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Images of quantum representations and finite index subgroups 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]
August 20, 2011
Abstract
We prove that, except for a few explicit roots of unity, the quantum image of any Johnsonsubgroup of the mapping class group contains an explicit free non-abelian subgroup. Our secondresult is that, in general, the images of quantum representations are not isomorphic to higherrank irreducible lattices in semi-simple Lie groups. In particular, when the the parameter p is anodd prime greater than or equal to 7, the images are subgroups of infinite index within the groupof unitary matrices with cyclotomic integers entries. The proof consists in finding an infinitenormal subgroup of infinite index in the image, which is impossible for higher rank irreduciblelattices, according to Margulis. The same method also shows that the quantum images havefinite index subgroups which surject onto Z. In particular, there exist finite index subgroups ofmapping class groups with infinite abelianization.
2000 MSC Classification: 57 M 07, 20 F 36, 20 F 38, 57 N 05.Keywords: Mapping class group, Dehn twist, triangle group, braid group, Burau representation,Johnson filtration, quantum representation.
1 Introduction and statements
The aim of this paper is to study the images of the mapping class groups by quantum represen-tations. Some results in this direction are already known. We refer the reader to [36] and [23] forearlier treatments of quantum representations. In [11] we proved that the images are infinite andnon-abelian (for all but finitely many explicit cases) using earlier results of Jones who proved in [20]that the same holds true for the braid group representations factorizing through the Temperley-Lieb algebra at roots of unity. Masbaum then found in [30] explicit elements of infinite order inthe image. General arguments concerning Lie groups actually show that the image should containa free non-abelian group. Furthermore, Larsen and Wang showed (see [26]) that the image of thequantum representations of the mapping class groups at 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 [4] 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 [26] 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
1
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 of order 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 [4],4.11). The set of admissible colors is {0, 1, 2, . . . , p
2 − 2}, for even p and is {0, 2, 4, . . . , p − 3} forodd p. Therefore the order of the image of a Dehn twist by ρp is p.
We will now consider 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 is well-known the Johnson filtration shows up within the framework of finite type invariants of3-manifolds (see e.g. [14]).
Our next result shows that the image is large in the following sense (see also Propositions 3.2 and3.5):
Theorem 1.1. 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’srepresentation (see [11]) 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 thelower central series of PB3 by Burau’s representation at roots of unity in order to prove Theorem1.1.
The analysis of the contribution of mapping classes supported on small sub-surfaces of a surface,which are usually holed spheres, to various subgroups of the mapping class groups was also used inan unpublished paper by T. Oda and J. Levine (see [27]) for obtaining lower bounds for the ranksof the graded quotients of the Johnson filtration.
Our construction also provides explicit free non-abelian subgroups (see Theorems 3.1 and 3.2 forprecise statements).
On the other hand, Gilmer and Masbaum proved in [15] that the mapping class group preservesa certain free lattice within the space of conformal blocks associated to the SO(3)-TQFT. Let usintroduce the following notation. For p ≥ 5 an odd prime we denote by Op the ring of integersOp = Z[ζp], if p ≡ −1(mod 4) and Op = Z[ζ4p], if p ≡ 1(mod 4) respectively, where ζp is a primitivep-th root of unity.
