Math 213a (Fall 2021) Yum-Tong Siu 1

ELLIPTIC FUNCTIONS (Approach of Weierstrass)

Three Basic Properties of General Elliptic Functions. Before we discuss the approach of Weierstrass to elliptic functions, we first look at some ba- sic properties of doubly periodic meromorphic functions. These functions are called elliptic functions. The Jacobian elliptic functions we have seen and the Weierstrass elliptic functions we are introducing are special cases of these general elliptic functions. The three basic properties of of general el- liptic functions are as follows. We denote two primitive periods of an elliptic function f by ω1 and ω2.

(i) The sum of the residues of the function inside a fundamental parallel- ogram is zero.

(ii) The number of zeroes of the function equals the number of poles inside a fundamental parallelogram.

(iii) Inside a fundamental parallelogram the sum of the coordinates of the zeroes equals the sum of the coordinates of the poles modulo a period. Math 213a (Fall 2021) Yum-Tong Siu 2

Figure 1: Fundamental Parallelogram

To prove (i) we integrate f(z) dz along the boundary of the fundamental parallelogram. By the residue theorem the integral is simply 2πi times the sum of the residues of f inside the parallelogram. On the other hand the integral is zero, because, by the periodicity of f, the integral over [a, a + ω1] equals the integral over [a + ω2, a + ω1 + ω2] and the integral over [a, a + ω2] equals the integral over [a + ω1, a + ω1 + ω2]. Property (ii) follows from integrating 1 f 0(z) dz 2πi f(z) over the boundary of the fundamental parallelogram and from the argument principle. Again the integral along the boundary of the fundamental paral- lelogram because of cancellation on account of the periodicity of f. The proof of Property (iii) is slightly more complicated. One integrates 1 zf 0(z) dz 2πi f(z) Math 213a (Fall 2021) Yum-Tong Siu 3 over the boundary of the fundamental parallelogram, but in this case the integral may not be zero, because 1 Z zf 0(z) 1 Z zf 0(z) dz − dz 2πi [a+ω1,a+ω1+ω2] f(z) 2πi [a,a+ω2] f(z) 1 Z f 0(z) = ω1 dz. 2πi [a,a+ω2] f(z) However, 1 Z f 0(z) dz 2πi [a,a+ω2] f(z) 1 equals 2πi times the difference of the value of log f(z) at a+ω2 and at a when z runs along [a, a + ω2]. Since f(z) has the same value at a as at a + ω2, the difference of the value of log f(z) at a + ω2 and at a when z runs along [a, a + ω2] must be 2πi times an integer. Therefore 1 Z f 0(z) dz 2πi [a,a+ω2] f(z) is an integer and 1 Z zf 0(z) 1 Z zf 0(z) dz − dz 2πi [a+ω1,a+ω1+ω2] f(z) 2πi [a,a+ω2] f(z) is a period of f. Likewise 1 Z zf 0(z) 1 Z zf 0(z) dz − dz 2πi [a,a+ω1] f(z) 2πi [a+ω2,a+ω1+ω2] f(z) is also a period of f. This concludes the proof of (iii).

Weierstrass Elliptic Functions to be Introduced as Infinite Sum of Par- tial Fractions. Though a Weierstrass elliptic function is to be introduced as an infinite sum of partial fractions which is doubly periodic as a meromorphic function on C, we will first discuss how the partial fractions in the infinite sum is motivated from the approach of Abel and Jacobi of introducing an elliptic function as the inverse function of an elliptic integral whose integrand is the reciprocal of the square root of a quartic polynomial. Our discussion of the motivation starts with the replacement of the quartic polynomial by a cubic polynomial. Math 213a (Fall 2021) Yum-Tong Siu 4

Reduction of Polynomial in Elliptic Integral from Quartic to Cubic. The approach of Abel and Jacobi inverts the indefinite integral Z dz , pF (z) where F (z) is a quartic polynomial, especially of the form (1 − z2)(1 − k2z2). Actually the case of deg F (x) = 3 and the case of deg F (x) = 4 are the same. Suppose F (x) is of degree 4, which without loss of generality can be written Q4 as ν=1(x − γν) with all four distinct points γν of C. Apply the M¨obius transformation aζ + b z = cζ + d with ad − bc 6= 0 to z. Then ad − bc dz = (cζ + d)2 and Z dζ Z dζ = (ad − bc) dζ pF (ζ) pQ(ζ) with 4 Y Q(t) = ((aζ + b) − γν(cζ + d)). ν=1

