Generating the mapping class group (an algebraic approach)

Publicacions Matem`atiques, Vol 40 (1996), 457–468. GENERATING THE MAPPING CLASS GROUP (AN ALGEBRAIC APPROACH) James McCool∗ Abstract We give an algebraic proof of the fact that a generating set of the mapping class group Mg,1(g≥3) may be obtained by replicating a generating set of M2,1. 1. Introduction. We denote by F(S) the free group generated by the subset Sof the set of symbols X=a1,b 1,a 2,b 2,..., and put Fk= F(Xk), where Xkconsists of the first kelements of X(all groups F(S) are considered as subgroups of F(X)). L(X), the set of letters, is defined to be X∪X−1, i.e. the set a1, a1,b 1, b1,..., where a1denotes a−1 1, etc.; for S⊂X, the set of letters L(S)ofF(S)isF(S)∩L(X). For w∈F(X), L(W) is the set of letters occurring in the reduced form of w. We put A(S) = Aut F(S), with Akfor A(Xk), and denote by Πgthe element of F2ggiven by Πg= g  i=1 [ai,b i], where [ai,b i]=aibiaibi. The group M(Πg) is defined by M(Πg)={θ∈A 2g;Π gθ=Π g}. Let ρgdenote conjugation in F2gby the element Πg. The subgroup Ng of M(Πg) generated by ρgis central, and the quotient Mg,1=M(Πg)/Ng may be described as an (orientation preserving) (algebraic) mapping class group. It was shown in [9] that Mg,1is finitely presented, though computation of an explicit presentation valid for all gwas beyond the scope of the results of [9]. Such a presentation of the geometric mapping class group was found by Wajnryb [11]. Since the geometric and algebraic mapping class groups are known to coincide (see, e.g., the remarks and ∗Research supported by a grant from the Natural Sciences and Engineering Research Council of Canada 458 J. McCool references in [3]), this provides a presentation for our Mg,1. Wajnryb’s work is geometrically based, as is earlier work on generating sets by Dehn [2], Lickorish [6] and Humphries [5]. The present paper has the modest object of providing a purely algebraic method for obtaining a generating set of M(Πg), and hence of Mg,1. Thus we define the groups M(r, g), for 1 ≤r≤g−1, by M(r, g)={θ∈A{ar,b r,a r+1,b r+1};[ar,b r][ar+1,b r+1]θ =[ar,b r][ar+1,b r+1]}. Clearly each M(r, g) is an isomorphic copy of M(Π2), and each M(r, g) is naturally embedded in M(Πg), as is each M(Πr), for r<g. We will show Theorem. Let Grbe a generating set of M(r, g),1≤r≤g−1. Then g−1  r=1 Gris a generating set of M(Πg). It only remains, in order to fulfil our objective, to find a generating set G2of M(Π2). We discuss this after the proof of the theorem. We assume below that the reader is familiar with the notation and results of [8] and [9] (see also [7]). In addition, we will need the following definition. Let S⊂L(Xg), and θ∈A g. We say that θinvolves only the letters of Sif, writing S1for S±1∩Xg, there exists ϕ∈A(S1) such that θand ϕagree on S1and θis the identity on Xg−S1. 2. Preliminary results. The following result was proved by Shenitzer in [10]. Lemma 1. Let Wbe a minimal element of Fkwith |W|>1.Let (A;a)be a T2in Ak, with a, a not in L(W)and with A∩L(W)not the empty set. Then |W(A;a)|≥|W|+2. As a consequence of this we have Corollary 2. (1) A product W1W2···Wrof disjoint minimal elements of Fkis minimal if, and only if, |Wi|≥2,1≤i≤r. (2) Two equivalent minimal words involve the same number of generators. (3) If Wis minimal, |W|>1and (A;a)=(x1,... ,x j,a;a)is a T2 such that A∩L(W)is non-empty and |W(A;a)|≤|W|, then W must contain a subword xiaor axifor some i,1≤i≤j. Generating the mapping class group 459 Proof: Parts (1) and (2) were proved by Shenitzer in [10]. An immediate consequence of these is the fact that for any S⊂Xand W∈F(S), Wis minimal in F(S) if, and only if, Wis minimal in F(X). Now suppose W,(A;a) satisfy the conditions of (3), and no subword of the desired form exists. Let Wbe the unreduced word obtained from Wby replacing each letter bin Wby b(A;a). It is known [4] that w(A;a) is obtained from Wby deleting all subwords of Wof the form aa. Since Wcontains no subword of the form xiaor axi, the aand a symbols in any subword aaof Wmust both be ‘new’. Now let W1be obtained from Wby replacing each a, aby x, x respectively, where xis a letter not in L(W)∪A∪A−1. From the above remark, it is clear that |W1(A;a)|=|W(A;a)|≤|W|. However, this contradicts Lemma 1, and so proves (3). It follows from (1) that Πgis minimal, since [ai,b i] is clearly minimal. We denote by m(Πg) the set of minimal equivalents of Πgin F2g.If V∈m(Πg) then we observe that Vmust contain exactly one occurrence of each letter in L(X2g). Now if V∈m(Πg) has a subword xy, where x, y ∈L(X2g), then it is clear, since Vcontains one occurrence of each of x, x, that V(x, y;y) belongs to m(Πg) (as does V(y,x;x)). Combining this observation with (3) of Corollary 2, we obtain Corollary 3. Let V∈m(Πg)and let (A;a)=(y1,... ,y r,a;a)∈A 2g be such that V(A;a)∈m(Πg). Then there is a permutation σ∈Srsuch that V(yσ(1),a;a)···(yσ(i),a;a)∈m(Πg) for 1≤i≤r. We next prove Lemma 4. Let r, k be positive integers with r<kand let Y=Xk− Xr.LetU, V, W be such that U, W ∈Fr,L(V)∪L(V−1)=Y∪Y−1 and Vis minimal. If β∈A kis such that xiβ=xi,1≤i≤r, where Xr={x1,... ,x r}, and (WV)β=UV, then U=Wand βinvolves only the letters of Y. Proof: Let W1=U−1W, so that (W1V)β=V. We put Z={x1,... ,x r,... ,x 1,... ,x r,W 1V}, where Zcontains Noccurrences of the r-tuple Z1=(x1,... ,x r), and N is chosen so N>|W1V|. Then Zis mapped by βto Z2={Z1,... ,Z 1,V}. 460 J. McCool Since |Z|≥|Z2|, there exists (see [8], [9]) a factorisation β=P 1···P s, where P 1,... ,P s∈W, and an integer t,1≤t≤s, such that (1) |ZP 1···P i|<|ZP 1···P i−1|,i≤t, and (2) |ZP 1···P i|=|Z2|,i≥t. Each P iwith i≤tmust be a T2. Now for any tuple Z3, type one Tand type two P,if|Z3|=|Z3(TP)|, then |Z3|=|Z3(TPT−1)|=|Z3(TPT−1)T|, and TPT−1∈T2. Using this observation, we can modify the original factorisation of βto obtain β=P1···PlTwhere P1,... ,P lare T2’s, Tis aT1(possibly the identity) and (1), (2) hold with P1,... ,P lin place of P 1,... ,P s. From the choice of Zit is easy to see that no Pican increase the length of any one of x1,... ,x r, and hence each Piand Tmust fix all of x1,... ,x r. If P1has multiplier from L(Y), then |(W1V)P1|=|W1(VP 1)|≥|W1V|, since Vis minimal and no cancellation occurs between W1and VP 1.In view of (1) it follows that W1= 1 in this case. If P1has multiplier from L(Xr), then by Lemma 1 |VP 1|≥|V|+2; moreover, in (W1V)P1=W1(VP 1), at most one cancellation can occur between W1and VP 1, so that |(W1V)P1|≥|W1V|and again we must have W1=1. Hence we have shown that W1= 1. It now follows from Lemma 1, as above, that P1cannot have multiplier from L(Xr), and the same argument shows, inductively, that no Pican have multiplier from L(Xr). Since each Pifixes Xrpointwise, so must T. This proves the lemma. Definition. Let