Free subgroups within the images of quantum representations
aa r X i v : . [ m a t h . G T ] S e p Free subgroups within the images of quantum representations
Louis Funar Toshitake Kohno
Institut Fourier BP 74, UMR 5582 IPMU, Graduate School of Mathematical SciencesUniversity of Grenoble I The University of Tokyo38402 Saint-Martin-d’H`eres cedex, France 3-8-1 Komaba, Meguro-Ku, Tokyo 153-8914 Japane-mail: [email protected] e-mail: [email protected]
November 21, 2018
Abstract
We prove that, except for a few explicit roots of unity, the quantum image of any Johnsonsubgroup of the mapping class group contains an explicit free non-abelian subgroup.2000 MSC Classification: 57 M 07, 20 F 36, 20 F 38, 57 N 05.Keywords: Mapping class group, Dehn twist, triangle group, braid group, Burau representation,Johnson filtration, quantum representation.
The aim of this paper is to study the images of the mapping class groups by quantum represen-tations. Some results in this direction are already known. We refer the reader to [29] and [19] forearlier treatments of quantum representations. In [8] we proved that the images are infinite andnon-abelian (for all but finitely many explicit cases) using earlier results of Jones who proved in [17]that the same holds true for the braid group representations factorizing through the Temperley-Lieb algebra at roots of unity. Masbaum then found in [23] explicit elements of infinite order inthe image. General arguments concerning Lie groups actually show that the image should containa free non-abelian group. Furthermore, Larsen and Wang showed (see [21]) that the image of thequantum representations of the mapping class groups at roots of unity of the form exp (cid:0) πi r (cid:1) , forprime r ≥
5, is dense in the projective unitary group.In order to be precise we have to specify the quantum representations we are considering. Recallthat in [2] the authors defined the TQFT functor V p , for every p ≥ A of order 2 p . These TQFT should correspond to the so-called SU (2)-TQFT, for even p and tothe SO (3)-TQFT, for odd p (see also [21] for another SO (3)-TQFT). Definition 1.1.
Let p ∈ Z + , p ≥ , such that p . The quantum representation ρ p is theprojective representation of the mapping class group associated to the TQFT V p for even p and V p for odd p , corresponding to the following choices of the root of unity: A p = − exp (cid:16) πip (cid:17) , if p ≡ − exp (cid:16) ( p +1) πip (cid:17) , if p ≡ . Notice that A p is a primitive root of unity of order p when p is even and of order p otherwise.Remark . The eigenvalues of a Dehn twist in the TQFT V p i.e., the entries of the diagonal T -matrix are of the form µ l = ( − A p ) l ( l +2) , where l belongs to the set of admissible colors (see [2],1.11). The set of admissible colors is { , , , . . . , p − } , for even p and is { , , , . . . , p − } forodd p . Therefore the order of the image of a Dehn twist by ρ p is p .We will now consider the Johnson filtration by the subgroups I g ( k ) of the mapping class group M g of the closed orientable surface of genus g , consisting of those elements having a trivial outeraction on the k -th nilpotent quotient of the fundamental group of the surface, for some k ∈ Z + .As is well-known the Johnson filtration shows up within the framework of finite type invariants of3-manifolds (see e.g. [11]).Our next result shows that the image is large in the following sense (see also Propositions 3.1 and3.4): Theorem 1.1.
Assume that g ≥ and p
6∈ { , , , , , } or g = 2 , p is even and p , , , , , } . Then for any k , the image ρ p ( I g ( k )) of the k -th Johnson subgroup by thequantum representation ρ p contains a free non-abelian group. The idea of proof for this theorem is to embed a pure braid group within the mapping classgroup and to show that its image is large. Namely, a 4-holed sphere suitably embedded in thesurface leads to an embedding of the pure braid group
P B in the mapping class group. Thequantum representation contains a particular sub-representation which is the restriction of Burau’srepresentation (see [8]) to a free subgroup of P B . One way to obtain elements of the Johnsonfiltration is to consider elements of the lower central series of P B and extend them to all of thesurface by identity. Therefore it suffices to find free non-abelian subgroups in the image of thelower central series of P B by Burau’s representation at roots of unity in order to prove Theorem1.1.The analysis of the contribution of mapping classes supported on small sub-surfaces of a surface,which are usually holed spheres, to various subgroups of the mapping class groups was also used inan unpublished paper by T. Oda and J. Levine (see [22]) for obtaining lower bounds for the ranksof the graded quotients of the Johnson filtration.Our construction also provides explicit free non-abelian subgroups (see Theorems 3.1 and 3.2 forprecise statements). Acknowledgements.
