# Hochschild homology of acyclic complex

Let $$A$$ be a differential graded algebra over a commutative ring $$R$$. Suppose that $$H_*(A)=0$$, i.e. $$A$$ is acyclic.

Question: Does this imply that the Hochschild homology $$HH_*(A)$$ also vanishes identically? If not, what hypotheses does one need to impose on $$R$$ and $$A$$? (In the case I am mostly interested in, $$R$$ is a polynomial ring in one variable.)

Remark: if we assume that $$R$$ is a field, then it's a general fact that one can cook up an $$A_{\infty}$$-algebra structure on $$H_*(A)$$ such that $$A \to H_*(A)$$ is an $$A_{\infty}$$ quasi-isomorphism. Moreover, $$A_{\infty}$$-algebra quasi-isomorphisms are invertible and $$HH_*(-)$$ is functorial. So it follows that that $$HH_*(A)=0$$ is $$H_*(A)=0$$. So the result should be true in this case.

• I guess the answer depends on how precisely define the Hochschild homology over a general ring $R$ (i.e., are you using the derived tensor product or not?). – Denis Nardin Apr 19 at 8:05
• I think I'd like to define HH_* just using the bar complex (i.e. the same way as I would defined it over a field). However, I'm happy to assume that A is flat over R, so I presume I don't have to derive anything? (In fact, in the case I'm interested in, A is a free associative algebra and R is a polynomial ring in one variable.) – user155668 Apr 19 at 8:26
• If you use the derived tensor product (which coincides with the classical one if $A$ is flat over $R$), then the definition of $HH_\ast(A)$ is manifestly invariant under quasi-isomorphism of algebras. In particular the map $A\to 0$ is a quasi-isomorphism... – Denis Nardin Apr 19 at 8:29
• @DenisNardin Sorry, but while it's clear that a morphism of dg-algebras $A \to B$ induces a morphism $HH_*(A) \to HH_*(B)$, it's not clear why a quasi-isomorphism of dg-algebras induces an isomorphism on Hochschild homology. Maybe I'm using a bad definition for $HH_*(-)$? Could you say a few words about why this is clear? – user155668 Apr 19 at 17:51
• Both derived tensor products and geometric realizations of simplicial objects preserve quasi-isomorphisms, and the bar complex is obtained by composing the two operations. I'm not sure I can say much more than what is in Weibel's Homological algebra. – Denis Nardin Apr 19 at 18:58

Hochschild homology is a derived invariant, and in particular a quasi-isomorphism invariant, since quasi-isomorphic algebras are derived invariant. Your algebra is non-unital, I assume, since $$1$$ is usually a non-trivial cycle in general.
In any case, however, you can consider the Hochschild cyclic complex of $$A$$, call it $$C_*(A)$$, which has an internal differential coming from $$A$$ and an external one coming from the usual Hochschild differential. This is a bicomplex and so there is a spectral sequence with second page $$HH_*(H_*(A))$$ that converges (perhaps you need to check this part carefully) to $$HH_*(A)$$. The Hochschild homology of $$k$$ is $$k$$, and hence so is that of $$A$$. If $$A$$ is non-unital and $$H_*(A) = 0$$, you would get that $$HH_*(A) = 0$$, too.