Symplectic submanifolds and first integrals - MathOverflow

Symplectic submanifolds and first integrals

2012-05-03T19:47:43Z

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.

Answer by Vladimir S Matveev for Symplectic submanifolds and first integrals

2012-05-03T20:10:17Z

I believe the conditions are that the poisson bracket of $C_1 $ and $ C_2$ is $\ne 0$ and the differentials $dC_1$ and $dC_2$ are linearly independent at all points of $\mathcal{N}$. The second condition implies that N is a submanifold. If the first condition is not fulfilled, $N$ is clearly not a sympletic submanifold because we could take the first two coordinates in the Darboux theorem to be $C_1,C_2$. If the first condition is fulfilled, we take the coordinate system such that the first two coordinates are $C_1, C_2$. In this coordinates the matrix of the symplectic form is blockdiagonal with the first $2\times 2$ block being nondegenerate matrix and all other elements such that the index contains ``2'' equal to $0$. Then, the second block is nondegenerate and it is more or less the same as the matrix of the restriction of the form to $\mathcal{N}$

Answer by Giuseppe for Symplectic submanifolds and first integrals

2012-05-04T14:31:59Z

I am posting here the answer that I gave to the same question when it was posted yesterday on MSE.

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 $\Sigma.$