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On images of quantum representations of mapping class groups∗
Louis Funar Toshitake KohnoInstitut Fourier BP 74, UMR 5582 IPMU, Graduate School of Mathematical Sciences
University of Grenoble I The University of Tokyo
38402 Saint-Martin-d’Heres cedex, France 3-8-1 Komaba, Meguro-Ku, Tokyo 153-8914 Japan
e-mail: [email protected] e-mail: [email protected]
December 21, 2010
Abstract
We consider subgroups of the braid groups which are generated by k-th powers of the stan-dard generators and prove that any infinite intersection with even k is trivial. This is motivatedby some conjectures of Squier concerning the kernels of Burau’s representations of the braidgroups at roots of unity. Further we show that the image of the braid group on 3 strands bythese representations is either a finite group, for a few roots of unity, or a finite extension of atriangle group, by using geometric methods. The second part of this paper is devoted to appli-cations of these results to qualitative characterizations of the images of quantum representationsof the mapping class groups. First, we prove that, except for a few explicit roots of unity, thequantum image of any Johnson subgroup contains a free non-abelian subgroup. Our main resultis that, in general, the images of quantum representations are not isomorphic to higher rankirreducible lattices in semi-simple Lie groups. In particular, when the level is an odd primegreater than or equal to 7 the images are subgroups of infinite index within the group of unitarymatrices with cyclotomic integers entries.
2000 MSC Classification: 57 M 07, 20 F 36, 20 F 38, 57 N 05.Keywords: Mapping class group, Dehn twist, Temperley-Lieb algebra, triangle group, braid,Burau representation, Johnson filtration, quantum representation.
1 Introduction and statements
The first part of the present paper is devoted to the study of groups related to the kernels ofthe Burau representations of braid groups at roots of unity. We consider two conjectures statedby Squier in [43] concerning these kernels. These conjectures were part of an approach to thefaithfulness of the Burau representations and it seems that they were overlooked over the yearsbecause of the counterexamples found by Moody, Long, Paton and Bigelow (see [36, 31, 3]) forbraids on n ≥ 5 strands.
Specifically, letBn denote the braid group on n strands with the standard generators g1, g2, . . . , gn−1.Squier was interested to compare the kernel of the Burau representation βq at a k-th root of unityq to the normal subgroup Bn[k] of Bn generated by gk
j , 1 ≤ j ≤ n − 1. Our first result answers astrengthened form of conjecture C2 from [43]:
Theorem 1.1. The intersection of Bn[2k] over any infinite set of integers k is trivial.
∗Preprint available electronically at http://www-fourier.ujf-grenoble.fr/˜funar
1
Our method does not give any information about the intersection of Bn[k] with odd k.
The proof uses the asymptotic faithfulness of quantum representations of mapping class groups, dueto Andersen ([1]) and independently to Freedman, Walker and Wang ([16]). The other conjecturestated in [43] is that Bn[k] is the kernel of the Burau representation. This is false because Burau’srepresentation at a generic parameter is not faithful for n ≥ 5 (see Proposition 2.4).
The main body of the first part of this paper is devoted to the complete description of the imageof the Burau representation of B3. We can state our main result in this direction:
Theorem 1.2. Assume that q is a primitive n-th root of unity and g1, g2 are the standard generatorsof B3. Then β−q(B3) has a presentation with generators g1, g2 and relations:
1. If n = 2k and k is odd:
Braid relation: g1g2g1 = g2g1g2,Power relations: g2k
1 = g2k2 = (g2
1g22)
k = 1,
2. If n = 2k and k is even:
Braid relation: g1g2g1 = g2g1g2,Power relations: g2k
1 = g2k2 = (g2
1g22)
2k = 1,
3. If n = 2k + 1:
Braid relation: g1g2g1 = g2g1g2,
Power relations: g2k+11 = g2k+1
2 = (g21g
22)
2(2k+1) = 1,
This result was also obtained independently by Masbaum in [34] in a slightly different context.Consider the 2-dimensional SO(3)-quantum representations of the mapping class group of thepunctured torus at a primitive 2p-th root of unity for odd p, with the puncture labeled by thecolor c = p−1
2 − 2. Then the result proved by Masbaum is that the kernel of this representationis normally generated by the p-th powers of the Dehn twists. However, these quantum represen-tations are covered by the Burau representations of B3, so the two results above are equivalent.The same arguments apply to the quantum representations of the mapping class group M0,4 of the4-holed sphere. Notice that 2-dimensional representations of B3 are equivalent either to abelianrepresentations, to some not completely reducible representations, or else to the Burau representa-tion. Another consequence of this theorem is the fact that the image of a pseudo-Anosov mappingclass in the mapping class group of the punctured (or holed) torus by the quantum representationsconsidered above is of infinite order for large enough p. This solves a particular case of a conjec-ture formulated by Andersen, Masbaum and Ueno in [2] and it was independently announced byMasbaum in ([20], p.4).
The proof of this algebraic statement has a strong geometric flavor. A key ingredient is Squier’stheorem concerning the unitarizability of Burau’s representation (see [43]). The non-degenerateHermitian form defined by Squier is invariant under the braid group but it is not always positivedefinite. First, we find whether it is positive definite, so that the representation can be conjugateinto U(2). On the other hand when this Hermitian form is not positive definite the representationcan be complex-unitarized, namely it can be (rescaled and) conjugated into U(1, 1).
We will focus then on the complex-unitary case. We show that that the image of some free subgroupof the pure braid group PB3 on three strands by the Burau representation is a subgroup of PU(1, 1)generated by three rotations in the hyperbolic plane. Here the hyperbolic plane is identified to theunit disk of the complex projective line CP 1. Geometric arguments due to Knapp, Mostow and
2
Deraux (see [24, 37, 13]) show that the image of PB3 is a discrete triangle group and thus wecan give an explicit presentation for it. Then an easy argument permits to describe the imageof the slightly larger group B3. In particular, we obtain a description of the kernel of the Buraurepresentation of B3 at roots of unity, thus proving Theorem 1.2. This first part is not only purelytechnical preparation for the second part of the paper. In fact, finding the image of the Buraurepresentation seems to be a difficult problem interesting by itself (see e.g. [8, 7, 35]).
The second half of the present article consists of applications of these results to the study of themapping class groups images by quantum representations. Some results in this direction are alreadyknown. In [17] we proved that the images are infinite and non-abelian (for all but finitely manyexplicit cases) using earlier results of Jones who proved in [27] that the same holds true for the braidgroup representations factorizing through the Temperley-Lieb algebra at roots of unity. Masbaumfound then in [33] explicit elements of infinite order in the image. General arguments concerningLie groups show actually that the image should contain a free non-abelian group. Further, Larsenand Wang showed (see [29]) that the image of the quantum representations of mapping class groupsat roots of unity of the form exp
(2πi4r
), for prime 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 [5] the authors defined the TQFT functor Vp, for every p ≥ 3 and a primitive root of unityA of order 2p. These TQFT should correspond to the so-called SU(2)-TQFT, for even p and tothe SO(3)-TQFT, for odd p (see also [29] for another SO(3)-TQFT).
Definition 1.1. Let p ∈ Z+, p ≥ 3, such that p 6≡ 2(mod 4). The quantum representation ρp is theprojective representation of the mapping class group associated to the TQFT V p
2for even p and Vp
for odd p, corresponding to the following choices of the root of unity:
Ap =
− exp(
2πip
), if p ≡ 0(mod 4)
− exp(
(p+1)πip
), if p ≡ 1(mod 2)
Notice that Ap is a primitive root of unity of order p when p is even and 2p otherwise.
Remark 1.1. The eigenvalues of a Dehn twist in the TQFT Vp, i.e. the entries of the diagonalT -matrix, are of the form µl = (−Ap)
l(l+2), where l belongs to the set of admissible colors (see [5],4.11). The set of admissible colors is {0, 1, 2, . . . , p
2 − 2}, for even p and {0, 2, 4, . . . , p− 3} for oddp. Therefore the order of the image of a Dehn twist by ρp is p.
We will consider now the Johnson filtration by the subgroups Ig(k) of the mapping class groupMg 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 it is well-known the Johnson filtration shows up within the framework of finite type invariantsof 3-manifolds (see e.g. [18]).
Our next result shows that the image is large in the following sense (see also Propositions 4.2 and4.5):
Theorem 1.3. Assume that g ≥ 3 and p 6∈ {3, 4, 8, 12, 16, 24} or g = 2, p is even and p 6∈{4, 8, 12, 16, 24, 40}. Then for any k, the image ρp(Ig(k)) of the k-th Johnson subgroup by thequantum representation ρp 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 PB3 in the mapping class group. Thequantum representation contains a particular sub-representation which is the restriction of Burau’s
3
representation (see [17]) to a free subgroup of PB3. One way to obtain elements of the Johnsonfiltration is to consider elements of the lower central series of PB3 and extend them to all of thesurface by identity. Therefore it suffices to find free non-abelian subgroups in the image of the lowercentral series of PB3 by the Burau representation at roots of unity in order to prove Theorem 1.3.
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 [30]) for obtaining lower bounds for the ranksof the gradings of the Johnson filtration.
Our construction provides also explicit free non-abelian subgroups (see Theorems 4.1 and 4.2 forprecise statements).
On the other hand Gilmer and Masbaum proved in [19] that the mapping class group preserves acertain free lattice within the conformal blocks associated to the SO(3)-TQFT. Let us introduce thefollowing notations. For p ≥ 5 an odd prime we denote by Op the ring of integers in the cyclotomicfield Q(ζp), where ζp is a primitive p-th root of unity. Thus Op = Z[ζp], if p ≡ −1(mod 4) andOp = Z[ζ4p], if p ≡ 1(mod 4).
The main result of [19] states that, for every odd prime p ≥ 5, there exists a free Op -latticeSg,p in the C-vector space of level p conformal blocks associated by the TQFT Vp to the genusg closed orientable surface and a non-degenerate Hermitian Op-valued form on Sg,p such that (acentral extension of) the mapping class group preserves Sg,p and keeps invariant the Hermitianform. Therefore the image of the mapping class group consists of unitary matrices (with respectto the Hermitian form) with entries from Op. Let PU(Op) be the group of all such matrices, up toscalar multiplication.
A natural question is to compare the image ρp(Mg) and the discrete group PU(Op). The mainresult of this article shows that the image is small with respect to the whole group:
Theorem 1.4. Suppose that g ≥ 4, p 6∈ {3, 4, 5, 8, 12, 16, 24, 40} and if p = 8k with odd k thenthere exists a proper divisor of k which is greater than or equal to 7. Then the group ρp(Mg) isnot an irreducible lattice in a higher rank semi-simple Lie group. In particular, if p ≥ 7 is an oddprime then ρp(Mg) is of infinite index in PU(Op).
The idea of the proof is to show that ρp(Mg) has infinite normal subgroups of infinite index. We usethe same method as in the proof of Theorem 1.3. Namely, consider a subgroup of a mapping classgroup whose intersection with some pure braid subgroup PB3 has the property that its image bythe Burau representation is infinite. Then the image of the quantum representation of the subgroupis infinite, as well. We are able to make this strategy working for the subgroup generated by thef -th powers of Dehn twists (for most f which divide p) and also for the second derived subgroup[[Kg,Kg], [Kg,Kg]] of the second Johnson subgroup Kg = Ig(2).
It is known that PU(Op) is an irreducible lattice in a semi-simple Lie group G obtained by theso-called restriction of scalars construction. Specifically, G is the product
∏σ PUσ, over all complex
valuations σ of Op. The factor PUσ is the (projective) unitary group associated to the Hermitianform conjugated by σ, thus corresponding to some Galois conjugate root of unity.
By a celebrated theorem of Margulis (see [32], chapter IV, [46], chapter 8) an irreducible lattice in ahigher rank group, in particular an irreducible lattice in G, cannot have infinite normal subgroupsof infinite index. Therefore ρp(Mg) is not isomorphic to a lattice in a higher rank semi-simple Liegroup. In particular, when p ≥ 7 is an odd prime the discrete subgroup ρp(Mg) should be of infinitecovolume in G and hence of infinite index in the lattice PU(Op).
Although the proof of Theorem 1.3 does not use the full strength of our results from the first part– a shorter proof along the same lines of reasoning but using a density result from [15, 28] and the
4
Tits alternative instead of the explicit description of the image of Burau’s representation of B3 isoutlined in subsection 4.2.4 – they seem essential for Theorem 1.4.
