As a first application let us prove a precise form of Chebotarev for a finite étale Galois covering of curves. Let \varphi : Y \to X be a finite étale Galois covering with group G. This corresponds to a homomorphism
Sketch.
Write
\varphi _*(\overline{\mathbf{Q}_ l}) = \oplus _{\pi \in \widehat{G}} \mathcal{F}_{\pi }
where \widehat{G} is the set of isomorphism classes of irred representations of G over \overline{\mathbf{Q}}_ l. For \pi \in \widehat{G} let \chi _{\pi }: G \to \overline{\mathbf{Q}}_ l be the character of \pi . Then
H^*(Y_{\overline{k}}, \overline{\mathbf{Q}}_ l) = \oplus _{\pi \in \widehat{G}} H^*(Y_{\overline{k}}, \overline{\mathbf{Q}}_ l)_\pi =_{(\varphi \text{ finite })} \oplus _{\pi \in \widehat{G}} H^*(X_{\overline{k}}, \mathcal{F}_\pi )
If \pi \neq 1 then we have
H^0(X_{\overline{k}}, \mathcal{F}_\pi ) = H^2(X_{\overline{k}}, \mathcal{F}_\pi ) = 0,\quad \dim H^1(X_{\overline{k}}, \mathcal{F}_\pi ) = (2g_ X - 2)d_\pi ^2
(can get this from trace formula for acting on ...) and we see that
|\sum _{x \in X(k_ n)} \chi _\pi (\mathcal{F}_ x)| \leq (2g_ X - 2) d_\pi ^2\sqrt{q^ n}
Write 1_ C = \sum _\pi a_\pi \chi _\pi , then a_\pi = \langle 1_ C, \chi _\pi \rangle , and a_1 = \langle 1_ C, \chi _1\rangle = \frac{\# C}{\# G} where
\langle f, h\rangle = \frac{1}{\# G}\sum _{g \in G} f(g)\overline{h(g)}
Thus we have the relation
\frac{\# C}{\# G} = ||1_ C||^2 = \sum |a_\pi |^2
Final step:
\begin{align*} \# \left\{ x \in X(k_ n) \mid F_ x \in C\right\} & = \sum _{x \in X(k_ n)} 1_ C(x) \\ & = \sum _{x \in X(k_ n)} \sum _\pi a_\pi \chi _\pi (F_ x) \\ & = \underbrace{\frac{\# C}{\# G} \# X(k_ n)}_{ \text{term for }\pi = 1} + \underbrace{\sum _{\pi \neq 1}a_\pi \sum _{x\in X(k_ n)}\chi _\pi (F_ x)}_{ \text{ error term (to be bounded by }E)} \end{align*}
We can bound the error term by
\begin{align*} |E| & \leq \sum _{\pi \in \widehat{G}, \atop \pi \neq 1} |a_\pi | (2g - 2) d_\pi ^2 \sqrt{q^ n} \\ & \leq \sum _{\pi \neq 1} \frac{\# C}{\# G} (2g_ X - 2) d_\pi ^3 \sqrt{q^ n} \end{align*}
By Weil's conjecture, \# X(k_ n)\sim q^ n.
\square
Comments (3)
Comment #867 by Emmanuel Kowalski on
Comment #868 by Pieter Belmans on
Comment #870 by Johan on