On the Schur multipliers of Coxeter groups

Robert B. Howlett

Department of Pure Mathematics, University of Sydney, N.S.W. 2006, Australia.

Abstract

The Schur Multipliers of all finite and certain infinite Coxeter groups have been calculated by Ihara and Yokonuma [5] and Yokonuma [7]. The purpose of this paper is to give a unified treatment of the Schur multipliers of all Coxeter groups of finite rank, avoiding the case by case arguments used in [5] and [7] and permitting speedy calculations of the Schur multipliers by inspection of the Coxeter diagram associated with a Coxeter group.

§1 Coxeter groups

Suppose that S is a finite set and

m: { (s, t) | s, t ∈ S and s ≠ t } ⟶ { n | n ∈ ℤ and n ≥ 2 } ∪ {∞} a function satisfying m(s, t) = m(t, s) for all s, t ∈ S. This information can be conveniently depicted by a graph D with vertex set S and edge set { {s, t} | m(s, t) ≥ 3 }, the edge {s, t} being assigned the weight m(s, t) if m(s, t) ≥ 4. Such a weighted graph is called a Coxeter diagram.

Given a Coxeter diagram D let F be the free group on the set S and let N be the smallest normal subgroup of F containing

{ s^{2} | s ∈ S } ∪ { (st)^{m(s,t)} | s, t ∈ S, s ≠ t, m(s, t) ≠ ∞ }.

The Coxeter group associated with the diagram D is the quotient group W = F /N. In other words, W is generated by a set \overline{S} in bijective correspondence with S, subject to defining relations

s̄^{2} = 1  (for all s ∈ S)

and

(s̄t̄)^{m(s,t)} = 1  (for s ≠ t and m(s, t) ≠ ∞).

§2 The main theorem

With D, W as above, let A_{2} = { {s, t} | m(s, t) = 2 }, the set of nonadjacent pairs of vertices. Let us write {s, t} ≈ {s, t′} if {s, t}, {s, t′} ∈ A_{2} and m(t, t′) is odd, and let ∼ be the equivalence relation on A_{2} generated by ≈. That is, {s, t} ∼ {s′, t′} if and only if there is a sequence {s_{0}, t_{0}}, {s_{1}, t_{1}}, …, {s_{k}, t_{k}} in A_{2} with {s_{0}, t_{0}} = {s, t}, {s_{k}, t_{k}} = {s′, t′} and {s_{i−1}, t_{i−1}} ≈ {s_{i}, t_{i}} for i = 1, 2, …, k.

Let us define

n_{1}(D) = number of vertices of D
n_{2}(D) = number of edges of D of finite weight
n_{3}(D) = number of equivalence classes of ∼ on A_{2}
n_{4}(D) = number of connected components of D′

where D′ is the graph obtained from D by deleting all edges of even weight and all edges of infinite weight. In practice these numbers are all easily calculated, at least if D is small. Our main theorem is the following:

Theorem A. The Schur multiplier of W is an elementary abelian 2-group of rank n_{2}(D) + n_{3}(D) + n_{4}(D) − n_{1}(D).

Note that the arguments used to prove that the multiplier has exponent 2 and rank at most n_{2}(D)+n_{3}(D)+n_{4}(D)−n_{1}(D) are due to Yokonuma and Ihara [5] (although rephrased slightly in §4 below). Thus the main objective of this paper is to prove that the bound is attained.

§3 Schur multipliers

Our treatment of Schur multipliers follows that of Curtis and Reiner [2]. Thus the Schur multiplier, Mult(G), of a group G is the group of all equivalence classes of factor sets G × G → ℂ^{×}, where ℂ^{×} is the multiplicative group of the complex field and

(i) a factor set is a function α satisfying

α(g_{1}, g_{2})α(g_{1}g_{2}, g_{3}) = α(g_{1}, g_{2}g_{3})α(g_{2}, g_{3})

for all g_{1}, g_{2}, g_{3} ∈ G, (ii) factor sets α and β are equivalent if there exists δ: G → ℂ^{×} such that

α(g_{1}, g_{2}) = β(g_{1}, g_{2})δ(g_{1})δ(g_{2})δ(g_{1}g_{2})^{−1}

for all g_{1}, g_{2} ∈ G,

(iii) multiplication of factor sets is pointwise multiplication of functions.

Note that this definition amounts to

Mult(G) = H^{2}(G, ℂ^{×})

regarding ℂ^{×} as a trivial G-module. Some authors define

Mult(G) = H_{2}(G, ℤ)

or

Mult(G) = ([F, F] ∩ N)/[F, N]

(where F is a free group and N a normal subgroup of F with F /N ≅ G). These latter two definitions are equivalent (see, for example, Hilton and Stammbach [4]), whereas

H^{2}(G, ℂ^{×}) ≅ Hom(H_{2}(G, ℤ), ℂ^{×})

could conceivably be different, but only if it is infinite.

§4 Preliminary facts

Throughout the remainder of this paper W will be a Coxeter group, and we will use all the notation of §1 and §2. Let K be the smallest normal subgroup of F containing { s^{2} | s ∈ S }, and let F̂ = F /K. Thus F̂ is the free product of groups of order 2 in bijective correspondence with the vertices of D. Observe that W ≅ F̂/N̂ where N̂ = N/K, and let g ↦ ĝ and g ↦ ḡ be the natural maps F → F̂ and F → W.

1 Proposition Let A be the group of all homomorphisms λ: N̂ → ℂ^{×} such that λ(x^{−1}yx) = λ(y) for all x ∈ F̂, y ∈ N̂, and let B be the group of all homomorphisms N̂ → ℂ^{×} which are restrictions of homomorphisms F → ℂ^{×}. Then Mult(W) ≅ A/B.

