and h(g11) =(h(g))12 for each gG1.(Hint. Use 3.1.2. to achieve the former statement.)3.1.5. Show that ifh : G1G2 is a group homomorphism, then h(G1) is a subgroup ofG2.

We are interested in the concept of (linear) representation of a group on a vector space.In order to state the corresponding definition, notice that, if V is a (not necessarily finite-dimensional) vector space, L(V|V) contains an important group. This is GL(V) which is theset of both injective and surjective elements ofL(V|V) equipped with the usual compositionrule of maps. We can give the following definition.

Def.3.1. (Linear group on a vector space.) If V is a vector space, GL(V) denotes thegroup of linear mappings f : V V such that f is injective and surjective, with group compo-sition rule given by the usual mappings composition. GL(V) is said the linear group on V.

Remarks.(a) IfV := Kn then GL(V) =GL(n,K).(b) If V= V it is not possible to define the analogue of GL(V) considering some subset ofL(V|V). (The reader should explain the reason.)

Def.3.2. (Linear group representation on a vector space.) Let(G, )be a group andV avector space. A (linear group) representation of Gon Vis a homomorphism : GGL(V).Moreover a representation: GGL(V) is said:(1) faithful if is injective,(2) free if for eachvV, there is no gG withg=e (the unit element ofG) and(g)v= v,(3) transitive if for each pairv, vV there isgG withv= (g)v.(4) irreducible if there is no proper vector subspace S

V which is invariant under the

action of(G), i.e., which satisfies(g)SS for allgG.

Equipped with the above-given definitions we are able to study the simplest interplay of ten-sors and group representations. We want to show that the notion of tensor product allow thedefinitions of tensor products of representations. That mathematical object is of fundamental

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importance in applications to Quantum Mechanics, in particular as far as systems with manycomponents are concerned.Consider a group G (from now on we omit to specify the symbol of the composition rule when-ever it does not produces misunderstandings) and several representations of the group i : G

GL(Vi), whereV1, . . . , V nare finite-dimensional vector spaces on the same field K = C or R. ForeachgG, we may define a multi-linear mapping [i(g), . . . , n(g)]L(V1, . . . , V n|V1 . . .Vn)given by, for all (i(g), . . . , n(g))V1 . . . Vn,

[i(g), . . . , n(g)] : (v1, . . . , vn)(1(g)v1) . . . (n(g)vn) .That mapping is multi linear because of the multi linearity of the tensor-product mapping andthe linearity of the operators k(g). Using the universality theorem, we uniquely find a linearmapping which we indicate by 1(g) . . . n(g) :V1 . . . VnV1 . . . Vn such that:

1(g) . . . n(g)(v1 . . . vn) = (1(g)v1) . . . (n(g)vn) .

Def.3.3. (Tensor product of representations.) Let V1, . . . V n be finite-dimensional vectorspaces on the same fieldK= C orR and suppose there aren representationsi : GGL(Vk)of the same group G on the given vector spaces. The set of linear maps

{1(g) . . . n(g) :V1 . . . VnV1 . . . Vn| gG} ,defined above is calledtensor product of representations1, . . . , n.

The relevance of the definition above is evident because of the following theorem,

Theorem 3.1. Referring to the definition above, the elements of a tensor product of represen-

tations1, . . . n of the group G on spacesV1, . . . , V n define alinear group representation ofG on the tensor-product space V1 . . . Vn.

Proof. We have to show that the mapping

g1(g) . . . n(g) ,is a group homomorphism from G to GL(V1 . . . Vn). Taking account of the fact that eachi is a group homomorphism, ifg, g

G, one has1(g

). . .n(g)(1(g). . .n(g)(v1. . .vn)) =1(g). . .n(g)((1(g)v1). . .(n(g)vn))and this is

(1(g g)v1) . . . (n(g g)vn) .The obtained result holds true also using a canonical basis for V1. . .Vn made of usualelementse1,i1 . . . en,in in place ofv1 . . . vn. By linearity, it means that

(1(g) . . . n(g))(1(g) . . . n(g)) =1(g g) . . . n(g g) ,

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for all g, gG. Notice that each mapping 1(g) . . . n(g)GL(V1 . . . Vn) because it islinear and injective by construction. The latter is a consequance of the proof given above whichimplies that:

(1(g1) . . . n(g1))(1(g) . . . n(g)) = (1(g) . . . n(g))(1(g1) . . . n(g1))equals

1(e) . . . n(e) =I .The last identity follows from the facf that 1(e) . . . n(e) Iis linear and vanishes on anycanonical basis (show it!). These results conclude the proof because we have just shown thatthe map

g1(g) . . . n(g) ,is a homomorphis from G to GL(V1 . . . Vn).

More generally, ifAk : Vk Uk are n linear mappings (operators), and all involved vectorspaces are finite dimensional and with the same field K = C or R, it is defined the tensor productof operators.

Def.3.4 (Tensor Product of Operators.) If Ak : Vk Uk, k = 1, . . . , n are n linearmappings (operators), and all the vector spacesUi, Vj are finite dimensional with the same fieldK = C orR, the tensor product ofA1, . . . , An is the linear mapping

A1 . . . An: V1 . . . VnU1 . . . Ununiquely determined by the universality theorem and the requirement:

(A1 . . . An) = A1 . . . An,

whereA1 . . . An: V1 . . . VnU1 . . . Un,

is the multi-linear mapping such that, for all(v1, . . . , vn)V1 . . . Vn:

A1 . . . An: (v1, . . . , vn)(A1v1) . . . (Anvn) .

