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

Question

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.

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

Answer 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.

Albanese varieties

Let us fix $k$ to be a perfect field, and $X$ a normal proper geometrically integral scheme over $k$. We also fix a $k$-point $x$ of $X$.

Definition: An Albanese variety for the pair $(X,x)$ is a morphism $f:(X,x)\to (A,e)$ of pointed $k$-schemes, satisfying the following conditions:

• $A$ is an abelian variety over $k$ and $e$ is its identity section,
• it is initial amongst maps to such pointed abelian varieties.

If such an Albanese variety exists it is clearly unique up to unique isomorphism, and we denote it by $\mathrm{Alb}(X,x)$.

We then have the following beautiful, classical result.

Theorem (e.g. see [Mochizuki, Theorem A.4 and Proposition A.6]): Let $(X,x)$ be as above. Then, an Albanese variety $$f\colon (X,x)\to (\mathrm{Alb}(X,x),e)$$ exists and, moreover $\mathrm{Alb}(X,x)\cong ((\text{Pic}^0_{X/k})_{\text{red}})^\vee$.

This notation needs some explanation. Consider the functor

$$\mathrm{Pic}_{X/k}\colon \mathbf{Sch}_k\to\mathbf{Grp},\qquad T\mapsto \left\{(\mathscr{L},\iota): \begin{aligned}(1)&\quad \mathscr{L}\text{ is a line bundle on }X_T\\ (2)&\quad \iota\colon x_T^\ast\mathscr{L}\xrightarrow{\approx}\mathcal{O}_T\end{aligned}\right\}/\text{iso.}$$

Here $x_T\colon T\to X_T$ is the base change of $x\colon \mathrm{Spec}(k)\to X$ along $T\to X$, and an isomorphism $(\mathscr{L},\iota)\to (\mathscr{L}',\iota')$ is an isomorphism $\alpha\colon \mathscr{L}\to\mathscr{L}'$ such that $\iota'\circ x_T^\ast\alpha=\iota$. Then, by the discussion after [BLR, §8.1, Proposition 4] and [BLR, §8.1, Theorem 3], $\mathrm{Pic}_{X/k}$ is represented by a locally of finite type group $k$-scheme.

Moreover, with our assumptions on $X$, we in fact have that the reduced subscheme $(\mathrm{Pic}_{X_k}^0)_\mathrm{red}$ of the identity component $\mathrm{Pic}^0_{X/k}$ (which is still a group $k$-scheme by [Milne, Corollary 1.39]) is an abelian variety. Note that as it is automatically smooth (see [Milne, Proposition 1.26 and Proposition 1.28]) it suffices to show that it's proper. If $X$ is smooth over $k$ then this follows from [BLR, §8.4, Theorem 3]. If one assumes that $X$ is merely normal this requires more work (see [Mochizuki, Theorem A.4]).

In any case, as $(\mathrm{Pic}^0_{X/k})_{\mathrm{red}}$ is an abelian variety over $k$, it makes sense to take its dual abelian variety in the sense of [BLR, §8.4, Theorem 5], and this is the object that appears in the statement of the above theorem.

Trivial Albanese varieties : vanishing $H^1(X,\mathcal{O}_X)$

The upshot of all of this is that if one wants to study maps from $X$ to an abelian variety then, by choosing a point $x$ in $X(k)$ (which can always be achieved after a finite extension), we are reduced to studying $\mathrm{Alb}(X,x)$. In particular, if $\mathrm{Alb}(X,x)$ is trivial, then every $k$-map to an abelian variety is constant. This is useful, as the Albanese variety has many appealing properties.

For instance, we may compute its Lie algebra of $\mathrm{Alb}(X,x)$ (and thus its dimension) quite concretely.

Fact 1(see [BLR, §8.4, Theorem 1]): Let $e$ be the identity section of $\mathrm{Alb}(X,x)$. Then, there is a canonical isomorphism of vector $k$-spaces $$T_e \,\mathrm{Pic}_{X/k}\cong H^1(X,\mathcal{O}_X).$$

We then immediately deduce the following corollary which shows that, in particular, $\mathbb{P}^n_k$ admits no non-constant $k$-maps to an abelian variety over $k$, let alone embeddings.

Corollary 1: Suppose that $H^1(X,\mathcal{O}_X)$ is trivial, then every map $X\to A$, where $A$ is an abelian variety, is constant.

Proof: If $X\to A$ is non-constant, then so is $X_{\overline{k}}\to A_{\overline{k}}$, and so we may assume that $k$ is algebraically closed. It then suffices to show that if $x$ is any point of $X(k)$ then $\mathrm{Alb}(X,x)$ is trivial. But, note that

$$\dim \mathrm{Alb}(X,x)=\dim \,(\mathrm{Pic}^0_{X/k})_\mathrm{red}=\dim_k T_e\, (\mathrm{Pic}_{X/k})_\mathrm{red}\leqslant \dim_k \mathrm{Pic}_{X/k},$$

(where the first equality follows from [EMvdG, (6.18) Theorem] and the second equality from the fact that $(\mathrm{Pic}_{X/k})_\mathrm{red}$ is smooth). The conclusion follows from the above fact. $\blacksquare$

Of course, this Albanese discussion only applied a priori to proper things. In particular, there is nothing that, a priori, precludes us from having a non-constant map $\mathbb{A}^n_k\to A$, for an abelian variety $A$ that just does not extend to $\mathbb{P}^n_k$. As it turns out though, this cannot happen.

