The twist subgroup of the mapping class group of a nonorientable surface
aa r X i v : . [ m a t h . G T ] S e p THE TWIST SUBGROUP OF THE MAPPING CLASSGROUP OF A NONORIENTABLE SURFACE
MICHA L STUKOW
Abstract.
Let T ( N ) be the subgroup of the mapping class groupof a nonorientable surface N (possibly with punctures and/or bound-ary components) generated by twists about two–sided circles. Weobtain a simple generating set for T ( N ). As an application wecompute the first homology group (abelianization) of T ( N ). Introduction
Let N ng,s be a smooth, nonorientable, compact surface of genus g with s boundary components and n punctures. If s and/or n is zerothen we omit it from the notation. If we do not want to emphasise thenumbers g, s, n , we simply write N for a surface N ng,s . Recall that N g is a connected sum of g projective planes and N ng,s is obtained from N g by removing s open disks and specifying the set Σ = { z , . . . , z n } of n distinguished points in the interior of N g .Let Diff( N ) be the group of all diffeomorphisms h : N → N suchthat h is the identity on each boundary component and h (Σ) = Σ.By M ( N ) we denote the quotient group of Diff( N ) by the subgroupconsisting of maps isotopic to the identity, where we assume that iso-topies fix Σ and are the identity on each boundary component. M ( N )is called the mapping class group of N . The mapping class group of anorientable surface is defined analogously, but we consider only orienta-tion preserving maps.If we assume that maps and isotopies fix the set Σ pointwise thenwe obtain the so–called pure mapping class group PM ( N ng,s ). If wealso require that maps preserve the local orientation around each ofthe punctures then we obtain the group PM + ( N ng,s ). It is an easy Mathematics Subject Classification.
Primary 57N05; Secondary 20F38,57M99.
Key words and phrases.
Mapping class groups, Nonorientable surfaces.Supported by the Foundation for Polish Science. Some part of the research wasdone while the author was visiting the Institute Mittag-Leffler, Djursholm, Sweden.Their hospitality and financial support is greatly appreciated. observation that PM ( N ng,s ) is the subgroup of index n ! in M ( N ng,s )and PM + ( N ng,s ) is of index 2 n in PM ( N ng,s ).Define also T ( N ) to be the twist subgroup of M ( N ), that is thesubgroup generated by Dehn twists about two–sided circles.By abuse of notation we will use the same letter for a map and itsisotopy class and we will use the functional notation for the compositionof diffeomorphisms.1.1. Background.
The study of algebraic properties of mapping classgroups of an orientable surface goes back to the work of Dehn [3] andNielsen [14–16]. Probably the best modern exposition of this reachtheory is a survey article by Ivanov [6]. On the other hand, the nonori-entable case has not been studied much. The first significant result isdue to Lickorish [11], who proved that the twist subgroup T ( N g ) is asubgroup of index 2 in the mapping class group M ( N g ). Moreover, M ( N g ) is generated by Dehn twists and a so–called “crosscap slide”(or a “Y–homeomorphism”). Later Chillingworth [2] found a finitegenerating set for the group M ( N g ), and Birman and Chillingworth[1] showed how this generating set can be derived from the knownproperties of the mapping class group of the orientable double coverof N g .These studies were continued much later by Korkmaz [10], who foundfinite generating sets for the groups M ( N ng ) and PM ( N ng ). Korkmaz[9, 10] also computed the first integral homology group of M ( N ng ), andunder additional assumption g ≥
7, of PM ( N ng ). He also showed [10]that T ( N ng ) is of index 2 n +1 n ! in M ( N ng ), provided g ≥ M ( N ng,s ) and PM ( N ng,s ), provided g ≥
3. Inparticular we obtained simple generating sets for these groups and wecomputed their abelianizations.Finally, let us mention that recently Wahl [20] proved some stabilitytheorems for the homology of mapping class groups of nonorientablesurfaces and using the ideas of Madsen and Weiss [13] she managed toidentify the stable rational cohomology of M ( N ).Another very promising project is the work of Szepietowski [18] whoshowed a method to obtain a presentation for the group M ( N ng,s ). Us-ing this technique he managed [19] to derive a presentation of the group M ( N ).1.2. Main results.