Remark. Employing the given definition and the same proof used for the relevant part of theproof ofTheorem 3.1, it is simply shown that ifAi : Vi

Ui and Bi : Ui

Wi, i= 1, . . . , n

are 2n linear maps and all involved spaces Vi, Uj , Wk are finite dimensional with the same fieldK = C or R, then

B1 . . . Bn A1 . . . An= (B1A1) . . . (BnAn) .

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3.2 A quantum physical example.

Physicsts are involved with Hilbert spaces whenever they handle quantum mechanics. A Hilbertspace is noting but a complex vector space equipped with a Hermitean scalar product (see the

next chapter) such that it is complete with respect to the norm topology induced by that scalarproduct. As far as we are concerned we need only the structure of vector space. Physicallyspeaking, the vectors of the Hilbert space represent the states of the considered physical system(actually things are more complicated but we do not matter). To consider the simplest casewe assume that the vector space which describes the states of the system is finite-dimensional(that is the case for the spin part of a quantum particle). Moreover, physics implies that thespace H of the states of a composit system Smade of two systems S1 and S2 associated withHilbert spaces H1 and H2 respectively, is the tensor product H = H1H2. There should beseveral remarks in the infinite dimensional case since our definition of tensor product works forfinite-dimensional spaces only, however suitable and well-behaved generalizations are, in fact,possible. Let the systemS1be described by a state

H1and suppose to transform the system

by the action of an element R of some physical group of transformations G (e.g. SO(3)). Thetransformed state is given by U(1)R where G RU(1)R is a representation ofGin terms oflinear transformations U

(1)R : H1 H1. Actually, physics and the celebrated Wigners theorem

in particular, requires that every U(1)R be a unitary(or anti unitary) transformation but this is

not relevant for our case. The natural question concerning the representation of the action ofGon the composit system S is:

If we know the representations G RU(1)R and G RU(2)R , what about the representationof the action ofGon Sin terms of linear transformations in the Hilbert space H= H1H2?The answer given by physics, at least when the systems S1 and S2 do not interact, is that

Ug :=U(2)g U(1)g .

3.3 Permutation group and symmetry of tensors.

We remind the definition of the group of permutations ofn objectsand give some know resultsof basic group theory whose proofs may be found in any group-theory textbook.

Def.3.5. (Group of permutations.) Consider the setIn :={1, . . . , n}, the group of per-mutations ofn objects, Pn is the set of the bijective mappings: InIn equipped with thecomposition rule given by the usual composition rule of functions. Moreover,(a)the elements ofPn are saidpermutations (ofn objects);(b)a permutation ofPn with n2 is said transposition if differs from the identity mappingand reduces to the identity mapping when restricted to some subset ofIn containingn

2 ele-

ments.

Comments.(1) Pn contains n! elements.(2) Each permutation Pn can be represented by a corresponding string ((1), . . . , (n)).

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(3) If, for instance n = 5, with the notation above (1, 2, 3, 5, 4), (5, 2, 3, 4, 1), (1, 2, 4, 3, 5) aretranspositions, (2, 3, 4, 5, 1), (5, 4, 3, 2, 1) are not.(3) It is possible to show that each permutation Pn can be decomposed as a product oftranspositions = 1

. . .

k. In general there are several different transposition-product

decompositions for each permutation, however it is possible to show that if = 1 . . . k =1 . . . r, where i and j are transpositions, then r+ k is even. Equivalently, r is even orodd if and only ifk is so. This defines the parity, {1, +1}, of a permutation , where, = +1 if can be decomposed as a product of an evennumber of transpositions and =1if can be decomposed as a product of an odd number of transpositions.(4) IfA= [Aij ] is a real or complex n n matrix, it is possible to show (by induction) that:

detA=

PnA1(1) An(n).

Alternatively, the identity above may be used to define the determinant of a matrix.

We pass to consider the action ofPn on tensors. Fix n > 1, consider the tensor algebraAK(K) and single out the tensor space V

n:= V . . . Vwhere the factor V appears n timesThen consider the following action ofPn on V

n:= V . . . Vwhere the factor V appears ntimes. For each Pn consider the mapping

: VnVn such that (v1, . . . , vn)v(1) . . . v(n).It is quite straightforward to show that is multi linear. Therefore, let

: VnVn,be the linear mapping uniquely determined by by means of the universality theorem.

Theorem 3.2. The above-defined linear mapping

with Pn, is a group representation ofPn onVn.

Proof. First we show that if, Pn then((ei1 . . . ein)) = ( )(ei1 . . . ein) ,

for some canonical basis of elements ei1 . . . ein. By linearity such an identity will hold truefor all arguments in V

n

. We have((ei1 . . . ein)) = ( )(ei1 . . . ein)

holds true because, from definition of , it can be re-written

((ei1 , . . . , ein)) = ( )(ei1, . . . , ein) ,

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ore((1)) . . . e((n))= ( )(ei1, . . . , ein) ,

which is trivially satisfied by the definition of . Hence, we have proven that

= ( ).The mappings are linear by constructions and are bijective because

= ( )

implies1= 1= e= I .

The last identity can be proven by noticing that e Iis linear and vanishes when evaluated onany canonical base ofVn. We have shown that GL(Vn) for all Pnand the mapping

is a homomorphism. This concludes the proof.

