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

I'm somewhat surprised to figure out that P^n's does not admits any non-constant morphism to a curve: in fact this is not fancy, for P^1 it follows by Riemann-Hurwitz, and for P^n, note that it doesn't admit any non-constant morphism to P^1, while any morphism to a curve can be composed with a branched cover of P^1, one sees that any map from P^n to curves are constant.

So my question is, given an projective variety, when can one find a non-constant morphism to P^1? And when can one find a non-constant morphism to a curve of genus greater than 0?

share|improve this question
2  
Your question seems to require some minor modification, since of course $\mathbb{P}^1$ admits plenty of nonconstant morphisms to $\mathbb{P}^1$, e.g. the identity map. –  Pete L. Clark Aug 13 '10 at 8:39
add comment

5 Answers

A few opposite-looking remarks, and the case of (minimal) surfaces.

If a variety $X$ admits a non-constant morphism to a curve, then it admits a non-constant morphism to $\mathbb{P}^1$ (just compose the given morphism with a morphism of the curve to $\mathbb{P}^1$). Thus, if you only care about existence of a non-constant morphism to a curve, you may as well restrict your attention to the question of existence of a non-constant morphism to $\mathbb{P}^1$. On the other hand, the existence of a morphism to a curve of genus at least one is a birational property: any such morphism factors through the Albanese variety of $X$, and the Albanese variety of $X$ is a birational invariant. The property of admitting a morphism to $\mathbb{P}^1$ is clearly not a birational invariant property, as any non-constant rational function determines a rational map to $\mathbb{P}^1$ (see Charles Matthews' answer).

Both questions appear quite hard, though. An equivalent formulation of the question is the following: does $X$ admit two disjoint effective non-zero nef divisors? The equivalence of the statements is almost tautological. An easy implication is that the rank of the Neron-Severi group of $X$ is at least two, and that there are effective non-zero nef divisors that are not big.

For minimal surfaces, the situation is as follows. A surface of negative Kodaira dimension (i.e. a ruled surface or $\mathbb{P}^2$, here we do not need the surface to be minimal) admits a morphism to $\mathbb{P}^1$ if and only if it is not isomorphic to $\mathbb{P}^2$. A surface of Kodaira dimension zero (i.e. a K3, Enriques, Abelian of bielliptic surface) not admitting a morphism to $\mathbb{P}^1$ is a non-elliptic K3 surface. Every surface of Kodaira dimension one (a properly elliptic surface) admits a morphism to $\mathbb{P}^1$ (and in fact a canonical morphism to a curve).

EDIT: Among surfaces of Kodaira dimension zero, also simple abelian surfaces (i.e. abelian surfaces that are not isogenous to a product of two elliptic curves) admit no morphism to a curve.

For all surfaces (including the surfaces of Kodaira dimension two), one thing you can say is that there is a Theorem of Castelnuovo and de Franchis characterizing surfaces with a morphism to a curve of genus at least two.

Thus, in conclusion, it seems that for minimal surfaces of special type, the only surfaces not admitting a morphism to a curve are $\mathbb{P}^2$ and non-elliptic K3 surfaces and simple abelian surfaces (which does not look so bad, after all!). Note that every irreducible component of the moduli space of polarized K3 surfaces contains elliptic surfaces and non-elliptic ones.

share|improve this answer
    
The theorem of Castelnuovo and de Franchis has been generalized to arbitrary dimension by Catanese, Moduli and classification of irregular Kähler manifolds (and algebraic varieties) with Albanese general type fibrations, Invent. Math. 104 (1991),263-289. –  Angelo Aug 13 '10 at 11:55
1  
The following statement is wrong: "An equivalent formulation of the question is the following: does X admit two disjoint effective non-zero nef divisor". Indeed, consider and Abelian surface A and take a generic non-trivial bundle L with c_1(L)=0. Then consider the projectivisation of L+O over A (O is the structure sheaf). Obviously this P^1 bundle has two disjoint sections that are effective and nef. It is easy to see that if A does not admit a map to an elliptic curve, then P(L+O)(A) does not admit a map to CP^1, contrary to what you are claiming. –  Dmitri Aug 13 '10 at 21:55
    
