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Given a knot in the 3-sphere in Bridge Position you can find a presentation for the fundamental group of the complement (a Wirthinger presentation) containing $n$ generators and $n-1$ relators, where $n$ is the number of bridges in the presentation.

The bridge index of the knot is the minimal number of bridges it takes to present the knot. Is that the same as the minimal number of generators among all presentations of the fundamental group of the knot complement?

I imagine this is well known but after a while of Googling around and looking through Neuwirth's "Knot Groups" book I haven't found where this question is addressed.

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up vote 11 down vote accepted

The (p,q) torus knot has a presentation with two generators, namely $\langle x,y \mid x^p = y^q\rangle$, but if $p,q>2$ then it's non-alternating and so it must have bridge index greater than 2.

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The existence of such examples is proved in a preprint of Jesse Johnson. He proves that for any $n$, there is a tunnel number 1 knot which has bridge number at least $n$. A tunnel number one knot complement has a Heegaard surface of genus 2, which splits the knot complement up into a genus 2 handlebody and a genus 2 compression body with a torus boundary corresponding to the boundary of a regular neighborhood of the knot. Thus tunnel number 1 knots are 2-generator (and in fact with 1 relation), so this answers your question.

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Thanks Steven and Ian. – Ryan Budney Jul 24 '12 at 22:10
Steven Sivek's answer is much more elementary than the reference I gave, in fact torus knots have bridge numbers $\to \infty$, yet are tunnel number 1. The point of Jesse's construction is that he gave hyperbolic examples with this property. – Ian Agol Jul 24 '12 at 22:10
Okay, I'll give Steven the checkmark then. – Ryan Budney Jul 24 '12 at 22:11
Where do the lower bounds on the bridge numbers of torus knots come from? – Ryan Budney Jul 24 '12 at 22:39
A little Googling later I found in Jennifer Schulten's paper "Brige numbers of torus knots" that the bridge number of the $(p,q)$ torus knot is the minimum of $p$ and $q$, nice. And this is due to Schubert. – Ryan Budney Jul 24 '12 at 22:43

It is not a problem, but this is a problem.

A question posed by Cappell and Shaneson asks whether the bridge number of a given link equals the minimal number of meridian generators of its group.

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Thanks for pointing that out. – Ryan Budney Jul 25 '12 at 4:32
This is also problem 1.11 in Kirby's list, attributed therein to Cappell and Shaneson. – dvitek Oct 9 '13 at 4:30
Chris Cornwell obtained lower bounds on the meridional rank of certain pretzel knots, showing they are equal to the bridge number: – Ian Agol Mar 31 '15 at 3:32

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