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Suppose $(M,\omega)$ is a symplectic manifold, $X$ and $X'$ are symplectic submanifolds, and $X$ can be smoothly deformed via symplectic submanifolds to $X'$. Is there a diffeomorphism generated by a Hamiltonian vector field that takes $X$ to $X'$?

Additional necessary conditions and partial answers are welcome. For example, this is true if $X$ is a finite set of points from https://mathoverflow.net/a/62018/50754. Is it true if $X$ is a sphere? Or for example, what if we know that $M$, $X$, $X'$ are Kaehler and the homotopy is via Kaehler submanifolds?

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2 Answers 2

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The question is answered by Proposition 4 in this paper by Auroux. The Proposition says that given a family of symplectic submanifolds $\{X_t:t \in [0,1]\}$ in a symplectic manifold $M$, there is a family of symplectomorphisms $\Phi_t:M \to M$ such that $\Phi_0=\operatorname{Id}_M$ and $\Phi_t(X_0)=X_t$. In the question I had asked for a Hamiltonian diffeomorphism. A sufficient condition to ensure this is if $M$ is simply connected. In that case the family $\Phi_t$ is generated by a Hamiltonian vector field.

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For a finite set of points in a connected symplectic manifold, this is Theorem A in a paper of W. Boothby. This was copied from the answer here on this quite similar question https://mathoverflow.net/a/62018/50754

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  • $\begingroup$ Actually I am interested in cases beyond a set of points. That is, when the submanifold $X$ is positve dimensional. I made this point clearer and I included your link in the question itself. $\endgroup$
    – Anon
    Jun 23, 2016 at 3:29

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