On Reduced Lantern Relations in Mapping Class Groups?

Chaohui ZHANG

1 Statement of Results

LetSbe a hyperbolic Riemann surface of type(p,n)with a finite area,wherepis the genus andnis the number of punctures ofS.Assume throughout that 3p+n>3.Let H denote the hyperbolic plane.By the uniformization theorem(see[8]),there is a holomorphic covering mapfrom which we can obtain a covering groupGwhich acts on H as isometries and is a torsion free,finitely generated Fuchsian group of the first kind.

Denote bythe surface obtained fromSwith one pointxremoved.Letdenote the subgroup of the mapping class group onthat consists of mapping classes isotopic to the identity asxis filled in.It is well-known(see[3,5])thatis the image ofGunder the socalled “Bers isomorphism”.In the literature,elements ofare called point-pushing mapping classes.

Letbe the subset ofconsisting of elements with formsorwherea,bare simple closed geodesics onandtcis the positive Dehn twist about a geodesicc.It is clear that ifand botha,bare non-trivial curves onS,then=.Here and hereafter,we use the symbols a andto denote the geodesics onShomotopic toaandb,respectively.In the case where the pair(a,b)fills,that is,a∪bintersects every simple closed geodesic on,then by Thurston’s theorem(see[14]),are pseudo-Anosov.The elementhofGcorresponding tois called an essential hyperbolic element(see[10]for more information).

The main purpose of this article is to clarify the situation when the product of two parabolic elements ofGcan be identified with an element of(the product always belongs to).To state our results,we need some geometric and topological terms related to the mapping class group.

Let D be a thrice punctured disk with three puncturesx′,y,z.Denote byDthe boundary of D.Leta,b,c?D be the boundaries of twice punctured disks enclosing{x′,y},{x′,z}and{y,z},respectively,such thata,bandcpairwisely intersect twice(see Figure 1).Then the classical lantern relation(see[6,11])is simplified to the relation

Usually,ifcontains at least three punctures,D can be embedded intoin such a way thatx′is identified withxand{y,z}can be identified with two other punctures ofS.Thusa,b,candDcan be considered simple closed geodesics on.In this situation we call these geodesicsa,b,candDgeometrically related in Figure 1.

Figure 1 A reduced lantern relation

Assume thatSis non-compact.ThenGcontains infinitely many parabolic elements andcontains at least two punctures.For eachh∈G,leth?denote the corresponding element in.By Theorem 2 of[10],h?is the Dehn twist(positive or negative)along a geodesiceonthat is the boundary of a twice punctured disk enclosingxif and only ifhis a primitive parabolic element.In this case,eis a trivial loop onSand is called a preperipheral geodesic.

We first prove the following result.

Theorem 1.1Let h1,h2∈G be parabolic elements such thatandAssumethatwhich allows us to writeasorfor some simple closedgeodesics c,D on.Thencontains at least three punctures(so S contains at least two punctures)and a,b,c,D are geometrically related in Figure1.

During a conference hosted by AMS in 2002,J.D.McCarthy asked a question about how to characterize geometric relations by means of algebraic relations among various Dehn twists.We use the symboli(a,b)to denote the geometric intersection number betweenaandb.To the best knowledge of the author,only the following relations are well-known(see[9]):

(1)if and only ifj=kanda=b,

(2)if and only ifi(a,b)=0,and

(3)anda≠bif and only ifj=k=±1 andi(a,b)=1.

Some results related to the classical lantern relation and the chain relation were found in Margalit[12]and Hamidi-Tehrani[7].Their results rely on the strong hypothesis that some words generated bytaandtbare multi-twists(defined by finite collections of disjoint simple closed geodesics).

The proof of Theorem 1.1 leads to the following result,which does not impose any condition on commutativity and disjointness among simple closed geodesics and thereby gives a partial solution to the problem posed by McCarthy.

Theorem 1.2Let S be a Riemann surface of type(p,n)with a finite hyperbolic area.Assume that3p+n>3and n≥1.Let a,b,c and D be simple closed geodesics on S.Then the relation(1.1)holds if and only if S contains at least three punctures and a,b,c,D are geometrically related in Figure1.

Here is the outline of this paper.Section 2 is dedicated to preliminaries which include some definitions and well-known facts.In Section 3 we investigate parabolic loops in the fundamental group ofS.In Section 4,we prove several related lemmas.In Section 5,we prove Theorem 1.1 and Theorem 1.2.Section 6 includes a technical lemma that handles the case whereis also represented by a product of two Dehn twists along non-preperipheral geodesics.

2 Background and Preliminaries

LetGbe a Fuchsian group of the first kind that acts on H as a group ofisometries so thatElements ofGare either hyperbolic or parabolic,and every hyperbolic elementgkeeps invariant a unique oriented geodesic axis(g)called the axis ofg.

