ARITHMETICITY OF THE BRAID GROUP AT ROOTS OF UNITYvenky/fullbraid.pdf · 2012-09-01 ·...

ARITHMETICITY OF THE BRAID GROUP AT ROOTSOF UNITY

T.N.VENKATARAMANA

Abstract. We show that the image of the pure braid group underthe monodromy action on the homology of a cyclic covering ofdegree d of the projective line is an arithmetic group provided thenumber of branch points is sufficiently large compared to the degreed.

Contents

1. Introduction 31.1. The Braid Group and the Burau Representation 51.2. The Burau representations ρn(d) at d-th roots of unity 71.3. Description of the Proof 81.4. Organisation of the Paper 92. Algebraic Groups 102.1. Examples of Algebraic Groups 102.2. Arithmetic Groups 102.3. Parabolic Subgroups 102.4. The Real Rank 112.5. A Criterion for Arithmeticity 122.6. Subgroups of Products of Higher Rank Groups 132.7. Some results on algebraic groups 143. Unitary Groups 153.1. Rank of a Unitary Group 153.2. The Heisenberg Group and the Group P 173.3. A Degenerate Hermitian Form 173.4. An Inductive Step for Integral Unitary Groups 194. Properties of the Burau Representations ρn and ρn(d) 204.1. Notation 204.2. Non-degeneracy of the representation (An, ρn(d)) 214.3. A Central element of the Braid Group 23

1991 Mathematics Subject Classification. primary: 22E40. Secondary: 20F36T.N.Venkataramana, School of Mathematics, Tata Institute of Fundamental Re-search, Homi Bhabha Road, Colaba, Mumbai 40005, INDIA.

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2 T.N.VENKATARAMANA

4.4. Constructing Unipotent Elements when n = kd− 1. 244.5. Constructing unipotents when n = kd with k ≥ 1 255. Proof of Theorem 3 and Theorem 2 275.1. Proof of Theorem 3 275.2. Proof of Theorem 2 286. Applications 286.1. Some Complex Reflection Groups 286.2. Application to monodromy of type nFn−1 297. Proof of Theorem 1 32References 33

ARITHMETICITY OF THE BRAID GROUP AT ROOTS OF UNITY 3

1. Introduction

Let n ≥ 1 and d ≥ 2 be integers. Fix integers k1, k2, · · · , kn+1 with1 ≤ ki ≤ d− 1, and such that the g.c.d. of k1, k2, · · · , kn+1 and d is 1.Given n + 1 distinct points a1, a2, · · · , an+1 in the complex plane, puta = (a1, · · · , an+1) ∈ Cn+1. Consider the curve Xa,k given by the pair(x, y) satisfying the equation

(1) yd = (x− a1)k1(x− a2)k2 · · · (x− an+1)kn+1

with y 6= 0 and x 6= a1, a2, · · · , an+1.Let C be the space of points in Cn+1 all of whose coordinates are

distinct; as the point a ∈ C varies, we get a family

F = {(y, x, a) ∈ C∗ × C× C : yd =n+1∏i=1

(x− ai)ki},

and the fibration F → C given by the projection map (y, x, a) 7→ a.The fibre over a point a ∈ C is the affine curve given in equation (1).

The curve Xa,k is a compact Riemann surface X∗a,k minus a finite setof punctures. We may consider, analogously, the family of CompactRiemann surfaces

F∗ = {(p, a) ∈ P2 × C : p ∈ X∗a,kwith a fibration ( projection map) F∗ → C; the fibre over a point a ∈ Cis the projective curve X∗a,k (X∗a,k is the normalisation of the projectivi-sation of Xa,k).

The fundamental group of the space C is well known to be the purebraid group Pn+1 on n + 1 strands); thus the fibration F∗ → C yieldsa monodromy representation

ρ∗M(k, d) : Pn+1 → GL(H1(X∗a,k,Z)),

of Pn+1 on the integral homology of the fibre X∗a,k. If N be the rank ofthe abelian group H1(X

∗a,k,Z), then the image Γ of Pn+1 is a subgroup

of GLN(Z).

Given a subgroup Γ ⊂ GLN(Z), we say that Γ is an arithmeticgroup if Γ has finite index in its integral Zariski closure G(Z) (i.e.suppose G ⊂ GLN is the Zariski closure of Γ; then Γ ⊂ G(Z), whichby definition, is G ∩ GLN(Z). We say that Γ is arithmetic, if Γ is asubgroup of finite index in G(Z)).

4 T.N.VENKATARAMANA

The following Theorem says that if the number of branch points issufficiently large compared to the degree d of the cyclic covering, thenthe monodromy is arithmetic. Precisely, we prove

Theorem 1. Suppose d ≥ 2 and n ≥ 1 are integers, k1, · · · , kn+1 areintegers with 1 ≤ ki ≤ d − 1 with g.c.d(ki, d) = 1 for each i. Supposethat

n ≥ 2d.

Then the image Γ = ρ∗M(k, d)(Pn+1) of the monodromy representationρ∗M(k, d) of Pn+1 is an arithmetic group.

Moreover, the monodromy group is a product of irreducible latticeseach of which has Q-rank at least two.

If all the integers ki are 1, then g.c.d(ki, d) = 1 for all i. Considerthe compactification X∗a of the affine curve

yd = (x− a1)(x− a2) · · · (x− an+1),

with y 6= 0 and x 6= a1, · · · , an+1. There is now the monodromy actionof the pure braid group Pn+1 (even of the full braid group Bn+1) onH1(X

∗a ,Z).

Theorem 2. If n ≥ 2d, then the image Γ of the monodromy repre-sentation ρ(d) : Bn+1 → GL(H1(X

∗a ,Z)) = GLN(Z) is an arithmetic

group.

Moreover, the monodromy is a a product of irreducible lattices, eachof which is a non-co-compact arithmetic group and has Q-rank at leasttwo.

Remark 1. If n+ 1 ≤ d then the integral points of the Zariski closureof the monodromy is a product of irreducible arithmetic lattices, someof which form co-compact lattices of their real Zariski closures.

A result of A’Campo says ([A’C]) that Theorem 2 holds even whend = 2.

Theorem 2 answers a question raised in [Mc] (see question (11.1) in[Mc]) in the affirmative when n ≥ 2d. When n ≤ d − 2, the mon-odromy is, in some cases, not arithmetic as is shown by the examplesof Deligne-Mostow (cf. the example of d = 18 and n = 3 of Corollary(11.8) of [Mc]).


In a different direction ([Al-Car-Tol]), the arithmeticity the imageof the braid group of type E6 into U(4, 1) is proved; in [1] representa-tions of the braid group on the homology of non-cyclic coverings areconsidered.

A special case of Theorem 1 is the following

Corollary 1. Suppose d is a prime, k1, · · · , kn+1 integers with 1 ≤ ki ≤d− 1 and n ≥ 2d. Then the monodromy group Γ, namely the image ofPn+1 under the representation ρ(k, d) : Pn+1 → GL(H1(X

∗a,k,Z)) is an

arithmetic group.

Remark 2. If d is not assumed to be prime, then the analogue of Corol-lary 1 is false in general, even when n is large. As an example, considerd = 2 × 18 and let n be arbitrary. Suppose a1, a2, · · · , an, b1, b2, b3, b4are distinct complex numbers. Consider the two equations

Cd(a, b) : y2×18 = (n∏i=1

(x− ai))18(x− b1) · · · (x− b4) and

C18(b) : w18 = (x− b1) · · · (x− b4).There is a map C2×18(a, b)→ C18(b) given by (x, y) 7→ (x,w) with

w =y2

(x− a1) · · · (x− an).

The monodromy of the family C2×18(a, b) (as a and b vary) on the firsthomology of the curves C2×18(a, b) maps onto the corresponding mon-odromy of the family of the curves C18(b) (as b varies).

The latter is non-arithmetic by the half integrality criterion of Mostow(in [Mos], see page 104, with N = 5 and µi = 7/18, and µ∞ = 8/18;one can easily check, from the list given there, that the monodromyis non-arithmetic i.e. not of finite index in its integral Zariski closure;however, it is a lattice in its projection to one of the real factors whichis U(2, 1), of the real points of the Zariski closure). Therefore, themonodromy of the family C2×18(a, b) is also non-arithmetic.

Theorem 2 is deduced from the arithmeticity of the images of theBurau representation of the braid group Bn+1 at d-th roots of unity(analogously Theorem 1 is deduced from the arithmeticity of the im-ages of certain representations (the Gassner Representations at rootsof unity) of the pure Braid Group Pn+1 evaluated at roots of unity).

1.1. The Braid Group and the Burau Representation.

6 T.N.VENKATARAMANA

1.1.1. Definition. The Braid Group Bn+1 on (n+ 1)-strands is the freegroup on n generators s1, s2, · · · , sn modulo the relations

sjsk = sksj (| j − k |≥ 2),

andsjsksj = sksjsk (| j − k |= 1).

