Following along a similar line to the question asked here: Is there an explicit bound on the number of tetrahedra needed to triangulate a hyperbolic 3-manifold of volume V?

Let $K$ be a (hyperbolic) knot in $S^3$. Let $n$ be the minimal number of crossings of any diagram of $K$ and let $M = S^3 \backslash K$ be its complement. By Moise’s theorem the 3-manifold $M$ can be triangulated by tetrahedra. But is there any known bound on the number of tetrahedra needed to triangulate $M$ as a function of $n$? I am particularly interested in any known lower bounds, but upper bounds would also be interesting.

  • $\begingroup$ There's an upper bound coming from the number of vertices in a planar knot diagram -- this comes from the algorithm SnapPea uses to construct a topological ideal triangulation of the complement. You can then subdivide appropriately to construct a proper triangulation. I believe the number of tetrahedra needed should be linear in n although I haven't thought it through precisely. $\endgroup$ Nov 15, 2010 at 21:42
  • $\begingroup$ Point of clarification -- do you want your triangulation to be a hyperbolic ideal triangulation? I think it's an open problem as to whether or not cusped hyperbolic manifolds admit hyperbolic ideal triangulations, isn't it? They have cusped polyhedral (Epstein-Penner) decompositions but they're generally not triangulations. $\endgroup$ Nov 15, 2010 at 21:45
  • $\begingroup$ And you can't have a lower bound of the number of tetrahedra needed in terms of $n$ since you can always unneccessarily complicate your knot diagram. Perhaps you want $n$ to be the minimal number of crossings in a diagram for the knot? $\endgroup$ Nov 15, 2010 at 21:46
  • $\begingroup$ Yes Ryan you are correct, by number of crossings of a knot K I mean the minimal number of crossings of ANY diagram of K. I'll change the question to reflect this. $\endgroup$
    – Mark Bell
    Nov 15, 2010 at 22:58

1 Answer 1



Let $t(K)$ be the minimal number of tetrahedra needed to triangulate a knot complement. Let $c(K)$ be the minimal crossing number of a knot.

As Ryan Budney points out in the comments, $t(K)\leq C c(K)$ for some constant $C$. One may show that this lower bound is optimal, indirectly using a result of Lackenby. For alternating knots, one obtains a linear lower bound on the hyperbolic volume of the knot complement, and therefore on the number of tetrahedra needed to triangulate the complement (as noted in Thurston's answer to the other question), in terms of the twist number of the alternating knot, by work of Lackenby. One may find hyperbolic alternating knots $K_i$ where the twist number is equal to the crossing number $c(K_i)$, and thus $ c(K_i)\leq c_1 Vol(S^3-K_i) \leq c_2 t(K_i)$ for some constants $c_1, c_2$. This shows that the linear lower bound is optimal.

For the other direction, of course, there is some sort of upper bound, since there are only finitely many knot complements with a triangulation by $\leq n$ tetrahedra, so just take the one with the maximal crossing number! We may attempt to make this relation more explicit, using a result of Simon King. I think I can show that $c(K)\leq e^{p(t(K))}$, for some polynomial $p(n)$. Inverting this inequality gives a lower bound on $t(K)$ in terms of $c(K)$. King shows that if one has a triangulation $\tau$ of $S^3$ with $m$ tetrahedra, and a knot $K$ in the 1-skeleton of $\tau$, then the crossing number $c(K)$ is bounded by $C^{m^2}$ for some $C$. Given a triangulation of the knot complement, one wants to estimate the crossing number. To apply King's result, one must extend the triangulation of $S^3-K$ to a triangulation of $S^3$. The difficulty is that the meridian of the knot may be a very complicated curve in the triangulation of the torus boundary of the knot complement. Work of Jaco and Sedgwick allow one to (in principle) estimate the combinatorial length of the meridian in the triangulation of the boundary torus. Their work also allows one to extend the triangulated knot complement along a solid torus to get a triangulation of $S^3$ with the knot in the 1-skeleton. I think one could obtain from their work an estimate on the number of tetrahedra of $\tau$ which is polynomial in the number of tetrahedra $t(K)$. Together with Simon King's result, this should give a bound $c(K)\leq e^{p(t(K))}$ ($p(n)$ a polynomial) on the crossing number in terms of the number of tetrahedra. I expect the answer to be exponential though.

Exponential lower bounds are realized, for example, by the torus knots. The crossing number is bounded linearly below by the genus. One may find sequences of torus knot complements triangulated by $n$ tetrahedra, but with genus growing exponential in $n$, and therefore crossing number growing exponentially. Estimates of the number of tetrahedra needed to triangulate Seifert fibered spaces were given by Martelli and Petronio, in Complexity of geometric three-manifolds. The point is that you can get a triangulation of a $(p,q)$ torus knot with the number of tetrahedra growing like the continued fraction expansion of $p/q$, which can be like $log(|p|+|q|)$, but the genus is $(p-1)(q-1)/2$.

  • $\begingroup$ The octahedral decomposition triangulates a diagram with $c$ crossings using $4c$ or $5c$ tetrahedra (there are two variants) so one explicitly has $t(K) \le 4c(K)$. $\endgroup$ Mar 14 at 22:00

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