Letπ1(S,x)denote the fundamental group ofS.Thenπ1(S,x)is isomorphic toG.Letε:G→π1(S,x)be an isomorphism.An elementg∈Gis hyperbolic if and only ifε(g)is represented by a non-trivial closed geodesic;g∈Gis parabolic if and only ifε(g)is represented by a loop around a puncture ofS.More precisely,a hyperbolic elementg∈Gis simple if and only if?(axis(g))is a simple closed geodesic;it is essential hyperbolic if and only if?(axis(g))is a filling closed geodesic(in the sense that every component ofS?(axis(g))is either a polygon or a once punctured polygon);it is non-simple and non-essential if and only if?(axis(g))is a non-simple and non-filling closed geodesic.

LetT(S)denote the Teichmüller space ofS.That is,T(S)is the space of all conformal structuresμ(S)onSquotient by an equivalent relation,where two conformal structuresμ:S→μ(S)andμ′:S→μ′(S)are equivalent if and only if there is a conformal mapc:μ(S)→μ′(S)such that(μ′)?1cμis isotopic to the identity.The equivalence class ofμis denoted by[μ].It is well-known thatT(S)is a complex manifold of dimension 3p+n?3.

LetV(S)be the fiber bundle overT(S)so that any fiber ofV(S)over[μ]∈T(S)is the Riemann surface representing[μ].ThenV(S)is also a complex manifold of dimension 3p+n?2,and its universal covering manifoldF(S)is called the Bers fiber space.The fiber over[0]∈T(S)(represented byS)is the central fiber whichis identified with the hyperbolic plane H.Thus the covering groupGnaturally acts onF(S)that preserves each fiber inF(S).A remarkable result of Bers[3]states that there exists an isomorphismφofF(S)ontoT(),which induces(by conjugation)an isomorphismφ?ofGonto.

By Theorem 2 of[10],g∈Gis a primitive parabolic element if and only ifg?is a simple Dehn twisttaalong the boundaryaof a twice punctured disk enclosingx;g∈Gis essential hyperbolic if and only ifg?is pseudo-Anosov(in the sense of[14]);g∈Gis a simple hyperbolic element if and only ifg?is a spin mapwherekis an integer and{c1,c2}are the boundary components of anx-punctured cylinder on.Finally,g∈Gis non-simple and non-essential if and only ifg?is a pure mapping class that has a unique pseudo-Anosov component onthat contains the puncturex.

LetQ(G)denote the group of quasiconformal automorphismswof H such thatwGw?1=G.Two such mapsw,w′∈Q(G)are said to be equivalent ifwgw?1=w′g(w′)?1for everyg∈G.It is well-known thatGcan be regarded as a normal subgroup ofQ(G)/～andφ?extends to an isomorphism ofQ(G)/～onto thex-pointed mapping class groupof.Let[w]denote the equivalence class of an elementw∈Q(G)and[w]?denote the image of[w]∈Q(G)/～under the isomorphism

Leta?be a simple closed geodesic that is non-trivial onSasxis filled in.Letdenote the(non-trivial)simple closed geodesic homotopic toaonS.Thus the positive Dehn twist? defines a special non-trivial reducible mapping class.Let?H be a geodesic so thatDenote bythe components ofThen,? and ?′are invariants under the action of a simple hyperbolic element ofG.The Dehn twist?can be lifted to a mapτa:H→H with respect to?,say,which satisfies the conditions

(i)

(ii)

In addition to(i)and(ii)above,τadefines a collectionof half planes in H in a partial order defined by inclusion.There are infinitely many maximal elements ofall maximal elements(? is one of them)ofare mutually disjoint,and the complement

is not empty.In fact,it is a convex region bounded by a collection of disjoint geodesicswithIt is clear thatcontains infinitely many maximal elements ofand the mapτaconstructed above keeps each maximal element invariant and has the property that

The mapτaso obtained depends on the choice of a geodesicwithbut it does not depend on the choice of a boundary component of ?a.Moreover,τadetermines an element[τa]∈Q(G)/～.By Lemma 3.2 of[15],we can properly choose(and ?)so thatis represented by the Dehn twisttaalonga.If we use ?′to acquire a lifting mapτa′of?,we havewherea0together withaforms the boundary of anx-punctured cylinder.See[15,18]for more details.In the rest of this paper we call the triplethe configuration corresponding toa.

3 Products of Parabolic Elements

Letx0=x,x1,···,xndenote the punctures of.Letbe the set of preperipheral geodesics enclosingxandxi.Let

Assume thata,Thenaandbare trivial loops onSas the puncturexis filled in.This is equivalent to thataandbare preperipheral and thus are the boundaries of twice punctured disksD(a)andD(b)that enclosex.