Acknowledgements. We are grateful to Norbert A’Campo, Jørgen Andersen, Jean-Benoıt Bost,Martin Deraux, Greg Kuperberg, Francois Labourie, Yves Laszlo, Greg McShane, Ivan Marin,Gregor Masbaum, Daniel Matei, Majid Narimannejad, Christian Pauly, Bob Penner, ChristopheSorger and Richard Wentworth for useful discussions and to the referees for a careful reading of thepaper leading to numerous corrections and suggestions. The first author was partially supported bythe ANR Repsurf:ANR-06-BLAN-0311. The second author is partially supported by Grant-in-Aidfor Scientific Research 20340010, Japan Society for Promotion of Science, and by World PremierInternational Research Center Initiative, MEXT, Japan. A part of this work was accomplishedwhile the second author was staying at Institut Fourier in Grenoble. He would like to thankInstitut Fourier for hospitality.
2 Braid group representations
2.1 Jones and Burau’s representations at roots of unity
In this section we recall the definition of the Jones and Burau’s representations of braid groupsand show that they are equivalent except at primitive roots of unity of order 1 and 3. Moreover wediscuss when they are unitarizable or complex-unitarizable. We start with the following classicaldefinition.
Definition 2.1. The Temperley-Lieb algebra Aτ,n, for τ ∈ C∗, n ≥ 2 is the C∗-algebra generatedby the projectors 1, e1, . . . , en−1 satisfying the following relations:
e∗j = e2j = ej
eiej = ejei, if |i− j| ≥ 2
ejej+1ej = ejej−1ej = τej
According to Wenzl ([45]) there exist such unitary projectors ej , 1 ≤ j ≤ n − 1, for every naturaln ≥ 2 if and only if τ−1 ≥ 4 or τ−1 = 4cos2
(πk
), for some natural k ≥ 3. However, for given n one
could find projectors e1, . . . , en−1 as above if τ−1 = 4cos2 (α), where the angle α belongs to somespecific arc of the unit circle.
Another definition of the Temperley-Lieb algebra (which is equivalent to the former, at least whenτ verifies the previous conditions) is as a quotient of the Hecke algebra:
Definition 2.2. The Temperley-Lieb algebra An(q) is the quotient of the group algebra CBn of thebraid group Bn by the relations:
(gi − q)(gi + 1) = 0
1 + gi + gi+1 + gigi+1 + gi+1gi + gigi+1gi = 0
where gi are the standard generators of the braid group Bn. The quotient obtained by imposing onlythe first relation above is called the Hecke algebra Hn(q).
It is known that An(q) is isomorphic to Aτ,n where τ−1 = 2 + q + q−1, and in particular when q isthe root of unity q = exp
(2πin
). We suppose from now on that τ−1 = 2 + q + q−1.
We will analyze the case where n = 3 and q is a root of unity, and more generally for |q| = 1. ThenAτ,3 and An(3) are nontrivial and well-defined for all q with |q| = 1 belonging to the arc of circlejoining exp
(−2πi
3
)to exp
(2πi3
). We will recover this result below, in a slightly different context.
5
Further Aτ,3 is semi-simple and splits as M2(C)⊕C, where M2(C) denotes the simple C-algebra of2-by-2 matrices. There is a natural representation of B3 into Aτ,3 which sends gi into qei − (1− ei).This representation is known to be unitarizable when τ−1 ≥ 4 (see [27]).
Proposition 2.1. Let q = exp(iα).
1. Assume that q is not a primitive root of unity of order 2 or 3. Then every completely re-ducible representation ρ of B3 into GL(2,C) which factors through A3(q) is equivalent to somerepresentation ρq,C defined by:
ρq,C(g1) =
(q 00 −1
), ρq,C(g2) =
(− 1
q+1 −(q + 1)C
−ǫq(q + 1)Cr2 q2
q+1
)
where C ∈ C − {0}, r2 = r(q, C)2 = |C|−2|q + 1|−4|q + q + 1| and ǫq is the sign of the realnumber q + q + 1, namely:
ǫq =
{1, if α ∈ (−2π
3 ,2π3 )
−1, if α ∈ (2π3 , π) ∪ (π, 4π
3 )
2. Let q be a primitive root of unity of order 2 or 3. Then completely reducible representations ρof B3 into GL(2,C) which factor through A3(q) are abelian with finite image and equivalentto:
ρq,0(g1) = ρq,0(g2) =
(q 00 −1
)
We may extend the definition of ǫq, r(q, C) to this exceptional case by setting ǫq = 1, ifα ∈
{−2π
3 ,2π3
}, ǫq = −1, if α = π and r(q, 0)2 = 1. In this case ρq,0 is both unitarizable and
complex-unitarizable.
3. If α ∈(−2π
3 ,2π3
), then the representation ρq,C is unitarizable if r(q, C)2 = 1.
4. If α ∈(
2π3 ,
4π3
)then the representation ρq,C is complex-unitarizable if r(q, C)2 = 1.
Proof. We can choose ρ(g1) =
(q 00 −1
)since completely reducible 2-dimensional representations
are diagonalizable and the eigenvalues are prescribed. Since g2 is conjugate to g1 in B3 we haveρ(g2) = Uρ(g1)U
−1, where, without loss of generality, we can suppose that U ∈ SL(2,C). Wediscard from now on q = −1 when the representation should be abelian, as ρ(g1) is scalar.
Set U =
(a bc d
), where ad − bc = 1. Then we have ρ(g2) =
(qad+ bc −(q + 1)ab(q + 1)cd −qbc− ad
).
Therefore ρ factors through A3(q), namely the second identity of Definition 2.2 is satisfied, if andonly if:
qad+ bc = −1
q + 1
If 1 + q + q2 6= 0 we obtain the solutions: d = q(q+1)2a , and c = − q2+q+1
(q+1)2b . This implies that:
ρ(g2) =
(− 1
q+1 −(q + 1)C
− (q2+q+1)q(q+1)3C
q2
q+1
)
which coincides with the matrix ρq,C(g2) from the statement of Proposition 2.1, where C = ab and
r2 = r(q, C)2 = |q+q+1||q+1|4|C|2
.
6
If q is a primitive root of unity of order 3 then we find d = q(q+1)2a
and either b = 0 and c arbitrary
or c = 0 and b arbitrary. But the representation ρ is completely reducible only when b = c = 0 andthis gives the second claim of the Proposition 2.1.
We re-scale the representation ρq,C in order to take values in SL(2, C). This amounts to replaceρq,C(gj) by ρq,C(gj) = λρ(gj), where λ satisfies λ2q = −1. Then the condition r2 = 1 is equivalent
to ρq,C(g2) =
(u v
−ǫv u
), where |u|2 + ǫ|v|2 = 1. In this case the representation ρq,C takes values
in U(2), when ǫ = 1 and respectively in U(1, 1), when ǫ = −1.
Remark 2.1. Notice that representations associated to the same q, |C|2 are pairwise conjugate.
The representation ρq,C of B3 that arises as above and for which the parameter C satisfies r(q, C)2 =1 will be called a Jones representations of B3 at q. By the previous remark the conjugacy class ofρq,C is uniquely defined by the value of q. We omit the subscript C in the sequel when the choiceof C is not relevant.
Proposition 2.2. Let ρ : B3 → SU(2) be a unitary Jones representation at q = exp(iα), forα ∈ (−2π
3 ,2π3 ). Let Q : SU(2) → SO(3) be the usual 2-covering. Then Q ◦ ρ(g1) and Q ◦ ρ(g2) are
two rotations of angle π + α, whose axes form an angle θ which is given by the formula:
cos θ =cosα
1 + cosα
Proof. The set of anti-Hermitian 2-by-2 matrices, namely matrices A =
(w + ix y + iz−y + iz w − ix
)with
real w, x, y, z, is identified with the space H of quaternions w+ix+jy+kz. Under this identificationSU(2) corresponds to the sphere of unit quaternions. In particular any element of SU(2) acts byconjugacy on H. Let R3 ⊂ H be the vector subspace given by the equation w = 0. Then R3
is SU(2)-conjugacy invariant and the linear transformation induced by A ∈ SU(2) on R3 is theorthogonal matrix Q(A) ∈ SO(3).
A simple computation yields the following explicit formula for Q:
Q
(w + ix y + iz−y + iz w − ix
)=
1 − 2(y2 + z2) 2(xy − wz) 2(xz + wy)2(xy + wz) 1 − 2(x2 + z2) 2(yz − wx)2(xz − wy) 2(yz + wx) 1 − 2(x2 + y2)
This shows that Q(ρ(g1)) is the rotation of angle π + α around the axis i ∈ R3 in the space ofimaginary quaternions. Instead of giving the cumbersome computation of Q(ρ(g2)) observe thatQ(ρ(g2)) is also a rotation of angle π+α since it is conjugate to Q(σ1). Let N be the rotation axisof Q(ρ(g2)). If N = ui+ vj + wk then cos θ = u.
A direct computation shows that the matrix of the rotation of angle π + α around the axis N hasa matrix whose first entry on the diagonal reads u2 + (1 − u2) cos(π + α). Therefore we have theidentity:
u2 + (1 − u2) cos(π + α) = 1 − 2(y2 + z2)
where y, z are the off diagonal entries of ρ(g2), namely:
y2 + z2 = | − λ(q + 1)C|2 =|q + 1 + q|
|q + 1|2
This gives |u| =∣∣∣ cos α1+cos α
∣∣∣. Identifying one more term in the matrix of Q(σ2) yields the sign of u.
We omit the details.
7
Remark 2.2. In [42] the authors consider the structure of groups generated by two rotations offinite order for which axes form an angle which is an integral part of π. Their result is that thereare only few new relations. However the previous Proposition shows that we cannot apply theseresults to our situation. It seems quite hard just to find those α for which the axes verify thecondition from [42].
Definition 2.3. The (reduced) Burau representation β : Bn → GL(n − 1,Z[q, q−1]) is defined onthe standard generators
βq(g1) =
(−q 10 1
)⊕ 1n−3
βq(gj) = 1j−2 ⊕
1 0 0q −q 10 0 1
⊕ 1n−j−2, for 2 ≤ j ≤ n− 2
βq(gn−1) = 1n−3 ⊕
(1 0q −q
)
Jones already observed in [27] that the following holds true for the principal roots of unity, i.e. forthe roots of unity of the form exp
(2πin
), n ∈ Z:
Proposition 2.3. The Burau representation of B3 at q is conjugate to the tensor product of theparity representation and the Jones representation at q, for all q which are not primitive roots ofunity of order 2 or 3.
Proof. Recall that the parity representation σ : B3 → {−1, 1} ⊂ C∗ is given by σ(gj) = −1. TheBurau representation for n = 3 is given by
βq(g1) =
(−q 10 1
), βq(g2) =
(1 0q −q
)
Take then V =
(a 1
(q+1)a
0 1a
), for q 6= −1, where a is given by (q+ 1)3Ca2 = 1+ q+ q2 and C 6= 0
is chosen such that ρq,C is unitarizable, namely |C|2 = |q + 1|−4|1 + q + q|. One verifies easily that(σ ⊗ ρq,C)(gj) = V −1βq(gj)V .
Remark 2.3. The definition of Aτ,3 in terms of orthogonal projections has a unitary flavor andthus it works properly only when the Burau representation is unitarizable, namely only for thoseq = exp(iα), where α ∈ (−2π
3 ,2π3 ).
2.2 Two conjectures of Squier and proof of Theorem 1.1
This section is devoted to the study of the kernels of the Jones and Burau’s representations at rootsof unity. Motivation comes from the following conjectures of Squier from [43]:
Conjecture 2.1 (Squier). The kernel of the Burau representation β−q for a primitive k-th root ofunity q is the normal subgroup Bn[k] of Bn generated by gk
j , 1 ≤ j ≤ n− 1.
The second conjecture of Squier, which is related to the former one, is:
Conjecture 2.2 (Squier). The intersection of Bn[k] over all k is trivial.
8
In order to prove Theorem 1.1, which shows that a strengthening of Conjecture 2.2 holds we willneed first a number of definitions and lemmas. Let Σ0,n+1 be a disk with n holes. The (pure)
mapping class group M(Σ0,n+1) is the group of framed pure braids PBn and fits into the exactsequence:
1 → Zn → PBn → PBn → 1
where Zn is generated by the Dehn twists along the boundary curves.
The extended mapping class group M∗(Σ0,n+1) is the group of mapping classes of homeomorphismsof the disk with n holes that fix point-wise the boundary of the disk but are allowed to permutethe remaining boundary components, which are suitably parameterized. Thus M∗(Σ0,n+1) is thegroup of framed braids on n strands and we have then an exact sequence:
1 → Zn →M∗(Σ0,n+1) → Bn → 1
Since the unit tangent bundle splits the exact sequence above has a non-canonical splitting, i.e.there exists a section s : Bn → M∗(Σ0,n+1), which we fix once for all. The restriction of s to the
subgroup PBn yields a section PBn → PBn. Let g1, . . . , gn−1 denote the standard generators ofBn.
