Are orbifold singularities canonical? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T06:13:18Z http://mathoverflow.net/feeds/question/36392 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36392/are-orbifold-singularities-canonical Are orbifold singularities canonical? Charles Siegel 2010-08-22T20:10:26Z 2010-08-23T10:01:53Z <p>This is a direct consequence of my previous question: <a href="http://mathoverflow.net/questions/36321/extending-group-actions-on-varieties" rel="nofollow">http://mathoverflow.net/questions/36321/extending-group-actions-on-varieties</a></p> <p>In his answer, inkspot said that group actions can be extended if the variety has ample canonical class and is smooth, but mentions that canonical singularities can be allowed. Now, my situation doesn't involve a smooth variety, but instead I have an orbifold.</p> <p>However, I know that for surfaces, the canonical singularities are the duVal singularities, and are all orbifold points (they're all $\mathbb{C}^2$ modulo a finite subgroup of $\mathrm{Sl}_2$.)</p> <p>Now, I've not studied general singular points of surfaces, so I could be wrong already with surfaces, but are orbifold singularities canonical?</p> http://mathoverflow.net/questions/36392/are-orbifold-singularities-canonical/36411#36411 Answer by Richard Borcherds for Are orbifold singularities canonical? Richard Borcherds 2010-08-22T22:19:51Z 2010-08-22T22:31:03Z <p>In 2 dimensions canonical singularities are the du Val singularities, which are quotients by finite subgroups of SL<sub>2</sub>, so quotients by finite subgroups of GL<sub>2</sub> that are not in SL<sub>2</sub> are a good place to look for counterexamples. (However the quotients by some of these subgroups can be nonsingular, so you have to be a bit careful.) Unless I've misunderstood what is going on, most of the cyclic quotient singularities are non-canonical (these are quotients of C<sup>2</sup> by a cyclic group acting as (x,y) &rarr; (x&zeta;,y&zeta;<sup>n</sup>) for &zeta; a root of 1). </p> <p>Reid's paper <a href="http://www.warwick.ac.uk/~masda/surf/more/cyclic.pdf" rel="nofollow">http://www.warwick.ac.uk/~masda/surf/more/cyclic.pdf</a> gives more details</p> http://mathoverflow.net/questions/36392/are-orbifold-singularities-canonical/36430#36430 Answer by Remke Kloosterman for Are orbifold singularities canonical? Remke Kloosterman 2010-08-23T10:01:53Z 2010-08-23T10:01:53Z <p>For quotient singularities there is the so-called Reid-Tai criterion to check whether the singularity is canonical or not. Suppose $G$ is a finite subgroup of $GL_n(\mathbb{C})$ without quasi-reflections. Let $m=|G|$ and fix a primitive $m$-th root of unity $\zeta$. Let $g\in G$ and let $0\leq a_i &lt; m$ be such that $\zeta^{a_1},\dots,\zeta^{a_n}$ are the eigenvalues of $g$. Then the Reid-Tai sum of $g$ is defined as $\Sigma(g):=1/m(\sum a_i)$. </p> <p>The Reid-Tai criterion states that $\mathbb{C}^n/G$ has a canonical singularity at 0 if and only if $\Sigma(g)\geq 1$ for all $g\in G, g\neq id$.</p> <p>(Taking a quotient by quasi-reflections does not yield a singularity, the Reid-Tai sum depends on the choice of $\zeta$, the criterion does not.)</p> <p>For more on this see e.g. M. Reid's Young person's guide to canonical singularities (Bowdoin 1985 proceedings).</p>