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.


