User noz - MathOverflow

Answer by Noz for Contact manifolds that are not cooriented

2011-02-18T22:11:43Z

This is not an answer to your question but here is a natural example where a non cooriented contact structure arise:

the space of contact elements (the grassmanian of hyperplane) on a manifold is isomorphic $P(T^*M)=T^*M\setminus M_0/\mathbb{R}$ . The local contact forms are locally given by the Liouville form on a local section of the quotient map by the action of $\mathbb{R}$. However you cannot define a global form (you can on a double cover which corresponds to the space cooriented contact elements).

This space is of interest, for instance submanifolds of $M$ give Legendrian submanifolds of $P(T^*M)$, and isotopy give Legendrian isotopy. Section of $P(T^*M)$ which are contact submanifolds are exactly the contact structure on $M$, etc.

Answer by Noz for Do there exist closed symplectic manifolds with Euler characteristic zero?

2011-01-03T22:15:36Z

You can find plenty of example of symplectic $4$-manifolds with $0$ Euler characteristic by taking $Y\times S^1$ where $Y$ is a fibered $3$-manifolds. See: http://www.zentralblatt-math.org/zmath/en/search/?q=an:0324.53031&format=complete for a proof (I couldn't find an on-line version of the paper, however the construction is outlined there: http://arxiv.org/abs/1001.0132).

For the vanishing of $\omega$ on $\pi_2$ just take the fiber to be something not a sphere (the fibers are symplectic so it couldn't work with the sphere as a fiber).

Those manifolds are non-trivial fibrations over the torus if you take the monodromy to be non trivial.

Answer by Noz for extension of $G$-bundles

2010-11-25T13:43:38Z

Just a quick idea on how I would proceed if the field was $\mathbb{C}$. Let see $S=S\setminus\mathcal{N}(D)\cup \mathcal{N}(D)$ where $\mathcal{N}(D)$ is a normal neighborhood of $D$. You have a $G$ principal bundle $\mathcal{F}$ over $S\setminus\mathcal{N}(D)$ and you want to extend it to all $S$.

Basically you could do so if there is a $G$ principal bundle, $P$, over $S$ such that $\pi^*(P)=\mathcal{F}\vert_{\partial\mathcal{N}(D)}$ where $\pi$ is the natural projection from $\partial\mathcal{N}(D)$ to $D$.

I have the feeling that the only obstruction is that $\mathcal{F}$ has to be trivial when restricted to the fibers of the unit normal neighborhood of $D$ in $S$.

As we are talking about surfaces then $S$ is of real dimension $4$, if $D$ is of real dimension $2$ then those fiber have dimension $1$. And since you assumed that $G$ was connected then it is necessarily trivial on the fibers...

This is just some thought maybe I'm wrong...

Answer by Noz for nowhere vanishing vector field on a manifold

2010-11-25T11:00:44Z

A bundle is orientable if and only if its first Stiefel-Whitney class is 0 (one can see the first Stiefel-Whitney class as the function $w_1: H_1(M)\rightarrow \mathbb{Z}_2$ which associate to a loop the sign of the determinant of the monodromy).

As mentionned by Ryan, if a line bundle is non-orientable then there is a two sheeted cover of $M$ which orients $L$ (the covering correspond exactly to the index two subgroup $ker(w_1)$) this implies that if an oriented bundle admits a 1-dimensionnal sub-bundle the its Euler class has to be $0$ (regardless if the bundle is the tangent bundle or not).

Finally one can easily see that $T(T^2)$ is the sum of two non-trivial line bundle:

The canonical line bundle $\gamma$ on $\mathbb{R}P^1$ is non trivial but $\gamma\oplus\gamma^*$ is (it is oriented, 2 dimensionnal and admit a section given by the trace map). Pulling back this bundle by the projection $T^2\rightarrow S^1$ you get a trivial bundle (hence the tangent bundle) on $T^2$ written as a sum of two non trivial line bundle ($w_1$ is non zero on each summand).

Comment by Noz

2012-05-08T18:27:09Z

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 $\Omega\vert_M=\omega$.

Comment by Noz

2012-05-08T18:15:32Z

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.

Comment by Noz

2012-05-08T18:14:09Z

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.

Comment by Noz

2011-02-04T18:08:48Z

Forget my comment, I was out of my mind. I erased it

Comment by Noz

2011-01-14T23:11:37Z

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).

Comment by Noz

2011-01-03T22:18:23Z

OK Mike Usher was quicker and clearer I guess...

Comment by Noz

2010-11-25T16:27:10Z

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.

Comment by Noz

2010-11-25T14:03:10Z

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...