Julia sets in parameter spaces

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Julia sets in parameter spaces

Buff, X.; Henriksen, C.

Published in:

Communications in Mathematical Physics

Link to article, DOI:

10.1007/PL00005568

Publication date:

2001

Document Version

Peer reviewed version

Link back to DTU Orbit

Citation (APA):

Buff, X., & Henriksen, C. (2001). Julia sets in parameter spaces. Communications in Mathematical Physics,

Julia sets in parameter spa es. ? XavierBu 1 ,Christian Henriksen 2 1

UniversitePaulSabatier,UFRMIG,LaboratoireE.Pi ard,31062ToulouseCedex,Fran e.

E-mail:bupi ard.ups-tlse.fr

2

Te hni alUniversityofDenmark,DepartmentofMathemati s,2800Lyngby,Denmark.

E-mail: hrismat.dtu.dk

Re eived:date/A epted:date

Abstra t: Given a omplexnumber of modulus 1, weshowthat the

bifur- ation lo us ofthe oneparameter familyff

b (z)=z+bz 2 +z 3 g b2C ontains

quasi- onformal opiesofthequadrati JuliasetJ(z+z

2

).Asa orollary,we

show that whenthe Juliaset J(z+z

2

)is notlo ally onne ted (forexample

when z 7! z+z

2

has a Cremer point at 0), the bifur ation lo us is not

lo- ally onne ted.Toourknowledge,thisistherstexampleof omplexanalyti

parameterspa e of dimension1, with onne tedbut non-lo ally onne ted

bi-fur ationlo us.Wealsoshowthat theset of omplexnumbersofmodulus1,

forwhi h atleastoneoftheparameterrayshasanon-triviala umulationset,

ontainsadenseG Æ subsetofS 1 . Contents 1. Introdu tion. . . 2 2. ConformalrepresentationofC nM  . . . 5

3. Copies ofquadrati Juliasetsin thedynami alplane. . . 11

4. Denition ofthewakeW

0 .. . . 16 5. Dynami soff b inthewakeW 0 . . . 22

6. Holomorphi motionof rays. . . 26

7. Thedyadi wakesW

#

. . . 28

8. Copies ofquadrati Juliasetsin theparameterplane. . . 37

9. Nonlo al onne tivityintheparameterplane. . . 40

?

o-1.Introdu tion.

Inthisarti le,westudytheoneparameterfamilyof ubi polynomials

f b (z)=z+bz 2 +z 3 ; b2C; where=e 2i

isaxed omplexnumberofmodulus1.We allK(f

b

)the

lled-inJuliaset ofthepolynomialf

b ,J(f

b

)itsJuliaset, andM



the onne tedness

lo usofthefamily:

K(f b )=  z2C   f Æn b (z)
n2N isbounded J(f b )=K(f b ); and M  =  b2C   J(f b )is onne ted : ThenotationsK b andJ b

arekeptforother purposes.

In se tions 2 and 3, we re all some lassi al results related to the study

of the dynami s of ubi polynomials. Those results an be found in [BH1℄.

In parti ular, weprove that the onne tedness lo us M



is onne tedand we

onstru tdynami allya onformalrepresentation



:CnM



!CnD ( ompare

with[Z1℄).Thisenablesus todenetheparameterraysR



(),2R=Z.

Inse tion4,weprovethattheparameterraysR

 (1=6)andR  (1=3)landat a ommonparameterb 0

. Thete hniqueswe usearenotnew. Theyaresimilar

to those developed by Douady and Hubbard in [DH1℄ to study the landing

propertiesof parameterraysin thequadrati familyfz7!z

2

+ g

2C

. Wethen

dene the wake W

0

asthe onne ted omponent of C n

 R  (1=6)[R  (1=3) 

that ontainstheparameter rayR



(1=4) (see gure5). Inse tion 5,westudy

the dynami al features of the polynomials f

b

when the parameter b ranges in

thewakeW

0 .

Matters get interesting in se tion 6. Let us dene   R=Z (respe tively

 0

R=Z) to be the Cantor set of angles that anbe written in base 3with

only0'sand1's(respe tivelywithonly1'sand2's).WedenotebyX

b

theset of

dynami alrayswhoseargumentsbelongto.Inse tion6,weprovethattheset

X b

movesholomorphi allyaslongastheparameterbremainsin thewakeW

0 .

Asa onsequen e,weshowthatforanyparameterb2W

0

,thelled-inJuliaset

K(f b

) ontainsaquasi- onformal opyofthelled-inJuliasetK(z+z

2

)(see

gure11).

TheoremA.Forany parameterb2W

0

andfor any2,thedynami al ray

R b

() does notbifur ate.Wedene X

b tobethe set X b = [ 2 R b (): WealsodeneJ b tobethesetJ b =X b nX b andK b

tobethe omplementofthe

unbounded onne ted omponentofC nJ

b

.Then,K

b

is ontainedinthe lled-in

Julia set K(f

b

), its boundary J

b

is ontained in the Julia set J(f

b

) and K

b is

quasi- onformally homeomorphi tothelled-in JuliasetK(z+z

2 ).

In the wake W

0

, one an see a opy M

0

of a Mandelbrot set (see Figure

M  K(z+z 2 ) Fig.1. ZoomsinM  for=e i( p 5 1) .

homeomorphi to theMandelbrot set. Thishas beendone in [EY℄ in the ase

6=1,andisnotknowninthe ase=1.However,weshowthattheboundary

of M

0

isequal to thea umulationset ofthe parameterraysR

 (=3), 2  0 (seegure12): M 0 =X 0 nX 0 ; where X 0 = [ 2 0 R  (=3):

Atthe sametime,weshowthat the onne ted omponents ofW

0

nX

0

anbe

indexed by dyadi angles# 2R=Z.The onne ted omponentW

#

is bounded

by two parameter rays R



(# ) and R

 (#

+

) landing at a ommon parameter

b #

2 M

0

. Theangles # and #

+

are two onse utiveendpointsof theCantor

set 

0

. Weprove that given any dyadi angle #=(2p+1)=2

k , wehave# + = # +1=(2
3 k +1

).Wethendenethesets X

#

, J

#

andK

#

inthefollowingway:

X # = [ R   # 3 +  3 k +1  ;

J # =X # nX #

,wherethe losureistakeninC, andK

#

isthe omplementofthe

unbounded onne ted omponentofC nJ

#

.Ourmain resultsare thefollowing

(seeFigure1).

Main Theorem. Let 2S

1

be a omplexnumber of modulus1 and#2R=Z

be a dyadi angle. The set K

# is ontained in M  \W # , its boundary J # is

ontainedin the boundary of M



andthe parameter b

#

belongsto J

#

.Besides,

thereexistsaquasi- onformalhomeomorphism denedinaneighborhoodof K

# , sendingK # toK(z+z 2 ).

CorollaryA.Forea h omplexnumberofmodulus1,thebifur ationlo usof

theoneparameterfamilyf

b (z)=z+bz 2 +z 3 ,b2C, ontainsquasi- onformal

opiesof the quadrati JuliasetJ(z+z

2 ).

CorollaryB.IftheJuliaJ(z+z

2

)isnotlo ally onne ted,thenthebifur ation

lo usM



isnot lo ally onne ted.

Wewouldliketo mentionthat onehasto be areful.Indeed, inthe ontext

of Newton's method of ubi polynomials,Pas aleRoes h [R℄ hasan example

ofalo ally onne tedJuliaset ontaininga opyofaquadrati Juliasetwhi h

isnotlo ally onne ted.Inour ase,this doesnoto urbe ausetheset M

 is full.

Observe that when t 2 RnQ does not satisfy the Bruno ondition, the

quadrati JuliasetJ(e

2it

z+z

2

)is knowntobenon-lo ally onne ted.Hen e,

thesetofvaluesof2S

1

forwhi hM



isnotlo ally onne ted ontainsadense

G Æ

subsetofS

1

.Lavaurs[La℄provedthat the onne tednesslo usof thewhole

family of ubi polynomials is not lo ally onne ted. In the parameter spa e

ofreal ubi polynomials,thebifur ationlo usis alsoknowntobenon-lo ally

onne ted (see [EY℄).Toourknowledge, we givethe rst exampleof omplex

parameterspa eofdimension1with onne tedbutnon-lo ally onne ted

bifur- ationlo us.

ShizuoNakanebroughttoourattentionthatwe ouldprovetheexisten eof

parameterrayswith anon-triviala umulationset.Hehasalreadyprovedthis

result in the family of real ubi polynomialsin ajointwork with Y. Komori

(see[NK℄).Tostatethenext orollary,weneedtointrodu esomenotations.

Givenany omplexnumberofmodulus1,wedeneP

 tobethequadrati polynomialP  (z) = z+z 2

. Forany angle  2 R=Z,we dene R

P 

() to be

thedynami alrayofthepolynomialP



ofangle.Wealso onsidertheCantor

map (or devil stair ase) 



:R=Z ! R=Z whi h is onstant onthe losure of

ea h onne ted omponentofR=Zn andisdened on by:

  0  X i1 " i 3 i 1 A = X i1 " i 2 i ; where " i 2f0;1g:

Corollary C. Given any omplex number  of modulus 1, any dyadi angle

R 

(# =3+=3

k +1

)is redu edto apointif andonly if the a umulation setof

the quadrati rayR

P (



())isredu edtoa point.