Proof. Let { x_{w} | w ∈ W } be a set of representatives of the cosets of N̂ in F. It can be checked that for any λ ∈ A the function α_{λ}: W × W → ℂ^{×} defined by

α_{λ}(v, w) = λ(x_{v}x_{w}x_{vw}^{−1})

is a factor set, and the function which maps λ to the equivalence class of α_{λ} is a homomorphism A → Mult(W). The kernel of this homomorphism consists of those λ such that for some μ: W → ℂ^{×},

(1.1)  λ(x_{v}x_{w}x_{vw}^{−1}) = μ(v)μ(w)μ(vw)^{−1}

for all v, w ∈ W. But (1.1) is precisely what is required for x_{w}y ↦ μ(w)λ(y) to be a homomorphism F̂ → ℂ^{×} extending λ; hence the kernel is B.

It remains for us to prove that every factor set β is equivalent to α_{λ} for some λ ∈ A. Given β let ρ: W → GL(V) be a projective representation with factor set β; for instance, let V be the set of all functions f: W → ℂ and for w ∈ W define ρ(w): V → V by

(ρ(w)f)(u) = β(u, w)f(uw)

for all u ∈ W. For each s ∈ S we have ρ(s̄)^{2} = β(s̄, s̄)ρ(¯1), and this is a scalar multiple of the identity transformation. Choose μ(s) ∈ ℂ^{×} such that ρ(s̄)^{2} = μ(s)^{2}1, and define σ(ŝ) = μ(s)ρ(s̄). Then σ(ŝ)^{2} = 1 for all s ∈ S, and so σ extends to a homomorphism σ: F̂ → GL(V). Now for each g ∈ F we have

(1.2)  \begin{aligned}σ(ĝ) & = σ(ŝ_{1}ŝ_{2} … ŝ_{l})  (for some s_{i} ∈ S) \\ & = μ(s_{1})μ(s_{2}) … μ(s_{l})ρ(\overline{s_{1}})ρ(\overline{s_{2}}) … ρ(\overline{s_{l}}) \\ & = μ(ĝ)ρ(ḡ) \end{aligned}

for some scalar μ(ĝ). In particular, if g ∈ N it follows that σ(ĝ) = λ(ĝ)1 for some scalar λ(ĝ), and since σ is a representation we deduce that λ ∈ A. Now for all v, w ∈ W,

β(v, w)1 = ρ(v)ρ(w)ρ(vw)^{−1}
= μ(x_{v})^{−1}μ(x_{w})^{−1}μ(x_{vw})σ(x_{v}x_{w}x_{vw}^{−1})
(since (1.2) gives σ(x_{u}) = μ(x_{u})ρ(u) for all u ∈ W)
= μ(x_{v})^{−1}μ(x_{w})^{−1}μ(x_{vw})α_{λ}(v, w)1

so that β is equivalent to α_{λ}, as required.

2 Proposition With the notation as in Proposition 1, the group A is an elementary abelian 2-group of rank at most n_{2}(D) + n_{3}(D).

Proof. Let λ ∈ A, let s, t ∈ S with m(s, t) = n < ∞, and let x ∈ F̂. Since λ is F̂-invariant we have

λ(x^{−1}(ŝt̂)^{n}x)^{2} = λ((ŝt̂)^{n})λ(ŝ(ŝt̂)^{n}ŝ)
= λ((ŝt̂)^{n})λ((t̂ŝ)^{n})
= 1.

Since N̂ is generated by the set N_{0} of all such elements x^{−1}(ŝt̂)^{n}x it follows that λ^{2} = 1. This holds for all λ ∈ A; so A has exponent 2.

Let {s_{1}, t_{1}}, {s_{2}, t_{2}}, …, {s_{k}, t_{k}} be the edges of D of finite weight, and let {s_{k+1}, t_{k+1}}, …, {s_{q}, t_{q}} be representatives of the ∼ equivalence classes on A_{2}. Thus q = n_{2}(D) + n_{3}(D), and to prove that A has rank at most n_{2}(D) + n_{3}(D) it suffices to prove that each λ ∈ A is uniquely determined by its restriction to { (s_{i}t_{i})^{n_{i}} | i = 1, 2, …, q }, where n_{i} = m(s_{i}, t_{i}) for each i.

Let λ ∈ A. We will prove the following claim:

(2.1)

If {r, s} and {t, u} are elements of A_{2} satisfying
{r, s} ∼ {t, u} then λ((r̂ŝ)^{2}) = λ((t̂û)^{2}).

From this it will follow that for each element x^{−1}(r̂ŝ)^{n}x in the generating set N_{0}

λ(x^{−1}(r̂ŝ)^{n}x) = λ((r̂ŝ)^{n}) = λ((ŝr̂)^{n}) = λ((ŝ_{i}t̂_{i})^{n_{i}})

for some i, so that the values λ((ŝ_{i}t̂_{i})^{n_{i}}) do determine λ.

To establish (2.1) it suffices to prove that λ((st)^{2}) = λ((sr)^{2}) whenever {s, t} ≈ {s, r}. So, assume that

m(s, r) = m(s, t) = 2 and m(r, t) = 2k + 1

for some k ∈ ℤ. Let σ be the representation of F̂ induced from the representation λ of N̂. We have σ(y) = λ(y)1 for all y ∈ N̂ (since λ is F -invariant), and so

(1)  σ(ŝ)σ(t̂) = λ((ŝt̂)^{2})σ(t̂)σ(ŝ)

and

(2)  σ(ŝ)σ(r̂) = λ((ŝr̂)^{2})σ(r̂)σ(ŝ).

Thus we find that

(3)  (σ(r̂)σ(t̂))^{k}σ(ŝ)(σ(t̂)σ(r̂))^{k} = εσ(ŝ)

with ε = ±1, while

(4)  (σ(r̂)σ(t̂))^{k}σ(r̂)(σ(t̂)σ(r̂))^{k} = ν σ(t̂)

where ν = λ((r̂t̂)^{2k+1}). Writing γ = (σ(r̂)σ(t̂))^{k} we have

γσ(ŝ)σ(r̂)γ^{−1} = εν σ(ŝ)σ(t̂)  (by (3) and (4))
= λ((ŝt̂)^{2})εν σ(t̂)σ(ŝ)  (by (1))
= λ((ŝt̂)^{2})γσ(r̂)σ(ŝ)γ^{−1}  (by (3) and (4))

and comparing with (2) gives λ((st)^{2}) = λ((sr)^{2}), as required.

