Hochschild cohomology: some methods for computations Mar´ ıa Julia Redondo * INMABB, Universidad Nacional del Sur Bah´ ıa Blanca, Argentina. Abstract We present some results on computing Hochschild cohomology groups. We describe the lower cohomology groups and provide several examples. In the particular case of hereditary algebras, radical square zero algebras and incidence algebras, we construct convenient projective resolutions that allow us to compute their cohomology groups. Finally, we show an inductive method to compute the Hochschild cohomology groups. 2000 Mathematics Subject Classification : 16E40, 16G20. Keywords : cohomology, Hochschild, finite–dimensional algebras. 1 Introduction These notes correspond to a series of three lectures given in the Workshop on Representations of Algebras that took place in S˜ ao Paulo in July, 1999, before the Conference on Representations of Algebras (CRASP). The purpose of these lectures was to present some results on computing Hochschild cohomology groups. Let A be a finite–dimensional k–algebra (associative, with unit) and let M be an A–bimodule. The Hochschild cohomology groups H i (A, M ) were introduced by Hochschild [19] in 1945. He considered the group of i–linear applications L i k (A, M ) and he defined a coboundary operator L i k (A, M ) → L i+1 k (A, M ) in analogy with the corresponding in algebraic topology. He proved that A is separable if and only if H i (A, M ) = 0 for any A–bimodule M , and that there is a one to one correspondence between H 2 (A, M ) and the set of equivalence classes of singular extensions of A by M . The low–dimensional groups (i ≤ 2) have a very concrete interpretation of classical algebraic structures such as derivations and extensions. Moreover, H 2 (A, A) has a close connection to algebraic geometry. It was observed by Gerstenhaber [15] that H 2 (A, A) controls the deformation theory of A, and it was shown that the vanishing of H 2 (A, A) implies that A is rigid, that is, any 1– parametric deformation is isomorphic to the trivial one [16]. The converse is not true in general, but it holds if we add the condition H 3 (A, A) = 0. There exists also a connection between Hochschild cohomology and the representation theory of finite–dimensional algebras. It is known that if A is of finite representation type (this means that there exists a finite number of non-isomorphic indecomposable A–modules) then A is simply connected if and only if A is representation-directed * The author is a researcher from CONICET, Argentina. 1
Transcript
Hochschild cohomology: some methods for computations
Marıa Julia Redondo !
INMABB, Universidad Nacional del SurBahıa Blanca, Argentina.
Abstract
We present some results on computing Hochschild cohomology groups. We describethe lower cohomology groups and provide several examples. In the particular case ofhereditary algebras, radical square zero algebras and incidence algebras, we constructconvenient projective resolutions that allow us to compute their cohomology groups.Finally, we show an inductive method to compute the Hochschild cohomology groups.
These notes correspond to a series of three lectures given in the Workshop on Representations ofAlgebras that took place in Sao Paulo in July, 1999, before the Conference on Representations ofAlgebras (CRASP).
The purpose of these lectures was to present some results on computing Hochschild cohomologygroups.
Let A be a finite–dimensional k–algebra (associative, with unit) and let M be an A–bimodule.The Hochschild cohomology groups Hi(A,M) were introduced by Hochschild [19] in 1945. Heconsidered the group of i–linear applications Li
k(A,M) and he defined a coboundary operatorLik(A,M) ! Li+1
k (A,M) in analogy with the corresponding in algebraic topology. He proved thatA is separable if and only if Hi(A,M) = 0 for any A–bimodule M , and that there is a one to onecorrespondence between H2(A,M) and the set of equivalence classes of singular extensions of Aby M .
The low–dimensional groups (i " 2) have a very concrete interpretation of classical algebraicstructures such as derivations and extensions. Moreover, H2(A,A) has a close connection toalgebraic geometry. It was observed by Gerstenhaber [15] that H2(A,A) controls the deformationtheory of A, and it was shown that the vanishing of H2(A,A) implies that A is rigid, that is, any 1–parametric deformation is isomorphic to the trivial one [16]. The converse is not true in general, butit holds if we add the condition H3(A,A) = 0. There exists also a connection between Hochschildcohomology and the representation theory of finite–dimensional algebras. It is known that if Ais of finite representation type (this means that there exists a finite number of non-isomorphicindecomposable A–modules) then A is simply connected if and only if A is representation-directed
!The author is a researcher from CONICET, Argentina.
1
and H1(A,A) = 0, see [18]. The importance of the simply connected algebras follows from the factthat usually we may reduce the study of indecomposable modules over an algebra to that for thecorresponding simply connected algebras, using Galois coverings.
Despite this very little is known about computations for particular classes of finite–dimensionalalgebras, since the computations of these groups by definition is rather complicated, and it hasbeen done only in particular situations where explicit formulas have been obtained. The aim ofthese notes is to show how some computations can be done in particular cases.
In Section 2 we provide an introduction to the subject, that is to say, given any associativek–algebra A with unit, with k a commutative ring, we define the Hochschild (co)–homology groupsof A with coe!cients in an A–bimodule M .
In Section 3 we consider the lower cohomology groups, that is, Hi(A,M) for i = 0, 1, 2. Thesegroups have a concrete interpretation in terms of classical algebraic structures such as derivationsand extensions. We provide several examples concerning algebras of the form kQ/I where Q is aquiver, kQ is its path algebra and I is an ideal of kQ. For basic information on this subject werefer the reader to [1], [21].
