2016
DOI: 10.1093/imrn/rnw136
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Generic Thinness in Finitely Generated Subgroups of SL$_n(\mathbb Z)$

Abstract: We show that for any n ≥ 2, two elements selected uniformly at random from a symmetrized Euclidean ball of radius X in SL n (Z) will generate a thin free group with probability tending to 1 as X → ∞. This is done by showing that the two elements will form a ping-pong pair, when acting on a suitable space, with probability tending to 1. On the other hand, we give an upper bound less than 1 for the probability that two such elements will form a ping-pong pair in the usual Euclidean ball model in the case where n… Show more

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Cited by 14 publications
(16 citation statements)
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“…A related strategy towards the same goal would be to show that ρ(Aff(O)) fits the assumptions of the ping-pong lemma (compare with the proof of Lemma 2•3). Nevertheless, it is not easy to implement this idea in general because the construction of "ping-pong subsets" might be somewhat tricky (see [FR,page 5387 and Subsections 3•2 and 3•3]).…”
Section: •3 Final Commentsmentioning
confidence: 99%
“…A related strategy towards the same goal would be to show that ρ(Aff(O)) fits the assumptions of the ping-pong lemma (compare with the proof of Lemma 2•3). Nevertheless, it is not easy to implement this idea in general because the construction of "ping-pong subsets" might be somewhat tricky (see [FR,page 5387 and Subsections 3•2 and 3•3]).…”
Section: •3 Final Commentsmentioning
confidence: 99%
“…We attempted to verify whether each group is arithmetic by expressing its generators as words in standard generators of SL(n, Z) and running a coset enumeration with the presentation from [30]. As the enumeration never terminated, we suspect that these groups are not arithmetic (note that a random finitely generated subgroup of SL(n, Z) is likely to be thin [13,27]). 4.2.4.…”
Section: The Fundamental Group Of the Figure-eight Knot Complement Ad...mentioning
confidence: 99%
“…where R 2 R 1 ≥ 1. As remarked in [10], in the context of selecting elements uniformly at random form SL N (Z), this is the case R 2 = 1/R 1 is analogous to bounding the condition number ||M || ||M −1 ||. We would like to compute vol D R1,R2 , which is defined as the invariant measure (2.2) integrated over D R1,R2 .…”
Section: Invariant Measure and Volumesmentioning
confidence: 99%