Let us pass to consider the abstract index notation and give a representation of the action of within that picture.

Theorem 3.3. If t is a tensor inVn AK(V) withn2 and Pn, then the componentsoft with respect to any canonical basis ofVn satisfy

(t)i1...in =ti1(1)...i1(n) .

Proof.

t=

(

tj1...jnej1

. . .

ejn) =

tj1...jnej(1)

. . .

ej(n)

.

Since : In In is bijective, if we define ik := j(k), it holds jk = i1(k). Using this identityabove we find

t= ti1(1)...i1(n)ei1 . . . ein.That is nothing but the thesis.

To conclude we introduce the concept of symmetric or anti-symmetric tensor.

Def. 3.6 (Symmetric and anti-symmetric tensors.) LetV be a finite-dimensional vectorspace with fieldK = C or R. Consider the space Vn AK(V) of tensors of order (n, 0) withn

2.(a) tVn is saidsymmetric if

t= t ,

for all of Pn, or equivalently, using the abstract index notation,tj1...jn =tj(1)...j(n) ,

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for all of Pn.(b) tVn is said anti symmetric if

t= t ,

for all of Pn, or equivalently using the abstract index notation:

tj1...jn =tj(1)...j(n) ,

for all of Pn.

Remark. Concerning the definitition in (b) notice that = 1 .

Examples 3.1.3.1.1. Supposen= 2, then a symmetric tensor sVV satisfiessij =sji and an antisymmetrictensor aV V satisfies a

ij

=aji

.3.1.2. Suppose n = 3, then it is trivially shown that P3 has parity 1 if and only if is a cyclic permutation, i.e., ((1), (2), (3)) = (1, 2, 3) or ((1), (2), (3)) = (2, 3, 1) or((1), (2), (3)) = (3, 1, 2).Hence, a tensor eV V V is anti symmetric if and only if

eijk = 0 ,

if (i,j,k) is nota permutation of (1, 2, 3) and, otherwise,

eijk =e123 ,

where the sign + takes place if the permutation ((1), (2), (3)) = (i,j,k) is cyclic and thesigntakes place otherwise. That relation between parity of a permutation and cyclicity doesnothold true for n >3.

Remarks.(1) Consider a generic tensor space S AK(V) which contains n 2 spaces V as factors.We may suppose for sake of simplicity S = S1 Vn S2 where S1 = U1 . . .Un1,S2= Un+1 . . . Um and Ui = V or Ui = V. Anyway all what we are going to say holds truealso if the considered n spaces Vdo not define a unique block Vn. We may define the actionof Pn on the whole space Sstarting by a multi linear mapping

: U1 . . . Uk Vn

Uk+1 . . . UmU1 . . . Uk Vn

Uk+1 . . . Um,such that reduces to the tensor-product mapping on U1 . . . Uk and Uk+1 . . . Um:

: (u1, . . . , uk, v1, . . . vn, uk, . . . , um)u1 . . . uk v(1) . . . v(n) uk+1 . . . um.

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Using the universality theorem as above, we build up a representation ofPnon S,whichacts on Vn only. Using the abstract index notation the action of is well represented:

: tA i1...in B

tA

i1(1)...i1(n) B .

This allows one to define and study the symmetry of a tensor referring to a few indices singledout among the complete set of indices of the tensors. E.g., a tensor tij k

r may be symmetric oranti symmetric, for instance, with respect to the indices i and j or j and r or ij r.(2) Everything we have obtained and defined may be similarly re-obtained and re-defined con-sidering spaces of tensors of order (0, n) with n 2, i.e. covariant tensors. In that case theantisymmetric tensorsof order (0, n) are said n-forms.(3) Notice that no discussion on the symmetry of indices of different kind (one covariant andthe other contravariant) is possible.

Exercises 3.2.

3.2.1. Let t be a tensor in Vn (or Vn). Show that t is symmetric or anti symmetric ifthere is a canonical basis where the components have symmetric or anti symmetric indices, i.e.,ti1...in =ti(1)...i(n) or respectively ti1...in =t

i(1)...i(n) for all Pn.Note. The result implies that, to show that a tensor is symmetric or anti symmetric, it issufficient to verify the simmetry or anti symmetry of its components within a singlecanonicalbasis.3.2.2. Show that the sets of symmetric tensors of order (n, 0) and (0, n) are vector subspacesofVn and Vn respectively.3.2.3. Show that the subspace of anti-symmetric tensors of order (0, n) (the space ofn-forms)in Vn has dimension

dimV

n

ifndimV. What about n > dimV?

3.2.4. Consider a tensor ti1...in , show that the tensor is symmetric if and only if it is symmetric

with respect to each arbitrary chosen pair of indices, i.e.

t...ik...ip... =t...ip...ik... ,

for all p, k {1, . . . n}, p=k.3.2.5. Consider a tensor ti1...in , show that the tensor is anti symmetric if and only if it is antisymmetric with respect to each arbitrarily chosen pair of indices, i.e.

t...ik...ip... =t...ip...ik... ,

for all p, k {1, . . . n}, p=k.3.2.6. Show that V

V =A

Swhere

denotes the direct sum and A and Sare respectively

the space of anti-symmetric and symmetric tensors in V V. Does such a direct decompositionhold if considering Vn with n >2?

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4 Scalar Products and Metric Tools.

This section concerns the introduction of the notion of scalar product and several applicationson tensors.

