Lower bound on number of tetrahedra needed to triangulate a knot complement - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T00:20:27Z http://mathoverflow.net/feeds/question/46149 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46149/lower-bound-on-number-of-tetrahedra-needed-to-triangulate-a-knot-complement Lower bound on number of tetrahedra needed to triangulate a knot complement qwerty1793 2010-11-15T20:18:32Z 2010-11-19T02:45:18Z <p>Following along a similar line to the question asked here: <a href="http://mathoverflow.net/questions/38082/is-there-an-explicit-bound-on-the-number-of-tetrahedra-needed-to-triangulate-a-hy" rel="nofollow">http://mathoverflow.net/questions/38082/is-there-an-explicit-bound-on-the-number-of-tetrahedra-needed-to-triangulate-a-hy</a></p> <p>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.</p> http://mathoverflow.net/questions/46149/lower-bound-on-number-of-tetrahedra-needed-to-triangulate-a-knot-complement/46204#46204 Answer by Ian Agol for Lower bound on number of tetrahedra needed to triangulate a knot complement Ian Agol 2010-11-16T06:50:54Z 2010-11-19T02:45:18Z <p><strong>Revision</strong></p> <p>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. </p> <p>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 <em>twist number</em> of the alternating knot, by <a href="http://www.ams.org/mathscinet-getitem?mr=2018964" rel="nofollow">work of Lackenby</a>. 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. </p> <p>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 <a href="http://www.ams.org/mathscinet-getitem?mr=1983086" rel="nofollow">Simon King</a>. 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 <a href="http://www.ams.org/mathscinet-getitem?mr=1958532" rel="nofollow">Jaco and Sedgwick</a> 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.</p> <p>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 <a href="http://front.math.ucdavis.edu/0303.5249" rel="nofollow">Martelli and Petronio</a>. 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 $(p-1)(q-1)/2$.</p>