intuition for hochschild homology

Question

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 \begin{equation} \displaystyle \notag C_{n}(A,M)=M\otimes_{k}A^{\otimes n} \end{equation} and for each $n$ the maps $d_{i}:C_{n}(A,M)\rightarrow C_{n-1}(A,M)$ as follows. Define \begin{equation} \displaystyle d_{0}(m\otimes a_{1}\otimes\cdots\otimes a_{n})=ma_{1}\otimes a_{2}\otimes\cdots\otimes a_{n} \end{equation} and \begin{equation} \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 \end{equation} and \begin{equation} \displaystyle d_{n}(m\otimes a_{1}\otimes \cdots \otimes a_{n})=a_{m}m\otimes a_{1}\otimes\cdots\otimes a_{n-1}. \end{equation} When $i<j$, one can check that $d_{i}d_{j}=d_{j-1}d_{i}$. We define a linear operator \begin{equation} \displaystyle b=\sum_{i=0}^{n}(-1)^{i}d_{i} \end{equation} and an $ A$-module \begin{equation} \displaystyle C_{*}(A,M)=\bigoplus_{n\geq 0}C_{n}(A,M). \end{equation} 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 $HH_{*}(A)$.

Answer 1

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.

Answer 2

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.