In this blog post by John Baez, he paints a (perhaps not original) picture of how one might expect that the minimal resolution of a Kleinian singularity $\mathbb{C}^2/G$ is given by Nakajima's equivariant Hilbert scheme of points. At the end though, he admits that its not clear how one could interpret the exceptional divisor, indeed the end of the post reads:

As I hope you see, these are certain ‘limits’ of 600-cells that have ‘shrunk to the origin’… or in other words, highly symmetrical ways for 120 points in C2 to collide at the origin, with some highly symmetrical conditions on their velocities, accelerations, etc.

That’s what I need to understand.

I am looking for some resource or a pointer to how one could formalize this further, indeed since the Hilbert scheme of points is typically very singular (right?) its a miracle to me that this is a resolution, but I would also like an interpretation of the exceptional divisor in terms of the language above, i.e. trajectories of polyhedra.

Thanks in advance.

Crosspost on MathOverflow

  • $\begingroup$ There is a geometric interpretation using K-theory by Gonzalez-Sprinberg and Verdier. Also for a smooth surface the corresponding Hilbert scheme is smooth. This is the symmetric product which is singular. $\endgroup$ Commented Jan 15, 2019 at 23:50
  • $\begingroup$ I do know of the K-theory interpretation, this is sort of where I'm heading anyway...I wasn't aware that the Hilbert scheme of points on a smooth surface is smooth, thanks. I was hoping for a geometric interpretation of the exceptional divisor besides just a bunch of $\mathbb{P}^1$'s though, I was hoping to see the Dynkin quiver from viewpoint of Hilbert scheme of points, as hinted at my Baez above. $\endgroup$
    – DKS
    Commented Jan 16, 2019 at 2:16
  • 1
    $\begingroup$ There seems to be an interpretation at the end of these lecture notes math.lsa.umich.edu/~idolga/McKaybook.pdf A really down-to-earth interpretation is for type $A_n$ is to use toric geometry so you just have to make a picture and everything is clear here :) $\endgroup$ Commented Jan 16, 2019 at 8:36


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