In Section 4 we consider finite dimensional algebras over an algebraically closed field k. Weprovide convenient projective resolutions of A over the enveloping algebra Ae, which allow us tocompute the Hochschild cohomology groups of hereditary algebras, radical square zero algebrasand incidence algebras.
In Section 5 we present an inductive method to compute the Hochschild cohomology groups oftriangular algebras. We use a result due to Happel that says that for one point extension algebrasA = B[M ] there exists a long exact sequence connecting the Hochschild cohomology groups of Aand B, see [18].
2 Definition of Hochschild cohomology groups
Let k denote a commutative ring with unit and let A be a k–algebra (associative with an identity).The enveloping algebra Ae is the k–algebra whose underlying k–module is A#k Aop with product(a# b)(a! # b!) = aa! # b!b. The following lemma shows the importance of the enveloping algebra:
Lemma 2.1 The category of A–bimodules is equivalent to the category of left (right) Ae–modules.
Proof: If M is an A–bimodule, we define a left (right) Ae–structure in the following way:
(a# b)m = amb, (m(a# b) = bma).
On the other hand, if N is an Ae–module, we define
am = (a# 1)m, mb = (1 # b)m.
The axioms are verified and this defines an equivalence. !
Example 2.2 The tensor product A"n = A #k · · · #k A of A n–times over k is an A–bimodule,with a(a1 # · · ·# an)b = aa1 # · · ·# anb . Hence A"n is an Ae–module.The map b!n#1 : A
"n+1 ! A"n, n $ 1, given by
b!n#1(a0 # · · ·# an) =n#1!
i=0
(%1)ia0 # · · ·# aiai+1 # · · ·# an
is a morphism of A–bimodules. Hence, it is a map of Ae–modules.
Since Im s generates A"n as an A–module and b! is a morphism of A–modules, then b!nb!n+1 = 0.!
Let M be an A–bimodule. If we apply the functor M #Ae . (respectively HomAe(.,M)) to theHochschild resolution of A, we get a complex whose homology (respectively cohomology) is theHochschild (co)–homology of A with coe!cients in M , Hi(A,M) (respectively Hi(A,M)).
Let us see in detail the definition of Hi(A,M). Applying the functor HomAe(.,M) to theHochschild resolution of A and using the isomorphism
HomAe(A"n,M) (= Homk(A"n#2,M)
given by f ! f , with f(x) = f(1# x# 1), we get the following isomorphism of complexes:
where the maps !n are defined so as to make all the squares commutative. It can be verifieddirectly that !n : Homk(A"n,M) ! Homk(A"n+1,M) is given by
(!nf)(a0 # · · ·# an) = a0f(a1 # · · ·# an)
+n#1!
i=0
(%1)i+1f(a0 # · · ·# aiai+1 # · · ·# an)
+ (%1)n+1f(a0 # · · ·# an#1)an.
Then Hi(A,M) (= Ker !i/ Im !i#1.
Remark 2.4
3
i) If A is k–projective, then A"n#2 is k–projective. Hence the Ae–module A"n is projective,for n > 1, since
A"n = A#k A"n#2 #k A (= A#k A
op #k A"n#2 = Ae #k A
"n#2.
Then (A"n, b!) is a projective resolution of A over Ae. So we may define the Hochschild(co)–homology of A with coe!cients in M in the following way:
Hi(A,M) = TorAe
i (M,A),
Hi(A,M) = ExtiAe(A,M).
It follows that, in this case, the Hochschild (co)–homology of A with coe!cients in M doesnot depend on the projective resolution we consider to compute it.
ii) Let D = Homk(., k) . For any A–bimodule M , we can define maps
" : Hi(A,D(M)) ! D(Hi(A,M)),
# : Hi(A,D(M)) ! D(Hi(A,M)).
If k is a field then " is an isomorphism, and if A is a finite dimensional k–algebra then # isalso an isomorphism, see [7, page 181].
iii) Assume that k is a field. If A is Ae–projective, by i) we have that Hi(A,M) = 0 for anyi > 0 and for any A–bimodule M . By ii), we deduce that Hi(A,M) = 0 for any i > 0 andfor any A–bimodule M .
In fact, the following conditions are equivalent:
a) A is Ae–projective,
b) Hi(A,M) = 0, &i > 0, for any A–bimodule M ,
c) Hi(A,M) = 0, &i > 0, for any A–bimodule M ,
d) A is separable.
So we are interested in determining when A is Ae–projective. Let µ : Ae ! A be the mapdefined by µ(a# b) = ab. Then µ is a morphism of A–bimodules.
Lemma 2.5 The Ae–module A is projective if and only if there exists an element e ) Ae suchthat µ(e) = 1 and ae = ea, for any a in A.
Proof: Assume that A is Ae–projective. Then the Ae–epimorphism
Ae µ! A ! 0
splits. Hence there exists an Ae–morphism $ : A ! Ae such that µ$ = idA. Let e = $(1). Thenµ(e) = µ$(1) = 1 and ae = a$(1) = $(a) = $(1)a = ea.The converse is immediate if we define the map $ by $(a) = ae. !
