Killing the automorphism group of a Riemann surface It is well known, that the automorphism group of a Riemann sphere with at least three marked points on it is trivial, whereby an automorphism means an automorphism of the curve preserving the marked points. The same is true for smooth genus 1 curves with 5 marked points.
Is it true that for any $g$ there exists $N$ such that any smooth genus $g$ complex curve with $n>N$ marked points does not admit non-trivial automorphisms?
 A: I'll show that if $f$ is an automorphism of a curve of genus $g>1$ with more than $2+2g$ fixed points, $f$ must be the identity. 
Let $X$ be a smooth complete curve of genus $g>1$, and let $f$ be a nontrivial automorphism of $X$. For $g>1$, a smooth complete curve of genus $g$ has at most $84(g-1)$ automorphisms, by Hurwitz. I won't need the actual number $84(g-1)$ but I'll use the fact that this is a torsion group, so $f$ is torsion. Recall that if $f$ has a fixed point $P$, then there is a number $i(f, P)$, the index of $f$ at $P$, which is the multiplicity of the zero at  of $g(f(Q)) - g(Q)$ at $Q=P$, where $g(Q)$ is a uniformizing parameter at $P$. (Remark that in this holomorphic setting, the index is always a positive integer, being the order of a zero.)
By the Lefschetz fixed point theorem,
$$\sum_{P \in \mathrm{Fix}(f)} i(f, P) = \mathrm{Tr}(f|_{H_0(X, \mathbb Q)}) - \mathrm{Tr}(f|_{H_1(X, \mathbb Q)}) + \mathrm{Tr}(f|_{H_2(X, \mathbb Q)})$$
where the sum is taken over the fixed points $P$ of $f$, and $i(f, P)$ is the index of $f$ at $P$. Hence, since $i(f, P) \geq 1$ for every fixed point $P$, we have
$$\#\mathrm{Fix}(f) \leq \sum_{P \in \mathrm{Fix}(f)} i(f, P) \leq |\mathrm{Tr}(f|_{H_0(X, \mathbb Q)})| + |\mathrm{Tr}(f|_{H_1(X, \mathbb Q)})| + |\mathrm{Tr}(f|_{H_2(X, \mathbb Q)})|$$
Remark that for each $i$, $\sigma = f|_{H_i(X, \mathbb Q)} \in GL(H_i(X, \mathbb Q))$ is a torsion element, since $f$ is itself torsion. So all eigenvalues of $\sigma$ are roots of unity, which implies $|\mathrm{Tr}(\sigma)| \leq \dim H_i(X, \mathbb Q)$. Therefore
$$\#\mathrm{Fix}(f) \leq \dim H_0(X, \mathbb Q) + \dim H_1(X, \mathbb Q) + \dim H_2(X, \mathbb Q) = 2+2g.$$
Hence, if $f$ is an automorphism of a curve of genus $g>1$ with more than $2+2g$ fixed points, $f$ must be the identity. 