The following paper is a natural continuation ofthe results mentioned above, especially it should be thought as a con-tinuation of [17]. Namely, we study basic algebraic properties of the
HE TWIST SUBGROUP OF THE. . . 3 twist subgroup of the mapping class group of a nonorientable surfaceof genus g ≥
3. The crucial observation which makes such a studypossible is that T ( N ng,s ) is a subgroup of index 2 in PM + ( N ng,s ) (henceof index 2 n +1 n ! in M ( N ng,s ) – see Corollaries 6.3 and 6.4). Using thisobservation, we obtain surprisingly simple generating set for the twistsubgroup – see Theorem 6.2. Moreover, we compute the first integralhomology group (abelianization) of this subgroup – see Theorem 8.1.2. Preliminaries
By a circle on N we mean an unoriented simple closed curve on N \ Σ, which is disjoint from the boundary of N . Usually we identify acircle with its image. Moreover, as in the case of diffeomorphisms, wewill use the same letter for a circle and its isotopy class. According towhether a regular neighbourhood of a circle is an annulus or a M¨obiusstrip, we call the circle two–sided or one–sided respectively. We saythat a circle is generic if it bounds neither a disk with less than twopunctures nor a M¨obius strip disjoint from Σ.Let a be a two–sided circle. By definition, a regular neighbourhoodof a is an annulus, so if we fix one of its two possible orientations,we can define the right Dehn twist t a about a in the usual way. Weemphasise that since we are dealing with nonorientable surfaces, thereis no canonical way to choose the orientation of S a . Therefore by atwist about a we always mean one of two possible twists about a (thesecond one is then its inverse). By a boundary twist we mean a twistabout a circle isotopic to a boundary component. It is known that if a is not generic then the Dehn twist t a is trivial. In particular a Dehntwist about the boundary of a M¨obius strip is trivial – see Theorem3.4 of [4].Other important examples of diffeomorphisms of a nonorientable sur-face are the crosscap slide and the puncture slide . They are defined asa slide of a crosscap and of a puncture respectively, along a loop. Thegeneral convention is that one considers only crosscap slides along one–sided simple loops (in such a form it was introduced by Lickorish [11];for precise definitions and properties see [10]).The following two propositions follow immediately from the abovedefinitions. Proposition 2.1.
Let N a be an oriented regular neighbourhood of atwo–sided circle a in a surface N , and let f : N → N be any diffeo-morphism. Then f t a f − = t f ( a ) , where the orientation of a regularneighbourhood of f ( a ) is induced by the orientation of f ( N a ) . (cid:3) MICHA L STUKOW
Proposition 2.2.
Let K be a subsurface of N which is a Klein bottlewith one boundary component ξ , and let y be a crosscap slide on K such that y = t ξ . Then for any diffeomorphism f : N → N , f yf − isa crosscap slide on f ( K ) such that ( f yf − ) = t f ( ξ ) , where the orientation of a regular neighbourhood of f ( ξ ) is induced via f by the orientation of a regular neighbourhood of ξ . (cid:3) One of our main tools in studying properties of mapping class groupsis the so–called lantern relation . The proof can be found in Section 4of [7].
Proposition 2.3.
Let S be a sphere with four holes embedded in asurface N \ Σ and let a , a , a , a be the boundary circles of S . Definealso a , , a , , a , as in Figure 1(i) and assume that the orientationsof regular neighbourhoods of these seven circles are induced from theorientation of S . Then t a t a t a t a = t a , t a , t a , . (cid:3) Figure 1.
Circles of the lantern and torus with a hole relations.Finally, recall the torus with a hole relation – see Lemma 3 of [12].It can be also thought as an instance of the so–called star relation [5].
Proposition 2.4.
Let S be a torus with one boundary component c embedded in a surface N \ Σ and let a and a be two two–sided circlesas in Figure 1(ii). If the orientations of regular neighbourhoods of a , a and c are induced from the orientation of S then ( t a t a ) = t c . (cid:3) HE TWIST SUBGROUP OF THE. . . 5 Two models for a nonorientable surface
Let g = 2 r + 1 for g odd and g = 2 r + 2 for g even. Representthe surface N = N ng,s as a connected sum of an orientable surface ofgenus r and one or two projective planes (one for g odd and two for g even). Figures 2 and 3 show this model of N – in these figures thebig shaded disks represent crosscaps, hence their interiors are to beremoved and then the antipodal points on each boundary componentare to be identified. The smaller shaded disks represent components of ∂N (we will call them holes ). Figure 2.
Circles a i , b i , c i and u i for g = 2 r + 1. Figure 3.
Circles a i , b i , c i and u i for g = 2 r + 2.It is well known that N = N ng,s can be also represented as a spherewith n punctures, s holes and g crosscaps – cf Figure 4. In order todistinguish this model from the previous one (provided by Figures 2and 3), let us denote it by e N . The goal of this section is to constructan explicit homeomorphism Φ : N → e N .Let a , . . . , a g and u , . . . , u s be two–sided circles on N as in Figures2 and 3 ( u i are the boundary circles). Define also v , . . . , v n + s to bethe arcs as in Figure 5(i) and observe that if we cut N along the circles MICHA L STUKOW
Figure 4.
