29 votes

intuition for hochschild homology

Slogan: Hochschild homology is a (derived) categorification of the trace. This means the identity at the end of John Pardon's answer is a categorification of the identity $\text{tr}(AB) = \text{tr}(...
Qiaochu Yuan's user avatar
22 votes
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intuition for hochschild homology

If you have a right $A$-module $M_A$ and a left $A$-module $_AN$, then you can form their tensor product $$M\otimes_AN:=\operatorname{coker}(M\otimes_kA\otimes_kN\xrightarrow{(m,a,n)\mapsto(ma,n)-(m,...
John Pardon's user avatar
  • 18.3k
12 votes

What is the negative cyclic homology of a smooth projective variety?

There are conceptually simple definitions, but they require a more symmetric definition of Hochschild homology. The Hochschild homology of $X/k$ (with coefficients in $\mathcal O_X$) is the homology ...
Marc Hoyois's user avatar
  • 8,642
11 votes
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Why is every deformation of the universal enveloping algebra of a complex semisimple Lie algebra trivial?

The article Deformation par quantification et rigidite des algebres enveloppantes by M. Bordemann, A. Makhlouf, T. Petit addresses these questions. They call Lie algebras $\mathfrak{g}$ with $HH^2(U(\...
Dietrich Burde's user avatar
11 votes

A simple proof of the Weyl algebra's rigidity.

Since the Weyl algebra is a deformation of the polynomial ring in two variables, there is a short transparent proof using deformation theory, cf. M. Gerstenhaber and A. Giaquinto, On the cohomology of ...
Murray Gerstenhaber's user avatar
9 votes
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B-model and Hochschild cohomology

The boundary conditions of the B-model, ie, the D-branes, are the objects in $\mathcal{D}^b(X)$. A little bit of playing with pictures gives that the space of closed string states must be in the ...
Aaron Bergman's user avatar
9 votes
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Isomorphism in cyclic cohomology vs isomorphism in Hochschild cohomology

This was something that used to puzzle me when I was first learning this stuff. Refreshing my memory just now, I think that the trick is to use the fact that the Connes–Tsygan sequence has some ...
Yemon Choi's user avatar
  • 25.4k
8 votes
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A precise definition of contractible Banach algebras

$A$ is contractible if $H^1(A,X)=0$ for all Banach $A$-bimodules $X$ (here $H^1$ denotes continuous Hochschild cohomology for Banach algebras, as defined in the works of Johnson or Helemskii). It is ...
Yemon Choi's user avatar
  • 25.4k
8 votes
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What is the topological Hochschild cohomology of $\mathbb{F}_p$?

Let me write $HH^S(B) = THH(B) = B \wedge_{B^e} B$ for topological Hochschild homology, and $HH_S(B) = F_{B^e}(B, B)$ for topological Hochschild cohomology, where $B^e = B \wedge_S B^{op}$. For $B$ ...
John Rognes's user avatar
  • 8,627
8 votes
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Hochschild cohomology of group ring of a free group

Hochschild cohomology of group rings is reducible to ordinary group cohomology. (Not really if you're interested in multiplicative structure, but that's another story). Everyting works over any ...
Denis T's user avatar
  • 4,309
8 votes
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How to compute the periodic cyclic homology of this algebra

You can use a derived version of the HKR theorem, i.e. $HH(A) = \text{Sym}_A^\bullet(\mathbb{L}_A[1])$ where $\mathbb{L}_A$ is the cotangent complex (over $k$). I'm not sure about a reference though. ...
Harrison Chen's user avatar
7 votes
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Multiplicativity twisted Hochschild Kostant Rosenberg isomorphism

I am far from being expert in this subject, but I will try to present my understading of there this multiplicativity comes from. I wiil refer to authors you mention but only to the parts which I hope ...
SashaP's user avatar
  • 6,952
7 votes

Lie algebra of a p-group

I'm not an expert but it looks like some interesting simple Lie algebras do appear this way, yet one does not know how many. If the base field $k$ is $\mathbb{F}_p$, where $p$ is odd, and $G=\mathbb{Z}...
Alexander Premet's user avatar
7 votes

Book on Hochschild (co)homology

Sarah Witherspoon's book on Hochschild cohomology for Algebras is now published. See https://www.ams.org/publications/authors/books/postpub/gsm-204
6 votes
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Morita equivalence and isomorphisms in cohomology theories

