According to this post [Intuition for group homology](https://mathoverflow.net/questions/10879/intuition-for-group-cohomology), 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 $HH\_{\*}(A)$.
> 
> 
> 



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
$$M\otimes\_AN:=\operatorname{coker}(M\otimes\_kA\otimes\_kN\xrightarrow{(m,a,n)\mapsto(ma,n)-(m,an)}M\otimes\_kN).$$
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.


