0
$\begingroup$

Let $(M,g)$ be a Riemannian manifold which admit a non vanishing vector field.(That is $\chi(M)=0$ when $M$ is a compact manifold). We pull back The symplectic structure of the cotangent bundle to the $2$-form $\omega$ on $TM$.

Is there necessarily a non vanishing vector field $X$ on $M$ for which the following submanifold of $(TM, \omega)$ would be a symplectic submanifold?

$$\{v_p\in TM \mid |v_p|=1, v_p \perp X(p)\}$$

where $v_p$ is a vector in $TM$ based at point $p\in M$.

The motivation for this question is the following:

We would like to find some symplectic submanifolds of $TM$ which are in the form of a sub vector bundle of the tangent bundle or sub fiber bundle of unite tangent bundle.

In the standard coordinate $(x_1,x_2,\ldots,x_n,y_1,y_2,\ldots,y_n)$, the elementary examples of symplectic submanifolds are $$(x_1,x_2,\ldots,x_k,0,0,\ldots,0,y_1,y_2,\ldots,y_k,0,0\ldots,0)$$

In such elementary example we loose the whole base space.

$\endgroup$
3
$\begingroup$

I probably don't understand your question correctly, because the answer to the boxed question seems to be: obviously $X$ never exists if $M$ is compact. More generally, there is no closed manifold $V$ and map $f : V \to T^*M$ such that $f^*\omega$ is symplectic. Otherwise you would get an exact symplectic form on a closed manifold, and Stokes forbids this.

I can mention something that looks like what you wrote below the box. A cooriented hyperplane field $\xi$ on $M$ defines a submanifold $S\xi = \{\lambda ; \ker \lambda = \xi\} \subset T^* M$ (equality of cooriented hyperplanes) which is half a rank 1 subbundle. It is symplectic if and only if $\xi$ is contact.

$\endgroup$
  • $\begingroup$ Thank you very much for your interesting answer and your attention to my question. $\endgroup$ – Ali Taghavi Sep 2 '18 at 19:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.