A NNALES DE L ’I. H. P., SECTION A
J. M. G AMBAUDO
S. V AN S TRIEN C. T RESSER
The periodic orbit structure of orientation preserving diffeomorphisms on D2 with topological entropy zero
Annales de l’I. H. P., section A, tome 50, n
o3 (1989), p. 335-356
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335
The periodic orbit structure of orientation preserving diffeomorphisms
onD2 with topological entropy
zeroJ. M. GAMBAUDO
S. VAN STRIEN
C. TRESSER
Dept. de Math., Univ. de Nice, France Math. Dept., Techn. Univ. of Delft, the Netherlands
Dept. de Phys., Univ. de Nice, France
Vol. 49, n° 3, 1989, Physique ’ théorique ’
ABSTRACT. - A set of
periodic
orbits of a map on the disc is called"hereditarily
rotationcompatible"
if the orbits are like those of rotationsor
repeated
rotations above rotations. We prove C 1 orientationpreserving embeddings
of the disk have zerotopological
entropy if andonly
if itsperiodic
set is of this form. Moreover a tree structure on this set is introduced: someperiodic
orbitsimply
the existence of others.RESUME. 2014 L’ensemble des orbites
periodiques
d’uneapplication
dudisque
dans lui-meme estappele
« hereditairementcompatible
a unerotation »,
si ces orbites sont semblables a celles des rotations ou de rotations successivement construites sur d’ autres rotations.On montre
qu’ un plongement C 1
dudisque préservant
1’ orientation a uneentropie topologique
nulle si et seulement si 1’ensemble de ses orbitesperiodiques
est de cette forme.De
plus,
on introduit dans cet ensemble une structure d’arbres. Certaines orbitesperiodiques impliquent
l’existence d’autres orbitesperiodiques.
Annales de l’Institut Henri Poincaré - Physique theorique - 0246-0211
Vol. 49/89/03/335/22/$4,20/(e) Gauthier-Villars
GAMBAUDO,
1. INTRODUCTION
In the last decade several papers
appeared giving
relations between thetopological entropy
of a map and the existence ofperiodic
orbits. Forcontinuous interval maps this
relationship
was first discoveredby
A. N. Sarkovski
[Sar],
see also forexample [B. G. M. Y. ].
In this case theexistence of for
example
aperiodic
orbit ofperiod
threeimplies positive topological entropy.
For orientationpreserving homeomorphisms
of thedisc the situation is much less clear: rotations of the disc have
topological entropy
zero and can havearbitrary periods.
Even so one cangive
necessary and sufficient conditions related to the
"linking"
of aperiodic
orbit which
guarantee
that it can occur as aperiodic
orbit of a home-omorphism
of D2 with zero entropy, see[Kob], [Bo 1].
In this paper we will consider orientation
preserving diffeomorphisms
of D2 with zero
entropy,
but rather thanjust considering single orbits,
asin
[Kob]
and[Bo 1],
we will describe a tree structure on the set of allperiodic
orbits. This will enable us togive
acomplete description
of allperiodic
structures realizableby
orientationpreserving diffeomorphisms
of the two-disc with zero
topological entropy.
Thisdescription
is rathergeometric
and in terms of a notion which we call"hereditarily
rotationcompatibility"
and a"parent-child" relationship.
The
simplest diffeomorphisms
of the disc arerigid
rotations. The main theorem in this paper states that the set ofperiodic
orbits of a C1diffeomorphism
of the disc with zero entropy has a "rotationcompatible"
structure, i. e., it can be described
by
a tree structure on the set ofperiodic
orbits so that the map is a
composition
of a number of "rotations builton to each other". In this tree the orbits which are lower in the tree circle around the mother orbits
(like
a discrete version of moonscircling
aroundplanets).
InFigure
1 we have drawn anexample
where the set ofperiodic
orbits consists of three orbits of three consecutive
generations. Later,
whenwe
give definitions,
thispicture
will be madeprecise.
The result
implies
inparticular
that if one or more of the closed curvesassociated to
periodic
orbits via thesuspension
off
are too "linked"(we
will
give
aprecise
definitionbelow),
then thetopological entropy
off
ispositive,
seeFig.
2. It is remarkable that this theoremgives
necessary and sufficient conditions: thetopological
entropyof f is
zero if andonly
if theset of
periodic
orbits are"hereditarily
rotationcompatible".