Using an a umulation theorem due to Douady (see [S℄), we then prove

that theset of omplexnumbersof modulus 1,forwhi h at leastoneof the

parameterraysR



()C nM



hasanon-triviala umulationset, ontainsa

denseG

Æ

subsetofS

1 .

We would like to make some omments about the hoi e of the family f

b .

We wanted to work with a family of ubi polynomials having a persistently

indierent xed point. We de ided to put this xed point at the origin.This

onditionis a hieved,sin ethe mapf

b

hasanindierentxedpointat 0with

multiplier . Thereason why we have hosen this parametrization is that the

polynomial f

b

is moni and thus, has a preferred Bott her oordinate. This

will be usefulto dene a onformalrepresentation 



: C nM



!C nD in a

dynami al way. This is importantsin e wewantto be ableto transferresults

fromthedynami alplanetotheparameterplane.However,oneshouldobserve

that the maps f

b

and f

b

are always onjugate by the aÆne map z 7! z.

Indeed, f b ( z)= ( z+bz 2 z 3 )=f b (z):

Thisexplainswhyparameterpi turesaresymmetri withrespe ttotheorigin.

The entralargumentweuseisinspiredfromte hniquesdevelopedbyTanLei

in[TL℄.There,sheprovesthattherearesimilaritiesbetweentheMandelbrotset

and ertainJuliasets.WewouldalsoliketomentionthatPiaWillumsenproved

theexisten eof opiesofthequadrati JuliasetJ(z

2

1)intheparameterspa e

ofawell- hosenfamilyof ubi polynomials.

Hubbardmadethesuggestionthat thetwodimensional onne tednesslo us

ofthespa eof ubi polynomialsmay ontainhomeomorphi opiesof theset

n ( ;z)j K(z 2 + )is onne tedandz2K(z 2 + ) o :

After weexposed ourresultsin Crete

1

,Lyubi h andM Mullen madethe

ob-servationthat pushingfurther ourarguments,we should beableto provethis

result. This would show the existen e of ubi polynomialsbeing in the same

ombinatorial lass, but not being topologi ally onjugate. Su h a result has

been onje turedbyKiwiinhis thesis[K℄.

2.Conformal representation ofC nM

 .

Inthisse tion,wewilluseresultsbyBrannerandHubbard[BH1℄toprovethat

M 

isfull, onne tedandhas apa ity3=

3 p

4.Wewill onstru t,in adynami al

way,theRiemannmapping

 :C nM  !C nD,thatistangenttob7!b
3 p 4=3

atinnity. AsimilarstudyhasalreadybeendonebyZakeri[Z1℄.Workingwith

the es aping riti al value, he denes an analyti map from C nM



to C nD

whi hturnsouttobea overingmapofdegree3.Wewillinsteadworkwiththe

es aping o- riti alpoint.Wewillneedthisapproa hlater,totransferdynami al

resultstotheparameterplane.In[Z2℄,Zakerialsogivesaninterestingproofof

the onne tivityofM



basedonTei hmullertheoryofrationalmaps.

1

2.1.Potential fun tions.. Re all that Fatou proved that the Julia set of any

polynomialis onne tedifandonlyiftheorbitofea h riti alpointisbounded.

Inour ase,the mapf

b

hastwo riti alpoints.However,f

b

has anindierent

xedpointat0.Hen e,there isalwaysone riti alpointwithaboundedorbit.

Indeed,thereareonlythreepossible ases:

 thexedpointisparaboli (2Q), andthereisatleastone riti alpointof

f b

in itsbasinofattra tion;

 the xedpointis linearizable (it ould be the aseevenif  is notaBruno

number), andtheboundaryoftheSiegel diskisa umulatedbytheorbitof

atleastone riti alpointoff

b ;

 thexedpointisaCremer pointandis ontainedin thelimitsetofat least

one riti alpointoff

b .

Remark. Wewillsaythatthis riti alpointis\ aptured"by0.

Inparti ular, when J(f

b

) isdis onne ted, there isexa tlyone riti al point

! 1

withboundedorbit,andonees aping riti alpoint!

2 .

Letusnowre allsome lassi alresultsthat anbefoundin[DH1 ℄and[BH1℄.

Denition1(Potential fun tions). Forany b2C, dene g

b :C ![0;+1[ by g b (z)= lim n!1 1 3 n log +   f Æn b (z)   ; wherelog +

isthe supremum oflog and0.Also dene the fun tionG:C !R

+ by G(b)= sup f! jf 0 b (!)=0g g b (!):

Remark. Whenthe Julia set J(f

b ) is onne ted,G(b) =0. Otherwise,G(b) = g b (! 2 ).

Proposition 1.Wehavethe followingproperties:

1.g

b

is ontinuousandsubharmoni onall ofC;

2.g b (f b (z))=3g b (z); 3.g b vanishesexa tly on K(f b )and isharmoni on C nK(f b );

4.the riti alpointsofg

b

inCnK(f

b

)arethepreimagesofthees aping riti al

point! 2 by aniterate f Æn b ,n0; 5.the mapping(b;z)7!g b

(z)isa ontinuousplurisubharmoni fun tion;

6.the fun tion Gis ontinuousandsubharmoni .

Remark. We will see that G vanishes exa tly on the set M



and is harmoni

outsideM

 .

Denition2(Equipotentials).Thelevel urveg

1 b

fgis alledthedynami al

equipotential of level .The level urve G

1

fgis alledthe parameter

equipo-tential oflevel .

When the Julia set is onne ted, the two riti al points are ontained in

K(f b

),andtheharmoni mapg

b

:C nK(f

b

)!R

+

hasno riti alpoint.Hen e,

everydynami alequipotentialoff

b

isareal-analyti simple losed urve.More

G(b)g, and everydynami al equipotential of level  > G(b) is a real-analyti

simple losed urve.

Theorthogonal urvestodynami al(respe tivelyparameter)equipotentials

willbe alled dynami al(respe tivelyparameter)rays.Wewillbemorepre ise

aboutthedenitionofraysbelow.

Figure2showsalled-inJuliasetwithtwodynami alequipotentialsoflevel

1=3and 1,togetherwithfoursegmentsofdynami alrays.

f b f b 4 4i 4 4i R b (9=12) ! 2 ! 0 2 4+4i R b (1=12) R b (1=4) 4+4i R b (5=12) U b U 0 b f b

Fig.2. Adis onne tedJuliaset;

 (b)=' b (! 0 2 )=e 1=3+2i=12 .

2.2.The Bott her oordinate atinnity.. Theve toreld

 b = 1 2 grad(g b )=jgrad(g b )j 2

isameromorphi ve toreld onC nK(f

b

),havingpolesexa tlyat the riti al

pointsofg b in C nK(f b ). Denition3.We dene S b

to be the union of the riti al points of g

b

in C n

K(f b

)andtheirstablemanifoldsfortheve toreld

b

.Foranyb2C,wedene

V b

tobethe openset C n(K(f

b

)[S

b ).

We have normalizedour ubi polynomials so that theyare moni . Hen e,

andtangenttotheidentityatinnity.Considerthe ow(z;)7!F b

(z;)ofthe

ve toreld 

b

,where  2Ris arealtime. Foranypointz2V

b

,we anextend

' b

atz usingtheformula'

b (z)=e  ' b (F b (z;));where 2[0;+1[is hosen

large enough so that F

b

(z;) 2 U

b

. The following proposition is then easily

derivedfrom theanalyti ityof

b

anditsanalyti dependen eonb.

Proposition 2(Bott her oordinate). There exists a unique analyti

iso-morphism '

b

dened in a neighborhood of innity, tangent to the identity at

innity, andsatisfying

' b Æf b Æ' 1 b (z)=z 3 : The mapping ' b

extends toan analyti isomorphism '

b :V b !C and satises logj' b j=g b

onthis set.Furthermore, '

b

dependsanalyti ally onb,i.e., the set

V=

[

b2C

fbgV

b

isopenandthemapping:V!C

2

denedby(b;z)=(b;'

b

(z))isananalyti

isomorphism fromV ontoitsimage.

Remark. An easy omputation showsthat near innity, wehave'

b (z) = z+ b=3+O(1=jzj): WhenJ(f b )is onne ted,V b =C nK(f b

)andtheBott her oordinate'

b is aunivalentmapping ' b :C nK(f b )!C nD; andonCnK(f b ),wehaveg b =logj' b

j.Inparti ular,thedynami alequipotential

oflevel istheset

' 1 b n e +2i j2R=Z o ;

i.e.,thepreimageby'

b

ofthe ir leofradiuse



enteredat0.

WhenJ(f

b

)isdis onne tedthispropertystillholdsforequipotentialsoflevel

>G(b),i.e.,intheregionfz2C jg

b

(z)>G(b)g.

Inboth ases,thepush-forward('

b )

 (

b

)istheradialve toreldw=w.In

parti ular, '

b

maps everytraje tory of theve toreld 

b

to asegment ofline

with onstant argument. Hen e, '

b (V

b

) is a star-shaped domain with respe t

to innity, i.e., for every angle 2 R=Z,there exists a radius r(b;)1 su h

that w2'

b (V

b

)and arg(w)=2 ifandonlyif jwj>r(b;). Finally, alonga

traje toryz()oftheve toreld

b ,wehaveg b (z())=g b (z(0))+.

