Question

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!

Answer 1

I think the key point is that given a Hamiltonian torus action, the components of the moment map are Morse-Bott functions which have even dimensional critical submanifolds all with even index.

For example, suppose you have a Hamiltonian circle action on X. If p is fixed by the action, the circle acts on the tangent space T_pX at p and so T_pX decomposes into eigenspaces each of which is necessarily even-dimensional. The spaces where the action is trivial are tangent to the fixed locus, those with negative weight are where the hessian is negative definite and those of positive weight are where the hessian is positive definite. It follows that each component of the critical locus has even dimension and even index.

To see why this implies that the fibres are connected, at least in the case of a circle action, one applies Mores-Bott theory. When t passes a critical level the level set mu^-1(t) changes by a certain type of surgery. The only surgeries which can alter the connectedness of the level-set are those of index or coindex 1, but we have just seen that this never happens for a Hamiltonian circle action. So all the fibres are connected.

To pass from a circle action to a torus action one can use induction. If you get stuck, searching for things like "Morse-Bott even index torus" should help you find the proof on Google somewhere. Alternatively, if you are feeling more traditional, you could look at Atiyah's original proof in your library: "Covnexity and commuting Hamiltonians" in Bulletin of the LMS 1982 14(1).

Answer 2

The first question is OK, i.e., the fiber of \eta is indeed connected for every regular value \eta, since it is like a product here - please refer to Ehresmann's Theorem and also a related post "Can connectedness of fibers of a smooth map be checked on a dense set?".

So the real issue is on the second problem!