Symplectic submanifolds and first integrals - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T19:27:40Z http://mathoverflow.net/feeds/question/95909 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95909/symplectic-submanifolds-and-first-integrals Symplectic submanifolds and first integrals Ivah 2012-05-03T19:47:43Z 2012-05-04T15:09:18Z <p>I was working with symplectic submanifolds when I posed the following question:</p> <p>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?</p> <p>Any help is welcome.</p> http://mathoverflow.net/questions/95909/symplectic-submanifolds-and-first-integrals/95912#95912 Answer by Vladimir S Matveev for Symplectic submanifolds and first integrals Vladimir S Matveev 2012-05-03T20:10:17Z 2012-05-04T11:45:18Z <p>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 $\mathcal{N}$ </p> http://mathoverflow.net/questions/95909/symplectic-submanifolds-and-first-integrals/95987#95987 Answer by Giuseppe for Symplectic submanifolds and first integrals Giuseppe 2012-05-04T14:31:59Z 2012-05-04T15:09:18Z <p>I am posting here the answer that I gave to the same question when it was posted yesterday on <a href="http://math.stackexchange.com/questions/140470/symplectic-submanifolds-and-first-integrals" rel="nofollow">MSE</a>.</p> <p>Let $f_1$ and $f_2$ be independent functions on a symplectic manifold $(M,\omega).$<br> Let us denote by $\Sigma$ the submanifold $f_1^{-1}(0)\cap f_2^{-1}(0)$ of codimension $2$ in $M$.<br> The tangent bundle of $\Sigma$ is <code>$$T \Sigma= (\ker df_1\cap \ker df_2) |_\Sigma=(\text{span}\{X_{f_1},X_{f_2}\})^\perp|_\Sigma.\tag{1}$$</code><br> So in the symplectic vector bundle <code>$(T_{\Sigma} M,\omega |_\Sigma)$</code> the vector sub-bundle $T\Sigma$ has orthogonal complement $$(T\Sigma)^\perp=\operatorname{span}\{X_{f_1},X_{f_2}\}|_\Sigma.\tag{2}$$</p> <p>By definition, $\Sigma$ is symplectic in $(M,\omega)$ if and only $T\Sigma\cap(T\Sigma)^\perp=0 (\leftarrow\text{the zero section of }\Sigma).$<br> By (1) and (2), this means : in any point of $\Sigma$ the linear system <code>$$\left\{\begin{array}{c}0=\langle df_1,c_1X_{f_1}+c_2X_{f_2}\rangle=\{f_1,f_2\}c_2, \\0=\langle df_2,c_1X_{f_1}+c_2X_{f_2}\rangle=-\{f_1,f_2\}c_1\end{array}\right.$$</code> has only the trivial solution $c_1=c_2=0.$<br> Therefore $\Sigma$ is symplectic iff $\{f_1,f_2\}$ has no zeroes on $\Sigma.$ </p>