Hochschild cohomology

Hochschild cohomology Seminar talk complementing the lecture Homological algebra and applications by Prof. Dr. Christoph Schweigert in winter term 2011. by Steffen Thaysen Inhaltsverzeichnis 9. Juni 2011 1 Simplicial methods in homological algebra 1 2 Hochschild homology and cohomology for algebras 2 2.1 Hochschild homology and cohomology in degree 0........... 3 2.2 Hochschild homology and cohomology in degree 1........... 4 3 Relative Ext 6

1 SIMPLICIAL METHODS IN HOMOLOGICAL ALGEBRA 1 1 Simplicial methods in homological algebra To make life a little bit easier I will give a short introduction to simplicial and cosimplicial objects and its associated (co-)chain complex. In fact, Hochschild cohomology can be described by the cohomology of a cochain complex associated to a cosimplicial bimodule, that is a cosimplicial object in the category of bimodules. Definition. Let be the category whose objects are the finite ordered sets [n] = {0 <... < n} for integers n 0 and whose morphisms are nondecreasing monotone functions. For any category A we call a contravariant functor from to A, that is A: op A, a simplicial object in A. For simplicity we write A n for A([n]). Similarly a covariant functor C: A is called a cosimplicial object in A and we write C n for C([n]). One can check, that every nondecreasing monotone function can be written as a composition of so called face maps ǫ i : [n] [n+1] and degeneracy maps η i : [n] [n 1], for i = 1,...n, given by ǫ i (j) = { j, if j < i j +1, if j i, These maps fulfil the simplicial identities: η i (j) = ǫ j ǫ i = ǫ i ǫ j 1 if i < j η j η i = η i η j+1 if i j ǫ i η j 1, if i < j η j ǫ i = identity if i = j,j +1 ǫ i 1 η j if i > j +1. { j, if j i j 1, if j > i. Therefore, to give a simplicial Object A, it is sufficient to give the objects A n and face operators i : A n A n 1 and degeneracy operators σ i : A n A n+1 fulfilling the simplicial identities. Under this correspondence i = A(ǫ i ) and σ i = A(η i ). Definition. Let A be a simplicial object in an abelian category A. The associated, or unnormalized, chain complex C = C(A) has C n = A n, and its boundary operator is given by d := ( 1) i i : C n C n 1. In the same way we may obtain a cochain complex for a cosimplicial obejct in an abelian category. There is also a normalized chain complex associated to a simplicial object, N(A), which is a subcomplex of the unnormalized chain complex C(A). By the Dold- Kan correspondence there is an equivalence of categories betweem the category of simplicial objects in A and the category of chain cochain komplexes C in A with C n = 0 for n < 0. We check that in C we have d 2 = 0, since j i = i j 1 we get

2 HOCHSCHILD HOMOLOGY AND COHOMOLOGY FOR ALGEBRAS 2 d 2 = d( n+1 = ( 1) i i ) n+1 j 1 = + j=1 n+1 j 1 = ( 1) i+j i j 1 + j=1 n+1 i 1 = 1 + = = i=1 = 0. i ( 1) i+j+1 j i + ( 1) i+j+1 j i + 2 Hochschild homology and cohomology for algebras To get in touch with Hochschild cohomology we need to fix a unitary, commutative ring k. We will write for k and R n for the n-fold tensor product R R. Let R be a unitary k-algebra and M a R-R bimodule. We obtain a simplicial k- module,i.e.asimplicialobjectinthecategoryk-mod,m R with[n] M R n (M R 0 = M) - these modules are even k k bimodules -, by declaring mr 1 r 2 r n, if i = 0 i (m r 1 r n ) = m r 1 r i r i+1 r n, if 0 < i < n r n m r 1 r n 1 if i = n σ i (m r 1 r n ) = m r 1 r i 1 r i+1 r n, where m M and r i R. These formulas are obviously k-multilinear, so the i and σ i are well-defined homomorphisms. One can check that these homomorphisms respect the simplicial identities. So it makes sense to take a look at the associated chain complex of R R bimodules C(M R ), which looks as follows: 0 M 0 1 M R d M R R d...