The main result of [15] states that, for every odd prime p ≥ 5, there exists a free Op -lattice Sg,p
in the C-vector space of conformal blocks Vp associated by TQFT to the genus g closed orientablesurface and a non-degenerate Hermitian Op-valued form on Sg,p such that (a central extension of)the mapping class group preserves Sg,p and keeps invariant the Hermitian form. Therefore the
2
image of the mapping class group consists of unitary matrices (with respect to the Hermitian form)with entries in Op. Let PU(Op) be the group of all such matrices, up to scalar 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 1.2. Suppose that g ≥ 4 and p 6∈ {3, 4, 5, 8, 12, 16, 24, 40}. Suppose moreover that in thecase p = 8k with k odd there exists a proper divisor of k which is greater than or equal to 7. Thenthe group ρp(Mg) is not an irreducible lattice in a higher rank semi-simple Lie group. In particular,if p ≥ 7 is an odd prime, 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. Weuse the same method as in the proof of Theorem 1.1. More precisely, we consider a subgroup of themapping class group whose intersection with some pure braid subgroup PB3 has the property thatits image by Burau’s representation is infinite. Then the image of the quantum representation ofthe subgroup is infinite as well. We are able to make this strategy work for the subgroup generatedby the f -th powers of Dehn twists (for most f which divide p) and also for the second derivedsubgroup [[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 [29], chapter IV, [39], 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 1.1 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 [10, 25] and theTits alternative instead of the explicit description of the image of Burau’s representation of B3 isoutlined in subsection 3.2.4 – they seem essential for Theorem 1.2.
This method also gives the following result:
Theorem 1.3. Suppose that g ≥ 4 and p ≥ 7 is prime. Then the finite index subgroup ρp([Kg,Kg])of ρp(Mg) has infinite abelianization.
This solves Problem 2.11 from Kirby’s list ([21]), due to Ivanov.
Recall (see [28]) that the group Γ has property τ with respect to the family of unitary representationsL if there exists some ε > 0 such that for every unitary representation ρ ∈ L of Γ into some Hilbertspace H without invariant nontrivial vector and every unit vector v ∈ H we have |ρ(s)v − v| > ε,for s ∈ S. Here | ∗ | denotes the norm in H. Then Kazhdan’s property T corresponds to thecase when L is the set of all unitary (irreducible) representations, while property τ corresponds tothe family Lfin of unitary representations that factor through finite quotients of Γ. It is clear thatKazhdan’s property T implies property τ and both properties are inherited to and from finite indexsubgroups. Thus any group having some finite index subgroup with infinite abelianization has notproperty τ . In particular, we obtain the following easy consequences of the previous Theorem:
Corollary 1.4. 1. The group ρp(Mg), for g ≥ 4 and prime p ≥ 7 has not property τ and hencenot Kazhdan’s property T.
3
2. For g ≥ 4 and prime p ≥ 7 the finite index subgroups
Ug,p = ker(Mg → ρp(Mg)/ρp([Kg,Kg]))
have infinite abelianization.
3. The groups Mg, for g ≥ 4 have not property τ and hence not Kazhdan’s property T.
Remark 1.2. The restriction to g ≥ 4 is an artefact of our method. It is very likely that one canchange the subsurface embedding in order to obtain the same result for g = 3. The case g = 2 isalready known from the work of Taherkhani (see [37]) and Korkmaz (see [24, 12]).
Remark 1.3. The fact that Mg, g ≥ 2 does not have Kazhdan property T was first proved byAndersen in [2], under the assumption that the combinatorial TQFT used here is equivalent tothe geometric TQFT defined by the monodromy of the Hitchin connection. The latter result wasannounced by Andersen and Ueno.
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 Burau’s representations of B3 and triangle groups
Let Bn denote the braid group on n strands with the standard generators g1, g2, . . . , gn−1. Squierwas interested to compare the kernel of Burau’s representation βq at a k-th root of unity q withthe normal subgroup Bn[k] of Bn generated by gk
j , 1 ≤ j ≤ n− 1. Recall that:
Definition 2.1. 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
).
The paper [13] is devoted to the complete description of the image of Burau’s representation ofB3 at roots of unity. Similar results were obtained in [25, 30, 32]. For the sake of completenesswereview here the essential tools from [13] to be used later.
Let us denote by A = β−q(g21) and B = β−q(g
22) and C = β−q((g1g2)
3). As 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.
4
It is simple 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 of thecomplex-unitary group PU(1, 1).
