Parametrized Braid Groups of Chevalley Groups

Parametrized Braid Groups of Chevalley Groups

Jean-Louis Loday¹ and Michael R. Stein²

Received: February 17, 2005 Communicated by Ulf Rehmann

Abstract. We introduce the notion of a braid group parametrized by a ring, which is defined by generators and relations and based on the geometric idea of painted braids. We show that the parametrized braid group is isomorphic to the semi-direct product of the Steinberg group (of the ring) with the classical braid group. The technical heart of the proof is the Pure Braid Lemma, which asserts that certain elements of the parametrized braid group commute with the pure braid group. This first part treats the case of the root system An; in the second part we prove a similar theorem for the root systemDn. 2000 Mathematics Subject Classification: Primary 20F36; Secondary 19Cxx; 20F55.

Keywords and Phrases: Braid group, Steinberg group, parametrized braid group, root system.

1 Introduction

Suppose that the strands of a braid are painted and that the paint from a strand spills onto the strand beneath it, modifying the color of the lower strand as in the picture below.

u

u v+au

v

a= coefficient of spilling

@@

@@

¡¡

¡¡

Figure 1

1Partially supported by Northwestern University in 2002.

2Partially supported by Universit´e Louis Pasteur in 1998-99.

This gives rise, for any ringA, to theparametrized braid groupBrn(A), which is generated by elements y_i^a, wherei is an integer, 1≤i≤n−1, and ais an element ofA, subject to the relations

(A1) yi^ayi⁰yi^b=y⁰iyi⁰y^a+b_i

(A1×A1) y^aiyj^b=y^bjy^ai if|i−j| ≥2, (A2) y_i^ay^b_i+1y^c_i =y^c_i+1y_i^b+acy^a_i+1

for anya, b, c∈A

A variation of this group first appeared in [L]. The choice of names for these relations will be explained in section 2 below. The derivation of these relations from the painted braid model can be seen in Figures 2 and 3 below.)

Observe that when A = {0} (the zero ring), one obtains the classical Artin braid group Brn, whose presentation is by generators yi, 1 ≤i ≤n−1, and relations

(A1×A1) y_iy_j=y_jy_i if|i−j| ≥2, (A2) y_iy_i+1y_i=y_i+1y_iy_i+1

A question immediately comes to mind: does Figure 1 correctly reflect the elements of the parametrized braid group? Up to equivalence, a picture would be completely determined by a braid and a linear transformation of the set of colors. This linear transformation lies in the subgroup En(A) of elementary matrices. Hence if the elements of the parametrized braid group correspond exactly to the pictures, then the group should be the semi-direct product of En(A) by Brn. We will show that this is almost the case: we only need to replaceEn(A) by the Steinberg groupStn(A) (cf. [St] [Stb]).

Theorem. For any ringAthere is an isomorphism Brn(A)∼=Stn(A)⋊Brn , where the action ofBrn is via the symmetric groupSn.

The quotient of Brn(A) by the relation y_i⁰y⁰_i = 1 is the group studied by Kassel and Reutenauer [K-R] (in this quotient group, our relation (A1) becomes y^aiy⁰iyi^b =y^a+b_i , which is exactly the relation used in [K-R] in place of (A1)).

They show that this quotient is naturally isomorphic to the semi-direct product Stn(A)⋊Sn of the Steinberg group with the symmetric group. So our theorem is a lifting of theirs.

The proof consists in constructing maps both ways. The key point about their existence is a technical result called the Pure Braid Lemma (cf. 2.2.1). It says that a certain type of parametrized braid commute with the pure braid group.

In the first part of the paper we give a proof of the above theorem which corresponds to the family of Coxeter groupsAn. In the second part we prove a similar theorem for the familyDn, cf. section 3. The Pure Braid Lemma in theDn case is based on a family of generators of the pure braid group found by Digne and Gomi [D-G]. We expect to prove similar theorems for the other Coxeter groups.

A related result has been announced in [Bon].

2 The parametrized braid group

We introduce the parametrized braid group of the family of Coxeter groups An−1.

Definition 2.0.1. Let A be a ring (not necessarily unital nor commutative).

The parametrized braid groupBrn(A)is generated by the elementsy^a_i, wherei is an integer, 1≤i≤n−1, andais an element ofA, subject to the relations

(A1) yi^ayi⁰y^bi =y⁰iyi⁰y^a+b_i

(A1×A1) y^a_iy^b_j =y^b_jy^a_i if|i−j| ≥2, (A2) yi^ay_i+1^b y^ci =y^c_i+1y_i^b+acy_i+1^a

for any a, b, c∈A.

The geometric motivation for the defining relations of this group, and its con- nection with braids, can be seen in the following figures in which u, v, w (the colors) are elements of A, and a, b, c are the coefficients of spilling. Relation (A1) comes from Figure 2.

Relation (A1×A1) arises because the actions ofy^ai and ofy^bj on the strands of the braid are disjoint when|i−j| ≥2, so that these two elements commute.

