Given a possibly singular, connected, symplectic algebraic variety with a torus action, every fiber of the moment map admits a torus action. Is each fiber of this moment map connected? Any examples or counter-examples? Thanks!



connectedness of fibers of torus-equivariant moment maps
In the category of symplectic algebraic varieties moment maps have in general disconnected fibers. Easy example go as follows: Let $T={\bf G}\_m$ act on the affine plane ${\bf A}^2$ by $t\cdot(x,y)=(tx,t^{-1}y)$. Then symplectic form $\omega\_0=dx\wedge dy$ is $T$-invariant. The corresponding moment map is $m\_0(x,y)=xy$.


1. Let $X\_1:={\bf A}^2\setminus\{(0,0)\}$. Then the zero-fiber of $m\_0$ becomes disconnected.
2. A less trivial example is as follows: Let $f(z)$ be an arbitrary non-constant polynomial with derivative $f'(z)$. Now rescale the symplectic form to $\omega=f'(xy)\omega\_0$. Then $\omega$ is nondegenerate on the open subset $X\_2=\{f'(xy)\ne0\}\subseteq{\bf A}^2$. The corresponding moment map is $m(x,y)=f(xy)$. So, unless $f$ is linear, the generic fibers of $m$ are disconnected.
3. A similar example can be obtained as follows: Start with any moment map $m:X\to\mathfrak t^\*$. It is dominant if the action is effective. Let $f:Y\to\mathfrak t^\*$ be any étale morphism. Then $\tilde X:=X\times\_{\mathfrak t}Y$ is Hamiltonian where the symplectic form is the pull-back from $X$. The moment map is the composition $\tilde X\to Y\to \mathfrak t^\*$. It will have disconnected generic fibers unless $f$ is birational.


