# Elliptic operators and Leibniz rule

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 $$D(\{f,g\})=\{D(f),g\}+\{f,D(g)\}$$?

• I don't understand why the second question is different. It makes no reference to the Poisson structure on $M$. Sep 18 '20 at 15:31

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.