Is there an algebraic reason why a torus can't contain a projective space?

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.

Solution 1:

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

Solution 2:

Sorry for reviving such an old question, but there is a purely algebraic reason to expect this which, while not as simple as Georges's answer, is what I believe to be the 'correct reason' and which helps you figure out in general, when a variety admits non-trivial maps to abelian varieties. Again, I'm not claiming that this is the 'best' way to see the solution to this particular problem, but that it is what is happening behind the scenes.

Namely, recall that if $X$ is a smooth proper geometrically integral variety over $k$ (a perfect field, say) and $x\in X(k)$ then an Albanese variety for the pair $(X,x)$ is a morphism of varieties $f:(X,x)\to (A,e)$ where $A$ is an abelian variety over $k$ and $e\in A(k)$ its identity section, and such that for any other abelian variety $(A',e')$ and a pointed morphism $g:(X,x)\to (A',e')$ there exists a unique morphism of abelian varieties $\alpha:(A,e)\to (A',e')$ such that $g=\alpha\circ f$. Since this variety is evidently unique we denote it by $\text{Alb}(X,x)$

We then have the following beautiful result:

Theorem: Let $(X,x)$ be as above. Then, $(X,x)$ admits a Albanese variety $f:(X,x)\to (A,e)$ and, in fact, $A^\vee\cong (\text{Pic}^0_{X/k})_{\text{red}}$.

Here, as you probably know, $\text{Pic}^0_{X/k}$ is the identity component of the relative Picard scheme (which, here, can be constructed as the moduli space of line bundles with a trivialization along the section determined by $x$).

Why does this make our question about maps $\mathbb{P}^n_k\to A$, where $A$ is an abelian variety, trivial? Well, without loss of generality we may assume that $e\in A(k)$ is the image of some $x\in\mathbb{P}^n_k(k)$ and thus we know from the theorem that the map $\mathbb{P}^n_k$ must factor through $\text{Alb}(\mathbb{P}^n_k,x)$. But, by the theorem we know that $\text{Alb}(\mathbb{P}^n_k,x)$ is dual to $\text{Pic}^0_{\mathbb{P}^n_k/k}$ (the reduced is unecessary since we're in characteristic $0$ for your question). But, as is classical, we have a canonical identification

$$ T_e \text{Pic}^0_{\mathbb{P}^n_k/k}\cong H^1(\mathbb{P}^n_k,\mathcal{O}_{\mathbb{P}^n_k})=0$$

and thus $\text{Pic}^0_{\mathbb{P}^n_k/k}$ and thus $\text{Alb}(\mathbb{P}^n_k,x)$ is the trivial abelian variety. Thus, the map $\mathbb{P}^n_k\to A$ must be constant.

In fact, I think we can give a nice generalization of this result suggested by Georges answer. Namely, suppose that $X$ is smooth proper (integral) over $\mathbb{C}$ (for now) and, most importantly, simply connected. Then, you can't have a non-constant map $X\to A$ where $A$ is an abelian variety. Indeed, as Georges notes this would then give a map $X\to\mathbb{C}^n$ which must be constant since $X$ is proper. But, how can we prove this algebraically--does it hold over an arbitrary field. The answer is yes.

Namely, let $k$ be a perfect field and $X/k$ a smooth (normal should be fine) proper (geometrically) integral variety which is simply connected in the \'{e}tale sense, then I claim that any map $X\to A$, where $A$ is an abelian variety, is constant.

It suffices to prove the claim when $k$ is algebraically closed, so let's assume that. Choose any $x\in X(k)$ and, without loss of generality, choose it so that it maps to $e\in A(k)$. Then, the map $(X,x)\to (A,e)$ induces a map $\text{Alb}(X,x)\to A$, and thus it suffices to prove that $\text{Alb}(X,x)$, or equivalently $\text{Pic}^0_{X/k}$, is zero.

But, note that since $\text{Pic}^0_{X/k}$ is an abelian variety, we know that its rank $r$ satisfies $$\text{Pic}^0_{X/k}[n](k)=(\mathbb{Z}/n\mathbb{Z})^{2r}$$ for all $n$ invertible in $k$. But, note that $\text{Pic}^0_{X/k}[n](k)\subseteq \text{Pic}(X)[n]$ and by Kummer theory we have an isomorphism (since we assumed that $X$ was integral and proper)

$$\text{Hom}(\pi_1^{\acute{e}\text{t}}(X,x),\mathbb{Z}/n\mathbb{Z})=H^1(X_{\acute{e}\text{t}},\mathbb{Z}/n\mathbb{Z})\cong \text{Pic}(X)[n]$$

and since $X$ was assumed simply connected, this left term is zero. Thus, $\text{Pic}[n]$ and so $\text{Pic}_{X/k}^0[n](k)$ is zero for all $n$ invertible in $k$ and thus $\text{Pic}^0_{X/k}$ is rank $0$, which implies that $\text{Alb}(X,x)$ is trivial as desired.

Note that the above does actually show more than claimed, and more than (I think?) Georges argument (using that tori are $K(\mathbb{Z}^{2n},1)$'s). Namely, it shows that if $X/\mathbb{C}$ is a smooth projective (integral) variety such that $\text{Hom}(\pi_1^\text{top}(X,x),\mathbb{Z}/n\mathbb{Z})=0$ for some $n$, then any map $X\to A$ where $A$ is an abelian variety is trivial. Maybe one can see this by lifting to the universal cover--I'm not sure.

Of course, one sees that the same idea works for showing there are no maps from such $X$ to curves, since the Abel-Jacobi map $C\to\text{Jac}(C)=\text{Alb}(C)$ is injective (this also works over any arbitrariy field).

EDIT: In case anyone has an answer to this, I wonder to what extent the above discussion can be extended to affine simply connected things? Intuitively, there shouldn't be a map $X\to A$, where $A$ is an abelian variety and $X$ is affine, since then it'd factor $X\to \mathbb{C}^n\to A$ and that can't be algebraic I believe since $\mathbb{C}^n\to A$ is highly transcendental.