V∈m(Πg), A⊂L(X2g), A∩A−1=∅,|A|=2rfor some integer r≥1. We say that Ais interlocked in Vif the “quotient word” V(A) obtained by deleting all letters in L(X2g)−(A+A−1) from Vis a minimal equivalent of Πr. Let V∈m(Πg) have reduced form V=QxRxS, for some letter x. Then there is y∈R(i.e. letter ywhich is a subword of R) such that y/∈R, for otherwise conjugation of the generators occurring in Rby x would reduce the length of V. Hence for each x∈Vthere is a y∈V such that x, y are interlocked in V. We now observe Generating the mapping class group 461 Lemma 5. Let Abe interlocked in Vand let θ∈A 2gbe such that Vθ ∈m(Πg)and Gθ =G, where Gis the normal closure in F2gof L(X2g)−(A+A−1). Then Ais interlocked in Vθ. Proof: For ease of notation we suppose that A=X2r. Let pbe the projection p:F2g→F2g/G =F2r. Since Gθ =G,θinduces an automorphism θ1of F2rand pθ1=θp, so that Vpθ 1=V(A)θ1=Vθp=(Vθ)(A). Now V(A)∈m(Πr) since Ais interlocked in V.Thus(Vθ)(A)isan automorphic image of Πrand so belongs to m(Πr), since it has length 4r. Hence Ais interlocked in Vθ. 3. The complex Kg.Let Kgbe the complex for Πgconstructed in [9]; i.e. K0 g=m(Πg), K1 gis K0 gwith a directed edge labelled (V1,V 2;P) joining vertex V1to V2whenever P∈Wis such that V1P=V2, and Kg is K1 gwith a finite set of 2-cells attached. It was shown in [9] that there is an isomorphism κ:π1(Kg,Πg)→M(Πg), and that the isomorphism is the natural one, i.e. is induced by the homomorphism κfrom the groupoid of paths in Kgto A2gwhose effect on a path pin Kg, p=(V1,V 2;P1),(V2,V 3;P2),... ,(Vs−1,V s;Ps−1), is given by pκ =P1P2···Ps−1. Let V∈m(Πg) be such that x, y are interlocked in V. Then there is a (unique) T∈T1with Tinvolving only xand ysuch that VT =V1= AxByCxDyE (where the expression given for V1is reduced). Now let Ehave reduced form x1x2···xt. Then V(y,x1;x1)(y,x2;x2)···(y,xi;xi)∈m(Πg),0≤i≤t. The product µ1=(y,x1;x1)···(y,xt;xt) maps V1to V2=AxBEyCxDy, and may be denoted by µ1:y→Ey, since µ1fixes each letter other than y,y. The factorisation given for µ1yields a path p1in Kgof length rfrom V1to V2, with p1κ=µ1. Now define µ2,µ 3and µ4by µ2:x→xBE, µ3:y→yCBE,µ4:x→DCBEx. Then V2µ2=V3=AxyCBExDy, V3µ3=V4=AxyxDCBEy,V4µ4=V5=ADCBExyx y. Each µihas a factorisation similar to that given for µ1, and a corresponding path pi in Kgwith piκ=µi. We put µ=Tµ1µ2µ3µ4and let pbe the path (V1,V 2;T), p1,p 2,p 3,p 4, so that pκ =µ. The µiare instances of the familiar ‘cut and paste’ operations, and we shall refer to both pand µ as the CP operation on x, y taking Vto ADCBExyx y. We note that µ 462 J. McCool moves only xand y.NowADCBE is minimal, involves exactly 2g−2 elements of X, and each of these occur once with exponent one and once with exponent minus one. It follows easily from this that there is a sequence of CP operations which involve only the generators occurring in ADCBE and which map ADCBE to Πg−1J, where J∈T1. We now observe Lemma 6. Let a, b be interlocked in V∈m(Πg),g≥2.Letx∈ L(X2g)be such that x/∈{a, a, b, b}. Then there is y∈Vsuch that {a, b, x, y}is interlocked in V. Proof: Let µbe the CP’s on a, b taking Vto V1=U[a, b]. Then x∈U and there is y∈Usuch that x, y are interlocked in U. Clearly {a, b, x, y} is interlocked in V1, and so by Lemma 5, is interlocked in V. We now specify for each V∈m(Πg) a path τVfrom Vto Πg.For g= 1 and V∈m(Π1), there exists a unique type one TV∈A 2such that VT V=Π 1; we define