3 Proposition The subgroup B of A has rank n_{1}(D) − n_{4}(D).

Proof. Since F̂ is the free product of n_{1}(D) groups of order 2 it follows that Hom(F̂, ℂ^{×}) is an elementary abelian 2-group of rank n_{1}(D). Restriction of functions gives a surjective homomorphism ϕ: Hom(F, ℂ^{×}) → B with kernel consisting of those λ ∈ Hom(F, ℂ^{×}) which are trivial on (st)^{m(s,t)} for all s, t such that m(s, t) < ∞. But if m(s, t) is even then

λ((st)^{m(s,t)}) = (λ(st))^{m(s,t)} = 1

since λ(st) = ±1, while if m(s, t) is odd then λ((st)^{m(s,t)}) = 1 if and only if λ(s) = λ(t). So λ is in the kernel of ϕ if and only if λ(s) = λ(t) whenever s and t are in the same connected component of the graph D′ (which has vertex set S and edge set { {s, t} | m(s, t) is odd }). Since this graph has n_{4}(D) components it follows that the kernel of ϕ has rank n_{4}(D), and hence that B has rank n_{1}(D) − n_{4}(D).

Note that, in view of Propositions 1, 2 and 3, to complete the proof of Theorem A it remains to show that the bound in Proposition 2 is attained.

§4 Coxeter group prerequisites

In this section we describe the various well known properties of Coxeter groups which are needed for our proof of Theorem A. For the proofs the reader is referred to Bourbaki [1], Steinberg [6] and Deodhar [3].

4 Proposition Let V be a vector space over the real field with a basis Π = { a_{s} | s ∈ S } in bijective correspondence with S, and define a bilinear scalar product on V by

⟨a_{s}, a_{t}⟩ = \{ \begin{matrix}1 & if s = t \\ − cos(π/m(s, t)) & if m(s, t) ≠ ∞ \\ −1 & if m(s, t) = ∞. \end{matrix}

For each a ∈ V with ⟨a, a⟩ = 1 define ρ(a): V → V by b ↦ b − 2⟨b, a⟩a. Then there exists a homomorphism ϕ from W to O(V), the orthogonal group on V, such that ϕ(s̄) = ρ(a_{s}) for all s ∈ S.

This is proved by checking that the transformations ρ(a_{s}) (for s ∈ S) are all involutions, satisfy the defining relations of W, and preserve the scalar product. For w ∈ W and a ∈ V we will write “w(a)” for the image of a under the action of ρ(w).

For each x ∈ F̂ there exists a unique integer l ≥ 0 and unique elements ŝ_{1}, ŝ_{2}, …, ŝ_{l} ∈ S such that x = ŝ_{1}ŝ_{2} … ŝ_{l} and ŝ_{i−1} ≠ ŝ_{i} (i = 2, 3, …, l). We call this the canonical expression for x and define l to be the length l(x) of x. We say that x is reduced if l(y) ≥ l(x) for all y such that ȳ = x̄. For w ∈ W define l(w) to be the minimum of l(x) over all x ∈ F̂ such that x̄ = w. Note that x ↦ (−1)^{l(x)} is a homomorphism F̂ → {±1} with kernel containing N̂.

The following proposition can be proved by induction on l(w). (See Deodhar [4], for instance.)

5 Proposition Let w ∈ W and s ∈ S such that l(ws̄) ≥ l(w). Then the coefficients arising when w(a_{s}) is expressed as a linear combination of the basis vectors { a_{t} | t ∈ S } are all nonnegative.

Define Φ = { w(a_{s}) | w ∈ W, s ∈ S }. It follows from Proposition 5 that Φ is the disjoint union of Φ^{+} and Φ^{−}, where

Φ^{+} = { a ∈ Φ | a = ∑_{t∈S} λ_{t}a_{t}, with all λ_{t} ≥ 0 }
Φ^{−} = { a ∈ Φ | a = ∑_{t∈S} λ_{t}a_{t}, with all λ_{t} ≤ 0 }.

Elements of Φ are called roots, elements of the basis Π are called simple roots, and elements of Φ^{+} and Φ^{−} are called positive roots and negative roots respectively.

For w ∈ W define

neg(w) = { a ∈ Φ^{+} | w(a) ∈ Φ^{−} }.

Using Proposition 5 we deduce

6 Proposition If w ∈ W and s ∈ S then

l(ws̄) = \{ \begin{matrix}l(w) + 1 & if a_{s} ∉ neg(w) \\ l(w) − 1 & if a_{s} ∈ neg(w). \end{matrix}

As a corollary of Proposition 6 we obtain

7 Proposition The representation of W described in Proposition 4 is faithful.

For each J ⊆ S let F_{J} be the subgroup of F generated by J and let N_{J} be the smallest normal subgroup of F_{J} containing

(7.1) R_{J} = { s^{2} | s ∈ J } ∪ { (st)^{m(s,t)} | s, t ∈ J, s ≠ t, m(s, t) ≠ ∞ }.

Thus F_{J}/N_{J} is a Coxeter group, and so we may construct (by Proposition 4) a homomorphism ϕ_{J}: F_{J} → O(V_{J}) which has kernel N_{J} (by Proposition 7). Comparison of the definitions of V and V_{J} yields that V_{J} may be identified with the subspace of V spanned by Π_{J} = { a_{s} | s ∈ J }; moreover this subspace is ϕ(F_{J})-invariant, and for each g ∈ F_{J} the transformation ϕ_{J}(g) equals the restriction to V_{J} of ϕ(g). In particular this shows that

N ∩ F_{J} = ker ϕ ∩ F_{J} ⊆ ker ϕ_{J} = N_{J}

and hence that N ∩ F_{J} = N_{J}, since N ∩ F_{J} is a normal subgroup of F_{J} containing R_{J} (see (7.1) above).

