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Let $D$ be a differential operator on the space of smooth functions on a manifold $M$. The symbol of $D$ can be considered as a Hamiltonian on the cotangent bundle $T^*M$. We call this Hamiltonian as "Corresponding symbol Hamiltonian"

Motivated by the above interesting linked question and this post and this one we ask the following question:

Is there an elliptic operator on a manifold whose corresponding symbol Hamiltonian has an isolated periodic orbit?

Note: We add the ellipticity condition since we learn from this answer that for differential operator associated with a vector field, which is a non elliptic operator, we do not have an isolated periodic orbit

Denis Serre
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Ali Taghavi
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1 Answers1

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I'm going to assume that you want an isolated periodic orbit on some fixed energy level. Pick your favorite Riemannian manifold $(M,g)$ such that there is an isolated closed geodesic. Then, the geodesic flow on the unit tangent bundle has a corresponding isolated period orbit.

The Laplace-Beltrami operator $\Delta_{g}$ is elliptic and has principal symbol

$\sigma(\Delta_{g})(\xi)=\lVert \xi \rVert_{g}^{2}.$

as a function on $T^{*}M.$ The Hamiltonian flow of $\sigma(\Delta_{g})$ is the geodesic flow of $g,$ and therefore the previous comments apply.

Andy Sanders
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  • Thank you very much for your attention to my question. I am really sory for not replying earlier. By isolated i mean globally isolated not restricted to a level. But I realy appreciate your interesting approach using isolated closed geodesic. BTW how can one define (and gurante its existence) an ISOLATED closed geodesic?My previos +1 to your kindly attention and your answer! – Ali Taghavi Sep 11 '20 at 13:10