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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.

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    $\begingroup$ To get $\operatorname{Hom}_A(U,V)$ you do need f.g.p. of $U$, although nothing of $V$. $\endgroup$ Commented Oct 29, 2018 at 9:03
  • $\begingroup$ 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? $\endgroup$ Commented Oct 29, 2018 at 9:23
  • $\begingroup$ 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 $\endgroup$ Commented Oct 29, 2018 at 9:34
  • $\begingroup$ Ok, I will see what I can do with that. $\endgroup$ Commented Oct 29, 2018 at 9:40
  • $\begingroup$ (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?) $\endgroup$
    – Qfwfq
    Commented Oct 29, 2018 at 13:40

1 Answer 1

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(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.)

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  • $\begingroup$ 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?) $\endgroup$ Commented Oct 29, 2018 at 17:37
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    $\begingroup$ @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! $\endgroup$ Commented Oct 29, 2018 at 21:10
  • $\begingroup$ If I am not mistaken finitely generated suffices for directed colimits but I am not sure which modules are directed (rather than just filtered) colimits of projectives $\endgroup$ Commented Oct 29, 2018 at 21:55

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