8 Proposition For each J ⊆ S the subgroup W_{J} of W generated by { s̄ | s ∈ J } is a Coxeter group.

Proof. W_{J} = F_{J}N/N ≅ F_{J}/(F_{J} ∩ N) = F_{J}/N_{J}.

For J ⊆ S define Φ_{J}^{+} = Φ^{+} ∩ V_{J}. The generators of W_{J} all act trivially on the quotient space V /V_{J}, and from this it follows that neg(w) ⊆ Φ_{J} for all w ∈ W_{J}.

9 Proposition (i) If J ⊆ S and W_{J} is finite then there exists an element w_{J} ∈ W_{J} such that neg(w_{J}) = Φ_{J}^{+}. Furthermore, there exists a permutation σ_{J} of Π_{J} such that w_{J}(a) = −σ(a) for all a ∈ Π_{J}.

(ii) For all w, v ∈ W, if neg(v) ⊆ neg(w) then l(wv^{−1}) = l(w) − l(v). (iii) If w ∈ W and J ⊆ S is such that Π_{J} ⊆ neg(w) then W_{J} is finite.

Proof. Parts (i) and (ii) are well known and follow from Proposition 6. For part (iii), observe first that J ⊆ neg(w) yields Φ_{J}^{+} ⊆ neg(w), and hence neg(v) ⊆ neg(w) for all v ∈ W_{J}. Now the number of elements in the set neg(w) is l(w) (by Proposition 6), and so neg(w) has only finitely many distinct subsets. Moreover, if neg(v_{1}) = neg(v_{2}) then (ii) gives l(v_{1}v_{2}^{−1}) = 0, and hence v_{1} = v_{2}. Thus W_{J} has at most one element for each subset of neg(w), and is therefore finite.

We will say that a subset J of S is of finite type if W_{J} is finite.

The classification of finite Coxeter groups is well known. Our proof of Theorem A makes use only of the classification of finite Coxeter groups of rank three. The specific facts we need are collected in the next proposition.

10 Proposition Suppose that J is a three element subset of S of finite type. Then there exist r, s ∈ J such that m(r, s) is even. Furthermore, if t ↦ t′ is the permutation of J defined by σ_{J}(a_{t}) = a_{t′} then either {r, s} = {r′, s′} or else {r, s} ∼ {r′, s′} ∈ A_{2}.

Proof. Since the representation of W on V is orthogonal it follows from Proposition 9 (i) that m(r, s) = m(r′, s′) for all r, s ∈ J. The classification in fact yields that m(r, s) = 2 for some r, s ∈ J; so we can certainly choose r and s so that m(r, s) is even. If {r, s} ≠ {r′, s′} let {t, u} be the other two element subset of J. Then m(r, s) = 2 (since a 3-vertex diagram of finite type has at most one edge of weight greater than three) and W_{J} is the direct product of a group of order 2 and a dihedral group of order 2m(t, u). For such a group the element w_{J} is central if m(t, u) is even; since {r, s} ≠ {r′, s′} this does not occur, and it follows that {r, s} ≈ {r′, s′}.

We also need the following well known results:

11 Proposition If J ⊆ S then each coset wW_{J} of W_{J} in W contains a unique element v such that l(vu) = l(v)+l(u) for all u ∈ W_{J}. In this situation neg(vu) ∩ Φ_{J} = neg(u).

12 Proposition If w ∈ W and r, s ∈ S satisfy l(wr̄) = l(s̄w) = l(w) + 1 and l(s̄wr̄) < l(w) + 2 then wr̄ = s̄w.

§5 Completion of the proof

The set S will no longer be directly involved in our discussion; so, changing our notation, we let S be the set previously denoted by Ŝ.

Let B = { (r, s) | r, s ∈ S and m(r, s) < ∞ }. It can be seen that the number of functions μ: B → {±1} with the property that μ(r, s) = μ(t, u) whenever (r, s) ∼ (t, u) ∈ A_{2} is equal to 2^{q}, where q = n_{2}(D) + n_{3}(D). Our strategy is to show that for every such function μ there exists λ ∈ A such that λ((rs)^{m(r,s)}) = μ(r, s) for all (r, s) ∈ B; this will imply that the rank of A is at least q and thus complete the proof of Theorem A. The bulk of the work is in the proof of the next lemma, in which the function ε_{i} can be thought of as the sign change in going from one expression for an element of W to another expression for the same element.

13 Lemma Suppose that μ is a function of the kind described above. For each nonnegative integer i there exists a function

ε_{i}: X_{i} = { (x, y) | x, y ∈ F̂, xy^{−1} ∈ N, l(x) + l(y) ≤ 2i } ⟶ {±1}

with the following properties.

(i) The restriction of ε_{i} to X_{i−1} coincides with ε_{i−1}. (ii) ε_{i}((rs)^{m}, 1) = μ(r, s) whenever {r, s} ∈ B and m(r, s) = m ≤ i. (iii) ε_{i}(x, y) = ε_{i}(y, x) whenever (x, y) ∈ X_{i}.

(iv) ε_{i}(xz, y) = ε_{i}(x, yz^{−1}) for all x, y, z ∈ F̂ with (xz, y) and (x, yz^{−1}) in

X_{i} and xzy^{−1} ∈ N.

(v) ε_{i}(x, y)ε_{i}(x, z)ε_{i}(y, z) = 1 whenever (x, y), (x, z), (y, z) ∈ X_{i}. (vi) ε_{i}(sx, y) = ε_{i}(x, sy) whenever s ∈ S and (sx, y), (x, sy) ∈ X_{i}.

Proof. Note first that if x, y ∈ F̂ with x̄ = ȳ then l(x) + l(y) is even; so every such pair (x, y) lies in some X_{i}.