We are grateful to Jørgen Andersen, Greg Kuperberg, Greg McShane andGregor Masbaum for useful discussions and to Ian Agol for pointing out a gap in the previousversion of this paper. The second author is partially supported by Grant-in-Aid for ScientificResearch 23340014, Japan Society for Promotion of Science, and by World Premier InternationalResearch Center Initiative, MEXT, Japan. A part of this work was accomplished while the secondauthor was staying at Institut Fourier in Grenoble. He would like to thank Institut Fourier forhospitality. B and triangle groups Let B n denote the braid group on n strands with the standard generators g , g , . . . , g n − . Squierwas interested to compare the kernel of Burau’s representation β q at a k -th root of unity q withthe normal subgroup B n [ k ] of B n generated by g kj , 1 ≤ j ≤ n −
1. Recall that:
Definition 2.1.
The (reduced) Burau representation β : B n → GL ( n − , Z [ q, q − ]) is defined onthe standard generators β q ( g ) = (cid:18) − q
10 1 (cid:19) ⊕ n − , q ( g j ) = j − ⊕ q − q
10 0 1 ⊕ n − j − , for 2 ≤ j ≤ n − ,β q ( g n − ) = n − ⊕ (cid:18) q − q (cid:19) . The paper [10] is devoted to the complete description of the image of Burau’s representation of B at roots of unity. Similar results were obtained in [20, 23, 25]. For the sake of completeness wereview here the essential tools from [10] to be used later.Let us denote by A = β − q ( g ) and B = β − q ( g ) and C = β − q (( g g ) ). As is well-known P B isisomorphic to the direct product F × Z , where F is freely generated by g and g and the factor Z is the center of B generated by ( g g ) .It is simple to check that: A = (cid:18) q q (cid:19) , B = (cid:18) − q − q q (cid:19) , C = (cid:18) − q − q (cid:19) . Recall that
P SL (2 , Z ) is the quotient of B by its center. Since C is a scalar matrix the homomor-phism β − q : B → GL (2 , C ) factors to a homomorphism P SL (2 , Z ) → P GL (2 , C ).We will be concerned below with the subgroup Γ − q of P GL (2 , C ) generated by the images of A and B in P GL (2 , C ). When β − q is unitarizable, the group Γ − q can be viewed as a subgroup of thecomplex-unitary group P U (1 , πm , πn , πp , so that m + n + p <
1. The extended triangle group∆ ∗ ( m, n, p ) is the group of isometries of the hyperbolic plane generated by the three reflections R , R , R with respect to the edges of ∆. It is well-known that a presentation of ∆ ∗ ( m, n, p ) isgiven by∆ ∗ ( m, n, p ) = h R , R , R ; R = R = R = 1 , ( R R ) m = ( R R ) n = ( R R ) p = 1 i . The second type of relations have a simple geometric meaning. In fact, the product of the reflectionswith respect to two adjacent edges is a rotation by the angle which is twice the angle between thoseedges. The subgroup ∆( m, n, p ) generated by the rotations a = R R , b = R R , c = R R is a normal subgroup of index 2, which coincides with the subgroup of isometries preserving theorientation. One calls ∆( m, n, p ) the triangle (also called triangular, or von Dyck) group associatedto ∆. Moreover, the triangle group has the presentation:∆( m, n, p ) = h a, b, c ; a m = b n = c p = 1 , abc = 1 i . Observe that ∆( m, n, p ) also makes sense when m, n or p are negative integers, by interpretingthe associated generators as clockwise rotations. The triangle ∆ is a fundamental domain for theaction of ∆ ∗ ( m, n, p ) on the hyperbolic plane. Thus a fundamental domain for ∆( m, n, p ) consistsof the union ∆ ∗ of ∆ with the reflection of ∆ in one of its edges. Proposition 2.1 ([10]) . Let m < k be such that gcd( m, k ) = 1 where k ≥ . Then the group Γ − exp ( ± mπi k ) is a triangle group with the presentation: Γ − exp ( ± mπi k ) = h A, B ; A k = B k = ( AB ) k = 1 i . n is odd n = 2 k + 1, then the group Γ − q is a quotient of the triangle group associated to ∆,which embeds into the group associated to some sub-triangle ∆ ′ of ∆. Proposition 2.2 ([10]) . Let < m < k + 1 be such that gcd( m, k + 1) = 1 and k ≥ . Thenthe group Γ − exp ( ± mπi k +1 ) is isomorphic to the triangle group ∆(2 , , k + 1) and has the followingpresentation (in terms of our generators A, B ): Γ − exp ( ± mπi k +1 ) = h A, B ; A k +1 = B k +1 = ( AB ) k +1 = 1 , ( A − B k ) = 1 , ( B k A k − ) = 1 i . Proof.