1.1.2. The (Reduced) Burau Representations ρn. Let R = Z[q, q−1] bethe Laurent polynomial ring in the variable q with integral coefficients.Let M = Rn be the standard free R-module of rank n with standardgenerators e1, e2, · · · , en. For each j define the operator Tj ∈ EndR(M)by the formulae

Tj(ej) = −qej, Tj(ej−1) = ej−1 + qej, T (ej+1) = ej+1 + ej,

andTj(ek) = ek (| k − j |≥ 2).

The map sj 7→ Tj defines a representation ρn : Bn+1 → GLn(R).Denote by Γn the image of ρn.ρn is the (reduced) Burau Representation in degree n.

1.1.3. A Hermitian Form on Rn. The ring R = Z[q, q−1] has an invo-lution f 7→ f given by f(q) = f(q−1). The sub-ring S of invariants inR under this involution is clearly Z[q + q−1].

There is a unique map h : Rn ×Rn → R, which is a bilinear map ofS-modules, so that for all v, w ∈ Rn and all λ, µ ∈ R we have

h(v, w) = h(w, v), h(λv, µw) = λµh(v, w),

andh(ej, ej) = 0(| j − k |≥ 2),

h(ej, ej+1) = −(q + 1), h(ej, ej) =(q + 1)2

q.

We denote this form by h = hn ( and when n is fixed, we drop thesubscript, and write h). Then Γn preserves the Hermitian form h onRn. The unitary group U(h) of the Hermitian form h is an algebraicgroup scheme defined over S and

U(h)(S) = {g ∈ GLn(R) : h(gv, gw) = h(v, w) ∀v, w ∈ Rn}.Remark 3. This Hermitian form is essentially ( up to a scalar multipleand equivalence of Hermitian forms) the one constructed by Squier([Sq]) where he uses a form with coefficients in a quadratic extensionR′ = Z[

√q,√q−1] of R. We have written it in this form since we need

an algebraic group defined over S and not over R


1.2. The Burau representations ρn(d) at d-th roots of unity. Ifa ⊂ R is an ideal stable under the involution f 7→ f , and A = R/ais the quotient ring, then on the A-module An we get a correspond-ing Hermitian form hA, and a corresponding representation ρn(A) :Bn+1 → GLn(A), which maps Bn+1 into U(hA)(B) where B is the quo-tient ring S/a ∩ S.

If Φd(q) denotes the d-th cyclotomic polynomial in q, and a is theprincipal ideal ((Φd(q)) generated by Φd(q) in R, then ( in view ofthe equality Φd(q

−1) = ±Φd(q)q−φ(d) where φ(d) is the number of

residue classes modulo d coprime to d), it is clear that a is stableunder the involution f 7→ f . Denote a by (Φd(q)). Then the quotientA = Rd = R/(Φd(q) is also an integral domain (the ring of integersin the d-th cyclotomic extension Ed of Q). The reduction modulo theideal (Φd(q)) of the representation ρn yields a representation ρn(d) ofthe Braid Group Bn+1. This is referred to as the representation ob-tained by evaluating the representation ρn at (all the) primitive d-throots of unity.

Denote by Γn(d) the image of the representation ρn(d). The imagegoes into the group U(h)(OK) where K = Kd is the (totally real) sub-field of Ed invariant under the involution f 7→ f : K = Q(2cos(2π

d))

and OK is the ring of integers in K.

A subgroup ∆ ⊂ U(h)(OK) is said to be arithmetic, if it is of finiteindex.

1.2.1. Statement of Results. The main result of the paper is

Theorem 3. If d ≥ 3 and n ≥ 2d, then the image Γn(d) of the Braidgroup Bn+1 under the reduced Burau representation ρn evaluated atprimitive d-th roots of unity is an arithmetic group [ namely the imageof representation ρn(d) : Bn+1 → GLn(Z[q, q−]/(Φd(q))) is an arith-metic group ] .

More precisely, if h is the Hermitian form on An which Γn(d) pre-serves, then Γn(d) is a subgroup of finite index in U(h)OK), whereK = Q(cos(2π

d)) is the totally real sub-field of the d-th cyclotomic ex-

tension of Q.

These arithmetic groups are of Q-rank at least two, and in particular,are non-co-compact lattices.

8 T.N.VENKATARAMANA

Combining Theorem 3 with the results of Deligne-Mostow ([Del-Mos],[Mc]) we get:

Theorem 4. If d = 3, 4, 6 then for all n, the image of the Braid groupBn+1 under the representation ρn evaluated at a primitive d-th root ofunity is an arithmetic subgroup.

More precisely, the image of ρn(d) is an arithmetic subgroup of of theintegral unitary group U(h)(OK), where K = Kd is the field Q(2cos(2π

d)),

as in the foregoing.

Remark 4. If d = 4 and n = 1, 2 then the unitary group U(h) iscompact and Γ is discrete, therefore, Γn is finite. If n ≤ 6, then thearithmeticity is a well known result of Deligne and Mostow.

If d = 3 and n ≤ 4 then the arithmeticity is again a result of Deligne-Mostow.

Theorem 4 answers a question of [Mc] (see question (5.6) of [Mc]))in the affirmative, when d = 3, 4, 6.

A Theorem of [A’C] and Theorem 3 together imply the following:

Theorem 5. Consider the Burau representation

ρ : Bn+1 → GLn(Z[q, q−1]/(1 + q + · · ·+ qd−1)).

Then the image of ρ is an arithmetic group for all d ≥ 2 and n ≥ 2d.

1.3. Description of the Proof. The proof of Theorem 3 is by show-ing that for n ≥ 2d the image Γn(d) contains a large number of unipo-tent elements (precisely, Γn(d) contains an arithmetic subgroup of theunipotent radical of a parabolic Q- subgroup). By using results ofBass-Milnor-Serre and Tits ([Ba-Mi-Se], [Ti]), and their extensions toother groups ([Ra], [Va], [Ve]) on unipotent generators for noncocom-pact arithmetic groups of R-rank at least two, one can then show thatsuch groups are arithmetic if n ≥ 2d.

The construction of sufficiently many unipotent elements is especiallyeasy to describe when the representation is the Burau representationof Bn+1 at d-th roots of unity and n = 2d. Let s1 be as before. Denoteby ∆′ the product element

∆′ = (s2s3 · · · sn)(s2s3 · · · sn−1) · · · (s2s3)(s2).It can be shown that ∆′2 is central in B′n, the braid group gener-ated by s2, s3, · · · , sn. Form the commutator u = [s1, (∆

′)2]. Con-sider the group U ⊂ Bn+1 generated by conjugates of u of the form


{huh−1 : h ∈ B′n}. It can be shown that the image of this group Uunder the Burau representation at d-th rots of unity, is an arithmeticsubgroup of the unipotent radical of a (maximal) parabolic subgroupof the unitary group U(h).

This is enough to prove that the image of Bn+1 under ρn(d) is arith-metic, by the criteria of section 2.

1.4. Organisation of the Paper. This paper is organised as follows.In Section 2, we recall some basic notions from algebraic groups andstate a criterion due to Bass-Milnor-Serre and others on unipotent gen-erators for higher rank non-uniform lattices. In Section 3, we will applythis criterion to certain integral unitary groups.

The main section of the paper is Section 4. In Section 4, we showthat the image of the braid group Bn+1 at a primitive d-th root ofunity contains many unipotent elements (more precisely, contains anarithmetic subgroup of the unipotent radical of a parabolic subgroupdefined over the field Kd)). The criteria of section 3 then imply thearithmeticity when n ≥ 2d for all d.

In Section 6, we first consider (in subsection 6.1), complex reflectiongroups corresponding to root systems of type A and show the arith-meticity of the images of the corresponding Artin groups An(q) whereq is a primitive d-th root of unity and n ≥ 2d; this is an immediateconsequence of Theorem 3. This answers a question (question (5.6) in[Mc]) raised in [Mc], in many cases.

In the next subsection (6.2), we show that Theorem 3 implies thearithmeticity of the monodromy of certain hypergeometric series oftype nFn−1, in some special cases.

Acknowledgements: I am very grateful to Madhav Nori for mention-ing to me this problem of arithmeticity of the image of the Braid Group( Theorem 3); his formulation of the problem in purely algebraic termswas most helpful. I also thank him for generously sharing his insightson many of the questions addressed here and for very helpful remarkson many of the proofs and intermediate results (especially the remarkthat an earlier proof for the arithmeticity of the image of the Buraurepresentation should also go through for the Gassner representation).

10 T.N.VENKATARAMANA

I thank Peter Sarnak for mentioning the monodromy problem ofsubsection 6.2 and for very interesting discusions on the material ofsubsection 6.2.

The support of the JC Bose fellowship for the period 2008-2013 isgratefully acknowledged.

2. Algebraic Groups

For the facts on algebraic groups stated in this section, we refer to[Bor-Ti].

2.1. Examples of Algebraic Groups. If K is a number field andG ⊂ GLn is a subgroup which is the set of zeroes of a collection ofpolynomials in the matrix entries of GLn, such that the coefficients ofthese polynomials lies in K, then G is said to be an algebraic groupdefined over K.