Definition 2.4. Let k be a positive integer. The subgroup Bn{k} of Bn is the normal subgroupgenerated by the following elements:
g2k1 , (g1g2g1)
3k, (g1g2g3g2g1)4k, . . . , (g1g2 · · · gn−2gn−1gn−2 · · · g2g1)
nk
Observe that Bn[2k] ⊂ Bn{k}.
Definition 2.5. For any compact orientable surface Σ (possibly with boundary) we set M(Σ)[k]for the normal subgroup of M(Σ) generated by the k-th powers of Dehn twists.
Lemma 2.1. We have s(Bn{k}) ⊂M(Σ0,n+1)[k].
Proof. Every normal generator of Bn{k} is a pure braid and hence Bn{k} ⊂ PBn. Further observethat δj = (g1g2 · · · gj−2gj−1gj−2 · · · g2g1)
j is a Dehn twist along a curve encircling the first j + 1punctures of the n-punctured disk. Let now γ be an embedded curve in the n-punctured disk whichencircles j + 1 punctures. Then the (right) Dehn twist Tγ along the curve γ is conjugate to δj bymeans of some homeomorphism of the n-punctured disk sending γ into the the curve encircling thefirst j+ 1 punctures. Thus Bn{k} is the group generated by the k-th powers of Dehn twists on then-punctured disk.
The lift of a Dehn twist Tγ ∈ PBn into the mapping class group M(Σ0,n+1) of the n-holed disk isof the form s(Tγ) = Tγ
∏i T
εici
, where εi = ±1 and Tciare Dehn twists along boundary components
of Σ0,n+1. Therefore s(T kγ ) ∈M(Σ0,n+1)[k]. This proves the claim.
Remark 2.4. In the same spirit we can identify Bn[2k] with the subgroup of Bn generated by Dehntwists along curves encircling precisely 2 punctures.
The main result of this section is the following one which implies immediately Theorem 1.1 fromthe Introduction:
Theorem 2.1. The intersection of Bn{k} over an infinite set of integers k is trivial.
Proof. When n = 2 the claim holds trivially. Assume henceforth that n ≥ 3. We embed Σ0,n+1
into the closed orientable surface Σn+1 of genus n + 1 by gluing a one-holed torus along each
9
boundary component. Let Mn+1 denote the mapping class group of Σn+1. According to [41] thehomomorphism i : M(Σ0,n+1) →Mn+1 induced by the inclusion of surfaces is injective.
We have obviously i(M(Σ0,n+1))[k] ⊂ Mn+1[k] and so Lemma 2.1 implies that i(s(Bn{k})) ⊂Mn+1[k].
Consider now the projective quantum representations ρp of Mn+1 from Definition 1.1. Accordingto [1, 16], for any infinite set of even integers A we have ∩p∈A ker ρp = 1. However the proof givenin [16] for the the SU(2)-TQFT extends without any essential modification to the SO(3)-TQFTsVp from [5]. Therefore ∩p∈A ker ρp = 1, for any infinite set of integers A. Recall that ρp was definedin Definition 1.1 only when p 6≡ 2(mod 4).
Now Remark 1.1 shows that Mn+1[p] ⊂ ker ρp, for any p. Then the previous results and theinjectivity of i ◦ s imply that ∩p∈ABn{p} = 1, for any infinite set A.
Remark 2.5. The weaker statement that ∩k∈Z−{0}Bn[k] = 1 has actually an elementary proof,which was independently observed by Ivan Marin. Consider a residually finite group G having afinite system of generators S. Let G[k] be the normal subgroup of G generated by sk, with s ∈ S.We claim that ∩k∈Z−{0}G[k] = 1. In fact, suppose that there exists 1 6= a ∈ ∩k∈Z−{0}G[k]. By theresidual finiteness of G there exists some finite group F and a morphism f : G→ F with f(a) 6= 1.Now f(s)k = 1, for every s ∈ S, where k is the order of the finite group F . This shows that ffactors through G/G[k], which implies that f(a) = 1, contradicting our assumption. This provesthe claim. In particular, this implies that
∩k∈Z−{0}Bn[k] = ∩k∈Z−{0}Bn{k} = 1, ∩k∈Z−{0}Mn[k] = 1
However it seems that the proof of the stronger claim of Theorem 2.1 uses in an essential way theasymptotic faithfulness of the quantum representations.
Proposition 2.4. Conjecture 2.1 is false for n ≥ 5, for all but finitely many q of even order.
Proof. One knows by results of Bigelow ([3]), Moody ([36]), Long and Paton ([31]) that for n ≥ 5the (generic i.e. for formal q) Burau representation β into GL(n − 1,Z[q, q−1]) is not faithful. Leta ∈ Bn be such a non-trivial element in the kernel of β.
Suppose that Conjecture 2.1 is true for infinitely many primitive roots of unity q of even order.Then a should belong to the intersection of kernels of all βq, over all roots of unity q.
By Theorem 1.1 we have ∩∞k=2Bn[2k] = 1. If ker βq = Bn[2k] for infinitely many roots of unity q of
even order 2k, it follows that a ∈ ∩∞k=2Bn[2k] = 1, which is a contradiction.
Remark 2.6. The proof of the asymptotic faithfulness in [16] is given for one primitive root of unityq of given order. However, this proof works as well for any other primitive root of unity, by usinga Galois conjugacy.
3 The image of Burau’s representation of B3 at roots of unity
3.1 Finite and exceptional quotients of B3
The aim of this section is to understand the image of the Burau representation β−q(B3) at smallroots of unity and in particular to find an explicit presentation of it. Notice that we will considerthe representation at the root −q, instead of q, for reasons that will appear later.
If one is interested to know whether β−q(B3) is discrete one should first analyze the case whenthe image can be conjugated into U(2), and after rescaling into SU(2). There the discretenessis equivalent to the finiteness of the image. The finiteness of the Jones representation of B3 was
10
completely characterized in [27]. Jones studied the case where the roots of unity −q have the form−q = exp
(2πik
), but Galois conjugation yields isomorphic groups so that the discussion in [27] is
complete. The only cases where the image of the Jones representation of B3 at −q is finite iswhen −q is a primitive root of unity of order 1, 2, 3, 4, 6 or 10. Moreover, Burau’s representationis equivalent to the Jones representation only when the root of unity is neither −1 nor a primitivethird root of unity. These excluded cases should be treated separately. For the sake of completenesswe sketch the proofs below.
Proposition 3.1. Let q be a primitive root of unity of order n ∈ {2, 3, 4, 5}. Then β−q(B3) is afinite group with the following group presentation:
〈g1, g2 | g1g2g1 = g2g1g2, gn1 = gn
2 = 1〉
Proof. Set Bk(n) = Bk/Bk[n]. The Burau representation β−q factors through B3(n) when q is aprimitive root of unity of order n.
Coxeter gave in [10] the exhaustive list of the groups Bk(n) which are finite, together with theirrespective description (see also [11, 12]). The finite ones are those for which (k − 2)(n − 2) < 4,namely when k = 3, the following list:
1. B3(2) is the symmetric group S3;
2. B3(3) is isomorphic to SL(2,Z/3Z) (or the binary tetrahedral group ∆(2, 3, 3), see section3.3 for definitions) and has order 24;
3. B3(4) is isomorphic to the triangle group ∆(2, 3, 4) and has order 96;
4. B3(5) is isomorphic to GL(2,Z/5Z) and has order 600.
Set N(n) ⊂ B3(n) for the group generated by the image of (g1g2)3, which is a generator of the
center of B3. By direct computation β−q((g1g2)3) = −q31 is a scalar matrix and thus β−q induces
a well-defined homomorphism β−q : B3(n)/N(n) → PGL(2,C). Furthermore we have the followingcommutative diagram:
1 → N(n) → B3(n) → B3(n)/N(n) → 1
↓ β−q ↓ β−q ↓1 → C∗ → GL(2,C) → PGL(2,C) → 1
Further β−q((g1g2)3) = −q31 has order o(n), where o(2) = 1, o(3) = 2, o(4) = 4, o(5) = 10. Since
the order of N(n) is also o(n) it follows that the restriction of β−q at N(n) is injective.
From the previously cited results of Coxeter we derive that:
B3(n)/N(n) =
S3, if n = 2A4, if n = 3S4, if n = 4A5, if n = 5
where Sm and Am denote the symmetric and respectively the alternating group on m elements.
A direct inspection shows that the image of β−q is neither trivial nor of order 2 and thus β−q shouldbe injective since alternating groups are simple. Alternatively, we can use directly the computationsmade by Jones in [27]. This implies that β−q is injective as well and in particular β−q(B3) has thegiven presentation, establishing the claim.
The two excluded cases which have to be treated separately are as follows:
11
Proposition 3.2. 1. If q = 1 then β−q(B3) is the subgroup SL(2,Z) of GL(2,C) with thepresentation:
〈g1, g2 | g1g2g1 = g2g1g2, (g1g2)6 = 1〉
2. If q is a primitive 6-th root of unity then the representation β−q of B3 is not completelyreducible and its image β−q(B3) has the presentation:
〈g1, g2 | g1g2g1 = g2g1g2, g61 = 1, g−2
1 g2 = g2g21〉
Proof. The group β−1(B3) is generated by the images of the generators, namely
(1 10 1
)and
(1 0−1 1
), and thus it coincides with SL(2,Z) and the presentation follows.
Let now q be a primitive 6-th root of unity, so that t = −q is a primitive third root of unity.
Let V =
(−t 01 1
). We denote by Γ the subgroup V −1ρt(B3)V of GL(2,C). Then the matrices
hi = V −1βt(gi)V are both upper triangular, namely:
h1 =
(1 −t2
0 −t
), h2 =
(1 00 −t
)
We have therefore:
h1h−12 =
(1 t0 1
), h−1
2 h1 =
(1 t+ 10 1
), h2h
−11 h−1
2 h1 =
(1 10 1
)
Since the diagonal of the generators hi is (1,−t) the group Γ is contained into the group of matrices:
Γ =
{(1 r + st0 (−t)m
), m ∈ Z/6Z, r, s ∈ Z
}⊂ GL(2,C)
Any matrix from Γ can be written as a product
hm2 (h1h
−12 )s(h2h
−11 h−1
2 h1)r
such that Γ coincides with Γ.
Observe now that the map p : Γ → Z/6Z defined by:
p
(1 r + st0 (−t)m
)= m ∈ Z/6Z
is a well-defined homomorphism. We obtain then an exact sequence:
1 → Z2 → Γ → Z/6Z
where the inclusion i : Z2 → Γ is given by i(1, 0) = h1h−12 and i(0, 1) = h2h
−11 h−1
2 h1. Thus Γ is apolycyclic group. Denote by u = h1h
−12 and v = h2h
−11 h−1
2 h1 the two generators of the kernel of p.We obtain an explicit presentation of Γ out of one of Z2 by adding the generator h2 of order 6 whoseimage generates p(Γ) and the relations which describe its action by conjugacy on Z2. Specifically,we have:
Γ = 〈u, v, h2|uv = vu, h62 = 1, h2uh
−12 = v−1, h2vh
−12 = uv〉
Now, in order to describe Γ as a quotient of B3 we add the redundant generator h1 and the braidrelation and express u, v in terms of the hi. The conjugacy relations are now consequences of thebraid relation while the commutativity relation is equivalent to h2h
21 = h−2
1 h2. This gives thedesired presentation for the image of β−q(B3).
12
3.2 Discrete subgroups of PU(1, 1)
The aim of this section is to find whether the image of the Burau representation β−q is a discretesubgroup in PU(1, 1). The main result of this section is the identification of the image of a freesubgroup of PB3 by Burau’s representation with a group generated by two rotations. Then someresults of Knapp, Mostow and Deraux ([13, 24, 37]) give necessary and sufficient conditions forsuch a subgroup being discrete.
Let us denote by A = β−q(g21) and B = β−q(g
22) and C = β−q((g1g2)
3). As it is well-known PB3 isisomorphic to the direct product F2 × Z, where F2 is freely generated by g2
1 and g22 and the factor
Z is the center of B3 generated by (g1g2)3.
It is simply to check that:
A =
(q2 1 + q0 1
), B =
(1 0
−q − q2 q2
), C =
(−q3 00 −q3
)
Recall that PSL(2,Z) is the quotient of B3 by its center. Since C is a scalar matrix the homomor-phism β−q : B3 → GL(2,C) factors to a homomorphism PSL(2,Z) → PGL(2,C).
