# 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 ring $A$ and a bimodule $M$. Define an $A$-module by setting $$\displaystyle \notag C_{n}(A,M)=M\otimes_{k}A^{\otimes n}$$ and for each $n$ the maps $d_{i}:C_{n}(A,M)\rightarrow C_{n-1}(A,M)$ as follows. Define $$\displaystyle d_{0}(m\otimes a_{1}\otimes\cdots\otimes a_{n})=ma_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}$$ and $$\displaystyle d_{i}(m\otimes a_{1}\otimes\cdots\otimes a_{n})=m\otimes a_{1}\otimes\cdots\otimes a_{i}a_{i+1}\otimes\cdots\otimes a_{n},\quad 1\leq i\leq n-1$$ and $$\displaystyle d_{n}(m\otimes a_{1}\otimes \cdots \otimes a_{n})=a_{m}m\otimes a_{1}\otimes\cdots\otimes a_{n-1}.$$ When $i<j$, one can check that $d_{i}d_{j}=d_{j-1}d_{i}$. We define a linear operator $$\displaystyle b=\sum_{i=0}^{n}(-1)^{i}d_{i}$$ and an $A$-module $$\displaystyle C_{*}(A,M)=\bigoplus_{n\geq 0}C_{n}(A,M).$$ One can verify that $b:C_{*}(A,M)\rightarrow C_{*}(A,M)$ is a differential. The differential complex $(C_{*}(A,M),b)$ is called a Hochschild complex. The homology theory defined by a Hochschild complex is denoted by $H_{*}(A,M)$ and called a Hochschild homology. If $M=A$, the Hochschild homology is also denoted by $HH_{*}(A)$.

• You want $M$ to be an $A$-$A$-bimodule. Otherwise $d_n$ (or $d_0$) makes no sense. When $M = A$, the Hochschild cohomology is the higher derived functors of the center of $A$. – John Klein Dec 26 '17 at 12:33
• @John Klein Ooops, Forgive me for the misunderstanding. – Zbigniew Dec 26 '17 at 15:34
• Have you ever done anything with Hochschild (co)homology? Computed it for a few algebras or seen what lower cohomology groups are? Interpreted lower cocycles in the classical sense of Hochschild? It is not clear what "intuition" should mean here. Sure, you can talk about relative Tor and Ext and if your "intuition" for that is good be happy, or talk about derived functors and whatnot. But there's a very concrete and tangible aspect to this that should be the starting point for any meaning you may give to the word "intuition". We shouldn't forget about the utilitarian side of the story! – Pedro Tamaroff Dec 30 '17 at 2:31
• The existing answers to this are great, but I'm left wondering the following: there's a well-known story how the low-dimension Hochschild cohomology correspond to low-order deformations of associative algebras (tangent spaces/jets to the moduli space, see e.g. first chapter in the Kontsevich-Soibelman draft book). How does the "derived/categorification" interpretations in the existing answers fit into this story? @John Pardon – zzz Jan 2 '18 at 3:35
• @bianchira Yes. To a certain extent yes. See e.g. my paper with Sean Tilson available here: front.math.ucdavis.edu/1410.7089 , especially Th. C and Cor. D. – John Klein Jan 2 '18 at 7:23

If you have a right $$A$$-module $$M_A$$ and a left $$A$$-module $$_AN$$, then you can form their tensor product $$M\otimes_AN:=\operatorname{coker}(M\otimes_kA\otimes_kN\xrightarrow{(m,a,n)\mapsto(ma,n)-(m,an)}M\otimes_kN).$$ There is also a "derived" version of the tensor product, i.e. a chain complex whose homology groups are the $$\operatorname{Tor}^i_A(M,N)$$ groups, given by $$M\otimes_A^{\mathbb L}N:=\bigoplus_{i=0}^\infty M\otimes_kA^{\otimes_ki}\otimes_kN.$$ Notice that the $$i=0$$ and $$i=1$$ terms are exactly those appearing in the definition of the ordinary tensor product. [As noted in the comments, if the base ring $$k$$ is not a field, then one should be careful with this definition if $$M$$, $$N$$, and $$A$$ are not all flat over $$k$$.]

