Download 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
9. Juni 2011
Inhaltsverzeichnis
1 Simplicial methods in homological algebra
1
2 Hochschild homology and cohomology for algebras
2.1 Hochschild homology and cohomology in degree 0 . . . . . . . . . . .
2.2 Hochschild homology and cohomology in degree 1 . . . . . . . . . . .
2
3
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 An 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 ǫi : [n] → [n + 1] and degeneracy maps ηi : [n] →
[n − 1], for i = 1, . . . n, given by
(
(
j,
if j < i
j,
if j ≤ i
ǫi (j) =
ηi (j) =
j + 1, if j ≥ i,
j − 1, if j > i.
These maps fulfil the simplicial identities:
ǫj ǫi = ǫi ǫj−1
if i < j
ηj ηi = ηi ηj+1
if i ≤ j


if i < j
ǫi ηj−1 ,
ηj ǫi = identity if i = j, j + 1


ǫi−1 ηj
if i > j + 1.
Therefore, to give a simplicial Object A, it is sufficient to give the objects An and
face operators ∂i : An → An−1 and degeneracy operators σi : An → An+1 fulfilling
the simplicial identities. Under this correspondence ∂i = A(ǫi ) and σi = A(ηi ).
Definition. Let A be a simplicial object in an abelian category A. The associated,
or unnormalized,P
chain complex C = C(A) has Cn = An , and its boundary operator
is given by d := (−1)i ∂i : Cn → Cn−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 DoldKan 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
Cn = 0 for n < 0.
We check that in C we have d2 = 0, since ∂j ∂i = ∂i ∂j−1 we get
2 HOCHSCHILD HOMOLOGY AND COHOMOLOGY FOR ALGEBRAS
2
d = d(
=
n
X
2
(−1)i ∂i )
i=0
n+1
n
XX
(−1)i+j ∂j ∂i
j=0 i=0
j−1
n+1 X
n X
n
X
X
=
(−1)i+j ∂j ∂i +
(−1)i+j ∂j ∂i
j=1 i=0
=
j=0 i=j
j−1
n+1 X
X
(−1)
j=1 i=0
=
i+j
n X
n
X
∂i ∂j−1 +
(−1)i+j ∂j ∂i
j=0 i=j
n+1 X
i−1
n X
n
X
X
(−1)i+j ∂j ∂i−1 +
(−1)i+j ∂j ∂i
i=1 j=0
=
=
j=0 i=j
n X
i
X
(−1)
i=0 j=0
n X
n
X
i+j+1
n X
n
X
∂j ∂i +
(−1)i+j ∂j ∂i
(−1)i+j+1 ∂j ∂i +
j=0 i=j
(−1)i+j ∂j ∂i
j=0 i=j
= 0.
2
j=0 i=j
n X
n
X
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 kmodule, i.e. a simplicial object in the category k-mod, M ⊗R⊗∗ with [n] 7→ M ⊗R⊗n
(M ⊗ R⊗0 = M ) - these modules are even k − k bimodules -, by declaring


if i = 0
mr1 ⊗ r2 ⊗ · · · ⊗ rn ,
∂i (m ⊗ r1 ⊗ · · · ⊗ rn ) = m ⊗ r1 ⊗ · · · ⊗ ri ri+1 ⊗ · · · ⊗ rn , if 0 < i < n


rn m ⊗ r1 ⊗ · · · ⊗ rn−1
if i = n
σi (m ⊗ r1 ⊗ · · · ⊗ rn ) = m ⊗ r1 ⊗ · · · ⊗ ri ⊗ 1 ⊗ ri+1 ⊗ · · · ⊗ rn ,
where m ∈ M and ri ∈ R. These formulas are obviously k-multilinear, so the ∂i
and σi 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:
0o
∂0 −∂1
M ⊗Ro
Mo
d
M ⊗R⊗Ro
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
Hn (R, M ) = Hn C(M ⊗ R⊗∗ ).
We also obtain a cosimplicial k-module with [n] 7→ Homk (R⊗n , M ) (Homk (R⊗0 , M ) =
M ) by declaring


