Finding the Laurent Expansion of $\wp_{\Gamma}(u)$ in terms of Eisenstein series

Question

In Serre's "A Course in Arithmetic"(page $84$) he mentions that the Laurent series of the Weierstrass function $$\wp_{\Gamma}(u)= \frac{1}{u^2} + \sum_{\gamma \in \Gamma}(\frac{1}{(u-\gamma)^2} - \frac{1}{\gamma^2})$$ is given by $$(1)\qquad \frac{1}{u^2} + \sum_{k=2}^{\infty}(2k-1)G_k(\Gamma)u^{2k-2}$$ where $G_k(\Gamma)$ is the Eisenstein series of weight $2k$, and $\Gamma$ a lattice in $\mathbf{C}$.

I am attempting to prove that this is in fact the case, and have considered expanding $\frac{1}{(u-\gamma)^2} - \frac{1}{\gamma^2}$ as $\frac{1}{\gamma^2}\frac{1}{(1-\frac{u}{\gamma})^2} - \frac{1}{\gamma^2} = \frac{1}{\gamma^2}(\sum_{k=1}^{\infty}(k+1)(\frac{u}{\gamma})^k) = \sum_{k=1}^{\infty}(k+1)(\frac{u^k}{\gamma^{k+2}})$. Hence I am left with:

$$\frac{1}{u^2} + \sum_{\gamma \in \Gamma}(\sum_{k=1}^{\infty}(k+1)(\frac{u^k}{\gamma^{k+2}}))$$ Now in attempting to make this look like (1), I feel like this involves some tricky manipulation of the series to get $G_k(\Gamma)$, but I am not sure how to proceed.

Any hints would be greatly appreciated.

Answer 1

What you can do is reverse the order of the sums since the series is absolutely convergent to obtain that:

$$\frac{1}{u^2} + \sum_{\gamma \in \Gamma}(\sum_{k=1}^{\infty}(k+1)(\frac{u^k}{\gamma^{k+2}})) = \frac{1}{u^2} + \sum_{k=1}^{\infty}(k+1)(\sum_{\gamma \in \Gamma}(\frac{u^k}{\gamma^{k+2}}))$$ then factor out the $u^k$ to get:

$$\frac{1}{u^2} + \sum_{k=1}^{\infty}(k+1)(\sum_{\gamma \in \Gamma}(\frac{1}{\gamma^{k+2}}))u^k = \frac{1}{u^2} + \sum_{k=1}^{\infty}(k+1)G_k(\Gamma)u^k$$ We know that modular forms of odd weight are zero, and hence our sum becomes:

$$\frac{1}{u^2} + \sum_{k=2}^{\infty}(2k-1)G_k(\Gamma)u^{2k-2}$$ as desired.

Answer 2

Use $$\wp_\tau(z) = \frac{1}{z^2} + \!\!\!\sum_{(m,n) \in \mathbb{Z}^2\setminus (0,0)} \frac{1}{(m+n\tau-z)^2}-\frac{1}{(m+n\tau)^2}, \quad \!\! G_{2k}(\tau)=\sum_{(m,n) \in \mathbb{Z}^2\setminus (0,0)} \frac{1}{(m+n\tau)^{2k}}$$ With $f(z) = \wp_\tau(z)- \frac{1}{z^2}$ for $k \ge 1$ $$f^{(k)}(z) = \sum_{(m,n) \in \mathbb{Z}^2\setminus (0,0)} \frac{(k+1)!}{(m+n\tau-z)^{k+2}}, \qquad f^{(2k)}(0) = (2k+1)! G_{2k+2}(\tau)$$ And since for $|z| < r$, $f(z)$ is analytic and even we have the Taylor series $$\wp_\tau(z) -\frac{1}{z^2}=f(0)+\sum_{k=1}^\infty z^{2k} (2k+1) G_{2k+2}(\tau)$$