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Let $M$ be a manifold. Does it necessarily admit an elliptic operator on $C^{\infty}(M)$ which satisfy Leibniz rule?

Let $M$ be a symplectic manifold with the standard Poisson structure on $C^{\infty}(M)$. Does it necessarily admit an elliptic operator $D$ which satisfies $D(\{f,g\})=\{D(f),g\}+\{f,D(g)\}$?

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    $\begingroup$ I don't understand why the second question is different. It makes no reference to the Poisson structure on $M$. $\endgroup$
    – Deane Yang
    Sep 18 '20 at 15:31
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Let me answer your first question:

No, in general not: Assume $D$ is an elliptic differential operator on $\mathcal C^\infty(M)$ satisfying the Leibniz rule $$D(fg)=f D(g)+D(f)g$$ for all $f,g\in\mathcal C^\infty(M).$ Then, $D$ must be a first order differential operator as the commutator with the multiplication operator is a zeroth order differential operator. Hence, $D= X+m$ where $X\in\mathcal X(M)$ and $m$ denotes the multiplication with $m\in\mathcal C^\infty(M)$, see for example Chapter 2 in the book Global Calculus by Ramanan. Thus, ellipticity of $D$ implies that $M$ must be 1-dimensional. Of course, any 1-dimensional manifold admits an elliptic operator on the space of smooth function which satisfies the Leibniz rule.

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