2008
DOI: 10.2140/agt.2008.8.1833
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Commensurability classes of (−2,3,n) pretzel knot complements

Abstract: Let K be a hyperbolic . 2; 3; n/ pretzel knot and M D S 3 n K its complement. For these knots, we verify a conjecture of Reid and Walsh: there are at most three knot complements in the commensurability class of M . Indeed, if n ¤ 7, we show that M is the unique knot complement in its class. We include examples to illustrate how our methods apply to a broad class of Montesinos knots. 57M25

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Cited by 16 publications
(18 citation statements)
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“…It has also been shown in that the (2,3,n) pretzel knot is slim provided that it is hyperbolic and that n is not divisible by 3. In fact, (2,3,n) is hyperbolic precisely when n{1,3,5}; see [, p. 1834]. We refer the reader to [, p. 2] for a discussion of these facts.…”
Section: Some Computationsmentioning
confidence: 99%
“…It has also been shown in that the (2,3,n) pretzel knot is slim provided that it is hyperbolic and that n is not divisible by 3. In fact, (2,3,n) is hyperbolic precisely when n{1,3,5}; see [, p. 1834]. We refer the reader to [, p. 2] for a discussion of these facts.…”
Section: Some Computationsmentioning
confidence: 99%
“…Their work provides criteria for checking whether or not a hyperbolic knot complement is the only knot complement in its commensurability class. Specifically, we have the following theorem coming from Reid and Walsh's work in [37,Section 5]; this version of the theorem can be found at the beginning of [23]. Theorem 7.1.…”
Section: Commensurability Classes Of Hyperbolic Pretzel Knot Complementsmentioning
confidence: 99%
“…Macasieb and Mattman use this criterion in [23] to show that for any hyperbolic pretzel knot of the form K 1 −2 , 1 3 , 1 n , n ∈ Z \ {7}, its knot complement S 3 \ K 1 −2 , 1 3 , 1 n is the only knot complement in its commensurability class. The main challenge in their work was showing that these knots admit no hidden symmetries.…”
Section: Commensurability Classes Of Hyperbolic Pretzel Knot Complementsmentioning
confidence: 99%
See 1 more Smart Citation
“…Aside from the figure-eight, there are no knots with hidden symmetries with at most fifteen crossings [6] and no two-bridge knots with hidden symmetries [11]. Macasieb-Mattman showed that no hyperbolic (−2, 3, n) pretzel knot, n ∈ Z, has hidden symmetries [8]. Hoffman showed the dodecahedral knots are commensurable with no others [7].…”
mentioning
confidence: 99%