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Is there any work on manifolds (perhaps contact) with symplectic boundary (not asking about the boundary of a symplectic manifold)?

Giuseppe
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asked May 8 2012 at 14:15

jim stasheff
1,386●2●10

	Perhaps it would be good to make your question more precise : are you asking about manifolds whose boudary can be endowed with a symplectic structure ? – Samuel Tinguely May 8 2012 at 14:54
	One way to start thinking about this would be to understand what type of structure $M\times\mathbb{R}$ can have if $M$ is equipped with a symplectic form. If $\omega=-d\theta$ is exact then $dt+\theta$ is a contact form on $M\times\mathbb{R}$, however this exclude compact $M$'s. On the other side, the good notion of boundary of contact manifolds seems to be that of convex hypersurfaces, which outside a dividing set are exact symplectic. – Noz May 8 2012 at 18:14
	For closed symplectic manifold, here springerlink.com/content/m080126681712458 some symplectic manifold arises as boundary of quasi-symplectic manifold which are manifold whith a closed $2$-form with one dimensionnal kernel (in the case of $M\times\mathbb{R}$ it is simply $\omega$. I don't know whether or not those type of manifold have been extensively studied. – Noz May 8 2012 at 18:15
	Maybe one could make a definition like this. A filling of symplectic manifold $(M,\omega)$ is a $2n+1$ manifold with boundary $M$ and with stable Hamiltonian structure $(\Omega,\lambda)$ such that $\Omega\vert_M=\omega$. – Noz May 8 2012 at 18:27

Eugene Lerman · Answer 1 · 2012-08-02 16:33:14Z UTC

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There is a book by Guillemin, Gizburg and Karshon: Moment Maps, Cobordisms, and Hamiltonian Group Actions.

answered Aug 2 at 16:33

Symplectic boundary