Let $M^{2n}$ be a symplectic manifold and let $M^{2n-2}$ be a symplectic submanifold. How to construct a non-trivial Hamiltonian $S^1$-action on $M^{2n-2}$ on a small neighborhood of $M^{2n-2}$, that would fix $M^{2n-2}$? (I think I saw a reference to this statement in some book, but can not find it now.)
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$\begingroup$ A reference is Seidel's biased view of symplectic cohomology. $\endgroup$– YHBKJCommented Apr 27, 2017 at 15:14
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$\begingroup$ Thanks YHBKJ, though there is no proof there "In the simplest case of a smooth D = K, this is quite elementary (it follows from the tubular neighbourhood theorem for symplectic submanifolds)." $\endgroup$– aglearnerCommented Apr 27, 2017 at 15:46
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1 Answer
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I don't have any reference, but I think that the idea is as follows: There is a tubular neighborhood theorem according to which a neighborhood of $M^{2n-2}$ is isomorphic to symplectic disc bundle over $M^{2n-2}$. Take a complex structure J compatible with the symplectic form in each fiber. This gives you a metric $\omega(\cdot, J\cdot)$ on each fiber with associated norm $\|\cdot\|$.
Now take the Hamiltonian function $H(x,y)=\pi\|y\|^2$, where $x$ is the variable in $M^{2n-2}$ and $y$ is the fiber variable. It generates a rotation flow in each fiber, and at time 1, it is back to identity.
EDIT: this answer does not work. See Tim Perutz's comments below for how to fill the gap.
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$\begingroup$ Could you please say what you mean by "symplectic disc bundle"? Do you take $J$ for which the bundle is almost complex? $\endgroup$ Commented Apr 27, 2017 at 1:34
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1$\begingroup$ Thanks! Unfortunately there is a problem. In your defintion the total space of a symplectic bundle does not necessarily have a structure of symplectic manifold. In general, I don't believe that answer is correct in the present form. Indeed, consider, the symplectic manifold $S^2\times S^2$ with a standard product symplectic form $\omega$. Now add to this form a small generic form $\omega_t$ (so that $\omega+\omega_t$ is still symplectic). I claim, that there is no Hamiltonian $S^1$ symmetry of $\omega+\omega_t$ preserving vertical fibration - because of symplectic curvature of this fibration $\endgroup$ Commented Apr 27, 2017 at 9:21
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2$\begingroup$ @aglearner You are correct to notice the distinction between a symplectic vector bundle and a symplectic form on its total space. But the gap can be filled by the following construction of (IIRC) Kostant. Let $p: L\to M$ be a hermitian line bundle over the symplectic manifold $(M,\omega_M)$. Let $\alpha$ be a connection 1-form on the principal $U(1)$-bundle of unit vectors in $L$. Let $\mu$ be the moment map for the rotation action of $U(1)$ on $\mathbb{C}$... $\endgroup$ Commented Apr 27, 2017 at 14:22
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2$\begingroup$ ...Then the form $d\langle \alpha, \mu \rangle + \omega_{\mathbb{C}}$ on $P\times \mathbb{C}$ descends to $P\times_{U(1)} \mathbb{C} = L$, and after adding $p^*\omega_M$ one gets a symplectic form near the zero-section of $L$ which restricts to $\omega_M$ on the zero-section and which has a hamiltonian $U(1)$ action. $\endgroup$ Commented Apr 27, 2017 at 14:22
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2$\begingroup$ I don't know such a reference. My expectation would be that typical proofs will either apply the neighborhood theorem, or use its standard method of proof (Moser's argument). My reference to Kostant was wrong, it should have been to Sternberg, PNAS vol. 74, 1977. $\endgroup$ Commented Apr 27, 2017 at 18:10