The universal surface bundle over the Torelli space has no sections
TThe universal surface bundle over the Torelli space has no sections
Lei ChenOctober 11, 2017
Abstract
For g >
3, we give two proofs of the fact that the
Birman exact sequence for the Torelli group1 → π ( S g ) → I g, → I g → UI g,n Tu (cid:48) g,n −−−−→ BI g,n (resp. UPI g,n Tu g,n −−−−→
BPI g,n )denote the universal surface bundle over the Torelli space fixing n points as a set (resp. pointwise). Wealso deduce that T u (cid:48) g,n has no sections when n >
T u g,n has precisely n distinct sections for n ≥ It is a basic problem to understand when bundles have continuous sections, and the corresponding grouptheory problem as to when short exact sequences have splittings. These are equivalent problems when thefiber, the base and the total space are all K ( π, section we mean continuous section.Let S g,n be a closed orientable surface of genus g with n punctures. Let Mod g,n (resp. PMod g,n ) bethe mapping class group of S g,n , i.e. the group of isotopy classes of orientation-preserving diffeomorphismsof S g fixing n points as a set (resp. pointwise). Mod g,n and PMod g,n act on H ( S g ; Z ) leaving invariantthe algebraic intersection numbers. Let I g,n (resp. PI g,n ) be the Torelli group (resp. pure Torelli group ) of S g,n , i.e. the subgroup of Mod g,n (resp. PMod g,n ) that acts trivially on H ( S g ; Z ). We omit n when n = 0.The following Birman exact sequence for the Torelli group provides a relationship between I g, and I g ; see[FM12, Chapter 4.2]. 1 → π ( S g ) point pushing −−−−−−−−→ I g, T π g, −−−−→ I g → . (1.1)The main theorem of this paper is the following: Theorem 1.1 ( Nonsplitting of the Birman exact sequence for the Torelli group).
For g > , theBirman exact sequence for the Torelli group (1.1) does not split. Remark 1.2.
Our proof needs the condition g >
3. By [Mes92, Proposition 4], I is a free group. So theBirman exact sequence for I splits. The case g = 3 is open.Let BPI g,n := K ( PI g,n ,
1) be the pure universal Torelli space fixing n punctures pointwise and let S g → UPI g,n T u g,n −−−−→
BPI g,n (1.2)1 a r X i v : . [ m a t h . G T ] O c t e the pure universal Torelli bundle . Surface bundle (1.2) classifies smooth S g -bundle equipped with a basisof H ( S g ; Z ) and n ordered points on each fiber. Since PI g,n fixes n points, there are n distinct sections { T s i | ≤ i ≤ n } of the universal Torelli bundle (1.2). Let BI g,n := K ( I g,n ,
1) be the universal Torelli space fixing n punctures as a set and let S g → UI g,n T u (cid:48) g,n −−−−→ BI g,n (1.3)be the universal Torelli bundle . This bundle classifies smooth S g -bundles equipped with a basis of H ( S g ; Z )and n unordered points on each fiber. Theorem 1.1 says that T u g, has no sections. For n ≥
0, we have thefollowing complete answer for sections of (1.3) and (1.2).
Theorem 1.3 ( Classification of sections for punctured Torelli spaces).
The following holds:(1) For n ≥ and g > , every section of the universal Torelli bundle (1.2) is homotopic to T s i for some i ∈ { , , ..., n } .(2) For n > and g > , the universal Torelli bundle (1.3) has no continuous sections. Let M g := K (Mod g , M g : S g → UM g −→ M g has no sections. This can be seen from the corresponding algebraic problem of finding splittings of the Birman exact sequence → π ( S g ) → Mod g, → Mod g → . The answer is no because of torsion, e.g. see [FM12, Corollary 5.11]. The key fact is that every finitesubgroup of Mod g, is cyclic. However, this method does not work for torsion-free subgroups of Mod g . Forany subgroup G <
Mod g , there is an extension Γ G of G by π ( S g ) as the following short exact sequence.1 → π ( S g ) → Γ G → G → . (1.4)We call (1.4) the Birman exact sequence for G since it is induced from the Birman exact sequence. We posethe following open question: Problem 1.4 ( Virtually splitting of the Birman exact sequence).
Does the Birman exact sequencefor a finite index subgroup of
Mod( S g ) always not split?Let the level L congruence subgroup Mod g [ L ] be the subgroup of Mod g that acts trivially on H ( S g ; Z /L Z )for some integer L >
1. Theorem 1.1 implies that a finite index subgroup of Mod g containing I g does notsplit; in particular, this applies to all the congruence subgroups Mod g [ L ]. Error in G. Mess [Mes90, Proposition 2]
In the unpublished paper of G. Mess [Mes90, Proposition 2], he claimed that there are no splittings of theexact sequence (1.1). But his proof has a fatal error. Here is how the proof goes. Let C be a curve dividing S g into 2 parts S (1) and S (2) of genus p and q , where p, q ≥
2. Let
U T S g be the unit tangent bundle ofgenus g surface. Then I g contains a subgroup A , which satisfies the following exact sequence1 → Z → π ( U T S p ) × π ( U T S q ) → A → . Mess’ idea is to prove that the Birman exact sequence for A does not lift. However, in Case a) of Mess’proof for [Mes90, Proposition 2], Mess claimed that if the Dehn twist about C lifts to a Dehn twist about2 (cid:48) on S g, and C (cid:48) bounds a genus p surface with a puncture and a genus q surface, then there is a lift from π ( U T S p ) to π ( U T S p, ). This is a wrong claim. Actually even A does have a lift. We construct a lift of A as the following. Let PConf ( S p ) be the pure configuration space of S p , i.e. the space of 2-tuples of distinctpoints on S p . Let PConf , ( S p ) = { ( x, y, v ) | x (cid:54) = y ∈ S g and v a unit vector at x } . We have the following pullback diagram: π (PConf , ( S p )) (cid:47) (cid:47) f (cid:15) (cid:15) (cid:74) π (PConf ( S p )) g (cid:15) (cid:15) π (UT S p ) (cid:47) (cid:47) π ( S p ) . (1.5)The lift of π ( U T S p ) should lie in π (PConf , ( S p )) instead of π ( U T S p, ) as Mess claimed. As long as wecan find a lift of f , we will find a section of A to I g, ∗ . By the property of pullback diagrams, a section of g can induce a section of f in diagram (1.5). To negate the argument of [Mes90, Proposition 2], we only needto construct a section of g . We simply need to find a self-map of S g that has no fixed point. For example,the composition of a retraction of S p onto a curve c and a rotation of c at any nontrivial angle does not havea fixed point. Therefore Mess’ proof is invalid and does not seem to be repairable. Strategy of the proof of Theorem 1.1
Let T a be the Dehn twist about a simple closed curve a on S g . Our strategy is the following: assumethat we have a splitting of (1.1). The main result of [Joh83] shows that all bounding pair map i.e. T a T − b for a pair of nonseparating curves a, b that bound a subspace generate I g . Firstly we need to understand thelift of T a T − b . We show that the lift of a bounding pair T a T − b has to be a bounding pair T a (cid:48) T − b (cid:48) for a (cid:48) , b (cid:48) on S g, . Moreover, the curve a (cid:48) does not depend on the choice of b , i.e. for any other curve c that formsa bounding pair with curve a , the lift of T a T − c is T a (cid:48) T − c (cid:48) for the same a (cid:48) . Therefore, we have a lift fromthe set of isotopy classes of curves on S g to the set of isotopy classes of curves on S g, . Then we use thelantern relation to derive a contradiction. Our main tool is the canonical reduction system for a mappingclass, which in turn uses the Thurston classification of isotopy classes of diffeomorphisms of surfaces. Thisidea originated from [BLM83]. Acknowledgement
The author would like to thank Nick Salter for discussing the content of this paper. She thanks MattClay and Dan Margalit for reminding me the fact that I is a free group. Lastly, she would like to extendher warmest thanks to Benson Farb for his extensive comments as well as for his invaluable support fromstart to finish. Let g >
3. We assume that the exact sequence (1.1) has a splitting which is denoted by φ such that F ◦ φ = id . The goal of this section is to prove Theorem 1.1 by contradiction. In all the figures in thissection, ∗ represents the puncture of S g, , the genus g surface with one puncture.3 .1 Background In this subsection we discuss some properties of canonical reduction systems and the lantern relation. Let S = S bg,p be a surface with b boundary components and p punctures. Let Mod( S ) (reps. PMod( S )) be the mapping class group (resp. pure mapping class group ) of S , i.e. the group of isotopy classes of orientation-preserving diffeomorphisms of S fixing the boundary components pointwise and the punctures as a set (resp.pointwise). By “simple closed curves”, we often mean isotopy class of simple closed curves, e.g. by “preservea simple closed curve”, we mean preserve the isotopy class of a curve.Thurston’s classification of elements of Mod( S ) is a very powerful tool to study mapping class groups.We call a mapping class f ∈ Mod( S ) reducible if a power of f fixes a nonperipheral simple closed curve. Eachnontrivial element f ∈ Mod( S ) is of exactly one of the following types: periodic, reducible, pseudo-Anosov.See [FM12, Chapter 13] and [FLP12] for more details. We now give the definition of canonical reductionsystem. Definition 2.1 ( Reduction systems). A reduction system of a reducible mapping class h in Mod( S ) is aset of disjoint nonperipheral curves that h fixes as a set up to isotopy. A reduction system is maximal if itis maximal with respect to inclusion of reduction systems for h . The canonical reduction system CRS( h ) isthe intersection of all maximal reduction systems of h .For a reducible element f , there exists n such that f n fixes each element in CRS( f ) and after cutting outCRS( f ), the restriction of f n on each component is either periodic or pseudo-Anosov. See [FM12, Corollary13.3]. Now we mention three properties of the canonical reduction systems that will be used later. Proposition 2.2.
CRS( h n ) = CRS( h ) for any n .Proof. This is classical; see [FM12, Chapter 13].For a curve a on a surface S , denote by T a the Dehn twist about a . For two curves a, b on a surface S , let i ( a, b ) be the geometric intersection number of a and b . For two sets of curves P and T , we say that S and T intersect if there exist a ∈ P and b ∈ T such that i ( a, b ) (cid:54) = 0. Notice that two sets of curves intersectingdoes not mean that they have a common element. Proposition 2.3.
Let h be a reducible mapping class in Mod( S ) . If { γ } and CRS( h ) intersect, then nopower of h fixes γ .Proof. Suppose that h n fixes γ . Therefore γ belongs to a maximal reduction system M . By definition,CRS( h ) ⊂ M . However γ intersects some curve in CRS( f ); this contradicts the fact that M is a set ofdisjoint curves. Proposition 2.4.
Suppose that h, f ∈ Mod( S ) and f h = hf . Then CRS( h ) and CRS( f ) do not intersect.Proof. By conjugation, we have that CRS( hf h − ) = h (CRS( f )). Since hf h − = f , we get that CRS( f ) = h (CRS( f )). Therefore h fixes the whole set CRS( f ). A power of h fixes all curves in CRS( f ). By Proposition2.3, curves in CRS( h ) do not intersect curves in CRS( f ).We denote the symmetric difference of two sets A , B by A (cid:52) B . Lemma 2.5.
