Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X\subseteq\mathbb{C}P^n$ be a smooth subvariety. Let $C\subseteq X$ be an algebraic curve and let $[C]\in H_2(X;\mathbb{R})$ be the homology class it determines. Does there exist a continuous map $u:\Sigma\longrightarrow X$ satisfying properties (1),(2) and (3) below? (1) $\Sigma$ is a genus zero nodal curve (that is, a tree of spheres); (2) $u$ is holomorphic on each irreducible component of $\Sigma$; (3) $u_*[\Sigma]=[C]$;

Thank you.

share|improve this question
7  
No, X may have no rational curves on it - for example it could be a smooth elliptic curve. –  mdeland Jun 1 '11 at 0:57
add comment

1 Answer 1

up vote 3 down vote accepted

A complex analytic space $X$ is called Brody hyperbolic if any holomorphic map $\mathbb C\to X$ is constant. There are many spaces that are Brody hyperbolic (see this paper for instance) and thus what you ask will not happen. (In fact, according to Kobayashi's conjecture all general hyperssurfaces of high enough degree are Brody hyperbolic).

In fact, this is way too much firepower for that. It seems to me that essentially what you are asking is whether the cone of effective curves is generated by rational curves over $\mathbb Z$. Even if you ask this over $\mathbb Q$ it does not happen often, even when there are plenty of rational curves around. Take for instance a variety that is Brody hyperbolic (say a high genus curve) and its product with a rational curve. You will never get a rational curve "bend" towards the other direction.

One case when this actually happens is if $X$ is Fano. Then by the Cone Theorem the cone of effective curves is generated by rational curves. Actually, I am not sure if you can do it with a connected $\Sigma$, but you can certainly do it with the disjoint union of a few spheres. However, you have to allow rational coefficients, but you only need finitely many actual maps. It's all in the coefficients.

share|improve this answer
    
Hi Sándor, in fact it is not known whether a general projective hypersurface of high degree is Brody hyperbolic or not, as soon as its dimension is at least four. On the other hand it is conjecturally true (Kobayashi, '70). What is known is that they do not contain any subvariety not of general type. –  diverietti Jun 15 '11 at 8:49
    
Hi Simone, you are absolutely correct and I actually knew that. I don't know why I put it that way. Laziness, probably. Anyway, to make the point one does not need to know the Kobayashi conjecture, just that there exist Brody hyperbolic spaces and that is certainly true. –  Sándor Kovács Jun 15 '11 at 10:27
    
Of course! Anyway, it seemed to me that the fact that generic projective hypersurfaces of high degree are "algebraically hyperbolic" (ie they do not contain any subvariety not of general type) was sufficient for your aims. –  diverietti Jun 15 '11 at 11:40
    
Thank you for the explanations. Assume that $C$ is a rational curve in $X$ [i.e. birational to $\mathbb{C}P^1$]. Why is it possible to find holomorphic curves $u_1,\ldots,u_k:\mathbb{C}P^1\longrightarrow X$ such that $$[C]=(u_1)_*[\mathbb{C}P^1]+\ldots+(u_k)_*[\mathbb{C}P^1]?$$ [in $H_2(X;\mathbb{Q}$] –  user15512 May 23 '13 at 20:03
1  
For smooth projective curves being birational is equivalent to being isomorphic. Take the normalization of $C$. If $C$ is rational, then its normalization is $\mathbb CP^1$. –  Sándor Kovács May 24 '13 at 20:09
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.