Recursive identity for elliptic lattice constants $\sum_{\lambda\in\Lambda\setminus0} \lambda^{-2k}$

Question

I am stuck on Exercise 3 in these notes. To keep this question self-contained: we have

1. $\displaystyle\Lambda=\langle\omega_1,\omega_2\rangle=\omega_1\Bbb Z+\omega_2\Bbb Z\subset\Bbb C,$

2. $\displaystyle \wp(z) =\frac{1}{z^2}+\sum_{\lambda\in\Lambda\setminus0}\left(\frac{1}{(z-\lambda)^2}-\frac{1}{\lambda^2}\right)=\frac{1}{z^2}+\sum_{k=2}^\infty(2k-1)G_{2k}z^{2k-2} ~~ (z\approx0), $

3. $\displaystyle G_k=\sum_{\lambda\in\Lambda\setminus0}\frac{1}{\lambda^k}, \quad G_{\rm odd}=0,$

4. $\dot{\wp}^2=4\wp^3-40G_4\wp-160G_6.$

(See the link for proof of $\wp$'s Taylor expansion and the differential equation.) I want to prove

$$(n-3)(2n+1)(2n-1)G_{2n}=\sum_{\substack{k+l=n \\ k,l\ge2}} (2k-1)(2l-1)G_{2k}G_{2l}$$

for $n\ge4$. Multiplying both sides by $z^{2n}$ and summing should result in a differential equation that will surely yield an unwieldy equation involving $z,\wp,\wp^2,\dot{\wp},\ddot{\wp}$. I have no how to derive this from the known differential equation $(4)$, or if perhaps I should try a route other than generating functions and differential equations to prove the identity. Perhaps I should augment the generating function because the identity is only being proven for $n\ge4$?

Answer 1

Alright I think I got it. Set $G_0:=-1$ and $G_2:=0$ for the sake of being able to write

$$\wp(z)=\sum_{n=0}^\infty (2n-1)G_{2n}z^{2n-2}. \tag{1}$$

Differentiate $\dot{\wp}^2=4\wp^3-60G_4\wp-140G_6$ and then divide by $2\dot{\wp}$ to obtain

$$\ddot{\wp}=6\wp^2-30G_4. \tag{2}$$

Plug $(1)$ into $(2)$, cancel poles and constant terms, equate coefficients, get

$$(2n-1)(2n-2)(2n-3)G_{2n}=6\sum_{\substack{k+l=n \\ k,l\ge0}} (2k-1)(2l-1)G_{2k}G_{2l} \tag{3}$$

for $n\ge3$. By taking out the $(n,0),(n-1,1),(1,n-1),(0,n)$ terms we have

$$\sum_{\substack{k+l=n \\ k,l\ge0}} (2k-1)(2l-1)G_{2k}G_{2l}=2(2n-1)G_{2n}+\sum_{\substack{k+l=n \\ k,l\ge2}} (2k-1)(2l-1)G_{2k}G_{2l} \tag{4}$$

Thus subtracting $6\cdot2(2n-1)G_{2n}$ from both sides of $(3)$ yields

$$2(2n-1)[(n-1)(2n-3)-6]G_{2n}=6\sum_{\substack{k+l=n \\ k,l\ge2}} (2k-1)(2l-1)G_{2k}G_{2l} \tag{5}$$

Simplifying,

$$(2n-1)(2n+1)(n-3)G_{2n}=3\sum_{\substack{k+l=n \\ k,l\ge2}} (2k-1)(2l-1)G_{2k}G_{2l}.$$

(So in fact there was a factor of $3$ missing in the exercise statement.)