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Unformatted text preview: obal Torelli Theorem for K3 surfaces due to I. Shafarevich and I. Pjatetsky-Shapiro [94]. Theorem 5.11. Let AX be the image of Aut(X ) in O(SX ). Let G be the subgroup + + of O(SX ) generated by WX and AX . Then G = WX AX , and its index in O(SX ) is finite. The difficult part is the finiteness of the index. Corollary 5.1. The following assertions are equivalent: + • WX is of finite index in O(SX ); • Aut(X ) is a finite group; • the (−2)-reflection group of SX admits a fundamental polytope with finitely many faces defined by the classes of smooth rational curves; 32 IGOR V. DOLGACHEV • SX (−1) is a 2-reflective lattice. The third property implies that the set of (−2)-curves is finite if Aut(X ) is finite, but the converse is not true. Figures 10, 11, and 12 show the even 2-reflective lattices of rank 17, 18 and 19, the Coxeter diagrams of the reflection groups Ref−2 (SX ), and the corresponding K3 surfaces. • •••••••••• • • • SX = U ⊥ E8 (−1) ⊥ E7 (−1) • • • • • gaed `fbc Figure 10. ••••••••••••••••• • • SX = U ⊥ E8 (−1)2 VV VV VV VV VV VV Ö ÖÖ VV ÖÖ VV ÖÖÖ ÖÖ VVÖÖ VÖÖ ÖÖVV V ÖÖVV ÖÖ VV V ÖÖ V VV ÖÖ VV ÖÖ Ö V ÖÖ ÖÖ ÖÖ ÖÖ ÖÖ Ö Figure 11. • • ••••••••••••• • • •  dd dd • • •  dddddddddddd •d 2 SX = U ⊥ E 8 ⊥ A 1 ` `` `` `` `` ÖÖ ÖÖ `` `` ÖÖÖ ÖÖ `` Ö `` ÖÖ Ö ` ÖÖ Ö``` ÖÖ ÖÖ ``` ÖÖ `` ÖÖ `` ÖÖ ` ÖÖ ÖÖ Ö Ö Ö ÖÖ ÖÖ Ö Figure 12. The K3 surface is birationally isomorphic to the surface obtained as the double cover of P2 branched along the curve of degree 2 drawn thick, followed by the double REFLECTION GROUPS IN ALGEBRAIC GEOMETRY 33 cover along the proper inverse transform of the remaining curve of degree 4 (the cuspidal cubic and its cuspidal tangent line). It follows from the classification of 2-reflective lattices (see section 4) that each of them is isomorphic to the lattice SX for some K3-surface X . A similar assertion for any even reflective lattice is not true for the following trivial reason. Replacing M by M (2k) for some k, we obtain a reflective lattice with discriminant group whose minimal number of generators s ≥ rankM . Suppose rankM ≥ 12 and M (k) is primitively embedded in LK 3 . Its orthogonal complement is a lattice of rank 22 − 12 ≤ 10 with isomorphic discriminant group generated by ≤ 10 elements. This is a contradiction. A more serious reason is the existence of an even reflective hyperbolic lattice M of rank 22. One can take M = U ⊥ D20 realized as the sublattice of I19,1 of vectors with even v 2 . The reflectivity of M was proven by R. Borcherds [12]. Since rank SX ≤ 20 for any complex K3 surface X , the lattice M cannot be isomorphic to SX . However, the lattice M is realized as the Picard lattice of a K3 surface over an algebraically closed field of characteristic 2 isomorphic to a...
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