It appears that jvp discussed a version of this counterexample together with a (nontrivial) correction below. –  Donu Arapura Aug 13 '10 at 22:02
    
Donu, you are right, I did not read the answer of jvp, before commenting –  Dmitri Aug 13 '10 at 22:16
    
@Dmitri, you are right. I will not edit the answer to reflect this, since the correction is already contained in the comments here and in jvp's answer. Nevertheless, the implication needed for the statements about the Neron-Severi group holds, and this is indeed obvious! –  damiano Aug 14 '10 at 0:34
add comment

The existence of a morphism of a compact Kähler manifold to curve of genus $g \ge 2$ is a purely topological fact. This was first proved by Siu, and rediscovered independently by Beauville. The precise statement is the following:

A compact complex Kähler manifold $M$ admits a non-constant morphism to curve of genus $g \ge 2 $ if and only if there exits a surjective homomorphism of the fundamental group of $M$ to the fundamental group of some curve of genus $g' \ge 2$.

This has been generalized by Catanese here, where one can also find Beauville's argument.

In general lines the argument goes as follows:

  1. the morphism of between fundamental groups ensures the existence of a big subspace in $H^1(M, \mathbb C)$;
  2. Hodge Theory allows one to recognize a space in $H^0(M,\Omega^1)$ of dimension at least $g$ isotropic with respect to the wedge product;
  3. to conclude one applies Castelnuovo-De Franchis argument.

The existence of morphism to a curve of genus $1$ is more subtle and depends on the analytic structure of $M$, not just on its topology. For instance a general abelian variety has no non-constant morphism to an elliptic curve, but there are many that have.


In a different take there is also the following result by Totaro:

If a projective variety $M$ admits three pairwise disjoint connected hypersurfaces with Chern classes lying on a line of $H^2(M, \mathbb R)$ then $M$ fibers over a curve.

It has to be noted that two pairwise disjoint connect divisors do not suffice to draw the conclusion as the example $M = \mathbb P ( \mathcal O_N \oplus \mathcal L)$ ( with $N$ a projective variety and $\mathcal L$ a line-bundle with zero Chern class on it ) shows.

There is also a generalization of Totaro's result to arbitrary compact complex varieties here.

share|improve this answer
    
As Donu Arapura points out in his answer there is more literature on the subject than this answer suggests. –  jvp Aug 13 '10 at 12:17
add comment

As Damiano pointed out, if a smooth projective $X$ surjects onto a curve, and $\dim X>1$ then the Picard number is at least $2$. This gives a clear obstruction, which explains why $\mathbb{P}^n$ won't satisfy it (when $n>1$). Here is a small refinement:

Lemma: Suppose that an $n$ dimensional variety $X$ surjects onto a $m$-dimensional variety with $0<m<n$. Then $$\dim im[CH^m(X)\otimes \mathbb{Q} \to H^{2m}(X)]>1$$ where $H^*(X)$ denotes singular or $\ell$-adic cohomology.

Proof: The class of the general fibre and $H^{n-m}$, with $H$ an ample divisor, are independent.

Cor: $\mathbb{P}^n$ does not surject onto a nontrivial lower dimensional variety.

(This can also be seen directly from Bezout's theorem.)

On the other hand, $X$ will map onto $\mathbb{P}^1$ after blowing up, as Charles observed, and this is often a very useful trick in practice.

The case of (*) $X$ mapping onto curves of genus two or more is actually something that has been studied a number of people*. From Castelnuovo-De Francis (see Damiano's answer) one can extract a number of topological criteria. Here's one: $X$ satisfies (*) if and only if the fundamental group admits a surjective homomorphism onto the fundamental group of such a curve. (In my original answer, I hadn't realized that JVP already discussed this.) Some of this probably described in the multi-author book on Kaehler groups. Also take a look at my note in the Bulletin from way back in the last century.

In case it wasn't clear the results in the last paragraph are characteristic $0$ only. The positive characteristic case has not been looked at seriously, as far as I know, and is potentially very interesting. Warning: the standard tecnhiques such as Castelnuouvo-De Franchis, will fail!

