Bounded and measurable common fundamental domains for two lattices

Abstract

Suppose that $L, M$ are two full-rank lattices in Euclidean space with $\text{vol}(L)=\text{vol}(M)$. We give a new proof on the existence of a bounded and Lebesgue measurable set that tiles $\mathbb{R}^d$ with both $L,M$ using the measurable Hall's Theorem which was proved by T.Ciésla and M. Sabok. This proof is direct and does not go through the intermediate results on cut-and-project sets involved in the proof given by S.Grepstad and M.Kolountzakis. We also show the existence of a bounded, set-theoretic (i.e., not necessarily measurable) common fundamental domain of $L,M$ assuming only that $\text{vol}(L)=\text{vol}(M)$. Combining these results we show the existence of a bounded and Lebesgue measurable common fundamental domain for any two full-rank lattices of equal volumes. Finally we show that a set-theoretic bounded, common fundamental domain cannot exist when $\text{vol}(L)\neq \text{vol}(M)$.

Figures (8)

• Figure 1: Two different fundamental parallelepipeds of the lattice ${\mathbb Z}^2$. $P_1$ with respect to the basis $\{(1,0), (0,1)\}$ (left) and $P_2$ with respect to the basis $\{(1,1),(0,1)\}$ (right).
• Figure 2: An example of two ${\mathbb Z}^2$- equidecomposable sets, $P_1= [-\frac{1}{2}, \frac{1}{2})^2$ appears in blue and $P_2=1011\left[-\frac{1}{2}, \frac{1}{2}\right) ^2$ in red. The disjoint pieces $S_1 =P_1 \cap P_2,S_2,S_3$ partition $P_1$ and $S_1, S_2'=S_2 -e_2, S_3 '=S_3 +e_2$ partitions $P_2$ where $e_2=(0,1)\in {\mathbb Z}^2$.
• Figure 3: $L+M$ equidecomposition of $P_L$ (in blue), $P_M$ (in red) when $L={\mathbb Z}^2$ and $M= <(1/2,0), (0,2)>_{\mathbb Z}$.
• Figure 4: The bounded common fundamental domain $F$ of $L$,$M$ in Figure \ref{['PLPM']}.
• Figure 5: The set $F$ in Figure \ref{['F']} tiles ${\mathbb R}^2$ by translations of $M= <(1/2,0),(0,2)>_{\mathbb Z}$ (left) and by translations of $L={\mathbb Z}^2$ (right), as a common fundamental domain of $L,M$. In the figure to the left $F_{n,m}=F + n(1/2, 0)+ m(0,2)$ and to the right $F_{n,m}= F + n(1,0)+ m(0,1)$, for $n,m\in {\mathbb Z}$.

Theorems & Definitions (20)

• Theorem 1.1
• Theorem 1.2
• Theorem 1.3
• Lemma 2.1
• proof
• Definition 2.1
• Lemma 2.2
• proof
• Definition 2.2: Equidecomposition up to measure zero
• Lemma 2.3