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 $TM$.



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> Is there necessarily a non vanishing vector field $X$ on $M$ for which the following submanifold of $(TM, \omega)$ would be a symplectic submanifold?
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> $$\{v\_p\in TM \mid |v\_p|=1, v\_p \perp X(p)\}$$
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> where $v\_p$ is a vector in $TM$ based at point $p\in M$.
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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.



Symplectic submanifolds of the tangent bundle $TM$ which have the form of a vector or fiber bundle
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.