2 HOCHSCHILD HOMOLOGY AND COHOMOLOGY FOR ALGEBRAS 3 Definition. The Hochschild homology H (R,M) of R with coefficients in M is defined to be the k-modules H n (R,M) = H n C(M R ). Wealsoobtainacosimplicialk-modulewith[n] Hom k (R n,m)(hom k (R 0,M) = M) by declaring r 0 f(r 1,...,r n ), if i = 0 ( i f)(r 0,...,r n ) = f(r 0,...,r i r i+1,...,r n ), if 0 < i < n f(r 0,...,r n 1 )r n if i = n (σ i f)(r 1,...,r n 1 ) = f(r 1,...,r i,1,r i+1,...,r n 1 ). The associated cochain complex look like this: 0 M 0 1 Hom k (R,M) d Hom k (R R,M) d... Definition. The Hochschild cohomology of R with coefficients in M is defind to be the k-moduls H n (R,M) = H n C(Hom k (R,M)). 2.1 Hochschild homology and cohomology in degree 0 Proposition 2.1. Let k be a commutative ring, R be a k-algebra and M a R R bimodule, then H 0 (R,M) = M/[M,R]. Proof. To calculate H 0 (R,M) we need to look at the chain complex C(M R ) and i.e. the image of 0 1, which is given by elements of the form thus ( 0 1 )(m r) = 0 (m r) 1 (m r) = mr rm, H 0 (R,M) = M/[M,R]. In particular, we get H 0 (R,R) = R/[R,R]. Proposition 2.2. Let k be a commutative ring, R be a k-algebra and M a R R bimodule, then H 0 (R,M) = {m M rm = mr, r R}. Proof. To calculate H 0 (R,M) we take a look at the kernel of 0 1 in the cochain complex C(Hom k (R,M). An element m M is inside the kernel if for all r R. So 0 = ( 0 1 )(m)(r) = ( 0 m)(r) ( 1 m)(r) = mr rm, r R H 0 (R,M) = {m M rm = mr, r R}. In particular, we get H 0 (R,R) = Z(R).

2 HOCHSCHILD HOMOLOGY AND COHOMOLOGY FOR ALGEBRAS 4 2.2 Hochschild homology and cohomology in degree 1 Now, to investigate H 1 (R,M), we take a look at the kernel of d: Hom k (R,M) Hom k (R R,M) which consists of k-linear function f: R M satisfying the identity f(rs) = rf(s)+f(r)s for all r,s R, since 0 = ( 0 f)(r s) ( 1 f)(r s)+( 2 f)(r s) = rf(s) f(rs)+f(r)s. Such a function is called a k-derivation and the k-module of all k-derivations is denoted by Der k (R,M). The image of d: M Hom k (R,M) is given by the k-linear functions f m : R M defined by f m (r) = rm mr, since ( 0 m)(r) ( 1 m)(r) = rm mr. These functions are also k-derivations, beacause rf m (s)+f m (r)s = r(sm ms)+(rm mr)s = rsm rms+rms mrs = (rs)m m(rs) = f m (rs) for all r,s R, this fact is also given by d 2 = 0. A k-derivation of this type is called a principal k-derivation and we denote the submodule of principal k-derivations by PDer k (R,M). Thus we obtain Proposition 2.3. H 1 (R,M) = Der k (R,M)/PDer k (R,M). Now suppose that R is commutativ. Definition. The Kähler differentials of a ring R over k is the R-module Ω R/k having the following presentation: There is one generator dr for every r R, with dα = 0 if α k. For each r,s R there are two relations: d(r+s) = (dr)+(ds), d(rs) = r(ds)+s(dr). The map d: R Ω R/k, r dr, is a k-derivation. Lemma 2.4. Let k be a commutative ring and R be a commutative k-algebra. Then for every R-module M. Der k (R,M) = Hom R (Ω R/k,M) Let M be a right R-module. If we make M into a R R bimodule by setting rm = mr (remember that R is still commutative), we have H 1 (R,M) = Der k (R,M) and H 0 (R,M) = M. For homology in degree one we get the following result:

2 HOCHSCHILD HOMOLOGY AND COHOMOLOGY FOR ALGEBRAS 5 Proposition 2.5. Let k be a commutative ring, R be a commutative k-algebra and M a right R-module. Making M into a R R bimodule by setting rm = mr, we have H 1 (R,M) = M R Ω R/k. Proof. To caculate H 1 (R,M) we need to look at kerd 1 and imd 2. Since rm = mr we have d 1 (m r) = 0 (r m) 1 (r m) = mr rm = 0, hence, kerd 1 = M R. In the quotient H 1 (R,M) = M R/imd 2 we have 0 = d 2 (m r 1 r 2 ) = 0 (m r 1 r 2 ) 1 (m r 1 r 2 )+ 2 (m r 1 r 2 ) = mr 1 r 2 m r 1 r 2 +r 2 m r 1, so there is a well defined map H 1 (R,M) M R Ω R/k, m r m R dr, since in M R Ω R/k we have mr 1 R dr 2 m R d(r 1 r 2 )+r 2 m R dr 1 = m R r 1 (dr 2 ) m R r 1 (dr 2 ) m R r 2 (dr 1 )+M R r 2 (dr 1 ) = 0. On the other hand we have a bilinear map M Ω R/k H 1 (R,M), (m,r 1 dr 2 ) mr 1 r 2, since and (rm,r 1 dr 2 ) rmr 1 r 2 = r(mr 1 r 2 ) (m,rr 1 dr 2 ) mrr 1 r 2 = rmr 1 r 2 = r(mr 1 r 2 ). (Since M R is a siplicial R-module via r (m r 1...) = (rm r 1...).) Therefore we have, by the universal property of, a homomorphism M R Ω R/k H 1 (R.M), m r 1 dr 2 mr 1 r 2. These two maps are inverse to each other: m r m R dr m r, m R r 1 dr 2 mr 1 r 2 mr 1 R dr 2 = m R r 1 dr 2. So we have H 1 (R,M) = M R Ω R/k.

