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This is essentially a duplicate of: Lower bound on number of tetrahedra needed to triangulate a knot complement

Suppose a knot $K\subseteq\mathbb S^3$ is such that the complement $\mathbb S^3\setminus K$ has an ideal triangulation with $n$ simplices. What is the best bound on the crossing number $\operatorname{cr}(K)$ in terms of $n$? [perhaps we want to restrict to hyperbolic knots]

I'm half expecting that there is no such bound (probably because of twisting around a fixed unknot in the complement of $K$). In such a case, I'd like to ask for the following weaker bound in the same spirit, which I strongly suspect exists (though I'm not sure what it should be):

Suppose a knot $K\subseteq\mathbb S^3$ is such that the complement $\mathbb S^3\setminus K$ has an ideal triangulation with $n$ simplices. This of course gives a triangulation of the boundary torus $\partial N_\epsilon(K)$, and I define $A(\text{an isotopy class of simple closed curve in }\partial N_\epsilon(K))$ to be the minimum number of geometric intersections between the curve and the $1$-simplices in the triangulation of $\partial N_\epsilon(K)$. What is the best bound on $\operatorname{cr}(K)+A(m)$ (where $m$ is the meridian of $K$) in terms of $n$?

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    $\begingroup$ Shouldn't there be finitely many arrangements of $n$ simplices, therefore finitely many complements, therefore finitely many knots, therefore a bound on the crossing number? $\endgroup$
    – Will Sawin
    Commented Jan 11, 2012 at 3:03
  • $\begingroup$ @ Will: Yes, by the knot complement problem. $\endgroup$
    – Ian Agol
    Commented Jan 11, 2012 at 5:34
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    $\begingroup$ This is essentially a duplicate of this question: mathoverflow.net/questions/46149/… Translating between ideal triangulations and triangulations with boundary is a linear procedure. $\endgroup$
    – Ian Agol
    Commented Jan 11, 2012 at 5:42

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