[Symplectic geometry] Show that a submanifold is Lagrangian

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1. Jun 7, 2017

RedTachyon

1. The problem statement, all variables and given/known data

Let $(M, \omega_M)$ be a symplectic manifold, $C \subset M$ a submanifold, $f: C \to \mathbb{R}$ a smooth function. Show that $L = \{ p \in T^* M: \pi_M(p) \in C, \forall v \in TC <p, v> = <df, v> \}$ is a langrangian submanifold. In other words, you have to show that $\omega |_L = 0$ and $dim\ L = \frac{1}{2} dim\ M$

2. Relevant equations

$\alpha: M \to T^*M$
$\alpha^*\theta_M = \alpha$ where $\theta_M$ is Liouville's form ( $d\theta_M = \omega_M$ )

3. The attempt at a solution

I'm trying to start with decomposing $p \in L$ like that: $p = df + \varphi$, where $\varphi \in TC^\circ$, since $p$ can differ from $df$ only by $\varphi:\ <\varphi,v> = 0\ \forall v \in TC$. But I think there could be several possible $\varphi$ for each point? There should be, since we know nothing about $dim\ C$. I also tried doing something with the pullback $P^*\theta_M$ where P would be a one-form that takes on the value $p$ at $\pi_M(p)$ but I can't really get any results.

Thanks in advance for any help!

2. Jun 12, 2017

PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.