# Existence of anti-symmetric hochschild homology representatives

Let $$A$$ be an associative algebra over a field $$k$$. Let $$A_{L}$$ be the Lie algebra of $$A$$ with commutator bracket. Then if $$M$$ is a bimodule for $$A$$ there is an associated representation of $$A_{L}$$ called the adjoint representation and denoted $$M^{adj}$$.

There is a naturally defined map of homology groups $$H^{Lie} _{*} (A_{Lie}, M^{adj}) \rightarrow HH_{*} (A, M).$$ This is defined by anti-symmetrizing a Lie chain.

I'd like to know when this map on homology is surjective. It is if $$A$$ is smooth commutative (this follows from HKR). Less obviously it is true for $$U\mathfrak{g}$$, as follows from a computation of Chevalley and Eilenberg. A sort of minimal possible generalization would perhaps be to an almost commutative algebra, ie an associative algebra with an increasing filtration whose associated graded is smooth commutative. I'd love if it were true in this generality but any partial results would be greatly appreciated.

• The map from Lie cohomology to Hochschild is induced from the map of operads $f : \mathsf{Lie} \to \mathsf{As}$, since one can always pull-back $(A,M)$ to $(f^*A,f^*M)$, obtaining $f_* : H_{\mathsf{Lie}}^*(f^*A,f^*M) \to H_{\mathsf {As}}^*(A,M)$. But I have never thought how to rephrase your two situations in terms of this viewpoint, that is, what propety would we need on $f$ and on $A$ so that this map is iso/epi, etc. Feb 6, 2019 at 15:38
• Thanks for the comment Pedro. It'd be interesting to know a formulation in the operadic language, but I'm inclined to believe that such a reformulation would only be useful if it turned out to be quite a "formal" result, most probably implying a canonical splitting of the surjection in quite wide generality. I don't believe such a thing should exist, just by considering the smooth commutative case.
– EBz
Feb 6, 2019 at 16:36
• I think one can make smoothness appear more naturally into the picture if you consider the map $f_*$ in the first comment before taking homology, i.e. talking about cotangent complexes? Note that $U\mathfrak g$ is Koszul, and any Koszul algebra is homologically smooth in the sense of Kontsevich and Soibelman. This includes smooth commutative algebras. But at any rate I agree that phrasing things in such generality may make things much less tractable. Feb 6, 2019 at 16:46