Before we proceed we make a short digression on triangle groups. Let ∆ be a geodesic trianglein the hyperbolic plane of angles π
m, π
n, π
p, so that 1
m+ 1
n+ 1
p< 1. The extended triangle group
∆∗(m,n, p) is the group of isometries of the hyperbolic plane generated by the three reflectionsR1, R2, R3 with respect to the edges of ∆. It is well-known that a presentation of ∆∗(m,n, p) isgiven 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 reflectionswith respect to two adjacent edges is a rotation by the angle which is twice the angle between thoseedges. The subgroup ∆(m,n, p) generated by the rotations a = R1R2, b = R2R3, c = R3R1
is a normal subgroup of index 2, which coincides with the subgroup of isometries preserving theorientation. One calls ∆(m,n, p) the triangle (also called triangular, or von Dyck) group associatedto ∆. Moreover, the triangle group has the presentation:
∆(m,n, p) = 〈a, b, c ; am = bn = cp = 1, abc = 1〉.
Observe that ∆(m,n, p) also makes sense when m,n or p are negative integers, by interpretingthe associated generators as clockwise rotations. The triangle ∆ is a fundamental domain for theaction of ∆∗(m,n, p) on the hyperbolic plane. Thus a fundamental domain for ∆(m,n, p) consistsof the union ∆∗ of ∆ with the reflection of ∆ in one of its edges.
Proposition 2.1 ([13]). Let m < k be such that gcd(m,k) = 1 where k ≥ 4. Then the groupΓ− exp(±2mπi
2k ) is a triangle group with the presentation:
Γ− exp(±2mπi2k ) = 〈A,B;Ak = Bk = (AB)k = 1〉.
If n is odd n = 2k + 1, then the group Γ−q is a quotient of the triangle group associated to ∆,which embeds into the group associated to some sub-triangle ∆′ of ∆.
Proposition 2.2 ([13]). Let 0 < m < 2k + 1 be such that gcd(m, 2k + 1) = 1 and k ≥ 3. Thenthe group Γ− 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〉.
Proof. Here is a sketch of the proof. Deraux proved in ([8], Theorem 7.1) that the group ∆(2k+12 , 2k+1
2 , 2k+12 ),
which is generated by the rotations a, b, c around the vertices of the triangle ∆ embeds into thetriangle group associated to a smaller triangle ∆′. One constructs ∆′ by considering all geodesics of∆ joining a vertex and the midpoint of its opposite side. The three median geodesics pass throughthe barycenter of ∆ and subdivide ∆ into 6 equal triangles. We can take for ∆′ any one of the 6
5
triangles of the subdivision. It is immediate that ∆′ has angles π2k+1 ,
π2 and π
3 so that the associatedtriangle group is ∆(2, 3, 2k + 1). This group has 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 2.1. 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 2.1 and 2.2 is the following abstract description of the imageof Burau’s representation:
Corollary 2.1. If q is not a primitive root of unity of order in the set {1, 2, 3, 4, 6, 10}, then Γq isan infinite triangle group.
Alternatively, we obtain a set of normal generators for the kernel of Burau’s representation, asfollows:
Corollary 2.2. Let n 6∈ {1, 6} and q a primitive root of unity of order n. We denote by N(G) thenormal closure of a subgroup G of 〈g2
1 , g22〉. Then the kernel ker β−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 Johnson subgroups and proof of Theorem 1.1
3.1 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 subgroupsdefined as follows.
Definition 3.1. 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).
6
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. [19, 33]), which is often denoted by Kg.
Proposition 3.1. 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 twists 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 [19]) 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 [18]). Now p-th powers of the BP-pairs (TγT−1δ )p = T p
γ T−pδ
have trivial images inρp(Tg)ρp(Kg) . But the classes TγT
−1δ also generate 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 particular,ρp(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 a surjective homomorphism from A3/pA3
and hence is finite.
Remark 3.1. Recently Dimca and Papadima proved in [9] that H1(Kg) is finitely generated forg ≥ 3. The above proof implies that ρp([Kg,Kg]) is of finite index in ρp(Kg).
