I was working with symplectic submanifolds when I posed the following question:


Suppose I have a Hamiltonian system with the phase space $\mathcal{M}$, a symplectic manifold with the standard symplectic form. Now assume that the Hamiltonian system has two first integrals $C\_1,C\_2$. Define the restricted phase space $\mathcal{N}$ of $\mathcal{M}$ by taking $C\_1$=constant,$C\_2$=constant. What kind of conditions does $C\_1$ and $C\_2$ need to satisfy such that $\mathcal{N}$ is a symplectic submanifold?


Any help is welcome.



Symplectic submanifolds and first integrals
I am posting here the answer that I gave to the same question when it was posted yesterday on [MSE](https://math.stackexchange.com/questions/140470/symplectic-submanifolds-and-first-integrals).


Let $f\_1$ and $f\_2$ be independent functions on a symplectic manifold $(M,\omega).$  

Let us denote by $\Sigma$ the submanifold $f\_1^{-1}(0)\cap f\_2^{-1}(0)$ of codimension $2$ in $M$.  

The tangent bundle of $\Sigma$ is 
$$T \Sigma= (\ker df\_1\cap \ker df\_2) |\_\Sigma=(\text{span}\{X\_{f\_1},X\_{f\_2}\})^\perp|\_\Sigma.\tag{1}$$  

So in the symplectic vector bundle $(T\_{\Sigma} M,\omega |\_\Sigma)$ the vector sub-bundle $T\Sigma$ has orthogonal complement $$(T\Sigma)^\perp=\operatorname{span}\{X\_{f\_1},X\_{f\_2}\}|\_\Sigma.\tag{2}$$


By definition, $\Sigma$ is symplectic in $(M,\omega)$ if and only $T\Sigma\cap(T\Sigma)^\perp=0 (\leftarrow\text{the zero section of }\Sigma).$  

By (1) and (2), this means :
in any point of $\Sigma$ the linear system $$\left\{\begin{array}{c}0=\langle df\_1,c\_1X\_{f\_1}+c\_2X\_{f\_2}\rangle=\{f\_1,f\_2\}c\_2, \\0=\langle df\_2,c\_1X\_{f\_1}+c\_2X\_{f\_2}\rangle=-\{f\_1,f\_2\}c\_1\end{array}\right.$$ has only the trivial solution $c\_1=c\_2=0.$  

Therefore $\Sigma$ is symplectic iff $\{f\_1,f\_2\}$ has no zeroes on $\Sigma.$ 


