3
$\begingroup$

Let a compact Lie group $G$ acts on a closed symplectic manifold $(M,\omega)$. If the action is Hamiltonian with $\mu$ the moment map, then the integral $$\int_M e^{i\mu (X)+\omega}$$ is equal to the first term in the stationary phase approximation.

My question is if there is a similar integral formula for a compact $M$ with boundary.

Improve this question
$\endgroup$
3
  • $\begingroup$ Yes, at least when the boundary is $G$-invariant and there are no fixed points on the boundary. $\endgroup$
    – Liviu Nicolaescu
    Commented Dec 17, 2015 at 23:36
  • $\begingroup$ @LiviuNicolaescu Can you give a reference for the case you mentioned? $\endgroup$
    – Qijun Tan
    Commented Dec 19, 2015 at 21:47
  • $\begingroup$ The usual proof uses the fact that, outside the fixed point set the integrand can be written as the differential of an explicit form $\theta$ of degree $2n-1$, $2n=\dim M$. Denote by $T_r$ the tube of radius $r$ around the fixed point set. The integral is the equal to the integral of $\theta$ over the boundary of $M$ minus the integral over the boundary of $T_r$. Next let $r\to 0$ and argue as in the boundaryless case. $\endgroup$
    – Liviu Nicolaescu
    Commented Dec 20, 2015 at 11:17

1 Answer 1

Reset to default
1
$\begingroup$

There is the paper by E. Prato and and S. Wu, 1993 which deals with (at least) a special case of this.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .