The Torelli group and congruence subgroups of the mapping class group 2 1 0 ∗ 2 Andrew Putman n a January 20, 2012 J 8 1 Introduction ] T Let Σ be a compact oriented genus g surface with n boundary components. The g,n G mapping class group of Σ , denoted Mod , is the group of orientation-preserving g,n g,n h. diffeomorphisms of Σg,n that restrict to the identity on ∂Σg,n, modulo isotopies that at fix∂Σg,n. ThegroupModg,n playsafundamentalroleinmanyareasofmathematics, rangingfromlow-dimensionaltopologytoalgebraicgeometry. Atleastfordegreesless m thanabout2g/3,thecohomologyofMod iswell-understoodduetotheresolutionof g,n [ theMumfordconjecturebyMadsenandWeiss[27](togetherwithHarer’sunpublished 1 improvedversionofhishomologicalstabilitytheorem;see[5]foranexpositionofthis). v However,the cohomologyof finite-index subgroupsof Mod remains a mystery. g,n 6 In these notes, we will focus on one low-degree calculation. Consider n∈{0,1}. For 4 an integer p, the level p congruence subgroup of Mod , denoted Mod (p), is the 9 g,n g,n 3 subgroupofModg,n consistingofmappingclassesthatacttriviallyonH1(Σg,n;Z/p). . Another description of Mod (p) is as follows. The action of Mod on H (Σ ;Z) 1 g,n g,n 1 g,n 0 preserves the algebraic intersection pairing. Since n ≤ 1, this is a nondegenerate 2 alternating form, so we obtain a representation Modg,n → Sp2g(Z). Classically this 1 representationwas knownto be surjective (see §1.1). Let Sp (Z,p) be the subgroup 2g v: ofSp2g(Z)consistingofmatriceswhichequaltheidentitymodulop. ThenModg,n(p) Xi is the pullback of Sp2g(Z,p) to Modg,n. These notes will discuss the calculation of H2(Mod (p);Z). One motivation for g,n r a this is the study of line bundles on the finite cover of the moduli space of curves associated to Mod (p), which is known as the moduli space of curves with level p g,n structures. The first Chern class of such a line bundle lies in H2(Mod (p);Z), and g,n the determination of H2(Mod (p);Z) is the heart of the paper [36], which gives a g,n complete classification of such line bundles. However, in these notes we will ignore this connection to algebraic geometry. Instead, we will use the computation of this cohomology group as an excuse to discuss a number of interesting topics related to the mapping class group. ∗SupportedinpartbyNSFgrantDMS-1005318 1 The universal coefficients exact sequence for H2(Mod (p);Z) takes the form g,n 0−→Ext(H (Mod (p);Z),Z)−→H2(Mod (p);Z) 1 g,n g,n −→Hom(H (Mod (p);Z),Z)−→0. 2 g,n The thirdandfourthlecturewillbe devotedto calculatingthe kernelandcokernelof thisexactsequence. Theywillbeproceededbytwolecturesonnecessarybackground. Let us now give a more detailed description of the four lectures. • Lecture1willbedevotedtotheTorelligroup. DenotedI ,thisisthesubgroup g,n ofMod consistingofmappingclassesthatacttriviallyonH (Σ ;Z). There g,n 1 g,n are short exact sequences 1−→I −→Mod −→Sp (Z)−→1 g,n g,n 2g and 1−→I −→Mod (p)−→Sp (Z,p)−→1, g,n g,n 2g and the structure of Mod (p) is a sortof mixture of the structure of I and g,n g,n Sp (Z,p). 2g • Lecture2willbedevotedtotheJohnson homomorphism. SetH=H (Σ ;Z). 1 g,n The Johnson homomorphism is a surjective homomorphism τ : I −→ ∧3H. g,1 A deep theorem of Johnson shows that the Johnson homomorphism gives the “rational part” of the abelianization of Ig,1. More precisely, H1(Ig,1;Z) ∼= W⊕∧3H,whereW consistsoftorsion(infact,2-torsion). Wewillalsoconstruct a “mod p” version of the Johnson homomorphism which takes the form τ : p Mod (p)−→H , where H =H (Σ ;Z/p). g,1 p p 1 g,n • Lecture 3 is devoted to calculating H (Mod (p);Z) for odd p. See the begin- 1 g,1 ning ofthat lecture for why we restrictto odd p and do notconsider the closed case. There are two basic pieces. The first comes from the mod p Johnson homomorphism and the second comes from the abelianization of Sp (Z,p). 2g • Lecture 4 is devoted to proving that H2(Modg(p);Q) ∼= Q. Of course, this implies that Hom(H2(Modg,n(p);Z),Z)∼=Z. The major work here is related to homologicalstability. Lecture 1 : The Torelli group The Torelli group was first considered by Nielsen and Magnus in the early 20th century. However, its study only really took off in the late ’70’s and early ’80’s thanks to work of several people, most especially Birman and Johnson. Johnson’s workhas provenparticularlyfundamental and influential, and his survey[25] cannot be recommended enough. Throughout this lecture, we will fix some n∈{0,1}. 2 y x y′ Figure 1.1: A separating twist Tx and a boundingpair map TyTy−′1 Thesymplecticrepresentation. RecallthatI isthekerneloftherepresentation g,n Mod → Sp (Z) arising from the action of Mod on H (Σ ;Z). We will need g,n 2g g,n 1 g,n the followingfactaboutthis action. If x is asimple closedcurveonΣ ,then let T g,n x denotetherightDehntwistaboutx. Also,leti (·,·)denotethealgebraicintersection a pairing on H (Σ ;Z). 1 g,n Lemma 1.1. Let x be a simple closed curve on x. Orient x in an arbitrary way, and let [x]∈H (Σ ;Z) denote the associated homology class. Then for v ∈H (Σ ;Z), 1 g,1 1 g,1 we have T (v)=v+i ([x],v)·[x]. x a Remark. The Dehn twist T does not depend on an orientation on x. As a sanity x check, you should verify that despite its appearance,the formula in Lemma 1.1 does not depend on the orientation of x. Exercise 1.2. Prove Lemma 1.1. Basic elements of Torelli. Lemma 1.1 allows us to construct some important elements of I . First, if [x] = 0, then T acts trivially on H (Σ ;Z). This will g,1 x 1 g,1 happen exactly when x bounds an embedded subsurface of x (see Figure 1.1). We will call such elements of Torelli separating twists. Next, the formula Lemma 1.1 only depends on the homology class of the simple closed curve. Thus if y and y′ are homologous,then Ty and Ty′ act the same on H1(Σg,1;Z), so TyTy−′1 ∈Ig,1. If y and y′ aredisjointandhomologous,thentheirunionboundsanembeddedsubsurface(see Figure 1.1). If in addition to being disjoint neither y nor y′ is separating, then we will call T T−1 ∈I a bounding pair map. y y′ g,1 Generating sets for Torelli. We have the following theorem. Theorem 1.3. For all g ≥1, the group I is generated by bounding pair maps and g,1 separating twists. For g ≥3, only bounding pair maps are needed. 3 Remarks. 1. ThefactthatI isgeneratedbyboundingpairmapsandseparatingtwistswas g,1 originally proven by Powell [31], using earlier work of Birman [3]. This proof dependedonsomeheroiccalculationsinthesymplecticgroupwhosedetailswere omitted fromthe published papers. Morerecently, Putman[32] gavea modern proof using the curve complex. Even more recently, Hatcher and Margalit [16] have given an even shorter proof. 2. The fact that for g ≥3 only bounding pair maps are needed is due to Johnson [19]. He later proved a remarkable theorem which says that I is finitely g,1 generated for g ≥ 3 (see [22]). The size of Johnson’s generating set grows exponentiallying. AnsweringaconjectureofJohnson,Putman[38]hasrecently constructedageneratingsetforI thatgrowscubicallying. Aswewilldiscuss g,1 below, the abelianization of I has rank cubic in the genus, so one cannot do g,1 better. 