2. STATEMENT OF RESULTS
Let
D2 = ~ (x, y); x2 + y2 1 ~. By
a disc inDc:D2
we mean the closedset bounded
by
asimple
closed Jordan curve.Let f be
ahomeomorphism
FIG. 1. - An hereditarily rotation compatible orbit of generation 3 and period 5 x 3 x 4.
of
D2. The f -orbit A, f (A), f 2 (A), ...
of some set Ac D2,
is denotedby (O (A); f )
or, if no confusion can arise0(A).
We say that the set A isperiodic of period
andA, f (A), ... , f n -1 (A)
aredisjoint.
Vol. 49, n° 3-1989.
GAMBAUDO,
If
A c D2 and f "(A) = A
’ then we denote " the restrictionof f n to A simply
FIG. 2. - A map f with a suspension as drawn on the left must have positive topological entropy. The situation on the right does not imply anything about the topological entropy.
One can formalise the
"linking"
of aperiodic
orbitusing
braids or,equivalently, by
theisotopy
class ofFirst we say when a
periodic
orbit is called o rotationcompatible.
Take . a
periodic region A c D2
ofperiod
o We say that(O(A);f)
is a rotation
compatible f -orbit
ifClearly
aperiodic
orbit of a rotation of D2 is rotationcompatible.
Moregenerally,
it will later turn out that A is rotationcompatible
iff thereexists ~~ 1
and g
suchthat f ,
g :D~20140(A) -~ D2-O(A)
areisotopic
andsuch that and
are
conjugate,
where R is a rotation of D2 of order n(this
meansR n = id)
and
Do
aperiodic
disc inD2 (also
ofperiod n).
In order to define the notion of
hereditarily
rotationcompatibility,
weconsider renormalizations.
Roughly speaking,
a renormalization is aniterate of a map restricted to some rotation
compatible
subset.Indeed,
wesay that
/:D~20140(/?)-~D~20140(p)
can be renormalized if theperiod
ofO(p)
is theproduct
of twointegers
r, s > 1 and if there exists aperiodic topological
discDo
in D2 ofperiod
r such that each of the discsf i (Do)
contains s
points
ofO(p)
in its interior and(O(Do); f) is
rotationcompatible.
If this holds then we say that
(0(/?)
nDo; f r I Do)
renormalizes(0(/?); f ).
FIG. 3. - A renormalizable orbit.
Now we define
inductively (0 (/?);/)
to behereditarily
rotationtible as follows,
(0(~)?/)
ishereditarily
rotationcompatible
ofgeneration
0if p
is afixed
point.
It hasgeneration
1(0(/?); f )
is rotationcompatible.
The orbit
(O ( p); f )
ishereditarily
rotationcompatible
ofgeneration
k > 1 if there exists a
homeomorphism g
such thatf , g : D2 - O ( p) -~ D~-0(~)
areisotopic
such that
Vol. 49, n° 3-1989.
GAMBAUDO,
-
(O ( p); g)
can be renormalized to Thatis,
up toisotopy f rotates
a discDo containing
more than onepoint
ofO(p)
inits interior.
-
(O ( p)
nDo)
ishereditarily
rotationcompatible
ofgeneration
k-l.
Let
htop (f)
be thetopological
entropyof f.
The next result tells us that each of theseperiodic
ishereditarily
rotationcompatible.
THEOREM A. - Let an orientation
preserving homeomorphism of D2. If
then
( * *)
everyperiodic
orbitof f
ishereditarily
rotationcompatible
If f is a
diffeomorphism,
then(* *) is equivalent to (* *).
Let us describe these definitions in words.
(O ( p); f )
is rotation compa- tible itf acts
as a rotation onO ( p). (0(~);/)
ishereditarily
rotationcompatible
ofgeneration k
if there exist a map g in theisotopy
class of/:D~20140(~)-~D~20140(~)
and ag-periodic
discDo containing
at leasttwo
points
ofO(p)
in its interior suchthat g
rotates this disc and such that the restriction toDo
ishereditarily
rotationcompatible
ofgeneration k -1,
seeFig.
2.Theorem A is not difficult to prove and is not the main result of this paper. This theorem was
essentially
knownby
P.Boyland,
T.Kobayashi
and J. Smillie and others. For
example,
for the same class ofmappings,
P.
Boyland [Bo 1]
hasproved
that ifh ( f ) = 0
and p is aperiodic point
off whose period
is an oddprime
number thenO(p)
is rotationcompatible.
