16
$\begingroup$

Let $X$ smooth curve over a finite field $\mathbb{F}_q$ of type $(g,n)$ - that is, $X$ is an open subscheme of its genus $g$ compactification obtained by removing $n$ points.

Any such curve determines an exact sequence of etale fundamental groups

$$1\rightarrow \pi_1(X_{\overline{\mathbb{F}}_q})\rightarrow \pi_1(X)\rightarrow\underbrace{\text{Gal}(\overline{\mathbb{F}}_q/\mathbb{F}_q)}_{G_{\overline{\mathbb{F}}_q}}\rightarrow 1$$ whence a representation $$\rho' : G_{\overline{\mathbb{F}}_q}\rightarrow\text{Out}(\pi_1(X_{\overline{\mathbb{F}}_q}))\rightarrow \text{Out}(\pi_1'(X_{\overline{\mathbb{F}}_q}))$$ where $\pi'$ denotes the prime-to-$p$ fundamental group. (Actually since the absolute galois group of a finite field is projective, the exact sequence is split, and so we get an actual representation into $\text{Aut}(\cdots)$)

Are there any examples where $X$ is hyperbolic (equivalently, $\pi_1'(X_{\overline{\mathbb{F}}_q})$ is nonabelian) such that the representation $\rho'$ is known?

This is equivalent to understanding the action of Frobenius on $\pi_1'(X_{\overline{\mathbb{F}}_q})$.

References would also be appreciated.

Improve this question
$\endgroup$
12
  • 1
    $\begingroup$ your exact sequence splits, so your outer action lifts to a real one. $\endgroup$
    – Niels
    Commented Feb 12, 2017 at 22:02
  • 2
    $\begingroup$ @oxeimon The absolute Galois group of a finite field is projective (actually free) profinite group, hence any epimorphism on it splits in the profinite category. It is not equivalent to existence of a rational point $\endgroup$
    – Lior Bary-Soroker
    Commented Feb 12, 2017 at 23:25
  • 1
    $\begingroup$ @LiorBary-Soroker Good point! Neat! $\endgroup$
    – Will Chen
    Commented Feb 12, 2017 at 23:31
  • 1
    $\begingroup$ @SaalHardali No problem! A hyperbolic curve over a finite field is a smooth separated geometrically connected scheme of dimension 1 over a finite field such that if $g$ denotes the genus of its smooth compactification, and $n$ denotes the number of points one needs to add to compactify it, then either $g \ge 2$, $n\ge 3$, or $g = n = 1$. $\endgroup$
    – Will Chen
    Commented Feb 20, 2018 at 2:15
  • 1
    $\begingroup$ @SaalHardali The terminology comes from the following: Given a finite type Riemann surface $X$ of type $(g,n)$ (that is, it is obtained by removing $n$ points from a compact Riemann surface of genus $g$), then there are 3 possibilities for the universal cover of $X$. Either the universal cover is $\mathbb{C}$ (the geometry is flat), or $\mathbb{P}^1_\mathbb{C}$ (the geometry is spherical), or it is the upper half plane $\mathbb{H}$ (the geometry is hyperbolic). $\endgroup$
    – Will Chen
    Commented Feb 20, 2018 at 2:19

0

Reset to default

You must log in to answer this question.