Kernel of $\omega^\#$ is $k$-dimensional

Question

Let $M$ be a smooth manifold with coordinates $\{q^i\}_{i=1}^n$ .The variables $(q^i,p_i)$ are coordinates on the cotangent space $\Omega=T^*M$. Any cotangent space carries a natural one-form $\tilde{\theta}=\sum_i p_idq^i$.

The surface $\Sigma$ can be specified by giving a function $H:\Omega \to \mathbb{R}^k$. $\Sigma$ is then defined by $H=0$ and has dimension $2n-k$.

Denote $\theta$ the restriction of $\tilde{\theta}$ to the surface $\Sigma$ and let $\omega=d\theta$.I want to show that the kernel of $\omega^\#$ defined by $\omega^\#(X):=\omega(X,-)$ is $k$-dimensional but I failed in proving it. I can't get my head around this, any help would be very much appreciated. Thanks.

Answer 1

No wonder you cannot prove that - your claim is false!

Let $i : \Sigma \hookrightarrow T^*M$ be the natural embedding $i(x) = x$. Notice that $\theta = i^* \tilde \theta$. Let $\tilde \omega = \textrm d \tilde \theta .$ Notice that

$$\omega = \textrm d \theta = \textrm d i^* \tilde \theta = i^* \textrm d \tilde \theta = i^* \tilde \omega ,$$

so $\omega$ is a closed, skew-symmetric $2$-form on $\Sigma$, possibly degenerate. The thing is that sometimes it is not degenerate, for instance when $\Sigma$ is a symplectic sub-manifold, and in these cases it follows that $\omega ^\sharp$ is a bundle-isomorphism between $T\Sigma$ and $T^* \Sigma$, therefore its kernel is trivial.

As a concrete example: if $M = \Bbb R^2$ with global coordinates $(x,y)$ then $T^* M \simeq \Bbb R^4$ with global coordinates $(x,y,p,q)$, and with symplectic form $\tilde \omega = \textrm d x \wedge \textrm d p + \textrm d y \wedge \textrm d q$; let $H : T^* M \to \Bbb R^2$ be given by $H(x,y,p,q) = (y,q)$. Notice that, at every point, $\textrm d H = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$, clearly of rank $2$, so $\Sigma = H^{-1} (\{0\})$ is $\{(x,0,p,0) \mid x,p \in \Bbb R \}$, a submanifold of dimension $2$, which can be naturally identified with $\Bbb R^2$ (with coordinates $(x,p)$), and the restriction of $\tilde \omega$ to $\Sigma$ is $\omega = \textrm d x \wedge \textrm d p$, so that $(\Sigma, \omega)$ is clearly a symplectic submanifold, therefore $\ker \omega ^\sharp = 0$.

To conclude, it is not even true in general that $\ker \omega ^\sharp$ is non-trivial - let alone the computation of its dimension.

Answer 2

I believe the kernel of the contraction $\omega^\#$ at a point $p\in \Sigma$ is dependent on the vector field $X$. For example, if we consider $X$ such that $X(p) \in (T_p\Sigma)^{\omega}$ for each $p\in \Sigma$ then the one form is trivial. We can do this with a non-zero vector field anytime the submanifold is not symplectic so that we have a non trivial intersection $(T_p\Sigma)^{\omega}\cap T_p\Sigma $. I believe that the kernel of $\omega^\#$ at a point will be k dimensional when $\Sigma $ is a coisotropic submanifold and $X(p)\notin (T_p \Sigma)^{\omega}$ but I'm sure more could be said with the added structure of the cotangent bundle with the tautalogical 1-form. I'm also not exactly sure if you mean the kernel of $\omega^\#$ at a point or if you mean for what vector fields will $\omega^\#$ be a trivial 1-form.

An easy target for such vector fields $X $ that would make $\omega^\#$ trivial are precisely the vector fields whose flow induces a foliation of isotropic leaves.