For example, the group SLn and Sp2g are algebraic groups definedover K. If E/K is a quadratic extension and H : En × En → E isa K bilinear form which is hermitian with respect to the non-trivialautomorphism of E/K, then the unitary group of the Hermitian formh is an algebraic group defined over K.

2.2. Arithmetic Groups. Let OK denote the ring of integers in K;The group G(OK) is by definition, the intersection GLn(OK) ∩ G. Asubgroup Γ ⊂ G(K) is said to be arithmetic if the intersection Γ ∩G(OK) is of finite index in Γ and in G(OK).

Definition 1. An algebraic group G is said to be K-isotropic if thereexists an injective morphism Gr

m → G of algebraic groups defined overK for some r ≥ 1. Gr

m is called the K-split torus of dimension r and theembedding is called a K-embedding. The K-rank of G is by definitionthe maximum of the dimensions of K-split tori which are K-embeddedin G.

For example, K-rank SLn is n− 1; the K-rank of Sp2g is g.

2.3. Parabolic Subgroups. Suppose now that G ⊂ GLn is an alge-braic group defined over a number field K; embed K in C and supposethat the Lie algebra of G(C) is a simple Lie algebra over C. Then G issaid to be absolutely almost simple.


An algebraic subgroup P ⊂ G is said to be a parabolic subgroupdefined over K, if P is defined over K and the quotient G/P is aprojective variety. For example a subgroup P ⊂ GLn = GL(V ) is aparabolic subgroup if and only if it is the subgroup preserving a partialflag

{0} ⊂ W1,⊂ W2 ⊂ · · · ⊂ Wr−1 ⊂ V,

where Wi form a sequence of subspaces of V with Wi ⊂ Wi+1.

An absolutely almost simple K-algebraic group G has positive K-rank if and only if G contains a proper parabolic subgroup P definedover K.

Let G be an absolutely almost simple algebraic group defined over K;suppose that r = K − rank(G) ≥ 1 and T a K-split torus defined overK of dimension r and K-embedded in G (T is then called a maximalK-split torus). Then there exists a parabolic K-subgroup P containingT . Furthermore, there exists a non-trivial maximal unipotent normalsubgroup U of P , called the unipotent radical of P . The Lie algebraof U is stable under the action of the group T and splits into characterspaces for the adjoint action of T .

The structure theory of parabolic subgroups says that there exists aparabolic subgroup P− defined over K containing T , with a unipotentradical U− such that the characters of T on Lie(U−) are the inversesof the characters of T on Lie(U). The group P− is said to be oppositeto P and U− is said to be opposed to U . The intersection M = P ∩P−is called a Levi subgroup of P and we have the Levi decompositionP = MU . The group M normalises both U and U−.

A K-parabolic subgroup P0 containing T is minimal if it is of thesmallest dimension among the parabolic subgroups containing T . IfP ⊃ P0 then we have the (reverse) inclusion of the unipotent radicalsU ⊂ U0. There exists a minimal parabolic subgroup P−0 opposed toP0, with unipotent radical U−0 .

2.4. The Real Rank. Suppose that G is an absolutely almost simplealgebraic group defined over a number field K. Write G∞ for the prod-uct group G∞ =

∏v archG(Kv) (which is a real semi-simple group),

where v runs over all the archimedean completions of K; we write

∞− rank(G) = R− rank(G∞) =∑v arch

Kv − rank(G),


and call this the real rank of G∞.

2.5. A Criterion for Arithmeticity. Bass, Milnor and Serre provedthat for any N ≥ 1, the N -th powers of the upper and lower triangularunipotent matrices in SLn(Z) generate a subgroup of finite index inSLn(Z) for n ≥ 3. A similar result holds for Sp2g(Z). In this subsec-tion, we describe a result which is an analogue of this result, for allhigher R-rank groups.

The following result is due to many people: for SLn(n ≥ 3 andSp2g (g ≥ 2), this is due to Bass, Milnor and Serre ([Ba-Mi-Se]). forsplit G (i.e. a maximal K -split torus is a maximal C-split torus), toTits ([Ti]); for classical groups to Vaserstein ([Va]) and to Raghunathan([Ra]) and [Ve] in general.

Theorem 6. Let G be an absolutely almost simple algebraic group de-fined over a number field K. Assume that K − rank(G) ≥ 1, R −rank(G∞) ≥ 2, and that P0, P

−0 are minimal parabolic K-subgroups of

G with unipotent radicals U0 and U−0 . Then the following hold.

[1] For every integer N ≥ 1, let ∆N be the subgroup of G(OK) gen-erated by N-th powers of the elements in U0(OK) and U−0 (OK). Then∆N is of finite index in G(OK).

[2] If ∆′N ⊂ ∆N is an infinite normal subgroup, then ∆′N is also offinite index in G(OK).

The above Theorem says that two maximal opposing unipotent sub-groups (i.e.the unipotent radicals of two minimal parabolicK-subgroups)generate a finite index subgroup if the∞-rank is at least two. We needa strengthening of this, where we assume only that the unipotent radi-cals of two opposing parabolic subgroups, not necessarily maximal, areinvolved. This is the following:

Corollary 2. Suppose G is absolutely almost simple and K−rank(G) ≥2. Let P and P− be two opposite parabolic subgroups containing a max-imal K-split torus, and U,U− their unipotent radicals.

For any integer N ≥ 1, the group ∆N(P±) generated by N-th powersof U(OK) and U−(OK) is of finite index in G(OK).

Proof. In the notation above, letM = P∩P−. ThenM(OK) normalisesU(OK) and U−(OK) and hence normalises the group ∆N(P ). Now thegroup generated by M(OK) and ∆N(P±) contains U0∩M(OK) and U0∩


U(OK)N = U(OK)N ; the decomposition P = MU shows that P (OK) =M(OK)U(OK) and hence U0(OK) = (U0(OK) ∩ M(OK))U(OK) (allequalities are up to finite index).

Therefore the group generated by M(OK) and ∆N(P±) contains N -th powers of elements of U0(OK) and U−0 (OK). Consequently, the group∆N(P±) is normalised by ∆N . By the second part of the above theo-rem, ∆N(P ) is of finite index in G(OK). �

Remark 5. The second part of Theorem 6 is true for any irreduciblelattice in a real semi-simple group of real rank at least two (this is theMargulis Normal Subgroup Theorem).

In the next section, we will state a special case of this Corollary forcertain unitary groups.

2.6. Subgroups of Products of Higher Rank Groups. In a latersection (section 5), we will prove the arithmeticity in a product ofthe form U(h)(Z[q]/(qd − 1)). The latter is a product of the groupsU(h)(Oe) for certain rings of integers Oe. To deal with this case, wenow prove a Lemma which is a simple consequence of the super-rigiditytheorem of Margulis.

Suppose thatKe are number fields for each element e in a finite index-ing set X. Suppose that Ge is an absolutely almost simple semi-simplealgebraic group defined over Ke, and suppose that ∞− rank(Ge) ≥ 2for all e ∈ X. Suppose that Γ ⊂

∏Ge(Oe) is a subgroup such that

the image of its projection to each Ge(Oe) has finite index in Ge(Oe).Suppose also that Γ ∩Ge(Oe) contains non-trivial unipotent elements,for every e ∈ X. Assume in addition, that either Ke 6= Kf or else, thegroups Ge and Gf are not Ke = Kf -isomorphic.

Lemma 7. Under these assumptions, the group Γ has finite index inthe product

∏e∈X Ge(Oe).

Proof. We prove this by induction on the number of factors. Fix apoint p ∈ X. Replacing the arithmetic groups Ge(Oe) by subgroupsof finite index, we may assume that these are torsion free and hencethat Γ is torsion free. By induction, the projection of Γ in the prod-uct

∏e∈X,e 6=pGe(Oe) under the projection map pr :

∏e∈X Ge(Oe) →∏

e∈X,e 6=pGe(Oe) has finite index. Suppose Kp is kernel of restrictionof this projection map to Γ.

If the kernel is non-trivial, since Γ is assumed to be torsion free, Kp

is infinite; the conjugation action of Γ on the p-th factor Gp(Op) factors


through its projection to the p-th factor; but the p-th projection maphas image of finite index, and hence Kp is normalised by a subgroup offinite index in Gp(Op); by the Margulis normal subgroup Theorem, Kp

has finite index in Gp(Op); therefore, Γ maps onto a subgroup of finiteindex in the product of the “away from p” factors, and intersects thepth factor in a subgroup of finite index. Therefore, Γ is of finite index.

If the kernel is trivial, then Γ maps injectively into the product∏e∈X,e 6=pGe(Oe) (and has image of finite index; therefore, Γ is an arith-

metic subgroup of the higher rank lattice∏

e6=pGe(Oe) and has a non-

trivial representation (projection) into the arithmetic group Gp(Op) ;this contradicts the Margulis super-rigidity and therefore, the kernelcannot be trivial. �

2.7. Some results on algebraic groups. Let U ⊂ SLn(C) be theunipotent algebraic group consisting of the set of matrices u of the form

u =

1 x2 · · · xn0 1 · · · 0· · · · · · · · ·0 0 · · · 1

.