Here is a sketch of the proof. Deraux proved in ([5], Theorem 7.1) that the group ∆( k +12 , k +12 , k +12 ),which is generated by the rotations a, b, c around the vertices of the triangle ∆ embeds into thetriangle group associated to a smaller triangle ∆ ′ . One constructs ∆ ′ by considering all geodesics of∆ joining a vertex and the midpoint of its opposite side. The three median geodesics pass throughthe barycenter of ∆ and subdivide ∆ into 6 equal triangles. We can take for ∆ ′ any one of the 6triangles of the subdivision. It is immediate that ∆ ′ has angles π k +1 , π and π so that the associatedtriangle group is ∆(2 , , k + 1). This group has the presentation:∆(2 , , k + 1) = h α, u, v ; α k +1 = u = v = αuv = 1 i , where the generators are the rotations of double angle around the vertices of the triangle ∆ ′ . Lemma 2.1.
The natural embedding of ∆( k +12 , k +12 , k +12 ) into ∆(2 , , k + 1) is an isomorphism.Proof. A simple geometric computation shows that: a = α , b = vα v = u α u, c = uα u . Therefore α = a k +1 ∈ ∆( k +12 , k +12 , k +12 ).From the relation αuv = 1 we derive a k +1 uv = 1, and thus u = a k v . The relation u = 1 readsnow a k ( va k v ) a k v = 1 and replacing b k by va k v we find that v = a k b k a k ∈ ∆( k +12 , k +12 , k +12 ).Further u = a k v = a − b k a k ∈ ∆( k +12 , k +12 , k +12 ). This means that ∆( k +12 , k +12 , k +12 ) is actually∆(2 , , k + 1), as claimed.It suffices now to find a presentation of ∆(2 , , k + 1) that uses the generators A = a, B = b .It is not difficult to show that the group with the presentation of the statement is isomorphicto ∆(2 , , k + 1), the inverse homomorphism sending α into A k +1 , u into A − B k A k and v into A k B k A k .A direct consequence of Propositions 2.1 and 2.2 is the following abstract description of the imageof Burau’s representation: Corollary 2.1. If q is not a primitive root of unity of order in the set { , , , , , } , then Γ q isan infinite triangle group. Alternatively, we obtain a set of normal generators for the kernel of Burau’s representation, asfollows:
Corollary 2.2.
Let n
6∈ { , } and q a primitive root of unity of order n . We denote by N ( G ) thenormal closure of a subgroup G of h g , g i . Then the kernel ker β − q : h g , g i → P GL (2 , C ) of therestriction of Burau’s representation is given by: ( N ( h g k , g k , ( g g ) k i ) , if n = 2 k ; N ( h g k +1)1 , g k +1)2 , ( g g ) k +1 , ( g − g k ) , ( g k g k − ) i ) , if n = 2 k + 1 . Johnson subgroups and proof of Theorem 1.1
For a group G we denote by G ( k ) the lower central series defined by: G (1) = G, G ( k +1) = [ G, G ( k ) ] , k ≥ Definition 3.1.
The k -th Johnson subgroup I g ( k ) is the group of mapping classes of homeomor-phisms of the closed orientable surface Σ g whose action by outer automorphisms on π/π ( k +1) istrivial, where π = π (Σ g ) . Thus I g (0) = M g , I g (1) is the Torelli group commonly denoted T g , while I g (2) is the group generatedby the Dehn twists along separating simple closed curves and considered by Johnson and Morita(see e.g. [16, 26]), which is often denoted by K g .The proof of Theorem 1.1 follows from the same argument as in [8], where we proved that theimage of the quantum representation ρ p is infinite for all p in the given range. The values of p which are excluded correspond to the TQFT’s V , V , V , V , V and V and it is known that theimages of quantum representations are finite in these cases.Before we proceed we have to make the cautionary remark that ρ p is only a projective representation.Here and henceforth when speaking about Burau’s representation we will mean the representation β q : B → P GL (2 , C ) taking values in matrices modulo scalars.We will first consider the generic case where the genus is large and the 10-th roots of unity arediscarded. This will prove Theorem 1.1 in most cases. Specifically we will prove first: Proposition 3.1.
