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I am looking for a reference to the following fact which seems to be true and which is probably well-known (at least to experts in resolution of singularities):

Consider an isolated cyclic quotient singularity ${\mathbb C}^n/G$, where $G$ is a finite cyclic group acting on ${\mathbb C}^n$ by linear complex transformations. Then there is a sequence of blow-ups of ${\mathbb C}^n$ with $G$-invariant smooth centers (always lying in the set of points with non-trivial stabilizers under the $G$-action) which yields a complex manifold $M$ so that $M/G$ is smooth and $M/G\to {\mathbb C}^n/G$ is a resolution of the original singularity.

Edit: Jason Starr's example below shows that the claim is actually false.

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    $\begingroup$ You should be able to deduce this from the explicit resolution of toric singularities by torus-equivariant blowups, e.g., as in Fulton's book. $\endgroup$ – Jason Starr Dec 24 '16 at 16:37
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Edit. As Michael Entov points out, the original example (below with strikethrough) is incorrect. (Sorry!) Here is a corrected example. It is not always possible to find an equivariant resolution satisfying all of the hypotheses you list. The simplest example I know of is as follows. Let $G$ be $\mu_5$, the group of fifth roots of unity. Let the action of $\mu_5$ on $\mathbb{C}^2$ be $\zeta\bullet(x,y) = (\zeta x,\zeta^3 y)$. The unique point with nontrivial stabilizer is the origin. One of the standard affine opens for the blowing up of the origin is $\text{Spec}\ \mathbb{C}[u,x]$ where $ux$ equals $y$. Now the action is $\zeta\bullet(u,x) = (\zeta^2u,\zeta x)$. Up to the automorphism $G\to G$ by $\zeta \mapsto \lambda = \zeta^2$, this action is $\lambda \ast(u,x) = (\lambda u,\lambda^3 x)$. Thus, we have the same type of action for the fixed point $(u,x)=(0,0)$ on the blowing up as for the origin on the original variety.

It is not actually possible to find an equivariant resolution satisfying all of the hypotheses you list. The simplest example I know of is when $G$ equals $\mu_3$, the group of cube roots of unity, and where the action of $\mu_3$ on $\mathbb{C}^3$ is by $\zeta\bullet(x,y,z)=(x,\zeta y, \zeta^{-1}z)$. The only point of $\mathbb{C}^3$ with a nontrivial stabilizer is the origin $(0,0,0)$. For the blowing up of the origin, $\nu:\widetilde{\mathbb{C}^3} \to \mathbb{C}^3$, for the point $p$ on the exceptional divisor corresponding to the tangent direction of the line $\text{Zero}(y,z)\subset \mathbb{C}^3$, there is an affine chart of $p$ in $\widetilde{\mathbb{C}^3}$ that is $\mu_3$-equivariantly isomorphic to the origin in $\mathbb{C}^3$. So it is too much to expect that you can resolve the singularities by only blowing up smooth centers inside the locus of points with nontrivial stabilizers.

What people usually do instead is allow the centers to be not-necessarily-smooth. This can still lead to a smooth manifold $M$ such that $M/G$ is a resolution of $\mathbb{C}^n/G$. The "canonical" reference for this is Section 4 of Miles Reid's article, paticularly the proof of Theorem 4.6.

MR0927963 (89b:14016) Reviewed
Reid, Miles(4-WARW)
Young person's guide to canonical singularities. Algebraic geometry, Bowdoin, 1985 (Brunswick, Maine, 1985), 345–414,
Proc. Sympos. Pure Math., 46, Part 1, Amer. Math. Soc., Providence, RI, 1987.
14E30 (14B05 14E05 14J10)

This is also discussed in Sections 2.2 and 2.4 of Fulton's book.

MR1234037 (94g:14028) Reviewed
Fulton, William(1-CHI)
Introduction to toric varieties.
Annals of Mathematics Studies, 131. The William H. Roever Lectures in Geometry.
Princeton University Press, Princeton, NJ, 1993. xii+157 pp.
14M25 (14-02 14J30)

Roughly the idea is to perform the blowings up in such a way that the locus of points with nontrivial stabilizer becomes divisorial in some $\widetilde{\mathbb{C}^n}$. In the (corrected!) example above, if you first blow up the ideal $\langle x,y^2 \rangle$, this makes the fixed locus divisorial, but introduces one ordinary double point. Now, if you blow up the ordinary double point, then you have a smooth $M$ such that also $M/G$ is smooth.

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    $\begingroup$ Thanks, but I think that in your example the original singularity is not isolated: the points $(x,0,0)$ in ${\mathbb C}^3$ all have non-trivial stabilizers. Also, what do you mean by $\textrm{Zero} (y,z)$? $\endgroup$ – Michael Entov Dec 25 '16 at 20:46
  • $\begingroup$ @Michael Entov: Thanks for catching the mistake. I will try to fix it. $\endgroup$ – Jason Starr Dec 25 '16 at 21:08

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