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Suppose that $X$ is a smooth manifold, whose $C^{\infty}$-functions we denote by $A$. Let $D_{poly}^*(A):=\bigoplus_{n\geq -1}Hom(A^{\otimes n+1},A)$ be the Lie algebra of polydifferential operators with the Hochschild differential and the Gerstenhaber bracket. A version of the Hochschild-Kostant-Rosenberg theorem shows the cohomology of $D_{poly}^*(A)$ is isomorphic to $\bigoplus_{n\geq -1}\wedge^{n+1}T_X$, polyvector fields. The graded vector space of polyvector fields is also a Lie algebra with the Schouten-Nijenhius bracket. However, the HKR-isomorphism is not a morphism of Lie algebras. The formality theorem of Kontsevich shows the HKR-isomorphism can be corrected to an $L_{\infty}$-morphism whose first term is the HKR-isomorphism. The upshot is this proves star-products on $C^{\infty}(X)$ correspond to formal Poisson structures.

In the article: M. de Wilde and P. B. A Lecomte: An homotopy formula for the Hochschild cohomology, Compositio Mathematica, tome 96, no. 1 (1995), the authors construct an explicit homotopy for the HKR-isomorphism. From this explicit homotopy the authors construct a star-product on $\frak g^*$, the dual of the Lie algebra $\frak g$. My question is the following: can one do the same thing for the general case of a Poisson manifold. In particular, is the explicit homotopy which induces Kontsevich's $L_{\infty}$-quasiisomorphism known? I believe that one has to exist since two $L_{\infty}$-algebras are quasiisomorphic if and only if they are homotopy equivalent. But, can you write down the explicit homotopy from the quasiisomorphism.

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The only way I know to construct a formality quasi-morphism for poy-vector fields out of an homotopy is via Tamarkins approach (i.e. $G_\infty$-formality). What Tamrakin does is

  1. prove that there is a suitable $G_\infty$-structure on Hochschild cochains

  2. prove that the obstruction to construct a $G_\infty$-formality step by step is unobstructed. At each step you have to make some choice, and the homotopy for the Hochschild complex of De Wilde-Lecomte gives you a way to do such choices.

But Tamarkin construction (part 1) involves the choice of an associator (i.e. choice of appropriate weights in Kontsevich's $L_\infty$-qausi-isomorphism)... so I think that this is hopeless to construct the formality out of an homotopy.

Moreover, even the other way I don't see how you could associate an homotopy for the Hochschild complex to an $L_\infty$-quasi-isomorphism from $T_{poly}$ to $D_{poly}$.

I might be wrong but I have the feeling that you are mixing two different notions of homotopy: that of higher homotopies in the $L_\infty$-morphism, and that of homotopy retract for the Hochschild cochain complex. In particular, what do you mean by a "homotopy equivalence" between $L_\infty$-algebras (whatever you mean, a homotopy for the Hochschild complex in the sens of De Wilde-Lecomte won't produce an example).

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    $\begingroup$ I agree with Damien. ;) Could you please let us know the definitions behind "I believe that one has to exist since two $L_\infty$-algebras are quasiisomorphic if and only if they are homotopy equivalent. But, can you write down the explicit homotopy from the quasiisomorphism." $\endgroup$
    – Bruno V.
    Commented May 27, 2011 at 12:21
  • $\begingroup$ Hi Damien, Thank you for your reply. The definition of homotopy equivalence that I had in mind was from Fukaya: \emph{Deformation theory, homological algebra, and mirror symmetry} page 44. Then when everything is in the $L_{\infty}$-category quasiisomorphism implies homotopy equivalence (But maybe the converse is not true). My other hope, maybe somewhat naive, was to choose a 'wise' enough homotopy at the chain level and then use an $L_{\infty}$ version of Markl:\emph{Transferring $A_{\infty}$ structures}. If the choice is 'wise' enough then this will imply that $T_{\poly}$ can't have any $\endgroup$
    – Jeremy Pecharich
    Commented May 27, 2011 at 16:03
  • $\begingroup$ higher brackets, an explicit $L_{\infty}$-quasiisomorphism, and an $L_{\infty}$-homotopy. But,'wise' might be too difficult to determine. $\endgroup$
    – Jeremy Pecharich
    Commented May 27, 2011 at 16:08

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