3 RELATIVE EXT 6 3 Relative Ext Fix an associative ring k and let k R be a ring map. For right R-modules M and N we get a cosimplicial abelian group [n] Hom k (M R n,n) with f(mr 0,r 1,...,r n ), if i = 0 ( i f)(m,r 0,...,r n ) = f(r 0,...,r i 1r i,...,r n ), if 0 < i < n f(r 0,...,r n 1 )r n if i = n (σ i f)(m,r 1,...,r n 1 ) = f(m,...,r i,1,r i+1,...,r n 1 ). Definition. If N is a right R-module we define the relative Ext groups to be the cohomology of the associated cochain complex C(Hom k (M R,N)): Ext n R/k(M,N) = H n C ( Hom k (M R,N) ). Proposition 3.1. Ext 0 R/k(M,N) = Hom R (M,N). Proof. The cochain complex C(Hom k (M R,N)) is given by 0 Hom k (M,N) 0 1 Hom k (M R,N)... ThereforeExt 0 R/k(M,N) = H 0 (Hom k (M R,N) = ker( 0 1 ).Letf Hom k (M,N), then ( 0 f)(m,r) ( 1 f)(m,r) = f(mr) f(m)r, which is zero iff f is an R-linear map. Thus Ext 0 R/k(M,N) = Hom R (M,N). To see the relation between Hochschild cohomology and the relative Ext groups we need to define the enveloping algebra, but first we need the following definition. Definition. Let R be a k-algebra. We define the k-algebra R op to be the algebra with the same underlying abelian group structure as R but multiplication in R op is the opposite of that in R, i.e. r s in R op is the same as sr in R. The nice thing about R op is that any right R-module M can be considerd as a left R op -module via r m = mr, where ist the multiplication in the left R op -module and the product mr is given by the right R-module operation. This module is welldefind since (r s) m = (sr) m = m(sr) = (ms)r = r (ms) = r (s m). Definition. Let k be a ring and R a k-algebra. We define the enveloping algebra by R e := R R op.

3 RELATIVE EXT 7 The main feature of the enveloping algebra is that any right R R bimodule M can be considered as a left R e -module via since (r s) m = rms, (r 1 s 1 )((r 2 s 2 ) m) = (r 1 s 2 ) r 2 ms 2 = r 1 r 2 ms 2 s 1 = (r 1 r 2 s 1 s 2 ) m = ((r 1 s 1 )(r 2 s 2 )) m. Similarily we may consider M as a right R e -module via m (r s) = smr. This means we may consider the category R-mod-R of R R bimodules as the category of left R e -modules or as the category of right R e -modules. For a commutative k-algebra R and an R R bimodule M we get the following results: Lemma 3.2. Hochschild cohomology ist isomorphic to relative Ext for the ring map k R e : H (R,M) = Ext R e /k(r,m). WewanttoknowhowtherelativeExtgroupExt 0 R e /k(r,m)isrelatedtotheabsolute Ext group Ext 0 R(R,M). Lemma 3.3. For absolute Ext we have Ext 0 R(R,M) = M. Proof. We know that Ext 0 R(R,M) = Hom R (R,M). We claim that the map Hom R (R,M) M, φ φ(1) is an isomorphism. Let φ: R M with φ(1) = 0, then φ(r) = φ(1 r) = φ(1)φ(r) = 0, thus injectivity is given. Let m M, the map r rm is a morphism with φ(1) = m, thus surjectivity is given. Corollary 3.4. For relative Ext we have Ext 0 R e /k(r,m) = Z R (M), where Z R (M) = {m M rm = mr, r R}. Proof. By 3.1 we have Ext 0 R e /k(r,m) = Hom R e(r,m) and since an R e -module is the same as an R R bimodule we have Hom R e(r,m) = Hom R bimod (R,M). Since every R R bimodulehomomorphism ist an R-linear map we may consider Hom R bimod (R,M) as a subgroup of Hom R (R,M) and thus as a subgroup of M. The claim follows since for φ Hom R bimod (R,M) we have φ(1) r = φ(1 r) = φ(r 1) = r φ(1). Recall that also H 0 (R,M) = Z R (M) by 2.2. This means that, in a way the Hochschild cohomology or the relative Ext group in degree zero is the center of the absolute Ext group.