We will be concerned below with the subgroup Γ−q of PGL(2,C) generated by the images of Aand B in PGL(2,C). When β−q is unitarizable the group Γ−q can be viewed as a subgroup ofthe complex-unitary group PU(1, 1). Specifically, consider the action of Γ−q on the projective line
CP 1. Let V be the matrix from the proof of Proposition 2.3, namely: V =
(a 1
(1−q)a
0 1a
), for
−q 6= −1, where a is given by (1 − q)3Ca2 = 1 − q + q2 and C 6= 0 is chosen such that ρ−q,C isunitarizable. Denote the conjugate V −1ZV by Z. We have then:
A =
(q2 00 1
), B =
(1+q2
1−q1+q3
(1−q)2a2
−q(1 + q)a2 − q+q3
1−q
), AB =
(q2−q4
1−qq2+q5
(1−q)2a2
−q(1 + q)a2 − q+q3
1−q
)
since AB =
(−q3 − q2 − q q2 + q3
−q − q2 q2
).
We know that V −1β−qV = σ ⊗ ρ−q,C and σ ⊗ ρ−q,C is unitarizable simply by rescaling. In factλ(σ ⊗ ρ−q,C) is complex-unitary (for those values of −q considered in Proposition 2.1) when λverifies the condition λ2q = 1. Since scalar rescaling does not affect the class of the matrix inPU(1, 1) we can work directly with the classes of the matrices A,B in PU(1, 1).
Definition 3.1. Let q = exp(iα), with α ∈(−π
3 ,π3
). The group Γ−q ⊂ PU(1, 1) is the subgroup
generated by the classes β−q(g21) and β−q(g
22), namely the classes of matrices A,B in PU(1, 1).
It appears that the search for discrete subgroups in the complex-unitary case is more interestingthan in the unitary case since we can find infinite discrete subgroups of PU(1, 1). The main resultin this section is the following:
Proposition 3.3. Let q = exp(iα), with α ∈(−π
3 ,π3
). Then the group Γ−q is a discrete subgroup
of PU(1, 1) if and only if q = exp(±2πi
n
), for n ∈ Z+ and n ≥ 7.
Proof. Recall that PU(1, 1) is a subgroup of PGL(2,C) which keeps invariant (and hence acts on)the unit disk D ⊂ CP 1. The action of PU(1, 1) on D is conjugate to the action of the isomorphicgroup PSL(2,R) on the upper half plane. The former is simply the action by isometries on thedisk model of the hyperbolic plane.
The key point of our argument is the existence of a fundamental domain for the action of Γ−q onD. We will look to the fixed points of the isometries A,B,AB on the hyperbolic disk D. We havethe following list:
13
1. A has fixed point set {0,∞} in CP 1, and thus a unique fixed point in D, namely its centerO.
2. B has fixed point set{− 1
(1−q)a2 ,−q2−q+1q(1−q)a2
}⊂ CP 1 and thus a unique fixed point in D,
namely P = − q2−q+1q(1−q)a2 . In fact, if cos(α+ π) ∈ [−1
2 ,−1], then∣∣∣∣q2 − q + 1
q(1 − q)a2
∣∣∣∣ = |1 − q||a|2 =√
|1 − q + q2| =√
1 + 2 cos(α+ π) ∈ [0, 1]
3. AB has fixed point set{− q
(1−q)a2 ,−q2−q+1(1−q)a2
}⊂ CP 1 and thus an unique fixed point in D,
namely Q = − q2−q+1(1−q)a2 .
Lemma 3.1. The elements A,B and AB of PU(1, 1) are rotations of the same angle 2α centeredat the three vertices of the equilateral geodesic triangle ∆ = OPQ in D, whose angles are equal toα.
Proof. We know from above that A,B and AB are elliptic elements of PU(1, 1). Actually all ofthem are rotations of angle ±2α:
1. A(z) = q2z and hence A is the counterclockwise rotation of angle 2α around O;
2. B is conjugate to A and thus is a rotation of angle ±2α around P .
3. AB has the eigenvalues −q3 and −q, which are distinct since q2 6= 1, and so is diagonalizable.Therefore AB is a rotation of angle ±2α around Q.
Consider now the geodesic triangle ∆ = OPQ in D. The angle POQ at O equals α since Q = qP .Since the argument of q is acute it follows that the orientation of the arc PQ is counterclockwise.Moreover, this shows that d(O,P ) = d(O,Q), where d denotes the hyperbolic distance in D and
hence the angles OPQ = OQP .
Let us introduce the element D = β−q(g2), which verifies D2 = B. Then D2
= B. We can computenext
D =
(1
1−qq2−q+1(1−q)2a2
−qa2 q2
1−q
)
We know that D is a rotation of angle ±α around P since is conjugated to β−q(g1). We cancheck that D(Q) = 0 and hence D is the counterclockwise rotation of angle α around P andd(P,Q) = d(P,O). Thus all angles of the triangle ∆ are equal to α. This also shows that B is thecounterclockwise rotation of angle 2α.
Since both A and B are counterclockwise rotations of angle 2α it follows that AB is also thecounterclockwise rotation of angle 2α.
We will use now a result of Knapp from [24], later rediscovered by Mostow (see [37]) and Deraux([13], Theorem 7.1):
Lemma 3.2. The three rotations of angle 2α in D around the vertices of an equilateral hyperbolictriangle ∆ of angles α > 0 generate a discrete subgroup of PU(1, 1) if and only if α = 2π
n , withn ∈ Z+ and n ≥ 7.
Proof. This is a particular case treated in ([13], Theorem 7.1), but also in [24]. Notice that theexistence of a hyperbolic triangle of angles equal to α requires that n ≥ 7.
This ends the proof of Proposition 3.3.
14
3.3 Triangle groups as images of a free pure braid subgroup
The aim of this section is to obtain finite presentations for the groups Γ−q. Discrete subgroupsof PU(1, 1) have explicit presentations in terms of a fundamental domain for their action on thehyperbolic disk D. This method leads us to an identification of Γ−q with a suitable triangle group.
Before we proceed we make a short digression on triangle groups. Let ∆ be a triangle in thehyperbolic plane of angles π
m ,πn ,
πp , so that 1
m + 1n + 1
p < 1. The extended triangle group ∆∗(m,n, p)is the group of isometries of the hyperbolic plane generated by the three reflections R1, R2, R3 inthe sides of ∆. It is well-known that a presentation of ∆∗(m,n, p) is given by
∆∗(m,n, p) = 〈R1, R2, R3 ; R21 = R2
2 = R23 = 1, (R1R2)
m = (R2R3)n = (R3R1)
p = 1〉
The second type of relations have a simple geometric meaning. In fact, the product of the reflectionsin two adjacent sides is a rotation by the angle which is twice the angle between those sides. Thesubgroup ∆(m,n, p) generated by the rotations a = R1R2, b = R2R3, c = R3R1 is a normalsubgroup of index 2, which coincides with the subgroup of isometries preserving the orientation.One calls ∆(m,n, p) the triangle (also called triangular, or von Dyck) group associated to ∆.Moreover, the triangle group has the following presentation:
∆(m,n, p) = 〈a, b, c ; am = bn = cp = 1, abc = 1〉
Observe that ∆(m,n, p) makes sense also 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 sides.
Proposition 3.4. Let m < k be such that gcd(m,k) = 1 where k ≥ 4. Then the group Γ− exp(±2mπi2k )
is a triangle group with the presentation:
Γ− exp(±2mπi2k ) = 〈A,B;Ak = Bk = (AB)k = 1〉
Proof. Denote by ∆(πα ,
πα ,
πα ) the group generated by the rotations of angle 2α around vertices of
the triangle ∆ of angles α, even if α is not an integer part of π. We saw above that Γ−q is isomorphicto ∆(π
α ,πα ,
πα).
When α = 2π2k the group ∆(π
α ,πα ,
πα ) is a triangle group, namely it has the rhombus ∆∗ as fundamen-
tal domain for its action on D. In particular Γ−q is the triangle group with the given presentation.
For the general case of α = 2πm2k where q is a primitive 2k-th root of unity the situation is however
quite similar. There is a Galois conjugation sending −q into − exp(±2πi2k
)and inducing an auto-
morphism of PGL(2,C). Although this automorphism does not preserve the discreteness it is anisomorphism of Γ−q onto Γ− exp(±2πi
2k ). This settles the claim.
If n is odd n = 2k+1 then the group Γ−q is a quotient of the triangle group associated to ∆ whichembeds into the group associated to some sub-triangle ∆′ of ∆.
Proposition 3.5. Let 0 < m < 2k+ 1 be such that gcd(m, 2k+ 1) = 1 and k ≥ 3. Then the groupΓ− exp(±2mπi
2k+1 ) is isomorphic to the triangle group ∆(2, 3, 2k + 1) and has the following presentation
(in terms of our generators A,B):
Γ− exp(±2mπi2k+1 ) = 〈A,B;A2k+1 = B2k+1 = (AB)2k+1 = 1, (A−1Bk)2 = 1, (BkAk−1)3 = 1〉
15
Proof. It suffices to consider m = 1, as in the previous Proposition. The discreteness proof of ([13],Theorem 7.1) shows that the group ∆(2k+1
2 , 2k+12 , 2k+1
2 ), which is generated by the rotations a, b, caround vertices of the triangle ∆ embeds into the triangle group associated to a smaller triangle∆′. One constructs ∆′ by considering all geodesics of ∆ joining a vertex and the midpoint of itsopposite side. The three median geodesics pass through the barycenter of ∆ and subdivide ∆ into6 equal triangles. We can take for ∆′ any one of the 6 triangles of the subdivision. It is immediatethat ∆′ has angles π
2k+1 ,π2 and π
3 so that the associated triangle group is ∆(2, 3, 2k+1). This grouphas the presentation:
∆(2, 3, 2k + 1) = 〈α, u, v ; α2k+1 = u3 = v2 = αuv = 1〉
where the generators are the rotations of double angle around the vertices of the triangle ∆′.
Lemma 3.3. The natural embedding of ∆(2k+12 , 2k+1
2 , 2k+12 ) into ∆(2, 3, 2k+1) is an isomorphism.
Proof. A simple geometric computation shows that:
a = α2, b = vα2v = u2α2u, c = uα2u2
Therefore α = ak+1 ∈ ∆(2k+12 , 2k+1
2 , 2k+12 ).
From the relation αuv = 1 we derive ak+1uv = 1, and thus u = akv. The relation u3 = 1 readsnow ak(vakv)akv = 1 and replacing bk by vakv we find that v = akbkak ∈ ∆(2k+1
2 , 2k+12 , 2k+1
2 ).
Further u = akv = a−1bkak ∈ ∆(2k+12 , 2k+1
2 , 2k+12 ). This means that ∆(2k+1
2 , 2k+12 , 2k+1
2 ) is actually∆(2, 3, 2k + 1), as claimed.
It suffices now to find a presentation of ∆(2, 3, 2k + 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, 3, 2k + 1), the inverse homomorphism sending α into Ak+1, u into A−1BkAk and v intoAkBkAk.
A direct consequence of Propositions 3.4 and 3.5 is the following abstract description of the imageof the Burau representation:
Corollary 3.1. If q is not a primitive root of unity of order from the set {1, 2, 3, 4, 6, 10}, then Γq
is an infinite triangle group.
Alternatively, we obtain a set of normal generators for the kernel of the Burau representation, asfollows:
Corollary 3.2. Let n 6∈ {1, 6} and q a primitive root of unity of order n. We denote by N(G) thenormal closure of the subgroup G of 〈g2
1 , g22〉. Then the kernel kerβ−q : 〈g2
1 , g22〉 → PGL(2,C) of the
restriction of Burau’s representation is given by:{N(〈g2k
1 , g2k2 , (g2
1g22)
k〉), if n = 2k
N(〈g2(2k+1)1 , g
2(2k+1)2 , (g2
1g22)
2k+1, (g−21 g2k
2 )2, (g2k2 g
2(k−1)1 )3〉), if n = 2k + 1
3.4 Proof of Theorem 1.2
In order to prove Theorem 1.2 we need some preliminary Lemmas explaining how to retrieve thekernel of Burau’s representation of B3 from known information on its restriction to the free subgroup〈g2
1 , g22〉 of PB3.
The case when q has odd order is particularly simple:
16
Lemma 3.4. If n = 2k + 1, k ≥ 3 the inclusion PB3 ⊂ B3 induces an isomorphism:
PB3
PB3 ∩ kerβ−q→
B3
ker β−q
Equivalently, we have an exact sequence:
1 → PB3 ∩ ker β−q → kerβ−q → S3 → 1
Proof. The induced map is clearly an injection. Observe next that g2k+11 , g2k+1
2 ∈ kerβ−q and thusfor every x ∈ B3 there exists some η ∈ ker β−q such that ηx ∈ PB3. Thus the image of the classηx is the class of x and this shows that the induced homomorphism is also surjective. The claimsfollow.
