C OMPOSITIO M ATHEMATICA
J AN M. A ARTS
L EX G. O VERSTEEGEN
The homeomorphism group of the hairy arc
Compositio Mathematica, tome 96, n
o3 (1995), p. 283-292
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283
The homeomorphism group of the hairy
arcJAN M.
AARTS1
and LEX G.OVERSTEEGEN2
1 Technical University Delft, Faculty Mathematics and Informatics, P.O. Box 5031, 2600 GA Delft,
The Netherlands
E-mail address:j.m.aarts@twi.tudelft.nl
2 University
of Alabama at Birmingham, Department of Mathematics, Birmingham,Alabama 35294, U.S.A.
E-mail address: overstee @vorteb.math. uab. edu
Received: 23 June 1993; accepted in final form 11 May 1994 Compositt’o Mathematica 96: 283-292,1995.
© 1995 Kluwer Academic Publishers. Printed in the Netherlands.
Abstract. In many pictures of planar Julia sets similar subsets appear which can be loosely described
as hairy objects, i.e., objects which contain Cantor sets of arcs such that each arc is a limit from both sides of other arcs. The Julia set of the exponential family À exp
z, 0 03BB (1/e), is a
prototype ofsuch objects. In [AO] it was shown that these objects are topologically unique and admit essentially only one embedding in the plane; the unique object is called the hairy arc. Since the hairy arc arises
in many non-conjugate dynamical systems, it is natural to study the homeomorphisms they admit. In
this paper we show that the group of homeomorphisms H of the hairy arc IHI is one-dimensional and
totally disconnected. We will also show that 1-l is dynamically much richer than the corresponding
group of homeomorphisms of the interval.
1. Introduction
In
[AO]
we defined thehairy
arc IHI and showed that it appears as the Julia set in theexponential family [D2].
Thehairy
arc istopologically unique.
In this paperwe shall
study
thehomeomorphism
group 1t of IHL In this section we shallpresent
some extensions and refinements of the results of
[AO].
In the next section weshow that 1t is one-dimensional and in the last section we
study
theconjugacy
classes of x.
A
special representation
of thehairy
arc is the sosha. First we shallgive
thedefinition of a
straight
one-sidedhairy
arc(abbreviated sosha)
in theplane.
Ahairy
arc is then defined as atopological image
of a sosha. The unit interval[0,1]
is denoted
by
E.1.1. DEFINITIONS. A sosha X is a
compact
subset of the unit squareII2
of thefollowing
form. There exists a function 1: U - E, called thelength function,
such that(1)
For all(03BE, ri)
inJI2
we have(e, ~)
~ X if andonly if 0 ~ 1(e),
(2)
The setsle e i |l(03BE)
>01
andte 1 1(e)
=01
are both dense in U andl(0)=l(1)=0.
(3)
Foreach e
in i withl(03BE)
> 0 there exist sequences(an)n
and~03B2n~n in I
suchthat an
~ 03BE, 03B2n ~ 03BE
andlim l(03B1n)
=lim l(03B2n)
=1(e).
Fig. 1. A prototype of the hairy arc IHL
The set i x
{0}
is called the base of the sosha.Throughout
the base of the sosha is denotedby
B. If x =(e, 0), 03BE
E(0, 1)
and1(e)
=0,
then we say that x is ahairless base
point.
Thepoints (0, 0)
and(1, 0)
are called endpoints
of the base.Note that the end
points
are not included in the set of hairlesspoints.
For eachx =
(03BE, 0)
in B with1(e)
>0,
the set{(03BE, q) |0
111(e)l
is called the hairat x and denoted
by hx;
thepoint x
is called the basepoint
of the hairhx
andex =
(03BE, 1(e»
is called theendpoint
of the hair.By
abuse oflanguage,
sometimeswe write
l(x)
in stead of1(e)
where x =(03BE, 0).
There is a natural order on thebase B :
(a, 0) ((3, 0)
if andonly if 03B1 03B2.
