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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.

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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 Þór Magnússon Feb 29 '12 at 10:07
up vote 21 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.

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 ($\mathbb P^1_k)=0$

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Well that was simple. Thank you Georges! – Gunnar Þór Magnússon Feb 29 '12 at 10:08

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