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Proceedings of the American Mathematical Society
Proceedings of the American Mathematical Society
ISSN 1088-6826(online) ISSN 0002-9939(print)

On Bieberbach's analysis of discrete Euclidean groups


Author: R. K. Oliver
Journal: Proc. Amer. Math. Soc. 80 (1980), 15-21
MSC: Primary 20H15; Secondary 22E40, 51M20
MathSciNet review: 574501
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Abstract: For a subgroup G of the euclidean group $ {E_n} = {O_n} \cdot {{\mathbf{R}}^n}$ (semidirect product) and a real number $ r > 0$, let $ {G^ \ast }$ denote the translation subgroup of G, $ {G_r}$ the group generated by all (A, a) in G with $ \left\Vert {1 - A} \right\Vert < r$ (operator norm), and $ {k_n}(r)$ the maximum number of elements of $ {O_n}$ with mutual distances $ \geqslant r$ relative to the metric $ d(A,B) = \left\Vert {A - B} \right\Vert$. We give an elementary, largely geometrical proof of the following results of Bieberbach: Let G be a subgroup of $ {E_n}$. (1) If G is discrete, then $ {G_{1/2}}$ is abelian, $ {G_{1/2}} \triangleleft G$, and $ [G:{G_{1/2}}] \leqslant {k_n}(1/2)$. (2) G is discrete if and only if $ G \subset {O_{n - k}} \times {E_k}$, where $ {p_2}G$ is discrete, $ {({p_2}G)^ \ast }$ spans $ {{\mathbf{R}}^k}$, and $ G \cap \ker {p_2}$ is finite. (Here $ {p_2}$ is the projection on the second factor.) (3) G is crystallographic if and only if G is discrete and $ {G^ \ast }$ spans $ {{\mathbf{R}}^n}$. Moreover, if G is crystallographic, then $ [G:{G^ \ast }] \leqslant {k_n}(1/2)$.


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Additional Information

DOI: http://dx.doi.org/10.1090/S0002-9939-1980-0574501-7
PII: S 0002-9939(1980)0574501-7
Keywords: Discrete euclidean groups, crystallographic groups
Article copyright: © Copyright 1980 American Mathematical Society