In this post we prove a general result that shows, in particular, that any map from a simply connected $X$ to a curve $C$ of genus at least $1$ is constant.

The key preliminary result

Let us begin by discussing a nice result which mirrors a classical result in topology. For completeness, let us recall this result:

Theorem 1: Let $X$ be a locally path connected, simply connected topological space and let $f:X\to Z$ be a continuous map. Then, if $p:Y\to Z$ is a covering map, then there exists a lift of $f:X\to Z$ to $\widetilde{f}:X\to Y$ .

This result is a standard result in a first course on topology, and underlies the key principle of simply connected spaces: they are the spaces which are unobstructed to lifting to coverings.

The same is true in the étale world for essentially the same reason. In particular we have the following:

Key lemma 2: Let $X$ be a conected, (étale) simply connected scheme and $f:X\to Z$ a morphism with $Z$ connected. Suppose that $\alpha: Y\to Z$ is a finite connected étale cover. Then, there is a morphism $\widetilde{f}:X\to Y$ lifting $f$ .

Proof: Consider the pull-back

$\begin{matrix}X\times_Z Y & \to & Y\\ \downarrow & & \downarrow\\ X & \to & Z\end{matrix}$

Since $X$ is simply connected we know that $X\times_Z Y$ is isomorphic to $\displaystyle \bigsqcup_i X$ (with $\deg(\alpha)$ copies of $X$ ) as an $X$ -scheme. In particular, we see that there is a morphism $g:X\to Y$ such that $\alpha\circ g=f$ as desired. $\blacksquare$

The general result

With this

Theorem(Main) 3: Let $X$ be a connected, (étale) simply connected scheme. Let $Y$ be a connected locally Noetherian scheme for which there are connected finite étale covers of arbitrarily high degree. Then, there is no flat proper map $X\to Y$ .

Proof: Suppose that $f:X\to Y$ is a flat proper map. Then, for any finite connected étale cover $\alpha:Z\to Y$ we know, by the Key Lemma, that we can factorize $f$ as $\alpha\circ g=f$ for some $g:X\to Z$ . Note that $g$ is necessarily flat and proper. Properness follows since

$X\times_Y Z=\displaystyle \bigsqcup_i X\to Z$

is proper (using the notation in the proof of the Key Lemma) and since the canonical inclusion

$\displaystyle X\to\bigsqcup_i X$

is a closed embedding their composition, $g$ is also proper. To see that $g$ is flat we note that since $\alpha\circ g=f$ , $f$ is flat and $\Delta_\alpha$ is an open embedding, that the cancellation lemma implies that $g$ is also flat.

Now, since $f$ is proper and flat (and $Y$ is locally Noetherian) we know that $f_\ast\mathcal{O}_X$ is a vector bundle. Similarly, we know that since $g$ is proper and flat (and $Z$ is locally Noetherian) that $g_\ast\mathcal{O}_X$ is a vector bundle. But, by construction

$\alpha_\ast(g_\ast\mathcal{O}_X)=f_\ast\mathcal{O}_X$

But, this implies that

$\mathrm{rank}(f_\ast\mathcal{O}_X)=\deg(\alpha)\mathrm{rank}(g_\ast\mathcal{O}_X)$

and so, in particular, $\mathrm{rank}(f_\ast\mathcal{O}_X)\geqslant \deg \alpha$ . Since we assumed we could find $\alpha$ of arbitrarily high degree this is a contradiction. $\blacksquare$

Applications to curves

As an application of the main theorem we may derive the following result:

Theorem 4: Let $X$ be a proper variety over a field $k$ such that $\pi_1^\mathrm{geom}(X):=\pi_1^{\mathrm{\acute{e}t}}(X_{\overline{k}})=0$ . Then, for any (smooth projective integral) curve $C/k$ of positive genus there are no non-constant maps $X\to C$ .

Here ‘variety’ means a separated, finite-type, and geometrically reduced $k$ -scheme. Of course, separatedness is not used in the proof.

Proof: Suppose that $X\to C$ is non-constant, then $X_{\overline{k}}\to C_{\overline{k}}$ is non-constant. Thus, it suffices to assume that $k=\overline{k}$ . But, then we see that $X\to C$ non-constant implies that’s surjective (by dimension considerations) and thus (since $X$ is reduced) a flat map. Since $X$ is proper and $C$ separated we know that $X\to C$ is automatically proper (by the cancellation lemma). Thus, we may apply the Main Theorem to arrive at a contradiction after noting that $C$ , being positive genus, has connected étale covers of arbitrarily high degree. $\blacksquare$

As a simple application of this, we see that there are no non-constant maps from a proper smooth rational (even rationally connected!) variety $X$ to a curve of positive genus $C$ . This follows since such varieties are smooth (see the discussion here).

