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Let a finite group $G\subset SL(n,\mathbb{C})$ act on $\mathbb{C}^{n}$ in a natural way. Assume there is a crepant resolution of $f:X\rightarrow \mathbb{C}^{n}/G$. When is it possible to write $X$ as $Y/G$ for some $Y$ and $G$-action on $Y$?

This is true for example $\pm id_{\mathbb{C}^{2}} \subset SL(2,\mathbb{C})$ acting on $\mathbb{C}^{2}$. Is it still true for example $$ \langle diag(e^{\frac{2\pi i}{3}},e^{\frac{2\pi i}{3}},e^{\frac{2\pi i}{3}})\rangle \cong \mathbb{Z}/3\mathbb{Z} \subset SL(n,\mathbb{C}) $$ acting on $\mathbb{C}^{3}$? FYI, this has toric crepant resolution.

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  • $\begingroup$ An obvious idea is take $Y$ to be a blowup of the fixed point set (or the set where $G$ does not act freely). I think this works for Kummer surfaces: if $A$ is an abelian surface then its Kummer surface $X$ is a crepant resolution of $A/i$ ($i$ is the involution $x\mapsto -x$ on $A$), but can also be described as $B/i$ where $B$ is the blowup of $A$ along the $2$-torsion subgroup (i.e. fixed points of $i$). $\endgroup$
    – Piotr Achinger
    Commented Aug 30, 2012 at 22:22
  • $\begingroup$ For each affine of $X$, you can take the Spec of the integral closure of the coordinate ring in $\mathbb C(x_1, \dots, x_n)$ and glue together. I'm not sure when this is smooth, a blow-up, etc. $\endgroup$
    – Will Sawin
    Commented Aug 31, 2012 at 0:12
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    $\begingroup$ The Kummer case is precisely Jerome's $SL(2,\mathbb{C})$ example, I think. $\endgroup$
    – Atsushi Kanazawa
    Commented Aug 31, 2012 at 7:12

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By definition $C^n/G = Spec(C[x_1,\dots,x_n]^G)$ and $X$ being a blow up of an ideal $I$ on $C^n/G$ can be written as $$ X = Proj_{Spec(C[x_1,\dots,x_n]^G)}(C[x_1,\dots,x_n]^G \oplus I \oplus I^2 \oplus \dots). $$ Now assume the ideal $I \subset C[x_1,\dots,x_n]^G$ can be written as $$ I = J \cap C[x_1,\dots,x_n]^G = J^G, $$ where $J \subset C[x_1,\dots,x_n]$ is a $G$-invariant ideal (the simplest thing to do is to take $J = C[x_1,\dots,x_n]\cdot I$). Take $$ Y = Proj_{C[x_1,\dots,x_n]}(C[x_1,\dots,x_n] \oplus J \oplus J^2 \oplus\dots). $$ Then $Y/G = X$.

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  • $\begingroup$ It is not clear that $(J^d)^G = (J^G)^d$ for any $d\in\Bbb N$. Certainly, "$\supseteq$" holds, but I think the other inclusion requires that $J$ is generated by $G$-invariants. I don't have an example though, maybe it works for some reason I don't see? $\endgroup$
    – Jesko Hüttenhain
    Commented Feb 13, 2017 at 9:56

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