Hochschild homology with coefficients in a certain bimodule

Question

Let $A$ be a finite-dimensional $k$-algebra and $U$ and $V$ two finite-dimensional projective $A$-modules (maybe neither the finiteness nor projectivity has to play a role, but these requirements are satisfied in my problem).

Now consider the $A$-bimodule \begin{align} M := \mathrm{Hom}_A (U,A) \otimes \mathrm{Hom}_A (A,V) \ , \end{align} where $\mathrm{Hom}_A (U,A)$ is a right $A$-module via right multiplication applied to $A$, and $\mathrm{Hom}_A (A,V)$ a left $A$-module by right action on $A$ in the source (again by right multiplication); this ensures $\mathrm{Hom}_A (A,V) =V$ as left $A$-modules.

I would like to compute the Hochschild homology $HH_*(A;M)$.

If you write down the corresponding Hochschild complex, there is an augmentation by $\mathrm{Hom}_A (U,V)$. Does that induce a quasi-isomorphism? Maybe such a result is known to experts?

Is there, more generally, a coefficient theorem I could use?

As usual, thank you for any hints.

Answer 1

(I'm assuming in the following that the base ring $k$ was a field.)

First, we note that for any $A$-bimodule of the form $M \otimes N$, where $M$ is a left $A$-module and $N$ is a right $A$-module, has an isomorphism $$ HH_*(A;M \otimes N) \cong Tor^A_*(N,M). $$ To see this, we note that there is an explicit simplicial isomorphism between the cyclic bar construction computing Hochschild homology and the two-sided bar construction computing Tor: $$ (M \otimes N) \otimes A^{\otimes p} \to N \otimes A^{\otimes p} \otimes M $$ In particular, the last map which moves the last factor of $A$ around and multiplies it on the left is carried simply to its left action on $M$.

Therefore, we find $$ HH_*(A;Hom_A(U,A) \otimes Hom_A(A,V)) \cong Tor^A_*(Hom_A(A,V), Hom_A(U,A)) \cong Tor^A_*(V, Hom_A(U,A)). $$ If $V$ is projective as a (right) $A$-module, then we find that the Tor-groups vanish for $* > 0$ and that the zero'th group is $$ V \otimes_A Hom_A(U,A). $$ The natural map augmentation that you are describing is then the natural map $V \otimes_A Hom_A(U,A) \to Hom_A(U,V)$, and (because $V$ is projective) this map is an isomorphism whenever $U$ is finitely presented. In particular, we don't need $U$ to be projective. (We could instead ask that $U$ is finitely generated projective and $V$ is arbitrary and get this result; this was stated in the comments already by მამუკა ჯიბლაძე.)