Denition4(Dynami al Rays). Forany b2C, the dynami al ray R

b () is denedas R b ()=' 1 b n re 2i j r>r(b;) o :

Remark. Theve toreld

b = 1 2 grad(g b )=jgrad(g b )j 2 anbe extended holomor-phi allytoC nK(f b

).Then,ithasasinkatinnityandthedynami alraysare

exa tlythestablemanifoldsofinnityfortheve toreld

b .

Whenr(b;)=1,thea umulationsetofadynami alrayis ontainedinthe

Juliaset J(f

b

).This istrueforanyangle 2R=Zwhen J(f

b )is onne ted. If thelimit z 0 =lim' 1 b (re 2i )

exists, we will say that the dynami al ray R b () lands at z 0 . When J(f b ) is

dis onne tedandwhenr(b;)>1,thenthelimit

z 0 = lim r&r(b;) ' 1 b (re 2i )

existsandisa riti alpoint! ofg

b

.Inthis ase,wewillsaythatthedynami al

rayR

b

()bifur ateson!.

If b

2

= 3, then there is unique riti al point. This riti al point annot

es ape(be ause0\ aptures"a riti alpoint),andb2M



.Ontheotherhand,

ifb 2 6=3andb2=M  ,thenf b (! 2 )hasapreimage! 0 2 6=! 2 . FollowingBranner

andHubbard,we allitthe o- riti alpointto!

2 .

Letus observethat '

b

iswelldened atthe o- riti alpoint!

0 2

.Indeed, !

0 2

annot be a riti al point of g

b

sin e it is not an inverse image of !

2

. Let us

onsiderthetraje toryz()denedbytheinitial onditionz(0)=!

0 2 .Wehave g b (z())=g b (! 0 2

)+. Inparti ular, sin ethe regionfz 2C j g

b (z)>g b (! 0 2 )g

doesnot ontain riti alpointsof g

b

wesee that thetraje tory z()is dened

on[0;+1[.Hen e,! 0 2 belongstoV b ,and' b (! 0 2 )iswelldened. Denition5.Givenb2C nM 

,thees aping riti alpointis alled!

2 andthe o- riti al pointto! 2 is alled! 0 2

.Wedene themapping 

 :C nM  !C by   (b)=' b (! 0 2 ):

Proposition 3(Branner-Hubbard [BH1℄ and Zakeri [Z1℄ [Z2℄). The set

M 

isfulland onne ted.Besides,themap



:CnM



!CnD isthe onformal

isomorphism whi h istangent tob7!b

3 p

4=3atinnity.

Proof. We haveseen that ifb

2 = 3, then b 2 M  . Now, if b 2 6= 3, the two

riti al points are the two distin t roots of the equation f

0 b

(z) = 0, and by

theimpli itfun tion theorem,we anfollowthemlo ally.Hen e,we anfollow

holomorphi allythetwo riti alpointslo allyoutsideM



.Forthesamereason,

we anfollowholomorphi allythetwodistin t o- riti al pointslo allyoutside

M 

.

Lemma1.Themapping 



:C nM



!C nD isanalyti .

Proof. Fix aparameter b

0 = 2 M  , let ! 2

be thees aping riti al pointand !

0 2

bethe o- riti al point to !

2

. There exist twoholomorphi maps dened in a

neighborhoodU ofb

0

thatfollowthetwo o- riti alpoints. Let!

0

:U !C be

theonewhi h oin idewith!

0 2 atb 0 .Theset W=f(b;z)2C 2 jz2C nK(f b )g

isthepreimageof℄0;+1[bythemapg(b;z)=g

b

(z)whi his ontinuousby5)

ofproposition1.Hen e,Wisopen.Thus,byrestri tingU ifne essary,wemay

assumethatforanyb2U,the o- riti alpoint!

0

(b)belongstoC nK(f

b

).This

showsthat,foranyb2U,thees aping o- riti alpointis!

0 (b).

Furthermore,the mapping (b;z) 7! '

b (z) is analyti in a neighborhood of anypoint(b 0 ;z 0 )su h thatz 0 2V b0

.Hen e,itisanalyti in aneighborhood of

(b 0

;! 0 2

).Itfollowsimmediatelythat

 (b)=' b (! 0 2 (b))isanalyti ina neighbor-hoodofb . 

Theproofthat 

isanisomorphismbetweenC nM



andC nD isan

appli- ationoftheprin iple:ananalyti mappingisanisomorphismifitisproperof

degree1.Weshalluseasimilarargumentforquasi-regularmappingsin se tion

8.

Lemma2.Outside M



, we have G(b) = logj



(b)j. Besides, the fun tion G

vanishes exa tlyon M  . Proof. Ifb62M  ,we anwrite: log     (b)    = 1 3 n log   ' b f Æn (! 0 2 )
   = 1 3 n log     f Æn (! 0 2 )+O  1 jf Æn (! 0 2 )j     =G(b): Sin e ' b

takes valuesoutsideD, sodoes



. Hen e,G is positive outsideM

 .

Besides, ifb 2M



, both riti alpoints arein the lled-in Juliaset K(f

b

). So,

theirorbitsareboundedandG(b)=0. 

We annowseethatM



isfull.Thisisanimmediate onsequen eofthefa t

thatsub-harmoni fun tionssatisfythemaximumprin iple.Thus,levelsetsare

full.

By Pi ard's theorem, the mapping 



: C nM



! C nD has a removable

singularity at innity. Hen e,we an extendit toinnity. Wene essarilyhave

 

(1)=1sin eotherwiseGwould beanon- onstantboundedsubharmoni

fun tiononP

1 .

Morepre isely,asimple omputationshowsthatwhenjbjtendstoinnity,

! 2 = 2b 3 +o(1); and ! 0 2 = b 2! 2 = b 3 +o(1): Then, f b (! 0 2 ) (! 0 2 ) 3 =4+O  1 jbj  ;

andforanyintegern1,wehave

f Æ(n+1) b (! 0 2 ) (f Æn b (! 0 2 )) 3 =1+O  1 jbj  : Hen e,weobtain   (b)=! 0 2 Y f Æ(n+1) b (! 0 2 ) (f Æn b (! 0 2 )) 3 ! 1=3 n+1 = 3 p 4 3 b+O  1 jbj  :

We willnowshow that 



: C nM



!C nD is apropermapping.Sin e G

is ontinuous, G(b)tendsto 0asb tendsto theboundaryofM



. Hen e,

 (b)

tendstoD whenbtendstoM

 fromoutsideM  .Sin e  isanalyti ,itisa propermapping.

We annally see that ithas degree1sin einnity hasonlyone preimage

ountedwithmultipli ity.Hen e,itisanisomorphismbetweenCnM



andCnD.

Wehavedenedparameterequipotentials.We annowdeneparameterrays

(seeFigure5).

Denition6(Parameter Rays). The parameterrayR

 ()isdenedas R  ()= 1  n e +2i j>0 o : If thelimit b 0 =lim r&1  1  (re 2i )

exists, wewillsaythat the parameterray R



() landsatb

0 .

3.Copies of quadrati Julia sets in the dynami alplane.

In this se tion, wewill rstre all a resultwhi h is essentiallydue to Branner

and Hubbard [BH2℄(see also [Br℄): when theparameter b is not in M



, there

existsarestri tionoff

b

whi hisaquadrati -likemapping.Thereaderwill nd

informationonpolynomial-likemappings andrelatedresultsin [DH2℄.

Denition7(Polynomial-like mappings). A polynomial-like mapping f :

U 0

! U of degree d is a ramied overing of degree d between two topologi al

disks U

0

and U, with U

0

relatively ompa t in U. One an dene its lled-in

JuliasetK(f)anditsJulia setJ(f)asfollows:

K(f)=fz2U 0 j(8n2N) f Æn (z)2U 0 g; and J(f)=K(f):

Apolynomial-like mappingof degree 2will be alledaquadrati -like mapping.

Let usre all theso- alled StraighteningTheorem due to Douadyand

Hub-bard.

Proposition 4(StraighteningTheorem).Iff :U

0

!U isapolynomial-like

mappingof degreed. Then thereexists

apolynomial P :C !C of degreed,

aneighborhood V ofthe lled-in Juliaset K(P)su h that the mappingP :

P 1

(V)=V

0

!V isapolynomial-like map,and

a quasi onformal homeomorphism ': U ! V with '(U

0

)=V

0

,su h that

'=0almost everywhere onK(f)andsu hthatonU

0

'ÆP =fÆ':

Moreover,if K(f) is onne ted, thenP isuniqueupto onformal onjuga y.

Denition8.Twopolynomial-likemappingsf andgaresaidtobehybrid

equiv-alentifthereisaquasi- onformalhthat onjugatesf andg,withh=0almost

everywhereon the lled-in JuliasetK(f).

Proposition 5.For any b 2 C nM



, let us denote by U

b

the open set fz 2

C j g

b

(z) <3G(b)g and U

0 b

the onne ted omponent of f

1 (U

b

) that ontains

the non-es aping riti al point !

1

. Then, the restri tion f

b : U 0 b ! U b is a 2

Figure2showsthedomainsU 0 b

andU

b

fortheparameter

1 

(e

1=3+2i=12 ).

