# Probabilistic knot theory

Take a smooth closed curve in the plane. At each self-intersection, randomly choose one of the two pieces and lift it up just out of the plane. (Perturb the curve so there are no triple intersections.) I don't really know anything about knot theory, so I don't even know if I'm asking the right questions here, but I'm wondering: What is the probability that this is the trivial knot? What can we say about how knotted this knot might be, and with what probabilities? (Measure "knottedness" in whatever way you like.) More generally, can we say anything about the probability of the various possible values in the usual invariants that people use to study knots?

I only have an idea of how to approach the first question, and even then it's only by brute force. I was just playing around with the easiest cases, and I think that with 0, 1, or 2 intersections, all knots are trivial, and with 3 intersections the knot is trivial with probability 75%.

A general analysis should presumably involve calculating the probability that we can simplify using various Reidemeister moves, but I don't know how to incorporate this. I'd imagine a computer could brute-force the first few cases pretty easily (I'm not so bold as to venture an order-of-magnitude guess on whether it's the first few hundred or the first few million)...

• I don't understand a thing. Do you want to know: A) given a knot with possible self intersection, what is the probability of getting the trivial knot varying the way of resolving the self intersections, or B) the probability that given a "random" knot it is the trivial knot? By being the trivial knot I mean, of course, homotopic to. Dec 18, 2020 at 22:25

One possible route to a model of random knots would be through the braid group. Every knot can be expressed (non-uniquely) as the closure of a braid. So, for example, you could apply the braid generators uniformly $n$ times across $k$ strands, close the braid using your favorite closure, and then ask this question sensibly. I don't think you can directly ask about the $n \to \infty$ limit for the braid group, though, because I don't think there is a notion of uniform measure for that group. Actually, perhaps I will post this as a separate question, but is the braid group amenable? I would wager that in this model, the probability of having the unknot decreases very quickly with $n$ and $k$.

To test if you have the unknot, it is conjectured that you just have to check the Jones polynomial. But even this is still hard in general, unless even if you happen to have a quantum computer. :)

(Edit: Thanks Greg Kuperberg, below, for the correction.)

• Even if you have a quantum computer. arxiv.org/abs/0908.0512 Nov 4, 2009 at 4:37
• Consider the subgroup consisting of braids where all but the last strand stand still, and the last strand winds around them. This subgroup is clearly free, being the fundamental group of an (n-1)-punctured disk, and so braid groups are not amenable. Nov 6, 2009 at 23:44
• As an aside, the process of "combing" a braid is given by filtering by the cosets of such subgroups. This exhibits the pure braid group as an iterated extension of free groups; this was used by Arnol'd in his beautiful computation of the cohomology ring of the pure braid group [MR242196]. (Arnold's paper is very readable, but the translation can be a bit hard to track down; for anyone who is interested, I have some handwritten notes on the cohomology of braid groups, including Arnold's proof, on my website.) Nov 6, 2009 at 23:45
• Interesting stuff! Remark: You can find this paper of Arnold in a new, competently done, English translation in volume 2 of his Collected Works. Apr 30, 2015 at 7:19
• The braid groups have nonabelian free subgroups, so are not amenable.
– YCor
Mar 19, 2019 at 17:31

The model you propose for random knots obviously depends on the curve you draw initially, so I'm not sure this is the most natural model to consider. People have certainly looked at various probability distributions of (various classes of) knots (or knot projections). One of the immediate problems is that even just doing computer simulations is hard since determining the knot type - or just unkottedness - of a given knot diagram is highly non-trivial.

A paper which does this with Vassiliev Invariants (a certain important class of polynomial-like invariants of knots) appears in the volume "Random Knotting and Linking", edited by Millett and Summers (look at the paper by Deguchi and Tsurusaki). Other papers in this volume may interest you, too.

