0
$\begingroup$

I've been trying to proof this using $U(n,\mathbb{C})$ action over all lagrangian subspaces of $\mathbb{R}^n\times \mathbb{R}^n$ but it didn't work. I mean, I got stuck and I didn't know what else to do. I found the excercise on Heckman's book symplectic geometry.

$\endgroup$
  • 1
    $\begingroup$ Um.... why doesn't it work? $U(n, \mathbb C)$ acts transitively, you need only to find which elements in $U(n, \mathbb C)$ fixes the standard Lagrangian $\mathbb R^n \times \{0\}$. $\endgroup$ – Arctic Char Sep 25 at 4:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.