[Math] Deformations of Nakajima quiver varieties

ag.algebraic-geometrydg.differential-geometryqa.quantum-algebraquivers

Are deformations of Nakajima quiver varieties also Nakajima quiver varieties ?

In case the answer to this is (don't k)no(w), here are some simpler things to ask for.

  1. (If you're a differential geometer) Is any hyperkahler rotation / twistor deformation of a Nakajima quiver variety also a Nakajima quiver variety ?

  2. (An example, for algebraic geometers) Consider the Hilbert scheme $H$ of $k$ points on the minimal resolution of $\mathbb C^2/(\mathbb Z/n)$. Or restrict to $n=2=k$ and consider $Hilb^2T^*\mathbb P^1$. Its exceptional divisor over the symmetric product defines a class in $H^1(\Omega_H)$ (despite the noncompactness). Using the holomorphic symplectic form, we get a deformation class in $H^1(T_H)$. (The corresponding deformation is not so far from the twistor deformation, and can be realised as a composition of a deformation of the ALE space followed by a twistor deformation.) Is there a Nakajima quiver variety in the direction of this deformation ?

For instance if I take the quiver $\bullet^{\ \rightrightarrows}_{\ \leftleftarrows}\bullet$, dimension vector (1,1), and an appropriate stability condition (or value of the real moment map) then I get $T^*\mathbb P^1$ as the moduli space over the value $0$ of the complex moment map, and the smoothing of the surface ordinary double point over nonzero values.

Now if I take dimension vector (2,2) I can presumably get $Hilb^2$ of these surfaces, for an appropriate stability condition. However, as I vary the value of the complex moment map I simply vary the surface that I take $Hilb^2$ of, rather than getting the deformation I'm after. (The hyperkahler rotation is not a $Hilb^2$, since the exceptional divisor disappears in this deformation.)

But is there another quivery way of producing this deformation ?

Best Answer

I do not know a general statement. I just want to give a comment:

Now if I take dimension vector $(2,2)$ I can presumably get $Hilb^2$ of these surfaces, for an appropriate stability condition.

No. You only get the symmetric product of $T^*P^1$ if you work on quiver varieties with the dimension vector $(2,2)$.

To get a $Hilb^2$ of the surface, one need to put the one-dimensional vector space $W $at the vertex 0, and take a suitable stability condition. (I hope you are familiar with convention for quiver varieties.)

Then we have two dimensional family of quiver varieties from the complex moment map deformation. Thus we get one more dimension from the deformation of the underlying surface.

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