To the best of my knowledge, there's no really good model of random knots for which the question "what is the probability that the knot is trivial" has a known answer, except that as the number of crossing tends to infinity this probability likely approaches 0 (as anyone who left a set of mobile headphones in his pocket for more than five minutes knows).

• Is it not clear which Reidemeister moves one should apply to simplify a knot? I guess I'd imagine it could be that it needs to get more complicated before it can get any simpler. Oct 30, 2009 at 5:59
• Yes, that is correct: it is not clear which move to apply to a given not diagram in order to simplify it (whatever "simplify" means), and some trivial knot diagrams have the property that they cannot be reduced to the unknot without introducing some additional intersections first. If this hadn't been the case, this whole beautiful theory would have been reduced to a simple algorithm... Oct 30, 2009 at 6:09
• In general, deciding which Reidemeister moves to do is a very difficult problem. While algorithms have been known for a long time (at least since the work of Haken in the '60's), they are very complicated and not at all practical. In particular, there are examples where you have to introduce a huge number of new crossings before your knot can start to be simplified. An accessible source of examples is Kauffman-Lambropoulou's paper "Hard Unknots and Collapsing Tangles", available on the arXiv here : arxiv.org/abs/math/0601525 Oct 30, 2009 at 6:09
• One approach that might be interesting and which avoids hard unknots is to represent a knot as a grid diagram and see whether it admits any series of commutation moves followed by a destabilization (these are some of the grid diagram analogues of Reidemeister moves). It's not clear what such a series of moves would be or how to figure this out efficiently, but a knot is the unknot iff you can repeat this until you get the trivial 2x2 diagram. See Dynnikov's paper "Arc-presentations of links. Monotonic simplification", arXiv:0208153. Oct 30, 2009 at 13:17

I suspect that answering this question would be very difficult. A more reasonable question would be to try to understand the distribution of the various numerical knot invariants. I don't know any references off hand, but I know I've heard talks on the subject.

If you want to try to make conjectures about this kind of thing, then I highly recommend Livingston's table of knot invariants, which contains an amazing amount of data.

You should look at the Knot Atlas, which contains lots of tabulated knot invariants, although often not in as convenient form as Livingston's site.

Really, though, you want to download the KnotTheory package (presupposing you have access to Mathematica), available at the Knot Atlas. With a bit of fiddling, you can easily run experiments of the type you describe. It can calculate many invariants from the presentation of a knot.

Best of all, you should go and think about "physically realistic" models of random knots, and then try to implement such a model using one of the many knot notations the KnotTheory package understands. There are some good papers written about this subject, and even some real life experiments with strings in boxes being shaken up and down! :-)

Thank you, APR, for alerting me to this post and for referencing my papers.

I am new to MathOverflow and so I cannot yet comment on all of the posts above, so here is a conglomeration of answers with some poor formatting.

1. To Steve Flammia's point: One problem I see with creating a model for random knots using the braid group is that you most often obtain a random link and not a random knot (see Ichihara and Yoshida.) Also see a list of four papers on random walks on the Cayley graph of the braid group in the introduction to my first paper on random knots on page 2.

2. To Alon Amit's last remark: My papers on random knots (the first mentioned above and also here) give the exact probability of a knot appearing in a very particular model. The way these papers are set up, not all knots appear in this model, but the model can be extended easily.

To his more general point: Cantarella, Chapman, and Mastin look at all knot projections of up to 10 "crossings'' and compute probabilities.

1. To Andy Putnam, Scott Morrison, and Sam Nead's remarks: Slides on Nathan Dunfield's website give some really great conjectures based on experimental data.

People studying the topology of DNA use various models of random knots. Most of them have some geometric input as DNA has an actual length and doesn't want to bend too much.

