Let \[ \omega=e^{2\pi i/3}, \qquad \Lambda=\mathbb Z\oplus\mathbb Zi, \qquad L=\mathbb Z\oplus\mathbb Z\omega. \] Prove that \[ \sum_{\lambda\in\Lambda\setminus\{0\}}\lambda^{-6}=0 \qquad\text{and}\qquad \sum_{\lambda\in L\setminus\{0\}}\lambda^{-4}=0. \]
Solution
Both sums converge absolutely, since in a rank-two lattice \[ \sum_{\lambda\ne0}|\lambda|^{-s}<\infty \qquad(s>2). \] Hence we may freely reindex them by lattice automorphisms.
1. For the square lattice,\[ \sum_{\lambda\in\Lambda\setminus\{0\}}\lambda^{-6}=0. \]
Multiplication by \(i\) permutes \[ \Lambda=\mathbb Z\oplus\mathbb Zi. \] Thus, writing \[ S_6=\sum_{\lambda\in\Lambda\setminus\{0\}}\lambda^{-6}, \] absolute convergence permits the change of variables \(\lambda=i\mu\): \[ \begin{aligned} S_6 &=\sum_{\mu\in\Lambda\setminus\{0\}}(i\mu)^{-6}\\ &=i^{-6}S_6\\ &=-S_6. \end{aligned} \] Therefore \[ 2S_6=0, \] and hence \(S_6=0\).
2. For the hexagonal lattice,\[ \sum_{\lambda\in L\setminus\{0\}}\lambda^{-4}=0. \]
Since \[ \omega^3=1, \qquad 1+\omega+\omega^2=0, \] multiplication by \(\omega\) preserves \[ L=\mathbb Z\oplus\mathbb Z\omega: \] indeed \(\omega\cdot1=\omega\) and \[ \omega^2=-1-\omega\in L. \] Set \[ S_4=\sum_{\lambda\in L\setminus\{0\}}\lambda^{-4}. \] Reindexing by \(\lambda=\omega\mu\) gives \[ S_4 =\omega^{-4}S_4 =\omega^2S_4. \] Since \(\omega^2\ne1\), \[ (1-\omega^2)S_4=0, \] so \(S_4=0\).