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184    9. Elliptic and Hypergeometric Functions

                                   1     1     λ − 1     λ
                       λ,  1 − λ,   ,       ,       ,       ,
                                   λ   1 − λ     λ     λ − 1
        and we will choose the one of these six λ with the property |λ|<1and |λ−1|<1. We
        will calculate ω 1 (λ) and ω 2 (λ) for λ in this open set. For this consider the invariant
        differential on E
                                 dx          dx
                             θ =    = √               .
                                 2y   2 x(x − 1)(x − λ)
        The period integrals take the following form:

                                                 dx
                                (      (
                        ω 1 (λ) =  θ =     √              .
                                          2 x(x − 1)(x − λ)
                                 C 1    L 1
                                                 dx
                                (      (
                        ω 2 (λ) =  θ =     √              .
                                          2 x(x − 1)(x − λ)
                                 C 2    L 2
        Now move ω 0 in the figure to 0 so that x(ω 0 ) goes to ±∞ on the real axis. The
        paths L 1 and L 2 are along the real axis −∞ to near 0 for L 1 and from near 1 to
        +∞. The cubic in the complex integrand of θ changes argument of 2π while turning
        around 0 for ω 1 (λ) or 1 for ω 2 (λ) which means that as the path L 1 deforms down to 0
        becoming two integrals between −∞ and 0 both integrals give half the contribution
        to the period. The same assertion applies to L 2 as it deforms down to 1 becoming
        two integrals between 1 and +∞. This analysis leads to the following expression for
        the periods:

                  (  0                              (  ∞
                             dx                                dx
           ω 1 (λ) =   √                and ω 2 (λ) =    √              .
                         x(x − 1)(x − λ)                  x(x − 1)(x − λ)
                    −∞                               1
        (6.1) Theorem. For a complex number λ satisfying |λ|, |λ − 1| < 1 the periods
        have the following form in terms of the hypergeometric function

                             1 1                         1 1
                ω 1 (λ) = iπ F  , , 1; 1 − λ ,  ω 2 (λ) = π F  , , 1; λ .
                             2 2                         2 2
        For the lattice L λ = Zω 1 (λ) + Zω 2 (λ) the complex tori C/L λ and E λ (C) are iso-
        morphic by a map made from an affine combination of the Weierstrass ℘-function
        and its derivative.
        Proof. The second statement will follow from the first and the above diagram and
        discussion. It remains to calculate the above integral expression for the periods. We
                                                          2
        do the integral for ω 2 (λ) by changing variables x = 1/t, t = s ,and s = sin θ
                        ∞       dx                     −dt/t
                      (                     (  0            2
              ω 2 (λ) =   √               =    √
                       1    x(x − 1)(x − λ)  1   (1/t)(1/t − 1)(1/t − λ)
                      (  1                  (  1
                                dt                     ds
                    =    √               = 2    +
                                                              2
                                                      2
                       0   t(1 − t)(1 − λt)  0   (1 − s )(1 − λs )
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