3. McCullough and Miller [28] proved that I is not finitely generated for n ∈ 2,n {0,1}. Later, in his thesis Mess [29] proved that I is an infinite rank free 2 group. 4. It is not known if I is finitely presentable for g ≥3. g,n The Birman exact sequence. We will need to make several calculations in I . g,n Forus,thekeytoolformakingsuchcalculationsis thefactthatI containsalarge g,n number of groups derived from surface groups. This follows from the Birman exact sequence,whichtakesthe followingform. LetUΣ be the unit tangentbundle ofΣ . g g For g ≥2, we then have an exact sequence 1−→π (UΣ )−→Mod −→Mod −→1. 1 g g,1 g The terms here have the following meanings. Let β be the boundary component of Σ . g,1 • ThemapMod →Mod comesfromgluingadisctoβ andextendingmapping g,1 g classes over this disc by the identity. • Thesubgroupπ (UΣ )ofMod isknownasthe“disc-pushingsubgroup”. The 1 g g,1 mapping class associatedto γ ∈π (UΣ ) “pushes” the boundary component β 1 g around the surface while allowing it to rotate. Of course, the original version of the Birman exact sequence goes back to work of Birman[2]. Theversionherefirstappearedin[22];see[12]foratextbooktreatment. The loop around the fiber. The fiber F of UΣ over the basepoint satisfies 0 g F0 ∼=S1. TheorientationonΣg determinesanorientationonF0,soitmakessenseto talkabout“clockwise”and“counterclockwise”directionsonF . The groupπ (UΣ ) 0 1 g contains a distinguished element δ which goes once around F in the clockwise di- 0 0 rection. The element of the disc-pushing subgroup of Mod corresponding to δ g,1 0 4 Figure1.2: Pushingtheboundarycomponentaroundasimpleclosedcurveinducesabound- ing pair map y3 γ3 x3 γ2 β x2 y2 γ1 x1 y1 Figure 1.3: Thelantern relation is γ˜1·γ˜2·γ˜3=δ0k. Tomakethefiguremoreattractive, the curves γi have what appears to be a singularity at the basepoint, but in reality one should imagine them rounded and smooth there. In terms of Dehn twists, the lantern relation is (Tx3Ty−31)(Tx2Ty−21)(Tx1Ty−11)=Tβ rotates the boundary component β by a full turn in the clockwise direction. Clearly this is simply T . Observe that T ∈I . β β g,1 Calculating in the disc-pushing subgroup. Consider some γ ∈ π (Σ ) that 1 g can be realized by a smoothly embedded simple closed curve. The derivative of a smoothsimplerepresentativeofγ isapathinthetangentbundleofγ whichdoesnot contain any zero vectors. For some fixed Riemannian metric on the surface, we can reparametrizeγ sothatitsderivativeisaloopγ˜ intheunittangentbundle. Ifγ 6=1, then the element γ˜ ∈π (UΣ ) is independent of the choice of a smoothly embedded 1 g representative of γ. Indeed, any two such realizations are smoothly homotopic (this can be proved using the techniques in [11]; to test your understanding, you should verify that this fails if γ =1). Let τ ∈ Mod be the element of the disc-pushing subgroup corresponding to γ g,1 γ˜ ∈ π (UΣ ). As is shown in Figure 1.2, the mapping class τ is a bounding pair 1 g γ map,andhence liesinI . Since the loopδ aroundthe fiberalsocorrespondstoan g,1 0 element of I , we deduce that the disc-pushing subgroup lies in I . This implies g,1 g,1 thatrelationsinπ (UΣ )yieldrelationsinI . Evenmorerelationscanbeobtained 1 g g,1 by embedding Ig,1 into Ig′,n via a subsurface inclusion Σg,1 ֒→Σg′,n. As anexample,considerthe relationγ ·γ ·γ =1in π (Σ )depicted inFigure 1 2 3 1 g,1 1.3. We have γ˜ ·γ˜ ·γ˜ =δk for some k ∈Z. 1 2 3 0 Exercise 1.4. Prove that k =1. The associated relation τ τ τ =δ in I is the lantern relation γ3 γ2 γ1 0 g,1 (T T−1)(T T−1)(T T−1)=T ; (1) x3 y3 x2 y2 x1 y1 β 5 x1 x3 x2 y2 y3 y1 Figure 1.4: x1x2x3 =1 and y1y2y3 =1 here the curves x and y are as depicted in Figure 1.3. i i Remark. The order of the terms in (1) is the opposite of what one might expect because elements in the fundamental group are composed left to right but mapping classes are composed right to left. Observe that if g′ ≥ 3, then this relation can be embedded in Ig′,n to express a separating twist as a product of bounding pair maps (c.f. Theorem 1.3). Killing off separating twists. The lantern relation gives numerous ways of ex- pressingT asaproductofboundingpairmaps. For1≤i≤3,letx ,y ∈π (Σ )be β i i 1 g the curves in Figure 1.4. Observe that x x x = 1 and y y y = 1, so we have two 1 2 3 1 2 3 different lantern relations T =τ τ τ and T =τ τ τ . β x3 x2 x1 β y3 y2 y1 These curves have the property that x is homologous to y−1 for 1 ≤ i ≤ 3. The i i group I acts on π (Σ ), and it is not hard to see that there exists some f ∈ I g,1 1 g i g,1 such that f (x )=y−1. To use this, we will need the following exercise. i i i Exercise 1.5. Ifγ ∈π (Σ )canberealizedbyasimpleclosedcurveandiff ∈Mod , 1 g g,1 then τ =fτ f−1. f(γ) γ Applying Exercise 1.5 severaltimes, we obtain the following relation in I . g,1 T2 =(τ τ τ )(τ τ τ ) β x3 x2 x1 y3 y2 y1 =(τ τ τ )(τ−1 τ−1 τ−1 ) x3 x2 x1 f3(x3) f2(x2) f1(x1) =(τ τ τ )(f τ−1f−1f τ−1f−1f τ−1f−1) (2) x3 x2 x1 3 x3 3 2 x2 2 1 x1 1 UponabelianizingI ,therighthandsideof (2)vanishes. Letting[T ]∈H (I ;Z) g,1 β 1 g,1 be the associated element of the abelianization, we obtain that 2[T ]=0. β If x is a separating curve on a surface of genus at least 3, then we can embed the above relation into the surface to get that T2 has to vanish upon abelianizing the x Torelli group. We have proven the following. 6 Lemma 1.6. Fix g ≥ 3 and n ∈ {0,1}. Let T be a separating twist in I . Then x g,n the image [T ] of T in H (I ;Z) satisfies 2[T ]=0. x x 1 g,n x Lemma 1.6 first appeared in [24]. The above is a version of Johnson’s proof. For an alternate exposition of that proof which arranges the details a little differently, see [35, §7.2]. A preview. In Lecture 2, we will constructthe importantJohnsonhomomorphism. Letting H=H (Σ ;Z), this is a surjective homomorphism 1 g,n τ :I −→∧3H. g,1 Thereisalsoaversionforclosedsurfaces,butwewillnotdiscussit. Thekeyproperty of the Johnson homomorphism is that its kernel is exactly the subgroup generated by separating twists. Lemma 1.6 will then allow us to deduce the following theorem of Johnson [24]. Theorem 1.7. For g ≥ 3, we have H1(Ig,1;Z) ∼= W ⊕∧3H, where W consists of 2-torsion. Remark. Johnson also calculated the 2-torsion W. The associated Z/2-quotients of I come from the Rochlin invariants of homology 2-spheres. They were originally g,1 constructedbyBirmanandCraggs[4]. Later,in[21]JohnsonpackagedallofBirman and Craggs’s homomorphisms together into a single homomorphism and determined exactly how many linearly independent quotients they had constructed. As a prologuefor the construction,we recommendperformingthe followingexercise, which explains the appearance of ∧3H in the Johnson homomorphism. Exercise 1.8. Let Tn denote the n-torus (S1)n. 1. ProvethatthecohomologyringH∗(Tn;Z)isisomorphictothe exterioralgebra ∧∗Zn. 2. Let G be an abelian topological group. Define a product H (G;Z)⊗H (G;Z)−→H (G;Z) i j i+j via the composition φ ψ H (G;Z)⊗H (G;Z)−→H (G×G;Z)−→H (G;Z), i j i+j i+j where φ is the map coming fromthe Ku¨nneth exact sequence andψ is induced by the group product G×G → G. Prove that with this product structure, H∗(G;Z) is a graded-commutative algebra. We remark that this product is known as the Pontryagin product. 