Remark 3.2. A natural question is whether ρp(Ig(k + 1)) is of finite index in 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) inIg(k)
Ig(k+1) by the Johnson homomorphism. Here Mg[p] denotes the normal
subgroup generated by the 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.
3.2 Proof of Theorem 1.1
The proof of Theorem 1.1 follows from the same argument as in [11], where we proved that theimage of the quantum representation ρp is infinite for all p in the given range. The values of pwhich are excluded correspond to the TQFTs V2,V3,V4,V5,V6,V8 and V12, and it is known thatthe images of quantum representations are finite in these cases.
Before we proceed we have to make the cautionary remark that ρp is only a projective representation.Here and henceforth when speaking about Burau’s representation we will mean the representationβq : B3 → PGL(2,C) taking values in matrices modulo scalars.
7
We will first consider the generic case where the genus is large and the 10-th roots of unity arediscarded. This will prove Theorem 1.1 in most cases and will also provide a construction whichwill be useful in the proof of Theorem 1.2 in the next section. Specifically we will prove first:
Proposition 3.2. 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
The pure braid group PB3 embeds into M0,4 using a non-canonical splitting of the surjectionM0,4 → PB3. Furthermore, M0,4 embeds into Mg when g ≥ 4, by using the homomorphisminduced by the inclusion of Σ0,4 into Σg as in the figure. Then the group generated by the Dehntwists a and b is identified with the free subgroup generated by g2
1 and g22 into PB3. Moreover, PB3
has a natural action on a a subspace of the space of conformal blocks associated to Σg as in [11],which is isomorphic to the restriction of Burau’s representation at some root of unity dependingon p. Notice that the two Dehn twists above are elements of Kg.
We will need the following Proposition whose proof will be given in section 3.2.1:
Proposition 3.3. 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 3.2 is the following Proposition which will be proved in 3.2.2:
Proposition 3.4. 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 3.3 implies that ρp(Ig(k)) contains a free non-abelian subgroup, which willcomplete the proof of Proposition 3.2.
We further consider the remaining cases and briefly outline in section 3.3 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 3.5. 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 3.2 and 3.5 above will prove Theorem 1.1.
8
3.2.1 Proof of Proposition 3.3
Choose the base point ∗ for 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 in 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, ∗).
Now, the action of B on the homotopy classes of loops of type I is trivial. The action of B on thehomotopy classes of loops of type II is completely described by the action of b ∈ PB3 on π1(Σ0,4, ∗).Specifically, let A : B3 → Aut(F3) be the Artin representation (see [3]). 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 3.1. 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 isa normal subgroup PBn,k of Bn. Furthermore, the central series of subgroups PBn,k verifies thefollowing (see e.g. [35]):
[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 the homotopy classes of loops of type III can be described in a similar way.Let a homotopy class a of this kind be represented by a loop λ−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 tosee 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 3.2. 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.
9
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).
3.2.2 Proof of Proposition 3.4
First we want to identify some sub-representation of the restriction of ρp to PB3 ⊂Mg. Specificallywe have:
Lemma 3.3. 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.
Proof. For even p this is the content of [11], Prop. 3.2. We recall that in this case the invariant2-dimensional subspace is the the space of conformal blocks associated to the surface Σ0,4 with allboundary components being labeled by the color 1. The odd case is similar. The invariant subspaceis the space of conformal blocks associated to the surface Σ0,4 with boundary labels (2, 2, 2, 2), whenp = 5 and (4, 2, 2, 2), when p ≥ 7 respectively. The eigenvalues of the half-twist can be computedas in [11].
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 Burau’srepresentation 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 the groupis dense in SO(3), then it contains a free non-abelian subgroup. Moreover, solvable subgroups ofSU(2) (and hence of SO(3)) are abelian. The finite subgroups of SO(3) are well-known. Theyare the following: cyclic groups, dihedral groups, tetrahedral group (automorphisms of the regulartetrahedron), the octahedral group (the group of automorphisms of the regular octahedron) and theicosahedral group (the group of automorphisms of the regular icosahedron or dodecahedron). Allbut the last one are actually solvable groups. The icosahedral group is isomorphic to the alternatinggroup A5 and it is well-known that it is simple (and thus non-solvable). As a side remark this groupappeared in relation with the non-solvability of the quintic equation in Felix Klein’s monograph[22].
