Timeline for Constructing a Hamiltonian $S^1$-action on a neighborhood of a symplectic divisor
Current License: CC BY-SA 3.0
13 events
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| Apr 28, 2017 at 7:55 | history | edited | Vincent H | CC BY-SA 3.0 |
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| Apr 28, 2017 at 7:52 | comment | added | Vincent H | Yes, I clearly missed something here. Thanks for filling the gap Tim. | |
| Apr 27, 2017 at 18:10 | comment | added | Tim Perutz | 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. | |
| Apr 27, 2017 at 14:53 | comment | added | aglearner | Tim, thanks for these comments! They fill the gap indeed. May I ask you, maybe you saw the statement in my question proven in some book? (I wonder if one can not construct this Hamiltonian $S^1$-action without using Weinstein neighbourhood theorem..) | |
| Apr 27, 2017 at 14:22 | comment | added | Tim Perutz | ...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. | |
| Apr 27, 2017 at 14:22 | comment | added | Tim Perutz | @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 $\mathbb{C}$... | |
| Apr 27, 2017 at 9:21 | comment | added | aglearner | 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 | |
| Apr 27, 2017 at 7:44 | history | edited | Vincent H | CC BY-SA 3.0 |
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| Apr 27, 2017 at 7:44 | comment | added | Vincent H | $J$ turns this bundle into a complex bundle (actually there is no need for the word "almost" here, and I removed it from my initial answer) Does that answer your question? | |
| Apr 27, 2017 at 7:42 | comment | added | Vincent H | This bundle is symplectic in the sense that there is a (natural) symplectic form on each fiber, which smoothly depends on the base point. Since the base is also a symplectic manifold, the total space carry a symplectic structure as well. | |
| Apr 27, 2017 at 7:38 | comment | added | Vincent H | Let me be more precise on the tubular neighborhood theorem: the bundle involved is the bundle whose fiber at $x\in M^{2n-2}$ is the symplectic orthogonal to $TM^{2n-2}$. This is proved in the book by McDuff and Salamon | |
| Apr 27, 2017 at 1:34 | comment | added | aglearner | Could you please say what you mean by "symplectic disc bundle"? Do you take $J$ for which the bundle is almost complex? | |
| Apr 26, 2017 at 19:55 | history | answered | Vincent H | CC BY-SA 3.0 |