Let h, f ∈ Mod( S ) be two reduced mapping classes such that hf = hf . Then CRS ( h ) (cid:52) CRS ( f ) ⊂ CRS ( hf ) . roof. Suppose that γ ∈ CRS( h ) and γ / ∈ CRS( f ). By Corollary 2.4, γ does not intersect CRS( f ). Thecanonical form of f has a component C that contains γ . From f hf − = h we know that f permutes CRS( h ),e.g. a power of f fixes γ . Since a pseudo-Anosov element does not fix any curve, a power of f is the identityon C .Since hf h − = f , we know that h permutes the components in the canonical form of f , e.g. h ( C ) isanother component in the canonical form of f . Since f permutes CRS( h ), a power of f fixes γ . This showsthat C and h ( C ) intersect, therefore we have that h ( C ) = C .Suppose that on the component C , the curve γ / ∈ CRS( hf ). This means that there is a curve γ (cid:48) ⊂ C such that ( hf ) n ( γ (cid:48) ) = γ (cid:48) for some integer n and i ( γ, γ (cid:48) ) (cid:54) = 0. A power of f is the identity on C , therefore f fixes γ (cid:48) . However no power of h fixes γ (cid:48) by Proposition 2.3. Therefore, no power of hf fixes γ (cid:48) . This is acontradiction, which shows that γ ∈ CRS( hf | C ).For any element e ∈ Mod( S ) such that the action on S can be broken into actions on components { C , ..., C k } , we have CRS( e | C ) ∪ ... ∪ CRS( e | C k ) ⊂ CRS( e ) . Therefore γ ∈ CRS( hf | C ) ⊂ CRS( hf ).Now, we introduce a remarkable relation for Mod( S ) that will be used in the proof. Proposition 2.6 ( The lantern relation).
There is an orientation-preserving embedding of S , ⊂ S andlet x, y, z, b , b , b , b be simple closed curves in S , that are arranged as the curves shown in the followingfigure.In Mod( S ) we have the relation T x T y T z = T b T b T b T b . Proof.
This is classical; see [FM12, Chapter 5.1].
Let { a, b } be a bounding pair as in the following figure, i.e. a, b are nonseparating curves such that a and b bounds a subsurface. Denote by T c the Dehn twist about a curve c . In this subsection, we determine φ ( T a T − b ). For two curves c and d , denote by i ( c, d ) the geometric intersection number of c and d . For acurve c (cid:48) on S g, , when we say c (cid:48) is isotopic to a curve c on S g , we mean that c (cid:48) is isotopic to c on S g .5igure 2.1: A bounding pair a, b Lemma 2.7.
Let { a, b } be a bounding pair as in Figure 2.1. Up to a swap of a and b , CRS ( φ ( T a T − b )) can be one of the following two cases. Moreover, either φ ( T a T − b ) = T a (cid:48) T − b (cid:48) as in case 1 or there exists aninteger n such that φ ( T a T − b ) = T na (cid:48) T − na (cid:48)(cid:48) T − b (cid:48) as in case 2. For a Dehn twist T s about a separating curve s ,there exists a pair of disjoint curves s (cid:48) and s (cid:48)(cid:48) such that they are all isotopic to s and φ ( T s ) = T ns (cid:48) T − ns (cid:48)(cid:48) forsome integer n . Figure 2.2: Case 1 Figure 2.3: Case 2
Proof.
Let T a T − b ∈ I g be a bounding pair map. Since the centralizer of T a T − b ∈ I g contains a copy of Z g − as a subgroup of I g , the centralizer of φ ( T a T − b ) ∈ I g, contains a copy of Z g − as well. Howeverby [McC82, Theorem 1], the centralizer of a pseudo-Anosov element is virtually cyclic group. g > g − >
3. Therefore φ ( T a T − b ) ∈ I g, is not pseudo-Anosov. For any curve γ (cid:48) on S g, , denote by γ thesame curve on S g . We decompose the proof into the following three steps. Claim 2.8 ( Step 1).
CRS ( φ ( T a T − b )) only contains curves that are isotopic to a or b .Proof. Suppose the opposite that there exists γ (cid:48) ∈ CRS( φ ( T a T − b )) such that γ is not isotopic to a or b .There are two cases. Case 1: γ intersect a and b . Since a power of φ ( T a T − b ) fixes γ (cid:48) , a power of T a T − b fixes γ . ByCRS( T a T − b ) = { a, b } and Lemma 2.3, we know that T a T − b does not fix γ . This is a contradiction. Case 2: γ does not intersect a and b . In this case by the change of coordinate principle, we canalways find a separating curve c such that i ( a, c ) = 0, i ( b, c ) = 0 and i ( c, γ ) (cid:54) = 0. Since T a T − b and T c commute in I g , the two mapping classes φ ( T a T − b ) and φ ( T c ) commute in I g, . This shows that a power of φ ( T c ) fixes CRS( φ ( T a T − b )); more specifically a power of φ ( T c ) fixes γ (cid:48) . However by Lemma 2.3, no powerof T c fixes γ . This is a contradiction. Claim 2.9 ( Step 2).
CRS ( φ ( T a T − b )) must contain curves that are isotopic to a and b .Proof. Suppose the opposite that CRS( φ ( T a T − b )) does not contain a curve γ (cid:48) such that γ is isotopic to a .Then by Step 1, CRS( φ ( T a T − b )) either contains one curve b (cid:48) isotopic to b or two curves b (cid:48) and b (cid:48)(cid:48) both6sotopic to b . After cutting CRS( φ ( T a T − b )), we have a component C that is not a punctured annulus. C homeomorphic to the complement of b in S g .If φ ( T a T − b ) is pseudo-Anosov on C , then the centralizer of φ ( T a T − b ) | C at most contains one copy of Z by [McC82, Theorem 1]. Combining with T b (cid:48) and T b (cid:48)(cid:48) , the centralizer of φ ( T a T − b ) at most contains one copyof Z as a subgroup. This contradicts the fact that the centralizer of φ ( T a T − b ) contains a subgroup Z g − as a subgroup because 2 g − >
3. Here we need to use g ≥
4. Therefore φ ( T a T − b ) is identity on C . Thiscontradicts the fact that T a T − b is not identity on C . Claim 2.10 ( Step 3).
Either φ ( T a T − b ) = T a (cid:48) T − b (cid:48) as in case 1 or there exists an integer n such that φ ( T a T − b ) = T na (cid:48) T − na (cid:48)(cid:48) T − b (cid:48) as in case 2.Proof. Suppose that φ ( T a T − b ) is pseudo-Anosov on a component C after cutting out CRS( φ ( T a T − b )) from S g, . Since g ( C ) ≥
1, there exists a separating curve s on C such that φ ( T s ) commutes with φ ( T a T − b ).Therefore φ ( T a T − b ) fixes CRS( φ ( T s )), which is either one curve or two curves isotopic to s . Thus a power of φ ( T a T − b ) fixes curves on C , which means that φ ( T a T − b ) is not pseudo-Anosov on C . Therefore, φ ( T a T − b )is not pseudo-Anosov on each of the components. By the canonical form of a mapping class, a power of φ ( T a T − b ) is a product of Dehn twists about CRS( φ ( T a T − b )). By the fact that φ ( T a T − b ) is a lift of T a T − b ,the lemma holds.The same argument works for T s the Dehn twist about a separating curve s .When n = 0 or n = 1, we have that ( T a (cid:48) ) n ( T a (cid:48)(cid:48) ) − n T b (cid:48) = T (cid:48) a (cid:48)(cid:48) T − b (cid:48) or ( T a (cid:48) ) n ( T a (cid:48)(cid:48) ) − n T b (cid:48) = T (cid:48) a (cid:48) T − b (cid:48) . There-fore, we can combine the results to get that φ ( T a T − b ) = ( T a (cid:48) ) n ( T a (cid:48)(cid:48) ) − n T b (cid:48) . In φ ( T a T − b ) = ( T a (cid:48) ) n ( T a (cid:48)(cid:48) ) − n T b (cid:48) ,denote ( T a (cid:48) ) n ( T a (cid:48)(cid:48) ) − n by the a component of φ ( T a T − b ). Notice that by symmetry, the b component of φ ( T a T − b ) could also be a product of Dehn twists. In the following lemma, we will prove that the a compo-nent of φ ( T a T − b ) does not depend on the choice of b . Lemma 2.11.
For two bounding pairs { a, b } and { a, c } , the a component of φ ( T a T − b ) is the same as the a component of φ ( T a T − c ) .Proof. If b, c are disjoint, φ ( T a T − b ) and φ ( T a T − c ) commute. By Lemma 2.5, we haveCRS( φ ( T a T − b )) (cid:52) CRS( φ ( T a T − c ) − ) ⊂ CRS( φ ( T a T − b )) φ ( T a T − c ) − ) . However φ ( T a T − b )) φ ( T a T − c ) − = φ ( T c T − b ), we have that CRS( φ ( T a T − b )) φ ( T a T − c ) − ) only containscurves that are isotopic to b or c . So CRS( φ ( T a T − b )) (cid:52) CRS( φ ( T a T − c ) − ) does not contain curves iso-topic to a . This shows that the a components of φ ( T a T − b ) and φ ( T a T − c ) are the same so that they cancancel each other through multiplication.When b, c intersect, there are a series of curves { b = b, b , ..., b n = c } such that i ( b i , b i +1 ) = 0 and i ( a, b i ) = 0 for all 1 ≤ i ≤ n −
1. This fact can be deduced from the connectivity of the complex ofhomologous curves, e.g. see [Put08]. Therefore, the a components of φ ( T a T − b ) and φ ( T a T − c ) are the same.We denote by the capital letter A the subset of curves in CRS( φ ( T a T − b )) that are isotopic to a . ByLemma 2.11, A only depends on the curve a . It can be a one-element set or a two-element set. Lemma 2.12. If i ( a, b ) = 0 , then A is disjoint from B . roof. Suppose that a, b are nonseparating. The case of separating curves are the same. If a, b bound, thenby Lemma 2.7, A and B are disjoint. If a, b do not bound, then there are curves c, d such that they form thefollowing configuration. g ≥ φ ( T a T − c ) and φ ( T b T − d ) commute, their canonical reduction systems do notintersect by Corollary 2.4. Therefore A and B are disjoint. Let D n be an n -punctured 2-disk. The n-strand pure braid group is denoted by P B n , i.e. the pure mappingclass group of D n fixing the n punctures pointwise. In this subsection, we prove a nonsplitting lemma forthe braid group that will be used in the proof of Theorem 1.1. Lemma 2.13.
Let F : P B → P B be the forgetful map forgetting the 4th punctures. There is no homo-morphism G : P B → P B such that Dehn twists map to Dehn twists, the center maps to the center and F ◦ G = id .Proof. Suppose the opposite that we have G : P B → P B such that Dehn twists map to Dehn twists, thecenter maps to the center and F ◦ G = id . Let c be a simple closed curve on D and we call c (cid:48) the lift on D such that G ( T c ) = T c (cid:48) . In the figure below, the lantern relation gives T a T b T c = T d ∈ P B . Therefore wehave T a (cid:48) T b (cid:48) T c (cid:48) = T d (cid:48) ∈ P B . Because G maps the center to the center, d (cid:48) is the boundary curve of D n . Since a, b, c do not intersect d , we have that T a , T b , T c commute with T d .If i ( a (cid:48) , b (cid:48) ) >
2, then T a (cid:48) T b (cid:48) is pseudo-Anosov on some subspace of D by Thurston’s construction, e.g. see[Che17a, Proposition 2.13]. Therefore, T a (cid:48) T b (cid:48) = T d (cid:48) T − c (cid:48)(cid:48) is not a multitwist. So i ( a (cid:48) , b (cid:48) ) = 2. Every curve in D n surrounds several points. For example, a ⊂ D surrounds 2 points. There are several cases we need toconcern about the number of surrounding points. 8igure 2.4: Case 1 Figure 2.5: Case 2 Figure 2.6: Case 3 Case 1: a (cid:48) bounds 2 points and b (cid:48) bounds 2 points. Then we have T a (cid:48) T b (cid:48) = T e (cid:48) T − c (cid:48) by thelantern relation as is shown in Figure 2.4. We also have the relation T a (cid:48) T b (cid:48) = T d (cid:48) T − c (cid:48) from the lift of the re-lation T a T b T c = T d ∈ P B . However CRS( T d (cid:48) T − c (cid:48) ) = { c (cid:48) } (cid:54) = CRS( T e (cid:48) T − c (cid:48) ) = { e (cid:48) , c (cid:48) } . This is a contradiction. Case 2: a (cid:48) bounds 2 points and b (cid:48) bounds 3 points. Then we have T a (cid:48) T b (cid:48) = T d (cid:48) T e (cid:48) T − c (cid:48) by thelantern relation as is shown in Figure 2.5. However T d (cid:48) T e (cid:48) T − c (cid:48) ) = 2 > T d (cid:48) T − c (cid:48) ). This is acontradiction. Case 3: a (cid:48) bounds 3 points and b (cid:48) bounds 3 points. Then we have T a (cid:48) T b (cid:48) = T d (cid:48) T e (cid:48) T − c (cid:48) by thelantern relation as is shown in Figure 2.6. We have T d (cid:48) T e (cid:48) T − c (cid:48) ) = 2 > T d (cid:48) T − c (cid:48) ). This is acontradiction. In this proof, we do a case study on the possibilities of φ ( T a T − e ) for a bounding pair map T a T − e . Case 1is when the a component is not a single Dehn twist. We reach a contradiction by the lantern relation. Case2 is when the component of every curve is a single Dehn twist, we use Lemma 2.13 to cause contradiction. Proof of Theorem 1.1.