Relation (A2) derives from Figure 3.

u v

@@

@@@

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¡¡ y^b_i

u v+bu

@@

@@@

¡¡

¡¡ y⁰_i

u v+bu

@@

@@@

¡¡

¡¡ y^a_i

u v+bu+au

=

u v

@@

@@@

¡¡

¡¡ y^a+b_i

v+ (a+b)u u

@@

@@@

¡¡

¡¡ y⁰_i

u v+ (a+b)u

@@

@@@

¡¡

¡¡ y⁰_i

v+ (a+b)u u Figure 2

u v w

@@

@@

¡¡

¡¡ yi^c

v+cu u w

@@

@@

¡¡

¡¡ y_i+1^b

v+cu w+bu u

@@

@@

¡¡

¡¡ y_i^a

w+bu

+a(v+cu)v+cu u

=

u v w

@@

@@

¡¡

¡¡ y_i+1^a u w+av v

@@

@@

¡¡

¡¡ y_i^b+ac w+av

+(b+ac)u u v

@@

@@

¡¡

¡¡ y_i+1^c w+av

+(b+ac)u v+cu u Figure 3

2.1 Technical lemmas

Let Φ be the root systemAn orDn and letAbe a ring (not necessarily unital) which is supposed to be commutative in the Dn case. Let ∆ be a simple subsystem of the root system Φ. Elements of ∆ are denoted by αorαi. The image ofb∈Br(Φ) in the Weyl group W(Φ) is denotedb. The parametrized braid group Br(Dn, A) is defined in 3.1.1.

Lemma2.1.1. The following relations inBr(Φ, A)are consequences of relation (A1):

(y_α⁰y_α⁰)y_α^a = y_α^a(y_α⁰y⁰_α), y^a_α(y_α⁰)⁻¹y_α^b = y_α^a+b,

y_α^−a = y_α⁰(y_α^a)⁻¹y⁰_α.

Proof. Replacingbby 0 in (A1) shows thaty⁰_αy⁰_αcommutes withy^a_α. From this follows

y_α^a(y_α⁰)⁻¹y^b_α = (y⁰_αy⁰_α)⁻¹y_α^a(y_α⁰y⁰_α)(y_α⁰)⁻¹y_α^b

= (y⁰_αy⁰_α)⁻¹y_α^ay_α⁰y_α^b

= (y⁰_αy⁰_α)⁻¹y_α⁰y_α⁰y_α^a+b

= y_α^a+b.

Puttingb=−ain the second relation yields the third relation.

Lemma2.1.2. Assume that the Pure Braid Lemma holds for the root systemΦ.

Suppose that α∈∆ and b∈Br(Φ) are such that¯b(α)∈∆ (where¯b∈W(Φ) denotes the image of b). Then for anya∈Aone has

byα^a(y⁰α)⁻¹b⁻¹=y¯^ab(α)(y¯b(α)⁰ )⁻¹ in Br(Φ, A).

Proof. First let us show that there existsb^′∈Br(Φ) such thatb^′y^a_αb^′−1=y_¯b(α)^a . The two roots αand ¯b(α) have the same length, hence they are connected, in the Dynkin diagram, by a finite sequence of edges withm= 3 (cf. [Car, Lemma 3.6.3]). Therefore it is sufficient to prove the existence of b^′ whenαand ¯b(α) are adjacent. In that case α and ¯b(α) generate a subsystem of type A2; we may assumeα=α1 and ¯b(α) =α2 ∈A2; and we can use the particular case of relation (A2), namely

y⁰_α

1y_α⁰

2y_α^a

1 =y^a_α

2y_α⁰

1y_α⁰

2

to show that

y⁰_α1y⁰_α2y_α^a₁(y⁰_α2)⁻¹(y_α⁰₁)⁻¹=y^a_α2. (Hereb^′=y_α⁰

1y⁰_α

2,α=α1=−ǫ1+ǫ2,b¯^′(α) =α2=−ǫ2+ǫ3.) To conclude the proof of the Lemma it is sufficient to show that

by_α^a(y_α⁰)⁻¹b⁻¹=y_α^a(y⁰_α)⁻¹

wheneverb(α) =α. According to [H, Theorem, p. 22], ¯bis a product of simple reflectionsσαi forαi∈∆ which are not connected toαin the Dynkin diagram of ∆. Hence we can write b as the product of an element in the pure braid group and generatorsyαi which commute withy^aαby relation (A1×A1). Since we have assumed that the Pure Braid Lemma holds for Φ, we can thus conclude that byα^a(yα⁰)⁻¹b⁻¹=yα^a(y⁰α)⁻¹ as desired.

2.2 Braid group and pure braid group

The group Brn(0) = Brn is the classical Artin braid group with generators yi,1≤i≤n−1,and relations

yiyj = yjyi, |i−j| ≥2, yiyi+1yi = yi+1yiyi+1.

The quotient ofBrn by the relations yiyi= 1,1≤i≤n−1 is the symmetric group Sn; the image of b ∈ Brn in Sn is denoted by ¯b. The kernel of the surjective homomorphism Brn → Sn is thepure braid group, denoted P Brn. It is generated by the elements

aj,i:=yjyj−1· · ·yiyi· · ·yj−1yj,

forn≥j≥i≥1, ([Bir]; see Figure 4 below).

HH HH HH H

©© ©©

©

©

i j

· · ·

Figure 4: the pure braid aj,i

Lemma 2.2.1 (Pure Braid Lemma for An−1). Let y_k^a be a generator of Brn(A) and letω ∈P Brn =P Brn(0). Then there existsω^′ ∈Brn, indepen- dent of a, such that

yk^a ω=ω^′ y^ak.