Lemma 3.4. If q is not a primitive root of unity of order in the set {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 2.1.
Now any infinite triangle group has a finite index subgroup which is a surface group of genus atleast 2. Therefore, each term of the lower central series of that surface group embeds into thecorresponding term of the lower central series of 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.
10
Lemma 3.4 shows that whenever p is as in the statement of Proposition 3.4, 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 3.4.
3.2.3 Explicit free subgroups
The main interest of the elementary arguments in the proof presented above is that the free non-abelian subgroups in the image are abundant and explicit. For instance we have:
Theorem 3.1. Assume that g ≥ 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.
We considered in Lemma 3.3 the root of unity qp with the property that βqp is a sub-representationof the quantum representation ρp. We derive from Lemma 3.3 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 ≥ 7;52 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 3.1 we will need the following result concerning the structure of commutatorsubgroups of triangle groups:
Lemma 3.5. The commutator subgroup ∆(r, r, r)(2) of a triangle group ∆(r, r, r), r ∈ Z−{0, 1, 2},is a 1-relator group with generators cij , for 1 ≤ i, j ≤ r − 1, and the 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−rb−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 relationin the statement of the lemma.
11
Proof of Theorem 3.1. Recall now the classical Magnus Freiheitsatz, which states that any subgroupof a 1-relator group which is generated by a proper subset of the set of generators involved in thecyclically reduced word relator is free.
Assume now that o(p) ∈ Z and o(p) ≥ 4. Then βqp([g21 , g
22 ]) and βqp([g
41 , g
22 ]) are the elements c11
and c21 of ∆(o(p), o(p), o(p))(2) respectively.
An easy application of the Freiheitsatz to the commutator subgroup of the infinite triangle group∆(o(p), o(p), o(p)) gives us that the subgroup generated by 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 3.1.
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:
Theorem 3.2. 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 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. Here thefunction f(n) is given by:
f(n) = |PSL(2,Z/nZ)| ·n− 6
6n
and, in particular, when n is prime by:
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 2 in PSL(2,Z/nZ) respec-tively. It follows that the normal subgroup K(2, 3, n) = kerψ is torsion free, because every torsionelement in ∆(2, 3, n) is conjugate to some power of one the generators α, u, v (see [16]). ThereforeK(2, 3, n) is a surface group, namely the fundamental group of a closed orientable surface whichfinitely covers the fundamental domain of ∆(2, 3, n). The Euler characteristic χ(K(2, 3, n)) of thisFuchsian group can easily be computed by means of the formula:
χ(K(2, 3, n)) = |PSL(2,Z/nZ)| · χ(∆(2, 3, n)),
12
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 3.2, it will suffice to show that the images of theelements s, t under Burau’s representation βqp are normal generators of the group K(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 already knows presentations for thisgroup (see [6], Lemma 1 and [17]) 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 α = ak+1, vα2v = b, v = akbkak (see the proof of Lemma 2.1).
3.2.4 Second proof of Proposition 3.2
We outline here an alternative proof which does not rely on the description of the image of Bu-rau’s representation in Corollary 2.1. This proof is shorter but less effective since it does notproduce explicit free subgroups and uses the result of [26] and the Tits alternative, which needmore sophisticated tools from the theory of algebraic groups.
The image ρp(Mg) in PU(N(p, g)) is dense in PSU(N(p, g)), if p ≥ 5 is prime (see [26]), whereN(p, g) denotes the dimension of the space of conformal blocks in genus g for the TQFT Vp. Inparticular, the image of ρp is Zariski dense in PU(N(p, g)). By the Tits alternative (see [38]) theimage is either solvable or else it contains a free non-abelian subgroup. However, if the image weresolvable, then its Zariski closure would be a solvable Lie group, which is a contradiction. Thisimplies that 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 satisfy this condition, then we have again to use the result of Proposition 3.4 for k = 1.This result can be obtained directly from the computations in [20] 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 3.2.