Surface N as sphere with crosscaps. Figure 5.
Circles e i and arcs v i . a , . . . , a g and arcs v , . . . , v n + s we obtain a polygon ∆ with sides a , . . . , a g − , a g , a g − , . . . , a , v , v , . . . , v n , v n ,v n +1 , u , v n +1 , . . . , v n + s , u s , v n + s , a , . . . , a g − , a g , a g − , . . . , a where the labels indicate from which circle/arc the edge came from.Identical polygon can be obtained by cutting the surface e N along thecircles a , . . . , a g and arcs v , . . . , v n + s indicated in Figure 4. Moreover,it is not difficult to see that the identification patterns required toreconstruct N and e N form ∆ are identical (cf Figures 6 and 7). This Figure 6.
Cutting N and e N for ( g, s, n ) = (3 , , N → e N which maps HE TWIST SUBGROUP OF THE. . . 7
Figure 7.
Cutting N and e N for ( g, s, n ) = (4 , , a , . . . , a g , u , . . . , u s and arcs v , . . . , v n + s in Figures 2, 3, 5(ii)to the circles/arcs with the same labels in Figure 4.The above geometric description of Φ is very convenient because itprovides a simple method for transferring circles between two modelsof N . In fact, if c is a circle on N then c becomes a collection of arcsin ∆. Moreover, up to isotopy we can assume that c does not passthrough any of the vertices of ∆. Since ∆ is simply connected each ofthis arcs is uniquely determined by the position of its endpoints. Toobtain the image Φ( c ) it is enough to reconstruct the surface e N from∆ keeping track of the collection of arcs composing c . In practise thiscan be easily done using pictures like Figures 6 and 7. Moreover, it isnot difficult to see that we can transfer not only the circles but alsothe orientations of their neighbourhoods (if it exists) – small plus andminus signs in Figures 6 and 7 indicate our choice of the orientation of∆. Of course the above procedure works as well in the other direction.Keeping in mind the above description, from now on we will transfercircles form N to e N and vice versa without further comments.4. Generators for the group PM + ( N ng,s )Let C = { a , . . . , a g − , b , . . . , b r , c , . . . , c r , e , . . . , e n + s − , u , . . . , u s } for g odd, and C = { a , . . . , a g − , b , . . . , b r +1 , c , . . . , c r , e , . . . , e n + s − , u , . . . , u s } for g even, where the circles a i , b i , c i , u i are as in Figures 2 and 3 and e i are as in Figure 5(ii). Moreover, these figures indicate the orientationsof local neighbourhoods of circles in C . We did it by indicating thedirection of twists about these circles. Therefore by a twist about oneof the circles in C we will always mean the twist determined by thisparticular choice of orientation (the general rule is that we consider MICHA L STUKOW right Dehn twists , that is if we approach the circle of twisting we turnto the right. Define also y to be a crosscap slide supported on a Kleinbottle cut of by the circle ξ indicated in Figure 8. To be more precise, Figure 8.
Circles ξ and λ .in terms of the model e N , let C g − and C g be crosscaps as in Figure 9.The same figure shows the circle ξ – it cuts off these two crosscaps. Figure 9.
Circle ξ .Hence we can define y to be a slide of the crosscap C g along the path µ indicated in that figure. In particular y = t ξ .Now we are ready to state the main theorem of this section, which isa simplification of the known generating set for the group PM + ( N ng,s ). Theorem 4.1.
Let g ≥ . Then the mapping class group PM + ( N ng,s ) is generated by { t l , y | l ∈ C} .Proof. By Theorem 5.2 of [17] and by Propositions 2.1 and 2.2, thegroup PM + ( N ng,s ) is generated by • { t l , y | l ∈ C ′ } if g is odd • { t l , ( t a g − t a g − ) − y ( t a g − t a g − ) , t − b r t λ t b r | l ∈ C ′ } if g is even,where λ is as in Figure 8(ii) and C ′ = C ∪ { f , . . . , f n + s } for f , . . . , f n + s as in Figure 10(i) (Figure 10(i) defines f i for g even; for g odd just forget about the second crosscap). HE TWIST SUBGROUP OF THE. . . 9
Figure 10.