Hence for any integer k and any element a ∈ A, the element y^a_k(y⁰_k)⁻¹ ∈ P Brn(A)commutes with every element of the pure braid groupP Brn. Notation. Before beginning the proof of Lemma 2.2.1, we want to simplify our notation. We will abbreviate

y⁰_k = k (y_k⁰)⁻¹ = k⁻¹

y_k^a = k^a Note thatk⁻¹ doesnotmeank^a fora=−1.

If there existω^′ andω^′′∈Brn such thatk^aω=ω^′j^aω^′′, we will write k^aω∼j^aω^′′.

Observe that∼isnotan equivalence relation, but it is compatible with multi- plication on the right by elements ofBrn: ifη∈Brn,

k^aω∼j^aω^′′⇔k^aω=ω^′j^aω^′′⇔k^aωη=ω^′j^aω^′′η⇔k^aωη∼j^aω^′′η For instance

k^a k k ∼ k^a by Lemma 2.1.1, k^a (k−1)k ∼ (k−1)^a by relation (A2), k^a (k−1) ∼ (k−1)^a k by relation (A2).

Proof of Lemma 2.2.1. It is clear that the first assertion implies the second one: since ω^′ is independent of a, we can set a = 0 to determine that ω^′ = y⁰_k ω (y⁰_k)⁻¹. Substituting this value of ω^′ in the expression y^a_k ω = ω^′ y^a_k completes the proof.

To prove the first assertion, we must show, in the notation just introduced, that k^aω∼k^a for every ω∈P Brn, and it suffices to show this whenω is one of the generatorsaj,i=j(j−1) · · ·i i· · · (j−1)jabove. Sincey^a_k commutes withy⁰_i for alli6=k−1, k, k+ 1, it commutes withaj,iwheneverk < i−1 or wheneverk > j+ 1. So we are left with the following 3 cases:

(j+1)^aaj,i ∼ (j+1)^a (1a)

j^aaj,i ∼ j^a (1b)

k^aa_j,i ∼ k^a i−1≤k≤j−1 (1c)

Our proof of case (1a) is by induction on the half-length of ω =aj,i. When i=k−1,ω= (k−1) (k−1), and we have

k^a (k−1) (k−1) = k^a (k−1)k⁻¹k(k−1)

∼ (k−1)^a k(k−1)

∼ k^a . and more generally,

k^a (k−1) (k−2)· · ·(k−2) (k−1)

∼ (k−1)^a k(k−2)· · ·(k−2)(k−1)

= (k−1)^a (k−2)· · ·(k−2)k(k−1)

∼ (k−1)^a k(k−1) (by induction),

∼ k^a.

The proof of case (1b) is also by induction on the half-length ofω=a_j,i. When i=k,ω=k k, and we have

k^ak k∼k^a by Lemma 2.1.1.

Then

k^ak(k−1)· · ·(k−1)k

= k^ak(k−1)k k⁻¹ (k−2)· · ·(k−2) (k−1)k

∼ k^a(k−1)k(k−1) (k−2)· · ·(k−2)k⁻¹(k−1)k

∼ (k−1)^a(k−1) (k−2)· · ·(k−2) (k−1)k(k−1)⁻¹

∼ (k−1)^ak(k−1)⁻¹ (by induction)

∼ k^a.

For case (1c) it is sufficient to check the cases ω = (k+1) (k+1)

(2a)

ω = (k+1)k k(k+1) (2b)

ω = (k+1)k(k−1)· · ·(k−1)k(k+1) (2c)

which are proved as follows:

k^a(k+1) (k+1) = k^a(k+1)k⁻¹k(k+1)

∼ (k+1)^a k(k+1)

∼ k^a Case (2a) k^a(k+1)kk(k+1) ∼ (k+1)^ak(k+1)

∼ k^a Case (2b)

k^a(k+1)k(k−1)· · ·(k−1)k(k+1)

∼ (k+1)^a(k−1)· · ·(k−1)k(k+1)

∼ (k+1)^ak(k+1)

∼ k^a Case (2c)

Proposition 2.2.2. For anyω ∈Brn(0) the element ωy^a_k(y⁰_k)⁻¹ω⁻¹ depends only on the class ω¯ of ω in Sn. Moreover if ω(j) =¯ k andω(j¯ + 1) =k+ 1, thenωy^a_k(y⁰_k)⁻¹ω⁻¹=y_j^a(y_j⁰)⁻¹.

Proof. The first statement is a consequence of Lemma 2.2.1 since the Weyl group W(An−1) =Sn is the quotient ofBrn byP Brn.

The second part is a consequence of relation (A1×A1) and the following computation:

2⁻¹1⁻¹2^a12 = 1^a2⁻¹1⁻¹2⁻¹12

= 1^a1⁻¹2⁻¹1⁻¹12

= 1^a1⁻¹ .

¤

2.3 The Steinberg group and an action of the Weyl group When Φ =An−1, the Steinberg group of the ringAis well-known and custom- arily denoted Stn(A). In that case it is customary to writex^a_ii+1 =xii+1(a) for the element xα(a), α = ǫi −ǫi+1 ∈ ∆, and, more generally, x^a_ij when α=ǫi−ǫj ∈An−1.

