Two-sided cells in type $B$ (asymptotic case)

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Two-sided cells in type B (asymptotic case)

Cédric Bonnafé

To cite this version:

Cédric Bonnafé. Two-sided cells in type B (asymptotic case). Journal of Algebra, Elsevier, 2006, 304,

pp.216-236. �hal-00004156v3�

ccsd-00004156, version 3 - 30 Jun 2006

TWO-SIDED CELLS IN TYPE (ASYMPTOTIC CASE)

C´EDRIC BONNAF´E

Abstract. We compute two-sided cells of Weyl groups of typeBfor the “asymptotic”

choice of parameters. We also obtain some partial results concerning Lusztig’s conjectures in this particular case.

Let W

be a Weyl group of type B

. The present paper is a continuation of the work done by L. Iancu and the author [3] concerning Kazhdan-Lusztig theory of W

for the asymptotic choice of parameters [3, §6]. To each element w ∈ W

is associated a pair of standard bi-tableaux (P (w), Q(w)) (see [19] or [3, §3]): this can be viewed as a Robinson-Schensted type correspondence. Our main result [3, Theorem 7.7] was the complete determination of the left cells: two elements w and w

are in the same left cell if and only if Q(w) = Q(w

). For the corresponding result for the symmetric group, see [12] and [1]. We have also computed the character afforded by a left cell representation [3, Proposition 7.11] (this character is irreducible).

In this paper, we are concerned with the computation of the two-sided cells. Let us state the result here. If w ∈ W

, write Q(w) = (Q

(w), Q

(w)) and denote by λ

(w) and λ

(w) the shape of Q

(w) and Q

(w) respectively. Note that (λ

(w), λ

(w)) is a bipartition of n.

Theorem (see 3.9). For the choice of parameters as in [3, §6], two elements w and w

are in the same two-sided cell if and only if (λ

(w), λ

(w)) = (λ

(w

), λ

(w

)).

Lusztig [17, Chapter 14] has proposed fifteen conjectures on Kazhdan-Lusztig theory of Hecke algebras with unequal parameters. In the asymptotic case, Geck and Iancu [9]

use some of our results, namely some informations on the preorder 6

(see Theorem 3.5 and Proposition 4.2), to compute the function a and to prove Lusztig’s conjectures P

, for i ∈ {1, 2, 3, 4, 5, 6, 7, 8, 11, 12, 13, 14}. On the other hand, Geck [8] has shown that Lusztig’s conjectures P

and P

hold. More precisely, he proved that the Kazhdan-Lusztig basis is cellular (in the sense of [11]). He also proved a slightly weaker version of P

(but his version is sufficient for constructing the homomorphism from the Hecke algebra to the asymptotic algebra J ).

The present paper is organized as follows. In Section 1, we study some consequences of Lusztig’s conjectures on the multiplication by T

, where w

is the longest element of

Date: June 30, 2006.

1991 Mathematics Subject Classification. According to the 2000 classification: Primary 20C08; Sec- ondary 20C15.

1

a finite Weyl group. From Section 2 to the end of the paper, we assume that the Weyl group is of type B

and that the choice of parameters is done as in [3, §6]. In Section 2, we establish some preliminary results concerning the Kazhdan-Lusztig basis. In Section 3, we prove the above theorem by introducing a new basis of the Hecke algebra: this was inspired by the work of Geck on the induction of Kazhdan-Lusztig cells [7]. Section 4 contains some results related to Lusztig’s conjectures. In Section 5, we determine which specializations of the parameters preserve the Kazhdan-Lusztig basis.

1. Generalities

1.A. Notation. We slightly modify the notation used in [3, §5]. Let (W, S) be a Coxeter group with |S| < ∞. We denote by ℓ : W → N = {0, 1, 2, . . . } the length function relative to S. If W is finite, w

denotes its longest element. Let 6 denote the Bruhat ordering on W . If I ⊂ S, we denote by W

the standard parabolic subgroup of W generated by I.