We can choose a, b, c, d so that a−γ1c = 0. Then b−γ1d 6= 0 from ad−bc 6= 0. Then the degree of Q(ζ) is 3 in ζ. Geometrically this means that the M¨obius transformation aζ + b z = cζ + d maps the point ζ = ∞ to the point z = γ1 and with respect to the ζ coordinate our integral comes from a polynomial of degree 3. An affine 3 variable change can reduce the polynomial to 4ζ − g2ζ − g3.

Elliptic Integral in Weierstrass Form and Principal Part of Weierstrass ℘ Function at 0. An indefinite elliptic integral of Weierstrass is of the form Z w dζ , p 3 ζ=∞ 4ζ − g2ζ − g3 Math 213a (Fall 2021) Yum-Tong Siu 5

where g2, g3 are complex numbers and the three roots e1, e2, e3 of the cubic 3 polynomial 4ζ − g2ζ − g3 are distinct, which means that its discriminant

Y 2 3 2 (ej − ek) = −16(g2 − 27g3) 1≤jRiemann surface.

The choice of ∞ as the initial point of integration means that the inverse function of the indefinite elliptic integral

(i) has a double pole at the origin and

(ii) is an even function.

We would like to see what the principal part at the origin should be chosen for the elliptic function so that its translates by the periods can be used in the infinite sum of partial fractions to define the Weierstrass elliptic function.

By the fundamental theorem of calculus, the elliptic function ℘(z) defined as the inverse function of Z w dζ z = ℘−1(w) = p 3 ζ=∞ 4ζ − g2ζ − g3 satisfies dz 1 = p 3 dw 4w − g2w − g3 which is the same as

d℘(z)2 = 4℘(z)3 − g ℘(z) − g . dz 2 3

If the principal part of ℘(z) at 0 is

−1 X c−ν , zν ν=−` Math 213a (Fall 2021) Yum-Tong Siu 6 equating the terms of most negative power on both sides yields `2c2 c3 −` = 4 −` , z2(`+1) z3` 2 2 3 which implies 2(`+1) = 3` and ` c−` = 4c−`, i.e., ` = 2, and c−2 = 1. By the evenness of ℘(z) (from the initial point of integration being a branch-point 1 of the Riemann surface), the principal part of ℘(z) at 0 must be z2 .

Rigorous Definition of Weierstrass ℘ Function. The above discussion involving the Riemann surface for the square root of a cubic polynomial with leading coefficient 4 and without second-highest degree term is just for the sake of giving the background motivation for choosing the principal part at 1 the origin to be z2 . The discussion helps to understand why the infinite sum of partial fractions to define the Weierstrass ℘ function is chosen to be of that particular form, but it is completely unnecessary for the rigorous definition of ℘(z) which we now give. We can actually start our treatment of the Weierstrass elliptic function ℘(z) from this point and forget about all the preceding material concerning elliptic functions. Let ω1, ω2 be two complex numbers which are R-linearly independent. Denote by L the lattice Zω1 + Zω2 generated by ω1, ω2. The Weierstrass ℘(z) for the lattice L is defined as the infinite series 1 X  1 1  ℘(z) = + − . z2 (z − `)2 `2 n∈L−{0} The reason for the infinite sum over ` ∈ L is the need to end up with a function for which every element of L is a period. The term 1 1 − (z − `)2 `2 is used instead of just 1 , (z − `)2 1 because without subtracting `2 the infinite sum fails to converge for the following reason. One is not able to conclude that for z in a compact subset of C the Cauchy partial sum X 1 (z − `)2 `∈L, p≤|`|≤q Math 213a (Fall 2021) Yum-Tong Siu 7 to go to 0 as p, q → ∞, because it is comparable to

q   q q X X 1 X 1 X 1 ∼ k = .  |`|2  k2 k k=p k−1≤|`|≤k+1 k=p k=p On the other hand, X  1 1  − (z − `)2 `2 `∈L, p≤|`|≤q is comparable to

q   q q X X 1 X 1 X 1 ∼ k = ,  |`|3  k3 k2 k=p k−1≤|`|≤k+1 k=p k=p which goes to 0 as p, q → ∞.

