As of late, I’ve been reading the proof due to Laumon of the Weil conjectures (a simplification of Deligne’s second proof) via the Fourier-Deligne transform. This is quite interesting, and I’d like to start a series of posts on it soon, based on the notes I’ve been taking. I don’t currently have the time to edit the notes, so I’ll devote this post to a curious fact I learned today.

One of the first results one proves when studying the classical fundamental group in topology is that the fundamental group of a topological group is abelian. As I learned today, the analogous result for the etale fundamental group fails.

Let $G_0$ be a smooth group scheme, defined over a finite field $\kappa$ of size $q$, and let $G$ be the base-change to the algebraic closure. Then there is the Lang map $L: G \to G$ sending $x \mapsto (Fx) x^{-1}$, for $F$ the Frobenius.

It is a theorem (of Lang) that this map $L$ is a surjective map. It is also finite, since the fiber over the identity is finite (the $\kappa$-rational points), and a morphism of homogeneous spaces for an algebraic group with finite fibers is finite. (Quick but probably unnecessarily non-elementary proof: a morphism of reduced homogeneous spaces over a smooth group scheme is faithfully flat, by generic flatness and a translation argument. As a result, it’s a quotient map. Since the fibers are finite, one can check that the map is closed by a similar translation argument. Then, proper plus finite fibers implies finite by Zariski’s Main Theorem.)

Also, the Frobenius induces the zero map on the tangent spaces. As a result, $x \mapsto (Fx) x^{-1}$, the Lang morphism, is smooth as the morphism on tangent spaces is a bijection. So, since the fibers are finite, we have an etale cover!

The claim is that it is a Galois cover, and the automorphism group is $G_0(\kappa)$: indeed, the right translations by the $\kappa$-rational points are automorphisms of the cover. Since this is the right number of translations for the degree of the cover, we have indeed a Galois cover with the appropriate Galois group. But, in general, this Galois group won’t be abelian. So the fundamental group, which surjects onto every Galois group, can’t be abelian either.

I initially tried to prove the false result by the Eckmann-Hilton argument (until I was told about this counterexample), but it seems not to be correct: perhaps the problem is that the etale fundamental group doesn’t commute with products! (It does for proper schemes over an algebraically closed field, but this is very nontrivial.)