This is the subgroup which preserves the partial flag

0 ⊂ Ce1 ⊂ Cn,

and acts trivially on successive quotients.

Proposition 8. Let H ⊂ SLn(C) be a reductive algebraic subgroupwhich contains the unipotent algebraic group U . Then H = SLn(C).

Proof. Denote by T the group of diagonals in SLn. The Lie algebraof SLn splits into eigenspaces for the action of T , and the eigenvectorsare Eij and Eii − Ejj where Eij is, in the usual notation, the n × nmatrix whose ij-th entry s 1 and all other entries are zero. Then theLie algebra u of U is spanned by E1i with 1 < i.

Let h be the Lie algebra of H. Write the direct sum decompositionsln(C) = h⊕ h′ as modules under the adjoint action of H (we use theassumption that H is reductive). Since U is unipotent, if h′ is non-zero,there exists an X ∈ h′, X 6= 0, which is fixed by U . This means thatthe linear transformation X commutes with U .

The centraliser z of U in sln(C) is stable under the action of thediagonals T , since U is T -stable. Hence z splits into eigenspaces for T .Since E1i ∈ Lie(U), the equation [E1i, Eij] = E1j shows that z cannot


contain Eij with i ≥ 2. It is also clear that the lie algebra of T actsfaithfully on Lie(U) under the adjoint action; therefore,

z = ⊕Cj≥2E1j = Lie(U) ⊂ h.

Hence X must lie in h; this is impossible and therefore, h′ = 0. �

Let V be an n-dimensional vector space over C and W,W ′ be twodistinct codimension one subspaces and suppose we are given a decom-position V = W ⊕Cv1 and V = W ′ ⊕Cv′. We may then view SL(W )and SL(W ′) as subgroups of SL(V ).

Lemma 9. The group SL(V ) is generated by SL(W ) and SL(W ′).

Proof. Put X = W ∩W ′ and write W = X ⊕Cw and W ′ = X ⊕Cw′.Write E = Cw⊕Cw′. Then sl(V ) = sl(X)⊕X⊗E∗⊕X∗⊗E⊕sl(E).If h is the sub-algebra generated by sl(W ) and sl(W ′), then h containsE ⊗X and X∗ ⊗ E as subspaces; the sub-algebra generated by thesesubspaces contains sl(E). Therefore h = sl(V ). �

3. Unitary Groups

Notation. Suppose that E/K is a quadratic extension, Y a finite di-mensional vector space over E, and h a non-degenerate hermitian formon Y with respect to this extension. Write E = K ⊕ K

√α for some

non-square element α in K and given z ∈ E, write z = x + y√α

accordingly. Then x, y are called the “real” and “imaginary” parts,respectively. Denote by z the element x−

√αy.

Given n ≥ 1, the vector space En may be viewed as a 2n dimensionalvector space over K. If h : En × En → E is a Hermitian form withrespect to E/K, then elements of the unitary group U(h) satisfy theproperty that they commute with scalar multiplication by elements ofE, viewed as K- linear endomorphisms of En; further, they preservethe real and imaginary parts of h. These properties characterise theelements of U(h) and in this manner, the group U(h) may be viewedas a K-algebraic subgroup of GLK(K2n) = GL2n(K).

3.1. Rank of a Unitary Group. Suppose E/K is a quadratic ex-tension whose non-trivial Galois automorphism is denoted x 7→ x. Leth2 : E2 × E2 → E be a form such that it is bilinear as a K-module.Suppose also that h2(v, w) = h2(w, v) ∀v, w ∈ E2. Then h is called aHermitian form with respect to E/K.


Suppose that the Hermitian form h2 can be written as

h2 =

(0 11 0

).

That is, if v, w are column vectors in E2 with entries v = (z1, z2) andw = (w1, w2), then h2(v, w) = z1w2 + z2w1. Then h is called a hyper-bolic form. The standard basis of E2 may be written v, v∗, and theform h2 is such that h(v, v) = h(v∗, v∗) = 0 and h(v, v∗) = 1.

Suppose h is a non-degenerate Hermitian form on En. Then (En, h)can be written as a direct sum of r copies of the hyperbolic form (E2, h2)and a form (En−2r, h0) which does not represent a zero in En−2r. Thenh0 is said to be anisotropic. Let U(h) be the unitary group of the Her-mitian form h. It can be shown that K − rank(U(h)) = r. We writeK − rank(En, h) = r.

With respect to this decomposition h = h2 ⊕ · · ·h2 ⊕ h0, write thestandard basis of the j-th copy of (E2, h2) as vj, v

∗j . we may rearrange

the basis of En in the form

v1, v2, · · · , vr, w1, · · ·ws, v∗r , · · · v∗2, v∗1,

where w1, · · · , ws is a basis of (En−2r, h0). Let Av1 ⊂ W ⊂ V = En,where W is the E-sub-module spanned by

v1, · · · , vr, w1, · · ·ws, v∗r , · · · v∗2.

In the terminology of the preceding section, if n ≥ 2, then SU(h)is an absolutely simple algebraic group defined over K; a subgroup ofSU(h) of the form P which preserves the flag Ev1 ⊂ W ⊂ V is a par-abolic subgroup defined over K.

Suppose (EN , h) is a Hermitian space which is a sum of at least twocopies of the two dimensional hyperbolic form (E2, h2) and a space(W,h0). Let Ev ⊂ Ev ⊕W ⊂ V be the partial flag considered earlierand P the parabolic subgroup of SU(h) preserving this flag and let Ube the subgroup which acts trivially on successive quotients of this flag.Then U is the unipotent radical of P .

The partial flag Ev∗ ⊂ Ev∗ ⊕W ⊂ V defines a parabolic subgroupP− of U(h), and U− is simply the subgroup of P− which acts triviallyon the successive quotients of this flag.


Corollary 3. With the preceding notation and hypotheses, the groupgenerated by the N-th powers of U(OK) and U−(OK) is an arithmeticsubgroup of SU(h)OK).

The same conclusion holds if

K − rank(SU(h)) = 1 but ∞− rank(SU(h)) ≥ 2.

Proof. We have already noted that K − rank(SU(h)) is the num-ber of hyperbolic 2-planes in the decomposition of h; therefore, K −rank(SU(h)) ≥ 2. Then the Corollary follows from Corollary 2 of thepreceding section.

The second part follows by Theorem 6 ((the group has higher realrank but Q-rank one). �

3.2. The Heisenberg Group and the Group P . Assume now that(V, h) is an n = m+1 dimensional E vector space with a non-degenerateHermitian form h. Assume that there exists a codimension two sub-space X of V such that h is non-degenerate on X; assume that thereexist isotropic vectors v, v∗ in V which are orthogonal to X, such thath(v, v∗) 6= 0. We have a filtration

0→ Ev → Ev ⊕X → V = Kv ⊕X ⊕Kv∗.

3.3. A Degenerate Hermitian Form. Suppose that m ≥ 1 is aninteger. Let (X, h) be a non-degenerate Hermitian space over E/K ofdimension m− 1. Let Ev be a one dimensional Hermitian space withthe zero Hermitian form. The direct sum W = Ev ⊕X has the prop-erty that the hermitian form on W is degenerate with W⊥ = Ev.

The quotient map W → W/Av = X preserves the Hermitian struc-ture on both sides, since Ev is orthogonal to W . This induces a sur-jective map U(W )→ U(X) of unitary groups, with kernel isomorphicto X∗ = HomE(X,Ev). We have a split short exact sequence

0→ X∗ → U(W )→ U(X)→ 1,

and we may write U(W ) = U(X)X∗. An element α of U(W ) mayaccordingly be written as a pair α = (g, x) with g ∈ U(X) and x ∈ X∗.If g ∈ U(X), then g gives a transformation on the dual X∗ defined byx 7→ xg (the transpose of g). With this notation, if β = (h, y) ∈ U(W )then

αβ = (gh, xh+ y).

Therefore, α−1 = (g−1,−xg−1).


The subgroup U(X) ⊂ U(W ) will be denoted M . Suppose thatm ∈ U(X) ⊂ U(W ) is of the form (λ, 0), which is the multiplicationby the scalar λ on X and acting by 0 on v. Given a, b ∈ U(h) denoteby [a, b] the commutator aba−1b−1.

Lemma 10. If α = (g, x) and m = (λ, 0) then the commutator [α,m]is of the form

[α,m] = (1, (1− λ−1)xg−1).In particular, if x 6= 0 and λ 6= 1, then the commutator [α,m] is a

non-zero element of the vector space V∗.

We denote by U0 the abelian vector group X∗ = Hom(X,Ev). Thenon-degenerate Hermitian form hX on X gives a Hermitian form on U0

as well, which we again denote hX .Now, V contains a codimension one subspace W = Ev⊕X ⊂ V such

that the restriction of h to W satisfies the conditions of the precedingsubsection. Consider the subgroup P of = U(h) = U((V, h)) whichpreserves the flag

Ev ⊂ W ⊂ V,

and U ⊂ P the subgroup which acts trivially on successive quotientsof this filtration. Then the group U is non abelian with centre equalto the one dimensional vector space Ga over K and we have an exactsequence

{0} → Ga ⊂ U(K)→ U0(K)→ {1}.Moreover, if x, y ∈ U ,and x∗, y∗ their images in U0, then the commu-tator [x, y] as an element of K is simply the imaginary part of h(x∗, y∗).