* Among them: Amoros, Beauville, Bressler, Campana, Catanese, Gromov, Green, Lazarsfeld, Ramachandran, Simpson Siu, and me. People coming from hyperplane arrangement theory have also looked at the question for open varieties, but here the literature is so large, I won't even attempt it.

share|improve this answer
add comment

This is a substantial revision of my previous answer based on the comments and the other answers. I am making it community wiki for two reasons: it incorporates ideas from my answer (for which I already have gained reputation), as well as ideas form the other posts (for which I do not deserve reputation!); people can edit it to correct mistakes and typos that are certainly present.

In what follows I will frequently not specify that a morphism is assumed to be non-constant; I will leave it to the reader to figure out when the word "non-constant" can be added! Moreover, I will only address the question of a variety of dimension at least two mapping to a curve: the case of curves can be made very difficult by specifying the question further, but I will not worry about this.

First, the question in the title asks about morphisms to a curve. We shall see that there are differences based on the Kodaira dimension of the target curve, the higher the dimension, the more rigid the situation: we go from negative Kodaira dimension (target of genus zero) where the property is neither a birational, nor deformation invariant; to zero Kodaira dimension (target is a curve of genus one) where the property is a birational invariant, but not deformation invariant; to positive Kodaira dimension (target of genus at least two) where the property is both a birational invariant and a deformation invariant (and more!). Let's start.

If a variety $X$ admits a non-constant morphism to a curve, then it admits a non-constant morphism to $\mathbb{P}^1$ (just compose the given morphism with a morphism of the curve to $\mathbb{P}^1$). Thus, if you only care about existence of a non-constant morphism to a curve, you may as well restrict your attention to the question of existence of a non-constant morphism to $\mathbb{P}^1$. The property of admitting a morphism to $\mathbb{P}^1$ is clearly not a birationally invariant property, as any non-constant rational function determines a rational map to $\mathbb{P}^1$ (see Charles Matthews' answer and think of $\mathbb{P}^2$). It is also not a deformation invariant property, as the example of a family of quartic surfaces in $\mathbb{P}^3$ containing a quartic with Picard number one and a quartic with a line clearly shows.

For this general case there is an easy necessary condition.

Lemma 1. Let $X$ be a smooth projective variety of dimension at least two and let $f \colon X \to \mathbb{P}^1$ be a non-constant morphism. Then there are on $X$ two non-zero disjoint effective linearly equivalent divisors $D_0,D_1$. In particular, there are on $X$ nef divisors that are not ample and hence the Picard number of $X$ is at least two.

Proof. Just let $D_0$ and $D_1$ be the fibers of the morphism $f$ above two distinct points. For the second statement, if the Picard number of $X$ were equal to one, then any non-zero effective divisor $H$ on $X$ would be ample, and in particular we would have $D_0 \cdot D_1 \equiv H^2 \neq 0$, contradicting the fact that $D_0$ and $D_1$ are disjoint. $\square$

As Donu Arapura remarks, a variation of Lemma 1 also works for non-constant morphisms to varieties $Y$ of dimension smaller than the dimension $X$. In this case we loose the fact that the divisors are linearly equivalent, but we can still find $\dim(Y)+1 \leq \dim(X)$ (Weil) divisors on $Y$ whose intersection is empty. Their pull-backs to $X$ give us $\dim(Y)+1$ divisors with empty intersection; again it follows that the Picard number of $X$ is at least two (see Donu Arapura's answer for a statement with Chow groups and cohomology).

As jvp remarks, there is a refinement of Lemma 1 by Totaro to an equivalence (see jvp's answer for more details).

Let us move on to the case of morphisms to smooth non-rational curves. The existence of a morphism to a curve of genus at least one is a birational property: any such morphism factors through the morphism between the Albanese variety of $X$ and the Jacobian of the curve, and we are done, since the image of $X$ in its Albanese variety is a birational invariant and is all we need to decide the existence of a morphism.

Also in this case, though, the property is not deformation invariant: already in the case of jacobians of genus two curves we can find examples of simple abelian surfaces and of abelian surfaces that are products of elliptic curves (or also just isogenous to products; see, for instance, this MO question).

