Why is "coisotropic" necessary in coisotropic reduction

Question

The setup for coisotropic reduction is as follows. Let $(M, \omega)$ be a symplectic manifold and $X \subset M$ a coisotropic submanifold (meaning $T_x X \subset T_x M$ is coisotripic i.e. $(T_x X)^\perp \subset T_x M$ where $W^\perp = \{ v \mid \forall w \in W : \omega(v, w) = 0 \}$).

Now we consider the null distribution $N \subset TX$ defined by $N_x = (T_x X)^\perp$ and check that it is involutive towards Frobenius' theorem. Since $\omega|_X$ descends to a non-degenerate form on $TX/N$, we can hope to parametrize the resulting foliation by a symplectic manifold.

My question is: ``where is $X$ being coisotropic actually necessary?''.

Starting at the level of linear algebra, I've seen it written that we need $W^\perp \subset W$ such that $\omega$ descends to a well-defined non-degenerate form on $W/W^\perp$. However, I think this works in general for $W/(W^\perp \cap W)$ where $W^\perp \cap W = \ker{\omega|_W}$ (considering the map $w \mapsto \omega|_W(w,-)$).

Given any submanifold $X \subset M$ I should be able to define a distribution $N \subset TX$ given by $N = \ker{\omega|_X}$ or explicitly $N_x = (T_x X)^\perp \cap T_x X$. It suffices to check that $N$ is involutive.

Let $\tilde{\omega} = \omega|_X$. Let $X, Y \in \Gamma(X, N)$ then by definition $\iota_X \tilde{\omega} = 0$ and $\iota_Y \tilde{\omega} = 0$. Now consider, $$ \iota_{[X, Y]} \tilde{\omega} = \mathcal{L}_X \iota_Y \tilde{\omega} - \iota_Y \mathcal{L}_X \tilde{\omega} = \mathcal{L}_X \iota_Y \tilde{\omega} - \iota_Y (\iota_X \mathrm{d}{\tilde{\omega}} + \mathrm{d}{\iota_X \tilde{\omega}}) = 0 $$ because $\mathrm{d} \tilde{\omega} = 0$ and $\iota_X \tilde{\omega} = 0$ and $\iota_Y \tilde{\omega} = 0$. Therefore $N$ is involutive and $\tilde{\omega}$ descends to a non-degenerate form on $TX / N = TX / \ker{\tilde{\omega}}$ so the leaves of the foliation generated by $N$ should also be parametrized by a symplectic manifold.

Is there an error in my argument? Where does $X$ being coisotropic come in?

Answer 1

As mentioned in the comments, the only other thing to consider is that the rank of $\ker \omega\rvert_X$ should be constant, so that this is a vector bundle, and the Frobenius theorem applies. Of course, if you want an actual symplectic manifold as the reduction, then you should also assume the leaf space is a manifold (though that's not necessary if you work with more general spaces).

On the other hand, I think it's worth mentioning that this isn't more general than coisotropic reduction in the following sense. The assumption that $\ker\omega\rvert_X$ has constant rank makes $(X,\omega\rvert_X)$ into a so-called presymplectic manifold. Gotay proved that any presymplectic manifold embeds as a coisotropic submanifold in a symplectic manifold, and so this "non-coisotropic" reduction becomes the usual coisotropic reduction from the point of view of the new ambient manifold.