Technique to prove basepoint-freeness - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T09:25:22Z http://mathoverflow.net/feeds/question/37570 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/37570/technique-to-prove-basepoint-freeness Technique to prove basepoint-freeness Moon 2010-09-03T02:16:21Z 2010-10-02T17:22:15Z <p>Let $X$ be a smooth projective variety over $\mathbb{C}$. And let $L$ be a big and nef line bundle on $X$. I want to prove $L$ is semi-ample($L^m$ is basepoint-free for some $m > 0$).</p> <p>The only way I know is using Kawamata basepoint-free theorem:</p> <p>Theorem. Let $(X, \Delta)$ be a proper klt pair with $\Delta$ effective. Let $D$ be a nef Cartier divisor such that $aD-K_X-\Delta$ is nef and big for some $a > 0$. Then $|bD|$ has no basepoints for all $b >> 0$.</p> <p><strong>Question. What other kinds of techniques to prove semi-ampleness or basepoint-freeness of given line bundle are?</strong></p> <p>Maybe I miss some obvious method. Please don't hesitate adding answer although you think your idea on the top of your head is elementary.</p> <p>Addition : In my situation, $X$ is a moduli space $\overline{M}_{0,n}$. In this case, Kodaira dimension is $-\infty$. More generally, I want to think genus 0 Kontsevich moduli space of stable maps to projective space, too. $L$ is given by a linear combination of boundary divisors. It is well-known that boundary divisors are normal crossing, and we know many curves on the space such that we can calculate intersection numbers with boundary divisors explicitely.</p> http://mathoverflow.net/questions/37570/technique-to-prove-basepoint-freeness/37612#37612 Answer by Henri for Technique to prove basepoint-freeness Henri 2010-09-03T13:17:11Z 2010-09-04T10:32:44Z <p>I don't think that your assertion is true; for example, Lazarsfeld gives an example (PAG, 2.3.3) of a big and nef divisor on a surface such that its graded algebra is not finitely generated, so that the divisor can't be semiample.</p> <p>But there are some close results for nef and big divisors, or even for good divisors (when the Kodaira dimensions equals the numerical dimension) as Mourougane and Russo showed : for example, Wilson's theorem asserts that for any nef and big divisor on an irreducible projective variety, there exists $m_0\in \mathbb N$ together with an effective divisor $N$ such that for all $m\geq m_0$, the linear system $|mD-N|$ has no base-point. (PAG, 2.3.9)</p> http://mathoverflow.net/questions/37570/technique-to-prove-basepoint-freeness/39226#39226 Answer by J.C. Ottem for Technique to prove basepoint-freeness J.C. Ottem 2010-09-18T16:54:00Z 2010-09-18T17:00:25Z <p>Numerical criteria for base-pointfreeness are known only in specific cases such as the Kawamata basepoint-free theorem and Reider's theorem (for $\dim X=2$). </p> <p>In the case you mention, $X=\mathcal{M}_{0,n}$, the problem of classifying semi-ample divisors is an important problem. It is slightly easier in positive characteristic, thanks to a theorem of Keel which says that a nef line bundle $L$ is semi-ample if and only if the restriction $L|_E$ is semiample, where $E$ is the exceptional locus of subvarieties $Z$ such that $L^{\dim Z}.Z=0$. If $f:X\to Y$ is a morphism with exceptional locus $E$, then $L$ is semi-ample if and only $L^r$ is the pullback of an ample line bundle on $Y$ for $r>0$. For the precise statements, you might want to take a look at</p> <p><a href="http://www.kurims.kyoto-u.ac.jp/EMIS/journals/Annals/149_1/keel.pdf" rel="nofollow">S. Keel, Basepoint freeness for nef and big line bundles in positive characteristic, Annals of Mathematics(1999)</a>.</p> <p>According to G. Farkas' <a href="http://www-irm.mathematik.hu-berlin.de/~farkas/seattle.pdf" rel="nofollow">article</a>, there are currently no known examples of nef divisors which are not semi-ample.</p>