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1 vote
0 answers
119 views

Cyclic homology with coefficients in a bimodule

I've recently been trying to understand Hochschild and cyclic co/homology better, and I've noticed that while it's common to define the Hochschild homology $\mathrm{HH}_{\bullet}(A;M)$ of an $R$-...
6 votes
2 answers
569 views

Is the exterior algebra intrinsically formal?

Following 4.6 and 4.7 of this paper by Seidel and Thomas, a graded algebra $A$ is called intrinsically formal if any two dgas with cohomology $A$ are quasi-isomorphic. There is a sufficient condition ...
8 votes
2 answers
771 views

How to compute the periodic cyclic homology of this algebra

Let $k=\mathbb{C}$ be the field of complex numbers. I consider the (DG) algebra $A:=k[x]/(x^2)$ such that $\deg(x)=-1$. My question is how to compute the periodic cyclic homology, Hochschild homology ...
9 votes
0 answers
366 views

A characterisation of symmetric algebras using Hochschild (co)homology

A finite dimensional (connected if needed) $K$-algebra $A$ over a field $K$ is called symmetric when $A \cong Hom_K(A,K)$ as $A$-bimodules. Symmetric algebras are Frobenius algebras and include for ...
2 votes
1 answer
450 views

Can we compute the Hochschild cohomology for $k[x]$ through the Hochschild complex?

For an algebra $A$ we can define its Hochschild cohomology $HH^{\bullet}(A,A)$ as in this wikipedia page. Now let $A=k[x]$ be the polynomial ring where $k$ is a field. It is well-known that $HH^{0}(A,...
14 votes
2 answers
2k views

Applications of topological chiral homology and factorization algebras (aka higher Hochschild cohomology)

I recently heard a talk about these topics and found them very interesting. The talk was centered on the formal structure and didn't really focus on examples. So my question is: what is your favorite ...
7 votes
0 answers
579 views

Why is Hochschild homology interesting if its cohomology groups are infinite-dimensional?

I am trying to understand Hochschild homology, in particular the Hochschild–Kostant–Rosenberg theorem. As far as I understand this result gives an isomorphism between the algebraic (Kähler) ...
7 votes
2 answers
884 views

How do you prove that Hochschild cohomology is Morita invariant?

I am simply trying to show that $HH^\bullet(A)= HH^\bullet(M_r(A))$ for any matrix ring of $A$. In Loday's book (Sect 1.5.6) the Morita invariance is explained as follows : it says that if $M$ is an ...
37 votes
7 answers
6k views

Book on Hochschild (co)homology

There is currently no treatise treating Hochschild (co)homology systematically. There is a chapter in Weibel's book, there's parts of Loday's and a few others... What should be covered by such a ...
11 votes
0 answers
930 views

Higher traces in Hochschild cohomology

Let $A$ be an associative algebra over a field $k$. Let $\rho:A \rightarrow \mathrm{End}(M)$ a left module, finite dimensional over $k$. Then the map $a \mapsto \mathrm{tr}_M \rho(a)$ is a well ...
21 votes
2 answers
4k views

intuition for hochschild homology

According to this post Intuition for group homology, I wonder what is the intuition for Hochschild homology. The Hochschild homology is defined as the homology of this complex chain. Given a ...
14 votes
2 answers
2k views

Relationship between Hochschild cohomology and Drinfeld centers

Let $HH_*(A,N)$ (or $HH^*(A,N)$) be the Hochschild homology (or cohomology) of an associative algebra $A$ with coefficients in an $A$-bimodule $N$. I was reading nlab's entry on Hochschild cohomology ...
7 votes
0 answers
416 views

Definitions of Hochschild Cohomology $HH^{\bullet}(A)$

Let $A$ be an associative unital $k$-algebra, and let $M$ be a bimodule of $A$. The Hochschild cohomology of $A$ with coefficients in $M$ can be defined as $$HH^{n}(A,\,M)=\mathrm{Ext}^{n}_{A^{e}}(A,\,...
7 votes
1 answer
2k views

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

Let X be a smooth and projective variety. Hochschild homology and cohomology have a very simple definition in terms of Ext groups of the diagonal of X. The Hochschild-Kostant-Rosenberg (HKR) theorem ...
4 votes
0 answers
211 views

Inner automorphisms acts as identity on Hochschild homology

Let $A$ be a unital algebra and $u \in A$ be an invertible element. Let us consider $u_n(a_0 \otimes a_1 \otimes ... \otimes a_n):=ua_0u^{-1} \otimes ua_1u^{-1} \otimes ... \otimes ua_nu^{-1}$. Then $(...
11 votes
2 answers
1k views

Relation between Gerstenhaber bracket and Connes differential

Let $C$ be an arbitrary algebra (more generally, a linear 1-category). The following structures are well-known: A degree-0 product on the Hochschild cohomology $HH^*(C)$ $$ HH^*(C) \otimes HH^*(C) \...
2 votes
1 answer
578 views

Interpretation of Hochschild Homology groups

In all the literature I've come across there are many concrete interpretations of the first few Hochschild Cohomology groups. For example $HH^1(A,M)\cong Derivation/Inner Derivations$ etc.... In ...
3 votes
1 answer
697 views

Hochschild cohomology and formal smoothness

Hochschild cohomology can be used to characterise formal smoothness of unital associative algebras; in that such an algebra $A$ is formally smooth if and only if it is of Hochschild cohomological ...
2 votes
0 answers
90 views

Poincaré Duality of a quasi-free algebra

I'm completely stumped on this one (yet I feel it is obviously true or obviously false) If $A$ is a quasi-free algebra, then must it satisfy Poincaré duality? All i need to find is a protective ...
1 vote
1 answer
274 views

Jacobi-Zariski exact sequence question

Denote by $HC(A,M)$ the Hochschild homological complex of an algebra $A$ with coefficients in an $A$-bimodule $M$, and let $B\rightarrow A$ be an $R$-flat extension of $R$-algebras, for some $CRing$ $...
5 votes
1 answer
592 views

When do Hochschild homology and cohomology agree? (Ambidexterity?)

Suppose $X$ is a smooth algebraic variety over a field of characteristic $0$. What are the most general conditions under which Hochschild homology and cohomology of $X$ agree? The existence of a ...
7 votes
1 answer
408 views

Hochschild H^1 (R,M) = 0 vs. H_1 (R,M) = 0 where R is a ring and M is an (R,R)-bimodule

Let $k$ be a commutative ring (with unity). Let $R$ be a $k$-algebra (with unity, but not of necessity commutative). Let $M$ be an $\left(R,R\right)$-bimodule where $k$ acts in the same way from the ...