Definition2.3.1([Stb]). The Steinberg group of the ringA, denotedStn(A), is presented by the generators x^a_ij,1 ≤ i, j ≤ n, i 6= j, a ∈ A subject to the relations

(St0) x^a_ijx^b_ij =x^a+b_ij

(St1) x^aijx^bkl=x^bklx^aij, i6=l, j6=k (St2) x^aijx^bjk=x^bjkx^abikx^aij, i6=k.

We should make two observations about this definition. First, it follows from (St0) that x⁰_ij = 1. Second, relation (St2) is given in a perhaps unfamiliar form. We have chosen this form, which is easily seen to be equivalent to (R2), because of its geometric significance (cf. [K-S] for the relationship with the Stasheff polytope), and for the simplification it brings in computation.

The Weyl groupW(An−1) is the symmetric groupSn. Its action on the Stein- berg group is induced by the formula

(3) σ·x^a_ij :=x^a_σ(i)σ(j), σ∈ Sn, a∈A 2.4 The main result

Theorem 2.4.1. For any (not necessarily unital) ringA the map φ:Brn(A)→Stn(A)⋊Brn

from the parametrized braid group to the semi-direct product of the Artin braid group with the Steinberg group induced byφ(y_i^a) =x^a_{i i+1}yi is a group isomor- phism.

Proof. Step (a). We show thatφis a well-defined group homomorphism.

• Relation (A1):

φ(y_i^ay_i⁰y^b_i) = x^a_{i i+1}yiyix^b_{i i+1}yi, since x⁰_{i i+1}= 1,

= yiyix^a_{i i+1}x^b_{i i+1}yi since yiyi = 1∈ Sn,

= yiyix^a+b_{i i+1}yi by (St0),

= φ(y⁰iyi⁰y^a+b_i ).

• Relation (A1×A1) follows immediately from (St1).

• Relation (A2) is proved by using the relations of Brn and the 3 relations (St0),(St1),(St2) as follows:

φ(y^a_iy_i+1^b y_i^c) = x^a_{i i+1}yix^b_i+1i+2yi+1x^c_{i i+1}

| {z } yi

= x^a_{i i+1}yi x^b_i+1i+2x^c_{i i+2}

| {z }

yi+1yi,

= x^a_{i i+1}yix^c_{i i+2}

| {z }

x^b_i+1i+2yi+1yi,

= x^a_{i i+1}x^c_i+1i+2

| {z }

yix^b_i+1i+2

| {z }

yi+1yi,

= x^c_i+1i+2x^ac_{i i+2}x^a_{i i+1}x^b_{i i+2}

| {z }

yiyi+1yi

| {z } ,

= x^c_i+1i+2x^ac+b_{i i+2}x^a_{i i+1}yi+1

| {z }

yiyi+1,

= x^c_i+1i+2x^ac+b_{i i+2}yi+1

| {z }

x^a_{i i+2}yiyi+1

| {z }

,

= x^c_i+1i+2yi+1

| {z }

x^ac+b_{i i+1}yi

| {z }

x^a_i+1i+2yi+1

| {z }

,

= φ(y_i+1^c y_i^b+acy^a_i+1).

Step (b). This is the Pure Braid Lemma 2.2.1 forAn.

Step (c). We construct a homomorphism ψ : Stn(A)⋊Brn →Brn(A). We first constructψ:Stn(A)→Kerπ, whereπis the surjectionBrn(A)→Brn, by setting

ψ(x^aij) := ω y_k^a(y_k⁰)⁻¹ω⁻¹

where ω is an element of Brn such that ¯ω(k) = i and ¯ω(k+ 1) = j (for instance, ψ(x^a₁₂) =y₁^a(y⁰₁)⁻¹ and ψ(x^a₁₃) =y⁰₂(y₁^a(y⁰₁)⁻¹)(y₂⁰)⁻¹). Observe that this definition does not depend on the choice of ¯ω(by Lemma 2.1.2), and does not depend on how we choose a liftingωof ¯ω(by the Pure Braid Lemma 2.2.1).

In order to show that ψ is a homomorphism, we must demonstrate that the Steinberg relations are preserved.

• Relation (St0): it suffices to show thatψ(x^a₁₂x^b₁₂) =ψ(x^a+b₁₂ ), ψ(x^a₁₂x^b₁₂) =y^a₁(y₁⁰)⁻¹y^b₁(y⁰₁)⁻¹=y₁^a(y₁⁰)⁻²y₁⁰y₁^b(y⁰₁)⁻¹

= (y₁⁰)⁻²y^a₁y⁰₁y₁^b(y₁⁰)⁻¹ by 2.2.1,

=y^a+b₁ (y⁰₁)⁻¹ by (A1),

=ψ(x^a+b₁₂ ).