if i = 0
r0 f (r1 , . . . , rn ),
(∂i f )(r0 , . . . , rn ) = f (r0 , . . . , ri ri+1 , . . . , rn ), if 0 < i < n


f (r0 , . . . , rn−1 )rn
if i = n
(σi f )(r1 , . . . , rn−1 ) = f (r1 , . . . , ri , 1, ri+1 , . . . , rn−1 ).
The associated cochain complex look like this:
0
/M
∂0 −∂1
/ Homk (R, M )
d
/ Homk (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(Homk (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
H0 (R, M ) = M/[M, R].
Proof. To calculate H0 (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
(∂0 − ∂1 )(m ⊗ r) = ∂0 (m ⊗ r) − ∂1 (m ⊗ r) = mr − rm,
thus
H0 (R, M ) = M/[M, R].
In particular, we get H0 (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(Homk (R⊗∗ , M ). An element m ∈ M is inside the kernel if
0 = (∂0 − ∂1 )(m)(r) = (∂0 m)(r) − (∂1 m)(r) = mr − rm, ∀r ∈ R
for all r ∈ R. So
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
2.2
4
Hochschild homology and cohomology in degree 1
Now, to investigate H 1 (R, M ), we take a look at the kernel of d : Homk (R, M ) →
Homk (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 Derk (R, M ). The image of d : M → Homk (R, M ) is given by the k-linear
functions fm : R → M defined by
fm (r) = rm − mr,
since
(∂ 0 m)(r) − (∂ 1 m)(r) = rm − mr.
These functions are also k-derivations, beacause
rfm (s) + fm (r)s = r(sm − ms) + (rm − mr)s
= rsm − rms + rms − mrs = (rs)m − m(rs) = fm (rs)
for all r, s ∈ R, this fact is also given by d2 = 0. A k-derivation of this type is called
a principal k-derivation and we denote the submodule of principal k-derivations by
PDerk (R, M ). Thus we obtain
Proposition 2.3. H 1 (R, M ) = Derk (R, M )/ PDerk (R, M ).
Now suppose that R is commutativ.
Definition. The Kähler differentials of a ring R over k is the R-module ΩR/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 → ΩR/k , r 7→ dr, is a k-derivation.
Lemma 2.4. Let k be a commutative ring and R be a commutative k-algebra. Then
Derk (R, M ) ∼
= HomR (ΩR/k , M )
for every R-module 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 ) = Derk (R, M )
and H0 (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
H1 (R, M ) ∼
= M ⊗R ΩR/k .
Proof. To caculate H1 (R, M ) we need to look at ker d1 and im d2 . Since rm = mr
we have
d1 (m ⊗ r) = ∂0 (r ⊗ m) − ∂1 (r ⊗ m) = mr − rm = 0,
hence, ker d1 = M ⊗ R. In the quotient H1 (R, M ) = M ⊗ R/ im d2 we have
0 = d2 (m ⊗ r1 ⊗ r2 ) = ∂0 (m ⊗ r1 ⊗ r2 ) − ∂1 (m ⊗ r1 ⊗ r2 ) + ∂2 (m ⊗ r1 ⊗ r2 )
= mr1 ⊗ r2 − m ⊗ r1 r2 + r2 m ⊗ r1 ,
so there is a well defined map
H1 (R, M ) −→ M ⊗R ΩR/k ,
m ⊗ r 7−→ m ⊗R dr,
since in M ⊗R ΩR/k we have
mr1 ⊗R dr2 − m ⊗R d(r1 r2 ) + r2 m ⊗R dr1
= m ⊗R r1 (dr2 ) − m ⊗R r1 (dr2 ) − m ⊗R r2 (dr1 ) + M ⊗R r2 (dr1 ) = 0.
On the other hand we have a bilinear map
M × ΩR/k −→ H1 (R, M ),
(m, r1 dr2 ) 7−→ mr1 ⊗ r2 ,
since
(rm, r1 dr2 ) 7→ rmr1 ⊗ r2 = r(mr1 ⊗ r2 )
and
(m, rr1 dr2 ) 7→ mrr1 ⊗ r2 = rmr1 ⊗ r2 = r(mr1 ⊗ r2 ).
(Since M ⊗ R⊗∗ is a siplicial R-module via r · (m ⊗ r1 ⊗ . . . ) = (rm ⊗ r1 ⊗ . . . ).)
Therefore we have, by the universal property of ⊗, a homomorphism
M ⊗R ΩR/k −→ H1 (R.M ),
m ⊗ r1 dr2 7−→ mr1 ⊗ r2 .
These two maps are inverse to each other:
m ⊗ r 7→ m ⊗R dr 7→ m ⊗ r,
m ⊗R r1 dr2 7→ mr1 ⊗ r2 7→ mr1 ⊗R dr2 = m ⊗R r1 dr2 .
So we have
H1 (R, M ) ∼
= M ⊗R ΩR/k .
6
3 RELATIVE EXT
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] 7→ Homk (M ⊗ R⊗n , N ) with