The proof proceeds by induction on i. Clearly X_{0} = {(1, 1)} and, defining ε_{0}(1, 1) = 1, it is obvious that (i)–(vi) all hold—vacuously in most cases. Suppose that the assertions of the lemma hold for all i < k. For notational simplicity we let ε = ε_{k−1} and define a function θ which will have

the properties required of ε_{k}. We will also use the notation “\overbrace{ab...}^{h}” to denote the product of the first h terms of the alternating sequence a, b, a, b, … and

“ \overbrace{...ba}^{h} ” for the reverse of this. Let (x, y) ∈ X_{k}. Case 1. If l(xy^{−1}) < 2k define θ(x, y) = ε(xy^{−1}, 1). Case 2. Suppose that l(xy^{−1}) = 2k and l(x) = l(y) = k. If x and y are not reduced choose z ∈ F̂ with l(z) < k and z̄ = x̄ = ȳ, and define θ(x, y) = ε(x, z)ε(z, y). The inductive hypothesis ensures that this is independent of the choice of z: if we have also that l(u) < k and ū = x̄ then

ε(x, z)ε(z, y) = (ε(x, u)ε(u, z))(ε(z, u)ε(u, y))

= ε(x, u)ε(u, y) (by Properties (iii) and (v).

Case 3. Suppose that l(xy^{−1}) = 2k and l(x) = l(y) = k with x and y reduced, and let x = x′r and y = y′s with r, s ∈ S. Observe that r ≠ s, since l(xy^{−1}) = l(x) + l(y). By Proposition 9 (iii) we deduce that {r, s} ∈ B, and if we let m = m(r, s) then by Proposition 11 there exists z ∈ F̂ with l(z) = l(x) − m and

z \overbrace{...rs}^{m}, z \overbrace{...sr}^{m} ∈ xN̂ = yN̂.

We define

θ(x, y) = ε(x′, z \overbrace{...rs}^{m−1})ε(y′, z \overbrace{...sr}^{m−1})μ(r, s).

Properties (iii), (iv) and (v) for ε show that this is independent of the choice of z. Case 4. Suppose finally that l(xy^{−1}) = 2k and l(x) ≠ l(y). Let z ∈ F̂ be chosen so that the canonical expression for xz consists of the first k factors of the canonical expression for xy^{−1}. Then we have l(xz) = l(yz^{−1}) = k, and we define θ(x, y) = θ(xz, yz^{−1}).

We must show that θ satisfies Properties (i)–(vi) for i = k. Properties (i) and (iii) are trivial. If m(r, s) = k then by Cases 4 and 3 of the definition we have

θ((rs)^{k}, 1) = θ(\overbrace{rs...}^{k}, \overbrace{...sr}^{k}) = μ(r, s).

If m(r, s) < k the inductive hypothesis yields the same result; hence θ satisfies Property (ii).

Suppose that xzy^{−1} ∈ N̂ and (xz, y), (x, yz^{−1}) ∈ X_{k}. If l(xzy^{−1}) < 2k then, by Case 1 above, θ(xz, y) and θ(x, yz^{−1}) are both defined to be equal to θ(xzy^{−1}, 1), while if l(xzy^{−1}) = 2k then Case 4 defines both to equal ε(xzu, yu^{−1}) where xzu consists of the first k factors in the canonical expression for xzy^{−1}. Hence θ satisfies Property (iv).

The proofs of Property (v) depends upon a reduction to finite Coxeter groups of rank three, and we need the following claim:

(13.1) Suppose that J = {r, s, t} ⊆ S is of finite type, and let h = m(r, s),

m = m(r, t), n = m(t, s). Suppose that u \overbrace{...rs}^{h}, v \overbrace{...rt}^{m}, w \overbrace{...ts}^{n} are reduced

elements of F̂_{J} contained in the coset of N̂ corresponding to w_{J}, the longest element of W_{J}, and assume that l(w_{J}) ≤ k. Then

ε(u \overbrace{...rs}^{h−1}, v \overbrace{...rt}^{m−1})ε(u \overbrace{...sr}^{h−1}, w \overbrace{...st}^{n−1})ε(v \overbrace{...tr}^{m−1}, w \overbrace{...ts}^{n−1})

= μ(r, s)μ(r, t)μ(s, t).

Proof of claim. Let u ↦ u′ be the permutation of J corresponding to the permutation σ_{J} of Π_{J}. Observe that since the canonical expression for w cannot end with t or s it must end with r, and so w = w′r with l(w′) = l(w)−1. Similarly we have u = t′u′ and v = s′v′ with l(u′) = l(u)−1 and l(v′) = l(v) − 1.

Since J has finite type we know from Proposition 10 that one of h, n or m is even. Without loss of generality we may assume that n is even. Then our assumptions concerning μ together with Proposition 10 yield that

μ(s, t) = μ(s′, t′). Note also that since w_{J} s̄ = \overline{s′}w_{J} and \overbrace{...s̄t̄}^{n}s̄ = s̄\overbrace{...s̄t̄}^{n} it

follows that \overline{ws} = \overline{s′w}, and likewise for t.

Repeated applications of Property (vi) combined with Property (iv) give that

(13.2)  ε(xzy, xz_{1}y) = ε(z, z_{1})

whenever z̄ = \overline{z_{1}} and l(z) and l(z_{1}) are both less than k − l(x) − l(y). Hence for all i with 1 ≤ i ≤ n − 1 we have

(13.3)  \begin{aligned}ε(ws, s′w) & = ε(\overbrace{...s′t′}^{i−1} ws \overbrace{ts...}^{n−i−1}, \overbrace{...s′t′}^{i−1} s′w \overbrace{ts...}^{n−i−1}) \\ & = ε(\overbrace{...s′t′}^{i−1} w \overbrace{st...}^{n−i}, \overbrace{...t′s′}^{i} w \overbrace{ts...}^{n−i−1}) \end{aligned}

and similarly

(13.4)  ε(wt, t′w) = ε(\overbrace{...t′s′}^{i−1} w \overbrace{ts...}^{n−i}, \overbrace{...s′t′}^{i} w \overbrace{st...}^{n−i−1}).