Now, P is a semi-direct product of U with M ' P/U and M isisomorphic to U(X) as in the preceding subsection. Moreover, thecentre Z = Ga of U is normal in P and P/Z is isomorphic to X∗. Wemay write, using the decomposition P = MU , an element α of P inthe form α = (g, x) with g ∈ U(W ) and x ∈ U . Let λ ∈ U(h) be ascalar transformation.

Lemma 11. If α = (g, x) with x ∈ U , x has non-zero image in U0

(i.e. does not lie in the centre of U), and m = (λ, 0) ∈ M , then thecommutator [α,m] is a non-central element of U .

The proof is immediate from Lemma 10 since the image of the com-mutator [α,m] in Uo = U/Z is already non-trivial by Lemma 10.

We now state another simple observation as a Lemma.


Lemma 12. If U → U0 is the quotient map, then a subgroup N ∈U(OK) is of finite index if and only if its image N0 is of finite index inU0(K).

If Ev ⊂ Ev ⊕X ⊂ V = Ev ⊕X ⊕ Ev∗ is as before, then the groupH(X) defined by the exact sequence

{1} → K ⊂ H(X)→ X

where on X the bracket is defined to be the imaginary part of h(x, y)is a subgroup of the unitary group U(h) = U(V ); this is the unipotentradical of a maximal parabolic subgroup.

3.4. An Inductive Step for Integral Unitary Groups. In thissubsection, we prove a result which will be used in the inductive proofof Theorem 3. This says that a subgroup of the integral unitarygroup which contains finite index subgroups of smaller integral uni-tary groups, is of finite index.

Notation. Let V = (V, h) be a nondegenerate hermitian space overE such that K − rank(V ) ≥ 2. Let W,W ′ be codimension one sub-spaces such that the restriction off h to W , W ′ and the intersectionW ∩W ′ are all non-degenerate. We denote by Uv the unitary groupU(h), and define similarly UW , UW ′ , UW∩W ′ . If Y is one of the subspacesW,W ′W ∩W ′, then by the non-degeneracy assumption, V = Y ⊕ Y ⊥and hence UY may be thought of as the subgroup of UV which actstrivially on Y ⊥.

Suppose that Γ ⊂ UV (OK) is a subgroup such that its intersectionwith UW (OK) (resp UW ′(OK)) has finite index in UW (OK) (resp. inUW ′(OK)).

Lemma 13. With the preceding notation, The group Γ has finite indexin UV (OK).

Proof. Since WeW ′ has co-dimension two in V and h is non-degenerateon W eW ′ it follows that V is the direct sum of W eW ′ and its or-thogonal complement (W ∩W ′)⊥ in V ; the orthogonal complement isalso a non-degenerate unitary space (of dimension two); since K-rankof V is at least two, it follows that W eW ′ contains an isotropic vectorv, say. The non-degeneracy of h on W ∩W ′ shows that there exists avector v∗ ∈ W eW ′ such that h(v, v∗) 6= 0; by replacing v∗ by v∗ + λvfor a suitable scalar λ if necessary, we may assume that v∗ ∈ W ∩W ′

is also an isotropic vector. Write V = (Ev + Ev∗)⊕X, an orthogonaldecomposition.


Consider the filtration

0 ⊂ Ev ⊂ E ⊕X ⊂ Ev ⊕X ⊕ Ev∗ = V.

Denote the corresponding integral Heisenberg group (the unipotentsubgroup of U(V ) which preserves the flag and acts by identity on suc-cessive quotients), by H(V ) = HV (OK). Define similarly, the smallerHeisenberg groups H(W ) and H(W ′).

By assumption, H(W ) ∩ Γ has finite index in H(W ); similarly forH(W ′). The two Heisenberg groups generate H(W ) up to finite index,since two distinct vector subspaces of codimension one, span the wholespace. We thus find the integral unipotent radical of a parabolic Ksubgroup in Γ.

Similarly, we find an opposite integral unipotent radical; therefore,Γ is arithmetic, by the Corollary 2 to Theorem 6. �

We keep the notation of Lemma 13. recall that E/K is a quadraticextension with respect to which unitary groups were defined. SupposeΓ ⊂ UV (OK) is a subgroup which operates irreducibly on V and W ⊂ Va codimension one E-subspace such that h is non-degenerate on W .Suppose K − rank(V ) ≥ 2.

Proposition 14. If Γ ⊂ UV (Od) is a subgroup which operates irre-ducibly on V ⊗ E such that the intersection Γ ∩ UW is an arithmeticsubgroup of UW (OK), then Γ is an arithmetic subgroup of UV (OK).

Proof. The irreducibility of the action of Γ implies that there exists aδ ∈ Γ such that δ(W ) 6= W ; by Lemma 9, the Zariski closure of Γcontains UV (K) = SL(V ). This is easily shown to imply that thereexists an element γ ∈ Γ such that the restriction of the Hermitian formh to the intersection γ(W ) ∩W is non-degenerate.

Write W = W and W ′ = γ(W ); then the assumption of Lemma 13are satisfied, and hence by Lemma 13, the group Γ is arithmetic. �

4. Properties of the Burau Representations ρn and ρn(d)

4.1. Notation. As observed in the Introduction, the representationρn : Bn+1 → GLn(Z[q, q−1]) preserves the hermitian form


h = hn =

(q+1)2

q−(1 + q) 0 · · · · · ·

−(1 + q−1) (q+1)2

q−(q + 1) · · · · · ·

0 −(1 + q−1) (q+1)2

q−(q + 1) · · ·

· · · · · · · · · · · · · · ·

.

Then the entries of the matrix h satisfy

hkl = 0 (| k − l |≥ 2), hk,k+1 = −(1 + q), hkk =(q + 1)2

q

(note that hkk is fixed under the involution q 7→ q−1). Denote by Dn

the determinant of hn.

Lemma 15. The determinant Dn of the matrix hn is

Dn = det(h) = (q + 1

q)n(

qn+1 − 1

q − 1).

Proof. Expanding the determinant of hn+1 using the first row, we seethat

Dn+1 =(q + 1)2

qDn − (1 + q)(1 + q−1)Dn−1.

Now an easy induction implies the Lemma. �

4.2. Non-degeneracy of the representation (An, ρn(d)). Considerthe ring A = R/(Φd(q)) ⊂ E = Q[q]/(Φd(q)) (q evaluated at a primi-tive root of unity). We then have corresponding Hermitian form on An

which we again denote by hn(d). We will say that elements v1, · · · vkare basis elements of An if there exist vectors vk+1, · · · , vn in An suchthat An is the free module generated by v1, · · · , vn. We have a repre-sentation ρn(d) : Bn+1 → U(h) ⊂ GLn(A) as before.

Lemma 16. Let A be as in the preceding. Assume that d ≥ 3, so thatq 6= ±1. Then

[1] The Hermitian form hn on the module An is non-degenerate ifand only if n is not congruent to −1 modulo d.

[2] If n is of the form kd− 1, then the vector

v = e1 + (q2 − 1

q − 1)e2 + (

q3 − 1

q − 1)e3 + · · · (q

kd−1 − 1

q − 1)ekd−1

is a basis vector and generates the orthogonal complement with respectto h, of the whole space V = An. Moreover, on the quotient module


V/Av, the form h is non-degenerate.

[3] The vector v is fixed by the elements sj under the representation

ρn(d). We therefore get the “quotient” representation ρn(d) of Bn+1 onthe quotient V/Av which again preserves the (non-degenerate) Hermit-ian form h on V/Av.

Proof. Part [1] is obvious because of Lemma 15: the determinant of hon An vanishes if and only if qn+1−1 = 0 where now q is a primitive d-throot of unity; therefore, this happens if and only if n+1 is divisible by d.

In part [2], the orthogonality of v with the vectors ej (j = 1, 2, · · · , kd−1 follows from an explicit computation; it is clear that v, e2, · · · , ekd−1form a basis of Akd−1. The matrix of the hermitian form hkd−1 onthe quotient Akd−1/Av with respect to the basis e2, · · · , ekd−1 is clearlyhkd−2; this is non-degenerate by part [1]. Thus part [2] follows.

Since v is orthogonal to ek it follows that sk fixed v: for any x ∈ Rn,we have

sk(x) = x− h(x, ek)

h(ek, ek)ek,

as can be easily checked by evaluating both sides of this equality onthe basis elements el. Therefore [3] follows.

�

Lemma 17. Suppose n is an integer not congruent to −1 modulo dand as before let Γn(d) be the image of Bn+1 under the representationρn(d) on An. If 0 6= W ⊂ V = An is a subgroup which is stable underΓn(d), then W contains λ(An) for some non-zero λ ∈ A; in particular,W is of finite index in An.

