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28 votes
6 answers
3k views

Why is there no symplectic version of spectral geometry?

First, recall that on a Riemannian manifold $(M,g)$ the Laplace-Beltrami operator $\Delta_g:C^\infty(M)\to C^\infty(M)$ is defined as $$ \Delta_g=\mathrm{div}_g\circ\mathrm{grad}_g, $$ where the ...
B K's user avatar
  • 1,942
3 votes
1 answer
366 views

When is the exterior derivation $d$ a Lie algebra morphism?

In this question we search for some conditions under which the exterior derivation $d:\Omega^i(M)\to \Omega^{i+1}(M)$ on a differentiable manifold $M$ is a Lie algebra morphism in a certain sense. We ...
Ali Taghavi's user avatar
2 votes
1 answer
243 views

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^{\...
Ali Taghavi's user avatar