Circles f i and h i .Therefore to complete the proof it is enough to show that if G = h t l , y | l ∈ Ci then t f , . . . , t f n + s ∈ G , and t λ ∈ G for g even.Let h , . . . , h n + s be circles as in Figure 10(ii) (as before, for g oddforget about the second crosscap). We claim that • f i = t − c r t − a g − y − t b r ( h i ) if g is odd, • f i = y − t − b r +1 t − a g − t λ t − a g − t − a g − t b r t a g − t b r ( h i ) if g is even.In fact, using the procedure of transferring circles between two mod-els N and e N described in Section 3, it is not difficult to check thatFigure 11(i) shows the circle t b r ( h i ) on e N and Figure 11(ii) shows y − ( t b r ( h i )). Then transferring this circle back to N easily leads tothe first of the above relations. The second one can be proved analo-gously. Figure 11.
Circles t b r ( h i ) and y − t b r ( h i ) for g odd.By Lemma 3.3 of [17], t h i ∈ G for i = 1 , . . . , n + s . Hence byProposition 2.1, t f i ∈ G provided t λ ∈ G . But this follows from therelation λ = t b r +1 y ( a g − ) . (cid:3) The action of M ( N ng,s ) on H ( N ng,s , R )It is known that H ( N, R ) has a basis in which every linear map f ∗ : H ( N, R ) → H ( N, R ), induced by a diffeomorphism f : N → N ,has a matrix with integral coefficients. Therefore we can define the determinant homomorphism D : PM + ( N ) → Z as follows: D ( f ) =det( f ∗ ) (we use the multiplicative notation for the group Z ). Lemma 5.1.
Let c be a two–sided nonseparating circle on N . Then D ( t c ) = 1 .Proof. It is an easy topological fact, that if c and c are two–sided,nonseparating circles in N such that both N \ c and N \ c are eitherorientable or nonorientable, then f ( c ) = c for some diffeomorphism of N . Let us stress the fact that f may not be the identity on ∂N , hencewe can not assume that f induces an element of the group M ( N ), how-ever this is of no importance to what follows. In particular, ( t c ) ∗ and( t c ) ∗ are conjugate in Aut(H ( N, R )), hence D ( t c ) = D ( t c ). More-over, if g is odd then there is no nonseparating two–sided circle c on N such that N \ c is orientable. Therefore, to prove the lemma it isenough to show that D ( t a ) = 1 and that D ( t b r +1 ) = 1 if g is even.This can be easily done – we skip the computations. (cid:3) Lemma 5.2.
Let c be a two–sided separating circle on N . Then ( t c ) ∗ : H ( N, R ) → H ( N, R ) is the identity map.Proof. Since c is two–sided, we can fix an orientation of a regular neigh-bourhood of c . Therefore for any circle a which is transversal to c , wecan define the algebraic intersection number I ( c, a ) in a usual way (wedo not claim that I ( c, a ) has any particular properties – it just countsthe points c ∩ a with appropriate signs). By the definition of a twist,it is obvious that [ t c ( a )] = [ a ] ± I ( c, a )[ c ]where [ x ] denotes the homology class of x . Moreover, it is clear that if c is separating then I ( c, a ) = 0 for any circle a . Therefore[ t c ( a )] = [ a ]and the lemma follows by the fact that homology classes of circles spanH ( N, R ). (cid:3) By Lemmas 5.1 and 5.2 we obtain
Proposition 5.3.
Let c be a two–sided circle on N . Then D ( t c ) = 1 . (cid:3) HE TWIST SUBGROUP OF THE. . . 11
The explicit definition of y made in the previous section easily leadsto the following Proposition 5.4.
Let D : PM + ( N ) → Z as above. Then D ( y ) = − . (cid:3) Generators for the twist subgroup
The main goal of this section is to find a simple generating set for thetwist subgroup T ( N ). Our main tool will be the following well knownfact from combinatorial group theory – see for example Chapter 9 of [8]. Proposition 6.1.
Let X be a generating set for a group G and let U be a left transversal for a subgroup H . Then H is generated by the set { uxux − : u ∈ U, x ∈ X, ux U } , where g = gH ∩ U for g ∈ G . (cid:3) Let e T ( N ) denote the kernel of the homomorphism D : PM + ( N ) → Z defined in Section 5. The reason for our choice of notation will becomeapparent after Corollary 6.3 below, where we will prove that in fact e T ( N ) = T ( N ) is the twist subgroup. Theorem 6.2.
Let g ≥ . Then the group e T ( N ) is generated by • { t l | l ∈ C ∪ { f , . . . , f n + s − , ξ }} if g = 3 , • { t l | l ∈ C ∪ { ψ, ξ }} if g ≥ odd, • { t l | l ∈ C ∪ { λ, ψ, ξ }} if g even,where ψ is the two–sided circle indicated in Figure 12, f , . . . , f n + s − are as in Figure 10(i), λ is as in Figure 8(ii) and ξ is as in Figure 8. Figure 12.
