Quotient singularities with no crepant resolution? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T13:45:09Z http://mathoverflow.net/feeds/question/66657 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66657/quotient-singularities-with-no-crepant-resolution Quotient singularities with no crepant resolution? Diego Matessi 2011-06-01T12:36:16Z 2011-06-03T09:49:48Z <p>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. </p> http://mathoverflow.net/questions/66657/quotient-singularities-with-no-crepant-resolution/66660#66660 Answer by Thomas Nevins for Quotient singularities with no crepant resolution? Thomas Nevins 2011-06-01T13:07:57Z 2011-06-01T13:07:57Z <p>There is a nice explicit description, in <a href="http://dx.doi.org/10.1090/S0002-9939-1984-0722406-4" rel="nofollow">this paper of Morrison and Stevens</a>, of four dimensional cyclic quotient singularities that are Gorenstein and terminal (cf. <a href="http://mathoverflow.net/questions/45818/resolution-of-singularities" rel="nofollow">this MO question</a>---terminal implies the non-existence of crepant resolutions). </p> http://mathoverflow.net/questions/66657/quotient-singularities-with-no-crepant-resolution/66694#66694 Answer by Jim Bryan for Quotient singularities with no crepant resolution? Jim Bryan 2011-06-01T22:41:07Z 2011-06-01T22:41:07Z <p>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. </p> http://mathoverflow.net/questions/66657/quotient-singularities-with-no-crepant-resolution/66702#66702 Answer by Sándor Kovács for Quotient singularities with no crepant resolution? Sándor Kovács 2011-06-02T02:18:20Z 2011-06-03T09:49:48Z <p>[<strong>EDIT:</strong> added proof that $\mathbb Q$-factoriality implies that the exceptional set of any resolution is a divisor.] </p> <p><em>Definition</em> A variety is called <em>$\mathbb Q$-factorial</em> if every Weil divisor on it is $\mathbb Q$-Cartier, i.e., some multiple of it is a Cartier divisor.</p> <p>The general statement you might want is that</p> <blockquote> <p><strong>Claim</strong> A $\mathbb Q$-factorial, terminal singularity does not admit a non-trivial crepant resolution.</p> </blockquote> <p><strong>Proof</strong> $\mathbb Q$-factoriality implies that the exceptional set of any resolution is a divisor, and being terminal implies that all the discrepancies are positive. $\square$</p> <blockquote> <p><strong>Remark</strong> 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 <em>Claim</em>.</p> </blockquote> <p>-</p> <blockquote> <p><strong>Example</strong> 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 $m=4$. </p> </blockquote> <p><strong>Addendum</strong> Here is a proof that $\mathbb Q$-factoriality implies that the exceptional set of any resolution is a divisor:</p> <blockquote> <p><strong>Claim</strong> 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 <em>set</em> 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 $Y$.</p> </blockquote> <p><strong>Proof</strong> Let $y\in E$ and suppose that <code>$\mathrm{codim}_YE\geq 2$</code> 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 <code>$C\not\subseteq H$</code>. 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.</p>