4.1 Scalar products.

First of all we give the definition of a pseudo scalar productand semi scalar productswhich dif-fer from the notion ofscalar productfor the positivity and the non-degenerateness requirementrespectively. In fact, a pseudo scalar product is a generalization of the usual definition of scalarproduct which has many applications in mathematical physics, relativistic theories in particular.Semi scalar products are used in several applications of quantum field theory (for instance inthe celebratedGNS theorem).

Def.4.1. (Pseudo Scalar Product.) LetVbe a vector space on the field eitherK = R orC.

(a) A pseudo scalar product is a mapping( | ) :V V K which is:(i) bi linear, i.e., for all u V both (u|) : v (u|v) and (|u) : v (v|u) are linear

functionals onV;(ii) symmetric, i.e., (u|v) = (v|u) for allu, vV;(iii) non-degenerate, i.e., (u|v) = 0 for allvV impliesu= 0.

(b)IfK = C, aHermitian pseudo scalar product is a mapping( | ) :V V Kwhich is:(i)sesquilinear, i.e., for alluV, (u|)and(|u)are a linear functional and an anti-linear

functional onV respectively;(ii) Hermitian, i.e., (u|v) = (v|u) for allu, vV;(iii) non-degenerate.

Def.4.2 (Semi Scalar Product.) LetVbe a vector space on the field eitherK = R orC.(a) IfK = R, a semi scalar product is a mapping( | ) :V V R which satisfies(ai),(aii)above and is

(iv) semi-defined positive, i.e., (u|u)0 for all uV .(b) If K = C, a Hermitian semi scalar product is a mapping (|) : VV K whichsatisfies(bi),(bii)above and is

(iv) semi-defined positive.

Finally we give the definition of scalar product.

Def.4.3. (Scalar Product.) Let Vbe a vector space on the fieldK = R (resp. C) endowed

with a pseudo scalar product (resp. Hermitian pseudo scalar product) ( | ). ( | ) is said scalarproduct(resp. Hermitian scalar product) if( | ) is also a semi scalar product, i.e., if it issemi-defined positive.

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Comments.(1) Notice that all given definitions do not require that Vis finite dimensional.(2)IfK = C and (|) is not Hermitian, in general, any requirement on positivity of (u|u) does notmake sense because (u

|u) may notbe real. If instead Hermiticity holds, we have (u

|u) = (u

|u)

which assures that (u|u)R and thus positivity may be investigated.(3) Actually, a (Hermitian) scalar product is positive defined, i.e,

(u|u)> 0 if uV\ {0} ,

because of Cauchy-Schwarz inequality

|(u|v)|2 (u|u)(v|v) ,

which we shall prove below for semi scalar products.(4) A semi norm on a vector space V with field K = C or R, is a mapping|| ||: V K suchthat:(i)||v|| R and in particular||v|| 0 for all vV;

(ii)||v||=||||v|| for all K and vV;(iii)||u + v|| ||u|| + ||v||for all u, vV.

A semi norm|| ||: V K is a normif(iv)||v||= 0 implies v = 0.

Notice that for semi norms it holds:||0||= 0 because of (ii) above. With the given definitions,it quite simple to show (the reader might try to give a proof) that ifV with field K = R (C) isequipped by a (Hermitian) semi scalar product then||v||:=

(v|v) for all vVdefines a semi

norm. Furthermore if ( | ) is a scalar product, then the associated semi norm is a norm.(5) If a vector space V is equipped with a norm|| || it becomes a metric spaceby defining thedistanced(u, v) :=||uv||for allu, vV. ABanach space(V, ||||) is a vector space equippedwith a norm such that the associates metric space is complete, i.e, all Cauchys sequences con-verge. A Hilbert space is a Banach space with norm given by a (Hermitean if the field isC) scalar product as said above. Hilbert spaces are the central mathematical objects used inquantum mechanics.

Exercices 4.1.4.1.1. Show that if (| ) is a (Hermitian) semi scalar product on V with field R (C) thenthe mapping on V, v ||v|| := (v|v), satisfies||u+ v|| ||u||+||v|| as a consequence ofCauchy-Schwarz inequality

|(u|v)|2 (u|u)(v|v) ,which holds true by all (Hermitian) semi scalar product.(Hint. Compute||u+v||2 = (u+v|u+v) using bi linearity or sesquilinearity property of ( | ),then use Cauchy-Schwarz inequality.)

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Theorem 4.1. (Cauchy-Schwarz inequality.) Let V be a vector space with field R (C)equipped with a (Hermitian) semi scalar product ( | ). Then, for allu, v V, Cauchy-Schwarzinequality holds:

|(u

|v)

|2

(u

|u)(v

|v) .

Proof. Consider the complex case with a Hermitian semi scalar product. Take u, vV. For allz C it must hold (zu+v|zu+v)0 by definition of Hermitean semi scalar product. Usingsesquilinearity and Hermiticity :

0zz(u|u) + (v|v) + z(u|v) + z(v|u) =|z|2(u|u) + (v|v) + z(u|v) + z(u|v) ,

which can be re-written as

|z

|2(u

|u) + (v

|v) + 2Re

{z(u

|v)

} 0 .