Example 2.6
4
i) Let A = Mn(k). The element
e =n!
i=1
ei1 # e1i
verifies the conditions of the previous lemma. Then Hi(Mn(k),M) = 0 = Hi(Mn(k),M)for i > 0 and for any Mn(k)–bimodule M .
ii) Let A = k[G], G a group with o(G) = n, such that n#1 ) k. Then the element
e =1
n
!
x%G
x#1 # x
verifies the conditions of the previous lemma. Hence Hi(k[G],M) = 0 = Hi(k[G],M) fori > 0 and for any k[G]–bimodule M .
iii) Let A = k[x]/ < xn >. We want to compute the Hochschild (co)–homology of A withcoe!cients in the A–bimodule A, Hi(A) = Hi(A,A) and Hi(A) = Hi(A,A). Since A isk–free, we may consider any projective resolution. Now,
· · · ! Ae dn! Ae ! · · · ! Ae d1! Ae µ! A ! 0
is a projective resolution of A over Ae, with d2i the multiplication by%n#1
i=0 xi # xn#1#i andd2i+1 the multiplication by 1# x% x# 1, see [17, page 54].If we apply the functors A #Ae . and HomAe(., A), using that A is commutative, we getcomplexes isomorphic to the following ones:
· · · ! Abn! A ! · · · ! A
b1! A ! 0,
0 ! Ab1
! A ! · · · ! Abn
! A ! . . .
where b2i = b2i is the multiplication by µ(%n#1
i=0 xi # xn#1#i) = nxn#1 and b2i+1 = b2i+1 isthe multiplication by µ(1# x% x# 1) = 0. Then
Hi(A) =
&
'
(
A, if i = 0,A/nxn#1A, if i is odd,Ann(nxn#1), if i is even and i > 0.
Hi(A) =
&
'
(
A, if i = 0,A/nxn#1A, if i is even and i > 0,Ann(nxn#1), if i is odd.
In particular, if 1n ) k, then
Hi(A) = Hi(A) =
"
A, if i = 0,A/nxn#1A, if i > 0.
If n = 0 in k then Hi(A) = Hi(A) = A for any i $ 0.In fact these computations may be generalized for any monic polynomial f ) k[x], see [17,page 54], and we get
Hi(A) =
&
'
(
A, if i = 0,A/f !A, if i is odd,Ann(f !), if i is even and i > 0.
5
Hi(A) =
&
'
(
A, if i = 0,A/f !A, if i is even and i > 0,Ann(f !), if i is odd.
Remark 2.7
i) Let A, B be k–algebras, M any A–bimodule and N any B–bimodule. Then, for any i $ 0,
Hi(A*B,M *N) = Hi(A,M)+Hi(B,N),
Hi(A*B,M *N) = Hi(A,M)+Hi(B,N),
see [22, page 305]. Hence, we may just consider indecomposable algebras.
ii) The Hochschild (co)–homology is invariant under Morita equivalence: given k–algebras Aand B such that modA is equivalent to modB, then Hi(A) (= Hi(B) and Hi(A) (= Hi(B),for any i $ 0, see [22, page 328]. Hence, we may just consider basic algebras.
3 Interpretation of the lower cohomology groups
Recall that the Hochschild cohomology of A with coe!cients in M is the cohomology of thefollowing complex:
0 ! M!0
! Homk(A,M)!1
! Homk(A"2,M)
!2
! Homk(A"3,M) ! . . .
3.1 The 0–Hochschild cohomology group
We have
H0(A,M) = Ker(!0)
= {m ) M : !0(m) = 0}
= {m ) M : !0(m)(a) = am%ma = 0, &a ) A}.
In particular, H0(A) = Z(A) the center of A.
3.2 The first Hochschild cohomology group
We have H1(A,M) = Ker(!1)/ Im(!0). Now,
Ker(!1) = {f ) Homk(A,M) : !1(f)(a# b) = af(b)% f(ab) + f(a)b = 0, &a, b ) A}
= Derk(A,M)
is the space of derivations of A in M , and
Im(!0) = {f ) Homk(A,M) : f = !0(m),m ) M}
= {fm ) Homk(A,M),m ) M : f(a) = am%ma}
= Der0k(A,M)
is the space of inner derivations of A in M . Then H1(A,M) (= Derk(A,M)/Der0k(A,M).
6
Example 3.1 Let A = k[x]/ < x2 >. The k–linear map ! : A ! A given by !(a + bx) = bx is aderivation. Since A is commutative, Der0k(A,A) = 0. Hence H1(A) (= Derk(A,A) ,= 0.
Example 3.2 Let A = kQ/J2, with J the ideal generated by the arrows. We want to show that ifH1(A) = 0 then Q is a tree (the underlying graph has no cycles).Assume Q0 = {1, . . . , n} and Q1 = {%1, . . . ,%r}. If Q is not a tree, there exists an arrow % ) Q1
such that Q \ {%} is connected. Suppose that % = %1. We may define a derivation ! : A ! A by!(ei) = 0 for i = 1, . . . , n,!(%1) = %1
!(%i) = 0 for i = 2, . . . , r, and we extend by linearity.Let us see that ! is not an inner derivation. If it were, there would exist x ) A such that ! = !xand !(a) = ax% xa for any a ) A. Let x =
%ni=1 &iei +
%rj=1 µj%j, &i, µj ) k. Then
%1 = !(%1) = %1x% x%1 = (&s("1) % &e("1))%1
0 = !(%i) = %ix% x%i = (&s("i) % &e("i))%i, i = 2 . . . , r.
