First of all we may restrict ourselves to minimal

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Unformatted text preview: f nonbasic rational surfaces are easy to describe, and they are rather dull. REFLECTION GROUPS IN ALGEBRAIC GEOMETRY 27 if N = 3. This embeds the lattice (5.7) EN = A2 ⊥ A1 E2,3,N −3 if N = 3, if N ≥ 4 in the lattice IN,1 with orthogonal complement generated by kN . The restriction ⊥ of φπ to kN defines an isomorphism of lattices 0 φπ : EN (−1) → SX . Let us identify the Coxeter group W (EN ) with the subgroup of O(I1,N ) generated by the reflections in the vectors ai from (5.6). A choice of a geometric basis in 0 SX defines an isomorphism from W (EN ) to a subgroup of O(SX ) generated by 0 reflections in vectors αi = φπ (ai ). It is contained in the reflection group Ref−2 (SX ). 0 Theorem 5.2. The image WX of W (EN ) in O(SX ) does not depend on the choice 0 of a geometric basis. The image of the homomorphism a : Aut(X ) → O(SX ) is contained in WX . Proof. To prove the first assertion it suffices to show that the transition matrix of two geometric bases defines an orthogonal transformation of I1,N which is the product of reflections in vectors ai . Let (e0 , . . . , eN ) and (e0 , . . . , eN ) be two geometric bases and e0 = m0 e0 − m1 e1 − . . . − mN eN . For any curve C on X , we have e0 · [C ] = π ∗ ([ ]) · [C ] = [ ] · [π (C )] ≥ 0. Intersecting e0 with ei we obtain that m0 > 0, mi ≥ 0, i > 0. We have 1 = e2 = m2 − m2 − . . . − m2 , 0 0 1 N −3 = e0 · kX = −3m0 + m1 + . . . + mN . (5.8) Applying the reflections in vectors αi = φπ (ai ), i > 1, we may assume that m1 ≥ m2 ≥ . . . ≥ mN . Now we use the following inequality (Noether’s inequality ): m0 > m1 + m2 + m3 if m0 > 1. (5.9) To see this we multiply the second equality in (5.8) by m3 and then subtract from the first one to get m1 ( m1 − m3 ) + m2 ( m2 − m3 ) − mi (m3 − mi ) = m2 − 1 − 3m3 (m0 − 1). 0 i≥ 4 This gives (m0 − 1)(m1 + m2 + m3 − m0 − 1) = (m1 − m3 )(m0 − 1 − m1 ) + (m2 − m3 )(m0 − 1 − m2 ) + mi ( m3 − mi ) . i≥ 4 The first inequality in (5.8) implies that m2 − m2 > 0; hence m0 − mi ≥ 1. Thus 0 1 the right-hand side is nonnegative, so the left-hand side is too. This proves the claim. Now consider the reflection s = rα1 . Applying it to e0 we get s(e0 ) = e0 + ((e0 − e1 − e2 − e3 ) · e0 )(e0 − e1 − e2 − e3 ) = (2m0 − m1 − m2 − m3 )e0 − (m0 − m2 − m3 )e1 − (m0 − m1 − m3 )e2 − (m0 − m1 − m2 )e3 . Using (5.9), we obtain that the matrix S · A, where S is the matrix of s, has the first column equal to (m0 , −m1 , − . . . , −mN ) with m0 < m0 and mi ≥ 0 for i > 0. 28 IGOR V. DOLGACHEV Since our transformations are isometries of SX , the inequalities (5.8) hold for the vector (m0 , −m1 , − . . . , −mN ). So, we repeat the argument in order to decrease m0 . After finitely many steps we get the transformation with first column vector equal to (1, 0, . . . , 0). Now the matrix being the orthogonal matrix of the quadratic form x2 − x2 − . . ....
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