Consider a complex $K3$ surface $X$ and take its group of automorphisms $Aut(X)$. It is a known fact that the action of $Aut(X)$ on the set of rational $-2$ curves of $X$ has only finite number of orbits.


**Questions.** What kind of ideas one has to use to prove this fact? Is there some nice exposition? Does this fact follows somehow from global Torelli theorem for $K3$'s?



Action of automorphisms of a $K3$ surface on its $(-2)$-curves
The group of symplectomorphisms $Aut(X)$ of a K3 is the group $O(\Lambda)$ of 
automorphisms of its period lattice $\Lambda=H^{1,1}(M,{\Bbb Z})$. For each 
(-2)-cohomology class $\eta\in H^{1,1}(M,{\Bbb Z})$, either $\eta$ or $-\eta$ is
represented by a curve (this follows from the Riemann-Roch formula). This curve
is unique, because its self-intersection is negative. Therefore, finiteness of the
number of orbits of $Aut(X)$ on curves is equivalent to the finiteness of the number of orbits of $O(\Lambda)$ on the set of (-2)-classes in $\Lambda$. This result follows from an answer to a question that I asked on Mathoverflow:
[orbits of automorphism group for indefinite lattices](https://mathoverflow.net/questions/145672/orbits-of-automorphism-group-for-indefinite-lattices)
(automorphism group $O(\lambda)$ of any lattice $(\Lambda, q)$acts on the set $R\_C$ of classes 
$\eta$ with $q(\eta, \eta)=C$ with finitely many orbits). 