3. The space Tn is an abelian topologicalgroup. Provethat the resulting graded- commutative ring H∗(Tn;Z) is isomorphic to the exterior algebra ∧∗Zn. 7 α1 α2 α3 β1 β2 β3 Figure 1.5: A geometric symplectic basis 1.1 Appendix to lecture 1 : the surjectivity of the symplectic representation Recall that the action of Mod on H (Σ ;Z) preserves the algebraic intersection g,1 1 g,1 pairing and thus gives a representation π :Mod → Sp (Z). In this appendix, we g,1 2g will give a sequence of exercises about the surjectivity of π. AsymplecticbasisforH (Σ ;Z)isabasis{a ,b ,...,a ,b }forH (Σ ;Z)such 1 g,1 1 1 g g 1 g,1 that i (a ,b )=δ and i (a ,a )=i (b ,b )=0 a i j ij a i j a i j forall1≤i,j ≤g. LetS bethesetofsymplecticbasesforH (Σ ;Z). Thefollowing 1 g,1 exercise should be straightforward. Exercise 1.9. Sp (Z) acts simply transitively on S. 2g If x and y are simple closed curves on Σ , then let i (x,y) be their geometric g,1 g intersection number; i.e. the minimal cardinality of x′ ∩y′ as x′ and y′ range over all simple closed curves homotopic to x and y, respectively. A geometric symplectic basis (see Figure 1.5) is a collection {α ,β ,...,α ,β } of simple closed curves on 1 1 g g Σ such that g,1 i (α ,β )=δ and i (α ,α )=i (β ,β )=0 g i j ij g i j g i j for all 1≤i,j ≤g. Let G be the set of geometric symplectic bases on Σ . We then g,1 have the following. Exercise 1.10. Mod acts transitively on G. Hint : given two geometric symplectic g,1 bases,proveusingthe Eulercharacteristicthatyougethomeomorphicsurfaceswhen you cut along them. The following lemma is the heart of the fact that π(Mod )=Sp (Z). g,1 2g Lemma 1.11. If {a ,b ,...,a ,b } is a symplectic basis for H (Σ ;Z), then there 1 1 g g 1 g exists a geometric symplectic basis {α ,β ,...,α ,β } on Σ such that [α ]=a and 1 1 g g g i i [β ]=b for 1≤i≤g. i i ProofsofLemma 1.11canbe foundin [32,Lemma A.3]and[12,3rdproofofLemma 6.4]; however, it is worthwhile to contemplate how one might prove it (though it is probably too hard for an exercise). Exercise 1.12. Combine Lemma 1.11 with Exercises 1.9 and 1.10 to deduce that π(Mod )=Sp (Z). g,1 2g 8 Lecture 2 : The Johnson homomorphism Let H = H (Σ ;Z). In this lecture, we will construct the Johnson homomorphism, 1 g which is a surjective homomorphism τ :I −→∧3H. g,1 This homomorphism can be constructed in a number of completely different ways. It was originally constructed in [20] by examining the action of I on the second g,1 nilpotenttruncationofπ (Σ ). Weexplainthisoriginalconstructioninanappendix. 1 g,1 In his survey [25], Johnson outlined several alternate constructions. We will use a definition in terms of mapping tori which was introduced in [25] and was first shown tobeequivalenttothe originaldefinitionbyHain[13]. Ourexpositionwillfollowthe paper [10] of Church and Farb, which gives a more direct proof of this equivalence. The construction. Consider f ∈ I . Though it is an abuse of notation, we will g,1 regard f as a homeomorphism of Σ . Glue a disc to the boundary component of g,1 Σ andextendf overthisdiscbytheidentitytoobtainahomeomorphismF ofΣ . g,1 g Let p ∈ Σ be the center of the glued-in disc, so F(p ) = p . Now let M be the 0 g 0 0 F mapping torus of F, i.e. the quotient Σ ×I/∼, where (x,1)∼ (F(x),0). Give M g F the basepoint q =(p ,0). 