There is an easy necessary criterion that everyone knows, but no one explicitly mentioned, so I will mention it here: if a variety $X$ admits a (generically smooth) morphism to a curve of genus $g$, then the irregularity of $X$ is at least $g$ (this means that the space of holomorphic one-forms has dimension at least $g$). The proof is obtained simply by pulling back forms from the curve to $X$. This rules out several possibilities, but is quite far from being sufficient - it is an "abelian version" of something that we will see later.

In this case we can state a criterion that might give a feel for what are the difficulties involved.

Lemma 2. Let $X$ be a smooth projective variety. The following are equivalent

  • $X$ admits a non-constant morphism to a curve of genus one;
  • the Albanese variety of $X$ is isogenous to a product of an elliptic curve and an abelian variety;
  • the Albanese variety contains an elliptic curve.

The equivalence of the statements above follows at once from the comment about the morphism factoring through the Albanese variety, as well as well-known statements about morphisms of abelian varieties. We could replace the Albanese variety in Lemma 2 by the Picard group (or the connected component of the identity), using the duality between the Albanese variety and the Jacobian. Thus deciding whether a variety admits a morphism to a curve of genus one reduces to the question of understanding the dimensions of the simple factors of its Albanese variety. This is tricky, but at least it feels like we made some progress, compared to the case of morphisms to curves of genus zero!

On to the case of morphisms to curves of genus at least two. We have already seen that the property is in this case a birational invariant; in fact the property is a deformation invariant (and more!). This follows from a beautiful characterization that only involves topological fundamental groups (over fields of characteristic zero). This characterization starts (in the case of surfaces) with the Theorem of Castelnuovo and de Franchis, later generalized (by Siu, Beauville and Catanese) to work for arbitrary smooth projective varieties. For more details, see the comment by Angelo to my previous answer, as well as the answers of jvp and Donu Arapura. As Donu mentions, it would be interesting to generalize this to varieties over general fields. For instance, what can be said of a variety whose étale fundamental group surjects onto the étale fundamental group of a curve of general type?

Finally, I will conclude this answer by analyzing the case minimal surfaces.

A surface of negative Kodaira dimension (i.e. a ruled surface or $\mathbb{P}^2$, here we do not need the surface to be minimal) admits a morphism to $\mathbb{P}^1$ if and only if it is not isomorphic to $\mathbb{P}^2$. Morphisms to curves of higher genus are also easy to analyze and quickly reduce to the case of morphisms between curves.

A surface of Kodaira dimension zero (i.e. a K3, Enriques, Abelian of bielliptic surface) not admitting a morphism to $\mathbb{P}^1$ is either a non-elliptic K3 surface or a simple abelian surface (i.e. an abelian surface that is not isogenous to a product of two elliptic curves). Morphisms to curves of genus at least one are ruled out for K3 surfaces by the irregularity; for abelian varieties every non-constant morphism to a curve factors through a morphism to an elliptic curve (and clearly there are no morphisms to curves of genus at least two).

Every surface of Kodaira dimension one (a properly elliptic surface) admits a morphism to $\mathbb{P}^1$; more precisely, the Stein factorization of the morphism determined by a sufficiently large multiple of the canonical divisor is a canonical morphism to a curve.

For surfaces of Kodaira dimension two, I do not know what else to say, besides what has been already remarked above!

Thus, in conclusion, for minimal surfaces of special type, the only surfaces not admitting a morphism to a curve are $\mathbb{P}^2$, non-elliptic K3 surfaces and simple abelian surfaces (which does not look so bad, after all!). Note that every irreducible component of the moduli space of polarized K3 surfaces contains elliptic surfaces and non-elliptic ones.

share|improve this answer
add comment

The question about the projective line seems more fundamental. And geometrically this seems to be about the graphs of rational functions. Certainly the function field of V provides morphisms from a dense open set in V to the line. The graph of such a morphism exists in the product of V and the line: take its closure. The projection onto the V factor exhibits this graph closure as some blowing up of V. The projection onto the line, in a sense, is the morphism we are seeking. So maybe you are asking for all varieties occurring as such graphs.

share|improve this answer
add comment

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.