The main result of
[AO]
is that any two soshas arehomeomorphic.
It followsthat any two
hairy
arcs arehomeomorphic.
Moreover under mild conditions all soshas areequivalently
embedded in theplane.
Thetopologically unique hairy
arcis denoted
by
IHLThe main
object
of this paper is thestudy
of thehomeomorphism
group 7y of E Besides theuniqueness
of thehairy
arc thefollowing
facts from[AO]
are relevantfor the discussion. The
length
function 1 of a sosha is upper semi-continuous.Consequently
1 attains a maximum on each closed subinterval[03B1, (3]
of i andassumes on
[a, 03B2]
all values between 0 and the maximum. The set ofendpoints
E of the
hairy
arc H is dense in H and E U B is a connected set. Without loss ofgenerality
we shall assume that a sosha X has aunique longest
hair M and thatM =
{(1 2, t)|0 t 1}. If X
and Y aresoshas, they
have the hair M in common285 and there is a
homeomorphism
of X to Y that is theidentity
on M. We write b ande instead of
( 2, 0)
and( 2, 1) respectively.
The first lemma states that the action of a
homeomorphism
of thehairy
arc isdetermined
by
its action on the base. The restriction of thehomeomorphism
F tothe subset A of the domain of F is denoted
by f A.
1.2. RIGIDITY LEMMA.
Suppose F,
G E 1t.If F j B
=G j B,
then F = G.Proof.
It iseasily
seen that for every H in 1t and every x in B we havel(x)
> 0 if andonly
ifl(H(x))
> 0. From this it follows thatH(ex)
=ejj(,,)
for every xwith
1(x)
> 0.Consequently,
ifF ~
B =G ~ B,
thenF(y)
=G(y)
for everyendpoint
y of H. As the set ofendpoints
isdense,
we have F = G.1.3. COROLLARY 1l is
totally
disconnected.Proof.
IfF,
G E 1t andF :1 G,
thenby
theRigidity
LemmaF(x) :1 G(x)
for some x in B. Without loss of
generality
we may assume thatF(x) G(x).
As the set of hairless base
points
isdense,
we may also assume that x is a hairlessbase
point.
Let c be a basepoint
betweenF(x)
andG(x)
withl(c)
> 0. Then{H
E111 H(x) c}
and{H
E1t H(x)
>c}
aredisjoint neighborhoods
of Fand G
respectively
whose union is li.The reader may wonder whether Lemma 1.2 can be
improved.
Inparticular
theRigidity
Lemmasuggests
thatF(b)
= b(and
henceF(M)
=M)
forces F tobe the
identity
on the hair M. Thefollowing
lemma shows that this is false andhighlights
the fact that the structure of 1t is morecomplicated
than thesubgroup
of
homeomorphisms
of i whichkeeps
the set of hairless basepoints
invariant.1.4. EXTENSION LEMMA.
Suppose f :
M ~ M is ahomeomorphism
suchthat
f (b)
= b. Then there exists an F in 1t such thatF M
=f.
Proof.
Let X be a sosha. Define G :JI2 ~ I2 by G(x, t) = (x, f(t)).
Fromthe definition of a sosha it
easily
follows thatG(X)
is a sosha.By
the remarkpreceding
Lemma 1.2 there exists ahomeomorphism
H ofG(X)
to X which isthe
identity
on M. Then F = H o G satisfies therequired
conditions.1.5 COROLLARY. There
are five homeomorphism
typesof points
inIHI, namely (1)
endpoints of
thebase,
(2)
endpoints of hairs, (3)
hairless basepoints,
(4)
basepoints
with hairsattached, (5)
interiorpoints of hairs.
Proof.
It iseasily
seen thatpoints
in distinct sets of the list aretopologically
different. It remains to show that for any two
points
listed under the same number there is ahomeomorphism
in 1l that sends the onepoint
to the other. This followseasily
from theuniqueness
of IHI for thepoints
listed under(1)
and(3).