We break our discussion into the following two cases.
Case 1: there is a bounding pair map T a T − e such that φ ( T a T − e ) = ( T a (cid:48) ) n ( T a (cid:48)(cid:48) ) − n T − e (cid:48) where n (cid:54) = 0 , There exist curves b, c, d such that a, b, c, d form a 4-boundary disk as Figure 2.7. We need g > S g Figure 2.8: On S g, f, g such that we have the lantern relation T a T − e T b T − f T c T − g T d = 1. Thelifts { b (cid:48) , c (cid:48) , d (cid:48) , f (cid:48) , g (cid:48) } of { b, c, d, f, g } do not intersect a (cid:48) , a (cid:48)(cid:48) as in Figure 2.8. After applying φ , we have( T a (cid:48) ) n ( T a (cid:48)(cid:48) ) − n T − e (cid:48) T b (cid:48) T − f (cid:48) T c (cid:48) T − g T d (cid:48) = 1 . Since T a (cid:48) T − e (cid:48) T b (cid:48) T − f (cid:48) T c (cid:48) T − g (cid:48) T d (cid:48) = 1 by the lantern relation on S g, , we have ( T a (cid:48) ) n ( T a (cid:48)(cid:48) ) − n = T a (cid:48) . It meansthat n = 1, which contradicts our assumption on n . This proof also works for the Dehn twist T s about aseparating curve s . Case 2: for any bounding pair map T a T − e , we have φ ( T a T − e ) = T a (cid:48) T − e (cid:48) and for any Dehntwist T s about a separating curve s , we have φ ( T s ) = T s (cid:48) Let S bg,p be a genus g surface with p punctures and b boundary components. In this case, firstly we wantto locate ∗ . Let us decompose the surface into pair of pants as the following Figure 2.9.Figure 2.9: A decomposition Figure 2.10: possibility 1 Figure 2.11: possibility 2The location of ∗ can be either in a pair of pants where all three curves are nonseparating like A or oneof them is separating like B . Suppose without loss of generality that ∗ lands on A or B . If ∗ lands on A ,we use Figure 2.10 to find four curves a, b, c, d and if ∗ lands on B , we use Figure 2.11 to find four curves a, b, c, d . The curves a, b, c, d that we find satisfy the following properties:1) d is separating and a ∪ b ∪ c ∪ d bounds a 4-boundary sphere S ≈ S ⊂ S g .2) The lifts a (cid:48) , b (cid:48) , c (cid:48) , d (cid:48) are 4 disjoint simple closed curves on S g, such that d (cid:48) is separating and a (cid:48) ∪ b (cid:48) ∪ c (cid:48) ∪ d (cid:48) bounds a 4-boundary sphere with ∗ in S (cid:48) ≈ S , ⊂ S g, . See the following figures.Figure 2.12: On S g Figure 2.13: On S g, Claim 2.14.
Let W be the subgroup of I g generated by bounding pair maps with curves on S . Let W (cid:48) bethe subgroup of I g, generated by bounding pair maps with curves on S (cid:48) such that one of the curves lies in a (cid:48) , b (cid:48) , c (cid:48) . We have that W ∼ = P B and W (cid:48) ∼ = P B . roof. W only acts nontrivially on S ∼ = S . After gluing punctured disks to the boundaries a, b and c , thereis a homomorphism µ : W → P B . Since every closed curve inside S , is isotopic to one of the boundarycomponents, every bounding pair map of W maps to a Dehn twist in P B under f . It is clear that φ issurjective. If f ∈ ker( µ ) as a mapping class on S , then f is either trivial or equal to a product of Dehntwists on a, b, c . However, we claim that a nontrivial product of Dehn twists on a, b, c is not in Torelli group,which shows that µ is injective. Suppose the opposite that f = T ma T nb T lc ∈ I g and l (cid:54) = 0. Let a (cid:48) , b (cid:48) betwo curves and denote by I ( a (cid:48) , b (cid:48) ) the algebraic intersection number of a (cid:48) and b (cid:48) . For x ∈ H ( S g ; Z ), wehave that T ma T nb T lc ( x ) = mI ( a, x ) a + nI ( b, x ) b + lI ( c, x ) c + x . The fact that T ma T nb T lc is in I g implies that mI ( a, x ) a + nI ( b, x ) b + lI ( c, x ) c = 0 for any x . Since a, b are independent, there exists an element x such that I ( x, a ) = 0 and I ( x, b ) = 1. Since a ∪ b ∪ c separate implying that a + b + c = 0, we have that I ( c, x ) = − mI ( a, x ) a + nI ( b, x ) b + lI ( c, x ) c = nb − lc = 0 because b, c are independent and l (cid:54) = 0. Forthe same reason W (cid:48) ∼ = P B .The lifts of elements in W is inside W (cid:48) and F : W (cid:48) → W is the forgetful map forgetting the puncture F : P B → P B . To conclude the proof of the theorem we only need to apply Lemma 2.13 that there is nosplitting of F satisfying our assumption. In this section, we discuss the “section problem” for the universal Torelli bundle with punctures.