Circle ψ for g odd and even. Proof.
By Proposition 5.4, D is onto, hence[ PM + ( N ) : e T ( N )] = 2and as a transversal for e T ( N ) we can take U = { , y } . By Theorem4.1 and Propositions 6.1, 5.3, 5.4, e T ( N ) is generated by { t l , yt l y − , y | l ∈ C} . Clearly the circles • u , . . . , u s if g = 3, • a , . . . , a g − , b , . . . , b r − , c , . . . , c r − , e , . . . , e n + s − ,u , . . . , u s if g ≥ • a , . . . , a g − , b , . . . , b r , c , . . . , c r , e , . . . , e n + s − ,u , . . . , u s if g even,are disjoint from the support of y (cf Figures 2, 3 and 8). Since yt l y − = t ± y ( l ) and y = t ξ , e T ( N ) is generated by • { t l | l ∈ C ∪{ y ( a ) , y ( b ) , y ( c ) , y ( e ) , . . . , y ( e n + s − ) , ξ }} if g = 3, • { t l | l ∈ C ∪ { y ( a g − ) , y ( a g − ) , y ( b r ) , y ( c r ) , ξ }} if g ≥ • { t l | l ∈ C ∪ { y ( a g − ) , y ( a g − ) , y ( b r +1 ) , ξ }} if g even.We defined y using e N , so it is convenient to transfer circles to thatmodel. Once this is done, it is not difficult to check that we haverelations: • y ( a ) = a , y ( c ) = b , y − ( e ) = τ , . . . , y − ( e n + s − ) = τ n + s − if g = 3, where τ i is as in Figure 13, • y ( a g − ) = ψ , y ( a g − ) = a g − , y ( c r ) = b r if g ≥ • y ( a g − ) = t − b r +1 ( λ ), y ( a g − ) = a g − , y − ( b r +1 ) = ψ if g even. Figure 13.
Circles τ , . . . , τ n + s − for g = 3.To complete the proof it is enough to show that the twists t τ , . . . , t τ n + s − can be replaced by twists t f , . . . , t f n + s − in case g = 3. In order to provethis, let ̺ i = t − c t − a ( τ i ). Then it is straightforward to check that τ i = t − ̺ i − t a t c ( f i ) for i = 2 , . . . , n + s − . HE TWIST SUBGROUP OF THE. . . 13
Therefore, using the relation τ = f , we can inductively replace each t τ i by t f i . (cid:3) Corollary 6.3.
Let g ≥ . Then the kernel of the determinant homo-morphism D : PM + ( N ) → Z is the twist subgroup of M ( N ) , that is e T ( N ) = T ( N ) .Proof. Clearly T ( N ) ≤ PM + ( N ) and by Proposition 5.3, T ( N ) ≤ ker D = e T ( N ) . On the other hand, by Theorem 6.2, e T ( N ) ≤ T ( N ). (cid:3) Corollary 6.4.
The twist subgroup T ( N ) has index n +1 n ! in M ( N ng,s ) .Proof. By the previous corollary, T ( N ) = ker D , so the conclusionfollows from the obvious equality[ M ( N ) : ker D ] = [ M ( N ) : PM ( N )] · [ PM ( N ) : PM + ( N )] ·· [ PM + ( N ) : ker D ] = n ! · n · . (cid:3) In the case of a closed surface N g , the above corollary was first provedby Lickorish [11]. Later Korkmaz [10] proved it for a punctured surface N ng under additional assumption g ≥ Homological results for the twist subgroup
For the rest of the paper, for any f ∈ T ( N ) let [ f ] denotes thehomology class of f in H ( T ( N ) , Z ). Moreover, we will use the additivenotation in H ( T ( N ) , Z ).7.1. Homology classes of non–peripheral twists.Lemma 7.1.
Let a and b be two two–sided circles on N such that I ( a, b ) = 1 and the orientations of regular neighbourhoods of a and b are induced from the orientation of a ∪ b . Then t a and t b are conjugatein T ( N ) . In particular [ t a ] = [ t b ] in H ( T ( N )) .Proof. The Lemma follows form the braid relation t b = ( t a t b ) t a ( t a t b ) − . (cid:3) Lemma 7.2.
Assume that g ≥ and let a and b be two nonseparatingtwo–sided circles on N such that N \ a and N \ b are nonorientable.Then t a is conjugate in T ( N ) either to t b or to t − b . Proof.
By Lemma 7.1, it is enough to prove that there exits a sequenceof two–sided circles p , . . . , p k such that p = a , p k = b and I ( p i , p i +1 ) =1 for i = 1 , . . . , k −
1. In other words, using the terminology of [10], wehave to prove that a and b are dually equivalent. For a nonorientablesurface with punctures this was proved in Theorem 3.1 of [10]. It isstraightforward to check that the same proof applies to the case of asurface with boundary. (cid:3) Lemma 7.3.
