Why are normal crossing divisors nice? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T07:29:44Z http://mathoverflow.net/feeds/question/23770 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/23770/why-are-normal-crossing-divisors-nice Why are normal crossing divisors nice? Makhalan Duff 2010-05-06T19:00:43Z 2010-10-08T00:02:16Z <p>This question is going to be extremely vague.</p> <p>It seems that wherever I go (especially about Grothendieck's circle of ideas) the higher-dimensional analogue of a curve minus a finite number of points is a scheme minus a normal crossing divisor.</p> <p>Why is that? What's so special about a normal crossing divisor that it simulates a curve minus a finite number of points better?</p> http://mathoverflow.net/questions/23770/why-are-normal-crossing-divisors-nice/23776#23776 Answer by Torsten Ekedahl for Why are normal crossing divisors nice? Torsten Ekedahl 2010-05-06T19:57:25Z 2010-05-06T19:57:25Z <p>It mostly has to do with finding nice compactifications. Compactifications of varieties are a good thing as they allow us to control what happens at "infinity". If the variety itself is smooth it seems a good idea (and it is!) to demand that the compactification also be smooth. However, you need the situation to be nice at infinity in order to make the study of asymptotic behaviour at infinity to be as easy as possible. The best behaviour at infinity would be if the complement were smooth but that is in general not possible. What is always possible is to demand that the complement be a divisor with normal crossings. In practice it works essentially as well as having a smooth complement: You have a bunch of smooth varieties intersecting in as nice a manner as possible.</p> http://mathoverflow.net/questions/23770/why-are-normal-crossing-divisors-nice/40984#40984 Answer by Sándor Kovács for Why are normal crossing divisors nice? Sándor Kovács 2010-10-04T04:59:59Z 2010-10-04T04:59:59Z <p>A related point is that if say $X\subset Y$ and you want an embedded resolution of $X$, then you can of course ask that you want a birational morphism $\pi:Z\to Y$ such that the strict transform of $X$ is smooth, but a better thing to ask is the the entire pre-image of $X$ is as nice as possible. Unfortunately (in general) you cannot make the preimage of $X$ smooth as the exceptional set will add additional components and where they meet is going to be a singular point. So, you can ask for the next best thing: normal crossings. You could even say that <em>normal crossings</em> is the reducible analogue of <em>smooth</em>. </p> <p>Anyway, this is the result of Hironaka, JS Milne referred to above: for any $X\subset Y$ (plus some reasonable assumptions) there exists a projective birational $\pi$ such that $Z$ is smooth and $\pi^{-1}X$ is a normal crossing divisor. If $Y\setminus X$ is smooth, then you may even require that $\pi$ is an isomorphism outside $X$.</p> <p>The compactification result is a simple consequence of this: if $U$ is open (say quasi-projective), pick a projective compactification $Y$ and let $X=Y\setminus U$. Perform Hironaka's embedded resolution of singularities and you get $U\subset Z$ with the complement being a normal crossing divisor.</p> <hr> <p>One, slightly independent word on <em>normal crossings</em>. There seems to be some confusion in the literature about what <em>normal crossings</em> mean. Or rather what the difference is between <em>normal crossings</em> and <em>simple normal crossings</em>. Well, the point is that (nowadays) the latter is understood in the Zariski topology while the former in the analytic or formal topology. In other words, <em>simple normal crossings</em> mean that each irreducible component is smooth and they meet transversally, while <em>normal crossings</em> allows for a component to meet itself transversally. In particular, a nodal curve has <em>normal crossings</em> but not <em>simple normal crossings</em>. </p> <p>In the above discussion and in the other answers before this one, you can always put <em>simple normal crossings</em> in place of <em>normal crossings</em> and the statements remain true. It is possible that back when Hironaka proved his famous theorem, this distinction had not been made so in older texts the meaning might be different. At the same time, according to Miles Reid, it was the Japanese who invented the term <em>simple normal crossings</em>.</p>