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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.

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1 Answer 1

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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 $$T\Sigma=(\ker df_1\cap\ker df_2)|_{\Sigma}=(\operatorname{span}\{X_{f_1},X_{f_2}\})^\perp|_{\Sigma}.\tag{1}$$
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 exactly that $df_1(X_{f_2})\equiv\{f_1,f_2\}$ has no zeroes on $\Sigma.$

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That is very helpful, Thank you! –  Novo May 4 '12 at 10:42
@user29751 You are welcome. It is gratifying to be useful to other mathematicians. By the way, how do you have met this question? –  Giuseppe Tortorella May 4 '12 at 12:08
I was reading an article on a KAM problem in celestial mechanics. In this article the author restricted the phase space by fixing two first integrals. In doing so a degeneracy in the frequency vector could be overcome. I was wondering why the resulting restricted phase was a symplectic submanifold since the author did not make a comment. –  Novo May 5 '12 at 9:50
It is interesting. And, in this situation, were you able to verify the non-vanishing condition for the Poisson bracket of the two first integrals on their common level-set? –  Giuseppe Tortorella May 5 '12 at 10:15
Yup, in the problem the integrals corresponded to total angular momenta $C_1,C_2,C_3$. I fix $C_1=C_2=0$ and $C_3 \neq 0$ to construct the symplectic submanifold. This works since the angular momenta satisfy the property: $\{C_1 ,C_2 \} = -C_3$. Although, ...now I am not sure anymore if $C_1=C_2=0$ and $C_3 \neq 0$ is a symplectic submanifold –  Novo May 6 '12 at 15:08

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