Fact 2 (cf. [BLR, §4.4, Theorem 1]): Let $V$ be a geometrically connected finite type smooth $k$-scheme, and $U$ a non-empty open subset of $V$. Then, for an abelian variety $A$ over $k$, any $k$-map $U\to A$ extends (necessarily uniquely) to a map $V\to A$.

For a separated finite type $k$-scheme $U$, let us call a proper $k$-scheme $X$ containing $U$ as a (topologically and scheme-theoretically) dense open subscheme a compactification of $U$. It is a deep theorem of Hironaka, that if $k$ is of characteristic $0$ and $U$ is smooth, then a smooth compactification of $U$ exists.

From the above discussion, we trivially see the following which implies that there are no non-constant $k$-maps $\mathbb{A}^n_k\to A$ for an abelian variety $A$ over $k$.

Corollary 1 (redux): If $U$ is a separated finite type smooth $k$-scheme which has a smooth compactification $X$ with $H^1(X,\mathcal{O}_X)$ then there exists no non-constant $k$-maps $U\to A$ for an abelian variety $A$ over $k$.

Vanishing Albanese varieties: 'small' fundamental groups

This approach using the Albanese variety is robust enough to go beyond cases shown in Corollary 1 (redux). To give examples of this, we have the following fact concerning the relationships between $\mathrm{Alb}(X,x)$ and the etale fundamental group of $X$.

Fact 3 ([Mochizuki, Proposition A.3]): Let $p$ be the characteristic of $k$. Then, for any geometric point $\xi$ of $X_{\overline{k}}$, the map $$\pi_1^\mathrm{et}(X_{\overline{k}},\xi)^{(p)}\to \pi_1^\mathrm{et}(\mathrm{Alb}(X,x)_{\overline{k}},\xi)$$ is surjective.

For a group $G$, we denoting by $G^{(p)}$ the inverse limit over all quotients of $G$ of order coprime to $p$.

Corollary 2: Suppose that either $U$ is either

• a geometrically integral normal proper $k$-scheme,
• a geometrically integral smooth $k$-scheme possessing a smooth compactification.

Suppose that for some geometric point $u$ of $U_{\overline{k}}$ one has that $\text{Hom}(\pi_1^\mathrm{et}(U_{\overline{k}},u),\mathbb{Z}/q\mathbb{Z})$ for some prime $q$ invertible in $k$. Then, any morphism of $k$-schemes $U\to A$ for an abelian variety $A$ over $k$ is constant.

Proof: We quickly reduce to the case when $k$ is algebraically closed. Note that in the second case, if $X$ is a smooth compactification of $U$, then by Fact 2, it suffices to show $X$ admits no non-constant $k$-maps to an abelian variety over $k$. But, as the natural map $\pi_1^\mathrm{et}(U,u)\to \pi_1^\mathrm{et}(X,x)$ is a surjection (cf. this), we quickly reduce to the first case. We then see from Fact 3 that

$$\mathrm{Hom}(\pi_1(\mathrm{Alb}(X,x)_{\overline{k}},u),\mathbb{Z}/q\mathbb{Z})=0.$$ But, if $n$ is the dimension of $\mathrm{Alb}(X,x)$, then

$$\pi_1(\mathrm{Alb}(X,x)_{\overline{k}},u)^{(p)}=\prod_{q\ne p}\mathbb{Z}_q^{2g},$$

(see [EvdGM, (10.37) Corollary]). Thus, from the assumption that $$\mathrm{Hom}(\pi_1(\mathrm{Alb}(X,x)_{\overline{k}},u),\mathbb{Z}/q\mathbb{Z})=0,$$ we see that $n=0$ and thus $\mathrm{Alb}(X,x)$ is trivial as desired. $\blacksquare$

As an application of this, if $X$ is a so-called singular Enriques surface over $k$, then $H^1(X,\mathcal{O}_X)\ne 0$ but $\pi_1^\mathrm{et}(X_{\overline{k}},x)=\mathbb{Z}/2\mathbb{Z}$. Thus, $X$ admits no non-trivial $k$-maps to an abelian variety over $k$.

References

[BLR] Bosch, S., Lütkebohmert, W. and Raynaud, M., 2012. Néron models (Vol. 21). Springer Science & Business Media.

[EvdGM] http://van-der-geer.nl/~gerard/AV.pdf

[Milne] Milne, J.S., 2017. Algebraic groups: the theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.

[Mochizuki] https://www.kurims.kyoto-u.ac.jp/~motizuki/Topics%20in%20Absolute%20Anabelian%20Geometry%20I.pdf