Given a branched cover with branch cycle description $(g_1,...,g_r)$, does $g_i$ generate some decomposition group?

2011-05-05T01:27:55Z

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...

Answer by JSE for Given a branched cover with branch cycle description $(g_1,...,g_r)$, does $g_i$ generate some decomposition group?

2011-05-05T15:09:56Z

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.

Given a branched cover with branch cycle description $(g_1,...,g_r)$, does $g_i$ generate some decomposition group? - MathOverflow

Given a branched cover with branch cycle description $(g_1,...,g_r)$, does $g_i$ generate some decomposition group?

Answer by JSE for Given a branched cover with branch cycle description $(g_1,...,g_r)$, does $g_i$ generate some decomposition group?