By Theorem 2 of[10]and Theorem 2 of[13],there exist primitive parabolic elementsTa,Tb∈Gsuch thatandUnder the isomorphismε:G→ π1(S,x),TaandTbcorrespond to parabolic loopseaandebpassing throughx,respectively,such thateagoes aroundx1,andebalso goes around a puncturexi.Note thatea,ebgo around the same puncture if and only ifTaandTbare conjugate to each other inG.

Letd(a)be the deformation retract ofD(a),that is,d(a)is a path onSconnectingxandx1so thatD(a)can be reconstructed from fatteningd(a).Likewise,letd(b)be the deformation retract ofD(b).Clearly,d(a)andd(b)determine the parabolic loopseaandebonSpassing throughx,respectively.Assume thatd(a)andd(b)intersect in a minimum number of points.We sayd(a)andd(b)are disjoint if they only meet atx.In this case,(d(a),d(b))forms a binary tree with two leavesx1andxi.IfD(a)andD(b)share both punctures,then by our convention,d(a)intersectsd(b).

Lemma 3.1Let[σ′]∈π1(S,x)correspond to the product TbTa.Then any representative of[σ′]is freely homotopic to a trivial or simple closed geodesic σif and only if d(a)and d(b)are disjoint.

ProofObviously,ifd(a)andd(b)are disjoint,i.e.,(d(a),d(b))forms a binary tree with two leavesx1andxi,thenea·ebis homotopic to the boundary of a twice punctured disk enclosingx1andxi.The converse can be proved by a geometric argument.Suppose thatd(a)andd(b)intersect at a minimum number of intersection points={ui;1≤i≤k}.Figures 2(a)–(b)show the first two such pointsu1andu2in two different situations.

Note that eachuicontributes four intersection points betweeneaandebwhich form vertices of a quadrilateralQi.Figure 3 illustrates some details of the curve concatenationea·ebatu1andu2,and atxbased on Figure 2(a).Verticesuij,1≤j≤4,of each quadrilateralQiare labeled counterclockwise.

Figure 2 Paths d(a)and d(b)and their intersection points

Figure 3 Fattenings of d(a)and d(b)produce D(a)and D(b)as well as the product ea·eb

Sinceσis freely homotopic toea·eb,ifσis a simple closed geodesic,then during the deformation,the points inand 1≤j≤4}are canceled in pairs,where at least one pair,according to the so-called bigon principle,constitutes vertices of a bigon.So it suffices to check if there exists a pointuijintogether with its neighboring point that forms vertices of a bigon.This can be done by examining each pointuifor 1≤i≤k.The case wherei=1 andj=4,is slightly different.Ifu14is the vertex of a monogonR,thenu1is not in,which contradicts thatd(a)andd(b)intersect at a minimum number of intersection points.

In the cases wherei>1,ori=1 andj≠4,each vertexuij,1≤j≤4,of the quadrilateralQiobtained fromuican not be canceled with any other vertex ofQi.Ifui2andu(i+1)1are also vertices of a bigon,thenuiandui+1are vertices of a bigon formed byd(a)andd(b).In this case,uiandui+1can be removed from.This contradicts thatd(a)andd(b)intersect at a minimum number of intersection points.After a finite number of steps,we see that there is no bigon in the complement ofea·eb,that is to say,no points incan be deleted.This leads to a contradiction.The case of Figure 2(b)can be handled in the same way.

Letza,zbdenote the fixed points ofTaandTb,respectively.Conjugating by a M¨obius transformation if necessary,we may assume without loss of generality thatzaandzbare south and north poles on S1,respectively.LetLandRdenote the left and right components of S1{za,zb},respectively.See Figure 4.

For each pointz∈R,one checks thatTbTa(z)≠z.Hence there are no fixed points ofTbTaonR.So the fixed point(s)ofTbTamust lie onL.The following lemma shows that there are actually two fixed points ofTbTaonL.

Lemma 3.2If3p+n>3,then TbTa∈G is hyperbolic.

ProofBy assumption,a,b∈(x).Ifd(a)andd(b)intersect,then by Lemma 3.1,TbTais not parabolic.Ifd(a)andd(b)are disjoint,then by Lemma 3.1 again,[σ′]is represented by a trivial or simple closed geodesicσ.Note that(p,n)≠(0,3),which impliesis not of type(0,4).Thusσis not trivial,which says thatσis a non-trivial simple geodesic.SoTbTais hyperbolic.

