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By K. O. Friedrichs

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TRANSFORMATIONS OF THETA FUNCTIONS Taking derivative in z at z = 0 we obtain ϑ(0; τ + 1)′1 1 = −Cϑ 1 1 (0; τ )′ . 14) that ∞ 1 ϑ 1 1 (0; τ )′ = −2πq 8 2 2 (1 − q m )3 . m=1 Since the substitution τ → τ + 1 changes q a into e2πa(τ +1) = q a e2πa we obtain C = eπi/4 . 8) 2 2 ϑ 1 0 (z; τ + 1) = −e πi/4 2 ϑ 1 0 (z; τ ). 9) 2 Now take M = ( 10 −1 0 ) . We have e−iπz 2 /τ ϑ00 (z/τ ; −1/τ ) = Bϑ00 (z; τ ) for some B depending only on τ . 10) where the square root takes positive values on τ ∈ iR. In particular, √ ϑ00 (0; −1/τ ) = −iτ ϑ00 (0; τ ).

So we have proved that any elliptic curve is isomorphic to a complex submanifold of the complex projective plane given by the Hesse equation. 2. Consider the affine part of the Hesse cubic where x0 = 0. It is isomorphic to the curve C ′ in C2 given by the equation 1 + x3 + y 3 + γxy = 0. 28) It follows that the functions Φ1 (z) = Θ1 (z; τ )3 , Θ0 (z; τ )3 Φ2 (z) = Θ2 (z; τ )3 Θ0 (z; τ )3 define a surjective holomorphic map C2 \ Z → C ′ whose fibres are equal to the cosets z + Z+ τ Z. Here Z is the set of zeroes of Θ 0 (z; τ )3 .

4 Show that the image of 2-torsion points on the Hesse cubic are the four points (0, 1, −1), (1, a, a), where a is a root of the cubic equation 2t3 + λt2 + 1 = 0. 5 Find the values of the parameter γ in the Hesse equation corresponding to the harmonic and anharmonic elliptic curve. 27) is equal to the following function in τ : λ=− ϑ(0; 3τ )3 + q 1/2 ϑ(t; 3τ )3 + q 2 ϑ(2τ ; 3τ ) . 2 in the case k = 2. Show that φ2 defines a holomorphic map E τ → P 1(C ) such that for all points x ∈ P1(C ) except four, the pre-image consists of 2 points and over the four points the pre-image consists of one point.

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