Then we pass to the polar representation ofz, z = rei with r, R arbitrarily and indepen-dently fixed. Decompose also (u|v), (u|v) =|(u|v)|ei arg(u|v). Inserting above we get:

F(r, ) :=r2(u|u) + 2r|(u|v)|Re[ei(arg(u|v) )] + (v|v)0 ,

for all r R when R is fixed arbitrarily. Since the right-hand side abive is a secon-orderpolynomial in r , the inequlity implies that, for all R,

2|(u|v)|Re[ei(arg(u|v) )]2 4(v|v)(u|u)0 ,

which is equivalent to

|(u|v)|2 cos(arg(u|v) ) (u|u)(v|v)0 ,

for all R. Choosing = arg (u|v), we get Cauchy-Schwarz inequality:

|(u|v)|2 (u|u)(v|v) .

The real case can be treated similarly, replacing z C with x R and the proof is essentiallythe same. .

Corollary. A bilinear symmetric (resp. sesquilinear Hermitian) mapping (|) : VV R

(resp. C

) is a scalar product (resp. Hermitian scalar product) if and only if it is positivedefined, that is(u|u)> 0 for alluV\ {0}.

Proof. Assume that ( | ) is a (Hermitian) scalar product. Hence (u|u) 0 by definition and( | ) is non-degenerate. Moreover it holds|(u|v)|2 (u|u)(v|v). As a consequence, if (u|u) = 0then (u|v) = 0 for all v V and thus u = 0 because ( | ) is non-degenerate. We have proven

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that (u|u) > 0 if u= 0. That is, a scalar product is a positive-defined bilinear symmetric(resp. sesquilinear Hermitian) mapping (|) : VV R (resp. C). Now assume that (|)is positive-defined bilinear symmetric (resp. sesquilinear Hermitian). By definition it is a semiscalar product since positive definiteness implies positive semi-definiteness. Let us prove that(|) is non-degenerate and this concludes the proof. If (u|v) = 0 for all v V then, choosingv= u, the positive definiteness implies u = 0.

4.2 Natural isomorphism between V and V and metric tensor.

Let us show that ifVis a finite-dimensional vector space endowed with a pseudo scalar product,V is isomorphic to V. That isomorphism is naturalbecause it is built up using the structureof vector space with scalar product only, specifying nothing further.

Theorem 4.2. (Natural (anti)isomorphism between V and V.) Let V be a finite-dimensional vector space with fieldK = R orC.

(a) IfV is endowed with a pseudo scalar product( | ) (also ifK = C),(i) the mapping defined onV, h: u(u|), where(u|) is the linear functional, (u|) :v

(u|v), is an isomorphism;(ii) (h(u)|h(v)) := (u|v) defines a pseudo scalar product onV.

(b) IfK = C andVis endowed with a Hermitean pseudo scalar product( | ),(i) the mapping defined onV, h: u(u|), where(u|) is the linear functional, (u|) :v

(u|v), is an anti isomorphism;(ii) (h(u)|h(v)) := (u|v) (= (v|u)) defines a Hermitian pseudo scalar product onV.

Proof. First consider (i) in the cases (a) and (b). It is obvious that (u|) V in both cases.Moreover the linearity or antilinearity of the mapping u

(u

|) is a trivial consequence of the

definition of pseudo scalar product and Hermitian pseudo scalar product respectively.Then remind the well-known theorem, dim(Kerf) +dimf(V) = dimV , which holds true forlinear and anti-linear mappings from some finite-dimensional vector space V to some vectorspace V. SincedimV = dimV, it is sufficient to show that h: V V defined by u (u|)has trivial kernel, i.e., is injective: this also assures the surjectivity of the map. Therefore, wehave to show that (u|) = (u|) implies u= u . This is equivalent to show that (u u|v) = 0for all v V implies u u = 0. This is nothing but the non-degenerateness property, whichholds by definition of (Hermitean) scalar product.Statements (ii) cases are obvious in both by definition of (Hermitian) pseudo scalar productsusing the fact that h is a (anti) isomorphism.

Remarks.(1) Notice that, with the definitions given above it holds also (u|v) = (h1u|h1v) and, forthe Hermitian case, (u|v) = (h1u|h1v) for all u, v V. This means that h and h1(anti)preservethe scalar products.(2) The theorem above holds also considering a Hilbert space and its topological dual space

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(i.e., the subspace of the dual space consisting of continuous linear functionals on the Hilbertspace). That is the mathematical content of celebrated Riesz representation theorem.

Exercise 4.2.4.2.1. Show that, for all u, vV:

(u|v) =u, h(v)= (h(v)|h(u)) ,no matter if ( | ) is Hermitian or not.

From now on we specialize to the pseudo-scalar-product case dropping the Hermitian case.Suppose ( |) is a pseudo scalar product on a finite-dimensional vector spaceV with field K eitherR or C. The mapping (u, v) (u|v) is bi linear on V Vand thus is a tensor g V V.Fixing a canonical basis in V V induced by a basis{ei}iI V, we can write:

g=

gij

e

i

ej ,

where, by Theorem 1.5,gij = (ei|ej) .

Def.4.4. (Metric Tensor.) A pseudo scalar product( |) =gVV on a finite-dimensionalvector spaceVwith fieldKeitherR orCis calledpseudo-metric tensor. IfK = R, a pseudo-metric tensor is calledmetric tensorif it defines a scalar product.