We first translate the “section problem” of the universal Torelli surface bundle into a group-theoretic state-ment. As is discussed in [Che17b, Chapter 2.1], we have the following correspondence when g > (cid:40) Conjugacy classes ofrepresentations ρ : π ( B ) → Mod g (cid:41) ⇐⇒ (cid:40) Isomorphism classes oforiented S g -bundles over B (cid:41) . (3.1)Let f : E → B be a surface bundle determined by ρ : π ( B ) → Mod g . Let f ∗ : π ( E ) → π ( B ) be the mapon the fundamental groups. By the property of pullback diagrams, finding a splitting of f ∗ is the same asfinding a homomorphism p that makes the following diagram commute, i.e. π g, ◦ p = ρ . π ( E ) (cid:47) (cid:47) f ∗ (cid:15) (cid:15) Mod g, π g, (cid:15) (cid:15) π ( B ) ρ (cid:47) (cid:47) p (cid:58) (cid:58) Mod g . (3.2)We have the following correspondence: (cid:40) Homotopy classes of continuoussections of S g → E f −→ B (cid:41) ⇐⇒ (cid:40) Homomorphisms p satisfying diagram (3.2) up toconjugacy by an element in Ker( π g, ) ∼ = π ( S g ) (cid:41) . (3.3)By the correspondence (3.3), we can translate Theorem 1.3 into the following group-theoretic statement.Let PI g,n T π g,n −−−−→ I g and I g,n T π (cid:48) g,n −−−−→ I g be the forgetful maps forgetting the punctures. Let I g,n T p g,n,i −−−−→ Mod g,
11e the forgetful homomorphism forgetting the fixed points { x , ..., ˆ x i , ..., x n } . Let P B n ( S g ) (resp. B n ( S g ))be the n -strand surface braid group , i.e. the fundamental group of the space of ordered (resp. unordered) n distinct points on S g . By the generalized Birman exact sequence (see e.g. [FM12, Theorem 9.1]), we havethat Ker ( T π g,n ) ∼ = P B n ( S g ) and Ker ( T π (cid:48) g,n ) ∼ = B n ( S g ). See [Che17b, Chapter 2.1] for more details. Wewill prove the following proposition in the next subsection. Proposition 3.1.
For g > and n ≥ . The following holds:1) Every homomorphism p satisfying the following diagram is either conjugate to a forgetful homomor-phism T p g,n,i by an element in PI g,n or factors through T π g,n , i.e. there exists f such that p = f ◦ T π g,n . → P B n ( S g ) (cid:47) (cid:47) R (cid:15) (cid:15) PI g,n T π g,n (cid:47) (cid:47) p (cid:15) (cid:15) I g (cid:47) (cid:47) = (cid:15) (cid:15) → π ( S g ) (cid:47) (cid:47) Mod g, π g, (cid:47) (cid:47) Mod g (cid:47) (cid:47) . (3.4)
2) For n > , every homomorphism p (cid:48) satisfying the following diagram factors through T π (cid:48) g,n , i.e. thereexists f (cid:48) such that p (cid:48) = f (cid:48) ◦ T π (cid:48) g,n → B n ( S g ) (cid:47) (cid:47) R (cid:48) (cid:15) (cid:15) I g,n T π (cid:48) g,n (cid:47) (cid:47) p (cid:48) (cid:15) (cid:15) I g (cid:47) (cid:47) = (cid:15) (cid:15) → π ( S g ) (cid:47) (cid:47) Mod g, π g, (cid:47) (cid:47) Mod g (cid:47) (cid:47) . (3.5) Proof of Theorem 1.3 assuming Proposition 3.1.
By Theorem 1.1, the short exact sequence1 → π ( S g ) → I g, π g, −−→ I g → The top exact sequence of diagram (3.4) gives us a representation ρ T : I g → Out(
P B n ( S g )). The followinglemma describes a property of ρ T . Let p i : P B n ( S g ) → π ( S g ) be the induced map on the fundamentalgroups of the forgetful map forgetting all points except the i th point. Lemma 3.2.
Let h > . For any surjective homomorphism φ : P B n ( S g ) → F h , there exists an element t ∈ I g such that t ( Ker ( φ )) (cid:54) = Ker ( φ ) .Proof. By Theorem [Che17b, Theorem 1.5], any homomorphism φ : P B n ( S g ) → F h factors through some p i . Thus we only need to deal with the case n = 1. We will prove the lemma by contradiction.Suppose the opposite that there exists a surjective homomorphism φ : π ( S g ) → F h such that for anyelement e ∈ I g , we have e (Ker( φ )) = Ker( φ ). Since φ is surjective, the induced map on H ( , Z ) is alsosurjective. Suppose that a , a , ..., a h ∈ π ( S g ) such that φ ( a ) , ..., φ ( a h ) generate F h . Since the cup product H ( F h , Z ) ⊗ H ( F h , Z ) cup −−→ H ( F h , Z ) is trivial, the image of φ ∗ : H ( F h ; Z ) → H ( S g ; Z ) is an isotropicsubspace with dimension at most g . Thus we can find b ∈ π ( S g ) such that φ ( b ) = 1 and [ b ] (cid:54) = 0 ∈ H ( S g ; Z ).12t is clear that [ b ] and { a , ..., a h } are linearly independent. Let π = π ( S g ), and π n +1 = [ π n , π ], we havethe following exact sequence. 1 → π /π → π /π → π /π → H := H ( S g ; Z ). Let ω = (cid:80) gj =1 a j ∧ b j . We know that π /π ∼ = ∧ H/ Z ω , where the identificationis given by [ x, y ] → x ∧ y . Notice that I g acts trivially on both π /π and π /π but nontrivially on π /π .The action is measured by the Johnson homomorphism τ : I g → Hom( H, ∧ H/ Z ω ); see [Joh80] for moredetails. Let t ∈ I g . For x ∈ H , let ˜ x ∈ π be a lift of x , i.e. ˜ x maps x under the map π → H . The Johnsonhomomorphism is defined by τ ( t )( x ) = t (˜ x )˜ x − ∈ π /π . It is standard to check that τ ( t ) does not dependon the choice of lift ˜ x .Johnson [Joh80, Theorem 1] proved that the image τ ( I g ) = ∧ H/H ⊂ Hom( H, ∧ H/ Z ω ). Thereforethere exists t ∈ I g such that τ ( t )( b ) = a ∧ a . By the definition of the Johnson homomorphism, we havethat t ( b ) b − = [ a , a ] T , where T ∈ π . Since φ ( b ) = 1, we have that φ ( t ( b )) = 1 by the assumption that t (Ker( φ )) = Ker( φ ). As a result, φ ([ a , a ]) φ ( T ) = 1.Let F h = [ F h , F h ] and F n +1 h = [ F nh , F h ]. We have that φ ( π n ) ⊂ F nh , which implies that φ ( T ) ∈ F h .However φ ([ a , a ]) (cid:54) = 1 ∈ F h /F h . This contradicts the fact that φ ([ a , a ]) = φ ( T ) − .We need the following lemma from [HT85, Lemma 2.2]. Lemma 3.3.
