MODULAR FORMS-page11

# MODULAR FORMS-page11 - 7 Assume γ = 0 Then the Moebius...

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Unformatted text preview: 7 Assume γ = 0. Then the Moebius transformation deﬁned by M −1 is the translation z → z + β and hence takes z out of the domain − 1 ≤ Re z ≤ 1 δ 2 2 unless β = 0 or β = ±1 and Re z = ± 1 . In the ﬁrst case M = ±I and f = f . In 2 1 ±1 the second case M = ± , f = ax2 ± axy + cy 2 and f = ax2 axy + cy 2 . 01 Assume γ = ±1. If γ = 1, then |z + δ | ≤ 1 implies (i) δ = 0, |z | = 1, or (ii) z = ρ := √ −1+ −3 2 and δ = 1. In case (i) we have M = ± −1 0 α 1 1 and M · z = α − z . This easily implies 01 and −1 0 ¯ M · f = cx2 − 2bxy + ay 2 . Since (c, b, a) ∈ Ω, we get a = c. Again f is of the form ax2 + 2bxy + ay 2 and is properly equivalent to ax2 − 2bxy + ay 2 . In the second case f = a(x2 + xy + y 2 ) and M f = a(x2 − xy + y 2 ). α α−1 Now, in case (ii), we get M = and M ·ρ = (αρ+(α−1))/(ρ+1) = 1 1 α + ρ. This implies α = 0, f = a(x2 + xy + y 2 ), M f = f . Finally, the case γ = −1 is reduced to the case γ = 1 by replacing M with −M . This analysis proves the following: α = 0 or (α, z ) = (−1, ρ), (1, −ρ2 ). So, in the ﬁrst case, M = Theorem 1.4. Let f = ax2 + 2bxy + cy 2 and f = a x2 + 2b xy + c y 2 be two properly reduced positive deﬁnite binary forms. Then f is properly equivalent to f if and only if f = f or f = ax2 ± axy + cy 2 , f = ax2 axy + cy 2 , or f = ax2 + 2bxy + ay 2 , f = ax2 − 2bxy + ay 2 . Moreover, M f = f for some M = ±I if and only if one of the following cases occurs: (i) f = a(x2 + y 2 ) and M = ± 0 −1 ; 10 (ii) f = a(x2 ± xy + y 2 ) and M = ± 1 −1 0 −1 ,± . 1 0 11 Deﬁnition. Let G be a group acting on a set X . A subset S of X is called a fundamental domain for the action of G on X if each orbit of G intersects S at exactly one element. ¯ The proof of Theorem 1.4 shows this enlarged set Ω contains a representative ¯ of each orbit of SL(2, Z). Moreover, two points (a, b, c) and (a , b , c ) in Ω belong to the same orbit of SL(2, Z) if and only if either a = c = a = c , b = −b or a = a, b = −b = a/2. Clearly ¯ Ω = R+ × D . ...
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