Remarks.(1)By Theorem 2.2, the isomorphismh : VVis represented by a tensor ofVVwhichacts on elements ofVby means of a product of tensors and a contraction. The introduction of

the pseudo-metric tensor allows us to represent the isomorphism h : V V by means of theabstract index notation determining the tensor representing h. Indeed, since h: u(u| ) and(u|v) = (ei|ej )uivj =gij viuj we trivially have:

(hu)i = gij uj .

Hence h is represented by g itself.(2)Notice that pseudo-metric tensors are symmetricbecause of the symmetry of pseudo scalarproducts:

g(u, v) = (u|v) = (v|u) =g(v, u) .

Components of pseudo-metric tensors with respect to canonical basis enjoy some simple butimportant properties which are listed below.

Theorem 4.3. (Properties of the metric tensor.) Referring to Def.4.4, the componentsof any pseudo-metric tensor g, gij := g(ei, ej) with respect to the canonical basis induced in

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V V by any basis{ei}i=1,...,nV, enjoy the following properties:(1) define a symmetric matrix [gij ], i.e,

gij =gji ;

(2) [gij ] is non singular, i.e., it satisfies:

det[gij ]= 0 ;(3) ifK = R andg is a scalar product, the matrix [gij ] is positive defined.

Proof. (1) It is obvious: gij = (ei|ej ) = (ej |ei) =gji .(2) Suppose det[gij ] = 0 and define n = dimV. The linear mapping K

n Kn determined bythe matrix [g] := [gij ] has a non-trivial kernel. In other words, there are n reals u

j , j = 1, . . . , ndefining a Kn vector [u] := (u1, . . . , un)t with [g][u] = 0 and [u]= 0. In particular [v]t[g][u] = 0for whatever choice of [v] Kn. Defining u := uj ej , the obtained result implies that there isu

V\{

0}

with (u|v) = (v

|u) = 0 for allv

V. This is impossible because (

|) is non degenerate

by hypothes.(3)The statement, (u|u)> 0 ifuV\{0}, reads, in the considered canonical basis [u]t[g][u]> 0for [u]Rn \ {0}. That is one of the equivalent definitions of a positive defined matrix [g].

The following theorem shows that a (pseudo) scalar product can be given by the assignmentof a convenient tensor which satisfies some properties when represented in some canonical bases.The important point is that there is no need to check on these properties foral lcanonical bases,verification for a single canonical basis is sufficient.

Theorem 4.4 (Assignment of a (pseudo) scalar product.) LetV be a finite-dimensionalvector space with fieldK either R orC. Supposeg

V

V

is a tensor such that there is acanonical basis of V V where the componentsgij ofg define a symmetric matrix [gij ] withnon-vanishing determinant. Theng is a pseudo-metric tensor, i.e. a pseudo scalar product.Furthermore, ifK = R and[gij ]is positive defined, the pseudo scalar product is a scalar product.

Proof. If g is represented by a symmetric matrix of components in a canonical basis then itholds in all remaining bases and the tensor is symmetric (see Exercise 3.2.1). This impliesthat (u|v) :=g(u, v) is a bi-linear symmetric functional. Suppose ( | ) is degenerate, then there isuV such that u=0 and (u|v) = 0 for all vV. Using notations of the proof of the item (2)ofTheorem 4.3, we have in components of the considered canonical bases, [u]t[g][v] = 0 for all[v] = (v1, . . . , vn)t Kn where n = dimV. Choosing [v] = [g][u], it also holds [u]t[g][g][u] = 0.

Since [g

] = [g

]

t

, this is equivalent to ([g

][u

])

t

[g

][u

] = 0 which implies [g

][u

] = 0. Since [u

]=0, [g

]cannot be injective anddet[g] = 0. This is not possible by hypotheses, thus (|) is non-degenerate.We conclude that (u|v) :=g(u, v) define a pseudo scalar product.Finally, ifK = R and [g] is also positive defined, ( | ) itself turns out to be positive defined, i.e.,it is a scalar product since (u|u) = [u]t[g][u]> 0 if [u]=0 (which is equivalent to u=0).

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Let us introduce the concept ofsignature of a pseudo-metric tensor in a vector space withfield R by reminding Sylvesters theorem whose proof can be found in any linear algebra text-book. The definition is interlaced with the definition of an orthonormal basis.

Theorem 4.5 (Sylvesters theorem.) LetA be a real symmetricn n matrix.(a) There is a non-singular (i.e., with non vanishing determinant) real nn matrix D suchthat:

DADt =diag(0, . . . , 0, 1, . . . , 1, +1, . . . , +1) ,where the reals 0, 1, +1 appear v 0 times, m 0 times and p 0 times respectively withv+ m +p= n.(b) the triple (v, m , p) does not depend on D. In other words, if, for some non-singular realn n matrix matrix E= D, EAEt is diagonal and the diagonal contains reals 0, 1, +1 only(in whatever order), then0, 1, +1 respectively appearv times, m times andp times.

Ifg V V is a pseudo-metric tensor on the finite-dimensional vector space v with field R,the transformation rule of the components ofg with respect to canonical bases (see Theorem2.1) induced by bases{ei}iI,{ej}jI ofV are

gpq = B i

p B jq gij.

Defining [g] := [gpq], [g] := [gij ], [B] := [B kh ], they can be re-written as

[g] = [B][g][B]t .