Let g = 2 r + 2 ≥ . Then t b r +1 is conjugate in T ( N ) to t − ψ , where ψ = y − ( b r +1 ) is as in Theorem 6.2.Proof. Figure 14(i) shows the circle b r +1 as a circle on e N . Using the Figure 14.
Circle b r +1 and crosscap slide y i , Lemma 7.3.notation from this figure, let y i for i = 1 , . . . , g −
1, be a slide of thecrosscap C i +1 along the loop µ i as shown in Figure 14(ii). In particular y g − = y . It is straightforward to check that y g − t a g − y g − t a g − · · · y t a y ( b r +1 ) = y − g − ( b r +1 ) = ψ. Moreover, y g − t a g − · · · y t a y , as a product of twists and even numberof crosscap slides, is in the kernel of the determinant homomorphism D : PM + ( N ) → Z defined in Section 5. Therefore, by Corollary 6.3, t b r +1 is conjugate in T ( N ) to either t ψ or t − ψ . Careful examination ofthe orientations of local neighbourhoods of b r +1 and ψ shows that infact t b r +1 is conjugate to t − ψ . (cid:3) Lemma 7.4.
Let g ≥ . Then [ t a ] = 0 in H ( T ( N )) .Proof. Figure 15(i) shows three two–sided nonseparating circles a, b and c such that I ( a, b ) = 1, I ( b, c ) = 1 and the complement of each of thesecircles in N is nonorientable. Hence by Lemma 7.1, all three twists t a , t b and t c are conjugate in T ( N ). Similarly, Figure 15(ii) shows that t a , t b ′ and t − c are conjugate. Hence t c and t − c are conjugate in T ( N ).By Lemma 7.2, the same is true for t a . (cid:3) HE TWIST SUBGROUP OF THE. . . 15
Figure 15.
Circles a, b, b ′ and c , Lemma 7.4. Remark . Observe that if a and b are two–sided nonseparating circleson N such that N \ a and N \ b are nonorientable, then Lemma 7.2gives us no hint if t a is conjugate to t b or maybe to t − b . However, byLemma 7.4, as long as g ≥ Lemma 7.6 (Lemma 6.7 of [17]) . Let g ≥ . Then [ t a ] = 0 . (cid:3) Lemma 7.7 (Lemma 6.6 of [17]) . Assume g = 2 r + 2 ≥ . Then [ t b r +1 ] = 0 . (cid:3) Lemma 7.8 (Lemma 6.12 of [17]) . Assume g ≥ . Then the boundarytwists t u , . . . , t u s are trivial in H ( T ( N )) . (cid:3) Lemma 7.9 (Lemma 6.14 of [17]) . Assume that g ≥ and let κ bea two–sided separating circle on N such that one of the componentsof N \ κ is a disk ∆ containing the punctures z , . . . , z n and the holes u , . . . , u s . Moreover, assume that the orientation of ∆ agrees with theorientations of neighbourhoods of u , . . . , u s and κ . Then [ t κ ] = [ t u · · · t u s ] = [ t u ] + · · · + [ t u s ] . (cid:3) Although each of the above lemmas was originally stated in terms ofthe group PM + ( N ng,s ), by Lemmas 7.1, 7.4 and by Remark 7.5, theirproofs work as well in the case of the twist subgroup T ( N ). Lemma 7.10.
Let g ≥ and let β be a separating circle on N suchthat one component of N \ β is Klein bottle with one hole and the secondcomponent is nonorientable. Then [ t β ] = 0 . In particular [ y ] = [ t ξ ] = 0 . Proof.
Figure 16 shows that there is a lantern configuration with onecircle β and all other twists either trivial or conjugate to t a . Hence[ t a t a ] = [ t β t a t a ] . (cid:3) Figure 16.
Lantern relation, Lemma 7.10.
Lemma 7.11.
Let g ≥ and let γ be a circle on N such that one ofthe components of N \ γ is a nonorientable surface of genus with onehole. Then [ t γ ] = 0 .Proof. Figure 17 shows that there is a lantern configuration with onecircle γ and all other circles either bounding M¨obius strips or satisfyingthe assumptions of Lemma 7.10. (cid:3) Figure 17.
Lantern relation, Lemma 7.11.7.2.
Homology classes of boundary twists.Lemma 7.12.
Let g = 4 . Then [ t u ] + [ t u ] + · · · + [ t u s ] = 0 . Proof.