When q has even order we will need an additional combinatorial argument:
Lemma 3.5. If n = 2k, k ≥ 4 then ker β−q ⊂ PB3. Thus the inclusion PB3 ⊂ B3 induces anexact sequence:
1 →PB3
PB3 ∩ ker β−q→
B3
kerβ−q→ S3 → 1
Proof. It suffices to show that β−q(g) 6∈ β−q(PB3) for g ∈ {g1, g2, g1g2, g2g1, g1g2g1}. Since noneof β−q(g), for g as above is a scalar matrix, this claim is equivalent to show that β−q(g) 6∈β−q(〈g
21 , g
22〉) = 〈A,B〉. We will conjugate everything and work instead with A, B. The trian-
gle group generated by A and B has a fundamental domain consisting of the rhombus ∆∗, whichis the union of ∆ with its reflection image Rj∆. The common edge of the two triangles of therhombus will be called a diagonal.
The image of gi is the rotation of angle α around a vertex of the triangle ∆. If this rotation werean element of ∆(k, k, k) then it would act as an automorphism of the tessellation with copies of∆∗. When the vertex fixed by gi lies on the diagonal of ∆∗ then a rotation of angle α sends therhombus onto an overlapping rhombus (having one triangle in common) and thus it cannot be anautomorphism of the tessellation, contradiction.
This argument does not work when the vertex is opposite to the diagonal. However, let us colorthe triangle ∆ in white and Rj∆ in black. Continue this way by coloring all triangles in black andwhite so that adjacent triangles have different colors. It is easy to see that the rotations of angle2α (and hence all elements of the group ∆(k, k, k)) send white triangles into white triangles. Butthe rotation of angle α around a vertex opposite to the diagonal sends a white triangle into a blackone. This contradiction shows that the image of the gi does not belong to ∆(k, k, k).
The last cases are quite similar. The images of g1g2 and g2g1 send ∆∗ into an overlapping rhombushaving one triangle in common. Eventually the image of g1g2g1 does not preserve the black andwhite coloring. This proves the Lemma.
We are able now to prove Theorem 1.2, which we restate here for reader’s convenience:
Theorem 3.3. Assume that q is a primitive n-th root of unity and g1, g2 are the standard generatorsof B3. Then β−q(B3) has a presentation with generators g1, g2 and relations:
1. If n = 2k and k is odd:
Braid relation: g1g2g1 = g2g1g2,Power relations: g2k
1 = g2k2 = (g2
1g22)
k = 1,
17
2. If n = 2k and k is even:
Braid relation: g1g2g1 = g2g1g2,Power relations: g2k
1 = g2k2 = (g2
1g22)
2k = 1,
3. If n = 2k + 1:
Braid relation: g1g2g1 = g2g1g2,
Power relations: g2k+11 = g2k+1
2 = (g21g
22)
2(2k+1) = 1,
Proof. When n ∈ {2, 3, 4, 5} this is already proved in Proposition 3.1. We suppose then n ≥ 7.
The strategy of the proof is to lift the triangle group presentation of Γ−q to β−q(〈g21 , g
22〉) and then
to β−q(PB3), by adding a central generator. We add further the standard generators g1, g2 of B3
and use the previous two Lemmas in order to obtain a presentation of β−q(B3) and then get rid ofredundant generators and relations.
Lemma 3.5 shows that ker β−q has the same normal generators as ker β−q ∩ PB3, when n is even.Lemma 3.4 states that for odd n = 2k + 1 a set of normal generators of ker β−q is obtained byadding the two elements g2k+1
1 and g2k+12 to a set of normal generators of ker β−q ∩PB3. This way
one produces a presentation of β−q(B3) from a presentation of β−q(PB3).
Further PB3 is the direct product of the free group 〈g21 , g
22〉 with the center of B3, which is generated
by (g1g2)3. Now β−q(g1g2)
3 is the scalar matrix −q31. The order of −q3 is 6k/(gcd(3, k)gcd(2, k+1))if q is a primitive 2k-th root of unity and respectively r = 6(2k + 1)/gcd(3, 2k + 1) when q is aprimitive 2k + 1-th root of unity. Therefore a presentation of β−q(PB3) can be obtained from apresentation of β−q(〈g
21 , g
22〉) by adjoining a new central generator (g1g2)
3 and the following centerrelations:
(g1g2)6k/gcd(2,k+1)gcd(3,k) = 1, for even n = 2k
(g1g2)6(2k+1)/gcd(3,2k+1) = 1, for odd n = 2k + 1
This new central generator will be redundant as soon as we pass to B3 with its standard generatorsg1, g2.
The group β−q(〈g21 , g
22〉) ⊂ GL(2,C) is a central extension of its image mod scalars Γ−q ⊂ PGL(2,C).
Thus we can obtain a presentation of it by looking at the lifts of the relations holding in Γ−q.
Let n = 2k. The lifts of the relations Ak = Bk = 1 in Γ−q are the relations g2k1 = g2k
2 = 1 inβ−q(〈g
21 , g
22〉). The eigenvalues of the matrix AB are −q3 and −q so that
β−q((g21g
22)
k) =
{−1, if k ≡ 0(mod 2)1, if k ≡ 1(mod 2)
Thus for odd k it is enough to add the relation (g21g
22)k = 1.
For even k the element (g21g
22)
k is central of order 2. On the other hand, one proves by recurrenceon m that the following combined relation holds true in B3:
(g1g2)3m = g2m
1 g2(g21g
22)
mg−12
Taking m = k and recalling that (g1g2)3 is central we find that g2k
1 = 1 implies that:
(g21g
22)
k = (g1g2)3k
Thus the fact that (g21g
22)
k is central is a consequence of the braid and powers relations. Thus itsuffices to add the power relation (g2
1g22)
2k = 1, in order to get a presentation of β−q(B3).
18
For odd n = 2k+1 the lifts of the relations An = Bn = 1 are g2n1 = g2n
2 = 1, which are consequencesof the power relations gn
1 = gn2 = 1. Further we verify that:
β−q((g21g
22)
2k+1) = −1
hence (g21g
22)
2k+1 is central of order 2. The argument used above for even k shows that g2k+11 = 1
and the braid relations imply that (g21g
22)
2k+1 is central, so it suffices to add the last power relation(g2
1g22)
2(2k+1) = 1. The remaining lifts of relations in Γ−q are redundant. In fact, powers and braidrelations give us:
(g−21 g2k
2 )2 = (g−21 g−1
2 )2 = (g1g2)−3
(g2k1 g2k−2
2 )3 = (g−11 g−3
2 )3 = (g1g2)−6
Eventually, a direct inspection shows that center relations are obtained from the combined relationabove along with the powers and braid relations.
4 Johnson subgroups and proof of Theorem 1.3
4.1 The first Johnson subgroups and their quantum images
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 ≥ 1
An interesting family of subgroups of the mapping class group is the set of higher Johnson sub-groups, defined as follows.
Definition 4.1. The k-th Johnson subgroup Ig(k) is the group of mapping classes of homeomor-phisms of the closed orientable surface Σg whose action by outer automorphisms on π/π(k+1) istrivial, where π = π1(Σg).
Thus Ig(0) = Mg, Ig(1) is the Torelli group commonly denoted Tg, while Ig(2) is the group generatedby the Dehn twists along separating simple closed curves and considered by Johnson and Morita(see e.g. [26, 38]), which is often denoted Kg.
Proposition 4.1. For g ≥ 3 we have the following chain of normal groups of finite index:
ρp(Ig(3)) ⊂ ρp(Kg) ⊂ ρp(Tg) ⊂ ρp(Mg)
Proof. There is a surjective homomorphism f1 : Sp(2g,Z) →ρp(Mg)ρp(Tg) . The image of the p-th power
of a Dehn twist in ρp(Mg) is trivial. On the other hand the image of a Dehn twist in Sp(2g,Z) is atransvection and taking all Dehn twist one obtains a system of generators for Sp(2g,Z). Using thecongruence subgroup property for Sp(2g,Z) (where g ≥ 2) the image of p-th powers of Dehn twistsin Sp(2g,Z) generate the congruence subgroup Sp(2g,Z)[p] = ker(Sp(2g,Z) → Sp(2g,Z/pZ)).Since the mapping class group is generated by Dehn twists the homomorphism f1 should factorthrough Sp(2g,Z)
Sp(2g,Z)[p] = Sp(2g,Z/pZ). In particular the image is finite.
By the work of Johnson (see [26]) one knows that when g ≥ 3 the quotientTg
Kgis a finitely generated
abelian group A isomorphic to∧3H/H, where H is the homology of the surface. Thus there is
a surjective homomorphism f2 : A →ρp(Tg)ρp(Kg) . The Torelli group Tg is generated by BP-pairs,
namely elements of the form TγT−1δ , where γ and δ are non-separating disjoint simple closed curves
bounding a subsurface of genus 1 (see [25]). Now p-th powers of the BP-pairs (TγT−1δ )p = T p
γ T−pδ
19
have trivial images inρp(Tg)ρp(Kg) . But the classes TγT
−1δ generate also the quotient A and hence the
classes (TγT−1δ )p will generate the abelian subgroup pA of those elements of A which are divisible by
p. This shows that f2 factors through A/pA which is a finite group because A is finitely generated.
In particularρp(Tg)ρp(Kg) is finite.
EventuallyKg
Ig(3) is also a finitely generated abelian group A3, namely the image of the third Johnson
homomorphism. Since Kg is generated by the Dehn twists along separating simple closed curves
the previous argument shows thatρp(Kg)
ρp(Ig(3)) is the image of an onto homomorphism from A3/pA3
and hence finite.
Remark 4.1. Recently Dimca and Papadima proved in [14] that H1(Kg) is finitely generated, forg ≥ 3. The proof from above implies that ρp([Kg,Kg]) is of finite index in ρp(Kg).
Remark 4.2. A natural question is whether ρp(Ig(k + 1)) is of finite index into ρp(Ig(k)), for everyk. The arguments above break down at k = 3 since there are no products of powers of commutingDehn twists in any higher Johnson subgroups. More specifically we have to know the image of the
group Mg[p] ∩ Ig(k) intoIg(k)
Ig(k+1) by the Johnson homomorphism. Here Mg[p] denotes the normal
subgroup generated by p-th powers of Dehn twists. If the image were a lattice inIg(k)
Ig(k+1) then we
could deduce as above thatρp(Ig(k))
ρp(Ig(k+1)) is finite.
4.2 Proof of Theorem 1.3
The proof of Theorem 1.3 follows the same lines as that from [17], where one proved that the imageof the quantum representation ρp is infinite for all p in the given range. The values of p which areexcluded correspond to the TQFTs V2,V3,V4,V5,V6,V8 and V12, whose quantum representationsare known to be finite.
Before we proceed we have to make the cautionary remark that ρp is only a projective representation.Most arguments can be carried on to the lift of ρp to a linear representation of some centralextensions of Mg, but this won’t be needed in the sequel. Here and henceforth when speakingabout the Burau representation we will mean the representation βq : B3 → PGL(2,C) takingvalues 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 1.3 in most cases and will also provide a construction whichwill be useful in the proof of Theorem 1.4 in the next section. Specifically we will prove first:
Proposition 4.2. Assume that g ≥ 4. Then the image ρp((〈g2
1 , g22〉)(k)
) contains a free non-abelian
group for every k and p 6∈ {3, 4, 5, 8, 12, 16, 24, 40}.
Proof. The first step of the proof provides us with enough elements of Ig(k) having their supportcontained in a small subsurface of Σg.
Specifically we embed Σ0,4 into Σg by means of curves c1, c2, c3, c4 as in the figure below. Then thecurves a and b which are surrounding two of the holes of Σ0,4 are separating.
1c
c3
c
c2
a b 4
20
The pure braid group PB3 embeds into M0,4 using a non-canonical splitting of the surjectionM0,4 → PB3. Further M0,4 embeds into Mg when g ≥ 4, by using the homomorphism induced bythe inclusion of Σ0,4 into Σg from the figure. Then the group generated by the Dehn twists a and bis identified with the free subgroup generated by g2
1 and g22 into PB3. Moreover, PB3 has a natural
action on a sub-block of the conformal block associated to Σg as in [17], which is isomorphic to therestriction of the Burau representation at some root of unity depending on p. Notice that the twoDehn twists above are elements from Kg.
We will need the following Proposition whose proof will be given in section 4.2.1:
Proposition 4.3. The above embedding of PB3 into Mg sends (PB3)(k) into Ig(k).
Recall now that 〈g21 , g
22〉 is a normal free subgroup of PB3. The second ingredient needed in the
proof of Proposition 4.2 is the following Proposition which will be proved in 4.2.2:
Proposition 4.4. Assume that g ≥ 4. Then the image ρp((〈g2
1 , g22〉)(k)
) contains a free non-abelian
group for every k and p 6∈ {3, 4, 5, 8, 12, 16, 24, 40}.
