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.


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.

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

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