Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Classically: Let $a_1,...,a_r$ be points in $\mathbb{P}^1_{\mathbb{C}}$, and let $\alpha_1,...,\alpha_r$ be simple loops around the $a_i$, all counterclockwise, and none touching (so $\alpha_1...\alpha_r=1$ in the fundamental group of the projective line minus those points). An (pointed, to be pedantic) unramified $G$-cover (meaning a normal covering space with deck transformations$=G$) of $\mathbb{P}^1_{\mathbb{C}}-a_1,...,a_r$ is given by a surjection $\pi_1(\mathbb{P}^1_{\mathbb{C}}-a_1,...,a_r) \rightarrow G$. Let $g_i$ be the image of $\alpha_i$. We say that this $G$-Galois branched cover has branch cycle description $(g_1,...,g_r)$ (note that this depends on our choice of the $\alpha_i$'s). This covering map of curves can be extended to a map of (smooth) projective groups. It can then be shown by a simple topological argument that $g_i$ generates the inertia group (=decomposition group in this case) of some point above $a_i$.

My question is whether (and if so, how?) this is also true for the $\overline{\mathbb{F}_p}$ case.

Let me be precise. It is known via Grothendieck that $\pi_1^{(p)}(\mathbb{P}^1_{\overline{\mathbb{F}_p}}-a_1,...,a_r)=\widehat{\langle \alpha_1,...,\alpha_r|\prod \alpha_i =1 \rangle}^{(p)}$ (the $^{(p)}$ indicates that we're taking the inverse limit of all prime-to-$p$ finite quotients). Since these $\alpha_i$'s are given in SGA1 through a rather mysterious method, I wonder if the phenomenon described in the first paragraph is still true.

My question, therefore, is: let $G$ be a prime-to-$p$ group, and let $X\rightarrow \mathbb{P}^1_{\overline{\mathbb{F}_p}}$ be a (pointed, to be pedantic) branched $G$-cover with branch points $a_1,...,a_r$. Let $\alpha_1,...,\alpha_r$ be such that $\pi_1^{(p)}(\mathbb{P}^1_{\overline{\mathbb{F}_p}}-a_1,...,a_r)=\widehat{\langle \alpha_1,...,\alpha_r|\prod \alpha_i =1 \rangle}^{(p)}$ (I'm almost positive that what I'm about to say is false if you're allowed to choose any such $\alpha_i$'s, so let's assume that we're taking the ones from Grothendieck's construction. If you see a better way of saying what the condition should be on the $\alpha_i$'s I would be very interested in that). Let the branch cycle description of this cover be $(g_1,...,g_r)$ (with respect to these $\alpha_i$'s). Is it true that $g_i$ generates the inertia group (=decomposition group in this case) of some point of $X$ above $a_i$?

The topological argument that we were able to use for the $\mathbb{C}$ case seems to no longer apply...

share|improve this question

1 Answer 1

I'm not sure I totally understand your question, but I think the answer is yes almost by definition. The procyclic group topologically generated by you alpha_i is the image of the map

pi_1(Spec O_i) -> pi_1(P^1 - a_1, ... a_r)

where O_i is the completed local ring of P^1 at the puncture a_i (so it's isomorphic to F_q((t)).) I guess the content here is that Grothendieck's comparison is compatible with the passage between local and global.

I was lazy and didn't write in base points; the choice of base point would induce the choice of WHICH decomposition group over a_i you get.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.