Thus the group ρp((PB3)(k)) contains ρp(〈g21 , g
22〉(k)) and so it also contains a free non-abelian
group. Therefore, Proposition 4.3 implies that ρp(Ig(k)) contains a free non-abelian subgroup,settling Proposition 4.2.
We further consider the remaining cases and outline briefly in section 4.3 the modifications neededto make the same strategy work also for small genus surfaces and for those values of the parameterp which were excluded above, namely:
Proposition 4.5. Assume that g and p verify one of the following conditions:
1. g = 2, p is even and p 6∈ {4, 8, 12, 16, 24, 40};
2. g = 3 and p 6∈ {3, 4, 8, 12, 16, 24, 40};
3. g ≥ 4 and p ∈ {5, 40}.
Then ρp(Mg) contains a free non-abelian group.
Then Propositions 4.2 and 4.5 above will prove Theorem 1.3.
4.2.1 Proof of Proposition 4.3
Choose the base point ∗ of the fundamental group π1(Σg) on the circle c4 that separates the sub-surfaces Σ3,1 and Σg−3,1. Let ϕ be a homeomorphism of Σ0,4 that is identity on the boundary andwhose mapping class b belongs to PB3 ⊂M0,4. Consider its extension ϕ to Σg by identity outsideΣ0,4. Its mapping class B in Mg is the image of b into Mg.
In order to understand the action of B on π1(Σg) we introduce three kinds of loops based at ∗:
1. Loops of type I are those included in Σg−3,1.
2. Loops of type II are those contained in Σ0,4.
3. Begin by fixing three simple arcs λ1, λ2, λ3 embedded in Σ0,4 joining ∗ to the three otherboundary components c1, c2 and c3, respectively. Loops of type III are of the form λ−1
i xλi,where x is some loop based at the endpoint of λi and contained in the 1-holed torus boundedby ci. Thus loops of type III generate π1(Σ3,1, ∗).
21
Now, the action of B on homotopy classes of loops of type I is trivial. The action of B onhomotopy classes of loops of type II is completely described by the action of b ∈ PB3 on π1(Σ0,4, ∗).Specifically, let A : B3 → Aut(F3) be the Artin representation (see [4]). Here F3 is the free groupon three generators x1, x2, x3 which is identified with the fundamental group of the 3-holed diskΣ0,4.
Lemma 4.1. If b ∈ (PB3)(k) then A(b)(xi) = li(b)−1xili(b), where li(b) ∈ (F3)(k).
Proof. 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 PBn,k of those purebraids b for which the length m Milnor invariants of their Artin closures vanish for all m ≤ k is anormal subgroup PBn,k of Bn. Further the central series of subgroups PBn,k verifies the following(see e.g. [40]):
[PBn,k, PBn,m] ⊂ PBn,k+m, for all n, k,m,
and hence, we have (PBn)(k) ⊂ PBn,k.
Now, if b is a pure braid then A(b)(xi) = li(b)−1xili(b), where li(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 ∈ PBn,k if and only if li(b) ∈ (Fn)(k).This proves the claim.
The action of B on homotopy classes of loops of type III can be described in a similar way. Let ahomotopy class a of this kind be represented by a loop λ−1
i xλi. Then λ−1i ϕ(λi) is a loop contained
in Σ0,4, whose homotopy class ηi = ηi(b) depends only on b and λi. Then it is easy to see that
B(a) = η−1i aηi
Let now yi, zi be standard homotopy classes of loops based at a point of ci which generate thefundamental group of the holed torus bounded by ci, so that {y1, z1, y2, z2, y3, z3} is a generatorsystem for π1(Σ3,1, ∗), which is the free group F6 of rank 6.
Lemma 4.2. If b ∈ (PB3)(k) then ηi(b) ∈ (F6)(2k).
Proof. It suffices to observe that ηi(b) is actually the i-th longitude li(b) of the braid b, expressednow in the generators yi, zi instead of the generators xi. We also know that xi = [yi, zi]. Letthen η : F3 → F6 be the group homomorphism given on the generators by η(xi) = [yi, zi]. Thenηi(b) = η(li(b)). Eventually, if li(b) ∈ (F3)(k) then η(li(b)) ∈ (F6)(2k) and the claim follows.
Therefore the class B belongs to Ig(k), since its action on every generator of π1(Σg, ∗) is a conju-gation by an element of π1(Σg, ∗)(k).
4.2.2 Proof of Proposition 4.4
First we want to identify some sub-representation of the restriction of ρp to PB3 ⊂Mg. Specificallywe have:
Lemma 4.3. Let p ≥ 5. The restriction of the quantum representation ρp at PB3 ⊂ M0,4 hasan invariant 2-dimensional subspace such that the corresponding sub-representation is equivalent tothe Jones representation ρqp,C , where the root of unity qp is given by:
qp =
A−4p = exp
(−8πi
p
), if p ≡ 0(mod 4)
A−35 = exp
(−3πi
5
), if p = 5
A−8p = exp
(−8(p+1)πi
p
), if p ≡ 1(mod 2), p ≥ 7
22
Proof. For even p this is the content of ([17], Prop. 3.2). We recall that in this case the invariant2-dimensional subspace is the conformal block associated to the surface Σ0,4 with all boundarycomponents being labeled by the color 1. The odd case is similar. The invariant subspace is theconformal block associated to the surface Σ0,4 with boundary labels (2, 2, 2, 2), when p = 5 andrespectively (4, 2, 2, 2), when p ≥ 7. The half-twist eigenvalues can be computed as in [17].
Thus the image ρp(PB3) of the quantum representation projects onto the image of the Jonesrepresentation ρqp,C(PB3).
Since qp is not allowed to be a root of unity of order 1 or 3 we can replace ρqp,C by the Buraurepresentation βqp in the arguments below. Up to a Galois conjugacy we can assume that βqp isunitarizable and after rescaling it takes values in U(2). Consider the projection of βqp((PB3)(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 thegroup is dense in SO(3) then it contains a free non-abelian subgroup. Moreover solvable groupsof 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) andthe icosahedral group (the group of automorphisms of the regular icosahedron or dodecahedron).All but the last one are actually solvable groups. The icosahedral group is isomorphic to thealternating group A5 and it is well-known that it is simple (and thus non-solvable). As a sideremark this group appeared first in relation with the non-solvability of the quintic equation in FelixKlein’s monograph [23].
Lemma 4.4. If q is not a primitive root of unity of order from {1, 2, 3, 4, 6, 10}, then (Γq)(k) is
non-solvable and thus non-abelian for any k. Moreover, (Γq)(k) cannot be A5, for any k.
Proof. If (Γq)(k) were solvable then Γq would be solvable. But one knows that Γq is not solvable.In fact if q is as above then Γq is an infinite triangle group, by Corollary 3.1.
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 Γq, 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 A5 either.
Lemma 4.4 shows that whenever p is as in the statement of Proposition 4.4 the group βqp((〈g2
1 , g22〉)(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 4.4.
4.2.3 Explicit free subgroups
The main interest of the low technology proof presented above is that the free non-abelian subgroupsin the image are abundant and explicit. For instance we have:
Theorem 4.1. Assume that g ≥ 4, p 6∈ {3, 4, 5, 12, 16} and p 6≡ 8(mod 16). Set x = ρp([g21 , g
22 ])
and y = ρp([g41 , g
22 ]). Then the group generated by the iterated commutators [x, [x, [x, . . . , [x, y]] . . .]
and [y, [x, [x, . . . , [x, y]] . . .] of length k ≥ 3 is a free non-abelian subgroup of ρp(Ig(k)).
It is well-known that the order of the matrix β−q(gi), i ∈ {1, 2} in PGL(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.
23
We considered in Lemma 4.3 the root of unity qp with the property that βqp is a sub-representationof the quantum representation ρp. We derive from Lemma 4.3 that the order of the root of unity−qp is 2o(p) where
o(p) =
p4 , if p ≡ 4(mod 8);p8 , if p ≡ 0(mod 16);p16 , if p ≡ 8(mod 16);p, if p ≡ 1(mod 2), p ≥ 752 if p = 5
Therefore βqp(〈g21 , g
22〉) is isomorphic to the triangle group ∆(o(p), o(p), o(p)). Notice that in general
o(p) ∈ 12 + Z and o(p) is an integer if and only if p 6≡ 8(mod 16), p 6= 5, as we suppose from now
on. Observe also that the order of βqp(g21) is a proper divisor of the order p of a Dehn twist ρp(g
21),
when p is even.
In the proof of Theorem 4.1 we will need the following result concerning the structure of commutatorsubgroups of triangle groups:
Lemma 4.5. The commutator subgroup ∆(r, r, r)(2) of a triangle group ∆(r, r, r), r ∈ Z−{0, 1, 2},is a 1-relator group with generators cij , for 1 ≤ i, j ≤ r − 1, and the following relation:
c11 · c21−1 · c22 · c32
−1 · · · cr r−1−1
· crr = 1
Proof. The kernel K of the abelianization homomorphism Z/rZ ∗ Z/rZ → Z/rZ × Z/rZ is thefree group generated by the commutators. Denote by a and b the generators of the two copies ofthe cyclic group Z/rZ. Then K is freely generated by cij = [ai, bj ], where 1 ≤ i, j ≤ r − 1. Thegroup ∆(r, r, r) is the quotient of Z/rZ ∗ Z/rZ by the normal subgroup generated by the element(ab)r a−r b−r, which belongs to K. This shows that ∆(r, r, r)(2) is a 1-relator group, namely the
quotient of K by the normal subgroup generated by the element (ab)r a−r 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 [ai, bj ]. This can be done as follows:
(ab)r a−r b−r = [a, b][b, a2][a2, b2] · · · [ar−1, br−1][br−1, ar][ar, br]
Therefore ∆(r, r, r)(2) has a presentation with generators cij , where 1 ≤ i ≤ j ≤ r, and the relationfrom the statement of the Lemma.
Proof of Theorem 4.1. 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 βqp([g21 , g
22 ]) and βqp([g
41 , g
22 ]) are the elements c11
and respectively c21 of ∆(o(p), o(p), o(p)).
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 βqp([g
21 , g
22 ]) and βqp([g
41 , g
22 ]) 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 ρp((PB3)(k)) ⊂ ρp(Ig(k)). This proves Theorem 4.1.
When p ≡ 8(mod 16), o(p) is a half-integer and βqp(〈g21 , g
22〉) is isomorphic to the triangle group
∆(2, 3, 2o(p)). If gcd(3, 2o(p)) = 1 then H1(∆(2, 3, 2o(p))) = 0, so the central series of this trianglegroup is trivial. Nevertheless the group ∆(2, 3, 2o(p)) has many normal subgroups of finite indexwhich are surface groups and thus contain free subgroups. In particular any subgroup of infiniteindex of ∆(2, 3, 2o(p)) is free. There is then an extension of the previous result in this case, asfollows:
24
Theorem 4.2. Assume that g ≥ 4, p 6∈ {8, 24, 40} and p ≡ 8(mod 16) so that p = 8n, for oddn = 2k + 1 ≥ 7. Consider the following two elements of 〈g2
1 , g22〉:
s = g2k1 g2k
2 g2(k−k2)1 g−2
2 g2k1 g−2
2 g2(k−k2)1 g2k
2 g2k1
andt = g2k
1 g2k2 g
2(k−k2)1 g−2
2 g2(k+1)1 g2
2g2(k+k2)1 g2k
2 g10k1
Let now N(s, t) be the normal subgroup generated by s and t in 〈g21 , g
22〉. Then for any choice of
f(n) elements x1, x2, . . . , xf(n) from N(s, t) the image ρp(〈x1, x2, . . . , xf(n)〉) is a free group. Herethe function f(n) is given by:
f(n) = |PSL(2,Z/nZ)| ·n− 6
6n
in particular, when n is prime:
f(n) =(n+ 1)(n − 1)(n − 6)
12
Then the group generated by the iterated commutators of length k ≥ 3 is a free subgroup of ρp(Ig(k)).