For thepoints
listed under(2)
and(4)
one argues as follows. Given a hair in a sosha there is asimple homeomorphism
which transforms this hair into thelongest
hair ofanother sosha. And we know
already
that any two soshas arehomeomorphic
undera
homeomorphism
that sends thelongest
hair to thelongest
hair. For thepoints
listed under
(5)
this follows from the Extension Lemma and(2).
2. Dimension of Ji
In this section we show that 7-f is one-dimensional. We shall use the
following
notation. The sum metric on
U2
is denotedby
d. The supremum metric on function spaces is denotedby
p. The function spaces we consider are 7-l and the space of allhomeomorphisms
of thelongest
hair M to itself both endowed with the supremum metric. For a sosha X we defineprojections
Jri : X - B and ?r2: X ~ Mby 7ri(e, q)
=(e, 0)
and03C02(03BE, q)
=(1, q).
Themapping
7ri is a monotoneretraction. For a =
(a, 0),
b =(03B2, 0)
in U with a03B2
we defineThe
following
lemmas are refinements of the Extension Lemma. The main feature is the control of thechange
in the second coordinate.2.1. LEMMA.
Suppose
that X and Y are soshas with commonunique longest
hair M.
Suppose
a, c are hairless basepoints of
X such that a b c and p, rare hairless base
points of
Y such that p b r.Then for
each - > 0 there existsa
homeomorphism
F : X ~ Y such thatProof.
Theproof
follows from theproof
of theuniqueness
of thehairy
arc in[AO,
Sect.3].
Thehomeomorphism
between two soshas isobtained by constructing
a sequence of
homeomorphisms
of the squaren2.
The firsthomeomorphism
of thesequence does not affect the second coordinate and can
easily
be made in such away that
(2)
is satisfied. For any otherhomeomorphism
of the sequence there is control over thechange
in the secondcoordinate;
thechange
can be madearbitrarily
small.
2.2. LEMMA.
Suppose
that X and Y are soshas. Letf :
M - M be ahomeomorphism with f (b)
= b.Then for each e
> 0 there exists ahomeomorphism
F: X ~ Y such that
F M
=f and |f
003C02(x) -
7T2 0F(x)| 03B5 for all
x.Proof.
Thehomeomorphism
F is obtained as thecomposition
H oG;
G is therestriction of the
homeomorphism
idx f : U2
~U2
to X and thehomeomorphism
H :
G(X) ~
Y is obtainedby applying
Lemma 2.1(with
the sames),
wherea = 0 = p and c = 1 = r.
287 2.3. LEMMA.
Suppose
that G is ahomeomorphism of the
sosha X to the sosha Ysuch that G(b)
= b.Then for every 03B5
> 0there exist hairless base points a
and cof
X with a b c such
that for
eachhomeomorphism f :
M ~ M withf (b)
= bthere exists a
homeomorphism
F : X - Y such that thefollowing
conditions aresatisfied
Proof. Using
thecontinuity
of G and 7r2, choose hairless basepoints
a and c suchthat a b c,
d(a, c)
-,d(G(a), G(c)) 03B5, and|Go03C02(x) - 03C02oG(x)| 03B5
for each x in a 0 c.
Define
F j ( (a 4l c)
=G (a
0c). Using
Lemma2.2,
define an extension F off
on a D c such thatF(a)
=G(a), F(c)
=G(c)
and(2)
as well as(4)
aresatisfied.
Then,
for each x in a 0 c, we haveSince
(3) holds, p(F, G) fi p( f, G M)
+ 3e. Thiscompletes
theproof
of thelemma.
2.4. THEOREM.
Suppose
that X is a sosha and1t is the groupof homeomorphisms of
X. For eachopen-and-closed
subset Uof1t
with id E U andfor
each y inMB{b, e} we have
Before
proving
the theorem we first discuss animportant corollary.