T.
Kobayashi, [Kob], proved
a theoremequivalent
to TheoremA,
but his result isphrased
in terms of a notion calledgraph
link. This notion is definedusing
thelanguage
of three dimensionaltopology.
J. Llibre and R. S.MacKay
announced a similar resultconsidering
also orientationreversing diffeomorphisms.
Remark 1. - We should also
point
out the connection with a paper of J.Morgan.
Let ’t be the solid torus D2 xst.
J. W.Morgan
hasproved
that if K c ’t is an
attracting
closed orbit for a Morse-Smale flow withoutsingular points
on ’t, then K is an iterated torus knot. Here a knot Kc:T is called an iterated torus knot if there is a sequence of solid tori ... ~T~ such that the core of ’to isunknotted,
the core of ii lieson a torus in which is
parallel
to for f~1,
and the coreof ir
isK. It is not hard to show that Theorem A
implies Morgan’s
result for flows which are thesuspension
of Morse-Smale maps.Remark 2. - We should note that the situation on other manifolds is
more
complicated.
Forexample
J. Smillie[Smi]
shows that for surfaceswith genus 145 there exists an
isotopy
class ofdiffeomorphisms
whichcontains two maps
f
i withtopological
entropy zero.f 1
has aperiodic point
ofperiod
5 andperiodic point
ofperiod 7,
but every map in thisisotopy
class which hasperiodic points
ofperiod
5 and ofperiod
7has
positive entropy.
Moreover thisisotopy
class contains no maps withzero
entropy
which have a fixedpoint.
3. A TREE STRUCTURE ON TH~ SET OF PERIODIC ORBITS A much more
precise description
of the set ofperiodic points
can begiven. Indeed,
there exists apartial ordering
on the set ofperiodic
orbits.Definition of
family relationship
(O(q);f)
is thepersistent
parentof (0(/?);/))
if for eachisotopy
~:D~-0(~)-~D~-0(~)
there exists a curve such that qo=q,and qt
is aperiodic point of ft of exactly period
and suchthat
- there
exist gt isotopic
toand a
topological
discDo, depending continuously
on t,containing preci- sely
onepoint
of(O (qt); ft)
and at least twopoints
of0(/?)
in itsinterior,
such that
-
Do
isgt-periodic
ofperiod
n =# O ( q)
-
Do
isconjugate
toa periodic
rotation ofD2.
If these conditions
only
hold forf ( instead
forft)
then(0(~):/)
iscalled the parent of
(O(~); f ).
For convenience we also call(O(~); f ~
thechild of
(O (q); f )
if(C~ (q); f )
is theparent
of(O ( p); f ).
Aperiodic
orbit(O (q); f )
is an ancestor of aperiodic orbit (0(/?); f )
if there is a chain ofperiodic
orbits , . ,so that
Oi
is aparent
ofo= + 1.
"Example. -
The orbit( Q ( p); f ) of period
15 inFigure
4 is the childof the orbit
(O (q); f )
ofperiod 5..
THEOREM B. - ~et
f
be aC1 diffeomorphism of D2
withtopological
entropy zero and isolated
periodic paint.s.
Then everyhereditarily
rotationcompatible
orbito, f’ generation
k hasa.persistent hereditarily
rotation com-patible
parentof generation
k - l; ,Vol. 49, nO 3-1989.
This theorem is useful in
particular
for bifurcationproblems.
In afamily
of
diffeomorphisms
if aperiodic
orbitpersists through
a bifurcation then itsparent
orbit must alsopersist.
Finally
we define a notion which expresses whether orbits can be consideredspatially separate
or not.Definition of
disjointness
of orbitsTwo orbits
(0(~);/)
and(0(~);/)
are calleddisjoint
if there are twodisjoint simple
closed curvesCt, C2 bounding
twodisjoint
discsDt, D2
containing O(p) respectively O(q)
and such that(In particular
two fixedpoints
aredisjoint.)
5. - Two disjoint orbits.
O(p)
andO(q)
lie nested if there exist a map g in theisotopy
class off:D2-O(p)~D2-O(p)
andg-periodic
discsD2~D1
such thatD 1 "D2
is ag-periodic annulus,
O(q)
is contained in the orbit of the interior of this annulus andO(p)
in the orbit of the interior of the discD 2’
seeFigure
6.THEOREM C. - Let
f
be a C1diffeomorphism of
D2 withtopological
entropy zero and isolatedperiodic points.