So &s("1) % &e("1) = 1 and &s("i) % &e("i) = 0 for i = 2, . . . , r. But this is a contradiction sinceQ \ {%1} connected implies that &i = &j, &i, j ) Q0.Hence ! is a derivation which is not inner, so H1(A) ,= 0.In fact, the following general result holds (see [18, page 114]): if A = kQ/J2, the followingconditions are equivalent
a) Hi(A) = 0, &i $ 1;
b) H1(A) = 0;
c) Q is a tree.
Example 3.3 Assume k has characteristic zero, A = kQ/I, I an homogeneous ideal (this meansthat I is generated by linear combinations of paths that have the same length). We want to showthat if H1(A) = 0 then Q has no oriented cycles.We may define ! : kQ ! kQ by !(w) = l(w).w, where l(w) is the length of the path w, and extendby linearity. A direct computation shows that ! is a derivation of kQ. Since I is homogeneous, !induce a derivation in A, ! : A ! A.Since H1(A) = 0, ! must be inner. Hence there exist a ) A such that !(x) = xa % ax for allx in A. Now !(ei) = l(ei)ei = 0 for all i, so aei = eia for all i. Hence a =
%
aiei + y, withy ) +ei(radA)ei.Take % ) Q1. Then
% = !(%) = %a% a% = (as(") % ae("))%+ y%% %y.
Since y%% %y ) rad2 A, we have that as(") % ae(") = 1 for any arrow % ) Q1.Assume that Q has an oriented cycle 1 ! 2 ! · · · ! n ! 1. Then
&
)
)
'
)
)
(
a2 % a1 = 1a3 % a2 = 1. . .a1 % an = 1
is a linear system that has no solution if k has characteristic 0.Using this result we get that for algebras A with rad3 A = 0, the vanishing of the first Hochschildcohomology group implies that its asociated quiver Q has no oriented cylces.
7
For some time it was suspected that the vanishing of the first Hochschild cohomology groupimplies that the corresponding quiver has no oriented cycles. Now it is known that this is not true(see [4]).
3.3 The second Hochschild cohomology group
Recall that H2(A,M) = Ker !2/ Im !1. Now,
Ker !2 = {f : A#A ! M : !2(f) = 0}= {f : A#A ! M : af(b# c)% f(ab# c) + f(a# bc)% f(a# b)c = 0}
and
Im !1 = {f : A#A ! M : f = !1(g), g ) Homk(A,M)}= {f : A#A ! M : f(a# b) = ag(b)% g(ab) + g(a)b, g ) Homk(A,M)}.
Definition 3.4 An extension of A is a k–algebra epimorphism " : B ! A that is k–split.
Let M be the kernel of ". Since M is a two–sided ideal of B, then M has an structure ofB–module. The product in B induces a product in M . If this product is such that M2 = 0 thisallows us to consider M as an A–bimodule in the following way:
a.m = b.m, if "(b) = a,
m.a = m.b, if "(b) = a. (1)
Observe that this is well defined since "(b) = "(b!) implies that b% b! ) M , so (b% b!)m = 0 sinceM2 = 0.On the other hand, if M has an structure of A–bimodule satisfying (1), then the product in Minduced by the product in B is zero.
Definition 3.5 Let A be a k–algebra, M an A–bimodule. An extension of A by M is a short exactsequence
0 ! Mi! B
#! A ! 0
with " an epimorphism of algebras that is k–split, i a monomorphism of k–modules such that
i("(b).m) = b.i(m),
i(m."(b)) = i(m).b, &b ) B,m ) M. (2)
Two extensions of A by M are said to be equivalent if there exists a commutative diagram
0 %%%%! Mi
%%%%! B#
%%%%! A %%%%! 0*
*
*F
#
#
$
*
*
*
0 %%%%! Mi
%%%%! B! #%%%%! A %%%%! 0
with F a morphism of algebras (necessarily isomorphism).
Remark 3.6 The conditions (2) are simply a translation of (1) when i is the inclusion M '! B.
8
Proposition 3.7 The set Ext(A,M) of isomorphic classes of extensions of A by M is in naturalbijection with H2(A,M).
Proof: Let0 ! M
i! B
#! A ! 0
be an extension of A by M , and let ( : A ! B be the k–linear map such that "( = idA. ThenB (= A+M as k–modules.If ( is an algebra morphism, B (= A!M as k–algebras, with (a,m).(b, n) = (ab, an+mb). In thiscase, B is said to be the trivial extension of A by M .In general, ( is not an algebra morphism. The failure of ( to be a morphism is measured by
f(a# b) = ((a)((b)% ((ab).
Since " is a morphism of algebras, we have
"(f(a# b)) = ab% ab = 0
and ( is k–linear, so f : A # A ! M . Now, B is completely determined by A, M and f as thek–module A+M with multiplication (a,m).(b, n) = (ab, an+mb+f(a#b)). We write B (= A!fM .Derived from the associative law, we have that f satisfies
f(a# b)c+ f(ab# c) = af(b# c) + f(a# bc).
This shows f to be in Ker !2.Hence we have a surjective map
Ker !2 ! Ext(A,M).
