# Can we prove $Aut(S_g) , g \geq 2$ is finite in the following way ?

I was trying to prove that $Aut( S_g$), g$\geq 2$ [ orientation preserving isometries ] is finite in the following way :

For fixed $M$ ( positive ) there are finitely many , say $k$ number of simple closed geodesics ( with repeated multiplicities ), say $c_1...c_k$ with length $\leq M$. Consider the group $Aut(S_g$) acting on this finite set $S$ of geodesics ( it acts, since an isometry preserve the length of geodesics ). Now the number of permutations of $S$ is $k!$ ( k-factorial ). So, if we can prove that :

Lemma : An automorphism of surface of genus $\geq 2$ is fully and uniquely determined by its action on finitely many ( say $k$ ,depending on genus g ) simple closed geodesics ( forming the set $S$ ),

then we can have only $k!$ permutations of $S$ and hence we would have only at most $k!$ distinct automorphism of $S_g$, if the Lemma is true . Is the lemma true and easily provable ? Or is there any other way to prove the main question ( $Aut(S_g)$ is finite ) ?

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The answer is yes. It's a little easier if we consider oriented geodesics (which is fine for your argument). Let $\alpha$ and $\beta$ be two oriented simple closed curves on $S_g$ that intersect once. Let $a$ and $b$ be geodesics that are homotopic to $\alpha$ and $\beta$, respectively. Then $a$ and $b$ only intersect once, say at a point $x \in S$. Now let two isometries $f_1,f_2 : S \rightarrow S$ be given. Assume that $f_1(a) = f_2(a)$ and $f_1(b) = f_2(b)$ (as oriented geodesics). I claim that $f_1=f_2$. Indeed, it is clear that $f_1(x)=f_2(x)$. Moreover, if $v$ is the unit tangent vector at $x$ pointing in the direction of $a$ (remember, $a$ has an orientation), then we also know that $f_1(v)=f_2(v)$. Since an element of $SO(2)$ is determined by its value at a single non-zero vector, this implies that both the values and the derivatives of $f_1$ and $f_2$ agree at $x$, and a standard result of Riemannian geometry (using the exponential map) says that then $f_1 = f_2$.
This implies that you can take $S$ to be the set of all geodesics whose lengths are at most the lengths of $a$ or $b$.
@ Dr Puttman : Thak you for your very clear answer . I would just take $M$ big enough such that there is at least a pair of once-intersecting geodesics. – Analysis Now Jan 30 '11 at 18:47