Note that to determine the convergence of an infinite series of partial fractions uniformly on a compact subset K of C, one removes first the finite number of terms with poles in K so that the convergence question is only for an infinite series of holomorphic functions instead of meromorphic functions. An unexpected difficulty in the verification of the periodicity property of ℘(z) arises with the use of the term 1 1 − (z − `)2 `2 for ` ∈ L − {0}, because the term for the case of ` = 0 is different due to the difficulty of dividing 1 by 0. In fact, ℘(z) so defined still has period ` for each ` ∈ L. One can verify this in one of the following ways. The first 1 way differentiates the infinite series of ℘(z) term-by-term to get rid of − `2 in each term for ` ∈ L − {0} and then integrate back. The vanishing of the constant of integration is guaranteed by the evenness of ℘(z).

More precisely X 1 ℘0(z) = −2 (z − `)3 `∈L which is justified from the convergence of the right-hand side absolutely and uniformly on compact subsets of C. Now clearly ℘0(z + `) = ℘0(z) Math 213a (Fall 2021) Yum-Tong Siu 8 for ` ∈ L. Integrating with respect to z yields

℘(z + `) = ℘(z) + C`

` for some constant C`. Evaluating at z = − 2 yields  `   `  ℘ = ℘ − + C , 2 2 ` which from the evenness of ℘(z) implies that C` = 0.

The second way of verifying the periodicity of ℘(z) is simply to rewrite the infinite sum as X 1 X 1 ℘(z) = 2 − 2 2 (z − n1ω1) n1ω1 n1∈Z n1∈Z−{0} X X  1 1  + 2 − 2 (z − n1ω1 − n2ω2) (n1ω1 − n2ω2) n2∈Z−{0} n1∈Z which clearly satisfies ℘(z+ω1) = ℘(z). For the same reason ℘(z+ω2) = ℘(z) holds. The key point of this second way of verification is that the two series of single summation

X 1 X 1 2 and 2 2 (z − n1ω1) n1ω1 n1∈Z n1∈Z−{0} converge.

Derivation of Differential Equation for ℘(z) by Eliminating Principal Part of Only Pole in Fundamental Parallelogram for Elliptic Function. We now derive the first-order differential equation for ℘(z) by eliminating the principal part of the only pole in a fundamental parallelogram for an elliptic function constructed as a polynomial of ℘(z) and ℘0(z). Since ℘0(z) as the derivative of an even function is odd, to work with even elliptic functions to minimize the number of terms in the principal part at the origin, we use ℘0(z)2 whose principal part at 0 is

4 a b + + z6 w4 w2 Math 213a (Fall 2021) Yum-Tong Siu 9 for some a, b ∈ C due to its evenness. To cancel the term 4 z6 in the principal part at 0 with the most negative power, we should use 4℘(z)3, but the process still leaves a principal part c d + z4 z2 for some c, d ∈ C. Thus, we know that

0 2 3 2 ℘ (z) = 4℘(z) + α2℘(x) + α1℘(x) + α0 for some constants α0, α1, α2. To determine the constants α0, α1, α2, we have to explicitly write down a few terms in the Laurent series expansion of ℘(z) at 0.

First Few Terms in Laurent Series Expansion of ℘(z) at 0. For n ≥ 3 Let X 1 s = . n `n `∈L−{0} Since 1 1 1 z z2 = = + 2 + 3 + ··· for |z| < |`|, (z − `)2 2 z
2 `2 `3 `4 ` 1 − ` it follows that

−2 2 4 ℘(z) = z + 3s4z + 5s6z + ··· , 0 −3 3 ℘ (z) = −2z + 6s4z + 20s6z + ··· , 0 2 −6 −2 ℘ (z) = 4z − 24s4z − 80s6 + ··· , 3 −6 −2 ℘(z) = z + 9s4z + 15s6 + ··· , and 0 2 3 ℘ (z) − 4℘(z) + 60s4℘(z) = −140s6 + ··· . Thus we have the differential equation

