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Suppose $\iota_0:L\xrightarrow{}M$ is a compact Lagrangian submanifold in a symplectic manifold $(M,\omega)$. Let $\iota_t:L\times I\xrightarrow{} M$ be an isotopy and denote $\iota_1(L)$ by $L’$. I wonder if there is a symplectomorphism $\varphi:M\xrightarrow{}M$ that takes $L$ to $L’$. Such a diffeomorphism exists by the Isotopy Extension Theorem, but I don’t see a way to turn that into a symplectomorphism. Any negative results are also appreciated.

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Here is a counter example for the claim:

Take the unit sphere $(S^2, d\theta \wedge dh) $. It is a 2-dimensional symplectic manifold, so every 1-dimensiononal submanifold is Lagrangian.

Let $L \subset S^2$ be the equator $\{h=0\}$. For every $-1<s<1$, the equator $L$ is isotopic to the circle $L_s := \{h=s\}$. Moreover, all of these circles are compact Lagrangian submanifolds.

However, these circles (for $s \ne 0$) cannot be images of $L$ through a symplectomorphism! $L$ divides the sphere into two parts of equal area, and this property must be preserved under a symplectomorphism. On the other hand, for $s \ne 0$, the circle $L_s$ divides the sphere into two parts of different areas.

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  • $\begingroup$ Thanks for the very geometric counterexample. It’s certainly quite nice. Are you aware of a more general result, maybe some kind of an obstruction? I’ll wait for a few days and accept your answer unless something more general comes up! $\endgroup$ Jul 3, 2023 at 1:31
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    $\begingroup$ I can't think of a general obstruction, but I think some similar claims can be proved using displacement energy. If a Lagrangian submanifold is non displacable (meaning it has infinite displacement energy), then it can't we mapped with a Hamiltonian diffeomorphism to any Lagrangian submanifold, if they are disjoint. Sadly, Hamiltonian diffeomorphisms are just a subgroup of the identity component of the group of symplectomorphisms, so it doesn't immediately generalize my answer to your case. $\endgroup$ Jul 3, 2023 at 5:50

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