MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $i$). – Piotr Achinger Aug 30 '12 at 22:22
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. – Will Sawin Aug 31 '12 at 0:12
The Kummer case is precisely Jerome's $SL(2,\mathbb{C})$ example, I think. – Atsushi Kanazawa Aug 31 '12 at 7:12

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 $Y/G = X$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.