Assume that g ≥ . Then the image ρ p ( (cid:0) h g , g i (cid:1) ( k ) ) contains a free non-abeliangroup for every k and p
6∈ { , , , , , , } .Proof. The first step of the proof provides us with enough elements of I g ( k ) having their supportcontained in a small subsurface of Σ g .Specifically we embed Σ , into Σ g by means of curves c , c , c , c as in the figure below. Then thecurves a and b which are surrounding two of the holes of Σ , are separating. c c cc a b The pure braid group
P B embeds into M , using a non-canonical splitting of the surjection M , → P B . Furthermore, M , embeds into M g when g ≥
4, by using the homomorphisminduced by the inclusion of Σ , into Σ g as in the figure. Then the group generated by the Dehntwists a and b is identified with the free subgroup generated by g and g into P B . Moreover, P B has a natural action on a a subspace of the space of conformal blocks associated to Σ g as in [8],which is isomorphic to the restriction of Burau’s representation at some root of unity dependingon p . Notice that the two Dehn twists above are elements of K g .We will need the following Proposition whose proof will be given in section 3.1.1:5 roposition 3.2. The above embedding of
P B into M g sends ( P B ) ( k ) into I g ( k ) . Recall now that h g , g i is a normal free subgroup of P B . The second ingredient needed in theproof of Proposition 3.1 is the following Proposition which will be proved in 3.1.2: Proposition 3.3.
Assume that g ≥ . Then the image ρ p ( (cid:0) h g , g i (cid:1) ( k ) ) contains a free non-abeliangroup for every k and p
6∈ { , , , , , , } . Thus the group ρ p (( P B ) ( k ) ) contains ρ p ( h g , g i ( k ) ) and so it also contains a free non-abelian group.Therefore, Proposition 3.2 implies that ρ p ( I g ( k )) contains a free non-abelian subgroup, which willcomplete the proof of Proposition 3.1.We further consider the remaining cases and briefly outline in section 3.2 the modifications neededto make the same strategy work also for small genus surfaces and for those values of the parameter p which were excluded above, namely: Proposition 3.4.
Assume that g and p verify one of the following conditions:1. g = 2 , p is even and p
6∈ { , , , , , } ;2. g = 3 and p
6∈ { , , , , , , } ;3. g ≥ and p = 40 .Then ρ p ( M g ) contains a free non-abelian group. Then Propositions 3.1 and 3.4 above will prove Theorem 1.1.
Choose the base point ∗ for the fundamental group π (Σ g ) on the circle c that separates the sub-surfaces Σ , and Σ g − , . Let ϕ be a homeomorphism of Σ , that is identity on the boundary andwhose mapping class b belongs to P B ⊂ M , . Consider its extension e ϕ to Σ g by identity outsideΣ , . Its mapping class B in M g is the image of b in M g .In order to understand the action of B on π (Σ g ) we introduce three kinds of loops based at ∗ :1. Loops of type I are those included in Σ g − , .2. Loops of type II are those contained in Σ , .3. Begin by fixing three simple arcs λ , λ , λ embedded in Σ , joining ∗ to the three otherboundary components c , c and c , respectively. Loops of type III are of the form λ − i xλ i ,where x is some loop based at the endpoint of λ i and contained in the 1-holed torus boundedby c i . Thus loops of type III generate π (Σ , , ∗ ).Now, the action of B on the homotopy classes of loops of type I is trivial. The action of B on thehomotopy classes of loops of type II is completely described by the action of b ∈ P B on π (Σ , , ∗ ).Specifically, let A : B → Aut( F ) be the Artin representation (see [1]). Here F is the free groupon three generators x , x , x which is identified with the fundamental group of the 3-holed diskΣ , . Lemma 3.1. If b ∈ ( P B ) ( k ) , then A ( b )( x i ) = l i ( b ) − x i l i ( b ) , where l i ( b ) ∈ ( F ) ( k ) . roof. This is folklore. Moreover, the statement is valid for any number n of strands instead of 3.Here is a short proof avoiding heavy computations. It is known that the set P B n,k of those purebraids b for which the length m Milnor invariants of their Artin closures vanish for all m ≤ k isa normal subgroup P B n,k of B n . Furthermore, the central series of subgroups P B n,k verifies thefollowing (see e.g. [28]): [