Furthermore, the group ρp(Tg) is of finite index in ρp(Mg), by Proposition 3.1, and hence it alsocontains a free non-abelian subgroup. Results of Morita (see [34]) show that for k ≥ 2 the groupIg(k + 1) is the kernel of the k-th Johnson homomorphism Ig(k) → Ak, where Ak is a finitelygenerated abelian group. This implies that [Ig(k), Ig(k)] ⊂ Ig(k+1), for every k ≥ 2. In particular,the k-th term of the derived series of ρp(Tg) is contained into ρp(Ig(k + 1)). But every term of thederived series of ρp(Tg) contains the corresponding term of the derived series of a free subgroupand hence a free non-abelian group. This proves Proposition 3.2.
Remark 3.3. Using the strong version of Tits’ theorem due to Breuillard and Gelander (see [5])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 the p-th powers of Dehn twists.
3.3 Proof of Proposition 3.5
If the genus g ∈ {2, 3}, then the construction used in the proof of Proposition 3.2 should bemodified. This is equally valid when we want to get rid of the values p = 5 and p = 40.
13
The proof follows along the same lines as Proposition 3.4, but the embeddings Σ0,4 ⊂ Σg are nowdifferent. In all cases considered below the analogue of Proposition 3.3 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.
Consider for even p the space of conformal blocks 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 equivalent to Burau’s representation βqp (see [11]). Therefore Proposition 3.4 shows thatρp(〈g
21 , 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 homomorphism M0,4 → M3 induced by this embedding is not injective but it also sends thefree subgroup 〈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 space of conformal blocks associatedto Σ0,4 with boundary 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 Burau’s representation βqp . Applying againProposition 3.4, 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[20]). 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
14
We consider the 3-dimensional space of conformal blocks associated to the surface Σ0,5 with theboundary labels (1, 1, 1, 1, 2), when p = 40 and the labels (2, 2, 2, 2, 2), when p = 5 respectively.This space of conformal blocks is ρp(i(PB4))-invariant. The restriction of ρp|PB4 to this invari-ant subspace is known (see again [11]) to be equivalent to the Jones representation of B4 at thecorresponding root of unity.
Now we have to use a result of Freedman, Larsen and Wang (see[10]) subsequently reproved andextended by Kuperberg in ([25], 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 in the proof of Proposition3.4 carry on to this setting and this proves Proposition 3.5.
Corollary 3.3. 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 the Dehn twist along separating simple closed curves.
4 Intermediary normal subgroups of ρp(Mg)
4.1 Normal subgroups for large divisors of p
We first show 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 f 6∈ {1, 2r} and divides 2r, when r is ahalf-integer, respectively.
The main result of this subsection is:
Proposition 4.1. 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 construct infinitesubgroups in triangle groups:
Lemma 4.1. Let r ∈ Z∪ 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 [16] 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 in [16] 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 [16] 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 2.1 forthe notations).
15
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 in section 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 2.2) in the hyperbolic disk D. By a direct computationwe see that the element α ∈ PGL(2,C) corresponds to the (projective) action of the matrix βq(g1)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 4.1. 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 , which proves 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 3.1 in order to derive that cff is of infinite order. Infact the image of cff in the abelianization H1(∆(r, r, r)(2)) of ∆(r, r, r)(2) also has infinite order.
The second step is to give some explicit subgroups of infinite index in triangle groups. We havefirst:
Lemma 4.2. 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 ]·
16
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 Burau’s 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’s representation βqp in Corollary 2.2 and the claim follows.
The case when f is odd is similar:
Lemma 4.3. 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 ]·
Proof. The arguments used in the proof of Lemma 4.2 work as well, but now we need to apply theodd case of Corollary 2.2.
