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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 |