The conceptual point is that all of these invariants are Morita invariant because they can be defined directly in terms of the category of modules. Explicitly: Starting from the category of modules $\...
Qiaochu Yuan's user avatar
6 votes
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Checking if Hochschild cohomology $\mathit{HH}^2(A)=0$

I believe that there is a 2-dimensional cocycle $g$ such that: $$ g(h_1 x \otimes h_2) = g(h_1 \otimes x h_2) = y h_1 x $$ and $g(a \otimes b) = 0$ for all other paths $a$ and $b$. To check that it's ...
Tyler Lawson's user avatar
  • 50.9k
6 votes

How to compute the periodic cyclic homology of this algebra

There's a spectral sequence starting with, for example, Hochschild cohomology of the cohomology of a DGA algebra $A$ and converging to the Hochschild cohomology of $A$, $E_2^{p,q}=H\!H^*H^*(A)\...
Dave Benson's user avatar
  • 11.2k
6 votes
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Hochschild cohomology of a group algebra

In general, if $A=K[G]$ for a group $G$ then an $A$-bimodule is the same as a vector space with action of $G\times G$. $HH^0(A,M)$ is the vector space of $A$-bimodule maps $A\to M$, which (in the case ...
Tom Goodwillie's user avatar
5 votes

How do you prove that Hochschild cohomology is Morita invariant?

It is even a derived invariant. Here is a proof in a special case, but Im not sure whether it works more general (with the same proof?) Let $A$ and $B$ two noetherian $K$-algebras for a commutative ...
Mare's user avatar
  • 25.5k
5 votes
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Can the Hochschild cochain complex be given the structure of a "homotopy BV algebra"?

Yes, this is the "cyclic Deligne conjecture", which also has several proofs by now. I believe the first one was Kaufmann's, using the cacti operad.
Dan Petersen's user avatar
4 votes
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A question on Hochschild cohomology

There are examples of non-semisimple algebras with trivial Hochschild cohomology: for example, the path algebra $C$ of a quiver whose underlying graph is a tree. Also, for finite dimensional algebras,...
Jeremy Rickard's user avatar
4 votes
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Hochschild cohomology of finite semisimple algebras

By corollary 18 of https://projecteuclid.org/journals/nagoya-mathematical-journal/volume-12/issue-none/On-the-dimension-of-modules-and-algebras-VIII-Dimension-of/nmj/1118799929.full we have (using the ...
Mare's user avatar
  • 25.5k
3 votes

Relation between the Hochschild cohomology of group algebras and groupoids

For every $\mathbb{Z}G$-bimodule $M$ there exists a $G$-module $U(M)$ such that the Hochschild cohomology of $\mathbb{Z}G$ with coefficients in $M$ is naturally isomorphic to the group cohomology of $...
Yonatan Harpaz's user avatar
3 votes

Hochschild cohomology of certain local algebras

The right term to look for is "truncated quiver algebras". There are two relevant references which I believe lead to a complete answer to your question in the non-commutative case. First, Section ...
Vladimir Dotsenko's user avatar
3 votes

Hochschild cohomology of an Azumaya algebra

It is true that $HH^*(A) \cong HH^*(Z)$ when $Z$ is a commutative $k$-algebra (i.e. in the setting of affine varieties). The following argument is due to Vadim Vologodsky, although any errors are mine:...
Joshua Mundinger's user avatar
3 votes

B-model and Hochschild cohomology

For correct attribution, one should at least mention the paper which the preprint of Moore and Segal itself quotes as the source for the particular case of the algebraic description of open-closed ...
amathematician's user avatar
3 votes

Two definitions of minimal models

The two definitions are the same. The thesis of Lefèvre-Hasegawa does not require the differential to be zero, it requires the component $m_1$ of the differential to be equal to zero: minimality ...
Vladimir Dotsenko's user avatar
3 votes
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Compute the cohomology of $\mathrm{Hom} (\Omega^*(M),\Omega^*(M))$

Since everything is linear over the field $\mathbb{R}$, $\Omega^*(M)$ is chain homotopy equivalent to $H^*_{dR}(M)$ endowed with the trivial differential, hence the cohomology of your complex $C$ is ...
Fernando Muro's user avatar
3 votes

Why is every deformation of the universal enveloping algebra of a complex semisimple Lie algebra trivial?

Means that as a $\mathbb C[[\hbar]]$ associative algebra $U_\hbar(\mathfrak g)$ is isomorphic to $U(\mathfrak g)[[\hbar]]$. Each element can be understood as a formal power series $\sum_{n=0}^\infty \...
Nicola Ciccoli's user avatar

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