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I was told that the following equation is true: Given a finitely generated Lie algebra $\mathfrak g$, there is a Gerstenhaber algebra isomorphism $$ HH(U\mathfrak g) \cong HH(\wedge^* \mathfrak g^\vee, \delta_{CE})$$ where $U\mathfrak g$ is the universal enveloping algebra, $\wedge^* \mathfrak g^\vee$ is the dual of exterior algebra of $\mathfrak g$ together with the differential $\delta_{CE}$ induced by Chevalley-Eilenberg differential, and bracket structure given by the Gerstenhaber bracket.

THE QUESTION is this:

  1. Is there an elementary way to prove that $$U\mathfrak g\rightarrow HOM_{(\wedge^*\mathfrak g^\vee,\delta_{CE})}(X,X)$$ is a quasi-isomorphism?

  2. In Lemma 6.5 (b) of Keller's paper 'DERIVING DG CATEGORIES', what does $RH_X\cong LT_{X^T}$ imply? Does it imply the essential surjectivity of $LT_X$? Or else, what does it mean?

In detail, I have tried to figure this isomorphism in detail, and found a guide line as follow:

The paper of B. Keller, 'DERIVED INVARIANCE OF HIGHER STRUCTURES ON THE HOCHSCHILD COMPLEX', says that in the above setting, if the derived categories of $U\mathfrak g$-Modules and ($\wedge^* \mathfrak g^\vee$, $\delta_{CE}$)-Modules [Modules over a dga] are derived equivalent, then the above isomorphism is achieved.

To show the derived equivalence of those two categories, we may find a $U\mathfrak g - (\wedge^* \mathfrak g^\vee,\delta_{CE})$ bimodule $$X=(U\mathfrak g \otimes \wedge^* \mathfrak g, d_{CE})$$ with the Chevalley-Eilenberg differential $d_{CE}$, so that two natural maps(induced by the actions) $$U\mathfrak g\rightarrow RHOM_{(\wedge^*\mathfrak g^\vee,\delta_{CE})}(X,X)$$ $$(\wedge^*\mathfrak g^\vee,\delta_{CE})\rightarrow RHOM_{U\mathfrak g}(X,X)$$ are quasi-isomorphisms.

Indeed, the second quasi-isomorphism is easily proved without the derived HOM. That is, $(\wedge^*\mathfrak g^\vee,\delta_{CE})\rightarrow HOM_{U\mathfrak g}(X,X)$ is a quasi-isomorphism. However, the first one does not seem to be easy.

I believe that $U\mathfrak g\rightarrow HOM_{\wedge^*\mathfrak g^\vee,\delta_{CE}}(X,X)$ (Notice that it is HOM rather than RHOM) is a quasi-isomorphism, and I have looked at another paper of Keller, 'DERIVING DG CATEGORIES' (to prove the case of derived hom), and the LEMMA 6.5 (b) seems to be telling something, but could not understand $RH_X\cong LT_{X^T}$ part so that whether it implies the essential surjectivity of the functor $LT_X$ or not.

Sorry for such a lengthy question. Please help me with this problem. Thanks.

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    $\begingroup$ Do you know a proof in some special cases? Even with abelian algebras this seems nontrivial and related somehow to the Koszul complex. $\endgroup$ – მამუკა ჯიბლაძე Jan 10 '18 at 6:20
  • $\begingroup$ I don't think this is true even in the case of a one dimension lie algebra---The right hand side is a completion of the left-hand side. In this case the right hand-side should be thought of as formal polyvector fields. This is elementary--- basically a "super" version of the HKR theorem. However the only written reference that comes immediately to mind is page 7 of arxiv.org/pdf/0812.1171.pdf. It will be true if g is a graded (dg) lie algebra concentrated in homologically positive degree (and Ug is suitably interpreted) by the Keller stuff. $\endgroup$ – Daniel Pomerleano Jan 10 '18 at 20:15
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    $\begingroup$ For clarity I should say that the problem is that the first map just below "so that two natural maps(induced by the actions)" is not a quasi-isomorphism. Even for the one dimensional Lie algebra you have to complete i.e. you have $k[[x]] \cong RHom_\Lambda(k,k)$ $\endgroup$ – Daniel Pomerleano Jan 10 '18 at 22:04
  • $\begingroup$ @DanielPomerleano My down to earth computation when $\mathfrak g$ is of dimension 1 and 2 gives the desired result. The zero'th cohomology of $HOM_{(\wedge^* \mathfrak g^\vee,\delta_{CE})}(U\mathfrak \otimes \wedge^* \mathfrak g, U\mathfrak \otimes \wedge^* \mathfrak g)$ together with the standard differential $\delta: f\mapsto d_{CE}f-(-1)^{|f|} fd_{CE}$ is a group of multiplication maps by an element of $U\mathfrak g$, and all the other degree vanishes. I am not sure why my computation conflicts with your statement. $\endgroup$ – sock Jan 11 '18 at 1:52
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    $\begingroup$ I suspect you are using a sum somewhere where you should be using a product. In any case, let's just do this for abelian $\mathfrak{g}$. In calculating $RHom_{Λ}(k,k)$ as a vector space, you only have to replace the first entry by your Koszul complex (basic homological algebra). If you do this, the differential on your complex will vanish and you can verify the assertion I made above additively. $\endgroup$ – Daniel Pomerleano Jan 14 '18 at 23:07

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