By Lemma 7.9,[ t u ] + [ t u ] + · · · + [ t u s ] = t κ , where κ is a circle on N bounding all the punctures and boundarycircles. On the other hand, Figure 18 shows that there is a lanternconfiguration with one circle γ and all other circles either boundingM¨obius strips or satisfying the assumptions of Lemmas 7.10 or 7.11. (cid:3) Lemma 7.13.
Let g ≥ . Then t u j ] = [ t u j ] = 0 for j = 1 , . . . , s . HE TWIST SUBGROUP OF THE. . . 17
Figure 18.
Lantern relation, Lemma 7.12.
Figure 19.
Lantern relation [ t u j t a t − a ] = [ t η j t − a t a ]. Proof.
Consider the lantern relation indicated in Figure 19. UsingLemma 7.1, it is easy to prove that all four twists about nonseparatingcircles in this figure are conjugate to t a , hence we have[ t u j t a t − a ] = [ t η j t − a t a ] . Therefore it is enough to show that f t η j f − = t − η j for some f ∈ T ( N ).The circle η j divides the surface N into a projective plane N ′ withtwo holes and a nonorientable surface N ′′ . Let h : N → N be a dif-feomorphism obtained as follows. On each of N ′ and N ′′ , h is a slideof η j along the core of a crosscap such that h is − id on η j . Clearly h ∈ PM + ( N ng,s ) and ht η j h − = t − η j . By Proposition 5.4, D ( y ) = − D : PM + ( N ) → Z is the determinant homomorphism. More-over yt η j y − = t − η j , hence by Corollary 6.3, either f = h or f = hy isthe required diffeomorphism. (cid:3) Homology classes of twists for g = 3 .Lemma 7.14. Assume that g = 3 . Then t a ] = [ t ξ ] = [ t u ] + [ t u ] + . . . + [ t u s ] . In particular t a ] = [ t ξ ] = 0 if s = 0 . Proof.
Applying Lemma 2.4 to the configuration shown in Figure 20,we obtain ( t a t a ) = t α . Hence by Lemmas 7.1 and 7.2,
Figure 20.
Relation ( t a t a ) = t α .(7.1) 12[ t a ] = [ t α ] . Using Lemma 7.1, one can check that all twists about nonseparatingessential circles indicated in Figure 21 are conjugate to t a . Therefore Figure 21.
Relation [ t a t a t κ ] = [ t a t a t α ].we have a lantern relation[ t a t a t κ ] = [ t a t a t α ] , where κ is as in Lemma 7.9. Together with (7.1) this yields(7.2) 12[ t a ] = [ t κ ] . On the other hand, by the lantern relation provided by Figure 22,we have [ t κ t a t − a ] = [ t ξ t − a t a ] . By (7.2), this gives 12[ t a ] = [ t κ ] = [ t ξ ]. Moreover, by Lemma 7.9,[ t κ ] = [ t u ] + [ t u ] + . . . + [ t u s ] . (cid:3) Lemma 7.15.
Assume that g = 3 . Then t a ] = 0 . HE TWIST SUBGROUP OF THE. . . 19
Figure 22.
Relation [ t κ t a t − a ] = [ t ξ t − a t a ]. Proof.
Figure 23 shows that there is a lantern relation t κ = t ξ t ξ t ξ , where κ is as in Lemma 7.14. Moreover by the proof of that lemma, Figure 23.
Lantern relation t κ = t ξ t ξ t ξ .[ t ξ ] = [ t ξ ] = [ t ξ ] = [ t κ ]. Hence[ t ξ ] = 3[ t ξ ] . Using once again Lemma 7.14, we have 24[ t a ] = 2[ t ξ ] = 0. (cid:3) Some special cases.Proposition 7.16.
Let N = N ng,s be a nonorientable surface of genus g with s holes and n punctures. Then H ( T ( N ) , Z ) = h [ t a ] i ∼ = Z for ( g, s, n ) = (3 , , , h [ t a ] i ∼ = Z for ( g, s, n ) = (3 , , , h [ t a ] , [ t b r +1 ] i ∼ = Z × Z for ( g, s, n ) = (4 , , .Proof. By Theorem 3 of [1], the group M ( N ) has a presentation M ( N ) = h t a , t a , y | t a t a t a = t a t a t a , yt a y − = t − a ,yt a y − = t − a , y = 1 , ( t a t a ) = 1 i . Using U = { , y } as a transversal for the subgroup T ( N ), it is straight-forward to obtain that T ( N ) = h t a , t a | t a t a t a = t a t a t a , ( t a t a ) = 1 i . This implies that H ( T ( N )) ∼ = h t a | t a = 1 i . The reasoning for the surfaces N , and N is similar, one has to usethe known presentations for the groups M ( N , ) and M ( N ) – seeTheorem 7.16 of [18] and Theorem 2.1 of [19]. (cid:3) Computing H ( T ( N ) , Z ) Theorem 8.1.
