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$X$ is smooth Poisson. Kontsevich formality theorem says that there is a $L_\infty$ quasi-isomorphism $$T_{\text{poly}}\xrightarrow{L_\infty}D_{\text{poly}},$$ where $T_{\text{poly}}:=(\bigwedge^\bullet_{\mathcal{O}_X}T^1(X))[1]$ is the dgla of (shifted) polyvector fields on $X$ and $D_{\text{poly}}:=C^\bullet(\mathcal{O}_X)[1]$ is the dgla of (shifted) Hochschild chains on $X$ (computing Hochschild cohomology of $\mathcal{O}_X$). The Poisson structure $\pi$ is an MC element on the LHS, we can form the MC twisting and get an $L_\infty$ quasi-isomorphism $$(T_{\text{poly}},d_\pi)\xrightarrow{L_\infty}(D_{\text{poly}},d_{\text{Hoch}_\hbar}),$$ where $d_{\text{Hoch}_\hbar}=d_{\text{Hoch}}+[\mu_\hbar-\mu,-]$, $\mu_\hbar$ is the Kontsevich quantisation.

My question is if this map is $\mathcal{O}_X$-linear. The question makes sense because $\mathcal{O}_X$ acts on both sides. It is clear it is $\mathbb{C}$-linear. It is not an algebra map, but I guess $\mathcal{O}_X$-linear sits somewhere in-between.

I am sorry if this is a trivial question.

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    $\begingroup$ Those aren't $\mathcal{O}_X$-linear complexes, because the differential $d_{\pi}$ isn't $\mathcal{O}_X$-linear. $\endgroup$ – Jon Pridham May 24 at 14:32
  • $\begingroup$ @Jon Pridham, Ah, yes. You are right. Indeed it was a stupid question. $\endgroup$ – FunctionOfX May 24 at 21:03
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$\DeclareMathOperator{\Br}{Br}$Willwacher proves (Theorem 4) in The Homotopy Braces Formality Morphism that there is a $\Br_\infty$-structure on $T_{poly}\mathbb R^n$ extending the $L_\infty$-algebra structure defined by the Schouten bracket for which the first $L_\infty$-morphism extends to a $\Br_\infty$-morphism, and sketches the generalization to arbitrary $X$ in Section 8.7. The $\Br_\infty$-structure is given explicitly in terms of graph integrals, and in particular it extends the Gerstenhaber algebra structure defined by the wedge product and Schouten bracket. The higher components of the $\Br_\infty$-morphism then give coherent nullhomotopies of the failure of the formality morphism being an algebra map. One may twist the $\Br_\infty$-formality morphism with a Maurer-Cartan element, i.e. a Poisson bitensor, to obtain similar coherent nullhomotopies extending your second $L_\infty$-morphism.

As Jon Pridham notes in the comments, your question does not have a definite answer as stated because the source and target of the second $L_\infty$-morphism do not carry $\mathcal O_X$-action. Maybe the following is a useful alternative question: The subspace $\mathcal O_X\subset T_{poly}$ is closed under the $\Br_\infty$-algebra structure defined by Willwacher, and the induced $\Br_\infty$-algebra structure on $\mathcal O_X$ is just given by the ordinary commutative product with vanishing shifted Poisson and higher brackets. I think the answer is yes, essentially since $\mathcal O_X$ are the invariants under the fiberwise rescaling of polyvector fields, for which all of these brackets have nonzero weight. This is of course no longer true in the twist by a Poisson bitensor, since $\mathcal O_X$ isn't even closed under the differential.

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