Why is there no symplectic version of spectral geometry? – mathoverflow.net

Why is there no symplectic version of spectral geometry? – mathoverflow.net

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Fist, 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 ...

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