Then(a)
two orbitsO1
and02
may have a common child but thenO1
and02
have the same
period
and the samegeneration
and any parentof O1
is alsoa parent
of 02.
(b)
LetO1
and02
beperiodic
orbits andO1 ~ 02.
Then there are threepossibilities:
they
aredisjoint,
orthey
lie nested and have a common ancestor, orone
of
these orbits is the parentof
the other orbit.Vol. 49, nO 3-1989.
GAMBAUDO,
Theorems
B
and Cgive
a veryprecise picture:
allperiodic
orbits of adiffeomorphism
ofD2
can beorganised
in a tree: eachorbit
ofperiod
~~ 2 has a
parent
orbit. ~ ~’ ’
.
It is not difficult to express the notion of rotation
compatibility,
child-parent
relationship
anddisjointness
in terms of braids. Inparticular
thereexists a finite
algorithm
to decide whether or not aparticular
orbit ishereditarily
rotationcompatible,
see[Bi].
’
Remark. -
Probably
thesetheorems
are also valid forC°
homeomor-phisms.
The reason we confine ourself toC1 diffeomorphisms
is that weneed a result from
[A.F.]
which isonly proved
forC1
maps. It seemslikely
however that the results in[A.F.] are
also valid forCO
maps.3.
BACKGROUND
Let us introduce some notation: We say that two
diffeomorphisms f ,
g : F ~ F of some
topological
space F areisotopic,
if thereexists
a continu-ous arc
Ft:
F -+ F ofdiffeomorphisms
such thatFo = f , F 1= g.
We abbrevi-ate this
(One
can prove that twohomotopic diffeomorphisms
onsurfaces are also
isotopic.)
The main
ingredient
of ourproof
will be thefollowing
theorem ofThurston. Assume that F is an
orientable
surface withnegative
genus which is closed withboundary,
or which is closed except that it has finite number of punctures. In this paper we considerhomeomorphisms
ofD2.
Since D2 is open, we can consider D2 as
S2 - { ~}
and ahomeomorphism
of
f : D2 ~
D2 induces ahomeomorphism
ofS2.
In this paper we will choose Fequal
to D2 minus a finite set; if this finite set consists of at least twopoints
then F hasnegative
genus. Take thehyperbolic
metricon F with constant curvature -1. A curve is essential if it is not
isotopic
to a
point,
nor ispuncture
orboundary parallel.
3.1. THEOREM
Nielsen, Thurston [Th],
see also[Ca]. - Let f:
F -+ F bean orientation
preserving homeomorphism of a
closed orientablesurface (with finite
numberof punctures).
Then(at least)
oneof
thefollowing
occurs.1.
f is periodic ;
i. e.isotopic to
theidentity
mapfor
some n.2.
f is reducible; i.e.
there exists ahomeomorphism that g
leaves some
finite
unionof disjoint simple
essential closed curvesinvariant.
3. there exists such
that for
any map has(In fact f is isotopic
to apseudo-Anosov map
but we will not ne~ed todefine
thisnotion in this
’ ’
The next result describes the
periodic
case and tells us thatif f is periodic
then
f
isisotopic
to ahisometry :
’ - . _3.2. THEOREM See
Kerckhoff [kerc]
an ’orientationpreserving homeomorphism. If fn
isisotopic
to id thenisotopic
to anisometry of F. -
-‘
~..
~
~
~
.
Finally
we will use thefollowing
theorem which will enable us to conclude that if onemap
has aperiodic
orbit then another map in thesame
isotopy
class will have an"isotopic" periodic
orbit. Moreprecisely,
let
fo, f 1:
F -+ F bediffeomorphisms
and po and plbe
fixedpoints
off o
and
f 1 respectively.
Thenf o) and (pl, f 1)
are in the same Nielsen class if there exist anisotopy ft between f0 and f 1
and a curve c :[0, 1]
-+ Fjoining
po and /)i such that(0, l)3f-~(~)) ~ homotopic
toc (t)
inpl ~. They
are in the same~,strong
Nielsen class if there exist suchan
isotopy ft and
a continuous curvec (t)
such for allt E [0, 1 ].
‘ .Vol. 49, n° 3-1989. - .