By Property (v) the product of the right hand sides of the equations (13.3) for odd values of i by the right hand sides of (13.4) for even values of i

is ε(w \overbrace{ts...}^{n−1}, \overbrace{...t′s′}^{n−1} w), while the product of the corresponding left hand sides is γε(ws, s′w), where γ = (ε(ws, s′w)ε(wt, t′w))^{(n/2)−1}. The same equality holds with s and t interchanged. Therefore, by Property (v) and (13.2),

ε(v \overbrace{...tr}^{m−1}, w \overbrace{...ts}^{n−1})γε(ws, s′w) = ε(v \overbrace{...tr}^{m−1}, \overbrace{...t′s′}^{n−1} w)

= ε(v′ \overbrace{...tr}^{m−1}, \overbrace{...t′s′}^{n−2} w)

= ε(v′ \overbrace{...rt}^{m}, \overbrace{...t′s′}^{n−2} wt) (13.5)

= ε(v′ \overbrace{...rt}^{m}, \overbrace{...s′t′}^{n−1} w)ε(t′w, wt).

By parallel reasoning we obtain also

(13.6)  ε(u \overbrace{...sr}^{h−1}, w \overbrace{...st}^{n−1})γε(wt, t′w) = ε(u′ \overbrace{...rs}^{h}, \overbrace{...t′s′}^{n−1} w)ε(s′w, ws).

Since ε(\overbrace{...rt}^{m}, \overbrace{...tr}^{m}) = μ(r, t) we have

(13.7)  \begin{aligned}μ(r, t)ε(v′ \overbrace{...rt}^{m}, \overbrace{...s′t′}^{n−1} w) & = ε(v′ \overbrace{...tr}^{m}, \overbrace{...s′t′}^{n−1} w) \\ & = ε(v′ \overbrace{...rt}^{m−1}, \overbrace{...s′t′}^{n−1} w′) \\ & = ε(v \overbrace{...rt}^{m−1}, \overbrace{...s′t′}^{n} w′) \end{aligned}

and similarly

(13.8)  μ(r, s)ε(u′ \overbrace{...rs}^{h}, \overbrace{...t′s′}^{n−1} w) = ε(u \overbrace{...rs}^{h−1}, \overbrace{...t′s′}^{n} w′).

However, since

ε(\overbrace{...s′t′}^{n} w, \overbrace{...t′s′}^{n} w) = ε(\overbrace{...s′t′}^{n}, \overbrace{...t′s′}^{n}) = μ(s,′ t′) = μ(s, t)

we have

ε(v \overbrace{...rt}^{m−1}, u \overbrace{...rs}^{h−1}) = μ(s, t)ε(u \overbrace{...rs}^{h−1}, \overbrace{...t′s′}^{n} w′)ε(v \overbrace{...rt}^{m−1}, \overbrace{...s′t′}^{n} w′)

and multiplying this by (13.5), (13.6), (13.7) and (13.8) gives the desired formula.

Returning to the proof of Lemma 13, our next objective is to prove that θ satisfies Property (v). Accordingly, let x, y, z ∈ F̂ with x̄ = ȳ = z̄ and l(x) + l(y), l(x) + l(z) and l(y) + l(z) all 2k at most; we must prove that

(∗)  θ(x, y)θ(x, z)θ(y, z) = 1.

Assume, without loss of generality, that l(xy^{−1}) ≤ l(xz^{−1}) ≤ l(yz^{−1}). There are several stages in the proof. (13.9) If l(yz^{−1}) < 2k then (∗) holds.

Assume that l(yz^{−1}) < 2k, and choose u of maximal length such that y = y_{1}u and z = z_{1}u with l(y) = l(y_{1}) + l(u) and l(z) = l(z_{1}) + l(u). Let r, s, t ∈ S be the last factors in the canonical expressions for y_{1}, z_{1} and xu^{−1} respectively. By maximality of u we know that r ≠ s. If r ≠ t then

l(xu^{−1}) + l(y_{1}) = l(xu^{−1}y_{1}^{−1}) = l(xy^{−1}) < 2k

so that (xu^{−1}, y_{1}) ∈ X_{k−1}, and similarly if s ≠ t then (xu^{−1}, z_{1}) ∈ X_{k−1}. Now by Property (iv) (which has already been proved for θ) we have that θ(x, y) = θ(xu^{−1}, y_{1}), and likewise θ(x, z) = θ(xu^{−1}, z_{1}). Hence if r, s and t are all distinct Case 1 of the definition gives

θ(x, y)θ(x, z)θ(y, z) = ε(xu^{−1}, y_{1})ε(xu^{−1}, z_{1})ε(y_{1}, z_{1})

which equals 1 by Property (v) for ε. So (∗) holds in this case. If r = t then writing xu^{−1} = x′t and y_{1} = y′t we have

θ(x, y)θ(x, z) = θ(xu^{−1}, y_{1})ε(xu^{−1}, z_{1}) = ε(x′, y′)ε(x′, z_{1}s)
= ε(y′, z_{1}s) = ε(y_{1}, z_{1}) = θ(y, z)

and hence (∗) holds. A similar argument holds in the case s = t. (13.10) Suppose that l(yz^{−1}) = 2k and l(y) = l(z). If y and z are not reduced then (∗) holds.

We have that l(y) = l(z) = k and that there exists w ∈ F̂ with w̄ = x̄ = ȳ = z̄ and l(w) ≤ k − 2. If l(x) < k then Case 2 of the definition immediately gives

θ(y, z) = ε(x, y)ε(x, z) = θ(x, y)θ(x, z).

