Ziggurats and rotation numbers
Danny Calegari (joint with Alden Walker)
September 12, 2012
Homeomorphisms of S 1
We denote by Homeo+ (S 1 ) the group of orientation-preserving
homeomorphisms of the circle. This is a topological group, with
the compact-open topology.
The dynamics of an element is often indicated informally by a
picture.
elements of Homeo+ (S 1 )
Rotation number
The most important dynamical invariant is rotation number. There
is a central extension
Z → Homeo+ (S 1 )∼ → Homeo+ (S 1 )
where Homeo+ (S 1 )∼ consists of homeomorphisms of R
commuting with integer translation.
For α ∈ Homeo+ (S 1 )∼ , define rot∼ (α) = limn→∞ αn (0)/n. This
descends to rot : Homeo+ (S 1 ) → S 1 .
Warning: rot and rot∼ are not homomorphisms.
An action of G on S 1 by homeomorphisms is the same as a
homomorphism ρ : G → Homeo+ (S 1 ).
Associated to ρ is a bounded Euler class e(ρ) ∈ Hb2 (G ; Z). This
has representative cocycles G 2 → {0, 1} defined as follows.
Fix p ∈ R, and for each g ∈ G let g ∼ be the unique lift to
Homeo+ (S 1 )∼ with g ∼ (p) ∈ [p, p + 1). Then define
ep (ρ)(g , h) = g ∼ h∼ (p) − (gh)∼ (p)
Different choices of p mod Z give homologous cocycles.
Theorem (Ghys): e(ρ) determines ρ up to (semi)conjugacy.
For any G there is an exact sequence
0 → H 1 (G ; S 1 ) → Hb2 (G ; Z) → Hb2 (G ; R)
coming from the coefficient exact sequence
0 → Cb∗ (G ; Z) → Cb∗ (G ; R) → C ∗ (G ; S 1 ) → 0
When G is amenable, Hb2 (G ; R) = 0 so rot is a homomorphism in
this case, and G is semiconjugate to a group of (linear) rotations.
Each g ∈ G determines Z → hg i → G which pulls back e(ρ) to an
element in Hb2 (Z; Z) = H 1 (Z; S 1 ) = S 1 . This pullback is rot(g ).
ROTATION NUMBER AND GHYS’ THEOREM
85
For each positive fixed b, the map a → rot(V ) is continuous and monotone,
Arnol’d tongues
but is locally constant on countably many open intervals, one for each ratioa,b
nal number. The interior in the a–b plane of the preimage of a rational number
" d tongue, which gets narrower as b → 0, limiting on the
p/q is called an
When we perturb
a Arnol
linear
action (i.e. an action by rotations) by
point ( p/q, 0 ). The bigger the denominator q is, the thinner the “tongue” and the
adding nonlinear
noise, the
phase
locking
creates
periodic
orbits.
sharper it approaches
a-axis; see
Fig. 2.4. Also
see § 4.11stable
and Theorem
4.58.
b=
1
2π
b=0
a=0
a=
1
2
a=1
F IG . 2.4. Arnol" d tongues for the family x → x + a + b sin(2 π x)
Rotation number of Fa,b : x → " x + a ++ b1 sin(2πx). Regions in the
We say that two elements α , α ∈ Homeo ( S ) are monotone equivalent if
a–b plane they
where
) ∈ Q actions
are in
define rot(F
monotone
of Zwhite.
on S1 (recall Definition 2.21). It
a,bequivalent
is clear that monotone equivalent α , α " have the same rotation number. The following converse was essentially known to Poincaré:
The widths of the tongues obey a power law width(p/q) ∼ q −2.292 .
Theorem 2.77. (Poincaré) rot is a complete invariant of the monotone equivalence
class of α .
We will prove a generalization of this theorem, due to Ghys, below.
For a group G, consider the short exact sequence of cochains:
Nonabelian actions
Fix F2 = ha, bi free of rank 2.
We would like to understand
X := Hom(F2 , Homeo+ (S 1 )∼ )/semiconjugacy
We think of rot∼ (w ), as w varies over F2 , as “characters” on X .
Fixing w1 , · · · , wk ∈ F2 and projecting X to Rk with coordinates
rot∼ (wi ) gives finite dimensional “shadows” of X .
Definition: Fix w ∈ F2 , and r , s ∈ R. Let X (w , r , s) denote the
set of values of rot∼ (w ) as we vary over all F2 → Homeo+ (S 1 )∼
with rot∼ (a) = r and rot∼ (b) = s.
Let R(w , r , s) = sup X (w , r , s). Reversing the orientation of S 1
sends rot∼ to −rot∼ , so inf X (w , r , s) = −R(w , −r , −s).
Lemma: X (w , r , s) is a compact interval.
Elementary properties of R(w , ·, ·)
For w ∈ F2 let ha (w ) count the signed number of a’s in w , and
similarly for hb (w ).
