Is there an obvious way for showing singularities are quotient? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T23:34:31Z http://mathoverflow.net/feeds/question/23091 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/23091/is-there-an-obvious-way-for-showing-singularities-are-quotient Is there an obvious way for showing singularities are quotient? Ariyan Javanpeykar 2010-04-30T10:07:43Z 2010-05-01T13:12:29Z <p>I'm stuck on a technicality concerning singularities. </p> <p>Basically, I have to show that the singularities of a $\mathbf{certain}$ normal projective variety over $\mathbf{C}$ are rational. (I won't bother you with the exact set-up. For, it's possible that I'm even wrong.) </p> <p>My idea was to use a theorem of Viehweg and show that all singularities are quotient. But I don't have a clue of how to do the latter. What are the standard techniques used in such a proof? That is, say you want to show that all singularities are quotient. What would be your first idea to apply? Do problems like this become easier in low dimensions or does it really not matter?</p> <p><strong>Note</strong>. Forgive me if the question is ill-posed/vague.</p> <p><strong>Added later</strong>. In view of Karl's remark, I decided to give the set-up.</p> <p>Let $X$ be a smooth projective variety over $\mathbf{C}$ (of any dimension). Let $\pi:Y\longrightarrow X$ be a finite morphism, where $Y$ is a normal projective variety over $\mathbf{C}$. We also are given a flat morphism $h:X\longrightarrow C$, where $C$ is a smooth projective curve. Of course, it can happen that $Y$ has singularities that are not rational. (Example?) But I would like to show that the singularities of $Y$ are rational in the situation I will describe now.</p> <p>Let $V\longrightarrow U$ be a connected finite etale covering of $U=X-D$, where $D$ is a simple normal crossings divisor on $X$. Define $Y$ to be the normalization of $X$ in the function field of $V$.</p> <p>So the set-up is quite general, i.e., I don't have any equations. Maybe one could try to apply arguments based on fundamental groups to show that the singularities of $Y$ are quotient? I think I can show that the singularities of $Y$ occur in the inverse image under $\pi$ of $D^{sing}$, where $D^{sing}$ is the singular locus of $D$.</p> <p>And Karl, what are the techniques in characteristic $p>0$ that you mention below?</p> http://mathoverflow.net/questions/23091/is-there-an-obvious-way-for-showing-singularities-are-quotient/23137#23137 Answer by Karl Schwede for Is there an obvious way for showing singularities are quotient? Karl Schwede 2010-04-30T19:45:44Z 2010-04-30T19:45:44Z <p>I'm not sure if I understand what you are doing correctly, but let me take a guess. </p> <p>Say you have $f : Y \to X$ which is finite and etale outside of a simple normal crossings divisor $D$ on $X$.</p> <p>Consider the pair $(X, (1-\varepsilon)D)$ where $\varepsilon > 0$ is a very small positive number. This pair is "Kawamata-log-terminal". See for example the book by Koll\'ar and Mori. </p> <p>It follows that $(Y, f^*(1-\varepsilon)D - Ram_f)$ is also Kawamata log terminal (see Proposition 5.20 in Kollar and Mori's book), here $Ram_f$ is the ramification divisor. Because $\varepsilon$ is very very small, $f^*(1-\varepsilon)D - Ram_f$ is effective (we are also using the fact that we are working in characteristic zero, so no wild ramification). It follows that $Y$ has rational singularities essentially by a theorem of Elkik (explicitly, see Theorem 5.22 in Kollar-Mori). </p> <p>If you'd like, I can give a slicker proof of klt (Kawamata log terminal) => rational too.</p> http://mathoverflow.net/questions/23091/is-there-an-obvious-way-for-showing-singularities-are-quotient/23139#23139 Answer by VA for Is there an obvious way for showing singularities are quotient? VA 2010-04-30T19:48:39Z 2010-05-01T12:48:20Z <p>Say $X=\mathbb A^n$ and $D_1,\dots, D_n$, the components of $D$, are the coordinate hyperplanes $x_i=0$, for simplicity. You can assume that WLOG, because your $(X,D)$ is isomorphic to this one in &eacute;tale topology. </p> <p>$\pi_1(X\setminus D) = \mathbb Z^n$. So the cover $V\to U$ corresponds to a finite quotient of $\mathbb Z^n$, which is a finite abelian group $G$. So the ring of regular functions on $V$ is generated by the <em>roots</em> of monomials in $x$. You can write these as $x^m$ for some $m\in \mathbb Q^n$. The lattice $H$ generated by $m_i$ contains $\mathbb Z^n$, and the quotient $H/\mathbb Z^n$ is the dual abelian group $G^{\vee}$.</p> <p>So what is $Y$ now? It is the normalization of the ring $k[x_1,\dots,x_n]$ in the bigger field $k(x^{m_i})$. So it is a toric problem now. The normalization is generated by monomials in the lattice $H$ which lie in the cone $(\mathbb R_{\ge0})^n$.</p> <p>So $Y$ is toric and simplicial, and every such singularity is an abelian quotient singularity.</p> <p>The condition $\pi_1(X\setminus D) = \mathbb Z^n$ fails in char $p$ (indeed, the fundamental group in that case is huge), so this argument and the statement both fail in char $p$.</p> http://mathoverflow.net/questions/23091/is-there-an-obvious-way-for-showing-singularities-are-quotient/23177#23177 Answer by Angelo for Is there an obvious way for showing singularities are quotient? Angelo 2010-05-01T13:12:29Z 2010-05-01T13:12:29Z <p>This is essentially Abhyankar's lemma.</p> <p>What VA says is correct. However, one can simply remark that the subgroups $(d\mathbb Z)^n$ for $d>0$ form a cofinal system of subgroups of finite index, and the normalization $Y_d$ of $X$ in the corresponding covering of $X \smallsetminus D$ is smooth. Since the fundamental group is abelian, the covering $Y \to X$ is Galois. If $G$ is the Galois group, the kernel $K$ of $\pi_1(X\smallsetminus D) \to G$ contains some $(d\mathbb Z)^n$; and then $Y$ is a quotient of $Y_d$ by the action of $K/(d\mathbb Z)^n$.</p>