• Relation (St1): it suffices to show that ψ(x^a₁₂x^b₃₄) = ψ(x^b₃₄x^a₁₂) and that ψ(x^a₁₂x^b₁₃) = ψ(x^b₁₃x^a₁₂). The first case is an immediate consequence of the Pure Braid Lemma 2.2.1 and of relation (A1×A1). Let us prove the second case, which relies on the Pure Braid Lemma 2.2.1 and relation (A2):

ψ(x^a₁₂x^b₁₃) = y^a₁(y₁⁰)⁻¹

| {z }

y₂⁰y^b₁(y⁰₁)⁻¹(y₂⁰)⁻¹

= (y₁⁰)⁻²y^a₁y₁⁰y⁰₂y^b₁

| {z }

(y₁⁰)⁻¹(y⁰₂)⁻¹

= (y₁⁰)⁻²y₁^ay₂^by₁⁰

| {z }

y⁰₂(y₁⁰)⁻¹(y⁰₂)⁻¹

| {z }

= (y₁⁰)⁻²y⁰₂y₁^by^a₂(y⁰₁)⁻¹(y₂⁰)⁻¹

| {z }

y⁰₁

= (y₁⁰)⁻²y⁰₂y₁^b(y₁⁰)⁻¹(y₂⁰)⁻¹y₁^ay₁⁰

| {z }

= (y₁⁰)⁻²y⁰₂y₁^b(y₁⁰)⁻¹(y₂⁰)⁻¹(y⁰₁)²

| {z }

y₁^a(y₁⁰)⁻¹

| {z }

= ψ(x^b₁₃)ψ(x^a₁₂).

• Relation (St2): it suffices to show thatψ(x^a₁₂x^b₂₃) =ψ(x^b₂₃x^ab₁₃x^a₁₂).

ψ(x^a₁₂x^b₂₃) = y₁^a(y₁⁰)⁻¹ y^b₂

|{z}

(y⁰₂)⁻¹

= y₁^a(y⁰₁)⁻¹y₂^by₁⁰

| {z }

(y⁰₁)⁻¹(y⁰₂)⁻¹

= y₁^ay⁰₂y₁^b

| {z }

(y₂⁰)⁻¹(y⁰₁)⁻¹(y⁰₂)⁻¹

| {z }

= y₂^b

|{z}y₁^ab

|{z}y^a₂(y₁⁰)⁻¹(y⁰₂)⁻¹(y⁰₁)⁻¹

= y₂^b(y₂⁰)⁻¹y⁰₂y₁^ab(y₁⁰)⁻¹y₁⁰y^a₂(y⁰₁)⁻¹(y₂⁰)⁻¹

| {z }

(y₁⁰)⁻¹

= y₂^b(y₂⁰)⁻¹

| {z }

y₂⁰y^ab₁ (y⁰₁)⁻¹(y₂⁰)⁻¹

| {z }

y₁^a(y⁰₁)⁻¹

| {z }

= ψ(x^b₂₃)ψ(x^ab₁₃)ψ(x^a₁₂),

as a consequence of relation (A2).

From 2.2.2 it follows that the action of an element of Brn by conjugation on Kerπ depends only on its class in Sn. The definition of ψ onStn(A) makes clear that it is anSn-equivariant map.

DefiningψonBrn byψ(yα) =y_α⁰ ∈Brn(0) yields a group homomorphism ψ:Stn(A)⋊Brn→Ker π⋊Brn=Brn(A).

The group homomorphismsφandψare clearly inverse to each other since they interchangey_α^a andx^a_αyα. Hence they are both isomorphisms, as asserted. ¤ Corollary 2.4.2 (Kassel-Reutenauer [K-R]). The group presented by generators y_i^a,1≤i≤n−1,a∈A, and relations

(y_i⁰)²= 1 y^a_i(y⁰_i)⁻¹y_i^b=y_i^a+b

yi^ayj^b=yj^by^ai if|i−j| ≥2 y^a_iy^b_i+1y_i^c=y_i+1^c y_i^b+acy^a_i+1

a, b, c∈A, is isomorphic to the semi-direct productStn(A)⋊Sn.

Observe that when the first relation in this Corollary is deleted, the second relation has several possible non-equivalent liftings. The one we have chosen, (A1), is what allows us to prove Theorem 2.4.1.

3 The parametrized braid group in the Dn case

In this section we discuss the parametrized braid groupBr(Dn, A) for a com- mutative ringA and prove that it is isomorphic to the semi-direct product of the Steinberg groupSt(Dn, A) by the braid groupBr(Dn,0).

3.1 The braid group and the parametrized braid group for Dn

Let ∆ ={α2, α2^′, α3, . . . , αn} be a fixed simple subsystem of a root system of type Dn, n ≥ 3. We adopt the notation of [D-G] in which the simple roots on the fork of Dn are labeled α2, α2^′. The system Dn contains 2 subsys- tems of type An−1 generated by the simple subsystems{α2, α3, . . . , αn} and {α2^′, α3, . . . , αn}, and, for n≥4, a subsystem of typeDn−1 generated by the simple subsystem{α2, α2^′, α3, . . . , αn−1}.

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3 4 n

•

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• • • . . . • •

Dynkin diagram ofDn

TheWeyl group W(Dn) is generated by the simple reflections{σi =σαi|αi∈

∆}, with defining relations

σ²_i = 1 (σiσj)^m(i,j) = 1 fori, j∈ {2,2^′,3, . . . , n}, where

m(i, j) =

(2 ifαi, αj are not connected in the Dynkin diagram, 3 ifαi, αj are connected in the Dynkin diagram

Since the only values form(α, β) are 1,2 and 3, the groupBr(Dn, A) involves only relations (A1),(A1×A1) and (A2).

Definition 3.1.1. The parametrized braid group of type Dn with parameters in the commutative ring A, denoted Br(Dn, A), is generated by the elements y^a_α, whereα∈∆ anda∈A. The relations are, for a, b∈A andα, β∈∆

(A1) y_α^ay⁰_αy^b_α=y⁰_αy_α⁰y_α^a+b

(A1×A1) y_α^ay_β^b =y^b_βy^a_α if m(α, β) = 2

(A2) y^a_αy_β^by^c_α=y^c_βy^b+ac_α y^a_β if m(α, β) = 3 andα < β Note that the simple roots inDn are ordered so thatα2^′ < α3.

