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

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.