We remind (see Theorem 2.1) that the non-singular matrices [B] are defined by [B] = [A]1t,where [A] = [Ai

j

] and em= Al

m

el

. Notice that the specification of [B] is completely equivalent

to the specification of [A] because [A] = [B]1t.Hence, since [g] is real and symmetric byTheorem 4.3, Sylvesters theorem implies that, start-ing from any basis{ei}iI V one can find another basis{ej}jI, which induces a canonicalbasis in VV where the pseudo-metric tensor is represented by a diagonal matrix. It issufficient to pick out a transformation matrix [B] as specified in (a) ofTheorem 4.5. In par-ticular, one can find B such that each element on the diagonal of [g] turns out to be either1or +1 only. The value 0 is not allowed because it would imply that the matrix has vanishingdeterminant and this is not possible because ofTheorem 4.3. Moreover the pair (m, p), where(m, p) are defined in Theorem 4.5, does not depend on the basis{ej}jI. In other words, itis an intrisic property of the pseudo-metric tensor: that is the signatureof the pseudo-metric

tensor.

Def.4.5 (Pseudo Orthonormal Bases and Signature). LetgVV be a pseudo-metrictensor on the finite-dimensional vector spaceV with fieldR.(a)A basis{ei}iI V is calledpseudo orthonormalwith respect to g if the components ofg

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with respect to the canonical basis induced inV V form a diagonal matrix with eigenvaluesin{1, +1}. In other words,{ei}iI is pseudo orthonormal if

(ei, ej ) =

ij .

If the pseudo-metric tensor is a metric tensor the pseudo-orthonormal bases are called orthonor-mal bases.(b)The pair(m, p), wherem is the number of eigenvalues1andp is the number of eigenvalues+1of a matrix representing the components ofg in an orthonormal basis is calledsignatureofg.(c) g and its signature are saidelliptic orEuclidean orRiemannian ifm= 0, hyperbolicifm >0 andp= 0, Lorentzian ornormally hyperbolic ifm= 1 andp= 0.(d) Ifg is hyperbolic, an orthonormal basis{ei}iI is saidcanonical if the matrix of the com-ponents ofg takes the form:

[g] =diag(1, ...,1, +1, ..., +1) .

Exercises. 4.2.4.2.1. Show that apseudo-metrictensorg is ametrictensor if and only if its signature is elliptic.

Remark. If{ei}iIis an orthonormal basis with respect to a hyperbolic pseudo-metric tensor g ,one can trivially re-order the vectors of the basis giving rise to a canonical orthonormal basis.

Comment. Let us consider a pseudo-metric tensor gin Vwith field R. Let (m, p) be the signatureofg and let Ng be the class of all of the canonical pseudo-orthonormal bases in Vwith respect

to g. In the following we shall indicate by [] the matrix diag(1, ...,1, +1,..., +1) whichrepresents the components ofg with respect to each basis ofNg. If [A] is a matrix correspondingto a change of basis in Ng, and [B] := [A]

1t is the associated matrix concerning change of basisin V, it has to hold

[] = [B][][B]t .

Conversely, each realn nmatrix [B] which satisfies the identity above determines [A] = [B]1twhich represents a change of basis in Ng. In particular each [B] which satisfies the identity abovemust be non singular. Indeed, taking the determinant of both sides in the identity above andtaking det[] = (1)m into account, we have that (det[B])2 = 1 and thus det[B] =1. Wherewe have also used det[B] =det[B]t.Noticing that [] = []1 and [A] = [B]1t, the identity above can be equivalently re-written interms of the matrices [A]:

[] = [A][][A]t .

The equation above completely determines the set O(m, p)GL(n,R) (n= m +p) of all realnon-singular n n matrices which correspond to changes of bases in Ng.It is possible to show that O(m, p) is a subgroup ofGL(n,R) called the pseudo orthogonal

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group of order (m, p). Notice that, if m = 0, O(0, p) = O(n) reduces to the usual orthogonalgroup of order n. O(1, 3) is the celebrated Lorentz groupwhich is the central mathematicalobject in relativistic theories.

Exercises. 4.3.4.3.1. Show that if [A]O(m, p) then [A]1 exists and

[A]1 = [][A]t[] .

4.3.2. Show that O(m, p) is a group with respect to the usual multiplication of matrices.Note. This implies that O(m, p) is a subgroup ofGL(n,R) with n = p + m.(Hint. You have to prove that, (1) the identity matrix Ibelongs to O(m, p), (2) if [A] and [A]belong to O(m, p), [A][A] belongs to O(m, p), (3) if [A] belongs to O(m, p), then [A]1 existsand belongs to O(m, p).)4.3.3. Show that SO(m, p) :={[A] O(m, p) | det[A] = 1} is not the empty set and is asubgroup ofO(m, p).Note. SO(m, p) is said the special pseudo orthogonal groupof order (m, p).4.3.4. Consider the special Lorentz group SO(1, 3) and show that the set

SO(m, p) :={[A]S O(1, 3)| A11>0}is a not empty subgroup.Note. SO(m, p) is said the special orthocronous Lorentz group.

4.3 Raising and lowering of indices of tensors.

Consider a finite dimensional vector space V with fieldK

=R

orC

endowed with a pseudo-metric tensor g. As we said above, there is a natural isomorphism h : V V defined byh: u(u|) =g(u, ). This isomorphism may be extended to the whole tensor algebra AK(V)using the universality theorem and Def.3.4.Indeed, consider a space S AK(V) of the form A V B, where A and B are tensor spacesof the form U1 . . . Uk, and Uk+1 . . . Um respectively, Ui being either V or V. We maydefine the operators:

h:= I1 . . . Ik h Ik+1 . . . Im : A V BA V B ,and

(h1):= I1 . . . Ik h1 Ik+1 . . . Im : A V BA V B ,whereIj :UjUj is the identity operator. UsingRemarkafter Def.3.4, one finds

(h1)h= I1 . . . Ik (h1h) Ik+1 . . . Im = I dAVB,and

h(h1)= I1 . . . Ik (hh1) Ik+1 . . . Im = I dAVB.

