All Questions
1 question
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 ...