MODULAR FORMS-page59

MODULAR FORMS-page59 - 55 6.4 We know that any elliptic...

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55 6.4 We know that any elliptic curve is isomorphic to a Hesse cubic curve. Let us give another cubic equation for an elliptic curve, called a Weierstrass equation. Its coefficients will give us new examples of modular forms. Recall that dimTh( k, Λ τ ) ab = k . Let use <,> to denote the linear span. We have Th(1 , Λ τ ) 1 2 1 2 = < ϑ 1 2 1 2 ( z ; τ ) > = < T > ; Th(2 , Λ τ ) = < T 2 ,X 0 >, Th(3 , Λ τ ) 1 2 1 2 = < T 3 ,TX 0 ,Y 0 >, for some functions X 0 Th(2 , Λ τ ) ,Y 0 Th(3 , Λ τ ) 1 2 1 2 . Now the following seven func- tions T 6 , T 4 X 0 , T 2 X 0 2 , X 0 3 , T 3 Y 0 , TX 0 Y 0 , Y 0 2 all belong to the space Th(6 , Λ τ ) . They must be linearly dependent and we have aT 6 + bT 4 X 0 2 + cT 2 X 0 2 + dX 0 3 + eT 3 Y 0 + fTX 0 Y 0 + gY 0 2 = 0 . (6.12) Assume g 6 = 0 ,d 6 = 0. It is easy to find X = αX + βT 2 , Y = γY 0 + δXT + ωT 3 which reduces this expression to the form Y 2 T - X 3 - AXT 4 - BT 6 = 0 , (6.13) for some scalars A,B . Let ( z ) = X/T 2 , 1 ( z ) = Y/T 3 . Dividing (6.13) by
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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.

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