MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 0 down vote accepted

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 $\Sigma.$

share|cite|improve this answer

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 $\mathcal{N}$

share|cite|improve this answer
Dear Vladimir S Matveev I have a silly doubt about your third paragraph. Let be $p\in\mathcal{N},$ in order to find coordinates $x_1,\ldot,x_n,y_1,\ldots,y_n$ around $p$ in $\mathcal{M}$ with $\omega=dx_i\wedge dy_i$ and such that $x_1=C_1,x_2=C_2$ we need that $\{C_1,C_2\}$ vanishes not only at $p,$ but in a neighborhood of $p$ in $\mathcal{M}.$ At least this is my comprehension of the the Caratheodory-Jacobi-Lie theorem. I would highly appreciate if could explain to me what I am missing. Thank you. – Giuseppe May 8 '12 at 15:44
Dear Guiseppe, you are right. To construct the Darboux coordinates we indeed need that the bracket of $C_1$ and $C_2$ vanish in a small neighborhood. Would the bracket of $C_1$ and $C_2$ vanish at the points of $\mathcal{N}$ only, one can find Poisson-commute functions $\tilde C_1$ and $\tilde C_2$ whose differentials at the point we consider coincides with that of $C_1$ and $C_2$, and construct Darboux coordinates starting from $\tilde C_1$ and $\tilde C_2$, which would be sufficient. – Vladimir S Matveev May 8 '12 at 16:49
But in your third paragraph, we need to consider even eventual isolated zeroes for $\{C_1,C_2\}=\Lambda(dC_1,dC_2).$ E.g. points $p\in\mathcal{N}$ such that $\{C_1,C_2\}(p)=0$ but $\{C_1,C_2\}(q)\neq 0,\ \forall q\in\mathcal{N}\setminus\{p\},$ And in such a case the same holds for $\{\overline{C}_1,\overline{C}_2\}$ independent from the choice of $\overline{C}_1,\overline{C}_2$). I hope to have clearly expressed my doubt, what I am missing? – Giuseppe May 8 '12 at 17:13
Guiseppe, I do not know what to tell you. You have asked whether you are missing something -- no, you are missing nothing and your explanation is more precise than mine one.One can make mine explanation also mathematically precise (because your can build darboux coordinate system such that the differential of the first two coordinates coincide with $dC_1$ and $dC_2$ provided the poisson bracket of $C_1$ and $C_2$ is zero at a point.) This observation will be enough to justify my answer, and you do more or less the same in your answer. – Vladimir S Matveev May 10 '12 at 15:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.