P B n,k , P B n,m ] ⊂ P B n,k + m , for all n, k, m, and hence, we have ( P B n ) ( k ) ⊂ P B n,k .Now, if b is a pure braid, then A ( b )( x i ) = l i ( b ) − x i l i ( b ), where l i ( b ) is the so-called longitude of the i -th strand. Next we can interpret Milnor invariants as coefficients of the Magnus expansion of thelongitudes. In particular, this correspondence shows that b ∈ P B n,k if and only if l i ( b ) ∈ ( F n ) ( k ) .This proves the claim.The action of B on the homotopy classes of loops of type III can be described in a similar way.Let a homotopy class a of this kind be represented by a loop λ − i xλ i . Then λ − i ϕ ( λ i ) is a loopcontained in Σ , , whose homotopy class η i = η i ( b ) depends only on b and λ i . Then it is easy tosee that B ( a ) = η − i aη i . Let now y i , z i be standard homotopy classes of loops based at a point of c i which generate thefundamental group of the holed torus bounded by c i , so that { y , z , y , z , y , z } is a generatorsystem for π (Σ , , ∗ ), which is the free group F of rank 6. Lemma 3.2. If b ∈ ( P B ) ( k ) , then η i ( b ) ∈ ( F ) (2 k ) .Proof. It suffices to observe that η i ( b ) is actually the i -th longitude l i ( b ) of the braid b , expressednow in the generators y i , z i instead of the generators x i . We also know that x i = [ y i , z i ]. Letthen η : F → F be the group homomorphism given on the generators by η ( x i ) = [ y i , z i ]. Then η i ( b ) = η ( l i ( b )). Eventually, if l i ( b ) ∈ ( F ) ( k ) , then η ( l i ( b )) ∈ ( F ) (2 k ) and the claim follows.Therefore the class B belongs to I g ( k ), since its action on every generator of π (Σ g , ∗ ) is a conju-gation by an element of π (Σ g , ∗ ) ( k ) . First we want to identify some sub-representation of the restriction of ρ p to P B ⊂ M g . Specificallywe have: Lemma 3.3.
Let p ≥ . The restriction of the quantum representation ρ p at P B ⊂ M , hasan invariant 2-dimensional subspace such that the corresponding sub-representation is equivalent tothe Burau representation β − q p , where the root of unity q p is given by: q p = − A − p = − exp (cid:16) − πip (cid:17) , if p ≡ − A − = − exp (cid:0) − πi (cid:1) , if p = 5; − A − p = − exp (cid:16) − p +1) πip (cid:17) , if p ≡ , p ≥ . Proof.
For even p this is the content of [8], Prop. 3.2. We recall that in this case the invariant2-dimensional subspace is the the space of conformal blocks associated to the surface Σ , with allboundary components being labeled by the color 1. The odd case is similar. The invariant subspaceis the space of conformal blocks associated to the surface Σ , with boundary labels (2 , , , p = 5 and (4 , , , p ≥ ρ p ( P B ) of the quantum representation projects onto the image of the Buraurepresentation β − q p ( P B ).Up to a Galois conjugacy we can assume that β − q p is unitarizable and after rescaling, it takes valuesin U (2). Consider the projection of β − q p (( P B ) ( k ) ) into U (2) /U (1) = SO (3).A finitely generated subgroup of SO (3) is either finite or abelian or else dense in SO (3). If the groupis dense in SO (3), then it contains a free non-abelian subgroup. Moreover, solvable subgroups of SU (2) (and hence of SO (3)) are abelian. The finite subgroups of SO (3) are well-known. Theyare the following: cyclic groups, dihedral groups, tetrahedral group (automorphisms of the regulartetrahedron), the octahedral group (the group of automorphisms of the regular octahedron) and theicosahedral group (the group of automorphisms of the regular icosahedron or dodecahedron). Allbut the last one are actually solvable groups. The icosahedral group is isomorphic to the alternatinggroup A and it is well-known that it is simple (and thus non-solvable). As a side remark this groupappeared in relation with the non-solvability of the quintic equation in Felix Klein’s monograph[18]. Lemma 3.4. If q is not a primitive root of unity of order in the set { , , , , , } , then (Γ q ) ( k ) is non-solvable and thus non-abelian for any k . Moreover, (Γ q ) ( k ) cannot be A , for any k .Proof. If (Γ q ) ( k ) were solvable, then Γ q would be solvable. But one knows that Γ q is not solvable.In fact if q is as above, then Γ q is an infinite triangle group by Corollary 2.1.Now any infinite triangle group has a finite index subgroup which is a surface group of genus atleast 2. Therefore, each term of the lower central series of that surface group embeds into thecorresponding term of the lower central series of Γ q , so that the later is non-trivial. Since the lowercentral series of a surface group of genus at least 2 consists only of infinite groups it follows thatno term can be isomorphic to the finite group A either.Lemma 3.4 shows that whenever p is as in the statement of Proposition 3.3, the group β − q p ( (cid:0) h g , g i (cid:1) ( k ) )is neither finite nor abelian, so that it is dense in SO (3) and hence it contains a free non-abeliangroup. This proves Proposition 3.3. The main interest of the elementary arguments in the proof presented above is that the free non-abelian subgroups in the image are abundant and explicit. For instance we have:
Theorem 3.1.
