Let $M$ be a smooth manifold with a Poisson bracket $\{-,-\}$. Kontsevich proved that there exists a deformation quantization of $M$, i.e. let $C^{\infty}(M)[[\hbar]]=C^{\infty}(M)\otimes_{\mathbb{R}}\mathbb{R}[[\hbar]]$ be the formal power series of smooth functions on $M$, there exists an associative star product $$ *_{\hbar}: C^{\infty}(M)[[\hbar]]\times C^{\infty}(M)[[\hbar]]\to C^{\infty}(M)[[\hbar]]$$ which is bilinear over $\mathbb{R}[[\hbar]]$ and for any $f, g\in C^{\infty}(M)$ we have $$f*_{\hbar}g=fg+\sum_{i=1}^{\infty}B_i(f,g)\hbar^i $$ where $B_1(f,g)=\{f,g\}$ is the Poisson bracket.

On the other hand we know on $M$ there is the de Rham algebra $(\Omega^{\bullet}(M),d,\wedge)$, which is a differential graded algebra, and $C^{\infty}(M)$ is the degree zero component of $(\Omega^{\bullet}(M),d,\wedge)$.

My question is: could we also deform $(\Omega^{\bullet}(M),d,\wedge)$ to a differential graded algebra $(\Omega^{\bullet}(M)[[\hbar]],d_{\hbar},\wedge_{\hbar})$ with $(C^{\infty}(M)[[\hbar]],*_{\hbar})$ as its degree zero component?

  • 1
    $\begingroup$ To write it more symmetrically $C^\infty ( T[1] M )$. $\endgroup$ – AHusain Jun 1 '17 at 22:50

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