by simply plugging the relevant functions into this differential equation. Near z = 0 ℘0(z) =− 2 z3 + 6G4z + 20G6z 3 +· · · , which is an elliptic function of order 3. If we square it, we get an elliptic function of order 6 since [℘0(z)]2 = 4 z6 − 24G4 z2 − 80G6 +· · · , where +· · · indicates a power series in z which vanishes at z = 0. If we cube ℘and multiple it by 4 we get 4℘3(z) = 4 z6 + 36G4 z2 + 60G6 +· · · . Therefore [℘0(z)]2 −4℘3(z) =− 60G4 z2 − 140G6 +· · · . Hence [℘0(z)]2 −4℘3(z) + 60G4℘(z) =−140G6 +· · · . The left hand side of the equation does not have a pole at z = 0 and hence has no poles in the period parallelogram. Therefore, it must be constant. That constant must be −140G6, and hence [℘0(z)]2 = 4℘3(z) −60G4℘(z) −140G6. We define invariants g2 = 60G4 and g3 = 140G6, which simplifies the differential equation for ℘ [1]. One last notation we need to cover before we get to the J-function is the discriminant, defined by ∆ =g3 2 −27g 2 3. This function, along with g2, is what we use to define the J-function. Note that this is not the only way to define ∆. If we look at the Euler function φ(q) = ∞ Y n=1 (1−qn), we see that if we multiply it by q 1 24 we get the function η(q). If we take (η(q))24 we get a function that many people call ∆. This is not quite the same as our ∆, but if we take (η(q))24(2π)12 we get our ∆ function. Although this may seem like a more natural way to define ∆, our more artificial construction of ∆ will help us as we define the J-function. Since ω =mω1 +nω2, we can consider g2, g3, and ∆ as functions of the two periods ω1 and ω2, and hence, g2 =g2(ω1, ω2), g3 =g3(ω1, ω2) and ∆ = ∆(ω1, ω2). By the Eisenstein series we see that g2 has degree −4 (since g2 = 60X ω6=0 (ω)−4) and g3 has degree −6 (since g3 = 140X ω6=0 (ω)−6). Both these functions are homogeneous. A function f is homogeneous Channels • 2022 • Volume 6 • Number 2 Page 39
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