Let p
be an isolatedperiodic point
ofperiod
n. Then(p, f )
is unremovableprovided
that forany g isotopic
tof
has aperiodic point q
such that(p, f )
and(q, g)
are in the same strong Nielsen class.3.3. THEOREM
(Asimov
andFranks, [A.F.]). - Let f:
F ~ F be adiffeo- morphism
with apoint p
which is theonly point in
its Nielsenclass with respect to the
map fn
and non-zeroLefschetz number, L (p, /")~0.
Then
(p, 1)
is unremovable.4. PROOF OF THEOREM A
Everywhere
in the remainder of this paper we will assume thatf is
an
diffeomorphism
of the disc with zerotopological
entropy. The ideaof the
proof
of Theorem A issimply
toapply
Theorem(3.1)
onf : D~20140(/?) -~
and use induction in the reducible case.4.1. LEMMA. - Let 0 be a
finite
subsetof D2 and R : D2 -
O ~D2 -
0an
isometry.
Then R extends to ahomeomorphism R : D2 ~ D2
which isconjugate
to a rotation ofD2.
Proof. -
SinceR:D~-0-~D~-0
is anisometry,
it can be extendedto a
homeomorphism R
ofD2.
The Lemma then follows from a theorem of Brouwer, see[Bro], [Kere]
and[Ei],
which says that ifg : D2 -~ D2
satisfies
then g
isconjugate
to a rotation.Q.E.D.
Remark. - Since R is an
isometry,
one can prove this lemma alsoeasily
withoutusing
Brouwer’s result: Denote p the fixedpoint
ofR :
D 2 ~D 2.
Let R’(/?): T p
F ~Tp
F be the derivative of R at p.Identify T p F
with R 2 and defineS(x)=/’(p)x.
Then S is a linear rotation ofD2.
Consider
geodesics
Y9 in Fthrough
phaving angle
e with some fixed linethrough p,
where8e[0, 7t].
Thesegeodesics
will either end in 0 or inaD2.
Then the
conjugacy
h between Rand S is definedby choosing
anappropri-
ate
homeomorphism h : D2 ~ D2
such that is contained in the line(through
theorigin 0~D2
which hasangle
ø with thex-axis).
Q.E.D.
4.2. LEMMA. - Let
: D2 ~ D2 topological diffeomorphism and O(p) be a periodic
orbitof period
n. Assumethat f: D2 - O ( p) ~
is periodic.
Thenl./":D~-0(~)-~D~-0(~)
and are iso-topic;
2.
fis isotopic to
anisometry (where
we take thehyperbolic
metric on
(0 (/?);/) if n >_ 2
and the usual metricifn= 1);
3.
D~20140(~)-~D~20140(~)M
rotationcompatible;
Proof. - Since f:D2-O(p)D2-O(p)
isperiodic, according
toTheorem
(3.2)
this map isisotopic
to anisometry
R ofD~20140(/?).
According
to Lemma(4.1),
R extends to a mapR : D2 ~ D2
which isconjugate
to a rotation onD2.
Since hasperiod n
it follows thatR"
fixes each
point
of0(/?)
and it follows thatRn=id.
This proves1,
2 and 3.Q.E.D.
4.3. LEMMA. - Assume
that f: D2 ->
D2 is ahomeomorphism of D2
withzero
topological
entropy. Thenevery periodic
orbitO (p) hereditarily
rotation
compatible.
Proof. -
Take aperiodic
orbit ofperiod
n. Ifn E ~ 1, 2}
thenO (p)
istrivially
rotationcompatible.
So let us assume that we haveproved by
induction that every
periodic
orbit ofperiod
n is rotationcompatible.
We will
apply
the theorems of section 3 ofF=D~20140(/?).
So take thehyperbolic
metric onD2 - O ( p).
Sinceh (f ) =0,
Theorem(3.1) implies
that
only
one of thefollowing
two cases can occur:Case 1. -
/:D~-0(~)-~D~-0(~)
isperiodic.
Then Lemma(4.1) gives
that(0(/?); f )
is rotationcompatible.
Case 2. -
f : D2 - O ( p) ~ D~-0(/?)
is reducible. Then there exist r~1,
a
simple
essential closed curveCo
and a map g in theisotopy
class of/:D~-0(~)-~-0(~)
such thatand
f=0,l,...,r20141
arepairwise disjoint.