Proof. We have seen that any dynami al equipotential of level  > G(b) is a

real-analyti simple losed urve. This applies to the dynami al equipotential

of level 3G(b). Thus, the set U

b

is a topologi al disk. Besides, it only

on-tains one riti al value of f

b

(the non-es aping one). The set f

1 (U b ) is the set fz 2 C j g b

(z) <G(b)g whi h is bounded by alemnis ate pin hing at the

es aping riti alpoint!

2

.Ea h onne ted omponentoff

1 (U

b

)isatopologi al

disk ompa tly ontainedin U

b

. Besides, therestri tionof f

b to the onne ted omponentof f 1 (U b

) ontainingthe non-es aping riti al point !

1

is a

ram-ied overing of degree 2, ramied at !

1

. This is pre isely the denition of a

quadrati -likemapping.

Next, to see that the hybrid lass of this quadrati -like mapping ontains

z7!z+z

2

,wewillusethefollowingresult.

Lemma3.Themultiplier of anindierent xedpointis aquasi- onformal

in-variant.

Remark. Naishul[Nai℄showsamu hbetterresultsin eheprovesthat the

mul-tiplierofanindierentxedpointisatopologi alinvariant.Perez-Mar o[PM℄

gaveanewproofofthisresultwhi hismu hsimpler.The aseofquasi- onformal

onjuga yiseasierto handle.R.Douadygaveaneasyproofbasedonthe

om-pa ity of the spa e of quasi- onformalmappings with bounded dilatation (see

[Y℄).Wewillpresentanewproofbasedonholomorphi motionsandthe

Ahlfors-Berstheorem.Thosetoolsaremore ompli atedthantheonesusedbyDouady,

buttheideaoftheprooftsverywellwithinthisarti le.

Proof. Assume that two germs f

0 : U 0 ! C and f 1 : U 1 ! C are

quasi- onformally onjugate.Call thequasi- onformal onjuga y.Then= =

is a Beltrami form invariant by f

0

. Integrating the Beltrami form 

" = ", "2D(0;1=jjjj

1

),weget afamilyofquasi- onformalhomeomorphisms

"

de-pendinganalyti allyon",andafamilyofanalyti germs

f " = " Æf 0 Æ 1 " :

We laimthat thisfamilyofgermsdepend analyti allyon" (thisisnot

imme-diate sin e

1 "

doesnot need to depend analyti ally on "; Douady explained

usageometri proof,andLyubi hexplainedusananalyti proofwhi hwegive

here).Sin ef " Æ " = " Æf 0

,foranyz2U we anwrite

f " "    "(z) + f " z
 " "    z + f " z
 " "    z =  " "    f0(z) : Sin ebothf " =z and "

="vanish,weseethat f

"

="vanishes.

In parti ular, themultiplier (") of thexed point depends analyti ally on

".Sin eit annotbe omerepellingorattra ting(allthegermsare onjugateto

f 0

whi hhasanindierentxedpoint),themodulusof(")is onstant.Hen e,

(")isa onstantfun tion,and(1)=(0). 

Thehybrid lassofthequadrati -likemapf

b :U 0 b !U b ontainsaquadrati

polynomialhavinganindierentxedpointwithmultiplier.Su hapolynomial

Denition9.For any parameter b 2 C nM 

, the lled-in Julia set of the

quadrati -like mapf b :U 0 b !U b is alledK b

anditsJuliasetis alledJ

b .

Wewillnowgivemoreinformationsaboutthedynami soff

b1

forthe

param-eter b

1

withpotential =1=3and external argument =1=4(we ouldhave

pi kedanyparameterwithpotential>0andexternalargument2℄1=6;1=3[).

Proposition 6.Letb 1 betheparameterb 1 = 1  (e 1=3+2i=4 ).If6=1,thetwo dynami al raysR b1 (0=1) andR b1

(1=2)bothland ata ommonxedpoint6=0

whi h is repelling. If  = 1, the rays R

b1

(0=1) and R

b1

(1=2) both land at the

paraboli xedpoint =0.

R b 1 (7=12) R b 1 (1=4) g 1 b 1 f1=3g R b1 (1=2) 4 5i 4+3i 4+3i 4 5i R b 1 ( 1=12) R b1 (0=1) g 1 b 1 f1g ! 0 2   !2 U 00 b 1 U b 1 U 0 b1

Fig.3. TheraysR

b1

(0=1)andR

b1

(1=2)bothlandata ommonxedpoint.

Proof. We still denote by U

b1 the set U b1 = fz 2 C j g b1 (z) < 3G(b 1 )g. Its preimage f 1 b 1 (U b1

) has two onne ted omponents. Note that U

0 b1 is the one ontaining! 0 2

in its boundary. Denote by U

00 b1

theother omponent(see gure

3). Remember that, f b1 : U 0 b 1 ! U b1

is a degree 2 proper mapping. Similarly

f b1 : U 00 b1 ! U b1

is a degree 1 proper mapping and sin e U

00 b1 is ompa tly ontained in U b1 , f b1

has exa tly one xed point in U

00 b

1

. This xed point is

repelling.Wewilldenoteitby.

Next,observethattheraysR

b1 ( 1=12)andR b1 (7=12)bifur ateon! 2 ,and

sin e 1=12 < 0 < 1=2 < 7=12, they separate  from the rays R

b 1 (0=1) and R b (1=2).

Sin ef b1 :U 0 b1 !U b1

isadegree2propermapping,andsin eU

0 b1

is ompa tly

ontainedinU

b1

,Rou he'sTheoremshowsthatf

b1

hasexa tlytwoxedpoints

in U

0 b 1

, ounted withmultipli ity. If 6=1, thosetwoxed points aredistin t.

Oneis0whi hisindierent,andhasmultiplier,theotheronewillbedenoted

by . A theorem due to Douady-Hubbard [DH1℄ and to Sullivan asserts that

everyxeddynami alraythat doesnotbifur ate, landsataxed pointwhi h

iseitherrepelling,orparaboli withmultiplier1.Sin ethetwoxeddynami al

rays R

b1

(0=1) and R

b1

(1=2) annot land at 0 (sin e the multiplier is neither

repelling norequalto 1), they mustbothland at thexed point . Sin e  is

thelandingpointofaray,eitheritisrepellingoritisamultiplexedpoint.But

sin ethereareonlytwoxedpointinU

0 b 1

ountedwithmultipli itytheformer

aseo urs.

On theother hand, if =1,there is only onexed point in U

0 b 1

: thexed

pointat0whi hisparaboli withmultiplier1.Hen ethetwoxedraysR

b 1 (0=1) andR b 1

(1=2)mustbothlandat0.

Wewill nowdes ribetheset ofraysthat a umulateon theJulia setJ

b 1

of

thequadrati -likemapf

b 1 :U 0 b 1 !U b 1 .

Denition10.Wedene R=Zto bethe set of angles su hthat for any

n0,3

n

2[0;1=2℄mod 1:

Remark. Thesetisthesetofanglesthat anbewritteninbase3withonly

0'sand 1's.Itis aCantor set andis forwardinvariantunder multipli ation by

3.

Figure 4showsthedynami alraysR

b 1

()for 2.Thefollowing

proposi-tionshowsthatthoseraysa umulateontheJuliasetJ

b1 ofthequadrati -like restri tionoff b1 . ! 2 R b 1 (1=2) R b 1 (0=1) R b 1 (1=3) R b 1 (1=6) R b 1 (4=9) R b 1 (1=18)

Fig.4. Thedynami alraysR

b1

(),2,a umulateontheJuliasetJ

b1 ofthe quadrati -likerestri tionoff b 1 .

Proposition 7.Let b 1 be the parameter b 1 =  1  (e 1=3+2i=4 ) and J b1 be the

Julia set of the quadrati -like mapping f

b1 : U 0 b1 ! U b1

.Then, for any  2,

the dynami al rayR

b1

()does notbifur ate.Besides, ifwedene

X b1 = [ 2 R b1 (); thenX b 1 nX b 1 =J b 1 :

Proof.Let us rstre all that therays R

b1

(0=1)and R

b1

(1=2) do notbifur ate

andlandatthesamexedpoint.Hen e,the urvefg[R

b1

(0=1)[R

b1 (1=2)

uts the plane in two onne ted omponents V

1 and V 2 . We all V 2 the one

ontainingthees aping riti al point!

2

.Observethat forany2[0;1=2℄,the

dynami alrayR

b1

()is ontainedin C nV

2

. Now,assumethat there exists an

angle 2 su h that thedynami alrayR

b 1

()bifur ates.Then, itbifur ates

on a preimage of the es aping riti al point !

2

and one of its forward image

bifur ateson!

2

.Butsin ebydenitionof,wehave3

k

2[0;1=2℄mod1,for

anyk0,theforwardorbitoftherayR

b1 ()is ontainedinC nV 2 .Hen e no forwardimageofR b1

() anbifur ateonthees aping riti alpoint!

2

2V

2 .

Sin ethesetis losed(itisaninterse tionof losedsets),X

b1 is losedin C nK(f b 1 ).Hen e, X b1 nX b1 J(f b1 ):

Wewill now showthat foranyangle 2, thea umulation setI of theray

R b1

()is ontainedintheJuliasetJ

b1

ofthequadrati -likemappingf

b1 :U 0 b1 ! U b 1

. Indeed,thea umulation set I is ontainedin theJulia set J(f

b 1 ) off b 1 ,

and its forward orbitis ontained in C nV

2

. Inparti ular, it annot enter the

regionU

00 b1

, and theforwardorbit ofI isentirely ontainedin U

0 b1 . This shows that I  K b 1

. Sin e I is ontained in the boundary of K(f

b 1 ), we see that IJ b1 , and X b 1 nX b 1 J b 1 :

Toprovethe reversein lusion,wewill usethefa t that the ba kwardorbit

of thexed point bythe quadrati -likemap f

b 1 : U 0 b 1 !U b 1 is densein J b 1 .

