# A characteristic-free proof that the action of a connected algebraic group $G$ on the fundamental group of a $G$-variety is trivial

Let $G$ be an algebraic group (not necessarily linear) defined over an algebraically closed field $k$, acting on a smooth integral $k$-variety $X$. Let $x_0\in X(k)$ and let $\pi_1(X,x_0)$ denote the étale (Grothendieck's) fundamental group of $X$. Assume that either $G$ fixes $x_0$ or the group $\pi_1(X,x_0)$ is abelian. In both cases $G(k)$ acts on $\pi_1(X,x_0)$. I need a proof that if $G$ is connected, then this action is trivial.

I know a proof in characteristic 0. In this case by the Lefschetz principle we may assume that $k=\mathbf{C}$, and we can consider the action of $G(\mathbf{C})$ on the topological fundamental group $\pi_1^{\mathrm{top}}(X(\mathbf{C}),x_0)$. Let $g\in G(\mathbf{C})$. Since $G$ is connected, we can connect $g$ with the unit element $e\in G(\mathbf{C})$ by a continuous path. We see that the automorphism $g_*\colon X(\mathbf{C})\to X(\mathbf{C})$ is homotopic to the identity automorphism. It follows that the induced automorphism of $\pi_1^{\mathrm{top}}(X(\mathbf{C}),x_0)$ is the identity.

I would like to see a proof that the action is trivial in arbitrary characteristic.

This is false as stated in positive characteristic. For example, suppose that the characteristic of $k$ is $p$; take $X = \mathbb A^1_k = \mathop{\rm Spec} k[t]$ and $G = \mathbb G_{\rm m}$. If $a \in k^*$ and $E$ is the standard étale cover $E = \mathop{\rm Spec} k[x,t]/(x^p - x - t) \to \mathop{\rm Spec} k[t]$ of $X$, its pullback through multiplication by $a$ is $\mathop{\rm Spec} k[x,t]/(x^p - x - at) \to \mathop{\rm Spec} k[t]$, which will not be isomorphic to $E$ in general.

However, it is true when $X$ is proper over $\mathop{\rm Spec} k$. The point is that in this case we have $\pi_1(X \times G, (x_0, 1)) = \pi_1(X, x_0) \times \pi_1(G, 1)$ (see, for example, Corollary 5.6.6 in Szamuely's wonderful book on the fundamental group); this uses the hypothesis that $G$ is connected. The embedding of $\pi_1(G, 1)$ corresponding in the decomposition above is induces by the embedding $G \subseteq X \times G$ defined by $g \mapsto (x_0, g)$. If $\alpha \colon X \times G \to X$ is the action, let us show that $\alpha_*\colon \pi_1(X \times G, (x_0, 1)) \to \pi_1(X, x_0)$ coincides with the homomorphism $\pi_1(X \times G, (x_0, 1)) \to \pi_1(X, x_0)$ induced by the first projection. From the formula above we see that it is sufficient to show that the composite $G \to X \times G \to X$, where the first map is $g \mapsto (x_0, g)$ and the second is the action, induces a trivial map on fundamental groups. But the map is in fact constant, because $G$ fixes $x_0$, so this is clear.

Now fix $g \in G(k)$. The composite $X \to X \times G \to X$, where the first map is $x \mapsto (x,g)$, and the second is the action, induces the action of $g$ on $X$. On the other hand, the composite $X \to X \times G \to X$, where the first map is $x \mapsto (x,g)$, and the second is the projection, induces the identity. From the previous fact, we have that these two maps induce the same homorphism on fundamental groups, and we are done.

• A related question: What is the fundamental group of $\mathbb A^1_k$ as a scheme (or, if it's not representable, as a functor)? Feb 28 '11 at 16:39
• @Angelo: Could you please add details? Why "if $\alpha \colon X \times G \to X$ is the action, we need to show that $\alpha_*\colon \pi_1(X \times G, (x_0, 1)) \to \pi_1(X, x_0)$ coincides with the homomorphism $\pi_1(X \times G, (x_0, 1)) \to \pi_1(X, x_0)$ induced by the first projection"? How is this related to the assertion that any element $g\in G(k)$ acts trivially on $\pi_1(X, x_0)$? Also, why "this is clear from the formula above, since the elements of $\pi_1(G, 1)$ go to the identity, because $G$ fixes $x_0$"? And where do you use the assumption that $G$ is connected? Feb 28 '11 at 16:47
• To André: I am not sure what you are asking. What's "the fundamental group as a functor"? To Mikhail: I will add more details. Feb 28 '11 at 17:04
• to André: if $char(k)=0$ then $\pi_{1}(\mathbb{A}_{k}^{1}=\pi_{1}(Spec(k))$. If $char(k)=p>0$ then the question is more delicate. Anyway it still exists a map $\pi_{1}(\mathbb{A}_{k}^{1} \rightarrow \pi_{1}(Spec(k))$ which is surjective, but not injective. So for example if the field k is of non-zero characteristic and separably closed, then $\pi_{1}(\mathbb{A}_{k}^{1}$ is non-trivial. Feb 28 '11 at 17:45
• @André: The precise structure of the fundamental group of an affine curve $X$ over an algebraically closed field $k$ of characteristic $p>0$ is not known. Assume that $X$ is the complement to $r\geq 1$ points in a projective connected smooth curve of genus $g$. Then, Abhyankar's conjecture, proved by Raynaud (for $X=\mathbf A^1_k$) and Harbater, states that a finite group $G$ is a quotient of $\pi_1(X)$ if and only its maximal prime-to-$p$ quotient $G'$ is the fundamental group of the analogous curve in characteristic zero, i.e., is generated by $2g-2+r-1$ elements.
– ACL
Mar 1 '11 at 8:29