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Let $\mathfrak{g}$ be a Lie algebra over a char. $0$ field and let $\iota: U\mathfrak{g}\rightarrow S\mathfrak{g}$ be the Poincaré-Birkhoff-Witt (PBW) isomorphism, inverse to that natural map from the symmetric algebra.

Let $C^\mathrm{Hoch}_{*} (U\mathfrak{g})$ be the graded vector space underlying the standard complex computing Hochschild homology of $U\mathfrak{g}$, ie forget the differential for now. Composition of $\iota$ with the Hochschild-Kostant-Rosenberg (HKR) morphism induces a map of graded vector spaces, $C^\mathrm{Hoch}_{*} (U\mathfrak{g})\rightarrow \Omega^{*} (S\mathfrak{g})$. This respects the natural filtrations on both sides.

What I'd like to know is whether there's a nice differential on the right hand side, such that our map intertwines the Hochschild differential on the LHS with this new differential. The leading order term (wrt the filtration) for this differential should be the lie derivative wrt the Poisson-Kirillov bracket on $S(\mathfrak{g})$.

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    $\begingroup$ I might be missing the point, but isn't it a consequence of Tsygan formality, see eg arxiv.org/abs/math/0010321 ? Ie there is in particular a quasi-iso from HH of $U(\mathfrak g)$ to Poisson homology of $\mathfrak g^*$. $\endgroup$
    – Adrien
    Commented Mar 6, 2019 at 16:27
  • $\begingroup$ @Adrien I agree that the existence should be implied by formality, I was hoping for a relatively simple formula for the diff on this. If I'm not mistaken in my calculations the map above doesn't intertwine the hochschild diff and the poisson one, and one needs to add further corrections of lower filt degree. Ofc these must be homologically insignificant as, like u mention, the two sides have the same homology. $\endgroup$
    – user108998
    Commented Mar 6, 2019 at 16:38
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    $\begingroup$ The assignment $U\mapsto CE^*(\Omega^*_c(U)\otimes\mathfrak g)$ (tensor with compactly supported forms to get a dgla, then take cohomology) defines a locally constant factorization algebra on manifolds of a given dimension. In particular, $F((a,b))$ is an ($E_1$-)algebra, and Owen Gwilliam proved you get $U\mathfrak g$ this way (people.math.umass.edu/~gwilliam/thesis.pdf Section 4.6). Then $HH_*(U\mathfrak g)\simeq F(S^1)$. To compute the latter, you may replace $\Omega_{(c)}^*(S^1)$ by its subalgebra of harmonic forms, so that $F(S^1)\simeq CE^*(\mathfrak g\ltimes \mathfrak g[1])$. $\endgroup$
    – Bertram Arnold
    Commented Mar 6, 2019 at 16:39
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    $\begingroup$ @EBz indeed, my point was somehow that the corrections you need to add are precisely the corrections you need to go from HKR to Kontsevich, so I doubt there will be a simple formula (basically in the "usual" formality picture you change the map and keep the same differential on both side, while you want to keep the map and change the differential, those are just two ways of doing the same thing). $\endgroup$
    – Adrien
    Commented Mar 6, 2019 at 17:06
  • $\begingroup$ @Adrien ok, that makes sense, I suppose I was being overly optimistic wrt the existence of anything simple. Thanks for the comments. $\endgroup$
    – user108998
    Commented Mar 6, 2019 at 17:26

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