Let usshowby indu tion onn that ifz 2J

b1

satises f

Æn b1

(z)=, thenthere

isan angle 2 su h that R

b 1

() landsat z.This is trueforn=0sin ethe

raysR

b1

(0=1)andR

b1

(1=2)landat. Now,iftheindu tionpropertyholdsfor

somen, let us show that it is truefor n+1. Given apointz 2 J

b1 satisfying f Æ(n+1) b 1 (z)=,its imagef b1

(z)satisesthe indu tionhypothesis. Thus,there

is an angle  2  su h that the ray R

b 1 () lands at f b 1 (z). Observe that, on

one hand,this ray annot ontain the es aping riti al value (indeed, the ray

ontaining the es aping riti al value has argument 3=4 2= ), and its three

preimages land atthe three preimages off

b1

(z).On the otherhand, there are

threeangles 1 , 2 and 3 su hthat3 i

=,i=1;2;3.Twoofthem,let'ssay

1

and

2

,arein, andthethirdone,

3

,is ontainedin℄2=3;5=6[mod1.Hen e,

therayR

b 1

( 3

)landsatthepreimage off

b 1 (z)whi his ontainedin U 00 b 1 .This

showsthatoneofthetworaysR

b1 ( 1 )orR b1 ( 2 )landsatz.

Remark. It is easy to see that noother dynami al ray an a umulate onJ

b 1

4.Denition ofthe wakeW 0

.

Wewillnowrestri tourstudytoaparti ularregionintheparameterplane:the

wakeW

0 .

Denition11.Thewake W

0

isdenedtobethe onne ted omponentof

C nR  (1=6)[R  (1=3)[R  (2=3)[R  (5=6)

that ontainstheparameter ray R

 (1=4).

Remark. In fa t, we will show that the parameter rays R



(1=6) and R

 (1=3)

landata ommonparameterb

0

whi h satisestheequationb

2 0

=4( 1).The

wakeW

0

istheregion ontainedbetweenthose tworays(seeFigure 5).

Thereareseveralwaysofprovingthelandingpropertyoftheparameterrays

R 

(1=6) and R



(1=3). We will use an argument similar to the one used by

Douady and Hubbardin [DH1℄. Wewill need to modify it slightlyin the ase

=1. R  (1=2) R  (1=3) R  (1=6) W0 R  (0=1) R  (2=3) R  (5=6) M 

Fig.5. TheparameterraysR



(1=6)andR



(1=3)landatb

0

,whereastheraysR



(2=3)and

R 

(5=6)landat b0.

Proposition 8.The parameter rays R



(1=6) and R



(1=3) land at the same

parameterb

0

satisfyingb

2 0

=4( 1).TheparameterraysR



(2=3)andR

 (5=6)

3 3i 3 3i 3+3i 3+3i R  (1=3) R  (2=3) R  (5=6) R  (1=6)

Fig. 6. The parameterspa e for  = 1. The four rays R

 (1=6), R  (1=3), R  (2=3) and R  (5=6)landat0.

Remark.When=1,wehaveb

0

=0andthefourrayslandat0(seeFigure6).

Proof.In the ase 6=1, we willshowthat for any parameterb

0

ontainedin

the a umulation set of therayR



(1=6), f

b0

has aparaboli xed pointwith

multiplier 1. The set of su h parameters is dis rete { in fa t b

2 0

= 4( 1).

Sin ethea umulationset ofanyrayis onne ted,thiswill provethat theray

R 

(1=6)lands. Asimilarargumentshowsthatthe raysR

 (1=3), R  (2=3)and R  (5=6)land at b 0 or b 0

. We will then haveto showthat therays R

 (1=6)

andR



(1=3)landatthesameparameter.

Inthe ase=1,wewillshowthattheonlyparameterinthea umulation

set of the rays R

 (1=6), R  (1=3), R  (2=3) and R  (5=6) is b 0 = 0. This will

on ludetheproofoftheproposition.

Lemma4.For any parameter b

0

ontained in the a umulation set of the ray

R  (1=6),R  (1=3),R  (2=3)orR  (5=6),thepolynomialf b0

hasaparaboli xed

pointwithmultiplier1.

Proof.LetusprovethislemmafortherayR



(1=6).Wewillpro eedby

ontra-di tion. Assume that f

b0

hasno paraboli xed pointwith multiplier 1. Sin e

b 0

2M



,thedynami alrayR

b0

(1=2)doesnotbifur ate.Itisaxeddynami al

ray. Hen e, it lands at a xed point , whi h is either repelling, orparaboli

We laimthat forb suÆ iently loseto b 0

,the ray R

b

(1=2) still lands ona

repellingxedpointoff

b

.Theproofis lassi aland anbefoundintheOrsay

Notes[DH1 ℄.

Thus,foranyb2U

1

,therayR

b

(1=2)doesnotbifur ateona riti alpoint.In

parti ular,thedynami alrayR

b

(1=6) annot ontainthe o- riti al point.But

thispre iselyshowsthattheparameterrayR



(1=6)omitstheneighborhoodU

1

ofb

0

whi hgivesthe ontradi tion. 

The xed points of the polynomial f

b

are 0 and the roots of the equation

 1+bz+z

2

=0. If 6= 1, there is a multiple root (i.e., aparaboli xed

pointwithmultiplier1)ifandonlyifthedis riminantiszero:b

2

4( 1)=0.

Hen e,when6=1,weseethattheparameterraysR

 (1=6),R  (1=3),R  (2=3) andR 

(5=6) anonlya umulate onb

0 or b 0 ,whereb 2 0 =4( 1). Sin ethe

a umulationset ofarayis onne ted, wehaveprovedthatthose rayslandat

b 0

or b

0 .

When=1,theoriginisapersistentlyparaboli xedpointwithmultiplier

1.Hen e,tobeableto on ludethattheparameterraysland,wemustimprove

ourlemma. Thefollowinglemma ompletestheproof oftheproposition in the

ase=1.

Lemma5.When  = 1 the parameter rays R

 (1=6), R  (1=3), R  (2=3) and R  (5=6)land atb 0 =0.

Proof. Let us prove this lemma for the parameter ray R



(1=6). The proof is

essentiallythesameasinlemma4.Wepro eedby ontradi tion,assumingthat

theparameterrayR



(1=6) a umulatesonb

0

6=0.

Ontheonehand,thedynami alrayR

b 0

(1=2) annotlandatarepellingxed

point, sin e otherwise there would be a neighborhood U

1 of b 0 in whi h the dynami alrayR b

(1=2)wouldnotbifur ate(asin lemma4).

Ontheotherhand,ifthedynami alrayR

b0

(1=2)werelandingataparaboli

xed point with multiplier 1 (i.e., the xed point 0) then we ould still show

thatthere existsaneighborhood U

1

inwhi hthedynami alrayR

b

(1=2)would

notbifur ate.Theideaoftheproofisthefollowing.

Sin e b

0

6=0,theparaboli xed point0is simple, i.e., f

00 b

0

(0)6=0. Wewill

showthatwe anfollow ontinuouslyarepellingpetalP

rep (b)inaneighborhood U 0 ofb 0

. Onthis repelling petal,the inverse bran hesf

1 b : P rep (b) !P rep (b)

are well dened and iterates of this inverse bran hes onverge to 0. We will

also showthat thedynami alray R

b 0

(1=2) enters the repellingpetal P

rep (b

0 ).

Consequently,thereexistsaneighborhoodU

1

ofb

0

su hthatforanyb2U

1

,the

dynami alrayR

b

(1=2) enters therepellingpetal P

rep

(b),and thusland at the

paraboli xedpoint0.

Let us ll in the details. Sin e we assume b

0

6= 0, there exists a

neighbor-hood U

0

of b

0

and a radius " > 0 su h that for any b 2 U

0

, f

b

restri ts to

an isomorphism between the disk V(b) entered at 0 with radius "=jbj and

f b

(V(b)). Now, observe that the hange of oordinates z 7! Z = 1=bz

on-jugatesf b :V(b)!f b (V(b))toanisomorphismF b : b V !F b ( b V),where b V =fZ2P 1 j1="<jZjg and F b (Z)=Z+1+O  1  :

1 i 1+i R b (1=2) R b (0=1) Prep(b) Patt(b)

Fig.7. Anattra tingpetalP

att

(b)andarepellingpetalP

rep

(b).Theattra tingpetalP

att (b)

is ontainedinK(f

b

)andtherayR

b

(1=2)eventuallyentersandstaysinPrep(b).

