# Irreducibility of holomorphic symplectic quotients

Let a connected algebraic group $G$ (over $\mathbb C$, say) act Hamiltonianly on an algebraic symplectic variety $M$, with moment map $\Phi: M\to \mathfrak g^*$. In the example I care about, vaguely related to Slodowy slices, $G$ is unipotent.

Is $\Phi^{-1}(0)$ necessarily irreducible?

Moreover $M$ contains an irreducible Lagrangian subvariety $L$.

Is $\Phi^{-1}(0) \cap L$ necessarily irreducible?

• As petit comment, In symplectic minimal model program, symplectic quotient is like flips and flops see arxiv.org/pdf/1508.01573.pdf,
– user21574
Commented Jan 16, 2017 at 21:02
• Question? 0 is the regular value of $\Phi$ and $G$ act freely on the level set $\Phi^{-1}(0)$ ? since if not symplectic quotient is the union of symplectic manifolds in general.
– user21574
Commented Jan 16, 2017 at 21:15
• If the momentum map is not proper, the reduced space $\Phi^{-1}(0)/G$ is not necessarily connected.
– user21574
Commented Jan 16, 2017 at 21:23
• My $M$ is going to be singular, so I don't really want to say "regular value" so much. What's an example of this reducibility? Commented Jan 16, 2017 at 21:32
• By Kirwan result fibers of Moment map are connected when choosen points are regular, but when we work on singular case I couldn't see a lot of papers related to that math.cornell.edu/~sjamaar/papers/sour.pdf. In fact when you take 0 not to be regular, it is as same as to take central fiber be singular, which need to impose some mild notion of singularities in the sense of minimal model program, like canonical singularites to work on that, and in fact to get the existence of flips and hence $\Phi^{-1}(0)$ as variety
– user21574
Commented Jan 16, 2017 at 21:39

Work of Gan and Ginzburg (https://arxiv.org/pdf/math/0409262v7.pdf) shows that one of our favorite examples, $M=T^*(\mathfrak{gl}_n\oplus \mathbb C^n)$ and $G=GL(n)$, has a 0 level of the moment map which has $n+1$ components and so is not irreducible (but for a given stability condition, all but one component is unstable).
EDIT: I'll just add that this situation can get quite bad; in the example above, at least things are equidimensional. Even this can fail, and $\Phi^{-1}(0)$ can have a larger dimension than you would expect from the GIT quotient, since only a component of smaller dimension has stable points; this happens for all quiver varieties corresponding to non-dominant weight spaces. For example, if $M=T^*\mathrm{Hom}(\mathbb C^n,\mathbb C^m)$ and $G=GL(n)$, then we have a pair of matrices $A$ and $B$ subject to $BA=0$ as the moment map condition. Obviously, the ranks of these matrices can sum to no more than $m$ or to $2n$. If $2n\leq m$, then there's a single component. If $m<2n$ (which is the non-dominant case; this is a $\mathfrak{sl}_2$ quiver variety), then there are $m-2n+1$ components, given by the closures of the sets where $A$ has rank $k$ and $B$ rank $m-k$. The dimension of this space is given by choosing a dimension $k$ subspace in $\mathbb C^m$, the surjective map of $\mathbb C^n$ to this space, and injective map from its quotient. This gives dimension $\binom mk+nm$. Thus, the largest component is when $k\approx m/2$, but the two different stability conditions are that $A$ is injective or that $B$ is surjective (so $k=n, m-n$); thus, if $n>m/2+1$, we're no longer equidimensional.