Proof. Since n 6= −1 (mod d), h is non-degenerate; since W 6= 0 W hasa nonzero vector x, and there exists a k such that h(x, ek) 6= 0. Theformula

sk(x) = x− h(x, ek)

h(ek, ek)ek,

shows that a nonzero multiple vk of ek lies in W ; applying s±1k to thismultiple of ek shows that (−q)±1vk ∈ W where q is a primitive rootof unity; hence the Z[q, q−1]- module generated by vk lies in W . ThusAvk ⊂ W .

Applying the elements sk±1 to Avk it follows that W contains a non-zero multiple vl of el for every l ≤ n and by the preceding paragraph,Avl ⊂ W for each l. Therefore the Lemma follows. �


Proposition 18. [1] The representation ρn(d) is irreducible unlessn ≡ −1(mod d).

[2] If n ≡ −1 (mod d) then ρn(d) contains a vector v fixed under Bn+1

and the quotient yields a representation ρn(d) of Bn+1. Restricted to

the subgroup of Bn+1 generated by s2, · · · , sn, the representation ρn(d)

is equivalent to ρn−1(d). Hence ρn(d) is irreducible.

[3] Suppose n is not congruent to −1 modulo d. Consider the rep-resentation ρn over an algebraically closed field E containing the ringS = Z[q+ q−1], and denote it ρn(E). Then ρn(E) is irreducible over E

Proof. In the proof of Lemma 17, we worked with the ring A ; instead,if we work with any field containing the quotient field K of A, then thesame proof works; since n is not congruent to −1 (mod d, we use thenon-degeneracy, to obtain the irreducibility.

The second part is obvious.

To prove the third part, The form hn has non-zero determinant byLemma 15. Therefore, again by the proof of Lemma 17, ρn(E) isirreducible. �

Let n ≡ −1 (mod d). Then the hermitian space V = (An, h) con-tains the Av as its orthogonal complement and V = V/Av has a non-degenerate hermitian form h induced by h on V . Let U(h) and Uh)denote the unitary groups. We have then a split exact sequence ofgroups

{0} → HomA(V ,Av)→ U(h)→ U(V )→ {1}.Here, the dual E of V may be identified with HomA(V ,Av). Thus

U(h) may be written as a semi-direct product U(h) = U(h)E with Enormal in U(h) and the conjugation action of U(h) on E is just thedual of the standard representation of U(h).

4.3. A Central element of the Braid Group. Consider the BraidGroup Bn+1 with generators s1, s2, · · · , sn. consider the element

∆ = (s1s2 · · · sn)(s1 · · · sn−1) · · · (s1s2)(s1),of Bn+1. It is elementary to check that ∆sn+1−k = sk∆. Hence ∆2 isin the centre of Bn+1.

If E is an algebraically closed field containing Z[q + q−1], and n isnot congruent to −1 mod d, then by part [3] of Proposition 18, ρn(E)is irreducible; therefore, by Schur’s Lemma, the element ∆2 acts by a


scalar delta, say. Since the determinant of each si is −q in the repre-sentation ρn(E), we see that the determinant of ∆2 is (−q)n(n+1) = δn.Therefore, (δ/qn+1)n = 1. On the other hand, the entries of ρn(E)(∆2)are Laurent polynomials in q with integral coefficients. Hence the scalarδ lies in the ring Z[q, q−1]. Therefore, the above equation means thatρn(∆2) = δ = (±1)qn+1.

Lemma 19. If n = kd− 2, then the element ∆2 acts by a scalar λ 6= 1on the representation (An, ρn(d)).

Proof. By Proposition 18, the representation ρkd−2 is irreducible. Bythe conclusion of the preceding paragraph, the central element ∆2 actsby a scalar λ = ±qn+1 = ±qkd−1 = ±q−1 since q is now a d-th root ofunity. which shows that λ 6= 1 since, otherwise, we get

q−2 = (±1)2 = 1.

This is impossible since d ≥ 3 and q is a primitive d-th root of unity.Hence λ 6= 1.

(It can be shown, by examining the action of the Braid Group ([Bir])on the free group on n + 1 generators, that the scalar in question isactually qn+1). �

4.4. Constructing Unipotent Elements when n = kd−1. Assumen = kd− 1. Consider the element

∆′ = (s2s3 · · · sn)(s2s3 · · · sn−1)(· · · )(s2s3)(s2)

which lies in the subgroup generated by s2, · · · , sn in Bn+1. By thepreceding lemma and Proposition 18, the element ∆′2 acts by a non-trivial scalar λ 6= 1 on ρkd−1(d).

Denote the element [s1,∆′2] of Bn+1 by u. We denote by [g, h] the

commutator ghg−1h−1 of g and h. The image of this element u underthe reducible representation ρkd−1(d), lies in the vector group An whereU(h) = U(h)×HomA(An, Av) is a semi-direct product as before.

Proposition 20. Let n = kd − 1. Under the representation ρn(d)the commutator element u = [s1, (∆

′)2] has the property that its imageu′ = ρkd−1(u) is a non-trivial unipotent element in the vector partAn−1 of U(h).

Proof. This follows from Lemma 10. Note that (An, h) is a degener-ate hermitian form with an isotropic vector v such that the quotientV/Av is non-degenerate; moreover, if X is the A-span of the vectorse2, e3, · · · , en then An is the direct sum Av ⊕X. Hence (V, h) satisfies


the hypotheses of Lemma 10.

By the preceding Lemma 19, the element m = ρn((∆′)2) acts (sinceDim(X) = kd− 2) by a scalar λ 6= 1 on X and acts trivially on v; theelement p = ρn(s1) takes the element e2 ∈ X into the element

−qe2 + v −n∑k=3

(qk − 1

q − 1)ek /∈ X,

which shows that p does not lie in M . therefore, Lemma 10 applies,and u′ is a non-zero vector in the vector part of U(h), and in particular,is a non-trivial unipotent element . �

Proposition 21. Suppose that n = kd− 1. The subgroup Un of Bn+1

generated by the conjugates huh−1 where h runs through the elementsof the group generated by s2, s3, · · · , sn has the property that under therepresentation ρn(d), it preserves the flag

Av ⊂< v, e2, · · · , en−1 >⊂ An,

and acts trivially on successive quotients of this flag. Here < v1, · · · , vk >denotes the sub-module generated by the vectors vi.

Further, if U0(OK) denotes the subgroup of U(h) which preserves theabove flag and acts trivially on successive quotients, then the image ofUn is a subgroup N0 of finite index in U0(OK).

Proof. We need only prove the last part, since the image of u lies in thenormal subgroup U0 and is preserved under conjugation by elementsof U(h). The conjugation action of M on U0 becomes the action ofM = U(X) on the dual X∗ = Hom(X,Av). By Lemma 17, N0 is offinite index provided it contains a non-zero element of U0; but it wasalready shown that the image of u is non-trivial in Uo. The Propositionfollows. �

4.5. Constructing unipotents when n = kd with k ≥ 1. We putn = kd − 1 = m + 1. The space V = An may be written as a directsum

V = Av ⊕X ⊕ Av∗, v = e1 +n−1∑k=2

xkek, v∗ = en +

n−1∑k=1

yken−k

as in the preceding. Then M = U(X) is the Levi part of the parabolicsubgroup P of U(h) which preserves the flag

0 ⊂ Av ⊂ Av ⊕X ⊂ V,


and U is the subgroup of P which acts trivially on successive quotientsof this flag.

Let ∆′ be as in the previous subsection; then m = ρn((∆′)2) lies inM since it acts trivially on v∗ and v (v∗ and v are orthogonal to all thevectors e2, e3, · · · , en−1, and hence are fixed by s2, · · · , sn−1); therefore,v, v∗ are fixed by ∆′, and ∆′ preserves the space X. By Lemma 19 ,(∆′)2 acts by a scalar λ 6= 1 on X.

Proposition 22. Assume n = kd. Then the following hold.[1] The element u = [s1, (∆

′)2] acts by a non-trivial unipotent ele-ment on An under the representation ρn(d).

More precisely, the element ρn(d)(u) lies in U(OK) \Z i.e. preservesthe flag

Av ⊂ Av ⊕X ⊂ V,

acts trivially on the successive quotients, and does not preserve thesubspace X.

[2] The group N generated by the conjugates huh−1 with h ∈ Γn−1(the group generated by the elements ρn(s2), · · · , ρn(sn−1)) is a sub-group of finite index in U(OK); is particular, Γn intersects the unipo-tent radical U(OK) in a subgroup of finite index.

Proof. The element u clearly preserves the flag of the Proposition, aswas verified in the preceding Proposition. Since u acts by a unipotentelement on W , and is a commutator, it follows that u acts trivially onthe one dimensional space V/W , and hence u acts unipotently on V .

By the preceding Proposition, u does not preserve the space X: ittakes the basis element e2 ∈ X into a sum of av and elements of X,with a 6= 0 a scalar. By Lemma 11, u does not lie in the centre of U .

Let B be the group generated in U(OK) by the conjugates huh−1 :h ∈ H where H is the group generated by s2, · · · , sn−1. Under thequotient map U → U0, B maps onto B0, and by the preceding Propo-sition, B0 has finite index in U0(OK).

