Hamiltonian vector field and fibers

Question

Let's assume that $(M,\omega)$ is a symplectic manifold, $N$ an arbitrary manifold and we have some differentiable map $\phi:M\to N$, with conneted fibers.

If we denote by $\phi^{-1}(p)$ the fiber corresponding to $p\in\text{Im}(\phi)$, is it true that $ X_{f\circ\phi}(x) \in T_x \phi^{-1}(q)$ for any $x\in M$, any $f\in C^\infty(N)$ and $q=\phi(x)$? I use $X_g$ to denote the hamiltonian vector field corresponding to $g\in C^\infty(M)$.

In other words, is it true that any vector field of the form $X_{f\circ\phi}$, for some $f\in C^\infty(N)$, sends any point $x\in M$ to a vector in the tangent space at $x$ of the fiber containing $x$ (which is $\phi^{-1}(\phi(x))$)?

Thank you.

Answer 1

This is not true and, in fact, it is the opposite case (in certain sense). Using that $f\circ\phi$ is constant along the fibers, we have, for all vector field $V$ mapping $M$ to the tangent space of some fiber: $$\mathcal{L}_V(X_{f\circ\phi})=0 \iff d(f\circ \phi)(V)=0\iff\omega(X_{f\circ\phi},V)=0. $$ By definition of the $\omega$-orthogonal, this proves that $X_{f\circ\phi}\in\Gamma((\ker(d\phi))^{\bot_\omega})$, where $\ker(d\phi)$ is the set of all the tangent spaces to the fibers.

Answer 2

Take $N$ to be $M$, and define $\phi:M \rightarrow M$ to be the identity map. Then every $f:M\rightarrow \mathbb{R}$ can be written as $f \circ \phi$. The level sets of $\phi$ are just points, so requiring $X_{f \circ \phi}$ to preserve the level sets of $\phi$ is equivalent to requiring it to not move any point, which is certainly wrong for every $f$ which is not locally constant.