Unformatted text preview: obal Torelli Theorem for
K3 surfaces due to I. Shafarevich and I. PjatetskyShapiro [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 2reﬂective 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 2reﬂective 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 2reﬂective lattices (see section 4) that each
of them is isomorphic to the lattice SX for some K3surface 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...
View
Full
Document
This note was uploaded on 02/24/2012 for the course MATH 285 taught by Professor Igordolgachev during the Fall '04 term at University of MichiganDearborn.
 Fall '04
 IgorDolgachev
 Algebra, Geometry, The Land

Click to edit the document details