and poles inside the cell. But f0/f is elliptic and has the same period as f, and the contour integral of an elliptic function around the boundary of any cell is zero, since the integrals along parallel edges cancel because of periodicity. So the difference between the number of zeros and the number of poles is 0, and hence the number of poles and the number of zeros for a given elliptic function are the same [1]. We now turn to the task of constructing a nonconstant elliptic function. The minimum order of such a function is 2, and so we start by constructing an elliptic function of that order. We will prescribe the periods and find an elliptic function of that order. We will start with a function of order 3 and then integrate it to get to a function of order 2. If we consider the series f(z) =X ω∈Ω 1 (z −ω)3 (where Ω ={ω =mω1 +nω2|m, n∈Z} is a lattice), we have an elliptic function with two periods, ω1 and ω2, and a pole of order 3 at each ω. We will use this function to construct an elliptic function of order 2 called the Weierstrass ℘ function. We do this by integrating f(z) term by term. To do this, we see that the integral of (z −ω)−3 is −(z −ω)−2/2, and hence we will have a summand of −(z −ω)−2/2 near each period. We can then multiply by −2 to simplify. Additionally, we can also consider starting our integration at the origin (ω = 0), and hence we will have an initial term of 1 z2. Basically, we can consider this as splitting the sum into two parts, ω = 0 and ω 6= 0, and integrating those two parts separately. Hence, ℘(z) = 1 z2 +Z z 0 X ω6=0 −2 (t −ω)3 dt = 1 z2 +X ω6=0 1 (z −ω)2 − 1 ω2 . Note that ℘0(z) =−2X ω∈Ω 1 (z −ω)3 , which is −2 times our original function. The Weierstrass function is incredibly important in the area of elliptic functions. It turns out that given some lattice, any elliptic function for that lattice is a rational function in ℘and ℘0 relative to that lattice [10]. This means that if you have the ℘function for some lattice, you can get whatever other elliptic function you want in terms of ℘ and ℘0. In the discussion that follows, we are going to refer to the series Gn =X ω6=0 1 ωn , which is called the Eisenstein series of order n. This series is used in the differential equation [℘0(z)]2 = 4℘3(z) −60G4℘(z) −140G6, which the Weierstrass ℘ function satisfies since we can form a linear combination of the powers of ℘and ℘0 which eliminates the pole at z = 0. Therefore, we can get a constant elliptic function, since it will have no poles and an elliptic function without poles is constant. We can verify that ℘ satisfies this differential equation Page 38 Riley • An Overview of Monstrous Moonshine
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