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

Now if I take dimension vector (2,2) $(2,2)$ I can presumably get Hilb2 $Hilb^2$ of these surfaces, for an appropriate stability condition.

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

To get a Hilb^2 $Hilb^2$ 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.
show/hide this revision's text 1

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 Hilb2 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.