Two extensions A!f1 M , A!f2 M are equivalent if and only if there exists a commutative diagram
0 %%%%! Mi
%%%%! A!f1 M#
%%%%! A %%%%! 0*
*
*F
#
#
$
*
*
*
0 %%%%! Mi
%%%%! A!f2 M#
%%%%! A %%%%! 0
with F a morphism of algebras.The commutativity of this diagram implies that F (a,m) = (a,m + g(a)), for g ) Homk(A,M).Now, F is a morphism of algebras if and only if
f1(a# b)% f2(a# b) = ag(b)% g(ab) + g(a)b, &a, b ) A.
This is just the condition for f1 % f2 to be in Im !1. !
Remark 3.8
1) The trivial extension of A by M corresponds to the zero element in H2(A,M).
2) If A = kQ/J2, then H2(A) = 0 if and only if Q does not contain loops, does not containnon–oriented triangles, and Q is not 1
&#%! 2, see [9, page 213]. This says that if Q satisfies
the hypothesis we have just mentioned, any extension of A by A splits.
9
4 Convenient projective resolutions of A over Ae
From now on A will denote a finite dimensional algebra over an algebraically closed field k. More-over, we will assume that A is basic and connected. For information on this subject see [1].
4.1 Minimal projective resolution
Let {e1, . . . , en} be a complete set of primitive orthogonal idempotents of A. Then {ei#ej}1'i,j'n
is a complete set of primitive orthogonal idempotents of Ae.So {P (i, j) = Ae(ei # ej) - Aei #k ejA} is a complete set of representatives from the isomorphismclasses of indecomposable projective Ae–modules.
Lemma 4.1 [18, page 110] Let
· · · ! Rm ! Rm#1 ! · · · ! R1 ! R0 ! A ! 0
be a minimal projective resolution of A over Ae. Then
Rm =+
i,j
P (i, j)dimk ExtmA (Sj ,Si).
Proof: Let Rm =,
i,j P (i, j)rij . Denote S(i, j) = top P (i, j) the corresponding simple Ae–module. Observe that S(i, j) - Homk(Sj , Si). Then by definition we have that
rij = dimk ExtmAe(A,S(i, j))
= dimk ExtmAe(A,Homk(Sj , Si))
= dimk ExtmA (Sj , Si)
The last equality follows from [7, Corollary 4.4, page 170]. !
The projective resolution constructed above allows us to get the immediate following conse-quences:
Proposition 4.2 pdAe A = gl.dimA.
Proposition 4.3 [8] Let A = kQ/I, Q with no oriented cycles. Then
Hi(A) =
"
k|Q0| if i = 0,0 if i ,= 0.
Proof: This follows from the fact that applying the functor A #Ae . to the minimal projectiveresolution given in the previous lemma, we may identify
A#Ae P (i, j) = A#Ae Ae(ei # ej) - ejAei.
But ExtmA (Sj , Si) ,= 0 for some m $ 1 implies that there is a path in Q from j to i. Since Q hasno oriented cycles, then ejAei = 0. Hence A#Ae Rm = 0 for all m $ 1. !
Proposition 4.4 [18, page 111] Let A be a basic indecomposable finite dimensional hereditaryalgebra, this means, A = kQ, Q connected without oriented cycles. Then
dimk Hi(A) =
&
'
(
1 if i = 0,0 if i > 1,1% n+
%
"%Q1dimk ee(")Aes(") if i = 1.
where n = |Q0| and ee(")Aes(") is the subspace of A generated by all the paths from s(%) to e(%).
10
Proof: Clearly H0(A) = Z(A) = k since Q is connected and has no oriented cycles.Since A is hereditary, we have that gl.dimA " 1, so Rm = 0 for all m $ 2. Hence Hi(A) = 0 fori $ 2 and
0 ! R1 ! R0 ! A ! 0
is the minimal projective resolution of A overAe, with R0 =,
i%Q0P (i, i) andR1 =
,
"%Q1P (e(%), s(%)),
because dimk Ext1A(Si, Sj) coincides with the number of arrows from i to j. Applying HomAe(., A)
to the previous exact sequence, we get
0 ! HomAe(A,A) ! HomAe(R0, A) ! HomAe(R1, A) ! 0.
ButHomAe(A,A) - k,
HomAe(R0, A) -+
i%Q0
eiAei - kn
andHomAe(R1, A) -
+
"%Q1
ee(")Aes(").
Thus dimk H1(A) = 1% n+%
"%Q1dimk ee(")Aes(") . !
Corollary 4.5 Let A = kQ, Q without oriented cycles. Then H1(A) = 0 if and only if Q is atree.
Remark 4.6
1) Locateli describes the minimal resolution considered in Lemma 4.1 in the particular case oftruncated algebras A = kQ/Jm, and she computes the corresponding Hochschild cohomologygroups [20].
2) Butler and King [6] and Bardzell [2] describe the morphisms of this minimal resolution inparticular cases (monomial algebras, truncated algebras, Koszul algebras).
3) The equation for the dimension of the first Hochschild cohomology group given in Proposition4.4 holds in a more general context. In fact,
dimk H1(A) = dimk Z(A)%
!
i%Q0
dimk eiAei +!