I believe there are a few known "random knotting" type results out there. Not the kind of results the original poster requested, but related. Take n points in R^3 generated by a random walk, join them up (cyclicly) by straight lines. That's generically a knot. And with probability 1 (as n gets large) it's non-trivial and has a trefoil knot summand. The paper by Deguchi and Tsurusaki in "Lectures at Knots '96" provides references for these results although I've never read them in detail.

edit: Andrew Rechnitzer also has quite a few results on random knotting. He takes an approach that uses knots on cubic lattices so his results have a different flavour than the Deguchi-Tsurusaki results. His results are more of the form, `what is the expectation of the number of prime summands (and of what type) on a random long knot with a certain number of edges?'.

• Something isn't right there! You must be thinking about some limit as n goes to the infinity? Nov 4, 2009 at 4:10

There are two papers on random knots I'm aware of which haven't been mentioned yet, one by Moshe Cohen and Sunder Ram Krishnan, and the other by Moshe Cohen, Chaim Evan-Zohar, and Sunder Ram Krishnan. The model is somewhat similar in spirit to what you've described, although the initial curve has to be drawn with some care, and this process is incorporated into the model. The references are:

“Random knots using Chebyshev billiard table diagrams,” with Sunder Ram Krishnan. Topology and its Applications. 194 (2015) 4-21. arXiv:1505.07681

“Crossing numbers of random two-bridge knots,” with Chaim Even-Zohar and Sunder Ram Krishnan. Topology and Its Applications. 247 (2018) 100-114. arXiv:1606.00277

Just to reply to comments above: if you stick to "random" diagrams with at most say 30 crossings, I am confident that SnapPea will give you answers essentially immediately.

Also, to second suggestions already made, the probabilities you get will depend very sensitively on the model you choose. (Which is why this question is not going to get a real answer!)

• SnapPea won't recognise torus knots, nor non-prime knots, or knots whose complements have incompressible tori. So if your random knots have a lot of prime summands (which is common to a lot of random knot generators), SnapPea will choke most of the time. Nov 10, 2009 at 3:16
• Burton, Rubinstein, Jaco and Tillmann are getting pretty close to having efficient algorithms for recognising such knots. Nov 10, 2009 at 3:19
• Sorry to disagree, but if the JSJ decomposition has a hyperbolic piece then SnapPea will present the splitting torus in the "splitting window". It does this by tracking the degeneration of the tetrahedra, and so finds the quad type (the speed of degeneration tells you the number of quads!) SnapPea can also sometimes guess at SL(2,R) representations (ie detect Seifert fibred spaces). Nov 10, 2009 at 3:24
• Re: Burton, Rubinstein, Jaco, Tillmann. I assume that their techniques will still be at least exponential time. SnapPea is not an algorithm, but it has the virtue of being fast! There are ways to kill SnapPea (eg feed it surface bundles where the monodromy is a high power and then ask it for a Dirichlet domain), but it is pretty hard to kill SnapPea with a hand-drawn knot... Nov 10, 2009 at 3:30
• I'm not sure how you're disagreeing with me. If your knot is a connect-sum of n trefoils, n=0,1,2,3,... does SnapPea ever say anything informative? Nov 10, 2009 at 3:41

Here is a proposal which would address your concern regarding both random trajectories in a configuration space (the complex plane Cartesian-producted with itself many times minus the space of degeneracies or collisions modded out by the symmetric group) and the statistical mechanical question of an infinite tangle (the limit as the particle number is very large for our configuration space). Consider a large matrix, say of dimension 1000, with entries all Brownian motions independent of one another. Usually the first constraint imposed on random matrices is that of symmetry or Hermiticity, but recent work examines the spectrum (or empirical spectral distribution, ESD) in the complex plane and, under certain conditions, obtains a uniform measure on the unit disk as a limit.

I haven't worked out what the SDE would be for this modification of Dyson Brownian Motion, but since the distribution on the eigenvalues still imposes a logarithmic potential (in the Gaussian case), it's likely that the property of almost surely no collisions should hold. Since we are now in the plane, the eigenvalues can move around each other as others have anticipated with prospective analysis of the statistics of the trajectories.