3.2 The Steinberg group of Dn and the main result

The roots of Dn are {±ǫi ±ǫj|1 ≤ i 6= j ≤ n}, [Car],[H]. The Weyl group W(Dn)∼= (Z/2)ⁿ⁻¹⋊Sn [Bour, p. 257, (X)] acts on the roots by permuting the indices (action ofSn) and changing the signs (action of (Z/2)ⁿ⁻¹). For the simple subsystem ∆ we takeαi=−ǫi−1+ǫifori= 2,· · ·, n, andα2^′ =ǫ1+ǫ2. If u and v are positive integers and α, β two roots, the linear combination uα+vβ is a root if and only if u = 1 = v, α = ±ǫi±ǫj, β ∓ǫj±ǫk and

±ǫi±ǫk 6= 0. In this case the definition of the Steinberg group is as follows.

Definition3.2.1([Stb][St]). The Steinberg group of typeDnwith parameters in the commutative ring A, denoted St(Dn, A), is generated by elements x^a_α, whereα∈Φanda∈A, subject to the relations (fora, b∈Aandα, β∈Φ)

(St0) x^a_αx^b_α=x^a+b_α

(St1) x^a_αx^b_β=x^b_βx^a_α if α+β6∈Dn andα+β6= 0, (St2) x^a_αx^b_β=x^b_βx^ab_α+βy^a_α if α+β∈Dn.

The Weyl group W(Dn) acts on St(Dn, A) by σ·x^aα = x^a_σ(α), and we can construct the semi-direct product St(Dn, A)⋊Br(Dn) with respect to this action.

Theorem 3.2.2. For any commutative ring Athe map φ:Br(Dn, A)→St(Dn, A)⋊Br(Dn) induced by φ(y^a_α) =x^a_αyα is a group isomorphism.

Corollary 3.2.3. The group presented by generators y^ai, i = 2^′,2,3,· · ·, n, a∈A and relations

(yi⁰)²= 1 y^ai(y⁰i)⁻¹yi^b=y_i^a+b

y_i^ay_j^b=y_j^by^a_i if |i−j| ≥2, ori= 2, j= 2^′, y^aiy^b_i+1yi^c=y_i+1^c y_i^b+acy^a_i+1 wherei+ 1 = 3when i= 2^′

fora, b, c∈A, is isomorphic to the semi-direct productSt(Dn, A)⋊W(Dn).

Proof of Corollary. For each simple rootαi∈Dn, writeyi^a foryα^ai. Proof of Theorem 3.2.2.

Step (a). Since the relations involved in the definitions of Br(Dn, A) andSt(Dn, A) are the same as the relations in the case of An−1, the mapφis well-defined (cf. Theorem 2.4.1).

Step (b). The proof of the Pure Braid Lemma in the Dn case will be given below in 3.3.

Step (c). Let π : Br(Dn, A) → Br(Dn) be the projection which sends each a∈A to 0 (as usual we identifyBr(Dn,0) withBr(Dn)). We define

ψ:St(Dn, A)⋊Br(Dn)→Br(Dn, A)∼= Kerπ⋊Br(Dn)

on the first component by ψ(x^a_α) = y^a_α(y_α⁰)⁻¹ ∈ Kerπ for α ∈ ∆. For any α∈ Dn there exists σ ∈W(Dn) such thatσ(α)∈∆. Let ˜σ∈ Br(Dn) be a

lifting of σ, and define ψ(x^a_α) = ˜σ⁻¹ψ(x^a_σ(α))˜σ∈Kerπ. This element is well- defined since it does not depend on the lifting of σby the Pure Braid Lemma forDn (Lemma 3.3.2), and does not depend on the choice ofσby Lemma 2.1.2.

In order to show that ψ is a well-defined group homomorphism, it suffices to show that the Steinberg relations are preserved. But this is the same verification as in theAn−1case, (cf. Theorem 2.4.1.)

The group homomorphismsφandψare inverse to each other since they inter- change y^a_αandx^a_αyα. Hence they are both isomorphisms. ¤

3.3 The Pure Braid Lemma for Dn

3.3.1 Generators for the Pure Braid Group of Dn

In principle, the method of Reidemeister-Schreier [M-K-S] is available to deduce a presentation ofP Br(Dn) from that ofBr(Dn). The details have been worked out by Digne and Gomi [D-G], although not in the specificity we need here.

From their work we can deduce that the group P Br(Dn) is generated by the elements yα², α ∈ ∆, together with a very small set of their conjugates. For example,P Br(D4) is generated by the 12 elements

2²,2^′2,3²,³2²,³2^′2,^322′3²,4²,⁴3²,⁴³2^′2,⁴³2²,^4322′3²,^4322′34²

where a prefixed exponent indicates conjugation: ^hg = hgh⁻¹. Here (and throughout) we use the simplified notations k^a =y^aαk and k=yα⁰k similar to those of 2.2.