So assume that l(x) = k and let r, s, t ∈ S such that x = x′r, y = y′s and z = z′t with l(x′) = l(y′) = l(z′) = k − 1. If r = s then by Case 1 of the definition we have

θ(x, y) = ε(x′, y′)
= ε(x′, ws)ε(y′, ws)  (by Property (v) for ε)
= ε(x, w)ε(y, w)  (by Property (iii)),

while if r ≠ s then this same equation comes directly from the definition of θ(x, y). Similarly we have

\begin{aligned}θ(x, z) & = ε(w, x)ε(w, z) \\ θ(y, z) & = ε(w, y)ε(w, z). \end{aligned}

and multiplying these expressions we see that (∗) holds. (13.11) Suppose that l(yz^{−1}) = 2k and l(y) = l(z) = k with y and z reduced. Then (∗) holds.

Since x̄ = ȳ and y is reduced we must have l(x) = k also. Write x = x′r, y = y′s and z = z′t with r, s, t ∈ S and l(x′) = l(y′) = l(z′) = k − 1. We know that s ≠ t, since l(yz^{−1}) = 2k, and the definition of θ(y, z) is

θ(y, z) = ε(y′, u \overbrace{...st}^{m−1})ε(z′, u \overbrace{...ts}^{m−1})μ(s, t)

where m = m(s, t) and u ∈ N̂y \overbrace{...st}^{m} is such that l(u) = k − m.

If r = s then the definitions of θ(x, y) and θ(x, z) are

\begin{aligned}θ(x, y) & = ε(x′, y′) \\ θ(x, z) & = ε(x′, u \overbrace{...st}^{m−1})ε(z′, u \overbrace{...ts}^{m−1})μ(s, t) \end{aligned}

Multiplying the expressions for θ(x, y), θ(x, z) and θ(y, z) and using Property (v) for ε gives (∗). A similar argument holds if r = t.

Suppose that r, s and t are all distinct. By Proposition 9 (iii) we have that J = {r, s, t} is of finite type. We will use the notation of (13.1). By

Proposition 11 there exists d ∈ F̂ such that du \overbrace{...rs}^{h}, dv \overbrace{...rt}^{m} and dw \overbrace{...st}^{n} lie in N̂x = N̂y = N̂z and

l(d) + l(u) + h = l(d) + l(v) + m = l(d) + l(w) + n = k.

The definition of θ and Property (vi) for ε give

θ(x, y) = ε(x′, du \overbrace{...rs}^{h−1})ε(y′, du \overbrace{...sr}^{h−1})μ(r, s)

= ε(d^{−1}x′, u \overbrace{...rs}^{h−1})ε(d^{−1}y′, u \overbrace{...sr}^{h−1})μ(r, s),

θ(x, z) = ε(x′, dv \overbrace{...rt}^{m−1})ε(z′, dv \overbrace{...tr}^{m−1})μ(r, t)

= ε(d^{−1}x′, v \overbrace{...rt}^{m−1})ε(d^{−1}z′, v \overbrace{...tr}^{m−1})μ(r, t),

θ(y, z) = ε(y′, dw \overbrace{...st}^{n−1})ε(z′, dw \overbrace{...ts}^{n−1})μ(s, t)

= ε(d^{−1}y′, w \overbrace{...st}^{n−1})ε(d^{−1}z′, w \overbrace{...ts}^{n−1})μ(s, t).

Using Property (v) we see that the product of the right hand sides of these equations is the quantity that was considered in (13.1), and is therefore equal to 1. Hence (∗) holds. (13.12) The equation (∗) holds in all cases.

The case l(yz^{−1}) < 2k was dealt with in (13.9), so we may assume that l(yz^{−1}) = 2k. We use induction on l(z) − l(y). Since (13.10) and (13.11) have dealt with all cases in which l(z) − l(y) = 0, assume that i ≥ 1 and l(x) ≤ l(y) = k−i and l(z) = k+i. Let z = z′t with t ∈ S and l(z′) = l(z)−1.

Then l(xt)+l(yt) ≤ 2k and l(xt)+l(z′) ≤ 2k. Now by the definition of θ(y, z) we have

θ(y, z) = θ(yt, z′)
= θ(xt, yt)θ(xt, z′)  (by the inductive hypothesis)
= θ(x, y)θ(x, z)  (by Property (iv)).

Hence (13.12) is established, and the proof that θ satisfies Property (v) is completed.

We embark now upon the proof of Property (vi). Replace x by sx if necessary and y by sy if necessary to ensure that l(sx) = l(x) + 1 and l(sy) = l(y) + 1. This either leaves (vi) unchanged or transforms it into

(vi)′  θ(sx, sy) = θ(x, y) whenever (x, y) and (sx, sy) are in X_{k}.

We deal with (vi)′ first; so for (13.13), (13.14) and (13.15) below we assume that x̄ = ȳ and l(sx)+l(sy) = l(x)+l(y)+s ≤ 2k. Without loss of generality we assume that l(x) ≤ l(y). (13.13) If l(xy^{−1}) < 2k − 2 then θ(sx, sy) = θ(x, y).

If l(xy^{−1}) < 2k − 2 we must have x = x′r and y = y′r with r ∈ S and l(x) − l(x′) = l(y) − l(y′) = 1. By Case 1 of the definition of θ and Property (vi) for ε we find that

θ(sx, sy) = ε(sx′, sy′) = ε(x′, y′) = θ(x, y).

(13.14) Assume that l(xy^{−1}) = 2k − 2 and l(x) = l(y) = k − 1. Then θ(sx, sy) = θ(x, y).

We may assume that k > 1 since the case k = 1 is trivial. If sx is not reduced then there exists z ∈ F̂ with z̄ = \overline{sx} = \overline{sy} and l(z) < k. The definition gives

θ(sx, sy) = ε(z, sx)ε(z, sy)
= ε(sz, x)ε(sz, y)  (by Property (vi) for ε)
= θ(x, y)  (by Property (v)).

