Let $G$ be a locally compact abelian group, and let $\Lambda$ be a lattice in $G$, i.e. a discrete subgroup such that the quotient group $G/\Lambda$ is compact.

A fundamental domain for $\Lambda$ in $G$ is a Borel set $B$ such that every $x \in G$ can be written uniquely as $x = b \lambda$ where $b \in B$ and $\lambda \in \Lambda$. A relatively compact fundamental domain will always exist (Lemma 2 of "Zeros of the Zak transform on locally compact abelian groups" by Kaniuth and Kutyniok).

My questions are the following:

  1. Given $G$ and $\Lambda$ as above, can we find always find a fundamental domain for $B$ in $\Lambda$ with non-empty interior?

  2. If 1 is not true in general, what kind of (preferably as general as possible) conditions can we place on $G$ and $\Lambda$ for it to hold?

One answer to 2 is that 1 is true in the case that $G$ is $\sigma$-compact and $\Lambda$ is countable (e.g. if $G$ is second countable). Since $G$ is $\sigma$-compact, page 35 of Margulis "Discrete subgroups of semisimple Lie groups" ensures the existence of a fundamental domain $B$ with $\mu(\partial B) = 0$. Since $\Lambda$ is countable, the identity $G = \bigcup_{\lambda \in \Lambda} \lambda B$ implies that $B$ cannot be nowhere dense, otherwise we would have a contradiction to the Baire category theorem. So let $U$ be a non-empty open subset of $\overline{B}$. Now if $B^{\circ} = \emptyset$, then $\overline{B} = \partial B$ so we have $U \subseteq \overline{B} = \partial B$. But then $\mu(U) \leq \mu(\partial B) = 0$ which contradicts the properties of a Haar measure.

  • 1
    $\begingroup$ Let $V$ be an open, $\sigma$-compact subgroup. Then $V+\Lambda$ has finite index in $G$. Hence enlarging $V$, we can suppose that $V+\Lambda=G$. Then choose a fundamental domain (with nonempty interior) of $\Lambda\cap V$ in $V$: this is a fundamental domain of $\Lambda$ in $G$. $\endgroup$ – YCor Sep 8 '18 at 1:07
  • $\begingroup$ That does the trick, feel free to post it as an answer and I will accept. Do you know if any fundamental domain will have nonempty interior, or are there counter-examples? $\endgroup$ – Ulrik Sep 10 '18 at 5:48

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