Proof. Observe that the map PSL(2,Z) → PSL(2,Z/nZ) factors through ∆(2, 3, n), namely wehave a homomorphism ψ : ∆(2, 3, n) → PSL(2,Z/nZ) defined by
ψ(α) =
(1 −10 1
), ψ(u) =
(1 −11 0
), ψ(v) =
(0 −11 0
)
The matrices ψ(α), ψ(u), ψ(v) are obviously elements of orders n, 3 and respectively 2 in PSL(2,Z/nZ).It follows that the normal subgroup K(2, 3, n) = kerψ is torsion free, because every torsion elementin ∆(2, 3, n) is conjugate to some power of one the generators α, u, v (see [21]). Therefore K(2, 3, n)is a surface group, namely the group of a closed orientable surface which finitely covers the funda-mental domain of ∆(2, 3, n). The Euler characteristic χ(K(2, 3, n)) of this Fuchsian group can beeasily computed by means of the formula:
χ(K(2, 3, n)) = |PSL(2,Z/nZ)| · χ(∆(2, 3, n))
where the (orbifold) Euler characteristic χ(∆(2, 3, n)) has the well-known expression:
−χ(∆(2, 3, n)) = 1 −
(1
2+
1
3+
1
n
)=n− 6
6n
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 4.2 it will suffice to show that the images of theelements s, t under the Burau representation βqp are normal generators of the groupK(2, 3, n). Thisis equivalent to show that these images correspond to the relations needed to impose in ∆(2, 3, n)in order to obtain the quotient PSL(2,Z/nZ). However one knows already presentations for thisgroup (see ([9], Lemma 1) and [22]), as follows:
PSL(2,Z/nZ) = 〈α, v, u | αn = u3 = v2 = 1, gvgv = gαg−1α−4 = 1〉
for odd n, where g = vαkvα−2vαk. The first three relations above correspond to the presentationof ∆(2, 3, n) and the elements gvgv and gαgα−4 correspond to the images of s and t in ∆(2, 3, n),by using the fact that (see the proof of Lemma 3.3) α = ak+1, vα2v = b, v = akbkak.
25
4.2.4 Second proof of Proposition 4.2
We outline here an alternative proof which does not rely on the description of the image of Burau’srepresentation from Corollary 3.1. This proof is shorter but less effective since it does not pro-duce explicit free subgroups and uses the result of [29] and the Tits alternative, which need moresophisticated tools from the theory of algebraic groups.
The image ρp(Mg) into PU(N(p, g)) is dense in PSU(N(p, g)), if p ≥ 5 is prime (see [29]), whereN(p, g) denotes the dimension of the conformal blocks in genus g for the TQFT Vp. In particularthe image of ρp is Zariski dense in PU(N(p, g)). By the Tits alternative (see [44]) the image iseither solvable or else it contains a free non-abelian subgroup. However if the image were solvablethen its Zariski closure would be a solvable Lie group, contradiction. This implies that ρp(Mg)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 pdoes not satisfies this condition then we have again to use the result of Proposition 4.4 for k = 1.This result can be obtained directly from the computations of [27] proving that the image of theJones representation of B3 is neither finite nor abelian for the considered values of p. This settlesthe case k = 1 of Proposition 4.2.
Further the group ρp(Tg) is of finite index in ρp(Mg), by Proposition 4.1, and hence it also containsa free non-abelian subgroup. Results of Morita (see [39]) show that for k ≥ 2 the group Ig(k+1) isthe kernel of the k-th Johnson homomorphism Ig(k) → Ak, where Ak is a finitely generated abeliangroup. This implies that [Ig(k), Ig(k)] ⊂ Ig(k + 1), for every k ≥ 2. In particular the k-th term ofthe derived series of ρp(Tg) is contained into ρp(Ig(k + 1)). But every term of the derived series ofρp(Tg) contains the corresponding term of the derived series of a free subgroup and hence a freenon-abelian group. This proves Proposition 4.2.
Remark 4.3. Using the strong version of Tits’ theorem due to Breuillard and Gelander (see [6])there exists a free non-abelian subgroup of Mg/Mg[p] whose image in PSU(N(p, g)) is dense. HereMg[p] denotes the (normal) subgroup generated by p-th powers of Dehn twists.
4.3 Proof of Proposition 4.5
If the genus g ∈ {2, 3} then the construction used in the proof of Proposition 4.2 should be modified.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 4.4, but the embeddings Σ0,4 ⊂ Σg are nowdifferent. In all cases considered below the analogue of Proposition 4.3 will still be true, namelythe image of the subgroup 〈g2
1 , g22〉(k) by the homomorphisms M0,4 →Mg will be contained within
the Johnson subgroup Ig(k).
If g = 2 we use the following embedding Σ0,4 ⊂ Σ2:
a
b
c3 c4
c1c 2
Although the homomorphism M0,4 → M2 induced by this embedding is not anymore injective, itsends the free subgroup 〈g2
1 , g22〉 ⊂ PB3 ⊂ M0,4 isomorphically onto the subgroup of M2 generated
by the Dehn twists along the curves a and b in the figure above.
26
Consider for even p the conformal block associated to Σ0,4 with boundary labels (1, 1, 1, 1). This 2-dimensional subspace is ρp(〈g
21 , g
22〉)-invariant and the restriction of ρp to this subspace is still equiv-
alent to the Burau representation βqp (see [17]). Therefore Proposition 4.4 shows that ρp(〈g21 , g
22〉(k)),
and hence also ρp(I2(k)), contains a free non-abelian group.
If g = 3 and p ≥ 7 is odd then we consider the following embedding of Σ0,4 ⊂M3:
c1 2
c
c 4ab
c 3
The homomorphismM0,4 →M3 induced by this embedding is not injective but it also sends the freesubgroup 〈g2
1 , g22〉 ⊂ PB3 ⊂ M0,4 isomorphically onto the subgroup of M3 generated by the Dehn
twists along the curves a and b in the figure above. The conformal block associated to Σ0,4 withboundary labels (2, 2, 2, 4) is a 2-dimensional subspace invariant by ρp(〈g
21 , g
22〉) and the restriction
of ρp to this subspace is equivalent to the Burau representation βqp . Applying again Proposition4.4 we find that ρp(〈g
21 , g
22〉(k)), and hence ρp(I3(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 cases p ∈ {5, 40}, when βqp(B3) is known to have finite image (see[27]). We will consider instead the representation ρp(i(PB4)), where PB4 embeds non-canonicallyinto M0,5 and M0,5 maps into M3 by the homomorphism i : M0,5 → M3 induced by the inclusionΣ0,5 ⊂ Σg drawn below:
c1 2
c c3 4c
c5
We consider the 3-dimensional conformal blocks associated to the surface Σ0,5 with boundary labels(1, 1, 1, 1, 2), when p = 40 and respectively the labels (2, 2, 2, 2, 2), when p = 5. These conformalblocks are ρp(i(PB4))-invariant. The restriction of ρp|PB4 to this invariant subspace is known (seeagain [17]) to be equivalent to the Jones representation of B4 at the corresponding root of unity.
Now we have to use a result of Freedman, Larsen and Wang (see[15]) subsequently reproved andextended by Kuperberg in ([28], Thm.1) saying that the Jones representation of B4 at a 10-throot of unity on the two 3-dimensional conformal blocks we chose is Zariski dense in the groupSL(3,C). A particular case of the Tits alternative says that any finitely generated subgroup ofSL(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 ρp(PB4), and hencealso ρp((PB4)(k)), contains a free non-abelian group. The arguments from the proof of Proposition4.4 carry on to this setting and this proves Proposition 4.5.
Corollary 4.3. For any k the quotient group Ig(k)/Mg[p]∩ Ig(k), and in particular Kg/Kg[p], forg ≥ 3 and p 6∈ {3, 4, 8, 12, 16, 24} contains a free non-abelian subgroup. Here Kg[p] is the normalsubgroup of Kg generated by the p-th powers of Dehn twist along separating simple closed curves.
27
5 Intermediary normal subgroups of ρp(Mg)
5.1 Normal subgroups for large divisors of p
We show first that there are several families of intermediary normal subgroups, at least when p haslarge divisors.
By abuse of language we will call in this section the integer f a proper divisor of r ∈ 12 + Z if
f 6∈ {1, r} and divides r, when r is an integer and respectively f 6∈ {1, 2r} and divides 2r, when ris a half-integer.
The main result of this subsection is the following:
Proposition 5.1. Assume that g ≥ 2 and f is a proper divisor of o(p) such that f 6≡ 2(mod 4),o(f) ≥ 4 if f is even, and o(f) ≥ 7
2 , when f is odd. Then ρp(Mg[f ]) is an infinite subgroup ofinfinite index in ρp(Mg).
We will need several preliminary Lemmas in the proof of the Proposition. First, we constructinfinite subgroups in triangle groups:
Lemma 5.1. Let r ∈ 12 + Z such that the triangle group ∆(r, r, r) is infinite and f be a proper
divisor of r. Then the subgroup 〈af , bf 〉 of ∆(r, r, r) is infinite.
Proof. If the subgroup 〈af , bf 〉 were finite, then by the results of [21] it would be actually finitecyclic. Thus there exists some c ∈ ∆(r, r, r) and integers d, e such that af = ce and bf = cd.
Consider first the case when r is an integer. By results from [21] any element c of finite order isconjugate to a power of one standard generator, namely wm, where w ∈ {a, b, ab} and 1 ≤ m ≤ r−1.Thus af is conjugate to wem and bf is conjugate to wdm. Consider the natural homomorphismξ : ∆(r, r, r) → H1(∆(r, r, r)) = Z/rZ × Z/rZ. We have then (f, 0) = ξ(af ) = emξ(w) and(0, f) = ξ(bf ) = dmξ(w). This leads to a contradiction since f 6≡ 0(mod r).
Consider now the case when r is a half-integer so that we work with the group ∆(2, 3, n), for oddn = 2r. Again using the above cited result of [21] the element c should be conjugate to a power ofone of the standard generators, namely wm, where w ∈ {α, u, v} (see the proof of Lemma 3.3 forthe notations).
If w = v then m = 1 and so e = d = 1, because v is of order 2 and af , bf are non-trivial elements.This implies that af = c = bf which is a contradiction. For instance compute the (triangular)matrices af and bf using the formulas from section 3.2.
If w = u then m ∈ {1, 2} and so e, d ∈ {1, 2}. Further af is conjugate to uem which is of order 3and thus α6f = a3f = 1 = αn, which implies that n = 3f , because n is odd. The abelianizationhomomorphism ξ : ∆(2, 3, n) → H1(∆(2, 3, n)) = Z/3Z is given by ξ(α) = −ξ(u) = 1, ξ(v) = 0.Thus ξ(af ) = 2f = −em ∈ Z/3Z, ξ(bf ) = ξ(vafv) = 2f = −dm ∈ Z/3Z. Since em 6= 0 it followsthat e = d. But this implies that af = bf which is a contradiction.
If w = α then we have c = hαmh−1, for some h ∈ ∆(2, 3, n). Thus α2f = af = ce = hαemh−1 andvα2fv = bf = cd = hαdmh−1. The order of αem is the same as the order of α2f , namely n/f , sincen and f are odd. Therefore em = λf , for some integral λ, with gcd(λ, n/f) = 1. In a similar wayone obtains dm = δf , for some integral δ, with gcd(δ, n/f) = 1.
Recall now that the element α ∈ ∆(2, 3, n) acts as a rotation of order n around a vertex of thetriangle ∆′ (see the proof of Proposition 3.5) in the hyperbolic disk D. By direct computation wesee that the element α ∈ PGL(2,C) corresponds to the (projective) action of the matrix βq(g1)
28
on CP 1. Therefore α2f is conjugate within PGL(2,C) to αem only if the absolute values of theirtraces coincide. Using the explicit form of βq on the generators it follows that:
|Tr(αs)| = |Tr(βq(gs1))| = |(−q)s + 1|, s ∈ Z
and hence we have the constraint:
|(−q)2f + 1| = |(−q)λf + 1|
This implies that λf = ±2f . In the similar way the fact that α2v is conjugate to αdm impliesthat δf = ±2f . Therefore we have either af = bf or else af = b−f , each alternative leading to acontradiction. This ends the proof of the Lemma.
Remark 5.1. Alternative proofs when r is integral can be given as follows. First we claim thataf b−f is not conjugate to a power of one of the generators a, b, ab and hence it is of infinite order.Assume first that 2f 6= r. Then the claim follows by inspecting their images in the abelianizationH1(∆(r, r, r)) = Z/rZ×Z/rZ. If 2f = r there is one possibility left, namely that afbf be conjugateto (ab)f , which is of order 2. By direct computation the matrix AfBf is of order 2 if and only ifq(1 + q + q2 + · · · q2f−1) = −2, where q is a primitive root of unity of order 4f , and this cannothappen for any f , proving the claim.
Second, the element cff = [af , bf ] belongs to the commutator subgroup ∆(r, r, r)(2). We can usethe Freiheitsatz as in the proof of Theorem 4.1 in order to derive that cff is of infinite order. Infact also the image of cff in the abelianization H1(∆(r, r, r)(2)) of ∆(r, r, r)(2) has infinite order.