For each
f=l,2,...,r,
is an essentialsimple
curve and therefore there exists s > 1 such that for each
~==0,... r -1
thecurve
C1
bounds a discD~ containing precisely s points (where
1 snbecause
C~
issimple)
and eachpoint
ofO(p)
is contained in one of the discsDo, Dt, ... , Dr _ 1.
Take Sincey (Co)
isisotopic
toCo
inD2 - O ( p),
it follows that all of thepoints
x,fr (x), f 2r (x),
... , arecontained in
Do.
From this and since each of the discsD~
containspoints
one has n=sxr. Notice that the discsDi, 0 _ i r,
arepairwise disjoint.
Choose a
’periodic’
essentialsimple
curveCo
inD2 - O (p)
as abovesuch that there is no
periodic
essentialsimple
curveCo bounding
morepoints
ofO(p)
thanCo.
Chooser and g corresponding
toCo.
We claim that
g)
is rotationcompatible. Indeed,
considerg : This map cannot be
isotopic
to apseudo-
Anosov map.
Indeed,
otherwise from Theorem(3 .1)
one gets that forVol. 49, n° 3-1989.
every essential curve r in has
~(T)~~(T)
for allThen f : DZ - O ( p) --~ D2 - O (p) would
also have thisproperty
and there-fore be
isotopic
to apseudo-Anosov
map. This isimpossible
since thetopological
entropyof f is
zero.Also g :
cannotbe
reducible,
since otherwise there would be an essentialsimple
curveCo
in
D2 - O (Do) c D2 - O (p)
which isperiodic up
toisotopy
andbounding
a disc
containing strictly
morepoints of O(p)
thanDo
nO ( p),
contradict-ing
the choice ofCo.
It follows from Thurston’s Theorem(3.1) that g : D2 - O(Do) --+ D2 - O(Do)
must beperiodic
and therefore there existsn >-1
such thatg",
id:D2 - O (Do)
->D2 - O (Do)
areisotopic.
As in Lemma
(4.1)
it follows that(O(Do); g)
is rotation compa- tible. Inparticular gr|D0: D0-O(p)~D0-0(p)
renormalizesf : D2-O(p) ~D2-O(p).
Since is aperiodic
orbitof gr |D0
of
period
1 s n and sinceh (gr ~ Do) = 0
it follows from the inductionassumption
thatgr|D0)
ishereditarily
rotationcompatible.
Therefore
by
definition(0(/?); f )
ishereditarily
rotationcompatible.
- Q.E.D.
4. 4. LEMMA. -
Let f be an orientation preserving C1+z diffeomorphism of D2
withisolated periodic points. If every periodic
orbitof f is hereditarily
rotation
compatible
thenf topological
entropy zero.Proof. -
Ifh ( f ) >
0 then a result of A. Katokimplies
that thereexists a
periodic
orbitO(p)
ofsaddle-type
with a transversal homoclinic intersection. Therefore there exists arectangle
R c D2 and n > 0 such thatfn R
has a horseshoe. It is well known that there are manyperiodic
orbitsin a horseshoe which are not rotation
compatible,
see[H. W.], [Fr], [Bo 1].
For
example
has aperiodic
orbitO(p)
of 5 as inFigure
7.Taking
the curve C as drawn in this
figure
it followsthat/"(C)
is notisotopic
toC in
D~20140(/?).
Hence4 ( p)
is not rotationcompatible.
Since 5 is aprime
number this orbit can also not behereditarily
rotationcompatible.
5.
PROOF OF
THEOREM B5.1. LEMMA. - Let
f :
D2 --+D2 C1 diffeomorphism
andO(p)
bea
periodic
orbitof period
n. Assume that/: D~20140(~)-~D~20140(~)
isperiodic.
Either p is afixed point
or there exists afixed point q of f
whichFIG. 7. - A periodic orbit of period 5 in the horseshoe map which generates positive entropy.
Proof. -
We may assume thatO(p)
is not a: fixedpoint.
Let R be the, isometry
ofD~ - O ( p) isotopic
tof : D~-0(~)-~D~-0(~)
and letR : D2 -+
D2 be the extension of R.Let p’
be the(unique)
fixedpoint
of,
R.
SinceO(p)
is not a fixedpoint,
onehas p’ ~ O ( p)
andp’
is a fixedpoint
of R. ~ ’ . ~ . ,.
,
Moreover,
sinceR
isconjugate
to a rotationthere exists
asimple
closedcurve
r
in D2 which goesthrough
each of thepoints
of0(p),
such that,
r
bounds a disccontaining p’
in its interior and such thatR (F)
=f.