τVto be (V1,Π1,T V). Now suppose that g>1 and that τVhas been defined for all V∈m(Πr), 1 ≤r<g. Let V∈m(Πg) and write V=AxByCxDy, where xis the first letter to the left of y in Vsuch that xand yare interlocked in V. Let θV=θ1θ2, where θ1is the type one interchanging bgand y, and θ2is the type one interchanging xθ1and ag. Then Vθ V=AagBbgCagDbg. We call θVthe correcting permutation on V. Now let µVbe the CP’s on ag,b gtaking Vθ Vto ADCBagbgagbg. From above, we know that ADCB ∈m(Πg−1) and so a path, call it γV, has already been defined from ADCB to Πg−1in Kg−1. Taking the obvious interpretation of γVas a path in Kg, we define τVto be the path (V,V θV;θV),µ V,γ V. We shall denote the images of the paths τV,µ V,γ Vunder κby the same symbols in what follows. Now it is clear that π1(Kg,Πg) is generated by the classes of the set of paths τ−1 V,e,τ V1, where Vranges over the points of Kgand e=(V,V1;P) ranges over the edges beginning at V. Moreover, it follows easily from Corollary 3 that we can restrict eto range over the edges (V,V1;P) where Pis a Nielsen automorphism, in fact either P∈T1or Pis of the form (a, b;b), where abor ba is a subword of V. It follows that M(Πg) is generated by all τ−1 VPτV1, i.e. by all γ−1 Vµ−1 Vθ−1 VPθV1µV1γV1, where here Vranges over m(Πg), Pranges over the Nielsen automorphisms described above, V1=VP and γV,γ V1 ,µ V,µ V1 , θV,θ V1are as defined above. We observe that if P∈T1then θVPθV1∈T1and does not involve ag or bg. Also, if P=(a, b;b), then θ−1 VPθV1=(aθV,bθ V;bθV)θ−1 VθV1=P1θV2θ, Generating the mapping class group 463 where P1=(aθV,bθ V;bθV), V2=Vθ VP1and θ=θ−1 V2θ−1 VθV1. The portion of Kgrelating to this will look like VPV1θV1 V4 θV V2θ V0 P1θV2V3µV1 µV0=µV µV2 θV > > > > > > > > > > where V0=Vθ V,V3=V2θV2,V4=V1θV1. We see that τ−1 VPτV1=τ−1 Vµ−1 Vθ−1 VPθV1µV1γV1 =(γ−1 Vµ−1 VP1θV2µV2γV2)(γ−1 V2µ−1 V2θµV1γV1) =(τ−1 V0P1τV2)(τ−1 V3θτV4). We note that θ∈T1and does not involve agor bg. From the above observations we see that M(Πg) is generated by the set of all k(V,N)=γ−1 Vµ−1 VNµV1γV1, where Vranges over the elements of m(Πg) with θV=1,Nis either a type one not involving agor bg (in which case θVN =1)orN=PθVP where Pis a type two Nielsen automorphism, and VN =V1. We say that a k(V,N)isnice if there is a set S={ag,b g,x,y}of letters such that Sis interlocked in Vand Ninvolves only the elements of S. We note that if k(V,N) is nice then, by Lemma 5, the corresponding set Sis interlocked in VN. The following is the key result in proving the theorem. Lemma 7. Let k(V,N)be nice. Then k(V,N)=k1hk2, where h∈ M(g−1,g)and k1,k 2∈M(Πg−1). Proof: We may assume that g≥3. Let Sbe a set such that S= {ag,b g,x,y}and Sis interlocked in V. Let V=AagBbgCagCbg,V1= A1agB1bgC1agD1bg. Then, by Lemma 5, x, y are interlocked in both ADCB and A1D1C1B1. Let ηbe the CP’s on x, y taking ADCB to (say) 464 J. McCool U 0[x, y], and η1the CP’s on x, y taking A1D1C1B1to (say) U 1[x, y]. Then h=η−1µ−1 VNµV1η1maps U 0[x, y][ag,b g]toU 1[x, y][ag,b g], and fixes each element of L(X2g)−S. Hence, by Lemma 4, hinvolves only x, y, agand bg, and U 0=U 1. Let τbe a type one not involving ag or bg, such that ag−1τ=xand bg−1τ=y. Let U 0τ−1=U0, so that {U 0[x, y]}τ−1=U0[ag−1,b g−1]. Clearly U0∈M(Πg−2). Choose λ∈ A2g−4such that U0λ−1=Π g−2. Now k(V,N)=γ−1 Vµ−1 VNµV1γV1 =(γ−1 Vητ−1λ−1)(λτhτ−1λ−1)(λτη−1 1γV1) =k1hk2 say. From their definition, it is clear that k1,k 2∈M(Πg−1). Since hinvolves only x, y, ag,b g, it follows that τhτ−1involves only ag−1,b g−1,a g and bg, and so commutes with λ. Hence h=τhτ−1∈M(g−1,g). 