"%Q1
dimk ee(")Aes(")
if A = kQ/I and
a) I = Jm, see [3, 20]; or
b) the ideal I is pregenerated, that is, eiIej = eikQej or ei(IJ + JI)ej for any i, j ) Q0,see [10, page 647] and [13]; or
c) A is schurian and semi–commutative, that is, dimk HomA(P, P !) " 1 for any indecom-posable projective modules P, P ! and if w,w! are two paths in Q sharing starting andending points, w ) I implies w! ) I, see [18, page 113].
Example 4.7
11
1) Let A = Tn(k) be the n * n–upper triangular matrices over k. Then A is an hereditaryalgebra and the ordinary quiver associated
Q : 1 ! 2 ! · · · ! n
is a tree. So
Hi(A) =
"
k if i = 0,0 if i ,= 0.
2) Let A = kQ, with Q0 = {1, 2} and Q1 = {%i : 1 ! 2}1'i'm. So
dimk Hi(A) =
&
'
(
1 if i = 0,0 if i > 1,1% 2 +
%mi=1 m = m2 % 1 if i = 1.
3) Let A = kQ/Jm, with Q the oriented cycle
1 ! 2 ! · · · ! n ! 1
Then dimk H1(A) = 1% n+ n = 1.
Recall that a left A–module T is called a tilting module if pdT < ., ExtiA(T, T ) = 0 for alli > 0 and there exists an exact sequence 0 ! A ! T 0 ! · · · ! T d ! 0 with T i ) addT .
Theorem 4.8 Let A be a finite dimensional k–algebra, T a tilting left A–module. Let B =EndA(T ). Then Hi(A) - Hi(B).
Proof: It is known that if B = EndA(T ), T a tilting A–module, then there is an isomorphismbetween the corresponding derived categories, "B,T,A : Db(A) - Db(B). Using this isomorphism,we may construct an isomorphism between the derived categories of the enveloping algebras Ae
and Be.In fact,
i) A#k T is a tilting A#k Bop–module and Ae - EndA"kBop(A#k T )
ii) T #k Bop is a tilting A#k Bop–module and Be - EndA"kBop(T #Bop)
So the map F : Db(Ae) ! Db(Be),
F = "#1Ae,A"kT,A"kBop "Be,T"kBop,A"kBop
is the desired isomorphism. Moreover, F (A) = B and F commutes with the shift. Hence
Hi(A) = ExtiAe(A,A) = HomDb(Ae)(A,A[i]),
HomDb(Be)(F (A), F (A[i])) = HomDb(Be)(F (A), F (A)[i]) = Hi(B)
and
HomDb(Ae)(A,A[i])F- HomDb(Be)(F (A), F (A[i])).
!
Corollary 4.9 Let A be a finite dimensional k–algebra, A = A0, A1, . . . , Am = kQ, Ti tilting Ai–modules, Ai+1 = EndAi(Ti), Q without oriented cycles. Then Hi(A) = 0 for all i $ 2, H0(A) = k,and H1(A) = 0 if and only if Q is a tree.
12
4.2 The resolution of the radical
The resolution we are going to construct in this section may be used to connect Hochschild coho-mology with simplicial cohomology.
Let A = kQ/I, E the subalgebra of A generated by the set of verticesQ0. Then E is semisimple,commutative and A = E + radA in the category of E–bimodules.
Lemma 4.10 Let A = kQ/I, A = E + radA. Then
· · · ! A#E (radA)"En #E Ab!! A#E (radA)"En#1 #E A ! · · · ! A#E radA#E A
b!! A#E A ! A ! 0
is a projective resolution of A over Ae, with b! the Hochschild boundary.
Proof: The boundary b! is well defined and (b!)2 = 0. The sequence is exact since the maps : A #E (radA)"En #E A ! A #E (radA)"En+1 #E A given by s(a # x) = 1 # a # x, forx ) (radA)"En #E A, a = e+ a ) E + radA, satisfies the equation b!s+ sb! = 1.On the other hand, A#E (radA)"En#EA - A#EAop#E (radA)"En, (radA)"En is E–projectiveand A#E Aop is A#k Aop–projective, hence A#E (radA)"En #E A is Ae–projective. !
4.3 Radical square zero algebras
The resolution above allows us to compute completely the Hochschild cohomology of radical squarezero algebras, that is, algebras of the form kQ/J2.
In fact, since rad2 A = 0, all the middle–sum terms of the boundary b! vanish, so
Theorem 4.11 [12, page 96] Let Q be a connected quiver, Q is not an oriented cycle. Then
dimk Hn(kQ/J2) =
&
'
(
1 + |Q1||Q0| if n = 0,|Q1||Q1|% |Q0||Q0|+ 1 if n = 1,|Qn||Q1|% |Qn#1||Q0| if n > 1,
where Qm is the set of paths in Q of length m and Qi||Qj = {((, (!) ) Qi*Qj : (, (!parallel paths}.
Corollary 4.12 [12, page 98] Let Q be a connected quiver, Q is not an oriented cycle. Then+n(0Hn(kQ/J2) is a finite dimensional vector space if and only if the quiver Q has no orientedcycles.If Q has an oriented cycle of length c, then Hcn+1(kQ/J2) ,= 0, for any n > 0.
Remark 4.13 The last result has been generalized by Locateli [20, page 660] for truncated algebras,that is, algebras of the form A = kQ/Jm.