Figure 4 The product of Taand Tbgives a hyperbolic element

By Lemma 3.2,g:=TbTa∈Gis hyperbolic,whose axis axis(g)meets1on the left componentLof1{za,zb}.We also know that the orientation of the axis is as shown in Figure 4.Otherwise,suppose that axis(g)takes an opposite orientation to the one shown in Figure 4.Thengandhave the same relative motion direction.By the same proof of Lemma 7.1 of[16],is hyperbolic,which would contradict thatandTais a parabolic element ofG.

4 Lantern Relation in a Reduced Form

In this section,we assume thata,b,candDare simple closed geodesics on,such that

wherea,b∈(x)(in the case wherethe discussion is the same).As usual,we letanddenote the geodesics homotopic toa,b,c,DonS,respectively.

Lemma 4.1With the above conditions,eitherandare trivial,orandare nontrivial.

ProofIfis trivial and ecis non-trivial,oris non-trivial and ecis trivial,then it quickly leads to a contradiction by filling in the puncturexin(4.1).

Lemma 4.2Assume that a,b∈(x)and satisfy(4.1).Then eitheroris non-trivial.

ProofSuppose that bothandare trivial.Thena,b,c,D∈(x)satisfy(4.1).We claim that this does not occur,and the contradiction will complete the proof of the lemma.

Indeed,there are primitive parabolic elementsTa,Tb,Tc,TD∈Gsuch thatandDenoteThese loops are parabolic inπ1(S,x).

By Lemma 3.2,TbTa∈Gis a hyperbolic element.Thus the curve concatenation

is homotopic to a non-trivial closed geodesicσ.As discussed in the proof of Lemma 3.1,during the homotopy fromea·ebtoσ,intersection points can be canceled only in pairs.Note that every interior intersection point betweend(a)andd(b)contributes four intersection points betweenaandb;nearx,aandbintersect twice.In addition,ifd(a)andd(b)intersect at the other endpointy,thenaandbintersect twice neary.We conclude thataandbintersect in an even number of points.Soσhas an even number of self-intersection points.

On the other hand,sincec,D∈(x),we havei(c,D)>0.Thus from the above argument,c=?D(c)andD=?D(D)have an even number of intersection points,but the curve concatenationhas an additional self-intersection point atx.So the number of self-intersection points ofis odd.During the homotopy fromto the geodesicσ,the self-intersection points could cancel only in pairs.We conclude that the number of selfintersection points ofσis odd.It follows thatThus via the Bers isomorphism,Similarly,we can prove

From Lemmas 4.1–4.2,we conclude that bothandare non-trivial.As a matter offact,more is true.

Lemma 4.3With the same conditions as in Lemma4.2,c and D are disjoint,and hence c and D are the boundary components of an x-punctured cylinder on.

ProofBy Lemma 6.1,we assert thati(c,D)=0.So eitherc=DorcandDare disjoint.Ifc=D,then from(4.1),tbtais trivial.But this is impossible sincea,b∈(x)and thusaandbintersect.We assume thatcandDare disjoint.By filling the puncturex,from(4.1),we see thatprojects to the trivial mapping class onS.But we know thatandare non-trivial geodesics.So=.It follows thatcandDare boundary components of anx-punctured cylinder on.

Lemma 4.4Under the same notations and conditions as above,D is disjoint from a and b,or equivalently,D is disjoint from d(a)∪d(b).

ProofLet?denote the component of Haxis(TbTa)that does not includezaandzb(as shown in Figure 4).Note thatWith the help of? one can construct a mapwhichis a lift of the Dehn twistwhere in factFrom Lemma 3.2 of[15],(see Section 2 for more details).

Letbe the configuration obtained from.By construction,?∈.Note thatTbTakeeps the set of maximal elements ofinvariant.If there is a maximal elementthat coverszabut notzb,thenand thuswhich says that ?0is not a maximal element of.If?0covers bothzaandzb,then either(i)is disjoint fromzbor(ii)coverszb.When(i)occurs,is disjoint from;when(ii)occurs,we haveTbTa(?0)??0.All these would imply thatTb(Ta(?0))is not a maximal element of.This contradiction tells us thatzacan not belong to any maximal element of.

Now by considering the inverseofTbTa,one can show thatzbcan not belong to any maximal element of.Hence bothzaandzb∈?∩S1.It follows that bothTaandTbcommute withτ.If[τ]?=tc,thentccommutes withtaandtb,and sotccommutes withtbta.But we havetctbta=tD,which implies thatcintersectsa∪b.This is absurd.We conclude thatThus bothtaandtbcommute withtD(buttaandtbdo not commute with each other).That is,Ddoes not intersecta∪b.

We now proceed to study the properties of conjugate parabolic elements and their products.Assume thata,b∈(x,x1),whichis equivalent to thatTaandTbare conjugate inG.

Lemma 4.5If Tbis conjugate to Tain G,then TbTa∈G is hyperbolic but not a simple hyperbolic element unless a=b.

