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seen Jun 23 '13 at 7:22

May
8
comment Symplectic boundary
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 $\Omega\vert_M=\omega$.
May
8
comment Symplectic boundary
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.
May
8
comment Symplectic boundary
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.
Apr
19
awarded  Critic
Apr
4
awarded  Fanatic
Feb
18
answered Contact manifolds that are not cooriented
Feb
4
comment Plurisubharmonic exhaustion functions without critical points at infinity
Forget my comment, I was out of my mind. I erased it
Jan
24
awarded  Enthusiast
Jan
14
comment Each element of fundamental group of a topological group represented by homomorphism?
It is probably far from what you're looking for, but you can find counter examples in symplectic geometry. Let $(M,\omega)$ be a symplectic manifold such that $M$ doesn't admit any circle action then there are no homomorphism from $S^1$ to $Ham(M,\omega)$ (the group of hamiltonian diffeomorphisms). However you can find plenty of $4$-dimensionnal example where $\pi_1(Ham(M,\omega)$ is non-trivial (blow ups of $K3$ surfaces for instance).
Jan
3
comment Do there exist closed symplectic manifolds with Euler characteristic zero?
OK Mike Usher was quicker and clearer I guess...
Jan
3
answered Do there exist closed symplectic manifolds with Euler characteristic zero?
Nov
25
comment nowhere vanishing vector field on a manifold
I agree that Stiefel-Whitney classes in general are designed for far more general problem. However in my answer I only talked about the first Stiefel-Whitney class which, correct me if I'm wrong, is specifically designed to address the orientability of vector bundle and is easily defined.
Nov
25
comment extension of $G$-bundles
Note that if you're still in the case of $\mathbb{C}$ and that dimension of $D$ is $0$ then the triviality of $\mathcal{F}$ on the fiber depends on $\pi_2(G)$ so if $G$ is a Lie group $\pi_2(G)$ is trivial, thus you can extend as well...
Nov
25
awarded  Supporter
Nov
25
answered extension of $G$-bundles
Nov
25
awarded  Teacher
Nov
25
answered nowhere vanishing vector field on a manifold