User ivah - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:33:01Z http://mathoverflow.net/feeds/user/23433 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100697/a-vector-field-on-a-symplectic-submanifold-intersecting-the-symplectic-complement A vector field on a symplectic submanifold intersecting the symplectic complement Ivah 2012-06-26T16:08:10Z 2012-06-26T16:08:10Z <p>Consider a 4-dim symplectic vector field $X$ on the symplectic manifold $(M, \omega)$ in $\mathbb{R}^4$ with $\omega= \sum_{i=1}^2 dy_i \wedge dx_i$. Moreover, the linear terms of $X$ are given by $y_1 \partial / \partial x_1 - x_1 \partial / \partial y_1$ (so not dependent on $(x_2,y_2)$). Let $X|_{\mathcal{N}}$ be the symplectic vector field restricted to</p> <p>$(\mathcal{N}, \omega|_{\mathcal{N}})$ a 2-dim symplectic submanifold of $(M, \omega)$. Furthermore, suppose that $T_0\mathcal{N}$ lies in the symplectic complement of the tangent space of the plane $(x_2,y_2)$. Does there exist a symplectic transformation such that in an open neighbourhood of zero the $(\mathcal{N}, \omega|_{\mathcal{N}})$ is given by the symplectic manifold $(\hat{\mathcal{N}}, \varpi)$ in $\mathbb{R}^2$ with</p> <p>$$\varpi= d\hat{y}_1 \wedge d\hat{x}_1<br>$$</p> <p>and such that the linear terms of $X|_{\hat{\mathcal{N}}}$ are given by<br> $$\hat{y}_1 \partial / \partial \hat{x}_1 - \hat{x}_1 \partial / \partial \hat{y}_1$$</p> <p>The existence of a 2-form follows by Darboux's theorem but I don't see how to construct a symplectic transformation such that the linear terms become as above.</p> <p>Any help is welcome.</p> <p>Thanks in advance.</p> http://mathoverflow.net/questions/95936/transversal-intersection-of-a-symplectic-manifold-with-a-plane Transversal intersection of a symplectic manifold with a plane Ivah 2012-05-03T23:36:43Z 2012-06-16T11:22:00Z <p>Perhaps this is stupid question. </p> <p>Let $\mathcal{M}$ be a symplectic manifold in $\mathbb{R}^4$ of codimension 2 with the symplectic 2-form $dx_1 \wedge dy_1 + dx_2 \wedge dy_2$. Suppose that $\mathcal{M}$ intersects the $(x_2,y_2)$-plane perpendicular in the origin. Does there exist a symplectic transformation $(x_2,y_2) \mapsto (\hat{x}_2,\hat{y}_2)$ such that for an open neighbourhood of the origin in $\mathcal{M}$ the $\hat{x}_2,\hat{y}_2$ are constant zero?</p> <p>Or is this complete nonsense?</p> 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>