2.5. COROLLARY. For each
open-and-closed
subset Uof 71, diam(U)
1.Hence, diam
H
1.Proof.
If idEU,
the resulteasily
follows from thepreceding
theorem. Let U be anopen-and-closed
subset of 1t and suppose F E U. Theright
translation G ~ G oF-1
is anisometry
of x onto itselfsending
U to anopen-and-closed
neighborhood
of id.Proof of Theorem
2.4. We may assume that Ji is the groupof homeomorphisms
of the sosha X. The
proof
isby
way of contradiction. Assume that there is anopen-and-closed neighborhood
of id and a y =( § , q)
inMB{b, e}
such thatLet gt :
M - M be apiecewise
linearhomeomorphism
such thatgt(b)
=b, gt(e)
= e andgt(y) = (1 t),
for t in(0, q].
Then there exists 6 > 0 such that for each extensionGt of gt
to ahomeomorphism Gt :
X - X we haveGt fi.
U forall t in
(0, 6).
We will construct a
Cauchy
sequence~Fn~n
in 7Y such thatF~
=lim Fn belongs
to theboundary
ofU,
which would contradict theassumption
that U isopen-and-closed.
Put
Fo
= id. SinceFo
E U there existsbo
> 0 such thatBao(id)
C U.To construct
Fi
weapply
Lemma 2.3 with G =Fo
and E = e 1minf 80, il
to find al and ci such that for each t in
(0, ~]
there exists ahomeomorphism Gt
satisfying
Since g~ =
id M, (5) implies
thatp(Fo, G1~) 3Ei bo,
whenceGI
E U. Lettl
=sup{t
E(0, q) |G1t ~ U}. Then t’
7î. Choose tl in(t’, q)
and sl in(0, t’1)
such that
P(9tl’ g.1) g
andG1s1 ~
U. PutFi = Gtl .
By (5)
we havep(Fo, F1) 03C1(g~, 9tl)
+3 . l
andby
anargument
similar tothat in the
proof
of Lemma 2.3 for each x in X weget
Hence p(Fi, H)V) 5 p(G’l) G1s1) 1.
SinceGIl
=FI
EU,
there is a61
> 0such that B03B41(F1)
C U.Inductively
for n >0,
construct basepoints
an and cn such that an bcn,
[an+1, Cn+1]
C(an, cn)
andlim(cn - an)
- 0 andhomeomorphisms Fn -
Gtn
in U with t 1 > t2 > ... > b such that289
Since
lim tn
=t~ 03B4
>0, lim gtn
= g~ is apiecewise
linearhomeomorphism
and the convergence is uniform. Hence
~gtn~n
is aCauchy
sequenceand, by (*), ~Fn~n
is aCauchy
sequence. The sequence~Fn~n
converges to a functionP~
in the space of all continuous functions of X to itself
(with
the supremumnorm).
By (* * *), F~ XBM
is one-to-one. Moreover, sinceFoo M
= 9~, it follows thatF~
is one-to-one and hence ahomeomorphism. By (**), P~ belongs
to theboundary
ofU,
asrequired.
To prove that dim
H
1 we use some results of[OT].
Atopological
space X is called almost zero-dimensionalprovided
there exists a basis B for the open sets such thatXBcl
B is the union ofopen-and-closed
sets for every B in B. There is thefollowing
sufficient condition for aseparable
metric space to be almost zero-dimensional. Denote the closed ball of radius - around x
by C03B5(x).
From Lemma3 in
[OT]
it follows that aseparable
metric space X is almost zero-dimensional wheneverXBC03B5(x)
is the union ofopen-and-closed
sets for each x in X and everyE > 0. A
uniquely
andlocally
arcwise connectedseparable
metric space is called aR-tree. It is known that a
non-degenerate
R-tree is one-dimensional. Thefollowing
theorem is Theorem 2 of
[OT].
2.6. THEOREM. Let X be an almost zero-dimensional space. Then X embeds in the set
of endpoints of an
R-tree. Inparticular,
X is at most one-dimensional.We use this result to show that 1t is at most one-dimensional.
