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I have the following question:

Let $(M,\omega)$ be a symplectic manifold and let $N_1$ and $N_2$ be submanifolds of $M$ such that there is a diffeomorphism $\psi:M \rightarrow M$ such that $\psi(N_1) = N_2$. Does there exist a symplectomorphism $\varphi:M \rightarrow M$ such that $\varphi(N_1) = N_2$?

Can anyone give me a hint? I'm really struggling. Thanks!

user353840
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1 Answers1

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As the simplest example I can think of, take a loop in $S^2$. There's a diffeomorphism taking any two loops to another; but the area of the two sides of the loop are preserved by a symplectomorphism (because symplectomorphisms are in particular volume-preserving).

More in line with my comment above, consider $T^*M$ with the standard form. The zero section is a Lagrangian submanifold, and you can easily check that a section (considered as a 1-form $\omega$) is Lagrangian if and only if $\omega$ is closed. But all sections are smoothly isotopic, whence there is a diffeomorphism taking one to another.

  • I don't get it. What is a 'loop' in S^2? Could you maybe give definitions or constructions? I'm not good with mental images of mathematical subjects I'm not familiar with. Also what do you mean by a section being Lagrangian? Sorry if these questions are stupid.. – user353840 Jul 20 '16 at 14:33
  • @user353840 An embedding of $S^1$ into $S^2$. For instance, any of the latitude curves. A submanifold of a symplectic manifold is Lagrangian if the pullback of the symplectic form vanishes identically on it. –  Jul 20 '16 at 14:45
  • So sections are submanifolds? – user353840 Jul 20 '16 at 14:48
  • Their images are. If you don't understand that part of the answer, don't worry about it. –  Jul 20 '16 at 14:48
  • Oh ok. How exactly does that work? If $\pi: E \rightarrow M$ is a vector bundle and $s:M \rightarrow E$ is a section, then $s: M \rightarrow s(M) \subseteq E$ is a bijection. Do when then equip $s(M)$ with the quotient topology or? – user353840 Jul 20 '16 at 15:04
  • subspace topology. – Thomas Rot Jul 21 '16 at 12:56