Proposition 3.3.1. Forn≥4,P Br(Dn)is generated by the elements

• a_j,i=j(j−1). . .(i+1)ii(i+1). . .(j−1)j, n≥j≥i≥2, and

• b_j,i=j(j−1). . .322^′3. . .(i−1)ii(i−1). . .32^′23(j−1)j, n≥j ≥i≥3, wherei+ 1 = 3wheni= 2^′.

Note. Since the notation can be confusing, let us be clear about the definition of these generators in certain special cases:

• Wheni=j,a_j,i=i².

• Wheni= 3,b_j,3=j(j−1). . .322^′332^′2. . .(j−1)j.

Proof of (3.3.1). We work in the case where W =W(Dn) in the notation of [D-G]. In the proof of [D-G, Corollary 2.7], we see that PW =Un⋊PW_In

−1; takingIn={s₁,s₂,s₂′, . . . ,s_n−1}, n≥4, as on [D-G, p. 10], we see that their PW is equal to (our)P Br(Dn) and theirPW_In

−1 is equal to (our)P Br(Dn−1).

It follows that a set of generators for P Br(Dn) can be obtained as the union of a set of generators forP Br(Dn−1) with a set of generators forUn. This sets the stage for an inductive argument, since D3 =A3 (with {α2, α3, α2^′} ⊂D3

identified with {α1, α2, α3} ⊂ A3). Because W(Dn) is a finite Weyl group, it follows from [D-G, Proposition 3.6], that Un is generated (not just normally

generated) by the elements ab,s, and a list of these generators in our case is given on [D-G, p. 10].

The calculations necessary to prove the Pure Braid Lemma forDnare simpler if we replace the Digne-Gomi generators by the equivalent set in which conju- gation is replaced byreflection; that is, we replace a generator^hg=hgh⁻¹ by hgh^′, where ifh=yi₁. . . yik,h^′=yik. . . yi₁. (We already used this trick in the case ofAn−1.) ForD4, this procedure yields as generators ofP Br(D4) the set

2²,2^′2,3²,32²3,32^′23,322^′3²2^′23,4²,43²4,432²34,432^′234, 4322^′3²2^′234,4322^′34²32^′234,

and, more generally, P Br(Dn), n≥ 4 is generated by the elements stated in the Proposition.

Lemma 3.3.2 (Pure Braid Lemma for Dn). Let αk ∈ Dn, lety^a_k be a generator ofBr(Dn, A), and letω∈P Br(Dn). Then there existsω^′∈Br(Dn), independent of a, such that

yk^a ω=ω^′ y^ak.

Hence for any integer k and any element a ∈ A, the element y_k^a(y_k⁰)⁻¹ ∈ Br(Dn, A)commutes with every element of the pure braid groupP Br(Dn).

Proof. Let us show that the first assertion implies the second one. Let ω ∈ P Br(Dn)⊂P Br(Dn, A). By the first assertion of the Lemma we have

y^a_kω=ω^′y^a_k

for some ω^′ ∈Br(Dn,0), independent of a. Setting a= 0 tells us that ω^′ = y⁰_kω(y⁰_k)⁻¹. Thus

yk^a(yk⁰)⁻¹ω=ωyk^a(yk⁰)⁻¹ as desired.

Before beginning the proof of the first assertion, we recall some notation intro- duced in 2.2. We abbreviate y_α^ak by k^a and y_α⁰k by k. Whenever there exist ω^′ andω^′′∈Br(Dn) such thatk^aω=ω^′j^aω^′′, we will writek^aω∼j^aω^′′. This is notan equivalence relation, but it is compatible with multiplication on the right by elements ofBr(Dn): ifη∈Br(Dn),

k^aω∼j^aω^′′⇔k^aω=ω^′j^aω^′′⇔k^aωη=ω^′j^aω^′′η⇔k^aωη∼j^aω^′′η From defining relations (A1),(A1×A1), and (A2) of 2, we can deduce the following:

k^akk ∼ k^a (4a)

k^a (k−1)k ∼ (k−1)^a (4b)

k^a (k−1)k⁻¹ ∼ (k−1)^a (4c)

k^a(k+1)k ∼ (k+1)^a (4d)

(k+1)^ak(k+1) ∼ k^a (4e)

(k+1)^ak(k+1)⁻¹ ∼ k^a (4f)

k^a(k+1)k⁻¹ ∼ (k+1)^a (4g)

k^a(k+1)⁻¹k⁻¹ ∼ (k+1)^a (4h)

(k+1)^ak⁻¹(k+1)⁻¹ ∼ k^a (4i)

k^ak(k−1) ∼ (k−1)^a(k−1)k⁻¹ (4j)

Proof, continued. In the notation just introduced, we must show, for every ω∈P Br(Dn), thatk^aω∼k^a, and it suffices to show this whenωis one of the generators a_j,ior b_j,i of 3.3.1. That is, we must show

k^aaj,i ∼ k^a n≥j ≥i≥2, 1≤k≤n (5a)

k^abj,i ∼ k^a n≥j ≥i≥3, 1≤k≤n (5b)

The proofs of (5a) for i≥3 are exactly the same as the corresponding proofs for An−1 (see section 2); the additional case i = 2^′ presents no new issues.

Thus we shall concentrate on proving (5b); the proof proceeds by induction on n.

