A couple of months ago I discussed a method to reduce a dynamical problem (computing the maximal rotation number of a prescribed element in a free group given the rotation numbers of the generators) to a purely combinatorial one. Now Alden Walker and I have uploaded our paper, entitled “Ziggurats and rotation numbers”, to the arXiv.
The purpose of this blog post (aside from continuing the trend of posts titles containing the letter “Z”) is to discuss a very interesting conjecture that arose in the course of writing this paper. The conjecture does not need many prerequisites to appreciate or to attack, and it is my hope that some smart undergrad somewhere will crack it. The context is as follows.
We let denote the group of orientation-preserving homeomorphisms of the circle, and let denote its universal cover, which is the group of orientation-preserving homeomorphisms of the real line which commute with integer translation. Poincaré’s rotation number is a class function which descends to . The function is a kind of “average translation distance”, defined by .
Let be a free group of rank 2 with generators and . An element is positive if it is a product of positive powers of the generators. Given a word and real numbers we let denote the supremum of under all
representations of into for which and .
The main theorems we prove are the following:
Rationality Theorem: If and are rational, and is positive, then is rational with denominator no bigger than the denominators of or .
Stability Theorem: If and are rational with denominators at most , and
is positive, there is some positive so that for all .
Both theorems can be proved rather easily by the combinatorial method described in my previous post. Roughly speaking, to compute look at all cyclic words in the alphabet with s and s, and for each one, compute a “combinatorial” rotation number associated to a discrete dynamical system. Then is the maximum of this finite list of rational numbers. A nice aspect of this proof is that it is effective, and gives the means to actually compute and draw a graph of it.
The graph of R(abaab,r,s) for r,s in
Now, although the function is nondecreasing as a function of it is discontinuous, and can jump up at a limit. We define to be the supremum of over . It is not hard to prove the following:
Lemma: is the supremum of under all representations of into for which and are conjugate to rigid rotations respectively.
Here the notation means the rotation . If we denote by the number of ‘s in , and by the number of ‘s in , then it is always true that , since we always have the representation for which and .
In contrast to the Stability Theorem, it turns out that there are words and points for which there is a strict inequality for all . We call such a point a slippery point for . The Slippery Conjecture is then the following:
Slippery Conjecture: If is positive, and is a slippery point for , then
How should one interpret this conjecture? One should think of the Rationality and Stability theorems as a kind of nonlinear analog of the phenomenon of Arnol’d tongues: when we perturb a linear system of circle rotations by adding nonlinear noise, phase locking tends to produce periodic orbits and therefore rational rotation numbers. In our context, the representation which is “maximally nonlinear” (i.e. for which differs from the most) tends to have a small denominator. If nonlinearity produces “rigidity”, then slippery phenomena should be associated with linearity.
The point (1/2,1/2) is slippery for abaab
Notice if is slippery for that must have arbitrarily large denominators as and . We can make a quantitative refinement of the Slippery Conjecture as follows:
Refined Slippery Conjecture: Let be positive, and suppose . Then
This conjecture says that the bigger the denominator of — i.e. the rotation number associated to the “maximally nonlinear” representation — the less nonlinear this maximal representation is. The Refined Slippery Conjecture implies the Slippery Conjecture.
Computer experiments support the Refined Slippery Conjecture, but we don’t have a good feel for why it might be true. But it can be translated into a purely combinatorial question, using cyclic -words, and maybe there is a clever combinatorial way to obtain the desired estimate.
Plot of against (the denominator of )