ProofFrom Lemma 3.2,TbTais hyperbolic.Ifa≠bandTbTais simple hyperbolic,thenTbTacorresponds to a simple closed geodesicγinπ1(S,x).

By assumption,there is an elementh∈Gsuch thatTb=hTah?1.ThusTbTa=hTah?1Ta.Note thatd(b)=h?(d(a))determines a parabolic loopeb,butebis also defined byhTah?1.We see thathTah?1Tadetermines a loopea·eb∈π1(S,x).Sinced(b)=h?(d(a)),d(a)andd(b)share both endpoints{x,x1}.This implies that the curve concatenationea·ebis homotopic to a geodesic with at least two self-intersection points(two of which are near the puncturex1).In other words,the axis ofhTah?1Taprojects to a non-simple closed geodesic.It follows from the definition thatTbTais not a simple hyperbolic element.

A mapping classMis called a multi-twist ifMis represented by a finite product of Dehn twists about disjoint simple closed geodesics.

Lemma 4.6If Tbis conjugate to Tain G,then(TbTa)?is not a multi-twist unless a=b,in which case Tb=Taand(TbTa)?is a power of a Dehn twist.

ProofAssume thatb≠aand(TbTa)?=Mis a multi-twist.SinceTbTa∈G,by Theorem 2 of[10],ifTbTais an essential hyperbolic element,or a non-simple non-essential hyperbolic element,then(TbTa)?can never be multi-twist.It follows that(TbTa)?is either parabolic or simple hyperbolic.By Lemma 3.2,(TbTa)?is not parabolic.So(TbTa)?must be simple hyperbolic.But this again contradicts Lemma 4.5.Ifa=b,thenTa=Tb.Soand hence

5 Proof of Theorems

Proof of Theorem 1.1We only handle the case where(TbTa)?is of the form(4.1).Suppose thatcontains only two puncturesxandx1.Thena,b∈(x,x1),and thusTa,Tbare conjugate inG.Sincea≠b,by Lemma 4.6,(TbTa)?is not a multi-twist,which implies thatcandDare not disjoint.On the other hand,since(TbTa)?∈0is of the form of(4.1),by Lemmas 4.1–4.2 and Lemma 6.1 in Appendix,we conclude thatcandDdo not intersect.This contradiction proves thatcontains at least three punctures.

Assume thata∈(x,x1)andb∈(x).By Lemmas 4.1–4.2,bothandare non-trivial.Lemma 4.4 then asserts thatcis disjoint fromDand{c,D}actually bounds anx-punctured cylinder onS.This implies thatis a multi-twist.By Theorem 2 of[10]and Theorem 2 of[13],there exists a simple hyperbolic elementh∈Gsuch thatButIt follows thath=TbTa,which tells us thatTbTais a simple hyperbolic element ofG.Hence by Lemma 4.5,Tais not conjugate(inG)toTb.As it turns out,b∈(x,xi)for somexi≠x1.Moreover,by Lemma 3.1,d(a)andd(b)are disjoint,which says that(d(a),d(b))forms a binary tree with two leavesx1andxi.

By Lemma 4.4,Dis disjoint fromd(a)∪d(b).This means thatDis disjoint froma∪b.Finally,to see thatDbounds a thrice punctured disk onS,we observe that the curve concatenationea·ebis homotopic to.But since(d(a),d(b))forms a binary tree with leaves{x1,xi},it is obvious thatea·ebbounds a twice punctured disk onSwhich encloses{x1,xi}.From the above argument,Dis disjoint froma∪b.IfDdoes not bound a thrice punctured disk,thenis not the boundary of any twice punctured disk,which leads to a contradiction.

We conclude thatDbounds a thrice punctured disk.Since{c,D}bounds anx-punctured cylinder onS,cbounds a twice punctured disk enclosing{x1,xi}.Thus Figure 1 has been reconstructed.This proves thata,b,candDare geometrically related by Figure 1.

To prove Theorem 1.2,we need some preliminary results.

Lemma 5.1Let a,b,c?be simple closed geodesics.We have the following claims:

(1)If tatbis trivial,then both a and b are trivial.

(2)Ifis trivial,then either a and b are trivial,or a=b.

(3)If tatb=tc,then either a,b and c are trivial,or a is trivial and b=c,or b is trivial anda=c.

(4)Ifthen either a,b and c are trivial,or a and b are non-trivial and c is trivial,or b is trivial and a=c,and

(5)Ifthen a,b and c are trivial.

Proof(1)Ifaandbare non-trivial,thenIfais trivial andbis non-trivial,orais non-trivial andbis trivial,thentatbis a simple Dehn twist that is also non-trivial.

(2)Ifis trivial,thenta=tb,which implies thata=bor bothaandbare trivial.