if i = 0
f (mr0 , r1 , . . . , rn ),
i
(∂ f )(m, r0 , . . . , rn ) = f (r0 , . . . , ri − 1ri , . . . , rn ), if 0 < i < n


f (r0 , . . . , rn−1 )rn
if i = n
(σ i f )(m, r1 , . . . , rn−1 ) = f (m, . . . , ri , 1, ri+1 , . . . , rn−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 (Homk (M ⊗ R⊗∗ , N )):
ExtnR/k (M, N ) = H n C Homk (M ⊗ R⊗∗ , N ) .
Proposition 3.1. Ext0R/k (M, N ) = HomR (M, N ).
Proof. The cochain complex C (Homk (M ⊗ R⊗∗ , N )) is given by
0
/ Homk (M, N ) ∂
0 −∂ 1
/ Homk (M ⊗ R, N )
/ ...
Therefore Ext0R/k (M, N ) = H 0 (Homk (M ⊗R⊗∗ , N ) = ker(∂ 0 −∂ 1 ). Let f ∈ Homk (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
Ext0R/k (M, N ) = HomR (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 Rop to be the algebra
with the same underlying abelian group structure as R but multiplication in Rop is
the opposite of that in R, i.e. r · s in Rop is the same as sr in R.
The nice thing about Rop is that any right R-module M can be considerd as a left
Rop -module via
r · m = mr,
where · ist the multiplication in the left Rop -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
Re := R ⊗ Rop .
7
3 RELATIVE EXT
The main feature of the enveloping algebra is that any right R − R bimodule M can
be considered as a left Re -module via
(r ⊗ s) · m = rms,
since
(r1 ⊗s1 ) ((r2 ⊗ s2 ) · m) = (r1 ⊗ s2 ) · r2 ms2 = r1 r2 ms2 s1
= (r1 r2 ⊗ s1 s2 ) · m = ((r1 ⊗ s1 )(r2 ⊗ s2 )) · m.
Similarily we may consider M as a right Re -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 Re -modules or as the category of right Re -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 → Re :
H ∗ (R, M ) ∼
= Ext∗Re /k (R, M ).
We want to know how the relative Ext group Ext0Re /k (R, M ) is related to the absolute
Ext group Ext0R (R, M ).
Lemma 3.3. For absolute Ext we have
Ext0R (R, M ) ∼
= M.
Proof. We know that Ext0R (R, M ) = HomR (R, M ). We claim that the map
HomR (R, M ) −→ M,
φ 7−→ 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 7→ rm is a morphism with φ(1) = m,
thus surjectivity is given.
Corollary 3.4. For relative Ext we have
Ext0Re /k (R, M ) = ZR (M ),
where ZR (M ) = {m ∈ M | rm = mr, ∀r ∈ R}.
Proof. By 3.1 we have Ext0Re /k (R, M ) = HomRe (R, M ) and since an Re -module
is the same as an R − R bimodule we have HomRe (R, M ) = HomR−bimod (R, M ).
Since every R − R bimodulehomomorphism ist an R-linear map we may consider
HomR−bimod (R, M ) as a subgroup of HomR (R, M ) and thus as a subgroup of M .
The claim follows since for φ ∈ HomR−bimod (R, M ) we have
φ(1) · r = φ(1 · r) = φ(r · 1) = r · φ(1).
Recall that also H 0 (R, M ) = ZR (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.