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§4. Elliptic Curves over the Real Numbers  287

        following basic modular transformations S(τ) =−1/τ and T (τ) =−1/(τ − 1).
        These two transformations have the following values on the corners.
           (1) For S it preserves the line Re(τ) = 0and S(ζ 4 ) = ζ 4 .Also S(ζ 3 ) = ζ 6 and
        S(ζ 6 ) = ζ 3 interchanging the lines Re(τ) =±1.
           (2) For T we have T (ζ 6 ) = ζ 6 , T (ζ 4 ) = (1/2) 1/2 ζ 8 ,and T (ζ 3 ) = (1/3) 1/2 ζ 12
        carries the circle |τ|= 1tothe line Re(τ) = 1/2 in the upper half plane.
        (4.6) The Modular Forms E 4 and E 6 . The transformation law

                     aτ + b           2k              ab

                 E 2k        = (cτ + d) E 2k (τ)  for      ∈ SL(2Z)
                     cτ + d                           cd
        specializes for the two transformations S(τ) and T (τ) to the following: E 2k (S(τ)) =
                                        2k
        τ  2k E 2k (τ) and E 2k (T (τ)) = (τ − 1) E 2k (τ).Wehavetwo casesfor thelines
        Re(τ) = 0 and 1/2.
           (1) For Re(τ) = 0: E 6 (ζ 4 ) = 0and J(ζ 4 ) = 1. Moreover, τ = it, t > 0, and q ≥
                                                        n
        0 in this case, and the expansion E 4 = 1+240  σ 3 (n)q shows that E 4 (it) ≥ 0
                                               n≥1
        on this line. In particular J(q) ≥ 1 on this line. The relation S(it) = i/t shows that
        the two parts of the positive imaginary axis i[1, ∞) in the fundamental domain and
        i(0, 1] are interchanged. The coefficients of the curve  (it) are transformed by the
        modular relations to give

                         i     4                i       6
                     E 4    = t E 4 (it)  and  E 6  =−t E 6 (it).
                         t                      t
           (2) For Re(τ) = 1/2: We have three distinguished values:

                τ = T (ζ 6 ) = ζ 6  T (ζ 4 ) = (1/2) 1/2 ζ 8  T (ζ 3 ) = (1/3) 1/2 ζ 12
            E 4 (τ) =   0             E 4 (i)> 0              0
            E 6 (τ) =  E 6 (ζ 6 )        0                −27 · E 6 (ζ 6 )
             J(t) =     0                1                    0

                                         6
        Here we use that E 6 (T (ζ 3 )) = (ζ 3 − 1) E 6 (ζ 6 ) =−27 · E 6 (ζ 6 ) for the last entry
        in the table. For τ = 1/2 + it, t ≥ 0, we have −1 < q ≤ 0and J(q) take values
        (−∞, 1], that is, the values not coming from Re(τ) = 0.

        (4.7) Two Real Forms. The two equations

                  2    3                    4    6     2    3     4     6
         E"α, β#: y = x − 3αx + 2β  and  E"t α, −t β#: y = x − 3αt x − 2t β
        over the real numbers are nonisomorphic, but over the complex numbers become
                                                          2
                                                              3
        isomorphic under the substitution which carries (x, y) to (−t x, it y) followed by
                    6
        dividing by −t . We can see how this can be carried out with the modular property
        of the coefficients.
        (4.8) Notation. Let C(ev) denote iR + compactified with a point, denoted 0 = i∞,
        and C(od) denote (1/2)+iR + compactified with a point, denoted 1/2 = 1/2+i∞.
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