Let ${X}$ be a variety over an algebraically closed field ${k}$. ${X}$ is said to be rational if ${X}$ is birational to ${\mathbb{P}_k^n}$. In general, it is difficult to determine when a variety in higher dimensions is rational, although there are numerical invariants in dimensions one and two.

• Let ${X}$ be a smooth projective curve. Then ${X}$ is rational if and only if its genus is zero.
• Let ${X}$ be a smooth projective surface. Then ${X}$ is rational if and only if there are no global 1-forms on ${X}$ (i.e., ${H^0(X, \Omega_{X/k}) = 0}$) and the second plurigenus ${H^0(X, \omega_{X/k}^{\otimes 2}) }$ vanishes. This is a statement about the negativity of the cotangent bundle (or, equivalently, of the positivity of the tangent bundle) which is a birational invariant and which holds for ${\mathbb{P}^2_k}$. The result is a criterion of Castelnuovo, extended by Zariski to characteristic $p$.

In higher dimensions, it is harder to tell when a variety is rational. An easier problem is to determine when a variety is unirational: that is, when there is a dominant rational map

$\displaystyle \mathbb{P}_k^n \dashrightarrow X;$

or, equivalently, when the function field ${k(X)}$ has a finite extension which is purely transcendental. In dimensions one and two (and in characteristic zero), the above invariants imply that a unirational variety is rational. In higher dimensions, there are many more unirational varieties: for example, a theorem of Harris, Mazur, and Pandharipande states that a degree ${d}$ hypersurface in ${\mathbb{P}^N}$, ${N \gg 0}$ is always unirational.

The purpose of this post is to describe a theorem of Serre that shows the difficulty of distinguishing rationality from unirationality. Let’s work over ${\mathbb{C}}$. The fundamental group of a smooth projective variety is a birational invariant, and so any rational variety has trivial ${\pi_1}$.

Theorem 1 (Serre) A unirational (smooth, projective) variety over ${\mathbb{C}}$ has trivial ${\pi_1}$.

The reference is Serre’s paper “On the fundamental group of a unirational variety,” in J. London Math Soc. 1959.

1. Finiteness of ${\pi_1}$

To prove this theorem, let’s fix a unirational (smooth, projective) variety ${X}$ over ${\mathbb{C}}$ of dimension ${n}$, and a dominant rational map

$\displaystyle f: \mathbb{P}^n_{\mathbb{C}} \dashrightarrow X.$

There is a subvariety ${Z \subset \mathbb{P}^n_{\mathbb{C}}}$ of codimension ${\geq 2}$ such that the map actually extends to an honest morphism ${\mathbb{P}^n_{\mathbb{C}} \setminus Z \rightarrow X}$ (via the valuative criterion).

Let’s form the universal cover ${\widetilde{X} \rightarrow X}$; here ${\widetilde{X}}$ is only a complex manifold, not necessarily an algebraic variety. The goal is to show that ${\widetilde{X} \rightarrow X}$ is actually an isomorphism. To start with, though, let’s show that it is a finite map: in other words, that ${\pi_1(X)}$ is finite. To see this, observe that the map ${f: \mathbb{P}^n_{\mathbb{C}} \setminus Z \rightarrow X}$ is generically finite and étale, and so we may choose an open subset ${U \subset X}$ such that

$\displaystyle f^{-1}(U) \rightarrow U$

is a finite covering map. We now have a diagram

Now ${U \times_X \widetilde{X}}$ is a cover of ${U}$, and since we’ve thrown away codimension one subsets of ${\widetilde{X}}$, it is a connected covering space. (Equivalently, the map ${\pi_1 (U) \rightarrow \pi_1(X)}$ is a surjection.)

The claim is that we can get a map of covering spaces ${f^{-1}(U) \rightarrow U \times_X \widetilde{X}}$ as in the diagram. In fact, we will do better: we will produce a lift in the diagram

We get this map simply by noting that ${\mathbb{P}^n_{\mathbb{C}} \setminus Z}$ is simply connected, since we have thrown away a subset of codimension ${\geq 2}$.

In view of this, we conclude that there is a map from the finite cover ${f^{-1}(U) \rightarrow U}$ to the connected cover ${U \times_X \widetilde{X} \rightarrow U}$ (where the latter is a ${\pi_1(X)}$-Galois cover). Since any map of connected covers is necessarily surjective, we must conclude that ${\pi_1(X)}$ is finite.

2. ${\pi_1(X)}$ is trivial

We need to go a bit further to conclude that ${\pi_1(X)}$ is actually trivial, though. First, since ${X}$ has finite fundamental group, the universal cover ${\widetilde{X} \rightarrow X}$ acquires the structure of an algebraic variety, and we get a finite étale map

$\displaystyle \widetilde{X} \rightarrow X.$

We’re going to show that in fact ${\widetilde{X}}$ is unirational. In fact, consider the diagram

where ${P}$ is the pull-back. Now ${P \rightarrow \widetilde{X}}$ is a dominant rational map. However, since ${P \rightarrow \mathbb{P}^n_{\mathbb{C}} \setminus Z}$ is a covering map, we conclude that ${P}$ is a disjoint union of copies of ${\mathbb{P}^n_{\mathbb{C}}\setminus Z}$ itself; taking any one produces the dominant rational map

$\displaystyle \mathbb{P}^n_{\mathbb{C}} \dashrightarrow \widetilde{X}.$

We conclude that ${\widetilde{X}}$ is also unirational as claimed, and from here will get a contradiction by looking at the arithmetic genus.

A basic property of a unirational variety (over ${\mathbb{C}}$) is that certain Hodge numbers vanish:

$\displaystyle h^p(X, \mathcal{O}_X) = 0, \quad p > 0,$

because Hodge theory gives equalities

$\displaystyle h^p(X, \mathcal{O}_X) = h^0(X, \Omega^p_X),$

and a nonzero holomorphic ${p}$-form on ${X}$ would pull back to one on ${\mathbb{P}^n_{\mathbb{C}}}$ (where there are none). In particular, we conclude that the holomorphic Euler characteristic is ${1}$:

$\displaystyle \chi(\mathcal{O}_X) = 1.$

Now, as we’ve seen, the holomorphic Euler characteristic of a unirational variety is one. Therefore,

$\displaystyle \chi(\mathcal{O}_X) = \chi(\mathcal{O}_{\widetilde{X}}) = 1.$

However, the Hirzebruch-Riemann-Roch formula allows one to compute the holomorphic Euler characteristic of a complex manifold via a polynomial in the Chern numbers and implies that it is multiplicative in finite covers. That is:

Proposition 2 Let ${\widetilde{Y} \rightarrow Y}$ be a finite cover of compact complex manifolds of degree ${n}$. Then

$\displaystyle \chi(\mathcal{O}_{\widetilde{Y}}) = n \chi(\mathcal{O}_Y).$

Putting this together, we conclude that the degree of the universal cover ${\widetilde{X} \rightarrow X}$ is equal to one, so that ${\pi_1(X) = 1}$ as desired.