(i) R(w , ·, ·) is lower semicontinuous.
(ii) R(w , r + n, s + m) = R(w , r , s) + nha (w ) + mhb (w ) for
n, m ∈ Z.
(iii) R(w , r , s) − rha (w ) − shb (w ) ≤ 2 · scl(w − aha (w ) − b hb (w ) )
(Milnor–Wood inequality).
The difference between the two sides in (iii) measures the extent to
which Hom(F2 , Homeo+ (S 1 )) fails to detect all of Hb2 (F2 ; Z).
A word w is positive if it contains a’s and b’s but no a−1 or b −1 .
Theorem (Rationality): If w is positive and r or s are rational,
R(w , r , s) is rational with denominator ≤ the minimum of the
denominators of r and s.
Theorem (Stability): If w is positive and r or s are rational,
R(w , ·, ·) is locally constant on [r , r + ) × [s, s + ) for some
positive . If R(w , r , s) = p/q then ≤ 1/q.
In other words, there is an open dense subset of the r –s plane
where R(w , ·, ·) is locally constant and takes values in Q.
The diameter of these locally constant regions goes to 0 as a
power law in the denominator of the constant value.
Question: Does the set where R(w , ·, ·) is locally constant have
full measure?
Example: The ziggurat (i.e. the graph of R) for w = ab.
Example: The ziggurat for w = abaab.
Example: The ziggurat for w = abaabaaab.
Example: The ziggurat for w = abbbabaaaabbabb.
Example: The “ziggurat” for w = abAB.
Theorem (Isobar): If w is positive, for any rational p/q the set of
r , s such that R(w , r , s) ≥ p/q is a finite sided rational polygon,
whose boundary consists of finitely many horizontal or vertical
segments.
The same is not true of the “level set” R(w , r , s) = p/q.
Let w be positive, and let R(w , r −, s−) denote the limit of
R(w , r 0 , s 0 ) as r 0 , s 0 approach r , s from below.
Lemma: R(w , r −, s−) is the supremum of rot∼ (w ) under
representations for which a, b are conjugate to rotations Rr , Rs
respectively.
Definition: (r , s) is slippery if R(w , r 0 , s 0 ) < R(w , r −, s−) for
r 0 , s 0 strictly less than r , s.
Slippery Conjecture: If (r , s) is slippery, then
R(w , r −, s−) = ha (w )r + hb (w )s
i.e. the “optimal” representation is conjugate to an action by rigid
rotations (“linear”).
Example: (1, t) and (t, 1) are slippery for every positive w .
(1/2, 1/2) is slippery for abaab, abaababb and abaabaaaabb.
Note: every representation for which a, b are conjugate to R1/2 is
semiconjugate to an action by rotations, because Z/2Z ∗ Z/2Z is
amenable.
Refined Slippery Conjecture: Suppose w = aα1 b β1 · · · b βm is
positive. If R(w , r , s) = p/q then there is an inequality
R(w , r , s) − ha (w )r − hb (w )s ≤ m/q
Note: this conjecture is true if the denominator of r or s is q; i.e.
if the denominator of R(w , r , s) is as big as possible.
Plot of q versus R(w , r , s) − ha (w )r − hb (w )s for w = abaab and
w = abaabbabbbababaab
The Rationality, Stability and Isobar theorems are proved by
reducing the computation of R(w , p1 /q1 , p2 /q2 ) to combinatorics.
a has a periodic orbit of order q1 , b has a periodic orbit of order
q2 . The configuration of these points in S 1 is encoded by a cyclic
word of length q1 + q2 in a two-letter alphabet {X , Y }, with q1 X s
and q2 Y s.
q1 = 5, q2 = 3; cyclic word XYYXXYXX
Suppose xi is the periodic orbit of a, so that a : xi → xi+p1 . Define
a+ by a+ : (xi−1 , xi ] → xi+p1 . Define b + similarly, and let w + be
obtained by substituting a+ , b + for a, b in w .
Since w is positive, rot∼ (w + ) ≥ rot∼ (w ). Conversely, a+ and b +
can be uniformly approximated by homeomorphisms with the same
periodic orbit, away from (xi , xi + ).
Hence R(w , p1 /q1 , p2 /q2 ) is the supremum of rot∼ (w + ) over all
XY -words of “type” q1 , q2 .
Example: We compute R(ab, 2/3, 1/2). There are two cyclic
XY -words of type 3, 2, namely XXXYY and XXYXY . An orbit
under successive applications of letters a, b is illustrated in green
and red respectively.
X Y Y |XXX
X Y Y |XXX
XYY · ··
· · · XXX
has rotation number 1, whereas
Y XY |XX
X YXY
Y |XXY X Y · · ·
· · · XXY X Y |XXY
has rotation number 3/2.
Hence R(ab, 2/3, 1/2) = 3/2.