The second step is to give some explicit subgroups of infinite index in triangle groups. We havefirst:
Lemma 5.2. Suppose that f ≡ 0(mod 4). We have then:
Mg[f ] ∩ 〈g21 , g
22〉 ⊂ N(〈g
2o(f)1 , g
2o(f)2 , (g2
1g22)
o(f)〉)
where N(G) denotes the normal closure of the subgroup G of 〈g21 , g
22〉. Consequently, the inclusion
homomorphism i : 〈g21 , g
22〉 →Mg induces an injection:
if :〈g2
1 , g22〉
N(〈g2o(f)1 , g
2o(f)2 , (g2
1g22)
o(f)〉)→
Mg
Mg[f ]
Proof. We know that 〈g21 , g
22〉 ∩Mg[f ] ⊂ 〈g2
1 , g22〉 ∩ ker ρf = ker ρf |〈g2
1 ,g22〉
because the image of the
f -th power of a Dehn twist by ρf is trivial. Since ρf |〈g21 ,g2
2〉contains the Burau representation βqf
as a sub-representation we obtain that ker ρf |〈g21 ,g2
2〉⊂ kerβqf
|〈g21 ,g2
2〉. But we identified the kernel
of Burau representation βqp in Corollary 3.2 and the claim follows.
The case when f is odd is similar:
Lemma 5.3. Suppose that f ≡ 1(mod 2). We have then:
Mg[f ] ∩ 〈g21 , g
22〉 ⊂ N
(⟨g4o(f)1 , g
4o(f)2 , (g2
1g22)
2o(f),(g−21 g
2o(f)−12
)2,(g2o(f)−12 g
2o(f)−31
)3⟩)
where N(G) denotes the normal closure of the subgroup G of 〈g21 , g
22〉. Consequently, the inclusion
homomorphism i : 〈g21 , g
22〉 →Mg induces an injection:
if :〈g2
1 , g22〉
N
(⟨g4o(f)1 , g
4o(f)2 , (g2
1g22)
2o(f),(g−21 g
2o(f)−12
)2,(g2o(f)−12 g
2o(f)−31
)3⟩) →
Mg
Mg[f ]
29
Proof. The arguments used in the proof of Lemma 5.2 work as well, but now we need to apply theodd case of Corollary 3.2.
Proof of Proposition 5.1. Let us show first that ρp(Mg[f ]) is infinite. We consider some inclusionΣ0,4 ⊂ Σg from the previous section. Recall that the induced homomorphism embeds the subgroup〈g2
1 , g22〉 within Mg. We have then
ρp(Mg[f ]) ⊃ ρp(PB3 ∩B3[f ]) ⊃ ρp(〈g2f1 , g2f
2 〉)
and thus it suffices to show that ρp(〈g2f1 , g2f
2 〉) is infinite. Since βqp is a sub-representation of
ρp|〈g21 ,g2
2〉it suffices to show that βqp(〈g
2f1 , g2f
2 〉) is infinite. Recall from Propositions 3.4 and 3.5
that βqp(〈g21 , g
22〉) is the triangle group ∆(o(p), o(p), o(p)). Observe that p 6≡ 8(mod16) is equivalent
to o(p) ∈ Z. This triangle group is generated by the elements a and b of order o(p) which are
corresponding to βqp(g21) and βqp(g
22). Now the image of 〈g2f
1 , g2f2 〉 by the Burau representation βqp
is the subgroup If (p) of ∆(o(p), o(p), o(p)) generated by the elements af and bf .
Lemma 5.1 implies then that If (p) and hence ρp(Mg[f ]) is infinite.
Let us show now that ρp(Mg[f ]) is of infinite index in ρp(Mg).
If f is even Lemma 5.2 implies that we have an injective homomorphism induced by if :
ρp(〈g21 , g
22〉)
ρp(N(〈g2o(f)1 , g
2o(f)2 , (g2
1g22)o(f)〉))
→ρp(Mg)
ρp(Mg[f ])
Further, since βqp is a sub-representation of ρp we have also an obvious injection:
βqp(〈g21 , g
22〉)
βqp(N(〈g2o(f)1 , g
2o(f)2 , (g2
1g22)
o(f)〉))→
ρp〈g21 , g
22〉)
ρp(N(〈g2o(f)1 , g
2o(f)2 , (g2
1g22)
o(f)〉))
Now we can identify the group of the left side of the last inclusion. The group βqp(〈g21 , g
22〉) is the
triangle group ∆(o(p), o(p), o(p)), while βqp(N(〈g2o(f)1 , g
2o(f)2 , (g2
1g22)
o(f)〉)) is its normal subgroup
generated by ao(f), bo(f), (ab)o(f), where a and b are the standard generators of the triangle groupfrom section 3.3. The quotient is therefore isomorphic to the triangle group ∆(o(f), o(f), o(f)).
Eventually ∆(o(f), o(f), o(f)) is infinite as soon as o(f) ≥ 4, in which case the group on the righthand side should also be infinite. This implies that ρp(Mg[f ]) is of infinite index in ρp(Mg), settlingtherefore the case when f is even.
Assume now that f is odd. As above, we have to show that the group:
βqp〈g21 , g
22〉
βqp
(N
(⟨g4o(f)1 , g
4o(f)2 , (g2
1g22)
2o(f),(g−21 g
2o(f)−12
)2,(g2o(f)−12 g
2o(f)−31
)3⟩))
is infinite. This is the quotient of the triangle group ∆(o(p), o(p), o(p)) by the normal subgroup
generated by the elements a2o(f), b2o(f), (ab)2o(f),(a−1b
2o(f)−12
)2,(b
2o(f)−12 a
2o(f)−32
)3. By the argu-
ments used at the end of the proof of Proposition 3.5 this group coincides with ∆(o(f), o(f), o(f)).If o(f) ≥ 7
2 then this group is infinite, which implies that ρp(Mg[f ]) is of infinite index in ρp(Mg).This proves Proposition 5.1.
Remark 5.2. The result of Proposition 5.1 is still valid when we replace Mg by the Torelli groupTg or by Kg and g ≥ 4, with the same proof.
30
5.2 The second derived subgroup of Kg and proof of Theorem 1.4
For a group G we denote by G(k) the derived central series defined by:
G(1) = G, G(k+1) = [G(k), G(k)], k ≥ 1
We can prove now:
Proposition 5.2. Assume that g ≥ 4, p 6≡ 8(mod 16) and p 6∈ {3, 4, 5, 12, 16}. Then the groupρp([[Kg,Kg], [Kg,Kg]]) is infinite and has infinite index in ρp(Kg).
Proof. We consider the embedding of Σ0,4 ⊂ Σg from subsection 4.2. This inclusion induces aninjective homomorphism M0,4 → Mg which restricts to an embedding PB3 → Kg as in [30]. Thisembedding further induces a homomorphism:
τ (k) :(PB3)
(k)
(PB3)(k+1)
→(Kg)
(k)
(Kg)(k+1)
It is known from [30] that τ (1) is an injective homomorphism. We want to analyze the higher
homomorphism τ (2). Since PB3 = Z × F2 we have (PB3)(2) = [F2,F2] is the free group generated
by the commutators dij = [ai, bj ], where i, j ∈ Z \ {0}. Let then dij denote the class of dij into(PB3)
(2)
(PB3)(3) = H1((PB3)
(2)). We are not able to show that τ (2) is injective. However a weaker statement
will suffice for our purposes:
Lemma 5.4. The image τ (2)(〈d11〉) in(Kg)(2)
(Kg)(3)is non-trivial.
Proof. Observe that the image of d11 in the second lower central series quotient(PB3)(2)(PB3)(3)
is non-zero
and actually generates(PB3)(2)(PB3)(3)
. We have then a commutative diagram:
(PB3)(2)(PB3)(3)
→(Kg)(2)(Kg)(3)
↑ ↑(PB3)
(2)
(PB3)(3) →
(Kg)(2)
(Kg)(3)
The vertical arrows are surjective. A result of Oda and Levine (see [30], Theorem 7) shows thatthe embedding PBg−1 ⊂ Kg induced from some subsurface Σ0,g ⊂ Σg such that Σ0,4 → Σ0,g
induces an embedding at the level of lower central series quotients(PBg)(2)(PBg)(3)
→(Kg)(2)(Kg)(3)
. Since the
homomorphism PBn+1 → PBn obtained by deleting a strand admits a section, the groups PBn areiterated semi-direct products of free groups. This implies that the inclusion PBn ⊂ PBn+1 induces
an embedding(PBn)(2)(PBn)(3)
→(PBn+1)(2)(PBn+1)(3)
. These two facts imply that the top homomorphism between
lower central series quotients arising in the diagram above, namely:(PB3)(2)(PB3)(3)
→(Kg)(2)(Kg)(3)
is injective.
This shows that the image of τ (2)(d11) into(Kg)(2)(Kg)(3)
is non-trivial and hence the claim follows.
In order to complete the proof it is enough to show that the image subgroup 〈ρp(d11)〉 into the
quotient groupρp((PB3)
(2))
ρp((PB3)(3))
is infinite. This is a consequence of the following:
Lemma 5.5. The image of the group 〈βqp(d11)〉 into the quotient groupβqp ((PB3)
(2))
βqp ((PB3)(3))
is infinite.
31
Proof. The quotient group in question is actually H1(∆(o(p), o(p), o(p))(2)). The hypothesis p 6=5, p 6≡ 8(mod 16) implies that o(p) ∈ Z. Let c11 denote the image of βqp(d11) in the groupH1(∆(o(p), o(p), o(p))(2)).
We obtained a presentation of the group ∆(r, r, r)(2) in Lemma 4.5. This presentation impliesthat, for integral r, the group H1(∆(r, r, r)(2)) is the quotient of the free abelian group generatedby the pairwise commuting classes cij (corresponding to the generators cij of ∆(r, r, r)(2)), for1 ≤ i, j ≤ r − 1, by the following relation:
r−1∑
i=1
(cii − ci+1i) = 0
Thus when r ≥ 3 the abelianization is infinite and c11 is of infinite order.
This means that c11 has infinite order in H1(∆(o(p), o(p), o(p))(2)), if o(p) ≥ 4 and this proves theclaim.
The last two Lemmas imply that the image of the element ρp(d11) in the quotient groupρp((Kg)(2))
ρp((Kg)(3))
is of infinite order and thus ρp([[Kg,Kg], [Kg,Kg]]) is of infinite index in ρp(Mg).
Eventually ρp([[Kg,Kg], [Kg,Kg]]) is infinite if g ≥ 4, p 6≡ 8(mod 16) and p 6∈ {3, 4, 5, 12, 16},because ρp(Kg) ⊃ ρp(Ig(3)) contains a free non-abelian subgroup, by Theorem 1.3. The Propositionfollows.
Remark 5.3. The condition p 6≡ 8(mod 16) is essential in our proof. When o(p) is a half-integer theresult of Lemma 5.5 does not hold. Indeed the corresponding group would be H1(∆(2, 3, 2o(p))(2))which is trivial, when gcd(3, 2o(p)) = 1, because ∆(2, 3, 2o(p)) is perfect.
We are able now to prove Theorem 1.4, which we restate here for reader’s convenience:
Theorem 5.1. Suppose that g ≥ 4, p 6∈ {3, 4, 5, 8, 12, 16, 24, 40} and if p = 8k with odd k thenthere exists a proper divisor of k which is greater than or equal to 7. Then the group ρp(Mg) isnot an irreducible lattice in a higher rank semi-simple Lie group. In particular, if p ≥ 7 is an oddprime then ρp(Mg) is of infinite index in PU(Op).
Proof. By a theorem of Margulis any normal subgroup in an irreducible lattice of rank at least 2is either finite or else of finite index. In particular Proposition 5.2 shows that ρp(Mg) is not anirreducible lattice.
Let now U be the complex unitary group associated to the Hermitian form on the conformal blocks.The Hermitian form depends on the choice of the root of unity Ap. Set G for the product
∏σ PUσ,
over all complex valuations of Op, or equivalently, over the Galois conjugates of Ap. The group Ghas rank at least 2, when the genus g ≥ 3. In fact one observes first that there exists at least oneprimitive root of unity Ap (p ≥ 7) for which the associated Hermitian form is not positive definitefor g ≥ 3. Then the multiplicative structure of conformal blocks shows that in higher genus g ≥ 4we have large sub-blocks on which the Hermitian form is negative (respectively positive) definite.Thus the rank of G is at least 2.
In particular, ρp(Mg) is not a lattice in G. Now ρp(Mg) is contained in the group PU(Op), whichembeds as an irreducible lattice into G. Since finite index subgroups in an irreducible lattice arealso irreducible lattices it follows that ρp(Mg) is of infinite index in PU(Op).
Remark 5.4. It seems that the results of this section could be extended to cover the case when p = 5.
In order to use similar arguments we need to understand whether the abelian groupβq((PB4)(2))
βq((PB4)(3))is
infinite, when q is a primitive root of unity of order 10.
32
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