Assume that g ≥ , p
6∈ { , , , } and p . Set x = ρ p ([ g , g ]) and y = ρ p ([ g , g ]) . Then the group generated by the iterated commutators [ x, [ x, [ x, . . . , [ x, y ]] . . . ] and [ y, [ x, [ x, . . . , [ x, y ]] . . . ] of length k ≥ is a free non-abelian subgroup of ρ p ( I g ( k )) . It is well-known that the order of the matrix β − q ( g i ), i ∈ { , } in P GL (2 , C ) is the order of theroot of unity q , namely the smallest positive n such that q is a primitive root of unity of order n .We considered in Lemma 3.3 the root of unity q p with the property that β − q p is a sub-representationof the quantum representation ρ p . We derive from Lemma 3.3 that the order of the root of unity q p is 2 o ( p ) where o ( p ) = p , if p ≡ p , if p ≡ p , if p ≡ p, if p ≡ , p ≥ . Therefore β − q p ( h g , g i ) is isomorphic to the triangle group ∆( o ( p ) , o ( p ) , o ( p )). Notice that ingeneral o ( p ) ∈ + Z and o ( p ) is an integer if and only if p β − q p ( g ) is a proper divisor of the order p of a Dehn twist ρ p ( g ),when p is even.In the proof of Theorem 3.1 we will need the following result concerning the structure of commutatorsubgroups of triangle groups: Lemma 3.5.
The commutator subgroup ∆( r, r, r ) (2) of a triangle group ∆( r, r, r ) , r ∈ Z − { , , } ,is a 1-relator group with generators f c ij , for ≤ i, j ≤ r − , and the relation: f c · f c − · f c · f c − · · · ^ c r r − − · f c rr = 1 . Proof.
The kernel K of the abelianization homomorphism Z /r Z ∗ Z /r Z → Z /r Z × Z /r Z is thefree group generated by the commutators. Denote by e a and e b the generators of the two copies ofthe cyclic group Z /r Z . Then K is freely generated by f c ij = [ e a i , e b j ], where 1 ≤ i, j ≤ r −
1. Thegroup ∆( r, r, r ) is the quotient of Z /r Z ∗ Z /r Z by the normal subgroup generated by the element( e a e b ) r e a − r e b − r , which belongs to K . This shows that ∆( r, r, r ) (2) is a 1-relator group, namely thequotient of K by the normal subgroup generated by the element ( e a e b ) r e a − r e b − r . In order to get theexplicit form of the relation we have to express this element as a product of the generators of K ,i.e., as a product of commutators of the form [ e a i , e b j ]. This can be done as follows:( e a e b ) r e a − r e b − r = [ e a, e b ][ e b, e a ][ e a , e b ] · · · [ e a r − , e b r − ][ e b r − , e a r ][ e a r , e b r ] . Therefore ∆( r, r, r ) (2) has a presentation with generators f c ij , where 1 ≤ i ≤ j ≤ r , and the relationin the statement of the lemma. Proof of Theorem 3.1.
Recall now the classical Magnus Freiheitsatz, which states that any subgroupof a 1-relator group which is generated by a proper subset of the set of generators involved in thecyclically reduced word relator is free.Assume now that o ( p ) ∈ Z and o ( p ) ≥
4. Then β − q p ([ g , g ]) and β − q p ([ g , g ]) are the elements f c and f c of ∆( o ( p ) , o ( p ) , o ( p )) (2) respectively.An easy application of the Freiheitsatz to the commutator subgroup of the infinite triangle group∆( o ( p ) , o ( p ) , o ( p )) gives us that the subgroup generated by β − q p ([ g , g ]) and β − q p ([ g , g ]) is free.This implies that the subgroup generated by x and y is free.Eventually the k -th term of the lower central series of the group generated by x and y is also a freesubgroup which is contained into ρ p (( P B ) ( k ) ) ⊂ ρ p ( I g ( k )). This proves Theorem 3.1.When p ≡ o ( p ) is a half-integer and β − q p ( h g , g i ) is isomorphic to the triangle group∆(2 , , o ( p )). If gcd(3 , o ( p )) = 1, then H (∆(2 , , o ( p ))) = 0, so the central series of this trianglegroup is trivial. Nevertheless the group ∆(2 , , o ( p )) has many normal subgroups of finite indexwhich are surface groups and thus contain free subgroups. In particular, any subgroup of infiniteindex of ∆(2 , , o ( p )) is free. There is then an extension of the previous result in this case, asfollows: Theorem 3.2.
