I’m in Melbourne right now, where I recently attended the Hyamfest and the preceding workshop. There were many excellent talks at both the workshop and the conference (more on that in another post), but one thing that I found very interesting is that both Michel Boileau and Cameron Gordon gave talks on the relationships between taut foliations, left-orderable groups, and L-spaces. I haven’t thought seriously about taut foliations in almost ten years, but the subject has been revitalized by its relationship to the theory of Heegaard Floer homology. The relationship tends to be one-way: the existence of a taut foliation on a manifold implies that the Heegard Floer homology of
is nontrivial. It would be very interesting if Heegaard Floer homology could be used to decide whether a given manifold
admits a taut foliation or not, but for the moment this seems to be out of reach.
Anyway, both Michel and Cameron made use of the (by now 20 year old) classification of taut foliations on Seifert fibered 3-manifolds. The last step of this classification concerns the case when the base orbifold is a sphere; the precise answer was formulated in terms of a conjecture by Jankins and Neumann, proved by Naimi, about rotation numbers. I am ashamed to say that I never actually read Naimi’s argument, although it is not long. The point of this post is to give a new, short, combinatorial proof of the conjecture which I think is “conceptual” enough to digest easily.
The conjecture concerns rotation numbers of circle homeomorphisms. Given an orientation-preserving homeomorphism of the circle , Poincaré defined the so-called rotation number of
as follows. Lift
to a homeomorphism
of the line, then define
. Then define
to be the reduction of
mod
.
In fact, the conjecture is about the real-valued rotation numbers of the lifts, and can be stated in the form of a question. Given homeomorphisms of the circle, and lifts
to homeomorphisms of the line with (real-valued) rotation numbers
, what is the maximum (real-valued) rotation number of the product
? We denote this maximum as
. For elementary reasons it satisfies
for any integers
so it suffices to restrict attention to
. It is also elementary to show that
is monotone nondecreasing (though not continuous) in both
and
; from the form of the answer it suffices to determine
for
rational.
In this language, what Jankins and Neumann conjectured, and Naimi proved, is the following:
Theorem: where the maximum is taken over all rational
and
.
We show how to turn this into a combinatorial problem, which can then be solved directly. Given homeomorphisms of the circle, with rotation numbers
and
respectively, we can choose periodic orbits
for
and
for
, so that
and
, indices taken mod
and
respectively. Denote the union of the
by
.
Now, in place of homeomorphisms and
consider (discontinuous) maps
and
defined by
for
, and similarly
for
. The point is that we can adjust the dynamics of
and
on the complement of the
and the
respectively without changing their rotation number. Replacing
and
with new
satisfying
and
for all
gives
. If successive elements of
are at least
apart, then providing
for
(and similarly for
) the powers of
and
have orbits that stay a bounded distance apart.
So in order to find it suffices to study the rotation numbers of
as above. Evidently, these rotation numbers depend (in a simple way, which we will now describe) only on the circular order of the points
. We encode the circular order of the
by a cyclic word
of
‘s and
‘s, one
for each
, and one
for each
. We define a dynamical system, whose states are the letters of
. The transformation
acts by moving to the right
‘s (including the
we start on, if we start on an
) and the transformation
acts by moving to the right
‘s (including the
we start on, if we start on a
). Any
with
‘s and
‘s is said to be admissible for
. For each admissible
the transformation
acting on
has an obvious rotation number, and
is the maximum of this rotation number over all admissible
. We illustrate this with an example:
Example: To compute the admissible
words are (up to cyclic permutation)
,
, and
. Starting on the last (cyclic) letter, and successively applying
gives in the first case a rotation number of
, in the second case a rotation number of
, and in the third case a rotation number of
, so
.
With this setup established, we now prove the theorem:
Proof: We prove the desired inequality for rational and
. Suppose
is an admissible
word, for which
has rotation number
, and suppose this is maximal over all
, so that
. We can decompose
(up to cyclic permutation) into
subwords
so that if
denotes the last letter of
, then
, indices taken mod
. We can similarly decompose a cyclic permutation of
into subwords
so that
, indices taken mod
. We can choose indices so that
and
. Let
be the subdivision of
generated by the
and
subdivisions. By the definition of the transformations
and
, each
is a letter
, and each
is a letter
, so the endpoints are distinct, and the
subdivision has exactly
elements. We may permute the letters within each
without changing the dynamics, providing we keep the last letter fixed. So we can assume that each
is either of the form
(if
for some
) or of the form
(if
for some
).
Now suppose some is entirely contained in some
. Hence
coincides with some
and therefore
. We claim we can move the
string to the left, past the rightmost string of
‘s in
(note that
). Since
ignores
‘s, we will still have
after this transformation. Moreover, since each interval
contains the same or fewer
‘s after this move, we have
after this transformation; i.e. for the new word we obtain, the rotation number of
is no smaller than it was for
. So without loss of generality, if
is entirely contained in some
then we can assume that
consists entirely of
‘s; similarly, any
contained entirely in
can be assumed to consist entirely of
‘s. But this means that
contains at most
consecutive strings of
‘s and
‘s, and therefore exactly
(since each
is an
and each
is a
), so each
is of the form
. This implies that the
and
alternate, so that there is a fixed
so that
for each
. Now,
so
. Since this is true for every
, and since
, we get an inequality
. Similarly, we have an inequality
. But
, as one can see by considering the dynamics of
and
on the word
. qed
This combinatorial language turns out to be quite flexible, and one can push the techniques substantially further; Alden Walker and I are busy writing this up at the moment. One of the nice aspects of this story is that it gives rise to attractive pictures; the graph of for
is the “ziggurat” appearing in the following figure. The vertical faces of the ziggurat correspond to places where
is not continuous as a function of
.
The Jankins-Neumann ziggurat (i.e. the graph of in the unit square)
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Maybe you could label the ziggurat diagram, explaining where the r and s axes are?
I should definitely put a better picture here at some stage. Let me explain in words for now: the r and s axes are “horizontal” in the picture, and R is “vertical”. Now, it turns out that R(r,s)=1 for r+s<1; those values of R should be represented by a "flat" triangle at the front of the figure (which has been omitted). The straight (horizontal) line at the front is r+s=1; note that R(p/q,(q-p)/q) = 1+1/q, so there is a vertical line of height 1/q at each point (p/q,(q-p)/q) (these end at the vertices of the "cubes"). There is an order 3 symmetry in the figure, coming from the order 3 symmetry of F_2 interchanging a,b,AB.
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