Now the Proposition follows, by appealing to the second part ofLemma 12.

�


5. Proof of Theorem 3 and Theorem 2

5.1. Proof of Theorem 3. We prove the main theorem (Theorem3) by induction on n ≥ 2d; we will prove Theorem 3 directly for alln which are multiples of d and are at least 2d. Then by induction,Theorem 3

5.1.1. Proof when n = kd, with k ≥ 2. The representation ρn−1 isnot irreducible. Let Av ⊂ Vn−1 be the subspace of invariants. Thequotient Vn−1/Av is an irreducible representation of the Braid groupB(s2, · · · , sn−1). Hence the commutator element

u = [s1,∆(s2, · · · , sn−1)2]

is a non-trivial unipotent element under ρn−1 and lies in vector groupHomA(Vn−1/Av,Av). Therefore, u acts on the flag 0 ⊂ Av ⊂ Vn−1 ⊂ Vand acts unipotently on V since u, being a commutator, has determi-nant one and is unipotent on Vn−1.

Therefore, by Proposition 22, the group generated by the elementshuh−1 with h ∈ B(s2, · · · , sn−1) generate a subgroup commensurablewith U(OK) where U is the unipotent group preserving the flag Av ⊂Vn−1 ⊃ V and acting trivially on successive quotients of the flag. Wehave therefore proved that Γn ⊃ U(OK)N for some integer N .

Similarly, Γn ⊃ U−(OK)N for some integer N , where U− is oppositeto U . Therefore, Γn is an arithmetic group, since n ≥ 2d and henceVn has K − rank at least two: the space Vd−1 spanned by e1, · · · , ed−1has an isotropic vector v by Lemma 16. The subspace V ′d−1 spanned byed+1, · · · , v2d−1 also has the same hermitian form hd−1 and contains anisotropic vector v′ by Lemma 16. Clearly, Vd−1 and V ′d−1 are mutuallyorthogonal, since the indices of the bases ej differ by at least two.Thus we have produced two mutually orthogonal independent isotropicvectors, and hence theK-rank of Vn is at least two. Therefore, Theorem3 follwos for n = kd from Corollary 3.

5.1.2. Proof that Theorem 3 for n − 1 implies Theorem 3 for n whenkd < n ≤ kd+d−2. In this case, ρn and ρn−1 are irreducible. Then Vncontains both the subspace Vn−1 and the span Wn−1 of e2, e3, · · · , en;moreover, both these subspaces Vn−1 and Wn−1 are non-degenerateunder h.

By the induction assumption, there exists an integer N such thatU(Vn−1)

N ⊂ Γn−1 and U(Wn−1)N ⊂ Γn−1 '< s2, · · · , sn >,. It follows

from Proposition 14 that Γn is arithmetic.


5.1.3. Proof that Theorem 3 for n − 1 implies Theorem 3 for n whenn = kd− 1. .

By the previous subsection, Γn−1 is arithmetic. Since n is congruentto −1 modulo d hn is degenerate. We claim that Γn intersects thevector part in an arithmetic group. This follows from Proposition 21).

Since the Hermitian form is degenerate, the unitary group U(Vn) ofVn is a semi-direct product of its reductive part U(Vn−1) and its unipo-tent part Hom(Vn−1, Av). Then, the decomposition U(Vn)(OK) =U(Vn−1)(OK)HomA(Vn−1, Av) that Γn is also arithmetic.

Putting the above subsections together, we obtain a proof of ourmain result, namely Theorem 3 for all n ≥ 2d.

5.2. Proof of Theorem 2. We note from Theorem (5.5) of [Mc] thatfor the compactified curve X∗a the action of the Braid group on the firstcohomology has the decomposition

H1(X∗a ,Q) ' ⊕e/dρn(e),

where ρn(e) is the reduced Burau representation evaluated at primitivee-th roots of unity and the bar denotes the quotient modulo the in-variant vector (if any). We have already proved (Theorem 3) that theimage of ρn(e) is arithmetic if n ≥ 2d ≥ 2e; we not appeal to TheoremLemma and the above result of [Mc] to conclude that the image of therepresentation on H1(X∗a) is arithmetic; by Poincare duality, the im-age of the representation on the first homology of X∗a is also arithmetic.This proves Theorem 2

6. Applications

6.1. Some Complex Reflection Groups. We will follow the nota-tion of Section 5 of [Mc]. Given the root system An (and thereforeits graph) the Artin group A(An) may be defined; given an a complexnumber q = e2πx with −1/2 ≤ x < 1/2, there exists a representation

ρq : A(An)→ GLn(C),

with image denoted An(q). The image preserves a hermitian form andis a subgroup of a unitary group U(r, s) ⊂ GLn(C). The image ofthe Braid group Bn+1 under the Burau representation ρn : Bn+1 →GLn(Z[q, q−1]) may be identified with the complex reflection groupAn(q). Question (5.6) of [Mc] asks when the image An(q) is a latticein U(r, s). In the notation of question (5.6) of [Mc], the image groupΓn = ρn(d)(Bn+1) is the group An(q) where q is a primitive d-th rootof unity; therefore, question (5.6) asks whether An(q) can be a lattice


in the real unitary group U(r, s); Theorem 3 answers this question ina large number of cases. We have the following Corollary of Theorem3 (and part [2] of the Corollary follows from [Del-Mos] and [Mc]).

Corollary 4. [1] If q = e2π/d then An(q) is a lattice in U(r, s) whend = 3, 4, 6 and n ≥ 2d.

[2] An(q) is always a lattice for all n when d = 3, 4, 6.[3] If d is not 3, 4, 6 and n ≥ 2d , then the image Γn under the

Burau representation is an irreducible lattice in a product U(h)Kd ⊗QR) '

∏U(r, s) where the product is over all the archimedean (real)

completions of the totally real field Kd, and the numer of factors is atleast two. Therefore, the projection of Γn to one of the factors is nevera lattice if d 6= 3, 4, 6.

Thus question (5.6) of [Mc] is open only if d 6= 3, 4, 6 and if n ≤ 2d−1.

I thank Curtis Mcmullen for helpful communications concerning ques-tion (5.6) of [Mc].

6.2. Application to monodromy of type nFn−1.

Definition 2. Let α = (α1, · · · , αn) ∈ Cn and β = (β1, · · · , βn) ∈ Cn

be complex numbers such that αj 6= βk (mod 1) for any j and k.

Denote by z a complex variable; write θ = z ddz

. Consider the differ-ential operator D = D(α1, · · · , αn, β1, · · · βn) given by

D = (θ + β1 − 1)(θ + β2 − 1) · · · (θ + βn − 1)− z(θ + α1) · · · (θ + αn).

The equationDu = 0,

is called the hypergeometric equation with parameters α, β.

This is a differential equation on C = P1\{0, 1,∞}. In can be shownthat the space of solutions is of dimension n. Denote by π1(C,

12), the

fundamental group of C based at the point 12. Then it acts on the space

(' Cn) of solutions u of the foregoing differential equation Du = 0 (u isanalytic on C). Denote by Γ = Γ(α, β) the image of the fundamentalgroup π1(C,

12) in the group GLn(C) of linear automorphisms of the

n-dimensional space of solutions. Denote by Mα,β the resulting repre-sentation of π1(C,

12).

If α, β ∈ Qn then it may be shown that Γ may be conjugatedinto GLn(O) where O is the ring of integers in a number field F . IfD = [F : Q] is the degree of F over Q, then Γ ⊂ GLnD(Z). In [Sar] the


following question is considered: determine the pairs α, β ∈ Qn suchthat the associated monodromy group is arithmetic or not arithmetic(i.e. has finite index in its integral Zariski closure in GLnD(Z)).

Fuchs, Meiri and Sarnak (see [Sar]) give an infinite family of exam-ples of pairs α, β for which the group Γ has a natural embedding inan integral orthogonal group On(Z) with Zariksi dense image and ofinfinite index. Thus the monodromy Γ is NOT arithmetic.

By using [A’C], an infinite family of examples can be constructed,where the monodromy is an arithmetic subgroup of Sp2m(Z). One can,using Theorem 3, give a more general formulation:

Let d ≥ 2 be an integer and 1 ≤ c ≤ d be an integer coprime to d.

Theorem 23. Suppose that

α = (c

d+

1

n+ 1, · · · , c

d+n− 1

n+ 1,c

d+

n

n+ 1),

β = (c

d+

1

n, · · · , c

d+n− 1

n, 1).

If d = 2, and n = 2m is even, then the monodromy group Γ (is anarithmetic group and) has finite index in the integral symplectic groupSp2m(Z).

If d ∈ {3, 4, 6} and n is arbitrary such that n + 1 is coprime to d,then the monodromy Γ is an arithmetic group and is of finite indexin the integral unitary group U(h)(Z) where h is a suitable Hermitianform defined over the rationals.