(3)Suppose that not alla,bandcare trivial.Ifais trivial,thentb=tcand thusb=c;otherwiseais non-trivial.Ifbis non-trivial,thentatbcan not be a single Dehn twist.It follows thatbis trivial.Thusta=tcand soa=c.

(4)Suppose that not alla,bandcare trivial.Ifbis trivial,thenta=tc,which saysa=c.Otherwise,bis non-trivial.Ifais also non-trivial,the only possibility is thatcis trivial anda=b.Ifais trivial,thenwhichis impossible.

(5)If only one ofa,bandcis trivial,thenIf any two ofa,bandcare trivial,the other one must also be trivial.If alla,bandcare non-trivial andaandbare disjoint,thentatbis multi-twist whileis a single Dehn twist.SoIfaandbintersect,thentatbcan not be a single Dehn twist either.

Proof of Theorem 1.2We first assume thatandare non-trivial(this is automatically true whenn=1;that is,Scontains only one puncture).Iforis trivial,then from(4.1),Lemma 5.1(3)and(5),we assert thatoror both are trivial.This is contradiction.Ifandare non-trivial,there are four subcases to consider:(i)i(c,D)=0,i(a,b)=0,(ii)i(c,D)>0,i(a,b)>0,(iii)i(c,D)=0,i(a,b)>0,and(iv)i(c,D)>0,i(a,b)=0.

Ifi(c,D)=0,thentbtais either the square of a positive Dehn twist or a multi-twist with two positive components.Clearly,(i)does not hold(sinceis either trivial or a multi-twist with one positive and one negative components).(iv)says thatcandDintersect.From Thurston’s theorem[14],we see that on the surface supported bycandD,is pseudo-Anosov.So(iv)can not happen either.

Figure 5 ?a∩?b=?

Figure 6 ?c∩?D=?

To handle the other two cases,we letbe the configurations corresponding toa,b,candD,respectively.

Suppose(ii)occurs with ?a∩?b≠?.Note thatτbτahas no fixed points on S1,whilehas two or infinitely many fixed points on S1.We see thaton S1.Now assume that ?a∩?b=?.There are maximal elementsandsuch thatWe refer to Figure 5 whereandlikewise,and(here and hereafter we denote by(AB)the minor arc on S1connecting two non-antipodal pointsAandBon1).

By examining the action ofτbτaon S1,we see that the fixed points forτbτa(if exist)must lie on the arc(D′C).LetQbe the fixed point ofτbτathat is closest toC.ThenQis also a fixed point ofIfthere are maximal elementsandsuch thatWe haveSee Figure 6.

For anyz∈1,letdenote the Euclidean length of the arc ofdetermined by the motion direction ofτbτaatz.Thenzis a fixed point ofτbτaif and only iffor an integerk.Similarly,we useto denote the Euclidean length of the arc ofdetermined by the motion direction ofatz.Since the motion directions ofandτDare opposite,zis a fixed point ofif and only ifNow we choose a sequencewithz0=C,andzn→Qfrom right.

Notice that 0L(whereLis the arc length ofWe conclude thatandIt follows thaton S1.

Similarly,we can handle the case where ?c∩?D≠?.

Suppose that(iii)occurs with ?a∩?b≠?.In this case,?c∩?D≠?.It is clear thatτbτahas no fixed points on S1,while there are infinitely many fixed points forThis is a contradiction.If ?a∩?b=?,a contradiction can also be derived by the similar argument as above(in this case,zis a fixed point ofif and only if

Note that for a surface with one puncture, ?,are automatically non-trivial.We conclude that there is no relation(4.1)onwhenn=1.

It remains to consider the case whereScontains two or more punctures andoror both are trivial.Suppose thata∈(x,x1).Our first claim is thatc/∈(x,x1).For otherwise,tcis conjugate totainand from(4.1),we obtain

Hence

Sincea∈(x,x1),we havetb(a)∈(x,x1).This implies thatcandtb(a)intersect.So ifi(b,D)=0,then the right side of(5.1)is a multi-twist or the identity,while the left side of(5.1)is neither the identity nor a multi-twist.This leads to a contradiction.We conclude thati(b,D)>0.But sincec,tb(a)∈(x),by Lemma 5.1,eitherandare trivial,or bothandare non-trivial and=.The former would contradict Lemma 4.2,and the latter would contradict Lemma 6.1.

Our next claim isb∈(x).Indeed,by assumption,a∈(x,x1).There are four cases to be considered.

Case 1andare both non-trivial.By filling the puncturex,from(4.1)we obtainBy Lemma 5.1,=andis trivial.That is,b∈(x).

Case 2is trivial andis non-trivial.By filling the puncturex,from(4.1)we obtainThis means that=id.By Lemma 5.1,this is impossible unless=is trivial.It follows thatb∈(x).