Letus hoose"suÆ ientlysmall,sothatjF

b (Z) Z 1j< p 2=2foranyb2U 0 andanyZ 2 b V.Then,denoteby b P att and b P rep these tors b P att =fZ2C   p 2=" Re(Z)<jIm(Z)jg; and b P rep =fZ2C   p 2="+Re(Z)<jIm(Z)jg: Besides,denotebyP att (b)andP rep (b)thesets P att (b)=fz2C  j 1=bz2 b P att g; and P rep (b)=fz2C  j 1=bz2 b P rep g: The set P att

(b) is alled an attra ting petal and the set P

rep

(b) is alled a

repelling petal (see Figure 7). One aneasily he kthat theassumptions on "

impliesthatforanyb2U

0 ,wehave (1) f b (P att (b))P att (b); (2) f Æn b

onvergesuniformlyon ompa tsubsetsofP

att

(b)to0;

(3) thereexistsaninversebran hf

1 b :P rep (b)!P rep (b); (4) [f 1 ℄ Æn

onvergesuniformlyon ompa tsubsetsofP

rep

LetusexpresstherayR b0

(1=2)asa ountableunionofsegments

S j =' 1 b 0 n e t j3 j t3 j+1 o ; j2Z; sothatf b0 (S j )=S j+1

.Clearly,weweethatP

att (b

0

)is ontainedinthelled-in

Juliaset K(f b 0 ). Thus, R b 0

(1=2) doesnotinterse t P

att (b

0

).Sin eweassumed

that the ray R

b 0

(1=2) lands at 0, there exists an integer j

0 su h that S j 0 is ontainedinP rep (b 0 ).Again,byshrinking U 0

ifne essary,wemayassumethat

U 0

fb2C jG(b)<3

j 0

g.This onditionimpliesthat foranyb2U

0

, theray

R b

(1=2)isdened uptopotentialatleast3

j 0 ,and S j (b)=' 1 b n e t j 3 j t3 j+1 o ; jj 0

is well dened. Finally, sin e f(z;b) j b 2 U

0

;z 2 P

rep

(b)g is open and sin e

' 1 b

depends ontinuously (even analyti ally) on b, we see that there exists a

neighborhood U 1  U 0 of b 0

, su h that for any b 2 U

1 the segment S j 0 (b) is ontainedin P rep (b).Hen e, S j 0 +k (b) =[f 1 b ℄ Æk (S j 0

(b)) iswelldened forany

k0,andtherayR

b

(1=2)landsat0.However,thisimpliesthattheparameter

rayR(1=6)doesnotinterse tU

1

. 

We still need to prove that when  6= 1, the parameter raysR



(1=6) and

R 

(1=3)landat thesameparameter.Rememberthat wedened thewakeW

0

asthe onne ted omponentof

C nR  (1=6)[R  (1=3)[R  (2=3)[R  (5=6)

that ontainstheparameterrayR

 (1=4).

Letus allb

0

thelandingpointoftheparameterrayR



(1=6).Wewillusethe

fa tthatthe onne tednesslo usM



issymmetri withrespe tto0(remember

that f

b

and f

b

are onjugateby z 7! z). The symmetry of M



showsthat

twoofthefourraysR

 (1=6),R  (1=3), R  (2=3)orR  (5=6)landat b 0 andthe

othertwolandat b

0

.Moreover,theparameterraysR

 (1=6)andR  (2=3)are symmetri ,sothatR  (2=3) annotlandatb 0 (6= b 0

).Hen e,iftheparameter

rayR



(1=3)were notlanding atb

0

,thentherayR



(5=6)would.Inthat ase,

thewakeW

0

would ontaintheparameterb =0(seeFigure 8). We willget a

ontradi tion by proving that for any parameter b 2 W

0

, the dynami al rays

R b

(0=1) and R

b

(1=2) land at the same point, whereas this is not the ase for

b=0.

Lemma6.For any parameter b 2 W

0

, the two dynami al rays R

b

(0=1) and

R b

(1=2) donot bifur ate.

Remark.Thislemmaandthefollowingoneareinfa ttrueassoonasbdoesnot

belongtooneoftheparameterraysR

 (1=6),R  (1=3), R  (2=3)orR  (5=6). Proof. If b 2= R  (1=3)[R 

(2=3); the dynami al ray R

b

(0=1) does not

bifur- ate. Indeed, if R

b

(0=1) were bifur ating, it would bifur ate on apreimage of

thees aping riti al point!

2

, i.e., there would beanon-negativen, su h that

f n b

(! 2

) belongs to therayR

b

(0=1).Sin e this isaxed ray,!

2 would belong totherayR b (0=1) and onsequently! 0 2

wouldlieoneitherR

b

(1=3)orR

b (2=3)

whi h ontradi tsthatb2=R



(1=3)[R



(2=3).Asimilarargumentshowsthatif

b2=R



(1=6)[R



(5=6);thedynami alrayR

b

b0 b0 0 b1 b 1 W0 R  (1=6) R  (5=6) R  (2=3) R  (1=3) R  (3=4) R  (1=4)

Fig.8. IftheparameterraysR



(1=6)andR



(1=3)werenotlandingatthesameparameter,

thewakeW

0

would ontaintheparameterb=0.

Lemma7.If6=1,thengivenanyb2W

0 ,theraysR b (0=1)andR b (1=2)both

landatthesamerepellingxedpoint(b)6=0.If =1,thengiven anyb2W

0 , the rays R b (0=1)and R b

(1=2)both land atthe paraboli xedpoint(b)=0.

Proof.Toprovethislemma, wewill useanideadueto PeterHassinskywhi h

hasbeenexplainedtousbyCarstenPetersen.Wehaveseenthatinthedomain

W 0

,thedynami alraysR

b

(0=1)andR

b

(1=2)donotbifur ate.Itfollowsthatthe

setX =R

b

(0=1)[R

b

(1=2)movesholomorphi allywithrespe ttotheparameter

b.Hen e,the-LemmabyMa~ne,SadandSullivan[MSS℄showsthatthe losure

ofX in P

1

movesholomorphi ally.Inparti ular,ifforsomeparameterb

1

2W

0

thetwodynami alraysR

b

(0=1)andR

b

(1=2)landatthesamexedpoint,they

dosoeverywhereinW

0

,i.e.,thereexistsaholomorphi fun tion(b)su hthat

(b)is axed pointoff

b

andisthelandingpointof thetworaysR

b

(0=1)and

R b

(1=2).Besides,themultiplierat(b)isaunivalentfun tion,thattakesvalues

inCnD. Hen e,eitherthemultiplieris onstantlyequalto1(whi h orresponds

to a persistently paraboli landing point)or it takes valuesin C nD (and the

landingpointremainsrepellinginallW

0 ).

Thus, wejust need toshowthat thereis aparameterb2W

0

forwhi h the

tworaysR

b

(0=1)andR

b

(1=2)landata ommonxedpoint,andthatthispoint

is repellingwhen 6=1,whereasit is paraboli with multiplier 1when =1.

Thisispre iselygivenbyproposition6fortheparameterb

1 = 1  (e 1=3+2i=4 ). 

To on ludetheproofoftheproposition,itisenoughtoseethatwhenb=0

and  6= 1, the two dynami al rays R

0

(0=1) and R

0

(1=2) annot land at the

samepoint.The polynomialf

0

(z)= z+z

3

is anodd polynomial. Thus, the

lled-in Julia set is symmetri with respe t to the origin. In parti ular, the

the two riti al orbits are symmetri , the Juliaset is onne ted, and the rays

land)thelandingpointsaresymmetri withrespe ttotheorigin.However,the

origin annot be the landing point of those rays be auseit is indierentwith

multiplier 6=1.Hen e,the twodynami alraysR

0

(0=1) and R

0

(1=2)land at

twosymmetri ,distin txedpoints.

We haveprovedthat in the wake W

0

, thetwodynami al raysR

b

(0=1)and

R b

(1=2)bothlandata ommonxedpoint(b)whi hdependsholomorphi ally

onb.If=1,wehaveseenthat (b)=0isadoublexedpoint,andthe ubi

polynomialf

b

hasonlyoneother xedpoint:(b)= b.If 6=1,themap f

b

hasthreedistin txedpoints:0,(b)and(b)= b (b).

Denition12.Foranyb2W

0

,we all(b)the landingpointofthedynami al

rays R

b

(0=1) and R

b

(1=2), and we all (b) = b (b) the xed point of f

b

whi h isneither0nor (b).

Remark. Sin ethefun tion isholomorphi inW

0

,thefun tionisalso

holo-morphi in W

0

. Infa t,sin eW

0

issimply onne tedanddoesnot ontainthe

parametersb

0

,itis learthatthethreexedpointsoff

b

depend

holomorphi- ally on b in W

0

, without usingthe fa t that (b) is thelanding parameter of

theraysR b (0=1)andR b (1=2). 5.Dynami s off b in the wake W 0 .

Wewillnowimproveourdes riptionofthedynami albehaviourofthe

polyno-mialf b , whenb2W 0 (seeFigure9). Proposition 9.For anyb2W 0

,the dynami s of the map f

b

isasfollows:

1.the two riti al points of f

b

are distin t and there exist two holomorphi

fun tions! 1 (b)and! 2 (b)denedinW 0

,su hthatforanyb2W

0 ,! 1 (b)and ! 2

(b) arethe two riti al pointsof f

b

,!

2

(b) beingthe es aping riti al point

wheneverb2W

0

nM



;the o- riti al pointsare!