Theorem 4.14 [12, page 98] If Q is the oriented cycle 1 ! 2 ! · · · ! m ! 1, and chark ,= 2then
i) if m = 1, A = k[x]/ < x2 > and
Hn(kQ/J2) =
"
A if n = 0,k if n > 0,
13
ii) if m > 1,
Hn(kQ/J2) =
"
k if n = 0, 2sm, 2sm+ 1 for any s ) N,0 otherwise.
Theorem 4.15 [12, page 98] If Q is the oriented cycle 1 ! 2 ! · · · ! m ! 1, and chark = 2then
i) if m = 1, A = k[x]/ < x2 > and Hn(kQ/J2) = A for any n $ 0,
ii) if m > 1,
Hn(kQ/J2) =
"
k if n = 0, sm, sm+ 1 for any s ) N,0 otherwise.
4.4 Incidence algebras
Let (P,") be a finite partially ordered set (poset). Without loose of generality, we may assumethat P = {1, 2, . . . , n}. Let I(P ) be the subalgebra of the square matrices over k, Mn(k), suchthat
I(P ) = {(aij) ) Mn(k) : aij = 0 if i " j}.
Then I(P ) is the so–called incidence algebra associated to the poset P .The ordinary quiver associated to an incidence algebra I(P ) is given as follows: the set of verticesQ0 is P , and there is an arrow i ! j in Q1 whenever i > j and there is no s ) P such thati > s > j. We say that two paths are parallel if they have the same starting and ending points.Then I(P ) = kQ/I, where I is the ideal generated by di"erences of parallel paths.
Example 4.16
1) The lower triangular square matrices algebra Tn(k) is an incidence algebra associated to theposet P = {1 " 2 " · · · " n}.
2) Let P = {1, 2, 3, 4} and 1 " 2 " 4, 1 " 3 " 4. So I(P ) = kQ/I with
Given any poset P , we may associate a simplicial complex #P = (Ci, di) with Ci = {s0 > s1 >· · · > si : sj ) P} the set of i–simplices. Let kCi be the k–vector space with basis the set Ci. Thecohomology of #P with coe!cients in k is the cohomology of the following complex:
Proof: Denote A = I(P ). We apply the functor HomAe(., A) to the resolution of the radical andwe use the following identification
HomAe(A#E (radA)"En #E A,A) - HomEe((radA)"En, A) - Homk(kCn, k).
These isomorphisms follow from the fact that the ideal I identifies parallel paths, and
radA = +s>tetAes - +s>tk
since dimk etAes = 1. Hence
(radA)"En = radA#E · · ·#E radA
= +s0>s1>···>snesnAesn"1#k esn"1
Aesn"2#k · · ·#k es1Aes0
= +s0>s1>···>snesnAes0
Moreover, these isomorphisms commute with the boundaries, hence we have a complex isomor-phism. !
Remark 4.18 The previous result says that the computation of the Hochschild cohomology is atleast as complicated as the computation of the cohomology of simplicial complexes.
Example 4.19 Consider the incidence algebra I(P ) given by the quiver
Then the corresponding simplicial complex is #P - Sn the n–sphere, and
Hi(I(P )) =
"
k if i = 0, n,0 otherwise.
To any poset (P,") we are going to associate a new poset P adding two new elements a, bsuch that a > u > b for any u ) P . If A = I(P ) and A = I(P ) are the corresponding incidencealgebras, and A = kQ/I, then A = kQ/I, where Q is the quiver Q with two new vertices a, b anda new arrow from a to each source vertex of Q and a new arrow from each sink vertex to b.
Theorem 4.20 [11, page 225] Hi(I(P )) - Exti+2A
(Sa, Sb) for any i $ 1.
Proof: SinceExti+2
A(Sa, Sb) - Hi+2(A,Homk(Sa, Sb))
15
and Homk(Sa, Sb) - ebAea as A–bimodules, we use the resolution of the radical to computeHi+2(A, ebAea). There is an isomorphism
HomEe((radA)"Ei+2, ebAea) - HomEe((radA)"E i, A) - Homk(kCi, k)
that follows from the fact that the paths in Q from a to b, that is, the i + 2–simplices a >s1 > · · · > si > b, are in correspondence with the paths in Q corresponding to the i–simplicess1 > · · · > si. Moreover, these isomorphisms commute with the corresponding boundaries, hencewe get the desired result using Theorem 4.17. !
Corollary 4.21 [14] If P is a poset with a unique maximal (minimal) element then Hi(I(P )) = 0,for all i $ 1.
Proof: Let x be the unique maximal element in P . Then
0 ! Px ! Pa ! Sa ! 0
is a projective resolution of Sa over A. So ExtjA(Sa, .) = 0 for any j $ 2. !
J.C. Bustamante has obtained a nice generalization of the previous result, see [5].
and I is the parallel ideal. Then Hi(A) = 0 for any i $ 1.
5 Inductive method to compute Hochschild cohomology of
triangular algebras
An algebra A is said to be triangular if the corresponding quiver has no oriented cycles. In thiscase, the quiver has sinks and sources, and this allows us to describe A as a one–point extension(co–extension) algebra.Let B be a finite dimensional k–algebra, M a left B–module. The one–point extension A = B[M ]of B by M is by definition the finite dimensional k–algebra
B[M ] =
-
k 0M B
.
with multiplication
-
a 0m b
.-
a! 0m! b!
.
=
-
aa! 0ma! + bm! bb!
.
where a, a! ) k, m,m! ) M
and b, b! ) B.