Assume that g ≥ , p
6∈ { , , } and p ≡ so that p = 8 n , for odd n = 2 k + 1 ≥ . Consider the following two elements of h g , g i : s = g k g k g k − k )1 g − g k g − g k − k )1 g k g k , and t = g k g k g k − k )1 g − g k +1)1 g g k + k )1 g k g k . et N ( s, t ) be the normal subgroup generated by s and t in h g , g i . Then for any choice of f ( n ) elements x , x , . . . , x f ( n ) from N ( s, t ) the image ρ p ( h x , x , . . . , x f ( n ) i ) is a free group. Here thefunction f ( n ) is given by: f ( n ) = | P SL (2 , Z /n Z ) | · n − n and, in particular, when n is prime by: f ( n ) = ( n + 1)( n − n − · Then the group generated by the iterated commutators of length k ≥ is a free subgroup of ρ p ( I g ( k )) .Proof. Observe that the map
P SL (2 , Z ) → P SL (2 , Z /n Z ) factors through ∆(2 , , n ), namely wehave a homomorphism ψ : ∆(2 , , n ) → P SL (2 , Z /n Z ) defined by ψ ( α ) = (cid:18) −
10 1 (cid:19) , ψ ( u ) = (cid:18) −
11 0 (cid:19) , ψ ( v ) = (cid:18) −
11 0 (cid:19) . The matrices ψ ( α ) , ψ ( u ) , ψ ( v ) are obviously elements of orders n, P SL (2 , Z /n Z ) respec-tively. It follows that the normal subgroup K (2 , , n ) = ker ψ is torsion free, because every torsionelement in ∆(2 , , n ) is conjugate to some power of one the generators α, u, v (see [13]). Therefore K (2 , , n ) is a surface group, namely the fundamental group of a closed orientable surface whichfinitely covers the fundamental domain of ∆(2 , , n ). The Euler characteristic χ ( K (2 , , n )) of thisFuchsian group can easily be computed by means of the formula: χ ( K (2 , , n )) = | P SL (2 , Z /n Z ) | · χ (∆(2 , , n )) , where the (orbifold) Euler characteristic χ (∆(2 , , n )) has the well-known expression: − χ (∆(2 , , n )) = 1 − (cid:18)
12 + 13 + 1 n (cid:19) = n − n . It is also known that any − χ ( G ) + 1 elements of a closed orientable surface group G generate a freesubgroup of G . Thus, in order to establish Theorem 3.2, it will suffice to show that the images of theelements s, t under Burau’s representation β − q p are normal generators of the group K (2 , , n ). Thisis equivalent to show that these images correspond to the relations needed to impose in ∆(2 , , n )in order to obtain the quotient P SL (2 , Z /n Z ). However, one already knows presentations for thisgroup (see [4], Lemma 1 and [14]) as follows: P SL (2 , Z /n Z ) = h α, v, u | α n = u = v = 1 , gvgv = gαg − α − = 1 i , for odd n , where g = vα k vα − vα k . The first three relations above correspond to the presentationof ∆(2 , , n ) and the elements gvgv and gαgα − correspond to the images of s and t in ∆(2 , , n ),by using the fact that α = a k +1 , vα v = b , v = a k b k a k (see the proof of Lemma 2.1). We outline here an alternative proof which does not rely on the description of the image of Bu-rau’s representation in Corollary 2.1. This proof is shorter but less effective since it does notproduce explicit free subgroups and uses the result of [21] and the Tits alternative, which needmore sophisticated tools from the theory of algebraic groups.The image ρ p ( M g ) in P U ( N ( p, g )) is dense in P SU ( N ( p, g )), if p ≥ N ( p, g ) denotes the dimension of the space of conformal blocks in genus g for the TQFT V p . In10articular, the image of ρ p is Zariski dense in P U ( N ( p, g )). By the Tits alternative (see [30]) theimage is either solvable or else it contains a free non-abelian subgroup. However, if the image weresolvable, then its Zariski closure would be a solvable Lie group, which is a contradiction. Thisimplies that ρ p ( M g ) contains a free non-abelian subgroup.If p is not prime but has a prime factor r ≥
5, then the claim for p follows from that for r . If p does not satisfy this condition, then we have again to use the result of Proposition 3.3 for k = 1.This result can be obtained directly from the computations in [17] proving that the image of theJones representation of B is neither finite nor abelian for the considered values of p . This settlesthe case k = 1 of Proposition 3.1.Furthermore, the group ρ p ( T g ) is of finite index in ρ p ( M g ), and hence it also contains a free non-abelian subgroup. Results of Morita (see [27]) show that for k ≥ I g ( k + 1) is the kernelof the k -th Johnson homomorphism I g ( k ) → A k , where A k is a finitely generated abelian group.This implies that [ I g ( k ) , I g ( k )] ⊂ I g ( k + 1), for every k ≥
