# How does the Frobenius act on the prime-to-$p$ $\pi_1(\mathbb{P}^1_{\overline{\mathbb{F_p}}}\setminus \{a_1,…,a_r\})$?

From Grothendieck's work we know that the prime-to-p fundamental group $\pi_1(\mathbb{P}^1_{\overline{\mathbb{F}_p}}\setminus\{a_1,...,a_r\})$ where $a_1,...,a_r \in \mathbb{F}_p$ is isomorphic to the prime-to-p part of the profinite completion of $\langle \alpha_1,...,\alpha_r|\alpha_1...\alpha_r=1\rangle$.

The question is: how does the Frobenius automorphism of $\overline{\mathbb{F}_p}$ act on the prime-to-p $\pi_1(\mathbb{P}^1_{\overline{\mathbb{F}_p}}\setminus\{a_1,...,a_r\})$ where $a_1,...,a_r \in \mathbb{F}_p$?

I don't actually expect an answer. I gather that this is not well understood.

Trivially?:) It seems that you will have an action only if your $a_i$ belong to the fixed field of the Frobenius. –  Mikhail Bondarko Dec 21 '10 at 0:59