Case 3is non-trivial andis trivial.Again by filling in the puncturex,we see thatThus=.Letanddenote the geodesics on∪{x1}homotopic toa,b,candDon∪{x1},respectively.By filling in the puncturex1,from(4.1)anda∈(x,x1),we obtain

Ifis trivial,then=andc∈(x,x1).This contradicts the fact thatc∈/(x,x1).

Assume thatis non-trivial.Then by Lemma 5.1 and(5.2),=and thusis trivial.This saysb∈(x1,x2)(b/∈(x,x1)sinceby assumption is non-trivial).Sincea∈(x,x1),we havea,b∈(x1).By switching the roles ofxandx1and by Lemma 6.1,we conclude that

Case 4BotheDand ecare trivial.In this case,by filling the puncturexonce again,from(4.1)we deduce thattebis trivial.Thusb∈(x).We are done.

We now use the same argument of Theorem 1.1 to complete the proof of Theorem 1.2.

6 Appendix

This section is devoted to the proof of a lemma which plays a key role in the proof of Theorem 1.1.With the same notations and terminology as in Section 4,we have the following Lemma.

Lemma 6.1Let a,b,c,D?be simple closed geodesics.Assume that a,b∈(x),and,are non-trivial on S as x is filled in.Thenand

ProofWe only prove thatSupposeBy assumption,a,b∈(x).Hence by filling the puncturex,we deduce thatis the identity.Sinceandare non-trivial,by Lemma 5.1(2),we have=.Nowandare well-defined non-trivial mapping classes onS.Letandbe the configurations corresponding tocandD,respectively(see Section 2 for an exposition).

Since=,all boundary geodesics of elements ofandare disjoint.Hence by Theorem 1.2 of[19],there exist maximal elementsandsuch thatandDenoteBy Theorem 1.2 of[19],is a hyperbolic element ofGwhose axis axisseparatesfrom(see Figure 7).

Figure 7 Both zaand zbare outside of?D

Figure 8 Only zais outside of?D

By assumption,the equality(4.1)holds.This particularly implies that axisaxis(whichis also denoted by).Since??cand??Dproject(underρ)to the simple closed geodesic=,is disjoint fromBy hypothesis,We conclude that axis=axis=geodesicconnectingAandB.

By combining Figure 4 and the remark thereafter,we deduce thatzaandzbmust lie on the right componentRof S1{A,B}and furthermore,zbis closer toAthanzais.In what follows,we denote by(P,Q)the minor on S1connecting two non-antipodal labeling pointsPandQonThere are several cases to be considered.

Case 1Bothzaandzblie in the arc(UW)(see Figure 8).In this case,ifTa(?D)coverszb,thenis not disjoint fromBut we know thatis disjoint fromThis is a contradiction.IfthenAgain this contradicts thatis disjoint fromIfthen one easily sees thatorboth of which would imply thatis not disjoint from

Case 2zb∈(AU)andza∈(UW)(see Figure 8).ThenHenceThis implies thatis not disjoint fromIfzb∈(UW)andza∈(BW),by considering the inverseofTbTaand by the same argument as above,we see that this case does not occur.

Case 3za,zb∈(AU)(see Figure 9).Noting thatbis preperipheral,sobbounds a twice punctured diskcontainingx.This implies thatis not of type(0,3).We can choose a non-trivial simple closed geodesicγonwhich can also be viewed as a geodesic onthat satisfies the conditions:(i)γis not preperipheral,(ii)γis disjoint fromb,and(iii)γintersectsDanda.Letbe the configuration corresponding toγ.By Lemma 2.2 of[19],and there exists a maximal element?∈so that?? crossesIt could be the case that?coversza,as shown in Figure 9.But it could also be the case that?does not coverza.Sincethere exists a maximal element,and call it?too,such that.In any case,zbis not contained inand?is disjoint from

Now we haveand thusButWe conclude that

So

Case 4and(see Figure 10).Assume without loss of generality that botha∪bandc∪Dfill.Since=,there existsh∈GsendingHencehis hyperbolic and its axis axis(h)separateszafromzb.By assumption,c∪DfillsFrom Lemma 2.2 of[21],axis(h)intersects at least one geodesicbetweenandUW.In general,we letbe the geodesics betweenand,wherek≥1.We redraw Figure 10 as Figure 11 and Figure 12.

