All Questions
3 questions
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 ...
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 ...
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^{\...