Proof of Proposition 4.1. Let us first show 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 2.1 and 2.2
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 Burau’s representation βqp is
the subgroup If (p) of ∆(o(p), o(p), o(p)) generated by the elements af and bf .
Lemma 4.1 implies then that If (p) and hence ρp(Mg[f ]) is infinite.
Now let us show that ρp(Mg[f ]) is of infinite index in ρp(Mg).
If f is even, Lemma 4.2 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 2. The quotient is therefore isomorphic to the triangle group ∆(o(f), o(f), o(f)).
17
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 2.2 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 4.1.
Remark 4.2. The result of Proposition 4.1 is still valid when we replace Mg by the Torelli groupTg or by Kg and g ≥ 4, with the same proof.
4.2 The second derived subgroup of Kg and proof of Theorem 1.2
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 4.2. 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 3.2. This inclusion induces aninjective homomorphism M0,4 → Mg which restricts to an embedding PB3 → Kg as in [27]. Thisembedding further induces a homomorphism:
τ (k) :(PB3)
(k)
(PB3)(k+1)
→(Kg)
(k)
(Kg)(k+1)
·
It is known from [27] 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 dij denote the class of dij in (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 4.4. 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)
18
The vertical arrows are surjective. A result of Oda and Levine (see [27], 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) in(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)〉 in the
quotient groupρp((PB3)
(2))
ρp((PB3)(3))
is infinite. This is a consequence of:
Lemma 4.5. The image of the group 〈βqp(d11)〉 in the quotient groupβqp((PB3)
(2))
βqp((PB3)(3))
is infinite.
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 3.5. This presentation impliesthat, for integer 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 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 1.1. The Propositionfollows.
Remark 4.3. The condition p 6≡ 8(mod 16) is essential in our proof. When o(p) is a half-integer theresult of Lemma 4.5 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 now able to prove Theorem 1.2, which we restate here for the reader’s convenience:
Theorem 4.1. 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).
19
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 4.2 shows that ρp(Mg) is not anirreducible lattice.
Now let U be the complex unitary group associated to the Hermitian form on the space of conformalblocks. 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. Thegroup G has rank at least 2, when the genus g ≥ 3. In fact one observes first that there exists atleast one primitive root of unity Ap (p ≥ 7) for which the associated Hermitian form is not positivedefinite for g ≥ 3. Then the multiplicative structure of the space of conformal blocks shows that inhigher genus g ≥ 4 we have large subspaces of the space of conformal blocks on which the Hermitianform is negative definite and also large subspaces on which the Hermitian form is negative definiterespectively. 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 4.4. 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.
4.3 Proof of Theorem 1.3
Notice first that ρp([Kg,Kg]) ⊂ ρ(Mg) is of finite index. This is the content of Remark 3.1. Wehave also the following alternative argument, which makes the proof independent of the result in[9]. The group ρp(Kg) is finitely generated since it is of finite index in the finitely generated groupρp(Mg). Thus ρp(Kg)/ρp(Kg,Kg) is abelian and finitely generated. Moreover ρp(Kg)/ρp([Kg,Kg])is generated by torsion elements of order p , since Kg is generated by Dehn twists along boundingsimple closed curves. Now, any finitely generated abelian group generated by order p elementsmust be finite.
Furthermore, we claim that the abelianization ρp([Kg,Kg])/ρp([[Kg,Kg], [Kg,Kg]]) of ρp([Kg,Kg])is infinite. In fact, from Proposition 4.2 ρp(Kg)/ρp([[Kg,Kg], [Kg,Kg]]) is infinite, but ρp([Kg,Kg])is of finite index in ρp(Kg) and the claim follows. Eventually, ρp([Kg,Kg]) is finitely generatedas being a finite index subgroup of a finitely generated group. But an infinite finitely generatedabelian group has an element of infinite order. This implies that ρp([Kg,Kg]) surjects onto Z, whichproves Theorem 1.3.
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