MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The knot group is the fundamental group of the knot complement in $S^{3} $. The Dehn presentation of the knot group is a particular group presentation obtained by looking at the regions and crossings of a knot (see Is it possible to look at an arbitrary group presentation and know whether or not it is a Dehn presentation of a knot group?

share|cite|improve this question

Let's call your candidate knot group $G$. Also, I will take "look at" to mean find easily computable invariants and or obstructions.

There are obvious obstructions that you might consider to help determine if $G$ could be a knot group. Many of these obstructions are more apparent from the Wirtinger presentation, which is described below the Dehn presentation in your reference.

The Wirtinger presentation is generated by conjugate elements or rather normally generated by one element. Since knot groups are torsion free, the abelianization of $G$ is $Z$. Furthermore, using the Wirtinger presentation, adding the relation that one generator is trivial makes the whole group trivial.

Those are two obvious necessary and often easily observable criterion for presentations involving few generators and short relations. However, if this group is showing up as the fundamental group of a 3-manifold with torus boundary more can be said. Refine our assumptions so $G$ be the fundamental group of a manifold $M$ with a single torus boundary component $T$ and let $\mu \in \pi_1(T)$. By Perelman's solution to the Poincaré conjecture and a simple argument using Dehn filling (see Rolfsen's Knots and Links for background on Dehn filling), $G/<<\mu>>$ is trivial if and only if $M$ is a knot exterior in $S^3$ (i.e. $M \cong S^3 - n(K)$, where $n(K)$ denotes the open neighborhood of an embedded knot).

Checking that $\mu \in \pi_1(T)$ is necessary here. The key point here is that there are lots of groups with abelianization $Z$ that are normally generated by one element that are not knot groups. A large family of examples of 3-manifold groups with this property can be constructed by taking 0/1 surgery along a (homologically framed) knot in $S^3$. With more work, one can find examples of such $\pi_1(M)$ where $M$ is a manifold with torus boundary having this property. Using Snappy, we can see the manifold 'm008' in the cusped census of hyperbolic 3-manifolds by Callahan, Hildebrand and Weeks has $\pi_1(m008)=\langle a,b | a^{2}b^{2}a^{-1}ba^{-1}b^{-2}\rangle$. Introducing the relation, $a=1$ makes the whole group trivial. In this case, $a$ does not correspond to an element in $\pi_1(\partial M)$ by trace considerations. To complete the argument, Snappy does not identify 'm008' as a CensusKnot, and so 'm008' has a fundamental group normally generated by one element such that $H_1(m008,Z)\cong Z$, but m008 is not a knot complement.

share|cite|improve this answer
Hi Neil, I took the question to be asking whether or not you can determine if the given presentation is a Dehn presentation of a knot group. This is quite different to determining whether or not the given presentation is isomorphic to a knot group. (Silly example: it's undecidable whether or not a given presentation presents the trivial group, but it's easy to decide if a given presentation is the trivial presentation.) – HJRW Apr 16 '13 at 13:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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