The case n= 3 is the case of the root system D3 =A3, which is part of the Pure Braid Lemma 2.2.1 forAn−1. Hence we may assumen≥4, and that (5b) holds wheneverj, k≤n−1. That is, we must prove (5b) in these cases:

k=n, j≤n−1; k≤n−1, j=n; k=n, j=n which further subdivide into the cases

k=n, j ≤n−2 (6)

k=n, j =n−1 (7)

k≤n−2, j =n (8)

k=n−1, j =n (9)

k=n, j =n (10)

• Case (6) k = n and j ≤ n−2. Since i ≤ j ≤ n−2, it follows from relation (A1×A1) thatn^a commutes with every generator which occurs in the expression forbj,i; hence

n^ab_j,i = n^aj(j−1). . .322^′3. . .(i−1)ii(i−1). . .32^′23(j−1)j

= bj,in^a

∼ n^a

as desired.

• Case (7) k=nandj =n−1.

Ifi≤n−2, then n^ab_n−1,i

= n^a(n−1)(n−2). . .322^′3. . .(i−1)ii(i−1). . .32^′23. . .(n−2)(n−1)

= n^a(n−1)n⁻¹

| {z }

n(n−2). . .322^′3. . .(i−1)ii(i−1). . .32^′23. . .(n−2)(n−1)

∼ (n−1)^an(n−2). . .322^′3. . .(i−1)ii(i−1). . .32^′23. . .(n−2)(n−1)

= (n−1)^a(n−2). . .322^′3. . .(i−1)ii(i−1). . .32^′23. . .(n−2)

| {z }

n(n−1)

∼ (n−1)^an(n−1)

∼ n^a as desired.

The casek=n, i=j=n−1, is considerably more complicated. We first prove some preliminary lemmas.

Lemma 3.3.3.

n^a(n−1)(n−2). . .322^′∼2^a32^′4. . .(n−2)(n−1)n Proof.

n^a(n−1)(n−2). . .322^′ = n^a(n−1)n⁻¹n(n−2). . .322^′

∼ (n−1)^an(n−2). . .322^′

= (n−1)^a(n−2). . .322^′n ...

∼ 3^a22^′4. . .(n−2)(n−1)n

= 3^a23⁻¹32^′4. . .(n−2)(n−1)n

∼ 2^a32^′4. . .(n−2)(n−1)n

¤ Lemma 3.3.4.

32^′4354. . .(n−3)(n−1)(n−2)nn=345. . .(n−1)nn2^′34. . .(n−2) Proof.

32^′4354. . .(n−3)(n−1)(n−2)nn

= 32^′4354. . .(n−3)(n−1)nn(n−2)

= 32^′4354. . .(n−1)nn(n−3)(n−2) ...

= 345. . .(n−1)nn2^′34. . .(n−2)

¤ We now complete the case k=n, i=j =n−1.

n^ab_n−1,n−1

= n^a(n−1)(n−2). . .322^′3. . .(n−2)(n−1)(n−1)(n−2). . . . . .32^′23. . .(n−2)(n−1)

∼ 2^a32^′4. . .(n−2)(n−1)n3. . .(n−2)(n−1)(n−1)(n−2). . .

. . .32^′23. . .(n−2)(n−1) (by Lemma 3.3.3)

= 2^a32^′4. . .(n−2)(n−1)3. . .(n−2)n(n−1)n⁻¹

| {z }

n(n−1)(n−2). . . . . .32^′23. . .(n−2)(n−1)

= 2^a32^′4. . .(n−2) (n−1)3. . .

| {z }

(n−2)(n−1)⁻¹n(n−1)n(n−1)(n−2). . . . . .32^′23. . .(n−2)(n−1)

= 2^a32^′4. . .(n−2)3. . .(n−1)(n−2)(n−1)⁻¹

| {z }

n(n−1)n(n−1)(n−2). . . . . .32^′23. . .(n−2)(n−1)

= 2^a32^′4. . .(n−2)3. . .(n−2)⁻¹(n−1)(n−2)

| {z }

n(n−1)n(n−1)(n−2). . . . . .32^′23. . .(n−2)(n−1)

...

= 2^a32^′434⁻¹

| {z }54. . .(n−1)(n−2)n(n−1)n(n−1)(n−2). . . . . .32^′23. . .(n−2)(n−1)

= 2^a32^′3⁻¹

| {z }4354. . .(n−1)(n−2)n(n−1)n(n−1)(n−2). . . . . .32^′23. . .(n−2)(n−1)

= 2^a2^′−132^′

| {z }4354. . .(n−1)(n−2)n(n−1)n(n−1)(n−2). . . . . .32^′23. . .(n−2)(n−1)

∼ 2^a32^′4354. . .(n−1)(n−2)n(n−1)n(n−1)

| {z }

(n−2). . . . . .32^′23. . .(n−2)(n−1)

= 2^a32^′4354. . .(n−1)(n−2)n n(n−1)n

| {z }

(n−2). . . . . .32^′23. . .(n−2)(n−1)

= 2^a345. . .(n−1)nn2^′34. . .(n−2)(n−1)n(n−2). . .

. . .32^′23. . .(n−2)(n−1) (by Lemma 3.3.4)

We now manipulate part of this expression so that we can apply induction.

2^a345. . .(n−1)nn2^′

= 2^a345. . .(n−1)nn(n−1). . .5433⁻¹4⁻¹5⁻¹. . .(n−1)⁻¹

| {z }

2^′

∼ 2^a3⁻¹4⁻¹5⁻¹. . .(n−1)⁻¹2^′