A knot in S^3 is small if its complement does not contain a closed incompressible surface. Is it a generic property for knots, meaning that among all knots with less than $n$ crossings, the proportion of small knots goes to 1 when $n$ goes to infinity?

2$\begingroup$ Questions like this are easy to ask but very hard to answer with the present technology, since crossing number of a knot $K$ is defined as the minimal crossing number of knot diagrams for $K$. A more realistic question would be to consider sets of all knot diagrams with $\le n$ crossings. I do not know the answer in this setting. But I do know that if you consider knot diagrams with edges on the rectangular grid, then with probability approaching $1$ (as $n\to \infty$), $K=K' \# T$, where $T$ is the trefoil and $K'$ is nontrivial, so the answer is strongly negative. $\endgroup$– MishaMay 21, 2013 at 12:19

$\begingroup$ I agree, this is a "cheap" question, but I am more interested in a feeling than in a proof. In particular, do you know if it is implemented in snappy so that I can check for small crossing number? Thanks also for your remark, I should add that the knot K is prime... $\endgroup$– Julien MarchéMay 21, 2013 at 13:12

1$\begingroup$ For prime knots, I believe this result (ams.org/mathscinetgetitem?mr=1304395) implies that (for a certain sense of random knots) most are satellite knots (have an incompressible torus). So I would guess you also want to restrict to hyperbolic knots. Even in that case my guess would be that with lots of crossings you get lots of incompressible surfaces generically. $\endgroup$– John PardonMay 21, 2013 at 15:56

$\begingroup$ I do not see how one can build "primality" condition into any feasible random model for knots; same for hyperbolicity. I do not know about Snappy knot census. In any case, that would cover only a finite number of knots and is likely to be misleading. $\endgroup$– MishaMay 21, 2013 at 16:39

$\begingroup$ I do not believe either in a nice random model producing only hyperbolic knots... that's why I will try to see if Snappy can detect smallness. After all, one can see in this way that most knots are nonalternating. $\endgroup$– Julien MarchéMay 22, 2013 at 7:59
2 Answers
In general, a knot is not small.
Abigail Thompson showed that if a thin position is not a bridge position, then there exists a closed incompressible, nonperipheral, surface in the knot complement (Corollary 3).
Thin position and bridge number for knots in the 3sphere
We need to show that in general, a knot in a thin position is not in a bridge position.
Even if it wasn't, FinkelsteinMoriah showed that there exists an essential meridional planar surface in the complement of a knot in a bridge position. Hence there exists a closed incompressible, nonperipheral, surface in the knot complement.

$\begingroup$ Yes, of course, but the question was about statistics of knots, not existence of large knots. For instance, one can ask for the probability of two natural numbers to be coprime (the correct answer is $6/\pi^2$, which is approximately $0.61$). Saying that 2 and 4 are not coprime numbers would not answer such question. $\endgroup$– MishaMay 22, 2013 at 4:23

1$\begingroup$ As far as I understand, the article by FinkelsteinMoriah shows that in a certain random model (using plat closure of braids) small knots are sparse. This model is not very geometric (I would prefer the closure of a random braid for instance) but this gives some information, thanks! $\endgroup$ May 22, 2013 at 7:52
Following up on the first comment by Misha: Your question is very sensitive to the model you choose. I vaguely remember a talk by Ken Millet (http://math.ucsb.edu/~millett/) in which he gave a natural model of knot generation where the generic knot seemed to be a connect sum of $O(n)$ copies of the trefoil. If you tweaked the model, then the generic knot was the unknot.
And to reply to another comment above: one could condition on the knot being hyperbolic. Then the model is more interesting to analyze. However such a model is "unusable in practice"  you can't actually generate knots this way because the waiting time is too long.