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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?

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    $\begingroup$ Yes, the Global Torelli theorem is the key ingredient. There's a sketch proof in Section 5 of this paper by Totaro: dpmms.cam.ac.uk/~bt219/algebraic.pdf. But Sterk's original paper is the only complete reference I know. $\endgroup$ – user5117 Oct 31 '13 at 13:06
  • $\begingroup$ Artie, thank you! This is indeed a good summary (meaning I can read it :) ) $\endgroup$ – aglearner Oct 31 '13 at 13:17
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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 (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).

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  • $\begingroup$ Misha, spasibo! :) $\endgroup$ – aglearner Oct 31 '13 at 15:58
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The standard reference is : H. Sterk, Finiteness results for algebraic K3 surfaces. Math. Z. 189 (1985), no. 4, 507--513. Of course it uses the global Torelli theorem for K3's, plus some reduction theory for arithmetic groups. The paper is quite short and shouldn't be too hard to read.

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