2008
DOI: 10.1080/10586458.2008.10129044
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Commensurators of Cusped Hyperbolic Manifolds

Abstract: Abstract. This paper describes a general algorithm for finding the commensurator of a non-arithmetic hyperbolic manifold with cusps, and for deciding when two such manifolds are commensurable. The method is based on some elementary observations regarding horosphere packings and canonical cell decompositions. For example, we use this to find the commensurators of all non-arithmetic hyperbolic once-punctured torus bundles over the circle.For hyperbolic 3-manifolds, the algorithm has been implemented using Goodma… Show more

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Cited by 28 publications
(51 citation statements)
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References 26 publications
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“…In fact, F is a simply asymptotic orthoscheme [4,4,3] in H 3 whose edge lengths satisfy tanh q 1 q 2 = 1/2 and tanh q 1 q 3 = 1/ √ 2 (see [18]). In a similar way, the glue facet F = F(P 2 ) for P 2 is a simply asymptotic orthoscheme [6,3,3] in H 3 whose corresponding edge lengths satisfy tanh q 1 q 2 = 1/2 and tanh q 1 q 3 = 1/ √ 3. Now, suppose that the groups 1 and 2 are commensurable.…”
Section: P N−1 Q ∞] Then Is a Subgroup Of Index 2 In θmentioning
confidence: 99%
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“…In fact, F is a simply asymptotic orthoscheme [4,4,3] in H 3 whose edge lengths satisfy tanh q 1 q 2 = 1/2 and tanh q 1 q 3 = 1/ √ 2 (see [18]). In a similar way, the glue facet F = F(P 2 ) for P 2 is a simply asymptotic orthoscheme [6,3,3] in H 3 whose corresponding edge lengths satisfy tanh q 1 q 2 = 1/2 and tanh q 1 q 3 = 1/ √ 3. Now, suppose that the groups 1 and 2 are commensurable.…”
Section: P N−1 Q ∞] Then Is a Subgroup Of Index 2 In θmentioning
confidence: 99%
“…Geometrically, a fundamental polyhedron for 1 2 is the Coxeter polyhedron arising by glueing together P 1 and P 2 along their common facet F. In this context, the following general result of Karrass and Solitar [17, Theorem 10] will be useful. 6 , (3, ∞, 4)], respectively (see also Fig. 3 and the "Appendix").…”
Section: Commensurable Hyperbolic Coxeter Groupsmentioning
confidence: 99%
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