If d ≥ 3 and n ≥ 2d is such that n + 1 is coprime to d, then Γ isan arithmetic group. Γ has finite index in an integral unitary group ofthe form U(h)(Od) where h is a non-degenerate hermitian form overthe totally real number field Kd = Q(cos(2π

d)), and Od is the ring of

integers in Kd.

Among these examples, only the case d = 2 (treated in [A’C]) has theproperty that the monodromy group Γ can be conjugated into GLn(Z)(in fact it already lies in Sp2m(Z)) with respect to the natural repre-sentation given in [Beu-Hec].

The theorem is an easy consequence of the fact that the image ofthe Burau representation ([Bir]) of the full braid group Bn+1 on n+ 1strands at d-th roots of unity is an arithmetic group in the cases stated


in the theorem. In case the image of the Burau representation is notarithmetic ([Del-Mos], [Mc]), the image of the monodromy representa-tion defines a thin group in the sense of Sarnak ([Sar]).

That the image of the Burau representation for d = 2 is a finiteindex subgroup of the integral symplectic group is a well known theo-rem of A’Campo ([A’C]). In the other cases, it follows from Theorem 3.

We now state the connection between Theorem 23 and the Buraurepresentation. Note that Bn+1is generated by the two elements h∞ =s1s2 · · · sn and h1 = s−1n . Write h0 = (s1s2 · · · sn−1)−1. Therefore,the braid group Bn+1 is a quotient of the fundamental group π1(C,

12).

Moreover, the image of h1 = sn under the Burau representation is acomplex reflection. One can show that the conditions of (definition(3.1)) of [Beu-Hec] are satisfied:

Definition 3. Let σ : ∆ = π1(C,12) → GLn(C) be a representation.

PutA = σ(h∞), B = σ(h−10 ),

where h∞, h0, h1 ∈ ∆ are small loops going counterclockwise around∞, 0, 1 respectively in C. Then ∆ is a free group on the generators h∞and h0 and h∞h1h0 = 1.We say that σ satisfies the Beukers-Heckman conditions ([Beu-Hec]),Definition (3.1)) if

[1] the characteristic polynomials Ch(t, A) and Ch(t, B) are also theminimal polynomials (are regular) and are of the form

Ch(t, A) =n∏j=1

(t− aj), aj = e2πiαj

Ch(t, B) =n∏k=1

(t− bk), bk = e2πiβk

with aj 6= bk for any j, k i.e. αj − βk /∈ Z for any j, k, and[2] the matrix C = A−1B fixes pointwise a codimension one subspace

W in Cn.

We claim that the Burau representation, which is a representationof the Braid group Bn+1 on n+ 1-strands, may be thought of as a rep-resentation of ∆; moreover, the resulting representation of ∆ satisfiesthe Beukers-Heckman condition.

The Burau representation is a representation of the braid group Bn+1

on n + 1 strands such that ρn(q) : Bn+1 → GLn(Z[q, q−1]). It can be


checked that under the standard basis ei of the Burau representation,the element h∞ has the matrix form

A = ρn(q)(h∞) =

0 0 0 · · · −qq 0 0 · · · −q0 q 0 · · · −q· · · · · · · · · · · · −q0 · · · · · · q −q

and its characteristic polynomial is of the form

Ch(t, A) =n∏j=1

(t− q e2πij/(n+1)).

It can also be shown that if B = ρn(q)(h−10 ) then the characteristicpolynomial is of the form

Ch(t, B) = (t− 1)n−1∏k=1

(t− q e2πik/n).

If q is specialised to be a primitive d-th root of unity e2πic/d, then theresulting monodromy representation image Γ is nothing but the imageof the Burau representation ρn(q). Therefore, the arithmeticity of The-orem 23 follows from Theorem 3.

We have proved (in Theorem 23) that a very special case of the rep-resentation Mα,β coincides with the monodromy of the Burau repre-sentation whose arithmeticity follows from Theorem 3 and from [A’C].Incidentally, the proof of Theorem 3 can be adapted to give a differentproof of the result of [A’C].

7. Proof of Theorem 1

We only sketch the proof (the proof is much more involved thanthat of Theorem 2). The proof of Theorem 2 used the properties ofthe Burau representation, which were established in Section 4. Theproof of Theorem 1 is quite similar, and uses properties of the Gassnerrepresentation (see [Bir], p.119 for the definition of the Gassner repre-sentation). The properties of Gassner representation which we will useare the following

Proposition 24. The reduced Gassner representation is has a non-degenerate skew hermitian form preserved by Pn+1. It is absolutelyirreducible. The centre of Bn+1 (which lies in Pn+1 and is generated by∆2) acts by scalars, and ∆2 acts by the scalar X1X2 · · ·Xn+1 on the


Gassner representation.

If we specialise Xi 7→ zi = qki where ki are coprime to d and q is agenerator of the cyclic group Z/dZ = qZ/qdZ , then the reduced Gassnerrepresentation evaluated at these d-th roots of unity is irreducible unlessz1z2 · · · zn+1 = 1 ( [Abd]).

The analogue of Theorem 3 is the following.

Theorem 25. Let gn(d) : Pn+1 → GLn(Od) denote the reduced Gassnerrepresentation evaluated at Xi = ωki where ki are coprime to d, and ωis a primitive d-th root of unity.

If n ≥ 2d, then the image of gn(d) is an arithmetic subgroup of asuitable unitary group.

The proof of Theorem 25 is similar to that of Theorem 3; the as-sumption n ≥ 2d implies, using a pigeon-hole principle argument, thatthe K rank of the associated unitary group is at least two.

The proof of Theorem 1 is deduced from Theorem 25. The proof isvery similar to that of Theorem 2 but much more involved and we omitthe details.

References

[Abd] M. Abdulrahim, Complex specialisations of the reduced Gassner Repre-sentation of the pure braid group, Proc. Amer. Math.Soc, 125 (1997), no.6, 1617-1624.

[Abd2] M. Abdurahim, A faithfulness criterion for the Gassner Representation ofthe pure braid group, Proc. Amer. Math. Soc 125, (1997), 1249-1257.

[A’C] N. A’Campo, Tresses, monodromie et groupes symplectique, Comment.Math. Helvetici, 54 (1987),318-327.

[Al-Car-Tol] D.Allcock, J. Carlson, D.Toledo, Complex Hyperbolic Geometry ofmoduli of cubic surfaces, J. Alg. Geometry 11, (2002), 659-724.

[1] D.Allcock, C. Hall, Monodromy Groups of Hurwitz type problems,Advances in math. 225 (2010), 69-80.

[Ba-Mi-Se] H.Bass, J.Milnor and J-P. Serre, Solution of the Congruence SubgroupProblem for SLn (n ≥ 3) and Sp2n (n ≥ 2), Publ. IHES, 33, (1967), 59-137.

[Beu-Hec] F.Beukers and G. Heckman, Monodromy for the hypergeometricfunction nFn−1, Invent. Math. 95 (1989), 325-354.


[Bir] J.Birman, Braids, Links and Mapping Class Groups, 82, Annals of Math.Studies, Princeton University Press, 1974.

[Bor-Ti] A.Borel and J.Tits, Groupes Reductifs, Publ. IHES.27, (1965), 55-150.

[Del-Mos] P.Deligne and G.D. Mostow, Monodromy of Hypergeometric Functionsand non-lattice integral monodromy, Publ.Math. IHES 63, (1986), 5-89.

[Long] D.D.Long, On linear Representations of Braid Groups, Transactions of theA.M.S. Vol 311, No.2, (1989), 535-560.

[Looi] E. Looijenga, Uniformization by Lauricella Functions, an overview of thetheory of Deligne-Mostow, 207-244, Porgr.Math.260 Birkhauser, Basel(2007).

[Mar] G.A.Margulis, Discrete Subgroups of semi-simple Lie groups, Ergebnisseder Mathematik und ihrer Grenzgebiete 3.Folge. Band 17, Springer-Verlag(1991).

[Mc] C. Mcmullen, Braid Groups and Hodge Theory, Math.Annalen, OnlineFirst, 27th March, (2012). http:// dx.doi.org/10.1007/s00208-012-0004-2 .

[Mos] G.D. Mostow, Generalised Picard Lattices arising from Half-IntegralConditions, Publ.Math. IHES, 63 (1986), 91-106.

[Ra] M.S. Raghunathan, A Note on Generators for Arithmetic Subgroups ofAlgebraic Groups, Pacific Journal of Mathematics, 152 (1991), 365-373.

[Sar] P. Sarnak, Notes on Thin Groups, MSRI hot topics workshop, Feb 2012).

[Sq] C.Squier, The Burau representation is unitary, Proc. Amer. Math. Soc. 90(1984), 199-202.

[Ti] J. Tits, Syst’eme‘s Generateurs de Groupes de Congruence, C.R.Acad.Sci. Paris, Serie A 283 (1976), 693.

[Va] L. Vaserstein, The Structure of Classical Arithmetic Groups of rankgreater than 1 (English Translation), Math. USSR, Sbornik 20 (1973),465-492.

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T.N.Venkataramana, School of Mathematics, TIFR, Homi BhabhaRoad, Colaba, Mumbai 400005, India

E-mail address: [email protected]

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