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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
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
SX deﬁnes an isomorphism from W (EN ) to a subgroup of O(SX ) generated by
reﬂections in vectors αi = φπ (ai ). It is contained in the reﬂection group Ref−2 (SX ).
Theorem 5.2. The image WX of W (EN ) in O(SX ) does not depend on the choice
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
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 ,
−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).
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
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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