# Integrabiliy of a distribution from a coisotropic submanifold

I just did the following exercise from the book "Introduction to Symplectic topology" and I wanted to make sure things are correct:

Let $$(M,\omega)$$ be a symplectic manifold and $$Q\subset$$ a coisotroipic sub manifold. Show that the distribution $$(TQ)^{\omega}\subset TQ$$ is integrable.

First let's see that the distribution is integrable.To do so we will use the Frobenius theorem. Since $$Q$$ is a submanifold of $$M$$, let's assume of codimension $$k$$, we know that locally,i.e., for every point $$p$$ there exists an open set $$U$$ containing it and local coordinates $$(x_1,...,x_n)$$, such that $$U\cap Q=\{x\in U:x_1(x)=...=x_k(p)=0\}$$.Now let´s consider the local functions $$x_1,...,x_k$$ that define $$Q$$ and the Hamiltonian vector fields $$X_{x_k}$$.We will have that these vector fields will span the distribution locally.This is due to the fact that $$X_{x_k}\subset TQ^{\omega}$$ since $$\omega(X_{x_k}(p),v)=d_px_k(v)=0$$ since $$v\in T_p Q$$.We also now that $$TQ^{\omega}$$ is a $$k$$-dimensional distribution, since $$Q$$ has codimension $$k$$, and now we just need to check that the vector fields $$X_{x_i}$$ with $$i=1,...,k$$ are linearly independent.But if this were note the case we would have that there exists an $$i$$ such that $$X_{x_i}=\sum_j a_jX_{x_j}$$ and consequently $$d_{x_i}=\sum_j a_j d_{x_j}$$ and this would contradict the fact that they are part of a coordinate system. \par Now since the distribution is locally generated by these vector fields we will have that it's smooth and now we just need to check that it's involutive. First we check that $$[X_{x_i},X_{x_j}]=0$$, and this is due to the fact that $$i[X_{x_i},X_{x_j}]\omega=dH$$ and that $$\omega$$ is non-degenerate , with $$H=\omega(X_{x_j},X_{x_i})=0$$, since $$Q$$ is a coisotropic submanifold. Now for general $$X,Y\in \mathfrak{X}(TQ^{\omega})$$ we will have that locally $$X=\sum_{j=1}^k a_j X_{x_j}$$ and $$Y=\sum_{i=1}^k b_iX_{x_i}$$, where $$a_j$$ and $$b_i$$ are smooth functions.And so we will have that

$$[X,Y]=\sum_{j,i=1}^ka_jX_{x_j}.b_i X_{x_i}-b_iX_{x_i}.a_j X_{x_j}+a_jb_i[X_{x_j},X_{x_i}]$$

where $$X_{x_j}.b_i$$ and $$X_{x_i}.a_j$$ are smooth functions, and since we know that $$[X_{x_j},X_{x_i}]=0$$ we get that this distribution is involutive. Since it's smooth and involutive frobenius theorem tells us that it's integrable.\par Now to check that it's isotropic notice that $$\omega$$ vanishes on $$TQ^{\omega}$$, since $$Q$$ is coisotropic and so the leaves of the corresponding foliation will be isotropic.

Your proof looks correct to me, assuming that $$Q$$ is an embedded (rather than immersed, say) submanifold - you use this when you find coordinates adapted to $$Q$$. There is a slightly quicker way of proving this: let $$\omega_Q$$ denote the restriction of $$\omega$$ to the submanifold $$Q$$ (i.e. $$\omega_Q = i_Q^*\omega$$, where $$i_Q:Q\to M$$ is the inclusion). Let $$X, Y$$ be smooth sections of $$(TQ)^\omega$$. Then $$i_X\omega_Q = i_Y\omega_Q = 0$$. Also, using Cartan's magic formula, $$\mathcal{L}_X\omega_Q = i_Xd\omega_Q + d(i_X\omega_Q) = 0$$ (using $$d\omega_Q = d(i_Q^*\omega) = i_Q^*d\omega = 0$$ for the first term). We wish to show that $$[X,Y]$$ is also a section of $$(TQ)^\omega$$, i.e. that $$i_{[X,Y]}\omega_Q = 0$$. This can then be shown using the identity $$i_{[X,Y]} = [\mathcal{L}_X,i_Y]$$, and the above facts. It follows that $$(TQ)^\omega$$ is involutive, hence integrable.