2. In particular, the k -th term of thederived series of ρ p ( T g ) is contained into ρ p ( I g ( k + 1)). But every term of the derived series of ρ p ( T g ) contains the corresponding term of the derived series of a free subgroup and hence a freenon-abelian group. This proves Proposition 3.1. Remark . Using the strong version of Tits’ theorem due to Breuillard and Gelander (see [3])there exists a free non-abelian subgroup of M g /M g [ p ] whose image in P SU ( N ( p, g )) is dense. Here M g [ p ] denotes the (normal) subgroup generated by the p -th powers of Dehn twists. If the genus g ∈ { , } , then the construction used in the proof of Proposition 3.1 should bemodified. This is equally valid when we want to get rid of the values p = 5 and p = 40.The proof follows along the same lines as Proposition 3.3, but the embeddings Σ , ⊂ Σ g are nowdifferent. In all cases considered below the analogue of Proposition 3.2 will still be true, namelythe image of the subgroup h g , g i ( k ) by the homomorphisms M , → M g will be contained withinthe Johnson subgroup I g ( k ).If g = 2 we use the following embedding Σ , ⊂ Σ : a bc c c c Although the homomorphism M , → M induced by this embedding is not anymore injective, itsends the free subgroup h g , g i ⊂ P B ⊂ M , isomorphically onto the subgroup of M generatedby the Dehn twists along the curves a and b in the figure above.Consider for even p the space of conformal blocks associated to Σ , with boundary labels (1 , , , ρ p ( h g , g i )-invariant and the restriction of ρ p to this subspace isstill equivalent to Burau’s representation β − q p (see [8]). Therefore Proposition 3.3 shows that ρ p ( h g , g i ( k ) ), and hence also ρ p ( I ( k )), contains a free non-abelian group.If g = 3 and p ≥ , ⊂ M :11 c c a b c The homomorphism M , → M induced by this embedding is not injective but it also sends thefree subgroup h g , g i ⊂ P B ⊂ M , isomorphically onto the subgroup of M generated by theDehn twists along the curves a and b in the figure above. The space of conformal blocks associatedto Σ , with boundary labels (2 , , ,
4) is a 2-dimensional subspace invariant by ρ p ( h g , g i ) andthe restriction of ρ p to this subspace is equivalent to Burau’s representation β − q p . Applying againProposition 3.3, we find that ρ p ( h g , g i ( k ) ), and hence ρ p ( I ( k )), contains a free non-abelian group.This also gives the desired results for any g ≥
3, and p as in the statement.Eventually we have to settle the case p = 40, when β − q p ( B ) is known to have finite image (see[17]). We will consider instead the representation ρ p ( i ( P B )), where P B embeds non-canonicallyinto M , and M , maps into M by the homomorphism i : M , → M induced by the inclusionΣ , ⊂ Σ g drawn below: c c c c c We consider the 3-dimensional space of conformal blocks associated to the surface Σ , with theboundary labels (1 , , , , p = 40 and the labels (2 , , , , p = 5 respectively.This space of conformal blocks is ρ p ( i ( P B ))-invariant. The restriction of ρ p | P B to this invari-ant subspace is known (see again [8]) to be equivalent to the Jones representation of B at thecorresponding root of unity.Now we have to use a result of Freedman, Larsen and Wang (see[7]) subsequently reproved andextended by Kuperberg in ([20], Thm.1) saying that the Jones representation of B at a 10-throot of unity on the two 3-dimensional conformal blocks we chose is Zariski dense in the group SL (3 , C ). A particular case of the Tits alternative says that any finitely generated subgroup of SL (3 , C ) is either solvable or else contains a free non-abelian group. A solvable subgroup has alsoa solvable Zariski closure. The denseness result from above implies then that ρ p ( P B ), and hencealso ρ p (( P B ) ( k ) ), contains a free non-abelian group. The arguments in the proof of Proposition3.3 carry on to this setting and this proves Proposition 3.4. Corollary 3.3.
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