For g > , a pseudo-Anosov element of Mod( S g,n ) does not fix any nonperipheral isotopy classof curves including nonsimple curves. Now we have all the ingredients to prove statement 1) in Proposition 3.1.
Proof of 1) in Proposition 3.1.
For any p : PI g,n → I g, , we have that for e ∈ PI g,n and x ∈ P B n ( S g ), R ( exe − ) = p ( e ) R ( x ) p ( e ) − . Denote by C e the conjugation by e in any group. This induces the following diagram: P B n ( S g ) C e (cid:47) (cid:47) R (cid:15) (cid:15) P B n ( S g ) R (cid:15) (cid:15) π ( S g ) C p ( e ) (cid:47) (cid:47) π ( S g ) . (3.6)By [Che17b, Theorem 1.5], a homomorphism R : P B n ( S g ) → π ( S g ) either factors through a forgetfulhomomorphism or has cyclic image. We break our discussion into the two cases. Case 1: Image ( R ) ∼ = Z In this case, the image is generated by x ∈ π ( S g ). By diagram (3.6), C p ( e ) preserves Image( R ) for any e . It is known that I g contains pseudo-Anosov elements; see [FM12, Corollary 14.3]. By Lemma 3.3, apseudo-Anosov element does not preserve Image( R ). Therefore R does not extend to p . Case 2: R factors through a forgetful homomorphism p i and does not have cyclic image In this case, we have a homomorphism S : π ( S g ) → π ( S g ) such that R = S ◦ p i . If S is a surjection, bythe same reason as in the proof of [Che17b, Theorem 2.4], we know that p is conjugate to p i . If S is not a13urjection, then Image( S ) is a noncyclic free group. By Lemma 3.2 and diagram 3.6, we know R does notextend to p .To prove statement 2) in Proposition 3.1, we need the following lemma. Lemma 3.4.
For n > , the image of any homomorphism B n ( S g ) → π ( S g ) is a free group.Proof. Suppose that there exists a homomorphism Φ : B n ( S g ) → π ( S g ) such that the image is not a freegroup. Then Image(Φ) ∼ = π ( S h ) where h ≥ g . After precomposing with the embedding i : P B n ( S g ) → B n ( S g ), we have a homomorphism Φ (cid:48) P B n ( S g ) → π ( S g ) with image a nontrivial finite index subgroup. ByTheorem [Che17b, Theorem 5], the map Φ (cid:48) factors through some p i , but there is no surjection from π ( S g )to a nontrivial finite index subgroup of π ( S g ). This is a contradiction. By the classification of subgroups of π ( S g ), the image of any homomorphism B n ( S g ) → π ( S g ) is a free group. Proof of 2) in Proposition 3.1.
By Claim 3.4, we know that R (cid:48) is not a surjection. Therefore, the imageof R (cid:48) is either cyclic or a noncyclic free group. For the cyclic image case, we use the same argument as inthe proof of 3.1 to show that R does not extend to p . In the case of noncyclic free group, by Lemma 3.2, weknow that R does not extend to p as well. In this subsection, we will prove the following corollary using Theorem 1.3.
Corollary 3.5.
For g > and m > n , the forgetful map F g,m,n : PI g,m → PI g,n forgetting the last m − n points does not have a section.Proof. We only need to show that the n sections of the bundle T u g,n have nontrivial self-intersection. Thenwe cannot find n + 1 disjoint sections on T u g,n . We restrict our attention to the subgroup
P B n ( S g ) of PI g,n .Let PConf n ( S g ) be the space of n-tuples of distinct points on S g . Since PConf n ( S g ) = K ( P B n ( S g ) , n ( S g ) is a subspace of BPI g,n . The bundle on PConf n ( S g ) is the trivial bundlePConf n ( S g ) × S g P n −−→ PConf n ( S g )with n sections s i ( x , ..., x n ) = ( x , ..., x n , x i ). By Poincar´e duality, the section is represented by a classin H (PConf n ( S g ) × S g ; Z ). So the self-intersection of a section is a class in H (PConf n ( S g ) × S g ; Z ). Let p i ( x , ..., x n ) = x i be the projection of PConf n ( S g ) to S g . We have the following pullback diagram such that s i is the pullback of the diagonal section for the trivial bundle P .PConf n ( S g ) × S g ( p i ,id ) (cid:47) (cid:47) P n (cid:15) (cid:15) (cid:74) S g × S gP (cid:15) (cid:15) PConf n ( S g ) (cid:47) (cid:47) S g . (3.7)Since s i is the pullback from the trivial bundle P , the self-intersection of s i is the pullback of the corre-sponding class γ in H ( S g × S g ; Z ). Let [ S g ] (resp. [ S g × S g ]) be the fundamental class of S g (resp. S g × S g ).It is classical that the class is γ = (2 − g )[ S g × S g ]. By the Gysin homomorphism, p ! ( p i , id ) ∗ (2 − g )[ S g × S g ] = (2 − g ) p ∗ i [ S g ] ∈ H (PConf n ( S g ); Z )which is nonzero by the computation in [Che17b, Lemma 3.4].14 Another proof of Theorem 1.1
We want to point out here that the punctured case can help us with the case of no punctures, i.e. Proposition3.1 can give us another proof of Theorem 1.1. Notice that the proof of Proposition 3.1 does not dependon Theorem 1.1. Let I bg,p be the Torelli group of S bg,p , i.e. the subgroup of Mod bg,p that acts trivially on H ( S g ; Z ). Second proof of Theorem 1.1.
