Suppose we have a Hamiltonian action of a torus T=T^m=R^m/Z^m on a compact, connected symplectic manifold M. According to the convexity theorem, we know every fiber of the momentum map \mu: M--->R^m is connected. My question here is about the proof.
We assume the Hamiltonian action is effective without loss of generality, i.e., only the zero point of T fixes M. I already know that the set of regular values of \mu is dense in \mu(M), also the set of points \eta in \mu(M) with (\eta_1, ..., \eta_(m-1)) a regular value for the reduced momentum map (\mu_1, ..., \mu_(m-1)) is dense in \mu(M). I also know that the fiber of \eta is connected whenever (\eta_1, ..., \eta_(m-1)) a regular value for the reduced momentum map. "Since the set of such points is dense in \mu(M) it follows by continuity that the fiber of \eta is connected for every regular value \eta." (in the book by McDuff and Salamon), and this is my first question. My second question is that then we can imply that every fiber of \mu is connected?
Notice that here the Hamiltonian action must play a special role, as the following is not true:
Suppose f: M--->N is a smooth map between two smooth manifolds, with M compact and connected, and suppose there is a dense subset of f(M) where each fiber is connected, then each fiber of f is connected. Example: consider a natural smooth surjection from S^1 to the figure eight. The fiber over the nodes of the figure eight has two points but every other fiber is a single point.
If anyone knows how to prove the connectedness part of the Convexity Theorem, could you please show us? Thank you very much!