16
Proposition 5.1 Let A be an algebra, Q its ordinary quiver. The following assertions are equiv-alent:
i) A is a one–point extension algebra;
ii) there is a simple injective A–module S;
iii) there is a vertex i ) Q0 which is a source.
Proof: i) ! ii) Assume that A = B[M ] is a one point extension algebra of B by M . ThenS = D(e11A) is a simple injective A–module, where
e11 =
-
1 00 0
.
ii) ! iii) Since the injective module S = D(eiA) is simple, then the corresponding vertex i is asource, that is to say, there is no arrow % ) Q1 ending at i.iii) ! i) Suppose there is a source i ) Q0 and let M = radAei and B = A/ < ei >, where < ei >denotes the two–sided ideal in A generated by the idempotent ei. Then M is a B–module, and itis easily checked that A - B[M ].
Example 5.2
1) Let B be the hereditary algebra with ordinary quiver
and let M be the B–module with representation M(1) = 0, M(2) = k, M(3) = k andM(4) = 0. Then A = B[M ], the one–point extension algebra of B by M , is the algebrakQA/IA with ordinary quiver QA
2) Tn(k) the algebra of n*n–upper triangular matrices over k is a one–point extension algebraof Tn#1(k) by the Tn#1(k)–module M = radTn(k)e11.
Theorem 5.3 [18, page 124] Let A = B[M ] be a one point extension algebra of B by M . Thenthere exists the following long exact sequence connecting the Hochschild cohomology of A and B
ii) The Ae–module I is projective, so ExtiAe(I, .) = 0 for all i > 0. Moreover, HomAe(I, B) = 0.So, applying HomAe(., B) to (3) we get that ExtiAe(A,B) = ExtiAe(B,B). But Be is a convexsubcategory of Ae, so
since I - Homk(S0, P0)) as A–bimodules, and the last equality follows from [7, Corollary4.4, page 170]. So,
a) applying the functor HomA(., P0) to (4) we get that Exti+1A (S0, P0) = ExtiA(M,P0) for
all i > 0 and Ext1A(S0, P0) = Homk(M,P0)/Homk(P0, P0).
b) since S0 is A–injective and HomA(M,S0) = 0, applying the functor HomA(M, .) to (4)we get that ExtiA(M,P0) = ExtiA(M,M) for all i $ 0. !
Example 5.4
i) Let A = kQ/I be the algebra with Q0 = {1, 2}, Q1 = {%,)}, where % : 1 ! 2, ) : 2 ! 2.Let I =< )2 >. Then A = B[M ], where B = k[x]/ < x2 > and M = radP1. Now, M isB–projective, HomB(M,M) = k2, H0(B) = k2 and H0(A) = k. So Hi(A) = Hi(B) for alli > 0, and we have already computed Hi(B) in Theorem 4.14.
and I is the ideal generated by )%. Then A = B[M ], B = A/ < e1 > an hereditaryalgebra, M the B–module with representation M(2) = k, M(3) = 0, M(4) = k, M(5) = k,M(!) = idk, M(*) = idk. Now,
0 ! P3 ! P2 ! M ! 0
is the projective resolution of M over B. Applying HomB(.,M) to this short exact sequence,we get the long exact sequence
But HomB(M,M) = k, HomB(P2,M) = M(2) = k and HomB(P3,M) = M(3) = 0. SoExtiB(M,M) = 0 for all i > 0. By the previous theorem, we have that Hi(A) = Hi(B) forall i $ 0. Since B is hereditary, we know that H0(B) = H1(B) = k and Hi(B) = 0 for alli > 1, see Proposition 4.4.
and I is the ideal generated by ()% % *)%. Then A = B[M ], B = A/ < e1 > an hereditaryalgebra, M the B–module with representation M(2) = k, M(3) = k, M(4) = k, M(5) = k,M()) = idk, M(() = idk, M(*) = idk, M(!) = idk. Now,
0 ! S5 ! P2 ! M ! 0
is the projective resolution of M over B. Applying HomB(.,M) to this short exact sequence,we get the long exact sequence
But HomB(M,M) = k, HomB(P2,M) = M(2) = k and HomB(S5,M) = k. So Ext1B(M,M) =k and ExtiB(M,M) = 0 for all i > 1. Since B is hereditary, we know, from Proposition 4.4,that H0(B) = H1(B) = k and Hi(B) = 0 for all i > 1. From Theorem 5.3 we have thatHi(A) = Hi(B) for i = 0 and i > 1, and we also have the exact sequence
0 ! H1(A) ! H1(B) ! Ext1B(M,M) ! H2(A) ! 0.
So H1(A) = H2(A) and dimk H1(A) " dimk H1(B) = 1. Hence H1(A) = H2(A) = 0 or k.In fact, A is a tilted algebra, that is, A - EndkQ(T ), with Q the quiver
that is a tree. So H1(A) = H1(kQ) = 0, see Theorem 4.8 and Corollary 4.5. This says thatthe non–inner derivations in B can not be extended to A.
19
3) Let A = I(P ) be the incidence algebra associated to the poset P . Let P = P /{a} be the posetsuch that a > u for all u ) P . Let A = I(P ) = A[M ], with M = radPa. Since Hi(A) = 0for all i > 0 (see Corollary 4.21) and Endk(M) = k, then Hi(A) = ExtiA(M,M).
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