Writey as the
disjoint
union of nsimple arcs i
in D2with efndpoints
inO ( p).
Let y, be thegeodesic (w. r. t.
to thehyperbolic
metric in
D2 - (O(p)~{p}) isotopic to i
inD2-(O(p)~{p})
and letr==
U Yf. Since anisometry
mapsgeodesics
ontogeodesies,
R(r)
= r. r is’Vol. 49, n° 3-1989. ~. ~ .
,
GAMBAUDO,
a
simple
closed curve and bounds a disc which contains the fixedpoint p’
of R in its interior.
As we remarked
before, p’
is theunique
fixedpoint
of R. Inparticular
this fixed
point
is theonly
fixedpoint
in its Nielsen class. Since R is anorientation
preserving isometry,
the Lefschetz numberL(//, R")
isequal
to one. It follows from Theorem
(3. 3)
thatQ/, R)
is unremovable. Hence there is anisotopy f
between R= fo
andf = fl
and a continuous curve c :[0,1]~D~-0(~)
such that~(c(t))=c(t).
Inparticular
forq = c ( 1)
one has
f (q) = q.
Take thehyperbolic
metric dt(resp. d)
on[resp. DZ - O ( p)]
and let it :D~B(0(/?) U {c(t)}
-+ be the canonical inclusion. Notice that for allt E [o,1]
and thatc(t) depends continuously
on t.Therefore we can
choose,
for eacht E [o,1],
ageodesic y~
inD~B(0(/?) U {c(t)}) (with
respect todt) isotopic
to y, in anddepending continuously
on t. Then is asimple
closed curve inU
{c(t)}.
Let rtdepends continuously
on t
[in
terms of the metric d onD~20140 (/?)].
Since for eachfe[0,l]
thisimplies
that for eachte[0,l],
rt is asimple
closed curvecontaining
each of thepoints
pi such that the disc Dt boundedby
rtcontains the
point
c(t)
in its interior. Since isisotopic
tor°
= rand f isotopic
toR, f (rt)
isisotopic
to rt inD2 - 4 ( p).
Q.E.D.
5. 2. LEMMA. - Assume
that f: D2 -+ 02
is aC1 diffeomorphism of D2
zero
topological
entropy. Then everyhereditarily
rotationcompatible periodic
orbit has apersistent
parent.Proof. -
We showby
induction that aperiodic
orbit(0 (/?);/)
off : D2
-+D2
which ishereditarily
rotationcompatible
ofgeneration k
hasa
persistent
parent. For k =1 this statement isproved
in Lemma(4. 2).
So assume that this induction statement is
proved
for allhereditarily
rotation
compatible
orbits ofgeneration ~k
and assume that(0 ( p); f )
is
hereditarily
rotationcompatible
ofgeneration
~+1.By
definition there exist a map g in theisotopy
classof f : D~20140(~)-~D~20140(~)
and ag-periodic
discDo
ofperiod
r such that(O (p)
nDo; gr I Do)
ishereditarily
rotation
compatible
ofgeneration
k. Consider theperiodic
orbitof
gr I Do
as aperiodic
orbit ofgr: D2 -+ D2.
Thenis also
hereditarily
rotationcompatible
ofgeneration
k and therefore
(O(p) n Do; fr)
ishereditarily
rotationcompatible
ofgeneration
k. It follows from the inductionassumption
thatDo; fr)
has apersistent
parent orbit. This orbit ispersistent
underisotopies
and thereforeunder
isotopies of f : D~-O(~) -~D~20140(~).
This means that there exist s eN andsimple
closed curves!=0,1, ...,s2014l,
suchthat q
is a
persistent periodic point
off’’
ofperiod
s, #(O ( p)
nDo)
=ro x s, for each i =0, 1,
... ,s -1, rir
bounds a disccontaining
qir =fir (q)
and ropoints
ofO ( p) n Do
and such thatf’ (rir)
isisotopic
to in02_{O(P)UO(q)}.
We can choose these curvesfir
such thatrircDo.
Then
ir
boundsprecisely
ropoints
ofO (p).
For k= 0,1,
... ,r-l,
chooserir+k -J k
and qi =f ~ (q).
It follows thatper Furthermore
ri
is asimple
curve,bounding
ropoints
ofO(p)
and q~ andf (ri)
isisotopic
to0393i+1
in02_{O(P)UO(q)}.
