3
$\begingroup$

Hi, everyone. I am interested in the complement of (1,1) bridge knot in a lens space, $S^{3}$. Is there one (1,1) bridge knot in $S^{3}$ or lens space such that its complement is hyperbolic?

Note: A knot $K$ in $S^{3}$ or Lens space is (1,1) if for the standard genus 1 Heegaard splitting of $S^{3}$ or lens space, $K$ intersects each solid torus only one arc which is boundary parallel.

$\endgroup$

2 Answers 2

4
$\begingroup$

All two-bridge knots in $S^3$ are $(1,1)$-knots in $S^3$. This is assigned as an exercise here. All two-bridge knots, other than the $(2,2k+1)$-torus knots, are hyperbolic.

$\endgroup$
0
3
$\begingroup$

Actually, there are a lot of these kinds of knots.

First, let's set some notation. Start with two manifolds $M_1$ and $M_2$ related by a Dehn surgery along an embedded curve $K$ in $M_1$. After removing a neighborhood of K, we glue in a solid torus $T$ along $\partial N(K)$ to obtain $M_2$. We say the core $K'$ of $T$ is the dual knot to $K$ in $M_2$. The dual knots to many Berge knots are (1,1) knots lens space. Ken Baker's work is a great place to find examples of this. Specifically, in this paper

  • Kenneth L. Baker, Surgery descriptions and volumes of Berge knots II: Descriptions on the minimally twisted five chain link, Journal of Knot Theory and Its Ramifications 17 No. 09 (2008) pp. 1099–1120, doi:10.1142/S021821650800652X, arXiv:math/0509055

all knots in families III-VI and VIII-XII are hyperbolic knots in $S^3$ that are dual to (1,1) knots in Lens spaces. Forcing the dual knots to be hyperbolic as well.

If you want a more concrete example. The (-2,3,7) pretzel knot is hyperbolic and admits 2 lens space surgeries. (This was first observed by Fintushel and Stern.) The lens spaces are (18,7) and (19,7) and the dual knots to the (-2,3,7) in these lens spaces are (1,1) knots.

$\endgroup$
1
  • $\begingroup$ Great, the answer is more than what I want. Thank you very much! $\endgroup$
    – yanqing
    Nov 16, 2012 at 0:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.