Let Γ be a totally ordered abelian group which will be denoted additively. The order on Γ will be denoted by 6 . If γ

∈ Γ, we set

Γ

= {γ ∈ Γ | γ < γ

}, Γ

= {γ ∈ Γ | γ 6 γ

}, Γ

= {γ ∈ Γ | γ > γ

} and Γ

= {γ ∈ Γ | γ > γ

}.

Let A be the group algebra of Γ over Z . It will be denoted exponentially: as a Z -module, it is free with basis (v

)

and the multiplication rule is given by v

v

= v

for all γ , γ

∈ Γ. If a ∈ A, we denote by a

the coefficient of a on v

, so that a = P

γ∈Γ

a

v

. If a 6= 0, we define the degree and the valuation of a (which we denote respectively by deg a and val a) as the elements of Γ equal to

deg a = max{γ | a

6= 0}

and val a = min{γ | a

6= 0}.

By convention, we set deg 0 = −∞ and val 0 = +∞. So deg : A → Γ ∪ {−∞} and val : A → Γ ∪ {+∞} satisfy deg ab = deg a + deg b and val ab = val a + val b for all a, b ∈ A.

We denote by A → A, a 7→ a ¯ the automorphism of A induced by the automorphism of Γ sending γ to −γ . Note that deg a = − val ¯ a. If γ

∈ Γ, we set

A

= ⊕

γ<γ₀

Z v

, A

= ⊕

γ6γ0

Z v

, A

= ⊕

γ>γ₀

Z v

and A

= ⊕

γ>γ0

Z v

.

We fix a weight function L : W → Γ, that is a function satisfying L(ww

) = L(w)+L(w

)

whenever ℓ(ww

) = ℓ(w)+ℓ(w

). We also assume that L(s) > 0 for every s ∈ S. We denote

by H = H(W, S, L) the Hecke algebra of W associated to the weight function L. It is the

associative A-algebra with A-basis (T

)

indexed by W and whose multiplication is determined by the following two conditions:

(a) T

T

= T

if ℓ(ww

) = ℓ(w) + ℓ(w

) (b) T

= 1 + (v

− v

)T

if s ∈ S.

It is easily seen from the above relations that (T

)

generates the A-algebra H and that T

is invertible for every w ∈ W . If h = P

w∈W

a

T

∈ H, we set h = P

w∈W

a

T

. Then the map H → H, h 7→ h is a semi-linear involutive automorphism of H. If I ⊂ S, we denote by H(W

) the sub-A-algebra of H generated by (T

)

.

Let w ∈ W . By [17, Theorem 5.2], there exists a unique element C

∈ H such that (a) C

= C

(b) C

∈ T

+

⊕

y∈W

A

T

. Write C

= P

y∈W

p

T

with p

∈ A. Then [17, 5.3]

p

= 1

p

= 0 if y 66 w.

In particular, (C

)

is an A-basis of H: it is called the Kazhdan-Lusztig basis of H.

Write now p

= v

p

. Then

p

∈ A

and the coefficient of p

on v

is equal to 1 (see [17, Proposition 5.4 (a)].

We define the relations 6

, 6

, 6

, ∼

, ∼

and ∼

as in [17, §8].

1.B. The function a. Let x, y ∈ W . Write C

C

= X

z∈W

h

C

, where h

∈ A for z ∈ W . Of course, we have

(1 . 1) h

= h

.

The following lemma is well-known [17, Lemma 10.4 (c) and formulas 13.1 (a) et (b)].

Lemma 1.2. Let x, y and z be three elements of W . Then deg h

6 min(L(x), L(y)).

Conjecture P

(Lusztig): There exists N ∈ Γ such that deg h

6 N for all x, y and z in W .

If W is finite, then W satisfies obviously P

. If W is an affine Weyl group, then it also satisfies P

[16, 7.2]. From now on, we assume that W satisfies P

, so that the next definition is valid. If z ∈ W , we set

a(z) = max

x,y∈W

deg h

.

Since h

= 1, we have a(z) ∈ Γ

. If necessary, we will write a

(z) for a(z). We denote by γ

∈ Z the coefficient of v

in h

. The next proposition shows how the function a can be calculated by using different bases.

