Let $\Gamma \subset\mathbb{R}^n$ be an integer lattice and $\Gamma^*$ its dual lattice (that is, $\langle\gamma,\gamma^*\rangle \in \mathbb{Z}$ for all $\gamma \in \Gamma$ and $\gamma^* \in \Gamma^*$, with the usual inner product).
Consider a linear map $M: \mathbb{R}^n \to \mathbb{R}^n$ that preserves the lattice, that is, $M\Gamma \subset \Gamma$ and all of its eigenvalues have $|\lambda| > 1$. If $\eta_1,...,\eta_m$ are the representatives of the cosets of the quotient group ${(M^*)}^{-1}\Gamma^* / \Gamma^*$, show that, for $\gamma \not\in M\Gamma$,
\begin{equation} \sum_{q=1}^m e^{2\pi i \langle\gamma,\eta_q\rangle} = 0 \end{equation}
First, I have proven that
- $|\det M| = m \in \mathbb{Z}_{>0}$, so $M$ is invertible (as is $M^*$);
- both quotient groups $\Gamma/M\Gamma$ and ${(M^*)}^{-1}\Gamma^* / \Gamma^*$ have order $m$;
so everything in the question is well-defined.
My hunch is that those exponentials are actually $m$-th roots of unity (or maybe some prime factor of $m$), for which I found strong evidence:
- $\langle \gamma, M^* \eta_q\rangle = n_q \in \mathbb{Z}$ for all $q = 1,...,m$. This comes from $\eta_q \in (M^*)^{-1}\Gamma^*$, so $M^* \eta_q \in \Gamma^*$.
- Even more, we can take $\eta_q$ such that $n_q \in \mathbb{Z}/m\mathbb{Z}$, since it's in an equivalence class.
- If the hunch is true, then the sum is equivalent to $\langle \gamma,\eta_q \rangle = \frac{n_q}{m}$, which is similar to the previous point.
This may be obvious for those that work with lattices, but I've been working with them for only about 2 weeks, in the context of multidimensional wavelets. It doesn't help that my algebra is quite rusty. Sorry if a similar question was asked, but I didn't find it.