4. Proof of the Theorem. The theorem follows immediately from Lemma 8. For each k(V,N)there exist k1,k 2∈M(Πg−1)and h∈ M(g−1,g)such that k(V,N)=k1hk2. Proof: Let V=AagBbgCagDbgand V1=A1agB1bgC1agD1bg. (1) Suppose that Ndoes not involve agor bg. Then k(V,N)=γ−1 Vµ−1 VNµV1γV1=γ−1 V(µ−1 VNµV1N−1)NγV1. Now µ−1 VNµV1N−1maps ADCB[ag,b g]to{(A1D1C1B1)N−1}[ag,b g] and fixes each element of Xg−1, so that, by Lemma 4, it must involve only agand bg. However, µ−1 VNµV1N−1fixes agand bgmodulo the normal closure of X2g−2in F2g, and so must be the identity. Hence k(V,N)=γ−1 VNγV1∈M(Πg−1). This disposes, in particular, of the case N∈T1. (2) We may now assume that N=(a, b;b)θVP =PθVP.IfNinvolves at most one other letter besides agand bg, then using Lemmas 5 and 6 it follows easily that k(V,N) is nice, and so the result holds by Lemma 7. We now consider a number of cases separately. Case 2.1. Pdoes not involve agor bg.IfθVP = 1, then this case is covered by (1) above. Otherwise, θVP must be ag↔cfor some letter Generating the mapping class group 465 c/∈{ag,b g, ag, bg}. Noting that ag,b gare interlocked in VP, we write VP =AagBbgCagDbg. Let µbe the CP’s on ag,b gtaking VP to ABCB[ag,b g], and let γ∈A 2g−2be such that (ADCB)γ=Π g−1. Then k(V,N)=γ−1 Vµ−1 VPθVPµV1γV1=(γ−1 Vµ−1 VPµγ)(γ−1µ−1θVPµV1γV1). Repeating the argument given in (1), we see that γ−1 Vµ−1 VPµγ∈M(Πg−1). Also, θVP involves only cbesides ag, so that, by (2), γ−1µ−1θVPµV1γV1 has a factorisation of the desired form. Hence, the result holds in this case. We may now assume that Pinvolves exactly one of ag,b g. We note, by Corollary 2, that Vmust contain a subword abor ba. Case 2.2. Pfixes each element of Xg−1. Then Pmust be one of (ag,b;b), (ag,b;b), (bg,b;b)or(bg,b;b), and so ag,b gare interlocked in VP. Suppose that one of the first three possibilities holds. The correcting permutation θVP in each of these cases is either trivial, or is ag↔bε, (ε=±1) (for example, if P=( ag,b;b) and bagis a subword of V, then V=AbagBbgCagDbg, where A=Ab, and VP =AagBbgCagbDbg,so that θVP is ag↔bif b∈B, and is the identity otherwise). Since only b and agare involved in N, the result holds. Suppose now that P=( bg,b;b). Then we have V=AagB1bbgCagDbg and VP =AagB1bgCagDbgb. Since θV= 1 we must have b∈A, so that V=A1bA2agB1bbgCagDbgsay, and then VP =A1bA2agB1bgCagDbgb. In order to describe θVP, we must choose the first letter cto the left of bin VP so that c, b are interlocked in VP.Thus cis in one of A2,B 1,C or D. The quotient words V(ag,b g,c,b) corresponding to these possibilities are cbcagbbgagbg,cbagcbbgagbg,cbagbbgc agbgand cbagbbgagcbgrespectively. Each of these is equivalent to Π2, so that {ag,b g,c,b}is interlocked in V.Thusk(V,N) is nice, and so the required result holds. This disposes of Case 2.2. The only remaining possibilities are that b∈{ag,b g, ag, bg}, and a∈ L(X2g−2). Case 2.3. b=agor b=ag. Here we note that the effect of Pon V is to shift the agor agin V, so that θVP must be the identity, or of the form ag↔c, for some letter c/∈{bg, bg}.IfθVP =1,orifc=a±1, then the result holds, since only agand aare involved in N. Otherwise, θVP is ag↔cand c=a±1. Then, for ε=±1. PθVP =(a, aε g;aε g)θVP =θVP{θ−1 VP(a, aε g;aε g)θVP} =θVP(a, cε;cε).