Eisenstein series for hexagonal lattice

Question

How I don't know how to prove that given the lattice $\Lambda=\lbrace m\rho+n \: : \: m,n\in\mathbb{Z}\rbrace$ with $\rho\in \mathbb{C}\setminus\lbrace 1\rbrace$ such that $\rho^3=1$, the Eisenstein series $$G_4(\Lambda):=\sum_{\omega\in \Lambda\setminus\lbrace0\rbrace}\dfrac{1}{\omega^4}=0.$$ Any help?

Answer 1

Otherwise, you can use the properties of the modular group and modular forms.

Note that $\scriptstyle \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\displaystyle e^{2i \pi / 3}=\frac{-1}{e^{2i \pi / 3}} = e^{2i \pi / 3}+1$

Since $G_4$ is a modular form of weight $4$ : $$G_4(\scriptstyle \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\displaystyle \tau) = \tau^4 G_4(\tau), \qquad G_4(\tau) = G_4(\tau+1)$$ and $$G_4(e^{2i \pi / 3}) = G_4(e^{2i \pi / 3}+1) = G_4(\scriptstyle \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\displaystyle e^{2i \pi / 3}) = e^{8 i \pi / 3}G_4(e^{2i \pi / 3})$$ i.e. $G_4(e^{2i \pi / 3}) = 0$

Answer 2

Since the sum converges absolutely, you can regroup the terms in group of threes $\{\omega ; \rho \omega ; \rho^2 \omega\}$. The sum on each group is $0$, and they partition the index set, so the whole sum is $0$.

Answer 3

$G_k$ is homogeneous of degree $k$, i.e., $$G_k(a\Lambda)=\frac{G_k(\Lambda)}{a^k}$$ for all nonzero $a\in \mathbb{C}$. If $a\Lambda=\Lambda$ then the homogeneity implies that $G_k(\Lambda)$ can only be nonzero if $a^k = 1$. Therefore $G_k(\Lambda)=0$ for the equilateral triangular lattice $\Lambda=\mathbb{Z}+\mathbb{Z}\rho$, unless $6\mid k$. In particular, $G_4(\Lambda)=0$.