0 0 There is a distinguished element t ∈ π (M ,q ) which traverses the embedded 1 F 0 loop p ×I/ ∼ in M in the positive direction. Fix a standard generating set S = 0 F {s ,...,s } for π (Σ ) that satisfies the surface relation 1 2g 1 g [s1,s2]···[s2g−1,s2g]=1. Since F(p ) = p , the map F acts on π (Σ ,p ). For 1 ≤ i ≤ 2g, let w be an 0 0 1 g 0 i expression for F∗(si) in terms of the generating set S. We then have a presentation π1(MF,q0)=hs1,...,s2g,t | [s1,s2]···[s2g−1,s2g]=1,tsit−1 =wi for 1≤i≤2gi. For γ ∈π (Σ ,p ), let [γ]∈H be the associatedelement of the abelianization. Since 1 g 0 F ∈ I , we have [s ] = [w ] for 1 ≤ i ≤ 2g. This implies that we can define a g i i homomorphism φ∗ :π1(MF,q0)→H such that φ∗(si)=[si] for 1≤i≤2g and such that φ∗(t)=0. The space M is clearly a K(π (M ),1). Let T2g be the 2g-torus. Fix an iden- F 1 F tification of π (T2g) with H. Since T2g is a K(H,1), the standard properties of 1 Eilenberg-MacLanespaces show that there is a canonicalhomotopy class of continu- ous maps φ:MF →T2g inducing the homomorphism φ∗. The space MF is a closed 3-manifold, so it has a canonical class [M ]∈H (M ;Z). Define F 3 F τ(f)=φ∗([MF])∈H3(T2g;Z)∼=∧3H. The last isomorphism here comes from Exercise 1.8 Summing up, we have constructed a map τ :I →∧3H. The following exercise g,1 is a good test of your understanding of the above construction. Exercise 2.1. Provethatτ isindependentofalltheabovechoicesexceptfortheiden- tificationofπ (T2g)with H (whichis fixed). Next, provethat τ is a homomorphism. 1 9 Effect on generators. The followinglemma calculatesτ onthe generatorsforI . g,1 Recall that if S is a genus h surface with at most 1 boundary component, then a symplectic basis for H1(S;Z) ∼= Z2h is a basis {a1,b1,...,ah,bh} for H1(S;Z) such that i (a ,b )=δ and i (a ,a )=i (b ,b )=0 a i j ij a i j a i j for all 1≤i,j ≤h. Here i (·,·) is the algebraic intersection form. a Lemma 2.2. 1. Let T ∈I be a separating twist. Then τ(T )=0. x g,1 x 2. Let T T−1 be a bounding pair map on I . Let S be the component of Σ cut x x′ g,1 g,1 alongx∪x′ thatdoesnotcontain∂Σg,1,soS ∼=Σh,2 forsomeh<g. LetS′ ⊂S be an embedded subsurface such that S′ ∼= Σh,1 and let {a1,b1,...,ah,bh} be a symplectic basis for H (S′;Z)⊂H (Σ ;Z). Then 1 1 g,1 τ(T T−1)=±[x]∧(a ∧b +···+a ∧b ). x x′ 1 1 h g We will discuss the proof of this lemma at the end of this section. Right now, we suggest doing the following two exercises. Exercise 2.3. Prove that the formula in Lemma 2.2 is independent of the choice of S′ and its symplectic basis. Exercise 2.4. Using Lemma 2.2, prove that τ is surjective. Johnson’s theorem. In [23], Johnson proved the following deep theorem, which is a sort of converse to part 1 of Lemma 2.2. For an alternate proof, see [35]. Theorem 2.5. The kernel of τ is generated by separating twists. As we indicated at the end of Lecture 1, this theorem together with Lemma 1.6 implies that H1(Ig,1;Z)∼=W ⊕∧3H for g ≥3, where W consists of 2-torsion. The Johnson homomorphism mod p. Set H = H (Σ ;Z/p). We wish to p 1 g,1 construct a “mod p” Johnson homomorphism τ :Mod (p)→H . p g,n p The construction goes exactly like the construction of the ordinary Johnson homo- morphism. Consider f ∈Mod (p). Regard f as a homeomorphism of Σ . Glue a g,1 g,1 disctotheboundarycomponentofΣ andextendf overthisdiscbytheidentityto g,1 obtain a homeomorphism F of Σ . Let p ∈Σ be the center of the glued-in disc, so g 0 g F(p )=p . LetM be the mapping torus of F and let q =(p ,0) be the basepoint 0 0 F 0 0 for M . F There is a distinguished element t ∈ π (M ,q ) which traverses the embedded 1 F 0 loop p ×I/ ∼ in M in the positive direction. Fix a standard generating set S = 0 F {s ,...,s } for π (Σ ) that satisfies the surface relation 1 2g 1 g [s1,s2]···[s2g−1,s2g]=1. 10