Letdenote the collection of components ofThen there exists a bijectionχbetweenand the set of geodesicsc0?with

Letbe contained in the region bounded byand be disjoint fromFor 1≤j≤k,we letbe contained in the region bounded byFinally,denote bythe component contained in the region bounded byand

It is clear thatA∈(U0U1)andB∈(WkW).From Figure 4,we haveza∈(BW)andIf(see Figure 11),we consider the component?1ofcontainingThenWriteIfwe are done.So we assume that

Figure 9 Both za,zbare inside of?c∩?D

Figure 10 za,zb∈?care separated by

Figure 11 zb∈(AU1)and za∈(BW)

Figure 12 zb∈(U1U)and za∈(BW)

By constructionandcovers the repelling fixed pointB(otherwise,we immediately see thatis shown as a shaded region in Figure 11.Letbe the configuration defined by

Letdenote the simple closed geodesic corresponding toBy Lemma 2.1 of[20],is disjoint fromc.Sinceby Lemma 2.2 of[19],intersectsa.Sinceby construction we know thatc′intersectstb(a),i.e.,intersectsa.It follows thatintersectsa.This tells us thatcontains a maximal element?which coversza(Lemma 2.2 of[19]).But sincezais disjoint from ??,? is disjoint from ??.

We claim that? does not cross axis(h).Otherwise,we note thatThis implies thatwould not be the last geodesic inthat lies in betweenandand crosses axis(h).Therefore,?is disjoint from both??and axis(h).?is shown in Figure 11 too.Sinceit is disjoint from(U0W0).But we know thatWe conclude thatis disjoint fromIt follows thatwhich in turn implies that

Ifzb∈(U1U)(see Figure 12),again,we let?1be the component of that contains(as shown in Figure 12).LetWe may also assume thatLet? be shown as in Figure 12.Then by the same argument as above,we conclude thatwhich implies that

This completes the proof of Lemma 6.1.

AcknowledgmentThe author is grateful to the referees for their careful reviews which helped to improve the paper in several aspects.

[1]Ahlfors,L.V.and Bers,L.,Riemann’s mapping theorem for variable metrics,Ann.Math.,72,1960,385–404.

[2]Beardon,A.,The Geometry of Discrete Groups,Springer-Verlag,New York,Heidelberg,Berlin,1983.

[3]Bers,L.,Fiber spaces over Teichmüller spaces,Acta Math.,130,1973,89–126.

[4]Bers,L.,An extremal problem for quasiconformal mappings and a theorem by Thurston,Acta Math.,141,1978,73–98.

[5]Birman,J.S.,Braids,Links and mapping class groups,Ann.Math.Studies,82,Princeton University Press,Princeton,1974.

[6]Dehn,M.,Papers on Group Theory and Topology,Springer-Verlag,New York,1987.

[7]Hamidi-Tehrani,H.,Groups generated by positive multi-twists and the fake lantern problem,Algebr.Geom.Topo.,2,2002,1155–1178.

[8]Farkas,H.M.and Kra,I.,Riemann Surfaces,Springer-Verlag,New York,Berlin,1980.

[9]Ivanov,N.V.and McCarthy J.D.,On injective homomorphisms between Teichmuller modular groups,I.Invent.Math.,135,1999,425–486.

[10]Kra,I.,On the Nielsen-Thurston-Bers type of some self-maps of Riemann surfaces,Acta Math.,146,1981,231–270.

[11]Johnson,D.,Homeomorphisms of a surface which act trivially on homology,Proc.Amer.Math.Soc.,75,1979,119–125.

[12]Margalit,D.,A lantern lemma,Algebr.Geom.Topo.,2,2002,1179–1195.

[13]Nag,S.,Non-geodesic discs embedded in Teichmüller spaces,Amer.J.Math.,104,1982,339–408.

[14]Thurston,W.P.,On the geometry and dynamics of diffeomorphisms of surfaces,Bull.Amer.Math.Soc.,19,1988,417–431.

[15]Zhang,C.,Pseudo-Anosov maps and fixed points of boundary homeomorphisms compatible with a Fuchsian group,Osaka J.Math.,46,2009,783–798.

[16]Zhang,C.,On products of pseudo-Anosov maps and Dehn twists of Riemann surfaces with punctures,J.Aust.Math.Soc.,88,2010,413–428.

[17]Zhang,C.,Pseudo-Anosov mapping classes and their representations by products of two Dehn twists,Chin.Ann.Math.Ser.B,30(3),2009,281–292.

[18]Zhang,C.,Invariant Teichmüller disks under hyperbolic mapping classes,Hiroshima Math.J.,42,2012,169–187.

[19]Zhang,C.,Pseudo-Anosov maps and pairs offilling simple closed geodesics on Riemann surfaces II,Tokyo J.Math.,36,2013.

[20]Zhang,C.,On the minimum of asymptotic translation lengths of point pushing pseudo-Anosov maps on one punctured Riemann Surfaces,preprint,2013.

[21]Zhang,C.,Point-pushing pseudo-Anosov mapping classes and their actions on the curve complex,preprint,2013.