All Questions
11 questions
21
votes
2
answers
4k
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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 ...
- 416
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 $HH^*(C)$
$$
HH^*(C) \otimes HH^*(C) \...
- 12.8k
9
votes
0
answers
366
views
A characterisation of symmetric algebras using Hochschild (co)homology
A finite dimensional (connected if needed) $K$-algebra $A$ over a field $K$ is called symmetric when $A \cong Hom_K(A,K)$ as $A$-bimodules. Symmetric algebras are Frobenius algebras and include for ...
- 26.5k
7
votes
1
answer
2k
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What is the negative cyclic homology of a smooth projective variety?
Let X be a smooth and projective variety. Hochschild homology and cohomology have a very simple definition in terms of Ext groups of the diagonal of X. The Hochschild-Kostant-Rosenberg (HKR) theorem ...
- 1,889
7
votes
1
answer
408
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Hochschild H^1 (R,M) = 0 vs. H_1 (R,M) = 0 where R is a ring and M is an (R,R)-bimodule
Let $k$ be a commutative ring (with unity). Let $R$ be a $k$-algebra (with unity, but not of necessity commutative).
Let $M$ be an $\left(R,R\right)$-bimodule where $k$ acts in the same way from the ...
- 33.8k
7
votes
0
answers
575
views
Why is Hochschild homology interesting if its cohomology groups are infinite-dimensional?
I am trying to understand Hochschild homology, in particular the Hochschild–Kostant–Rosenberg theorem. As far as I understand this result gives an isomorphism between the algebraic (Kähler) ...
- 1,144
7
votes
0
answers
416
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Definitions of Hochschild Cohomology $HH^{\bullet}(A)$
Let $A$ be an associative unital $k$-algebra, and let $M$ be a bimodule of $A$. The Hochschild cohomology of $A$ with coefficients in $M$ can be defined as
$$HH^{n}(A,\,M)=\mathrm{Ext}^{n}_{A^{e}}(A,\,...
- 661
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 $(...
- 9,330
2
votes
1
answer
448
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,...
- 9,019
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 ...
- 5,405
1
vote
1
answer
273
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 $CRing$ $...
- 5,405