• Another comment, equally infantile in its development as the above reply, would construct measures on knots from a process which was self-avoiding in the Riemann surfaces of non-zero genera -- thoughts, anyone? Can we build an SLE on the torus? Jun 30, 2015 at 15:48

You might be interested in this paper "The closure of a random braid is a hyperbolic link" by Jiming Ma which introduced me to a probabilistic look at knots. caveat: paywall

There could be a very hands-down, but still general, approach to this problem, that to a certain extent permits to calculate the distribution of some invariants.

Consider the configuration space of n points in the plane $$X_n := Conf_n (\mathbb{R}^2)$$: a point here consist of a tuple $$x_1, \ldots, x_n$$ of distinct points in $$\mathbb{R}^2$$. The topology here is given by being a subspace of some $$\mathbb{R}^N$$.

Now to a point in $$X_n$$ you can associate a closed curve in the plane in the following way: go from $$x_i$$ to $$x_{i+1}$$ linearly, and then go back from $$x_n$$ to $$x_1$$. This is not smooth, but any smooth immersion sufficiently close to this will do the job; in other words, if you take any lift to the space of this piecewise linear knot, the smoothly perturbed one will lift to a homotopic knot.

You can also go in the other direction: given a closed immersed curve $$C$$ in the plane, you can define the n-th truncation as the configuration of points of values of the curve at $$0, 1/n, 2/n, \ldots, (n-1) /n$$. As above, this could be not defined because points could be not distincts; but perturbing slightly some of the values $$k/n$$ will result in a (homotopic) configuration of points, since the curve is embedded. Maybe one could be even more formal about configurations by speaking directly about configurations obtained in the space by lifts of the curve, but this is heuristics.

The thing is that now we are somehow convinced that $$X_n:= Conf_n(\mathbb{R}^2)$$ is an approximation of the space of immersed curves, and it is finite dimensional. Since the (homotopic type of the) knot associated to a configuration is invariant by translations and positive dilatations (of both coordinates at the same time), we can equivalent study $$Y_n := X_n /\{ translations, dilatations\}$$. The cool fact is that now this space is finite dimensional it has a natural probability measure $$\mu_n$$, because it will be something like a subset of a sphere. Note that the subsets we wanted to measure in $$X_n$$ (configurations of points such that the associated knot has some homotopy type) are invariant by translations and dilatations, so they descend to subset of $$Y_n$$ and we can measure them with $$\mu_n$$.

The process I propose would be the following. Say you want the probability that some lifts of a plane curve as a homotopic invariant property $$H$$ and call $$A_H$$ the subsets of curves that has such homotopy type. Set $$A^{(n) }_H$$ as the subset of $$Y_n$$ of configurations (up to translations and dilatations) that yields knots in $$A_H$$. We'd like to set

$$p( C \text{ has } H) = \lim_{n \to \infty} \mu_n(A^{(n)}_H)$$

Even though this could seem very abstract, there is a possibility that one could compute $$\mu_n(A^n_H)$$ by induction on $$n$$. Indeed, the $$(n+1)$$-th knots are obtained by the $$n$$-th knots by allowing the last edge to have one "flexibility" point. By inductive hypothesis, you roughly know that the rest of the knot has 'arrangement' $$a_1$$ with probability $$p_1, \ldots,$$ arrangement $$a_k$$ with probability $$p_k$$. Now you insert this new point that forms a "V" with first and n-th point and mess up somehow the knot, and you sum up this messing up by taking into account the probability that such a new point falls there.

I know that this is very heuristic, but in practice it just means: hey let's suppose that the knots are piecewise linear".

Bye! Let me know what you think. The approximation with configuration spaces in the space is not completely justified, because goodwill calculus only works for dimension greater than 4; but I still think that this could give a cool approximation to the OP problem. If time permits, I'll try to approach the Alexander Polynomial distribution case and see what happens.