P¹ is simply connected

This is a cute proof that I ran into of the simple connectedness of \mathbb P^1. It does not use Riemann–Hurwitz or differentials, and instead relies on a purely geometric argument.

Lemma. Let k be an algebraically closed field. Then \mathbb P^1_k is simply connected.

Proof. Let f \colon C \to \mathbb P^1 be a finite étale Galois cover with Galois group G. We have to show that f is an isomorphism. The diagonal \Delta_{\mathbb P^1} \subseteq \mathbb P^1 \times \mathbb P^1 is ample, so the same goes for the pullback D = (f \times f)^* \Delta_{\mathbb P^1} to C \times C [Hart, Exc. III.5.7(d)]. In particular, D is connected [Hart, Cor. III.7.9].

But D \cong C \times_{\mathbb P^1} C is isomorphic to |G| copies of C because the action

    \begin{align*} G \times C &\to C \times_{\mathbb P^1} C\\ (g,c) &\mapsto (gc,c) \end{align*}

is an isomorphism. If D is connected, this forces |G| = 1, so f is an isomorphism. \qed

The proof actually shows that if X is a smooth projective variety such that \Delta_X is a set-theoretic complete intersection of ample divisors, then X is simply connected.

Example. For a smooth projective curve C of genus g \geq 1, the diagonal cannot be ample, as \pi_1(C) \neq 0. We already knew this by computing the self-intersection \Delta_C^2 = 2-2g \leq 0, but the argument above is more elementary.


[Hart] Hartshorne, Algebraic geometry. GTM 52, Springer, 1977.