2.7. THEOREM. 1t is almost zero-dimensional.
2.8. COROLLARY. 1t is one-dimensional.
Proof of
Theorem 2.7. We may assume that li is thehomeomorphism
group of the sosha X. It suffices to show that for allF,
G in 71 withp(F, G)
> E there exists aclopen
set U in 1t such that G E U and U flC,(F)
=0.
Since the set of
endpoints
E of X isdense,
there exists a hairhx
withendpoint
ex and an il > 0 such that
d(F(ex), G(ex))
= e + ~. Note thatF(x) 0 G(x)
forthe base
point x
ofhx.
Suppose,
for a moment, thatJr2(F(ex)) j 03C02(G(ex)).
It iseasily
seen thatC03B5+(~/2)(F(ex)) ~ hG(x) = ~.
Hence it ispossible
to choose a and c such thatB03B5+(~/2)(F(ex))
flG(a
0c)
=QJ.
In this way it follows that there exist a and cwith a x c such that
F(a
oc)
~G(a
CIc)
=0
andeither
B03B5+(~/2)(F(ex))
flG(a 0 c)
=~
orB03B5+(~,/2)(G(ex))
flF(a 0 c) = ~.
Notice that in the first case
d(F(ex), G(y)) 03B5 + 2
for all y in a 0 c, while in the second cased(F(y), G(ex)) 03B5 + il
for all y in a 0 c.In the final
part
of theproof
we make use of the linear order ofB,
inparticular
the usual interval notations. Choose
points
p and q in B and open intervals P andQ
in B such that(5)
theendpoints
of the intervals P andQ
are hairless basepoints.
Hence
p(F, H)
> e andC,(F)
fl U =0.
In
[KOT]
it is shown that the set of endpoints
of thehairy
arc ishomeomorphic
to the set E of
points
in Hilbert space all of whose coordinates are irrational. It is not known whether any twohomogeneous
almostzero-dimensional, topologically complete,
one-dimensional sets arehomeomorphic.
Inparticular,
thefollowing problem
is open.2.9. PROBLEM. Is 1t
homeomorphic
to the setof
endpoints of H?
3.
Conjugacy
classes of 1tLet X be a sosha and let N be the subset of B of hairless base
points.
Let I be thesubgroup
of the groupof homeomorphisms
of the interval B thatkeeps
N invariant.It is easy to see that 1 is zero-dimensional.
By
theRigidity
Lemma the continuous function 03A8: H ~I,
definedby BII(H)
=H [ B
for H inH,
is one-to-one. Since li and I have differentdimension, T
is not atopological embedding.
It is well known that any two
homeomorphisms
in I withexactly
two fixedpoints
are
topologically conjugate.
For each n, there areonly finitely
manytopological conjugacy
classes ofhomeomorphisms
with n fixedpoints.
We shall establish asimilar result for
homeomorphisms
in 1t withexactly
two fixedpoints,
but inall other cases the
analogy disappears.
That thehomeomorphisms
F and G aretopologically conjugate
is denotedby
F ~ G.3.1. THEOREM.
Suppose
X is a sosha andF,
G : X ~ X arehomeomorphisms
with
exactly two fixed points. Suppose
thatF B and G [
B areorder preserving.
Then F ~ G.
Proof.
Theproof
is similar to that for the interval(Note
that ahomeomorphism
of the interval with
exactly
two fixedpoints
isautomatically
orderpreserving).
Without loss
of generality
we may assume that bothF B
andG B
areincreasing.
We define a
conjugacy H
of F to G in thefollowing
way.H (0, 0) = (0, 0)
andH(1, 0) = (1, 0).
Let xo = yo be a hairless basepoint.
For n in Z we define291 let Xn =
Fn(x0) and yn = Gn(y0).
Thehairy
arcs x o 0 x 1 and yo 0 yi are thefundamental domains
(cf. [D 1 ]).