Suppose that sx is reduced let x = x′r and y = y′t with r, t ∈ S and l(y) − l(y′) = l(x) − l(x′) = 1. Then by Proposition 9 we know that m = m(r, t) is finite, and for an appropriate u,

θ(sx, sy) = ε(sx′, su \overbrace{...rt}^{m−1})ε(sy′, su \overbrace{...tr}^{m−1})μ(r, t)

= ε(x′, u \overbrace{...rt}^{m−1})ε(y′, u \overbrace{...tr}^{m−1})μ(r, t)

= ε(x, u \overbrace{...tr}^{m})ε(y, u \overbrace{...rt}^{m})ε(\overbrace{...rt}^{m}, \overbrace{...tr}^{m})

= ε(x, y).

(13.15) If l(xy^{−1}) = 2k − 2 then θ(sx, sy) = θ(x, y).

We use induction on l(y)−l(x). The case l(y)−l(x) = 0 has been dealt with in (13.14); so assume that l(y) > l(x) and let y = y′r with r ∈ S and l(y′) = l(y) − 1. Then by definition

θ(sx, sy) = θ(sxr, sy′)
= ε(xr, y′)  (by the inductive hypothesis)
= θ(x, y)  (by Property (iv)).

Turning now to (vi), assume that \overline{sx} = ȳ with l(sx)+l(y) ≤ 2k. Assume as before that l(sx) = l(x) + 1 and l(sy) = l(y) + 1, and, without loss of generality, that l(y) ≥ l(x). (13.16) If l(xy^{−1}) < 2k − 1 then θ(sx, y) = θ(x, sy).

Assume that l(xy^{−1}) < 2k − 1. Then there exist r ∈ S and x′, y′ ∈ F̂ with x = x′r and y = y′r and l(x) − l(x′) = l(y) − l(y′) = 1. We have

θ(sx, y) = ε(sx′, y′) = ε(x′, sy′) = θ(x, sy).

by Case 1 of the definition and Property (vi) for ε. (13.17) Assume that l(xy^{−1}) = 2k − 1 and l(sx) = l(y) = k. If y is not reduced then θ(sx, y) = θ(x, sy).

If y is not reduced then there exists z ∈ F̂ with z̄ = ȳ = \overline{sx} and l(z) < k. If l(sz) > l(z) then ε(z, y) = θ(sz, sy) by (13.13)–(13.15) above, while if l(sz) < l(z) then Property (vi) for ε gives ε(z, y) = ε(sz, sy). So in either case,

θ(sx, y) = ε(z, sx)ε(z, y)  (by definition of θ(sx, y))
= ε(sz, x)θ(sz, sy)
= θ(x, sy)  (by Property (v)).

(13.18) Assume that y is reduced, l(xy^{−1}) = 2k−1, and l(y) = l(sx) = k. Then θ(sx, y) = θ(x, sy).

If k = 1 then y ∈ S and x = 1. Now \overline{sx} = ȳ forces y = s, contradicting l(sy) = l(y) + 1. So k > 1. Let x = x′r and y = y′t with r, t ∈ S and l(x) − l(x′) = l(y) − l(y′) = 1. By definition we have θ(x, sy) = θ(xt, sy′). If xt is not reduced then, for some z with l(z) < k,

θ(xt, sy′) = ε(z, xt)ε(z, sy′)  (by definition)
= θ(sz, sxt)ε(sz, y′)

(either by (13.13)–(13.15) if l(sz) > l(z) or Property (vi) for ε otherwise)

= θ(sxt, y′)  (by Property (v))
= θ(sx, y)  (by Property (iv))

as required. Hence we may assume that xt is reduced.

We now have the situation that sx and xt are both reduced but sxt is not (since it maps onto the same element of W as does y′). By Proposition 12 it follows that \overline{sx} = \overline{xt}. We obtain

θ(x, sy) = θ(xt, sy′)
= θ(sx, xt)θ(sx, sy′)  (by Property (v))
= θ(sx, xt)ε(x, y′)  (by (13.13)–(13.15))
= θ(sx, xt)θ(xt, y)  (by Property (iv))
= θ(sx, y)  (by Property (v))

as required. (13.19) The equation θ(sx, y) = θ(x, sy) holds in all cases.

The case l(xy^{−1}) < 2k − 1 was dealt with in (13.16); so we may assume that l(xy^{−1}) = 2k − 1. We use induction on l(y) − l(x), noting that the case l(y) − l(x) = 1 has been dealt with in (13.17)–(13.19). Assume then that l(y) − l(x) > 1 and let y = y′r with r ∈ S and l(y′) = l(y) − 1. By definition of θ(x, sy) we obtain

θ(x, sy) = θ(xr, sy′)
= θ(sxr, y′)  (by the inductive hypothesis)
= θ(sx, y)  (by Property (iv))

as required. This completes the proof of Property (vi) and the proof of Lemma 13.

We continue with the notation of Lemma 13 and its preamble. For all x, y ∈ F̂ with x̄ = ȳ define ε(x, y) to be the common value of ε_{i}(x, y) for all i > l(x) + l(y), and define λ: N̂ → ℂ^{×} by λ(x) = ε(x, 1) for all x ∈ N̂.

For all x, y ∈ N̂ we have, by Properties (iv) and (v),

λ(x)λ(y) = ε(x, 1)ε(y, 1) = ε(x, 1)ε(1, y^{−1})
= ε(x, y^{−1}) = ε(xy, 1) = λ(xy)

so that λ is a homomorphism. Properties (vi) and (iv) give, for all s ∈ S and all x ∈ F̂,

λ(sxs) = ε(sxs, 1) = ε(xs, s) = ε(x, 1) = λ(x)

so that λ ∈ A. Finally, Properties (ii) and (iv) give λ((rs)^{m(r,s)}) = μ(r, s) for all (r, s) ∈ B. Since we have constructed a homomorphism λ with these properties for each μ it follows, as explained at the beginning of this section, that the rank of A is n_{2}(D) + n_{3}(D), and this completes the proof of Theorem A.

References

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4. P. J. Hilton and U. Stammbach, A Course in Homological Algebra, (Springer, New York, 1971).

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