0 i

(b)= b 2!

i (b);

2.thedynami alraysR

b

(1=6)andR

b

(1=3)donotbifur ateandbothlandata

preimage 1 (b)6=(b)of(b);theraysR b (2=3)andR b (5=6)do notbifur ate

and land at the other preimage 

2 (b)2= f(b); 1 (b)g; we dene V i tobe the onne ted omponentof C nR b (0=1)[R b (1=2)that ontains i (b);

3.ea h of the four onne ted omponents of C n

S

fj62Zg

R b

() ontains

ex-a tly oneof the four points!

1 (b), ! 2 (b), ! 0 1 (b)or! 0 2 (b); we allU i ,i=1;2,

the one ontaining !

i

(b) andU

0 i

,i=1;2,the one ontaining!

0 i (b); 4.themapf b :U 0 i !V i

,i=1;2,isanisomorphismandthemapf

b :U i !V i ,

i=1;2, isaramied overing of degree2ramiedat!

i (b).

Proof.Wewillrstshowthat we anfollowthetwo riti alpoints

holomorphi- allywhenb2W

0 .

Lemma8.Foranyb2W

0

,the two riti al pointsoff

b

aredistin t.Moreover,

thereexisttwoholomorphi fun tions !

1 (b)and! 2 (b) denedin W 0 ,su hthat

 1 (b) R b (1=2) R b (0=1) R b (1=3) R b (1=6) R b (5=6) R b (2=3) ! 0 2 (b) !1(b) 0 (b) ! 2 (b) ! 0 1 (b)  2 (b) (b) U 0 1 U 1 U 2 U 0 2

Fig.9. Thedynami alpi tureofthepolynomialf

b

whentheparameterbbelongstoW

0 .

Proof. When the two riti al points of f

b

are distin t, i.e., b

2

6= 3, we an

lo ally followthem. Sin eW

0

issimply onne ted,theproofof thelemmawill

be ompletedon e wehaveprovedthatfor anyb2W

0

, thetwo riti al points

off

b

aredistin t.

Wewillpro eedby ontradi tionandassumethatforsomeparameterb2W

0 ,

the polynomial f

b

has a unique riti al point !. The polynomial f

b

is then

onjugate by theaÆne hange of oordinate z 7! w =z ! to apolynomial

ofthe formw7!w

3

+ .TheJulia setof su h apolynomialis invariantunder

therotationw7!e

2i=3

w.ThisshowsthattheJuliasetoff

b

isinvariantunder

the rotation of angle 1=3 around !. In parti ular, the dynami al ray R

b (1=3)

(respe tivelyR

b

(2=3))istheimageofthedynami alrayR

b

(0=1)bytherotation

ofangle1=3(respe tively2=3)of enter! (seeFigure10).Forthesamereason,

thedynami al rayR

b (5=6) (respe tivelyR b (1=6))is obtainedfrom R b (1=2) by

rotating with angle 1=3 (respe tively 2=3) around !. We will show that the

dynami alraysR

b

(0=1)andR

b

(1=2) annotlandat thesamepoint(b).

Indeed,when b2W

0

, thetwodynami alraysR

b

(0=1)and R

b

(1=2) landat

(b).Byrotatingwithangle1=3,weseethat thetworaysR

b (1=3)andR b (5=6) land at e 2i=3

(b). Sin e those two rays are separated by the urve f(b)g[

R b

(0=1)[R

b

(1=2),they anonlymeetat(b).Hen e,(b)=e

2i=3

((b) !)+

2+2i 2 2i R b (5=6) R b (2=3) R b (1=2) ! R b (0=1) R b (1=6) R b (1=3)

Fig.10. TheJuliasetoff

b forb

2

=3.Thereisaunique riti alpoint!andtheJuliaset

isinvariantbyrotationofangle1=3around!.

Then, itis notdiÆ ultto he kthat the o- riti al points!

0 i (b) aredened by! 0 i (b)= b 2! i

(b), i=1;2.We stillhavea hoi eon whi h riti al point

will be labelled !

1

and whi h onewill be labelled !

2

. To omplete the proof

of(1), weneedto provethat we an hoose!

2 su h that ! 2 (b)is thees aping riti alpointoff b foranyb2W 0 nM 

.Thiswillbedonelaterandwewillnow

fo usontheproofof(2).

Lemma9.For any b 2 W

0

the dynami al rays R

b

(1=6) and R

b

(1=3) do not

bifur ate. They both land at a preimage 

1 (b) 2 f 1 b f(b)gnf(b)g of (b). The raysR b (2=3) and R b

(5=6) do notbifur ate and land atthe other preimage

 2 (b)2f 1 b f(b)gnf(b); 1 (b)g.

Proof.LetusassumethatR

b

(1=6)bifur atesforsomeparameterb2W

0

.Thenit

bifur atesonapreimageofthees aping riti alpoint!

2

,andoneofitsforward

imagebifur ateson! 2 .Sin ef b (R b (1=6))=R b

(1=2)isxed,thismeansthat!

2

belongstotherayR

b

(1=6)ortotherayR

b

(1=2).Ontheonehand,thelatter ase

isnotpossiblesin etherayR

b

(1=2)doesnotbifur ate.Ontheotherhand,sin e

b2W

0

,thees aping o- riti alpoint!

0 2

belongstoadynami alrayR

b

(),with

2℄1=6;1=3[.Hen e,theraysbifur atingon!

2

haveangle  1=32℄ 1=6;0[

and+1=32℄1=2;2=3[.Thus,therayR

b

(1=6) annotbifur ateon!

2 .

AsimilarargumentshowsthattheraysR

b (1=3),R b (2=3)andR b (5=6)donot bifur ate.

To ompletetheproofofthelemma,itisenoughtoprovethat(b)hasthree

(b) is not a riti alvalue of f b

. Indeed, we an then argue that sin e f

b

is a

lo al isomorphismin a neighborhood of 

i

(b) and sin e the rays R

b

(0=1) and

R b

(1=2) land at (b), twoof the rays R

b (1=6), R b (1=3), R b (2=3) and R b (5=6) landat 1

(b)andtwoofthemlandat

2

(b).TheonlypossibilityisthatR

b (1=6)

andR

b

(1=3)landat thesamepreimage,letussay

1

(b), andtheraysR

b (2=3)

andR

b

(5=6) landattheotherpreimage(b).

Toseethat(b)isnota riti alvalueoff

b

,wewillpro eedby ontradi tion.

Hen e,weassumethatforsomeparameterb2W

0

,one riti alpoint!ismapped

byf

b

to(b).Then,sin e(b)iseitherrepellingorparaboli withmultiplier1,

we see that ! 6=(b). Besides, we haveseenthat thetwo riti alpointsof f

b

aredistin t.Hen e,inaneighborhoodof!,themapf

b

isatwo-to-oneramied

overing,andthefourraysR

b (1=6),R b (1=3),R b (2=3)andR b (5=6)havetoland

at !.Butthis isnotpossiblesin etheraysR

b (1=6), R b (2=3) areseparatedby R b (0=1)and R b (1=2). 

Wewillnowprove(3)usingaholomorphi motionargument.

Lemma10.The set

X b =f! 1 (b);! 2 (b);! 0 1 (b);! 0 2 (b)g[ 0  [ fj62Zg R b () 1 A

undergoes aholomorphi motion asb movesin W

0 .

Proof. Thefun tions !

i

(b) and !

0 i

(b), i =1;2,are holomorphi when b 2 W

0 .

Besides, wehaveseenthat thedynami al raysR

b

(), 62Z,do notbifur ate

when b 2 W

0

, and thus, move holomorphi ally when b 2 W

0

. To prove the

lemma,weneedtoprovetheinje tivity onditionofholomorphi motions.Sin e

wealreadyknowthatthe riti alpointsaredistin t,weonlyneedtoshowthat

foranyb2W

0

,the riti alpointsand o- riti alpoints annotbelongtoanyof

theraysR

b

() ,62Z.Butthisis learsin eotherwise,oneofthoserayswould

haveto bifur ateona riti alpoint. 

Thedynami alpi tureforthepolynomialf

b1

hasbeenstudied in se tion3,

anditisnotdiÆ ultto he kthatea h onne ted omponentof

C n [ fj62Zg R b 1 ()

ontainsexa tlyoneofthefourpoints!

1 (b 1 ),! 2 (b 1 ),! 0 1 (b 1 )or! 0 2 (b 1 ).Noneof

thefourpointsare ontainedinthe set

S fj62Zg R b1 ()for anyb2W 0 : Sin e

the four points and

S

fj62Zg

R b1

() moves ontinuously when b hanges and

W 0

is onne ted,statement(3)follows.

We an now omplete the proof of (1). We hoose the fun tions !

1 (b) and ! 2 (b)sothat ! 2 (b 1

)isthees aping riti alpointoff

b 1

.Then,theboundaryof

U 0 2

(b 1

)istheunionofthetwodynami alraysR

b 1

(1=6),R

b 1

(1=3)andtheir

land-ingpoint

1 (b

1

).Usingtheholomorphi motion,weseethatthesameproperty

holds for U

0 2

(b), i.e., the boundaryof U

0 2

(b) is the union of thetwo dynami al

rays R

b

(1=6), R

b

(1=3) and theirlanding point 

1

(b). In parti ular, the region

U 0

(b) ontainsthedynami alraysR

b

know that when b2 W 0

nM



,the es aping o- riti alpoint belongsto one of

those rays. Hen e,for anyb 2W

0

nM



thees aping o- riti al point belongs

to the region U

0 2

(b). Thus the es aping o- riti al point is !