Let H* be anorder preserving homeomorphism
ofxo
0 Xl
1 toYo 0
yl . The definition of H iscompleted by defining
H from xn Il xn+1 to yn CI yn+1by H
=Gn o H* o F-n, for n
E Z.Itiseasily
seen that H isconjugacy
of F to G.
Now we discuss a situation of
exactly
four fixedpoints.
The result shows that the structure of 1t is much richer that of JI. The set of all countable ordinals is denotedby
!1.3.2. THEOREM. Let X be a sosha and let
f :
M ~ M be ahomeomorphism
withexactly two fixed points.
Then there exist extensionsFa :
X - Xof f,
0 cx03A9,
such
that for
distinct a and(3
we have( Fa r B) ~ (F03B2 t B),
butF03B1 ~ F03B2.
Proof.
Afterreplacing f by f -1 if necessary,
we may assume03C02(f(1 2, t))
t. We construct for each 03B1 in
03A9/{0}
a soshaXa
and ahomeomorphism Ga : X03B1 ~ X a
which extendsf.
Since all soshas arehomeomorphic,
there exists for each a ahomeomorphism
pa :Xa - X
suchthat ~
M = id.By choosing dynamically
distincthomeomorphisms Ga
onX03B1,
the inducedhomeomorphisms Fa
= pa oGa
o~-103B1
on X are notconjugate.
By
symmetry it suffices to constructX a only
in[0, 1]
x[0, 1].
For each a in flthere exists a
compact
zero-dimensional subsetE03B1 of (1 8, 1 4) {1}
such that a isthe smallest ordinal with the property that the derived set of order a is
empty (See [K]).
It is not difficult to construct for each a a soshaYa
such thatE03B1 ~ {(1 2, 1)} is
exactly
the setof endpoints
of hairsof length
1 and(1 8, 0)
and(1 4, 0)
are hairlessbase
points.
LetYj
be the subset(1, 0)
Il(1 4, 0)
ofY,,*.
Defineg(x) = 2 3x + 1 6.
Notice that
g(1 8) = 4
andlimn~~gn(x)
=1.
PutThen
L a
= M~ {(0, 0)}
U~{Yn03B1|n ~ Z}
is therequired
left half of the soshaXa.
DefineGa : La - La by
Clearly (G03B1 B) ~ (Gb B).
Notice that for all(x, y )
with(x, y) ~ (1 2, 1)
we have
limn~~03C02
oGn03B1(x, y)
= b if andonly
ifGn03B1(x, y)
nEa = 0
for alln in Z. It follows that
limn~~Gn03B1(x, 1)
= e for all(x, 1)
inEa.
From theseconsiderations we conclude that the set
Za = {e}
U~{Gn03B1(E03B1)|n ~ Z}
can becharacterized
dynamically
as the set ofpoints z
such thatlimn~~Gn03B1(z)
= e. Asa is the smallest number such that the derived set of order a + 1 of
Za
isempty,
itfollows that no
Fa
isconjugate
to anyF03B2
whena =1 /3.
Acknowledgements
This paper was written while the second author was
visiting
the Technical Univer-sity Delft;
he would like to thank theDepartment
of Mathematics for itshospital- ity.
Part of the results of this paper were announced at the AMS
Meeting
inKnoxville, Tennessee, University
ofTennessee,
March26-27, 1993.
References
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[D2] Devaney, R. L.: ez, Dynamics and bifurcations, International Journal of Bifurcation and
Chaos 1 (1991) 287-308.
[K] Kuratowski, C.: Topologie I et II, PWN, Warszawa.
[KOT] Kawamura, K., Oversteegen, L. G. and Tymchatyn, E. D.: On homogeneous totally discon-
nected 1-dimensional spaces (Preprint).
[OT] Oversteegen, L. G. and Tymchatyn, E. D.: On the dimension of certain totally disconnected spaces (Preprint).