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Let $G$ be a locally compact abelian group. A lattice in $G$ is a discrete subgroup $\Lambda$ such that the quotient $G / \Lambda$ is compact. A Borel fundamental domain of a lattice $\Lambda$ in $G$ is a Borel set $F \subseteq G$ that intersects each coset in $G / \Lambda$ in exactly one point, i.e., $F$ is a set of coset representatives.

My question is now the following:

Given two lattices $\Lambda_1, \Lambda_2 \subseteq G$ such that $\Lambda_1 \subseteq \Lambda 2$, is it always possible to take corresponding fundamental domains $F_i$ of $\Lambda_i$, where $i = 1,2$, such that $F_2 \subseteq F_1$?

By looking at special cases in, say, $G=\mathbb{R}$ or $G = \mathbb{Z}$, it seems that this is always possible.

Any comment or reference is highly appreciated.

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  • $\begingroup$ According to your definition, the empty set or a single point would do for both lattices! :) Do you know a better definition/construction? Say, for $\mathbb{R}^n$, the standard construction via intersection of half-spaces would answer your question in the affirmative. $\endgroup$ – Alex Degtyarev Nov 8 '17 at 19:23
  • $\begingroup$ I am sorry, in the definition "intersecting each coset at most once" should of course be "intersecting each coset in exactly once point". I have edited the question. $\endgroup$ – jvnv Nov 8 '17 at 19:29
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The image of $\Lambda_2$ in $G/\Lambda_1$ is finite. Take a set of representatives $s_1, s_2, \ldots s_n$ in $\Lambda_2$. Now consider $\cup s_i F_2$. I claim this is a Borel fundamental domain for $F_1$.

It is Borel as it is a finite union of Borel sets. Let $g\Lambda_1$ be a coset of $\Lambda_1$. Then $g\Lambda_2$ intersects $F_2$ in a single point $x$. Now, $\Lambda_2 = \cup s_i \Lambda_1$, and so $x$ is in some $s_i$ translate of $F_2$, and hence in $F_1$. It is unique as otherwise we have $s_i \lambda_1 = s_j \lambda_1'$, which would mean our set of representatives was too big in the first place.

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