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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