Hochschild homology with coefficients in a certain bimodule

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.

• To get $\operatorname{Hom}_A(U,V)$ you do need f.g.p. of $U$, although nothing of $V$. – მამუკა ჯიბლაძე Oct 29 '18 at 9:03
• So the homology is $\mathrm{Hom}_A(U,V)$ (concentrated in degree zero) if $U$ is finitely generated projective? Can you provide an argument or a reference? – Lukas Woike Oct 29 '18 at 9:23
• No no I don't know that much. I only wanted to say that the canonical map $\operatorname{Hom}_A(U,A)\otimes_AV\to\operatorname{Hom}_A(U,V)$ is an isomorphism for all $V$ if and only if $U$ is fgp, nothing more – მამუკა ჯიბლაძე Oct 29 '18 at 9:34
• Ok, I will see what I can do with that. – Lukas Woike Oct 29 '18 at 9:40
• (From how the OP is formulated, something tells me that for non-commutative rings $\mathrm{Hom}_A(A,V)$ is not isomorphic to $V$. Is that right?) – Qfwfq Oct 29 '18 at 13:40

(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.)
• Out of curiosity - Tor's would also vanish for $V$ just flat, and then also I think $V\otimes_A\operatorname{Hom}_A(U,A)\to\operatorname{Hom}_A(U,V)$ will be iso for $U$ f.g., since a flat $V$ is a filtered colimit of projectives while $\operatorname{Hom}_A(U,-)$ commutes with filtered colimits if $U$ is f.g. (or does one still need f.p. for that?) – მამუკა ჯიბლაძე Oct 29 '18 at 17:37
• @მამუკაჯიბლაძე You're absolutely correct about flatness on $V$. I think that you do need $U$ finitely presented to commute it across filtered colimits. E.g. if $U = A / (x_0, x_1, x_2, \dots)$ then you may have an element in $V$ which is annihilated by all the $x_i$ but any lift to the directed system is annihilated by only finitely many. This in fact is a characterization of being finitely presented. Life is much easier if you're Noetherian! – Tyler Lawson Oct 29 '18 at 21:10