MODULAR FORMS-page11

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 . ...
View Full Document

Ask a homework question - tutors are online