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

Screenshot-65

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

Screenshot-66

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

Screenshot-67where {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.

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