# Why is "coisotropic" necessary in coisotropic reduction

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?

• you're right, coisotropy is not really needed, though it does imply that the rank of the kernel is constant, which you would have to impose if you were using arbitrary submanifolds. Say in the usual symplectic reduction at a non-zero moment value one uses a non-coiso submanifold to get the reduced space. Jun 19, 2021 at 16:41

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.