0 2

(b) and for any

b2W

0

nM



,thees aping riti alpointis!

2 (b).

Wenallyprove(4).Wehave alledV

1

(b)andV

2

(b)thetwo onne ted

om-ponentsofC nR

b

(0=1)[R

b

(1=2). Sin ethepreimagesoftheraysR

b

(0=1)and

R b

(1=2) are the rays R

b

(), 6 2 Z, the onne ted omponents of f

1 b

(V i

),

i=1;2,arethe onne ted omponentsofC n

S

fj62Zg

R b

().LetU beoneof

them.Sin e thepolynomialf

b

:C !C is aramied overing,therestri tionof

f b

toU isaramied overingontoits image.Sin e U issimply onne ted,the

Riemann-Hurwitzformulashowsthatthedegreeoftherestri tionoff

b

to U is

n+1where n is thenumberof riti alpointsof f

b

in U, ountedwith

multi-pli ity. Hen e,to nishtheproofof(4),weonlyneedto showthatf

b (U i )=V i andf b (U 0 i )=V i fori=1;2.

Lemma11.For any b 2 W

0

, the omponent U

0 2

ontains the two dynami al

raysR

b

(2=9) andR

b

(5=18)thatbothland atapreimage of

2 (b).

Proof.Wehaveseenpreviouslythat foranyb2W

0 ,theregionU 0 2 ontainsthe dynami alraysR b (),2℄1=6;1=3[.Sin e 2=92℄1=6;1=3[and5=182℄1=6;1=3[,

therstpartofthelemmaisproved.

Next,wehaveseenthatf

b

isanisomorphismbetweenU

0 2

anditsimage.Sin e

U 0 2

ontainsthe twodynami al raysR

b

(2=9) and R

b

(5=18),its image ontains

thetwodynami alraysf

b (R b (2=9))=R b (2=3)andf b (R b (5=18))=R b (5=6)that bothland at 2 (b)2V 2

andthelemma isproved. 

Sin e f

b

maps the rays R

b (2=9) and R b (5=18) whi h are in U 0 2 to the rays R b (2=3) andR b (5=6)whi hland at 2 (b)2V 2 ,wesee that f b (U 0 2 )=V 2 . Sin e ! 2 (b)and! 0 2

(b)havethesameimage,weimmediately obtainthatf

b (U 2 )=V 2 . Hen e,f b :U 0 2 !V 2 isanisomorphismandf b :U 2 !V 2 isaramied overing ofdegree2,ramiedat! 2

.Sin ethepolynomialf

b

hasdegree3,the omponent

V 2

hasnootherpreimage, andf

b (U 1 )=f b (U 0 1 )=V 1

.This nishestheproof of

theproposition.

6.Holomorphi motionof rays.

Intherestofthisarti le,wewillworkinthewakeW

0

.Wewill onstantlyhave

todealwiththe riti alpoint!

2

(b),b2W

0

.Thus,thereadermustkeepinmind

that thefun tion !

2

isaholomorphi fun tion dened throughoutallthewake

W 0

, and that for any parameterb 2W

0 nM  , the point ! 2 (b)is the es aping riti alpoint.

TheoremA.Forany parameterb2W

0

andfor any2,thedynami al ray

R b

() does notbifur ate.Wedene X

b tothe set X b = [ 2 R b (): WealsodeneJ b tobethesetJ b =X b nX b andK b

tobethe omplementofthe

Julia set K(f b

), its boundary J

b

is ontained in the Julia set J(f

b

) and K

b is

quasi- onformally homeomorphi tothelled-in JuliasetK(z+z

2 ).

Figure11showsthesetK

b

andthesetofdynami alraysX

b foraparameter b2M  \W 0 . R b (1=18) R b (0=1) R b (1=2) R b (4=9)

Fig.11. ThesetK

b

andthesetofsetofdynami alraysX

b foraparameterb2M  \W 0 .

Proof. Let us rst provethat for any parameter b 2 W

0

and any  2 , the

dynami alrayR

b

()doesnotbifur ate.Wewillmimi theproofofproposition

7.

Forany b 2 W

0

, we have dened V

2

(b) to be the onne ted omponent of

CnR b (0=1)[R b (1=2)that ontains 2

(b).Sin ethetwodynami alraysR

b (2=3) andR b (5=6)landat 2

(b),theyare ontainedin V

2

(b),andforany2[0;1=2℄,

the dynami al ray R

b

() is ontained in C nV

2

(b). Sin e for any  2 , we

have3

k

2[0;1=2℄mod 1,foranyk0,theforwardorbit oftheray R

b

()is

ontained in C nV

2

(b).Next, for any b 2 W

0

, we laim that the riti al point

! 2

(b){whi h isthees aping riti alpointwhenb2W

0 nM  {belongsto the region V 2

(b). Indeed, we have seen that (b) annot be a riti al value of f

b .

Hen e,theset f!

2 (b); 2 (b)g[R b (0=1)[R b

(1=2)movesholomorphi allywhen

b 2 W 0 . Hen e, ! 2 (b) and  2

(b) are always in the same onne ted omponent

of C nR

b

(0=1)[R

b

(1=2). Now, assume that there exists an angle  2  su h

thatthedynami alrayR

b

()bifur ates.Then,itbifur atesonapreimageofthe

Butthis ontradi tsthefa tthattheforwardorbitoftherayR b

()is ontained

inC nV

2

(b)whi hdoesnot ontain!

2 (b).

Next, observe that the mapping h : W

0 X b 1 ! X b dened by h(b;z) = ' 1 b Æ' b1 (z)is aholomorphi motionof X b1 parametrizedby b 2W 0 .The

-LemmabyMa~ne,SadandSullivan[MSS℄showsthathextendstoaholomorphi

motionofthe losureX

b1

ofX

b1

inC. Sin eW

0

isasimply onne tedRiemann

surfa e,S`odkowski'sTheorem(see[Sl℄[D2℄)showsthatone anin fa textend

h to a holomorphi motion of the whole omplex plane C, still parametrized

by b2 W

0

. We will keep thenotation hfor this extension. Themappingz 7!

h b

(z) = h(b;z) is a K(b)-quasi- onformalhomeomorphism, where K(b) is the

exponentialofthehyperboli distan e betweenb

1

andb inW

0

.Itmapstheset

of dynami al rays X

b 1

to the set of dynami al rays X

b , and h b X b 1 nX b 1
= X b nX b

: Sin e  is losed, theset J

b

is ontainedin theJulia setJ(f

b

). Sin e

K(f b

)isfull, theset K

b

is ontainedin thelled-inJuliaset K(f

b

)Finally, h

b

providesaquasi- onformalhomeomorphismbetweenK

b

andK

b1

,andsin eK

b1

is quasi- onformallyhomeomorphi to thequadrati Juliaset K(z+z

2

) (see

propositions5and7),TheoremA isproved.

Observethat themapping h

b

onjugatesthepolynomialsf

b1 and f b on the setofraysX b1

,i.e.,foranyz2X

b1 wehaveh b Æf b1 =f b Æh b .By ontinuityof h b

,thispropertyholdsonthe losureX

b 1 andinparti ularonJ b 1 .

Observe also that the xed point0 never belong to the set X

b

so that the

setX

b

[f0gmovesholomorphi allywhenb movesin W

0

.Inparti ular,we an

hoose theextension hsothat h(b;0)=0foranyb2 W

0

. Sin e 02K

b1

,this

showsthatforanyb2W

0

,0belongstoK

b .

Wenally would liketo mentionthatwe ould hoose theextensionofhso

that h

b

onjugatesthe polynomialsf

b 1

and f

b

onthe whole set K

b 1

, and su h

thatthedistributionalderivativeh

b

=zvanishesonK

b1

.Butthiswouldrequire

extraworkand wewilljust mentiontheideaoftheproof.We ouldrstprove

that for anyb 2 W

0 , there is a restri tionof f b : U 0 b !U b to a neighborhood of K b

whi h is aquadrati -likemap. We ould then prove asin proposition 5

that the hybrid lass of this quadrati -like restri tion ontains the quadrati

polynomialz7!z+z

2

.Inparti ular,foranyb2W

0

,thepolynomial-likemaps

f b :U 0 b !U b and f b 1 :U 0 b1 !U b 1

wouldbehybrid onjugate, i.e.,there would

existaquasi- onformalhomeomorphismh

b :U b1 !U b su hthath b Æf b1 =f b Æh b onU 0 b1

andsu hthat thedistributionalderivativeh

b

=zvanishesonK

b1

.We

wouldnallyhavetoprovethattherestri tionofthemapping(b;z)7!h

b (z)to W 0 K b1

givesaholomorphi motionK

b1

extendingh.

7.The dyadi wakes W

# .

Observethatin thewakeW

0

wesee a opyM

0

of aMandelbrotset,with root

pointatb

0

.Inthisse tion,wewillexplainwhywesee su ha opy,andwewill

determine aCantor set 

0

su h that theboundary of M

0

is the a umulation

setoftheparameterraysR



(),2

0 .

The reason why su h a opy appears is that for any b 2 W

0 , the mapping f b : U 2 ! V 2

is a ramied overing of degree 2, ramied at !

2

. The sets U

2

and V

2

are topologi al disksand U

2

V

2

, and thefamily (f

b : U 2 !V 2 ) b2W 0