Suppose that in phase space $\mathbb{R}^{2n}$ we have a skew-symmetry metric: $\Omega=\sum_{i}dx^{i}\wedge dp_{i}$. We are given $n$ independent function $f_{1}(x,p) , f_{2}(x,p) , ... ,f_{n}(x,p)$ (i.e. having linearly independent vector dradients) which pairwise commute: $\left\{f_{i},f_{j}\right\}=0$ (the poisson bracket). Show that the surface $\Gamma$ in $\mathbb{R}^{2n}$ defined by $f_{1}=0,f_{2}=0,...,f_{n}=0$ is Lagrange (i.e. $\Omega|_{\Gamma}=0$).
I have no idea of this problem, my temp is to find the basis of the tangent space of the surface and shoe that $\Omega(v_{i},v_{j})=0$.... But I don't know how to find the vector.
Thanks in advance.
The gradient of the given $f_1,...,f_n,$ that is, $\{\nabla f_1,...,\nabla f_n\}$ will form a basis of the $n-$dim surface(check!). Then it remains to check how the form $\Omega$ acts these vectors.