Let N = N ng,s be a nonorientable surface with n punc-tures and s holes. Then H ( T ( N ) , Z ) = Z if g = 3 and s = 0 , Z × Z s − if g = 3 and s ≥ , Z × Z if g = 4 and s = 0 , Z s × Z if g = 4 and s ≥ , Z if g = 5 , , if g ≥ .Proof. By Theorem 6.2, Corollary 6.3, Lemmas 7.2 and 7.3, H ( T ( N ))is generated by • [ t a ] , [ t ξ ] , [ t u ] , . . . , [ t u s ] if g is odd, • [ t a ] , [ t b r +1 ] , [ t ξ ] , [ t u ] , . . . , [ t u s ] if g is even.Moreover • [ t a ] = 0 if g ≥ • [ t b r +1 ] = 0 if g ≥ • [ t u ] = · · · = [ t u s ] = 0 if g ≥ • [ t ξ ] = 0 if g ≥ • [ t u ] + · · · + [ t u s ] = 0 if g = 4 (Lemma 7.12), • [ t ξ ] = [ t u ] + · · · + [ t u s ] = 12[ t a ] if g = 3 (Lemma 7.14).Hence H ( T ( N )) is generated by • [ t a ] , [ t u ] , . . . , [ t u s − ] if g = 3, • [ t a ] , [ t b r +1 ] , [ t u ] , . . . , [ t u s − ] if g = 4, • [ t a ] if g = 5 , ( T ( N )) = 0 if g ≥
7. In particular this concludes the proof inthe case g ≥
7. Therefore in what follows we assume that g ≤ HE TWIST SUBGROUP OF THE. . . 21
We also know that2[ t a ] = 0 if g ≥ t u ] = · · · = 2[ t u s − ] = 0 if g ≥ t a ] = 0 if g = 3 and s = 0 (Lemma 7.14),(8.3) 24[ t a ] = 0 if g = 3 (Lemma 7.15).(8.4)Hence it is enough to prove that every relation in the abelian groupH ( T ( N )) is a consequence of the relations (8.1)–(8.4) above.Let g = 3 and suppose that(8.5) α [ t a ] + ε [ t u ] + · · · + ε s − [ t u s − ] = 0 . If s > N ′ to be a surface of genus 3 obtained from N by forget-ting all the punctures and gluing a disk to each boundary componentbut one, say u s . If s = 0 define N ′ by forgetting about the puncturesin N . We have a homomorphismΦ : T ( N ) → T ( N ′ ) . Clearly Φ( t u i ) = 0 for 1 ≤ i ≤ s −
1, hence equation (8.5) yields α [ t a ] = 0 in H ( T ( N ′ )).By Proposition 7.16, we have 24 | α if s > | α if s = 0. Thereforeby relations (8.3) and (8.4), equation (8.5) becomes(8.6) ε [ t u ] + · · · + ε s − [ t u s − ] = 0 . This concludes the proof if s ≤
1, hence assume that s ≥
2. Now let b N j for j = 1 , . . . , s −
1, be the surface obtained from N by forgetting thepunctures, gluing a cylinder to the circles u j and u s and finally gluinga disk to each of the remaining boundary components. Then b N j is aclosed nonorientable surface of genus 5. LetΥ j : T ( N ) → M ( b N j )be the homomorphism induced by inclusion. Since Υ j ( t u i ) = 0 for i = j and i = s , equation (8.6) gives us ε j [ t u j ] = 0 in H ( M ( b N j )) . Since u j is a nonseparating two–sided circle on b N j , by Theorem 1.1of [9], we have 2 | ε j . By relation (8.2), equation (8.6) becomes 0 = 0,which completes the proof for g = 3.The proof for g = 4 is analogous. If we assume that α [ t a ] + β [ t b r +1 ] + ε [ t u ] + · · · + ε s − [ t u s − ] = 0 the we can show that α = β = 0 in the same manner as in the case g = 3, namely by mapping N into a closed surface of genus 4 andusing Proposition 7.16. Similarly, we can show that ε = . . . ε s − = 0by mapping N into a closed surface of genus 6 and using Theorem 1.1of [9].If g = 5 or g = 6 the proof is even simpler, it is enough to map N into a closed surface and use Theorem 1.1 of [9], we skip the details. (cid:3) References [1] J. S. Birman. and D. R. J. Chillingworth. On the homeotopy group of a non–orientable surface.
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