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Let $X$ be a compact Riemann surface of genus $g>1$, $f\in Aut(X)$, a biholomorphism of $X$ onto itself, $x\in X$ a fixed point of $f$. Since tangent map of a holomorphic map (on the real tangent space) is of the form

\begin{bmatrix} r\cos(\theta) & -r\sin(\theta) \\ r\sin(\theta) & r\cos(\theta) \end{bmatrix}

The local Lefschetz number is always $1$, unless the tangent map is identity, which will impose $f$ to be the identity map. This fact can be easily seen when $f$ is lifted to the universal covering of $X$.

Can we carry on this idea to say something more about the group of automorphism of a compact Riemann surface of genus $g>1$? Recall that on such a Riemann surface, we have exactly $g(g-1)(g+1)$ Weierstrass points, counted with weight, and a biholomorphism must take Weierstrass points to Weierstrass points.

Thank you!

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I am sorry. I am a newcomer and I'm not familiar with typing math symbols yet. The row vector should have been a 2 by 2 matrix, I don't know what goes wrong. –  Andrew Dec 7 '11 at 16:03
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The software interprets backslashes in a funny way (it processes Markdown before LaTeX). You need to use three instead of two between lines. –  Qiaochu Yuan Dec 7 '11 at 16:22
    
I'm a little confused by your question. Automorphisms of Riemann surfaces need not have fixed points, and they can be any finite group... so what kind of flavour of result are you hoping for? -- why would you hope for an interesting connection here? –  Ryan Budney Dec 10 '11 at 10:46
    
I am just beginning to learn Riemann surfaces. I was trying to prove that the automorphisms of a compact Riemann surface is finite. Since there are only finitely many Weierstrass points on a compact Riemann surface and automorphisms take Weierstrass points to Weierstrass points, if the group is large, it must fix some Weierstrass points. I hope this can tell us something. It is just a very vague idea. –  Andrew Dec 11 '11 at 10:20

2 Answers 2

I will use this fact to prove that the group of automorphisms of a compact Riemann surface $C$ of genus $g>1$ is always faithfully represented in its action on the first homology $H_1(C)$.

Theorem: Every nontrivial automorphism $\varphi$ acts nontrivially on $H_1(C)$.

Proof: Assume otherwise, so that $\varphi$ is a nontrivial automorphism acting trivially on $H_1(C)$. Since $\varphi$ is holomorphic, it is orientation-preserving and thus acts trivially on $H_2(C)$ as well, and of course every self-map of $C$ acts trivially on $H_0(C)$. Thus the Lefschetz number of $\varphi$ is

$$L(\varphi)=\text{tr}(\varphi|H_0(C))-\text{tr}(\varphi|H_1(C))+\text{tr}(\varphi|H_2(C))=1-2g+1=2-2g.$$

The local description you give around a fixed point implies that every fixed point of $\varphi$ is isolated and has index 1. Thus by the Lefschetz fixed point formula, the Lefschetz number is equal to

$$L(\varphi)=\sum_{\text{fixed points}}1=\text{the }\#\text{ of fixed points of }\varphi.$$

But if $g>1$ the Lefschetz number $L(\varphi)=2-2g$ is negative, which is a contradiction.


I will now give a slightly different and perhaps more advanced application of this fact. I believe this argument goes back to Chevalley-Weil. Let $D\to C$ be an unramified Galois cover (aka a normal finite-sheeted covering space), and let $G=\text{Gal}(D/C)$ be the deck group, or Galois group. The finite group $G$ acts freely on the Riemann surface $D$ (with quotient $C$). We thus get an action of the group $G$ on the first homology $H_1(D;\mathbb{Q})$, and my goal is to describe this homology group as a representation of $G$. As before, let $g$ be the genus of $C$.

Theorem (Chevalley-Weil): There is an isomorphism $H_1(D;\mathbb{Q})\approx \mathbb{Q}^{\oplus 2}\oplus \mathbb{Q}G^{\oplus 2g-2}$ as representations of $G$.

Proof: A representation $V$ of a finite group over a field of characteristic 0 is determined by its character, the function $\chi_V(g)=\text{tr}(g|V)$. Thus it suffices to show that the character $\chi_{H_1(D)}$ coincides with the character $2\chi_{\mathbb{Q}}+(2g-2)\chi_{\mathbb{Q}G}$. Note that in the regular representation $\mathbb{Q}G$ every nontrivial $g\in G$ permutes the basis elements freely, so we have $\chi_{\mathbb{Q}G}(g)=0$ for $g\neq 1$ (and $\chi_{\mathbb{Q}G}(1)=|G|$). The character $\chi_{\mathbb{Q}}$ of the trivial representation is just the constant function 1.

First, the character $\chi_V(1)$ of the identity is just the dimension $\dim V$. If $h$ is the genus of $D$, multiplicativity of Euler characteristic gives $2-2h=|G|(2-2g)$, or $h=|G|(g-1)+1$. In particular, we have $\chi_{H_1(D)}(1)=2h=|G|(2g-2)+2$.

Now take $g\neq 1\in G$, and consider $g$ as an automorphism of $D$. Since $g$ acts trivially on $H_0(D)$ and $H_2(D)$, we have $$L(g)=1-\text{tr}(g|H_1(D))+1=2-\chi_{H_1(D)}(g).$$ But $G$ acts freely on $D$, so $g$ has no fixed points and the Lefschetz fixed point theorem gives $L(g)=0$. We conclude that $\chi_{H_1(D)}(g)=2$ for every nontrivial element $g\in G$, as desired. This shows that $\chi_{H_1(D)}=2\chi_{\mathbb{Q}}+(2g-2)\chi_{\mathbb{Q}G}$, which proves the theorem.

It is not hard to extend this argument to apply to ramified covers, since you can still describe the Lefschetz number of an automorphism by counting its ramification points (and this is what Chevalley-Weil actually did).

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Thanks so much for your detailed answer. –  Andrew Dec 11 '11 at 10:58

One consequence of Tom Church's answer, incidentally, is Hurwitz's theorem that the group of automorphisms of a Riemann surface of genus $g \geq 2$ is at most $84(g-1)$; see lecture 9 of these notes.

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