Is there any rational curve on an Abelian variety? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T14:45:51Z http://mathoverflow.net/feeds/question/9066 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9066/is-there-any-rational-curve-on-an-abelian-variety Is there any rational curve on an Abelian variety? Fei YE 2009-12-16T03:32:00Z 2010-01-10T13:29:11Z <p>Is that true that there is no rational curves contained in an Abelian variety? If it's true, is that because abelian varieties are not uniruled? How do I know whether an abelian variety is not uniruled?</p> http://mathoverflow.net/questions/9066/is-there-any-rational-curve-on-an-abelian-variety/9067#9067 Answer by jvp for Is there any rational curve on an Abelian variety? jvp 2009-12-16T03:43:14Z 2009-12-16T11:03:41Z <p>Yes, an abelian variety $A$ contains no rational curves.</p> <p>Suppose not and let $f: \mathbb P^1 \to A$ be a non-constant morphism. </p> <p>If $f$ is inseparable then it must be the composition of some power of Frobenius of $\mathbb P^1$ with a non-constant separable map $g: \mathbb P^1 \to A$. Thus we may assume that $f$ is separable, i.e., $df : T \mathbb P^1 \to f^{\ast} T A$ is not the zero morphism. Therefore the general $1$-form $\omega \in H^0(A,\Omega^1)$ will give rise to a non-zero $1$-form $f^{\ast} \omega$ on $\mathbb P^1$. Contradiction.</p> <p><hr /></p> <p>Remarks: </p> <ol> <li>Above, I have expanded the original answer "If $C$ is a curve on an abelian variety then $C$ has regular $1$-forms coming from $1$-forms on $A$" in order to incorporate Voloch's comment about positive characteristic.</li> <li>The same argument shows that non-algebraic compact complex tori contain no rational curves. </li> <li>Since over $\mathbb C$ the Albanese variety of a compact Kahler manifold $X$ is usually defined as $H^0(X, Omega^1)^{\ast} / H_1(X, \mathbb Z)$, the argument above is essentially the same as Voloch's when the characteristic zero.</li> <li>Let $X$ be a smooth projective variety and $f:X \to A$ be a morphism. If $df$ has maximal rank then $H^0(X,{\Omega^i}) \neq 0$ for every $i \le \dim X$. Thus an abelian variety contains no subvarieties without regular forms in any particular degree.</li> </ol> http://mathoverflow.net/questions/9066/is-there-any-rational-curve-on-an-abelian-variety/9068#9068 Answer by Charles Siegel for Is there any rational curve on an Abelian variety? Charles Siegel 2009-12-16T03:47:11Z 2009-12-16T03:47:11Z <p>There is not one. Reference is <a href="http://www.jmilne.org/math/CourseNotes/AV.pdf" rel="nofollow">Milne's notes</a>, Prop 3.9. More is true, Prop 3.10 in the same notes is that any rational map from a unirational variety to an abelian variety is constant.</p> http://mathoverflow.net/questions/9066/is-there-any-rational-curve-on-an-abelian-variety/9069#9069 Answer by Felipe Voloch for Is there any rational curve on an Abelian variety? Felipe Voloch 2009-12-16T03:49:12Z 2009-12-16T03:49:12Z <p>There are no rational curves in an abelian variety, this is much stronger than not being uniruled. If there is a map $P^1 \to A$, $A$ abelian, the map would factor through the Albanese variety of $P^1$, by definition. However, for curves, the Albanese is the Jacobian (from general theory of the Jacobian) and the Jacobian of $P^1$ is a point.</p> http://mathoverflow.net/questions/9066/is-there-any-rational-curve-on-an-abelian-variety/9090#9090 Answer by Georges Elencwajg for Is there any rational curve on an Abelian variety? Georges Elencwajg 2009-12-16T08:31:27Z 2009-12-16T08:31:27Z <p>Over $\mathbb C$ you can argue as follows. Suppose you have a morphism $\mathbb P^1(\mathbb C) \to A$ ($A$= abelian variety ). Since $\mathbb P^1(\mathbb C)$ is simply connected , the morphism lifts to the universal cover of $A$, affine space $\mathbb C^n$. But since $\mathbb P^1(\mathbb C)$ is complete and connected, the lift to affine space must be constant and hence the original morphism is constant too.</p> <p>The answers by Charles, Felipe and jvp are better because they work over arbitrary fields, but since the argument just given is so ridiculously elementary (introductory topology), I thought it might still be of some interest ( also it works in the holomorphic category if $A$ is a complex torus, maybe not algebraic). </p> http://mathoverflow.net/questions/9066/is-there-any-rational-curve-on-an-abelian-variety/9110#9110 Answer by Hao Sun for Is there any rational curve on an Abelian variety? Hao Sun 2009-12-16T15:51:00Z 2009-12-16T15:51:00Z <p>How about when the rational curve is singular?</p> http://mathoverflow.net/questions/9066/is-there-any-rational-curve-on-an-abelian-variety/11323#11323 Answer by norondion for Is there any rational curve on an Abelian variety? norondion 2010-01-10T13:29:11Z 2010-01-10T13:29:11Z <p>See also Cornell-Silverman, p. 107.</p>