Two simple examples which pop-up from this are the fact that there are no non-constant maps $\mathbb{P}^n_k\to C$ or $(\mathbb{P}^1_k)^n\to C$ for $C$ a positive genus curve. These, of course, both follow from elementary considerations, but this gives a conceptual way for showing the result.

The complex setting

If one is willing to consider a $\mathbb{C}$ -analogue one can prove Theorem 4 in a slightly different way—this was pointed out to my by my friend Alex Sherman. Namely, let $X/\mathbb{C}$ be a smooth projective (integral) variety which is topologically simply connected (i.e. $\pi_1^\mathrm{top}(X^\mathrm{an})=0$ ) and let $X\to C$ be a map. Then, by composing $C\to\mathrm{Jac}(C)$ (by choosing some base-point) we get a map $X\to\mathrm{Jac}(C)$ . Then, by the universal property of the Albanese variety one obtains a map $\mathrm{Alb}(X)\to \mathrm{Jac}(C)$ making the following diagram commute:

$\begin{matrix}\mathrm{Alb}(X) & \to & \mathrm{Jac}(C)\\ \downarrow & & \downarrow\\ X & \to & C\end{matrix}$

But, as a complex torus

$\mathrm{Alb}(X)=H^1(X,\mathcal{O}_X)/H^1_\mathrm{sing}(X,\mathbb{Z})$

and by Hodge theory we know that

$H^1_\mathrm{sing}(X,\mathbb{C})=H^1(X,\mathcal{O}_X)\oplus H^0(X,\Omega^1_\text{hol})$

But, since $X$ is simply connected we know that $\pi_1(X)=0$ and thus $H_1(X,\mathbb{Z})=0$ . Then, since $H^1_\mathrm{sing}(X,\mathbb{Z})=\left(H_1(X,\mathbb{Z})\right)^\vee$ we conclude that $H^1_\mathrm{sing}(X,\mathbb{Z})=0$ and thus $H^1_\mathrm{sing}(X,\mathbb{C})=0$ . Consequently, $H^1(X,\mathcal{O}_X)=0$ and so $\mathrm{Alb}(X)=0$ .

So, we see that the composition

$X\to\mathrm{Alb}(X)\to \mathrm{Jac}(C)$

is evidently constant. But, it’s equal to the composition

$X\to C\to\mathrm{Jac}(C)$

and since $C\to\mathrm{Jac}(C)$ is injective, it follows that $X\to C$ must have been constant as well.

This approach should be adaptable to work over characteristic $0$ fields besides $\mathbb{C}$ , namely. I haven’t checked the details but it seems like the following should work. The map $C\hookrightarrow \mathrm{Jac}(C)$ is still injective. Moreover, if $X$ is connected (and sufficiently nice) we know that $\mathrm{Alb}(X)$ exists and has dimension $H^1(X,\mathcal{O}_X)$ . So, we want to show that if $\pi_1^\mathrm{\acute{e}t}(X)=0$ then $H^1(X,\mathcal{O}_X)$ . Using standard techniques one reduces this to the case of $X=\mathbb{C}$ . Now, we claim that $\pi_1^\mathrm{\acute{e}t}(X)=0$ implies that $H_1(X^\mathrm{an},\mathbb{Z})$ is trivial. Indeed, if not then $H_1(X,\mathbb{Z})$ and thus $\pi_1^\mathrm{top}(X)$ would have a finite quotient. This then should imply that $\pi_1^\mathrm{\acute{e}t}(X)\ne 0$ which is a contradiction. Then, we have that $H_1(X,\mathbb{Z})$ so $H^1_\mathrm{sing}(X,\mathbb{C})=0$ . One then deduces that $H^1(X,\mathcal{O}_X)=0$ by Hodge theory. One could also prove this last part by something like, say, $p$ -adic Hodge theory. The rest of the argument then goes the same.

This seems doomed to work in general though. In positive characteristic it’s not true that $\pi_1^\mathrm{\acute{e}t}(X)=0$ implies that $H^1(X,\mathcal{O}_X)=0$ as an Enriques surface shows.

One comment

mayorliatmath says:

October 26, 2016 at 10:08 am

I don’t know much about Albanese in characteristic p, is it true that pi_1=0 would imply vanishing of Albanese (because Albanese map induces isomorphism between l-adic H^1_ét, where l is prime to p?)

Hard Arithmetic

Maps from simply connected projective varieties to curves

The key preliminary result

The general result

Applications to curves

The complex setting

One comment

Leave a Reply Cancel reply

The key preliminary result

The general result

Applications to curves

The complex setting

Share this:

Like this:

One comment

Leave a Reply Cancel reply