Quotient singularities with no crepant resolution?

2011-06-01T12:36:16Z

I read that in dimension $\geq 4$ there are Gorenstein abelian quotient singularities that have no crepant resolutions. What is the simplest example? I immagine that there should be toric examples. Is it the case that the cone does not admit a suitable subdivision in simplicial cones, corresponding to a crepant resolution? In your answers, please consider that I am only an amateur algebraic geometer, but I know some toric geometry.

Answer by Thomas Nevins for Quotient singularities with no crepant resolution?

2011-06-01T13:07:57Z

There is a nice explicit description, in this paper of Morrison and Stevens, of four dimensional cyclic quotient singularities that are Gorenstein and terminal (cf. this MO question---terminal implies the non-existence of crepant resolutions).

Answer by Jim Bryan for Quotient singularities with no crepant resolution?

2011-06-01T22:41:07Z

The simplest example is $\mathbb{C}^4/\pm1$ where $-1$ acts diagonally. I'm sure there is an elementary proof that this does not have a crepant resolution, maybe via toric geometry. Someone else on MO might know a reference. The proof I know uses a result of Yasuda which says that if $X$ is a Gorenstein orbifold and $Y\to X$ is a crepant resolution, then $H^*_{orb}(X) = H^*(Y)$ as graded vector spaces. In the case at hand, this would imply that the exceptional fiber of a crepant resolution $Y\to \mathbb{C}^4/\pm1$ would have cohomology in degree 0 and in degree 4 only, which is impossible for a projective variety.

Answer by Sándor Kovács for Quotient singularities with no crepant resolution?

2011-06-02T02:18:20Z

[EDIT: added proof that $\mathbb Q$-factoriality implies that the exceptional set of any resolution is a divisor.]

Definition A variety is called $\mathbb Q$-factorial if every Weil divisor on it is $\mathbb Q$-Cartier, i.e., some multiple of it is a Cartier divisor.

The general statement you might want is that

Claim A $\mathbb Q$-factorial, terminal singularity does not admit a non-trivial crepant resolution.

Proof $\mathbb Q$-factoriality implies that the exceptional set of any resolution is a divisor, and being terminal implies that all the discrepancies are positive. $\square$

Remark One might think that a Gorenstein terminal singularity does not admit a non-trivial crepant resolution. Here is an example that this is not true. Consider a cone over a smooth quadric surface in $\mathbb P^3$. This is a hypersurface in $\mathbb A^4$, so it is clearly Gorenstein. Blowing up the vertex and a simple calculation using adjunction shows that this is a terminal singularity. However, blowing up a divisor that is a cone over a line on the quadric surface gives a small resolution which will be crepant by being an isomorphism in codimension $1$. This shows that it is necessary to add the $\mathbb Q$-factoriality condition for the above Claim.

Example For the example in Jim Bryan's answer, $\mathbb C^4/\pm$, or more generally, $\mathbb C^{m}/\pm$, the point to notice is that this is just the cone over the Veronese embedding of $\mathbb P^{m-1}$. Isolated quotient singularities are $\mathbb Q$-factorial and an easy computation shows that the discrepancy of the single exceptional divisor of the blow up of the vertex is $\dfrac m2-1$. This implies that it is terminal as soon as $m>2$, but for $m$ odd it will not be Gorenstein, so the first example of the desired kind is for $m=4$.

Addendum Here is a proof that $\mathbb Q$-factoriality implies that the exceptional set of any resolution is a divisor:

Claim Let $X$ be a $\mathbb Q$-factorial variety and $f:Y\to X$ a proper birational morphism. Let $E=\mathrm{Exc}(f)$ denote the exceptional set of $f$, i.e., the largest (closed) subset of $Y$ such that $f|_{Y\setminus E}:Y\setminus E\to X\setminus f(E)$ is an isomorphism. Then $E$ is of pure codimension $1$ in $Y$.

Proof Let $y\in E$ and suppose that $\mathrm{codim}_YE\geq 2$ in a neighborhood of $y$. Let $C\subseteq E$ be an arbitrary proper curve such that $f(C)$ is a point and $y\in C$ and let $H\subseteq Y$ be an effective divisor such that $y\in H$, but $C\not\subseteq H$ . This implies that $H\cdot C>0$. Consider the Weil(!) divisor $f_*H$ on $X$ (the push-forward is meant as a cycle). As $X$ is $\mathbb Q$-factorial, some multiple of $f_*H$ will be Cartier, so replacing $H$ with that multiple we may assume that actually $f_*H$ is Cartier. Then it makes sense to pull it back (as a Cartier divisor). So we get a (Cartier) divisor $f^*f_*H$ which agrees with $H$ on $Y\setminus E$. In particular, if $\mathrm{codim}_YE\geq 2$ in a neighborhood $U$ of $y$, then $H|_U=(f^*f_*H)|_U$. Now by construction $y\in C\cap U\neq\emptyset$, so along $C$, $f^*f_*H=H+F$ where $F$ is an effective (exceptional) divisor that does not contain $C$. Finally, this leads to a contradiction, because we get that $$ 0=f^*f_*H\cdot C \geq H\cdot C>0$$ since $f(C)$ is a point.

Quotient singularities with no crepant resolution? - MathOverflow

Quotient singularities with no crepant resolution?

Answer by Thomas Nevins for Quotient singularities with no crepant resolution?

Answer by Jim Bryan for Quotient singularities with no crepant resolution?

Answer by Sándor Kovács for Quotient singularities with no crepant resolution?