Restriction on the number of component functions if Poisson brackets vanish? Problem: Let $\omega$ be a symplectic form on a manifold $M$, where $\dim(M) = 2n$, and let $F: (f_1, \ldots, f_k) : M \rightarrow \mathbb{R}^k$ be a submersion.
If $H$ is a function such that $\left\{H, f_i \right\} = 0$ for all $i$, then for all $p \in M$ the integral curve of $X_H$ through $p$ is contained in the submanifold $F^{-1} (F(p))$. 
Prove the following: if $\left\{f_i, f_j \right\} = 0$ for all $i, j$, then we necessarily have $k \leq n$. 
Attempt: I was not sure how to show this. By the remark, I know that for all $p \in M$ the level set $F^{-1} (F(p))$ is non-empty (because those integral curves are contained in it), hence it is a submanifold of $M$ of dimension $2n - k$. 
I also know that rank$[ \frac{ \partial f_i}{\partial x_j} ] = k$ because $F$ is a submersion. However, why cannot we have $k > n$? 
 A: In general, without using the remark you mentioned, one can show that $F^{-1}(q)$ is a coisotropic submanifold of $M$, for each $q$ lying in the image of $F$. Since coisotropic submanifolds are at least of middle dimension, we get
$$
2n-k=\dim F^{-1}(q)\geq n,
$$
and therefore $k\leq n$. To show that $F^{-1}(q)$ is coisotropic, you can proceed as follows:
1) Note that $T_{x}F^{-1}(q)=\ker d_{x}F=\ker d_{x}f_{1}\cap\cdots\cap\ker d_{x}f_{k}$.
2) Since $F$ is a submersion, we have that $d_{x}f_{1},\ldots, d_{x}f_{k}$ are linearly independent at each point $x$. By 1) we then get that the annihilator of $T_{x}F^{-1}(q)$ is given by
$$
\left(T_{x}F^{-1}(q)\right)^{0}=\left(\ker d_{x}F\right)^{0}=\text{Span}\{d_{x}f_{1},\ldots, d_{x}f_{k}\}.
$$
3) We can now show that $F^{-1}(q)$ is coisotropic by computing the symplectic orthogonal
\begin{align}
\left(T_{x}F^{-1}(q)\right)^{\omega_{x}}=\left(\omega_{x}^{\flat}\right)^{-1}\left(T_{x}F^{-1}(q)\right)^{0}&=\text{Span}\{\left(\omega_{x}^{\flat}\right)^{-1}(d_{x}f_{1}),\ldots,\left(\omega_{x}^{\flat}\right)^{-1}(d_{x}f_{k})\}\\
&=\text{Span}\{X_{f_{1}}(x),\ldots,X_{f_{k}}(x)\}.
\end{align}
Since $0=\{f_{i},f_{j}\}$, we have $d_{x}f_{i}(X_{f_{j}})=0$, so part 1) implies that
$$
\left(T_{x}F^{-1}(q)\right)^{\omega_{x}}\subset T_{x}F^{-1}(q).
$$ 
Using your remark, there is a simpler proof. The remark implies that the Hamiltonian vector fields $X_{f_{1}},\ldots,X_{f_{k}}$ are tangent to $F^{-1}(F(p))$. But since $F$ is a submersion, we have that $df_{1},\ldots,df_{k}$ are linearly independent everywhere, hence also $X_{f_{1}},\ldots,X_{f_{k}}$ are everywhere linearly independent. So we obtain
$$
2n-k=\dim F^{-1}(F(p))\geq k.
$$
