Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be an abelian variety. As abelian varieties are projective then $X$ contains lots and lots of subvarieties. Why can't one of them be a projective space?

If $X$ is defined over the complex numbers, then there is a relatively painless way to see this (modulo lots of painful differential geometry, depending on your tastes). Indeed, if $Z$ is a (say smooth) subvariety of any space $X$, then we have an exact sequence

$$ 0 \longrightarrow T_Z \longrightarrow T_X \longrightarrow N_{Z/X} \longrightarrow 0.$$

We can put a metric $\omega$ on $X$. By restriction, this gives a metric on $Z$. One can now calculate that the curvature of the metric on $Z$ is no more than that of the metric $\omega$ on $X$.

A torus admits flat metrics, that is Kähler metrics of zero curvature. If a torus could admit a projective space $\mathbb P^k$, we would then get a Kähler metric of non-positive curvature on $\mathbb P^k$. This cannot happen, for example, because then its Ricci curvature would be negative, in contradiction to the Ricci form representing the positive anticanonical bundle of $\mathbb P^k$.

Question: Is there an algebraic way of seeing this?

I'm interested because I absolutely don't know. I have little intuition for algebraic methods and would like to try to change that, a simple example like this might be a good place to start.

share|improve this question
    
A complex torus of dimension 1 (i.e., an elliptic curve) cannot contain a projective space because the function field of projective space would have to equal the function field of the complex torus. (I assume you want your complex torus contain a projective space as an open subset.) The same argument also holds in higher dimension. I don't know if it is "algebraic: enough though. –  Hoedan Feb 28 '12 at 22:12
    
No, I want the torus to contain the projective line as a proper subvariety (i.e. of lower dimension). An elliptic curve, or any curve, only contains points, so nothing happens there. –  Gunnar Magnusson Feb 29 '12 at 10:07

1 Answer 1

up vote 15 down vote accepted

1) Consider a complex torus $T$ of dimension $N$ over $\mathbb C$ (algebraic or not).
Theorem Every holomorphic map $f: \mathbb P^n (\mathbb C)\to T$ is constant.
The proof is very easy, without any "painful differential geometry":
Proof: Since $\mathbb P^n (\mathbb C)$ is simply connected , $f$ lifts to the universal cover of $\pi: \mathbb C^N \to T$, namely there exists a morphism $\tilde f:\mathbb P^n (\mathbb C) \to \mathbb C^N$ with $f=\pi\circ \tilde f$.
Since $\tilde f:\mathbb P^n (\mathbb C) \to \mathbb C^N$ is constant (by compactness of $\mathbb P^n $) , so is $f$.

2) In the purely algebraic case, if $A$ is an abelian variety over the field $k$, it contains no projective space.
It is clearly enough to show that every morphism $g: \mathbb P^1_k\to A$ is constant.
And this is Proposition 3.9 of Milne's Abelian Varieties, freely available here.

Edit
Here is a self-contained proof that there is no closed immersion $g:\mathbb P^1_k \hookrightarrow G$ to any algebraic group $G$ over the field $k$.
Indeed, $g$ would induce a tangent map $T_pg:T_p(\mathbb P^1_k)\hookrightarrow T_p(G)$ which would be non-zero at any $p\in \mathbb P^1_k$.
But then , since $\Omega _{G/k}$ is a trivial bundle, there would exist a differential form $\omega \in \Gamma(G,\Omega _{G/k})$, non-zero on the image of $T_p(\mathbb P^1_k)$ in $T_p(G)$ and thus $\omega $ would restrict to a non-zero differential form $res(\omega)\neq 0\in \Gamma(\mathbb P^1_k,\Omega _{\mathbb P^1_k})$ , contradicting $dim_k\Gamma(\mathbb P^1_k,\Omega _{\mathbb P^1_k})=$ genus ($\mathbb P^1_k)=0$

share|improve this answer
    
Well that was simple. Thank you Georges! –  Gunnar Magnusson Feb 29 '12 at 10:08

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.