Fourier expansion of Hilbert Eisenstein series

Question

Suppose $F=\mathbb{Q}(\sqrt{D})$ is a real quadratic field with class number 1. In J. Bruiner's article "Hilbert modular forms and applications", the Fourier series of Eisenstein series $$ G_{k, \mathcal{O}_{F}}(z_{1}, z_{2})=\sum_{(c,d)\in (\mathcal{O}_{F}\times \mathcal{O}_{F})/ \mathcal{O}_{F}^{\times}}\frac{1}{(cz_{1}+d)^{k}(c'z_{2}+d')^{k}} $$ for even $k>2$ is given by $$ \zeta_{F}(k)+\frac{(2\pi i)^{2k}}{(k-1)!^{2}}D^{1/2-k}\sum_{\nu\in \mathfrak{d}^{-1}, \nu>>0}\sigma_{k-1}(\mathfrak{d}\nu)e(tr(\nu z)) $$ where $$ \sigma_{k-1}(\mathfrak{l})=\sum_{\mathfrak{c}|\mathfrak{l}}N(\mathfrak{c})^{k-1}. $$ The author said that we can derive it in the same way as in case of elliptic modular forms. I know that for elliptic case, we use the Lipschitz's formula $$ \sum_{n\in \mathbb{Z}}\frac{1}{(z+n)^{k}}=\frac{(-2\pi i )^{k}}{(k-1)!}\sum_{r=1}^{\infty}r^{k-1}q^{r} $$ for $z\not\in \mathbb{Z}$. How can we apply this formula for Hilbert case?

Answer 1

Lipschitz's formula is a special case of the more general Poisson summation formula. In general, if $ h \in \mathcal{L}^1(\mathbb{R}^l) $ is any function such that the sum $ \sum_{d \in \mathbb{Z}^l} h(x+d)$ converges absolutely and uniformly on compact sets and is infinitely differentiable as a function of $x$, then $$ \sum_{d \in \mathbb{Z}^l} h(x+d) = \sum_{m \in \mathbb{Z}^l} \hat{h} (m) e^{2 \pi i \langle m, x \rangle} $$ (see e.g. Diamond and Shurman, "A first Course in Modular Forms", p. 144. Exercise 4.9.4 shows how the Lipchitz formula is a special case).

In this case, consider the function $$ h(x_1, x_2) = \begin{cases} N(x_1 + x_2 \phi)^{k-1} e^{\frac{2 \pi i}{\sqrt{5}} (x_1 z_1 - x_2 z_2) } & x_1 \ge 0, x_2 \ge 0 \\ 0 & else \end{cases} $$

Note that this is the function we are summing over on the right hand side. By applying the Poisson summation formula and some algebraic manipulation, one obtains the above equation, as mentioned in van der Geer, "Hilbert Modular Surfaces", p. 19.