Again let g >
3. We assume that the exact sequence (1.1) has a splittingwhich is denoted by φ such that F ◦ φ = id .By Lemma 2.7, the image φ ( T s ) of T s the Dehn twist about a separating curve s is T ns (cid:48) T − ns (cid:48)(cid:48) where s (cid:48) and s (cid:48)(cid:48) are curves on S g, that are isotopic to s . Let U T S g be the unit tangent bundle of genus g surface.Let s be a separating curve that separates S g into two parts C ∼ = S p and C ∼ = S q such that p, q ≥
2. Thecombination of torelli groups of C and C gives us a subgroup G of I g satisfying the following short exactsequence. 1 → Z ( T s ,T − s ) −−−−−−→ I p × I q → G → π ( U T S p ) → I p , i.e. see [FM12, Page 118]. The disk pushing subgroups of C and C give us a subgroup A of G satisfying the following short exact sequence.1 → Z ( T s ,T − s ) −−−−−−→ π ( U T S p ) × π ( U T S q ) → A → Claim 4.1. φ ( T s ) = T s (cid:48) for a curve s (cid:48) on S g, that is isotopic to s .Proof. We have already proved this result in the proof of Theorem 1.1, Case 1. Here we give another proofusing the Euler class. By Lemma 2.7, we have that φ ( T s ) = T ns (cid:48) T − ns (cid:48)(cid:48) . We only need to prove that CRS( φ ( T s ))only contains one curve.Suppose the opposite that CRS( φ ( T s )) contains two curves s (cid:48) and s (cid:48)(cid:48) such that they are isotopic to s . Then φ ( A ) is in the centralizer of of φ ( T s ). The centralizer φ ( T s ) is the subgroup of I g, that fixes s (cid:48) and s (cid:48)(cid:48) . Since φ ( A ) also satisfies the fact that it maps to A after forgetting ∗ . We know that φ ( A ) ⊂ π ( U T S p ) × π ( U T S q ).Since by computation,dim H ( U T S p × U T S q ; Q ) = dim H ( S p × S q ; Q ) = dim H ( A ; Q ) , we know that the Euler class of (4.1) is nonzero. Therefore (4.1) does not split which proves the claim.Since T s commutes with each element of G , we have that T s (cid:48) commutes with each element of φ ( G ).Therefore, φ ( G ) is a subgroup of the centralizer of T s (cid:48) . The centralizer C I g, ( T s (cid:48) ) of T s (cid:48) is the subgroup of I g, that fixes s (cid:48) . Since the two components of S g − s (cid:48) are not homeomorphic, any element in C I g, ( T s (cid:48) ) hasto fix the two components. Therefore C I g, ( T s (cid:48) ) satisfies the following exact sequence1 → Z −→ I p × I q, → C I g, ( T s (cid:48) ) → . Therefore we have a section of F : I q, → I q which maps T s to T s (cid:48) . This section gives a section of F q, , inthe following commutative diagram.1 (cid:47) (cid:47) Z (cid:47) (cid:47) (cid:15) (cid:15) I q, (cid:47) (cid:47) F (cid:15) (cid:15) PI q, (cid:47) (cid:47) F q, , (cid:15) (cid:15) (cid:47) (cid:47) Z (cid:47) (cid:47) I q (cid:47) (cid:47) I q, (cid:47) (cid:47) . (4.2)15owever, we already prove that F q, , does not have a section in Corollary 3.5, this implies that F does nothave a section. The statement follows. References [BLM83] J. Birman, A. Lubotzky, and J. McCarthy. Abelian and solvable subgroups of the mapping class groups.
Duke Math. J. , 50(4):1107–1120, 1983.[Che17a] L. Chen. Section problems for configuration spaces of surfaces. Pre-print, https://arxiv.org/abs/1708.07921,2017.[Che17b] L. Chen. The universal n-pointed surface bundle only has n sections.
Journal of Topology and Analysis ,pages 1–17, 2017. doi: 10.1142/S1793525319500134.[FLP12] A. Fathi, F. Laudenbach, and V. Po´enaru.
Thurston’s work on surfaces , volume 48 of
Mathematical Notes .Princeton University Press, Princeton, NJ, 2012. Translated from the 1979 French original by Djun M.Kim and Dan Margalit.[FM12] B. Farb and D. Margalit.
A primer on mapping class groups , volume 49 of
Princeton Mathematical Series .Princeton University Press, Princeton, NJ, 2012.[HT85] M. Handel and W.P. Thurston. New proofs of some results of nielsen.
Advances in Mathematics , 56(2):173–191, 1985.[Joh80] D. Johnson. An abelian quotient of the mapping class group I g . Math. Ann. , 249(3):225–242, 1980.[Joh83] D. Johnson. The structure of the Torelli group. I. A finite set of generators for I . Ann. of Math. (2) ,118(3):423–442, 1983.[McC82] J.D. McCarthy. Normalizers and centralizers of pseudo-anosov mapping classes. Pre-print, 1982.[Mes90] G. Mess. Unit tangent bundle subgroups of the mapping class groups. MSRI Pre-print, 1990.[Mes92] G. Mess. The Torelli groups for genus 2 and 3 surfaces.
Topology , 31(4):775–790, 1992.[Put08] A. Putman. A note on the connectivity of certain complexes associated to surfaces.
Enseign. Math. (2) ,54(3-4):287–301, 2008.Dept. of Mathematics, University of ChicagoE-mail: [email protected],54(3-4):287–301, 2008.Dept. of Mathematics, University of ChicagoE-mail: [email protected]