0 2 3 ℘ (z) = 4℘(z) − g2℘(z) − g3, Math 213a (Fall 2021) Yum-Tong Siu 10 where X 1 X 1 g = 60s = 60 and g = 140s = 140 . 2 4 `4 3 6 `6 `∈L−{0} `∈L−{0} From the differential equation it follows that w = ℘(z) is the inverse function of the indefinite integral Z w dζ w 7→ z = . p 3 ζ=∞ 4ζ − g2ζ − g3 The inversion step of using the fundamental group of the Riemann surface of the denominator of the integrand has been replaced by the elimination of the principal part at 0 of a polynomial in the doubly periodic functions ℘(z) and ℘0(z) when ℘(z) is defined as the infinite sum of partial fractions.

Addition Theorem for Weierstrass ℘ Function. The addition theorem for the Weierstrass ℘ function is obtained by using Property (iii) of doubly periodic functions. Let x = ℘(w) and y = ℘0(w) and, for some complex numbers a 6= 0 and b to be determined later, we consider the doubly periodic function y + ax + b. This doubly periodic function has a pole of order 3 at the origin and no other poles inside a fundamental parallelogram. So the sum of its three zeroes must be zero modulo a period. We are free to choose a and b. We can choose a and b so that the doubly periodic function y + ax + b vanishes at w1 and w2. Then y + ax + b must also vanish at −(w1 + w2). On the other hand we have the equation 2 3 y = 4x − g2x − g3. So by solving the two equation y + ax + b = 0 and 2 3 y = 4x − g2x − g3, we would get the values of x and y at −(w1 + w2). Since one equation is a linear equation and the second one is a cubic equation, we expect to get 3 solutions for (x, y). Math 213a (Fall 2021) Yum-Tong Siu 11

The other two solutions are the values of (x, y) at w1 and w2. Knowing these two solutions makes getting the third solution very easy, because one can use the fact that for a cubic equation with unit leading coefficient the sum of the three roots is the negative of the second coefficient.

We now carry out the details to get our addition theorem. From

0 ℘ (w1) + a℘(w1) + b = 0 0 ℘ (w2) + a℘(w2) + b = 0 we get ℘0(w ) − ℘0(w ) a = − 1 2 . ℘(w1) − ℘(w2) (As we see later we do not need to solve for b.) From the equation

2 3 (ax + b) = 4x − g2x − g3 we obtain (by the evenness of ℘(w))

2  0 0 2 a 1 ℘ (w1) − ℘ (w2) ℘(w1) + ℘(w2) + ℘(w1 + w2) = = . 4 4 ℘(w1) − ℘(w2) Thus we have the addition formula

 0 0 2 1 ℘ (w1) − ℘ (w2) ℘(w1 + w2) = −℘(w1) − ℘(w2) + . 4 ℘(w1) − ℘(w2)

Indefinite Integral Formulation of Addition Theorem for ℘. The addi- tion theorem for the Weierstrass ℘ function can be formulated in terms of indefinite integrals as

Z ξ dζ Z η dζ + p 3 p 3 ζ=∞ 4ζ − g2ζ − g3 ζ=∞ 4ζ − g2ζ − g3 √ √ !2 4ξ3−g ξ−g − 4η3−g η−g −ξ−η+ 1 2 3 2 3 Z 4 ξ−η dζ = , p 3 ζ=∞ 4ζ − g2ζ − g3 Math 213a (Fall 2021) Yum-Tong Siu 12

because when we let ξ = ℘(w1) and η = ℘(w2), we have

−1 −1 −1 ℘ (ξ) + ℘ (η) = w1 + w2 = ℘ (℘(w1 + w2))  0 0 2! −1 1 ℘ (w1) − ℘ (w2) = ℘ −℘(w1) − ℘(w2) + 4 ℘(w1) − ℘(w2)  !2 1 p4ξ3 − g ξ − g − p4η3 − g η − g = ℘−1 −ξ − η + 2 3 2 3 .  4 ξ − η 

TO BE CONTINUED ...