This proves that(O (q); f )
is apersistent
parent orbit of
( O ( p); f ).
This finishes theproof
of Lemma( 5 . 2).
Q.E.D.
6. PROOF OF THEOREM C
6.1. LEMMA. - Assume that
f : D2
-+02
is aC1 diffeomorphism of D2
with zero
topological
entropy. Then thefollowing
statements hold.1.
If
two orbitsO(p)
andO ( p’)
have a common childO(q)
thenO(p)
and O
( p’)
have the sameperiod,
the samegeneration
and any parentof O(p)
is also a parentof
O( p’).
"2.
If
two orbitsO(p)
andO ( p’)
aredisjoint
thenthey
have no common3.
If
two orbitsO(p)
andO ( p’)
aredisjoint
and have a common childO(q)
thenO(p)
andO ( p’)
are bothfixed points.
Proof. -
Consider threeperiodic
orbitsO ( p), O (p’)
andO(q)
as in1,
2 or 3. The Lemma is
trivially
true if Solet us assume that we have
proved by
induction that statements 1 and 2are true for all
periodic
orbits with ConsiderSince
h(/)=0, according
to Thurston’s theorem there are at most twopossibilities.
Case 1. -
f
isisotopic
to aperiodic
map and thereforeisotopic
to amap
which extends
continuously
to a map onD2
which isconjugate
to arotation of
D2.
In this case statements1,
2 and 3 followimmediately.
VoL49,n°3-1989. -
GAMBAUDO, S. VAN STRIEN AND C. TRESSER
Case . 2. -
f is
reducible. Thenthere
exist and asimple
essentialclosed
curveCo,
and a map g in theisotopy
class of/:
such that
~(Co)=Co,
and such that the curvesCi = g (Co),
~=0,1,
..., r -1 arepairwise disjoint.
There exists s> 1 such that for eachi = 0,
... , r -1 the curveC~
bounds a discOi containing precisely
spoints
of(where
becauseCi
issimple).
Choose a
"periodic"
essentialsimple
curveCo
inas above such that there is no
periodic
essential
simple
curveCo bounding
morepoints
U0(//)
UO(q)
than
Co.
Chooser and g corresponding
toCo.
Precisely
as in Lemma(4 . 3)
it follows that(O(Do); g)
is rotationcompatible
andthat g : D2 - O (Do)
-+ isperiodic.
So we mayassume
that g : D2 - O (Do)
-;D2 - O (Do) continuously
extends to a map R on D2 which isconjugate
to a rotation ofD2. Therefore,
thefollowing properties
are satisfied:,
(i)
if one of the orbitsO ( p), O (p’)
orO(~),
say0(/?),
is inD2 - O (Do)
and has children then
O(p)
is the(unique)
fixedpoint
of RandO(D)
has
period ~2;
,(ii)
ifO(q)
is inD2 - O (Do)
andO(p)
is a parent ofO(q)
then r =1(~. e. O(Do) = Do), O(p)
is a fixedpoint
andO(p) c Do;
(iii)
if one of the orbitsO ( p), O (p’)
orO(q)
is in thenthis orbit cannot be
disjoint
from one of the other orbitsO ( p), O (p’)
orO (q)
unlessR = id;
(iv) O(q)
is the parent ofO(p)
thenO(q)cO(Do)
and r= 1[i. e.
O(Do) =Do]
andO(q)
is a fixedpoint.
Let us first assume that both orbits
O ( p), O (p’)
are inSo from
(i)i
ifO(p)
andO (p’)
have a common child thenO(p)=O(p’)
is a fixed
poirit.
This proves 1. From(iii) they
cannot bedisjoint
unlessR ~= id. So statements 2 and 3 are
trivially
true.Next assume that
(precisely)
one of the orbits0(/?), 0(//),
sayO(p), is contained
in If0(/?)
and0(//)
have a commonchild,
then from (ii) above,
r =,1and O ( p),
acontradiction.
This proves 1
in
this case. From(iii) ~O ( p)
andO ( p’)
cannot bedisjoint
unless R = id and therefore 2 and 3
trivially
satisfied... So we may assume that both have at least one
point
in common withDo.
ThenO ( p), O(p’)cO(Do).
First assume thatThen if
O(p)
andO (p’)
are parents ofO(q) then,
from