MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

A fibered knot is a knot with a homeomorhism on compact surface with one boundary component. On the contrary, for a given homeomorohism on a compact surface with one boundary component, is there any way to determine whether it is a monodromy of a fibered knot?

share|cite|improve this question
You might be interested in qwerty1793's answer here:… . – Marco Golla May 29 '12 at 13:17

Form the mapping torus $M$, check whether $H_1(M;\mathbb{Z})=1$ and is generated by a loop on the torus boundary. If not, it isn't a fibered knot complement. If so, do Dehn filling to produce a closed 3-manifold. Use sphere recognition algorithms to check if the result of Dehn filling is homeomorphic to the 3-sphere.

--- Edit ---

As emphasized by the comments of Agol and Igor Rivin, this is not an algorithm until one specifies exactly how to compute which Dehn filling curves in $\partial M$ need to be checked. I'll say how to do this when the monodromy $\phi : S \to S$ is pseudo-Anosov, which corresponds to checking whether $M$ is a fibered hyperbolic knot complement (something similar will work to check whether $M$ is a satellite knot whose torus decomposition has a hyperbolic component incident to $\partial M$; something different needs to be done when $M$ is Seifert fibered or is a satellite knot complement whose torus decomposition has a Seifert fibered component incident to $\partial M$).

First compute the longitude and the degeneracy curves $\ell,\delta \in H_1(\partial M;\mathbb{Z})$.

The longitude $\ell$ is the generator of the kernel of $H_1(\partial M;\mathbb{Z}) \to H_1(M;\mathbb{Z}) \approx \mathbb{Z}$.

To compute $\delta$, first compute an invariant train track $\tau \subset S$ such that some component $C$ of $S \setminus \tau$ is a crown surface containing $\partial S$; one can use standard train track algorithms to do this, such as the Bestvina-Handel algorithm. Then suspend $\tau$ to get a branched surface $B$ in $M$. On $B$, take the suspension curve of a cusp of $C$. Isotope that curve through an annulus$\times [0,1]$ component of $M \setminus B$ to get $\delta \subset \partial M$. From a more invariant but less algorithmic point of view, if one suspends the stable geodesic measured lamination $\lambda \subset S$ of $\phi$ to get an essential lamination $\Lambda$ in $M$, the curve $\delta$ is the unique one which is isotopic to a curve in a leaf of $\Lambda$.

Now compute the three curves on $\partial M$ whose geometric intersection with $\ell$ equals $1$ and whose geometric intersection with $\delta$ equals $0$ or $1$. Those are the only three possible curves along which Dehn filling can give a 3-sphere: every Dehn filling along a curve which intersects $\ell$ either $0$ times or $\ge 2$ times has nontrivial 1st homology with $\mathbb{Z}$ coefficients; and every Dehn filling along a curve which intersects $\delta$ more than $1$ time has an essential lamination.

share|cite|improve this answer
I think you want to do Dehn filling along a slope which is either the loop generating the monodromy on the boundary, or a curve intersecting it once. But there are still only finitely many choices of slope by restricting to Dehn fillings which are homology spheres, since the slope must also intersect the longitude just once. – Ian Agol May 29 '12 at 14:36
How many is "finite"? Is there nan effective bound given OP's data (the monodromy), with the homology constraint? – Igor Rivin May 29 '12 at 18:26
@ Igor: 3, since the slope has to have intersection number $\leq 1$ with both the longitude and degeneracy slope, which can therefore be at most 3 curves. – Ian Agol May 29 '12 at 21:12
@Ian, I can see only one intrinsecally well-defined curve in the torus boundary, the boundary of the surface (is this what you call the longitude?). How can you tell among the infinitely many slopes intersecting it once, which all produce homology spheres (assuming H_1(M)=Z), those that give you S^3? – Bruno Martelli May 30 '12 at 15:37
@Bruno: the longitude is the generator of the kernel of $H_1(\partial M;Z) \to H_1(M;Z)$. The degeneracy slope is defined under the assumption that the mapping class is pseudo-Anosov, so in $M$ there is an essential lamination $\Lambda$ which is the suspension of the stable measured lamination, and there is an embedded annulus $A \subset M$ which intersects $\Lambda$ in one component of $\partial A$ and intersects $\partial M$ in the other component of $\partial A$ which by definition is the degeneracy slope. – Lee Mosher May 30 '12 at 19:00

This answer gets around some of the pesky issues in Lee Mosher's answer, but it is "morally" the same his initial approach. However, this argument doesn't intrinsically use the fiberedness of the manifold, incorporating this condition might be a way to improve it.

First, build a triangulation with one torus boundary component from the monodromy information.

Jaco and Sedgwick exhibit the $S^3-K$ decision problem (Algorithm (fancy) S in the reference below) for any cusped 3-manifold with real torus boundary (for ideal triangulations one can use Jaco-Rubinstein's inflation methods to reduce to this case). First they check that this manifold is irreducible and not a solid torus, then Jaco and Sedgwick enumerate the finitely many possible $S^3$-fillings and call upon $S^3$ recognition.

William Jaco and Eric Sedgwick, MR 1958532 Decision problems in the space of Dehn fillings, Topology 42 (2003), no. 4, 845--906. arxiv version

share|cite|improve this answer

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.