Maps from simply connected projective varieties to curves

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


But, this implies that


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


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

  1. 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?)

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