Exceptional loci are covered by rational curves: easy case - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T19:14:46Z http://mathoverflow.net/feeds/question/70747 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70747/exceptional-loci-are-covered-by-rational-curves-easy-case Exceptional loci are covered by rational curves: easy case Artie Prendergast-Smith 2011-07-19T15:48:53Z 2011-07-21T05:52:44Z <p>It's well-known that the exceptional locus of a birational morphism is covered by rational curves, in various degrees of generality. The best result I know in this direction is the following:</p> <p><strong>Theorem (Hacon&ndash;McKernan):</strong> Let (X,&Delta;) be a dlt pair, and f: Y -> X a birational morphism. Then the fibres of f are rationally chain connected.</p> <p>I'm curious about the following (very!) special case of that result:</p> <p><strong>Special case:</strong> Let f: Y -> X be a birational morphism where dim(X)=dim(Y)=3, X is terminal, Y is smooth, and the exceptional locus Exc(f) has codimension 2. Then Exc(f) is a union of rational curves. </p> <p>My main question is then the following:</p> <blockquote> <p>What's the simplest proof of this special case?</p> </blockquote> <p>One proof I can imagine is to use the fact that there exists an effective divisor D such that -D is f-ample, and apply the relative Cone Theorem to (Y/X, &epsilon;D) for small &epsilon;. (I haven't worked this out in full, but I think it can be made into a complete proof.) But that still uses the Cone Theorem, which seems like it might be overkill. So, is there a simpler proof? </p> <p>(Note: the assumptions on singularities in the special case come from a paper I was reading; there is no reason to consider them essential.)</p> <p>I've added the reference-request tag in case this is a completely standard argument written down somewhere; unfortunately I can't get my hands on books at the moment, so I can't check the usual places.</p> http://mathoverflow.net/questions/70747/exceptional-loci-are-covered-by-rational-curves-easy-case/70777#70777 Answer by Sándor Kovács for Exceptional loci are covered by rational curves: easy case Sándor Kovács 2011-07-19T20:00:37Z 2011-07-21T05:52:44Z <p>Artie, here is a sketch of an argument that I think should work.</p> <p><em>Definition</em> (just in case someone needs this): A threefold singularity is called <em>comopound Du Val</em> or <em>cDV</em> if a general hyperplane section through the singular point has Du Val singularities (a.k.a. rational double point).</p> <p><em>Remark</em> Since a rational double point surface singularity is Gorenstein, it follows that so is a cDV singularity. A typical cDV singularity is a one-parameter deformation of a rational double point.</p> <p>A terminal threefold singularity is a cyclic quotient of an isolated cDV singularity. Say $X'\to X$ is such that $X'$ has cDV singularities. Suppose we already know the statement for cDV singularities. Let $Y''$ denote a resolution of $Y'=X'\times_X Y$. I think $Y'$ should be smooth over a general point of the exceptional locus of $f:Y\to X$, so it should only have isolated singularities, so a general point of the exceptional locus of $f$ is covered by the codimension $2$ part of the exceptional locus of $Y''\to X$. In other words, if we know that that is covered by rational curves, then we are done. (Admittedly I did not work out this part, but it seems reasonable).</p> <p>Now for the cDV case, it should be easy. A general hyperplane section through the singular point has Du Val singularities, so the exceptional locus over that point can only contain rational curves forming a tree.</p> <p>I don't know if this is written down anywhere, I just made it up. (So it might not be air-tight).</p> <p>The same statement for $X$ smooth is known as <em>Abhyankar's lemma</em> and is (1.3) in Kollár-Mori. Interestingly, it is used in the proof of Bend-and-Break (although only for surfaces) which is used in the Cone Theorem, at least the original proof. And I think B&amp;B is actually enough to prove what you want, so you don't need the Cone Theorem, although that is of course also a big gun.</p> <p><strong>Addendum</strong> To answer Artie's question in the comments: The proof of this characterisation of terminal threefold singularities is actually not easy. However, it seems to me that it is not that far from what you want to be proven, so if that can't be proven easily, then probably neither can your statement.</p> <p>Here are some thoughts to support that pseudo-claim:</p> <p>1) The cyclic cover part is easy, since one takes the index-one cover, which will be an index-one terminal singularity. I will assume this from this point.</p> <p>2) A terminal singularity is rational and hence CM, so an index-one terminal singularity is Gorenstein. Therefore any hyperplane section is also Gorenstein. </p> <p>3) In the situation of the question a general hyperplane section through the singular point may not be resolved by $f$, but if it were, then the exceptional divisor of that resolution would be just the exceptional locus of $f$. That would imply that the exceptional locus of the resolution of the hyperplane section consists of rational curves and since it is equal to the exceptional locus of $f$, if it contained a loop, it would imply that $R^1f_*\mathscr O_Y\neq 0$ (I think I can prove this. It is relatively easy, but not super-easy.) In other words, since $X$ has rational singularities it follows that so does this general hyperplane section. That implies that it has Du Val singularties and hence $X$ has (isolated) cDV.</p> <p><strong>Conclusion</strong> Even if the addendum is not a full proof of the used characterisation of terminal threefold singularities, it seems to suggest that you question is about as hard as that is.</p>