An elementary combinatorial argument proves:
ab Theorem:
R(ab, r , s) =
sup
(p1 + p2 + 1)/q
p1 /q≤r , p2 /q≤s
This theorem was conjectured in 1985 by Jankins–Neumann, and
proved in 1994 by Naimi; it is the last step in the classification of
taut foliations of Seifert fibered spaces.
Note: the optimal XY word of type q, q for ab is (XY )q .
Proof of ab Theorem: Let W be a hypothetical XY word of type
q1 , q2 maximizing R(ab, p1 /q1 , p2 /q2 ). Every a hop ends on an X ,
and every b hop ends on a Y . Look at a periodic orbit (in red and
green):
Y XYYXYXY
Y XY X YYYXXY
Y YXXY X XXY · · ·
· · · YXXY X XYXYXY
Strings complementary to the orbit can be rearranged however we
like, since the dynamics doesn’t see their order:
X XXXYY Y XXXYYY Y YXX
X XXYYY Y YYXXX
X XXY · · ·
· · · YYXXX
Y or X Y for Y X can only increase rot∼ (w + ):
Swapping Y X for XY
X XXXXXXYY Y YYY Y YXX
X XXYYY Y YYXXX
X XXY · · ·
· · · YYXXX
If there is a string Y Y mY then an a hop starting at either Y will
land on the same X ; similarly if there is a string X X mX . So W
must have exactly q consecutive strings of X s and Y s where
rot∼ (w ) = p/q, each of which contains exactly one X or Y .
If si are the lengths of successive X strings,
p1 ≥ si+1 + si+2 + · · · + si+j
P
for every i and for some fixed j. Since
si = q1 , we get an
inequality p1 /q1 ≥ j/q. Similarly, p2 /q2 ≥ (p − j − 1)/q. This
completes the proof.
Nonpositive words
If w is arbitrary and r , s are rational, there is some positive word
w 0 (w , r , s) and an integer N(w , r , s) so that
R(w , r −, s−) = R(w 0 , r −, s−) − N
Since we can compute the right hand side (numerically, at least),
we can compute the left hand side.
The Slippery Conjecture implies the following:
Conjecture: For any w and rational r , s, R(w , r −, s−) is rational.
The interval game
Unlike the positive case, adjusting the dynamics of a in the
complement of a period orbit can make things both better and
worse.
26
DANNY CALEGARI AND ALDEN WALKER
a−1
wi
a
| {z }
I
| {z }
wi−1 (I)
Figure 6. Adjusting the dynamics of a on I.
This suggests the following (one-person) dynamical game, called
The Interval Game.
We are given a finite subset ϕi of Homeo+ (S 1 ) (belonging to the
enemies) and a single φ ∈ Homeo+ (S 1 ) (belonging to the player).
The player tries to find a winning interval I . This means a closed
interval I ⊂ S 1 and a positive integer n so that
(i) φn (I + ) ∈ I ; but
(ii) φj (I + ) is not in ϕi (I ) for any i and any 0 ≤ j ≤ n.
In our context φ = w , a word in F2 ending in a, and the ϕi are the
initial subwords wi ending in a−1 .
The dynamics of a can be adjusted on a winning interval to create
a period orbit for w which escapes from the influence of the enemy
a−1 s.
The key case to consider is when φ = Rθ , an irrational rotation.
Proposition: Suppose θ satisfies a certain Diophantine condition,
and ϕi are all C 1 and there is some p such that ϕ0i (p) 6= 1 for all i.
Then there is a winning interval.
Morally speaking, the interval game can be won provided the
enemies are “sufficiently nonlinear”. The Diophantine condition on
θ is that it is approximated “well” from below by infinitely many
rationals.
Theorem: In the case of a single enemy ϕ there is always a
winning interval unless ϕ is a rigid rotation Rψ and (ψ, θ) is in the
following subset of [0, 1] × [0, 1] (whose complement is open and
dense):
Corollary: If w is of the form va−1 b n a for some v containing no
a−1 , and r is rational, then either R(w , r , s) is rational, or
(R(w , r , s), ns) is in the “bad subset” of [0, 1] × [0, 1].
Unwinnable instances of the interval game are highly constrained;
these constraints force subwords to commute on big subsets of S 1 ,
which will tend to make rot∼ (w ) small. Therefore we make the
following:
Conjecture: For any w and rational r , s, R(w , r , s) is rational.
Conjecture: For any w ∈ [F , F ], maxr ,s R(w , r , s) is rational.
Comments and Questions
rot∼ (ababaabbaBBAABABAbababaabbABBAABABAB) = 0 for
every F2 → PSL(2, R).
Question: is there a nontrivial w ∈ F2 such that rot∼ (w ) = 0 for
every F2 → Homeo+ (S 1 )?
Every t in the interior of X (w , r , s) is realized by a C ω
representation.
Question: is R(w , r , s) realized by a C ω representation?
Reference: D. Calegari and A. Walker, Ziggurats and rotation
numbers; J. Mod. Dynamics 5 (2011), no. 4, 711–746