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Thereforeh is an isomorphism with inverse (h1).

The action ofh and (h1) is that of loweringandraising indices respectively. In fact, inabstract index notation, one has:

h: tAiB tA j B :=tAiBgij ,

and(h1): uA i B uAjB :=tA i B gij .

Above gij represents the pseudo-metric tensor as specified in Remark 1 after Def.4.4. Whatabout the tensor gV V representing h1 via Theorem 2.2?

Theorem 4.6. Leth: VV be the isomorphism determined by a pseudo scalar product, i.e.a pseudo-metric tensorg on the finite-dimensional vector spaceV with fieldK = R orC.

(a) The inverse mapping h1

: V

V

is represented via Theorem 2.2 by a symmetrictensorgV V such that, if{ei}iI is a basis ofV, grs := g(er, es) andgij :=g(ei, ej), thenthe matrix [g] := [gij ] is theinverse matrix of [g] := [gij ].(b) The tensorg coincides with the pseudo-metric tensor with both indices raised.

Proof. (a) By Theorem 2.2, h1 determines a tensor g V V with h1(u) = g(u, ). Incomponents (h1u)i =ukg

ki. On the other hand it must be

h(h1u) =u

or,ukg

kigir = ur ,

for all uV. This is can be re-written

[u]t([g][g] I) = 0 ,

for all Kn vectors [u] = (u1, . . . , un). Then the matrix (g[g]I)t is the null matrix. This implies

that[g][g] =I ,

which is the thesis. [g] is symmetric because is the inverse of a symmetric matrix and thus alsothe tensor g is symmetric.(b) Let g ij be the pseudo-metric tensor with both indices raised, i.e.,

gij :=grk gkj gri .

By (a), the right-hand side is equal to:

jr gri = gji = gij .

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That is the thesis.

Remark. Another result which arises from the proof of the second part of the theorem is that

gji =ji .

Comments.(1) When a vector space is endowed with a pseudo scalar product, tensors can be viewed asabstract objects which may be represented either as covariant or contravariant concrete tensorsusing the procedure of raising and lowering indices. For instance, a tensor tij ofV V may beviewed as a covariant tensor when represented in its covariant form tpq :=gpigqj t

ij . Also, itcan be viewed as a mixed tensor tp

j :=gpitij or ti q :=gqj t

ij .(2) Now consider a finite-dimensional vector space on R, V, endowed with a metric tensor g,i.e., with elliptic signature. In orthonormal bases the contravariant and covariant componentsnumerically coincides because gij =ij =g

ij . This is the reason because, using the usual scalar

product of vector spaces isomorphic to Rn and working in orthonormal bases, the differencebetween covariant and contravariant vectors does not arise.Conversely, in relativistic theories where a Lorentzian scalar product is necessary, the differencebetween covariant and contravariant vectors turns out to be evident also in orthonormal bases,since the diagonal matrix [gij ] takes an eigenvalue1.

4.4 Square Roots of operators and Polar Decomposition Theorem.

We racall some basic definitions and results which should be known by the reader from elemetarycourses of linear algebra.

IfA : VVis a (linear) operator on any finite-dimensional vector spaceV with field K = Ror C, an eigenvalueK ofA is a scalar such that

(A I)u= 0

for some u V\ {0}. In that case u is called eigenvector associated with . The set (A)containing all of the eigenvalues ofA is called the spectrumofA. TheeigenspaceE associ-ated with (A) is the subspace ofV spanned by the eigenvectors associated with .

Proposition 4.1. Let V be a real (complex) finite-dimensional equipped with a (resp. Her-mitean) scalar product (|). For every operator A : V V there exists exactly one of operatorA : VV, called theadjoint operatorofA, such that

(Au|v) = (u|Av) ,for allu, vV.

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Proof. Fix u V, the mapping v (u|Av) is a linear functional and thus an element of V.By Theorem 4.2 there is a unique element wu,AV such that (u|Av) = (wu,A|v) for all vV.Consider the map uwu,A. It holds, ifa, b are scalars in the field ofV and u, uV

(wau+bu,A|v) = (au+bu|Av) =a(u|Av)+b(u|Av) =a(wu,A|v)+b(wu,A|v) = (awu,A+bwu,A|v).Hence, for all vV:

(wau+bu,A awu,A bwu,A|v) = 0 ,The scalar product is nondegenerate by definition and this implies

wau+bu,A= awu,A+ bwu,A.

We have obtained that the mapping A : u wu,A is linear, in other words it is an operator.The uniqueness is trivially proven: if the operator B satisfies (Bu|v) = (u|Av) for all u, v V,it must holds ((B

A)u

|v) = 0 for all u, v

V which, exactly as we obtained above, entails

(B A)u= 0 for all uV. In other words B = A.

There are a few simple properties of the adjoint operator whose proofs are straightforward.Below A, B are operators in a real (complex) finite-dimensional vector space V equipped witha (resp. Hermitean) scalar product (|) and a, b belong

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