Let $X = \operatorname{Spec}(R)$ be a conical Poisson variety over $\mathbb{C}$. This means that $R$ is a non-negatively graded Poisson algebra over $\mathbb{C}$ with $R_0 = \mathbb{C}$ and that the Poisson bracket is homogeneous of negative degree. Assume that $X$ admits a conical symplectic resolution. Examples to keep in mind are:

  • the nilpotent cone in a simple Lie algebra
  • a Slodowy slice to a nilpotent orbit inside the nilpotent cone
  • the symmetric scheme of $n$ points on a Kleinian singularity
  • a hypertoric variety
  • a quiver variety.

Let $S$ be a symplectic leaf of $X$.

Q: Is it possible for the fundamental group of $S$ to be infinite?

In the first four examples listed above, the fundamental group of every symplectic leaf is finite. I don't know whether or not this property holds for quiver varieties, or for other examples.


1 Answer 1


I do not know the answer. I believe that it is so. See http://arxiv.org/abs/1301.1008 for algebraic fundamental groups.

If you consider quiver varieties of affine types, they are moduli spaces of instantons on ALE spaces. If you replace the base space by $\mathbb R^4$ and assume rank is $2$, then the finiteness follows from Atiyah-Jones conjecture for rank $2$. The Atiyah-Jones conjecture actually says much stronger: there is a homotopy equivalence between moduli spaces and the infinite dimensional space of all connections up to dimension explicitly given by the instanton number. The conjecture was proved by Boyer et al.

There are lots of generalizations of the Atiyah-Jones conjecture. So it probably applies to quiver varieties of affine types.

  • $\begingroup$ Thanks, your reference to Namikawa's paper is very helpful! In fact, finiteness of the algebraic fundamental group is sufficient for the application that I have in mind, so this is perfect. $\endgroup$ Mar 18, 2013 at 5:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.