Now, if you have an $$(A,A)$$-bimodule $$_AB_A$$, then you can "tensor together the right and left actions of $$A$$ on $$B$$" (recovering the above for $$_AB_A={}_AN\otimes_kM_A$$). The derived version of this is precisely the Hochschild homology chain complex. You can think of $$HH_\bullet(A,B)$$ as "$$B\otimes_A^{\mathbb L}{}$$" written on annular paper so that to the right of $$\otimes_A^{\mathbb L}$$ is again (the same copy of) $$B$$. This makes apparent identities such as $$HH_\bullet(R,M\otimes_S^{\mathbb L}N)=HH_\bullet(S,N\otimes_R^{\mathbb L}M)$$ for bimodules $$_RM_S$$ and $$_SN_R$$. It also means that $$HH_\bullet(A,A)$$ has a "circle action" in a homotopical sense.

• Thanks for this nicely explanation but I have some confusion: What did you mean by $B\otimes_A^{\mathbb L}{}$ I do not meet this notation before. – Zbigniew Dec 27 '17 at 10:46
• It means the derived tensor product: stacks.math.columbia.edu/tag/09LP – Qiaochu Yuan Dec 30 '17 at 0:24
• John is omitting the differential. Also I think this complex only computes the right thing if $M, N, A$ are all flat over the base commutative ring. – Qiaochu Yuan Dec 30 '17 at 1:11
• @QiaochuYuan One usually defines Hochschild (co)homology using Hochschild's complex, which ends up computing a relative Tor (or Ext). When the algebra is flat (or projective) over the base, you get usual Tor (or Ext) --no hypothesis should be necessary on the modules. I guess it merely depends on what you want to call HH, but I would follow Hochschild's original definition. (...) – Pedro Tamaroff Dec 30 '17 at 2:45
• This boils down to the fact that if $A$ is $k$-projective, then the bar resolution is a honest projective resolution of $A$ as a bimodule, and so doing $M\otimes B(A,A)\otimes N$ works to compute Tor. There is a silly cancellation (two $A$s for the two tensors over $A$) that gives the complex in this post. – Pedro Tamaroff Dec 30 '17 at 2:45

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 $\text{tr}(AB) = \text{tr}(BA)$.

To see this the idea is to think of bimodules as categorifications of linear maps. One way to think about this is in terms of the Eilenberg-Watts theorem, which identifies $(A, B)$-bimodules with cocontinuous functors $\text{Mod}(A) \to \text{Mod}(B)$: every such functor is tensoring with some $(A, B)$-bimodule. Composition is given by composition of cocontinuous functors, or tensoring bimodules.

A particularly nice special case is where $A = k^n, B = k^m$ for $k$ a field (and we consider bimodules over $k$): then $(A, B)$-bimodules are $m \times n$ "matrices" of vector spaces over $k$, and composition / tensor product is "matrix multiplication." This really makes the analogy to linear maps particularly explicit. For more details see this blog post.

Bimodules form part of a 2-category whose objects are rings (or more generally algebras over a base commutative ring), morphisms are bimodules, and 2-morphisms are morphisms of bimodules. This 2-category is symmetric monoidal with monoidal product given by the tensor product of rings, which satisfies a universal property analogous to the universal property of the tensor product of vector spaces: namely, functors $\text{Mod}(A) \times \text{Mod}(B) \to \text{Mod}(C)$ which are cocontinuous in each variable can be identified with functors $\text{Mod}(A \otimes B) \to \text{Mod}(C)$ (and hence with $(A \otimes B, C)$-bimodules).

Now, in any symmetric monoidal (higher) category one can define dualizable objects and traces of endomorphisms of dualizable objects; see this blog post for details and pictures, which among other things explains what taking traces has to do with circles. In this case

• every ring $A$ is dualizable with dual $A^{op}$,
• endomorphisms correspond to $(A, A)$-bimodules $M$, and
• the trace turns out to be the zeroth Hochschild homology $HH_0(A, M)$.

(In particular you can verify that when $A = k^n$, so that an $(A, A)$-bimodule is an $n \times n$ matrix of vector spaces, the trace is the direct sum of the diagonal entries, just as expected by analogy.)

The full Hochschild homology is obtained by deriving this whole story, so that now morphisms are given by derived categories of bimodules and composition is given by the derived tensor product.

All of this can be thought of as an elaboration of some particularly interesting special cases of 1-dimensional topological field theory, where traces correspond to circles, and higher-dimensional topological field theory features generalizations of Hochschild homology involving more complicated manifolds.

• This is really helpful! Question about the closing comment: are you saying 1d-TFT has something to do with this derived version rather than the non-derived linear algebra story? If so is there a reference for this? (Maybe Baez-Dolan is the right place to look?) – zzz Jan 1 '18 at 4:03
• @bianchira: you can talk about 1d TFT with values in any kind of target, either derived or underived. I don't know a reference off the top of my head for this exact story. – Qiaochu Yuan Jan 1 '18 at 9:16