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Let $G$ be a finite subgroup of $\mathrm{SL}(2, \mathbb{C})$ and let $X$ be the quotient surface $X = \mathrm{Spec}(\mathbb{C}[x, y]^{G})$. Denote by $\mathrm{Hilb}^r([X])$ the equivariant Hilbert scheme on $\mathbb{C}^2$: the connected component of the $G$-fixed loci of $\mathrm{Hilb}^{rg}(\mathbb{C}^2)$ dominating $\mathrm{Sym}^r(X)$, where $g$ is the order of $G$. Write this map

$f: \mathrm{Hilb}^r([X]) \to \mathrm{Sym}^r(X)$.

(The case where $r = 1$ is the $G$-Hilbert scheme of Ito-Nakamura, and the map $f: G-\mathrm{Hilb}(\mathbb{C}^2) = \mathrm{Hilb}^1([X]) \to \mathrm{Sym}^1(X) = X$ is the crepant resolution.)

It would be great if someone can point out some literature in understanding this may $f$ in general. My specific question, might be known to the community is:

Is the map $f$ semismall?

Instead of taking $\mathrm{Hilb}^r([X])$ if one takes the usual Hilbert scheme of $r$ points on the smooth surface $G-\mathrm{Hilb}(\mathbb{C}^2)$ that would give a symplectic resolution of $\mathrm{Sym}^r(X)$. In particular, that is semismall. Maybe one can see that the map $f$ is also a symplectic resolution?

Thanks!

This is a second part of the question, which I try to convince myself:

Q: Is the zero-fiber, the pre-image of the 0-cycle of multiplicity $d$ supported at the singularity of $X$ a Lagrangian subvariety? (I don't think it is irreducible in general though.)

Thanks!

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    $\begingroup$ Assuming that the characteristic of $k$ does not divide the order of $G$, the $G$-invariant element $\beta = dx\wedge dy$ in $H^0(\mathbb{A}^2,\omega_{\mathbb{A}^2/k})$ induces an everywhere nondegenerate element $\beta_n$ on the Hilbert scheme $\text{Hilb}^{P_r}_{[\mathbb{A}^2/G],k}$, where $P$ is the Hilbert polynomial of $r$ times the $G$-representation $k[G]$. Because $\text{Hilb}^{P_r}_{[\mathbb{A}^2/G],k}$ is smooth over $k$, it suffices to check that $\beta_r$ is nondegenerate at every codimension $1$ point, and this is the same computation as nondegeneracy of $\beta_1$. $\endgroup$
    – Jason Starr
    Commented Mar 12, 2017 at 23:24
  • $\begingroup$ @JasonStarr: Thanks! This is what I had in mind. But I am not comfortable to say that $\beta = dx \wedge dy$ is $G$-invariant, since it is only so up to a scalar multiple. $\endgroup$
    – Xudong
    Commented Mar 13, 2017 at 17:11
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    $\begingroup$ "But I am not comfortable to say that $\beta = dx \wedge dy$ is $G$-invariant, since it is only so up to a scalar multiple." If $G$ is a subgroup of $SL_2(\mathbb{C})$, then $G$ preserves $\beta$. $\endgroup$
    – Jason Starr
    Commented Mar 13, 2017 at 17:46
  • $\begingroup$ @JasonStarr: Can I say that the form $\beta_r$ is restricted from the one on $\mathrm{Hilb}^{rg}(\mathbb{A}^2)$ onto this $G$-invariant smooth subvariety? $\endgroup$
    – Xudong
    Commented Mar 13, 2017 at 22:46
  • $\begingroup$ "Can I say ..." Yes, for a finite quotient stack such as $[\mathbb{A}^2/G]$, you can construct $\beta_r$ in that way (up to dividing through by an integer). For a smooth, $2$-dimensional Deligne-Mumford stack with Gorenstein coarse moduli space that is not necessarily a finite quotient stack, it is better to use traces of $p$-forms as in my article with de Jong about rational curves on cubic threefolds. $\endgroup$
    – Jason Starr
    Commented Mar 13, 2017 at 23:59

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