MODULAR FORMS-page9

MODULAR FORMS-page9 - are properly equivalent if they...

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

View Full Document Right Arrow Icon
5 is the upper half-plane . Let us see how the group GL(2 , Z ) acts on the both sets. We have Mf = a (( αx + βy ) - z ( γx + δy ))(( αx + βy ) - ¯ z ( γx + δy )) = a ( x ( α - γz 1 ) - y ( - β + δz ))( x ( α - γ ¯ z ) - y ( - β + δ ¯ z )) = a | α - | 2 ( x - - β + δz α - γz y )( x - - β + δ ¯ z α - γ ¯ z y ) . Let us consider the action of GL(2 , Z ) on C \ R by fractional-linear transforma- tions (also called Moebius transformations ) defined by the formula ± α β γ δ ² · z = αz + β γz + δ . (1.5) Notice that Im M · z = Im αz + β γz + δ = Im ( αz + β )( γ ¯ z + δ ) | γz + δ | 2 = αδ - βγ | γz + δ | 2 Im z. (1.6) This explains why the transformation is well-defined on C \ R . Also notice that M - 1 = det M ± β - β - γ α ² . Thus the root z is transformed to the root z 0 = M - 1 · z and we obtain, for any M GL(2 , Z ), M - 1 · f = a | γz + δ | 2 ( x - M · z )( x - M · ¯ z ) . 1.4 Until now we considered binary forms up to the equivalence defined by an invertible integral substitution of the variables. We say that two binary forms
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: are properly equivalent if they differ by a substitution with determinant equal to 1. In other words, we restrict ourselves with with the subgroup SL(2 , Z ) of GL(2 , Z ). Since GL(2 , Z ) = SL(2 , Z ) ∪ ± 1-1 ² SL(2 , Z ) and ± 1-1 ² ( ax 2 + 2 bxy + cy 2 ) = ax 2-2 bxy + cy 2 we obtain that each f is properly equivalent to a form ax 2 + 2 bxy + cy 2 , where ( a,b,c ) ∈ ¯ Ω and ¯ Ω = { ( a,b,c ) ∈ R 3 : | 2 b | ≤ c ≤ a,a,ac-b 2 > } . Definition. We shall say that f = ax 2 + 2 bxy + cy 2 is properly reduced if ( a,b,c ) ∈ ¯ Ω....
View Full Document

This note was uploaded on 01/08/2012 for the course MATH 300 taught by Professor Ontonkong during the Fall '09 term at SUNY Stony Brook.

Ask a homework question - tutors are online