Proposition 1.3. Let (X

)

and (Y

)

be two families of elements of H such that, for every w ∈ W , X

− T

and Y

− T

belong to ⊕

A

T

. For all x and y in W , write

X

Y

= X

z∈W

ξ

C

. Then, if x, y, z ∈ W , we have:

(a) deg ξ

6 min{L(x), L(y)}.

(b) ξ

∈ γ

v

+ A

. In particular,

a(z) = max

x,y∈W

deg ξ

. Proof - Clear.

1.C. Lusztig’s conjectures. Let τ : H → A be the A-linear map such that τ (T

) = δ

if w ∈ W . It is the canonical symmetrizing form on H (recall that τ (T

T

) = δ

). If z ∈ W , let

∆(z) = − deg p

= − deg τ (C

).

Let n

be the coefficient of p

on v

. Finally, let D = {z ∈ W | a(z) = ∆(z)}.

Conjectures (Lusztig): With the above notation, we have:

P

. If z ∈ W , then a(z) 6 ∆(z).

P

. If d ∈ D and if x, y ∈ W satisfy γ

6= 0, then x = y

.

P

. If y ∈ W , then there exists a unique d ∈ D such that γ

6= 0.

P

. If z

6

z, then a(z) 6 a(z

). Therefore, if z ∼

z

, then a(z) = a(z

).

P

. If d ∈ D and y ∈ W satisfy γ

6= 0, then γ

= n

= ±1.

P

. If d ∈ D, then d

= 1.

P

. If x, y, z ∈ W , then γ

= γ

.

P

. If x, y, z ∈ W satisfy γ

6= 0, then x ∼

y

, y ∼

z

and z ∼

x

. P

. If z

6

z and a(z

) = a(z), then z

∼

z.

P

. If z

6

z and a(z

) = a(z), then z

∼

z.

P

. If z

6

z and a(z

) = a(z), then z

∼

z.

P

. If I ⊂ S and z ∈ W

, then a

(z) = a

(z).

P

. Every left cell C of W contains a unique element d ∈ D. If y ∈ C, then γ

6= 0.

P

. If z ∈ W , then z ∼

z

.

P

. If x, x

, y, w ∈ W are such that a(y) = a(w), then X

y^′∈W

h

⊗

h

= X

y^′∈W

h

⊗

h

in A ⊗

A.

Lusztig has shown that these conjectures hold if W is a finite or affine Weyl group and L = ℓ [17, §15], if W is dihedral and L is any weight function [17, §17] and if (W, L) is quasi-split [17, §16].

1.D. Lusztig’s conjectures and multiplication by T

0

. We assume in this subsection that W is finite. We are interested here in certain properties of the multiplication by T

for n ∈ Z . Some of them are partially known [14, Lemma 1.11 and Remark 1.12]. If y ∈ W and n ∈ Z , we set

T

C

= X

x∈W

λ

C

.

Note that λ

= 0 if x 66

y.

Proposition 1.4. Assume that W is finite and satisfies Lusztig’s conjectures P

, P

and P

. Let n ∈ Z and let x and y be two elements of W such that x 6

y. Then:

(a) If n > 0, then deg λ

6 n(a(x) − a(w

x)). If moreover x <

y, then deg λ

<

n(a(x) − a(w

x)).

(b) If n 6 0, then deg λ

6 n(a(y) − a(w

y)). If moreover x <

y, then deg λ

<

n(a(y) − a(w

y)).

(c) If n is even and if x ∼

y, then λ

= δ

v

.

Proof - If n = 0, then (a), (b) and (c) are easily checked. Let us now prove (a) and (b).

By [17, Proposition 11.4],

T

= X

u∈W

(−1)

p

C

. Consequently,

λ

= X

u∈W x6_Ru

(−1)

p

h

.

But, by P

, we have deg p

6 −a(w

u). If moreover x 6

u, then w

u 6

w

x and so

−a(w

x) > −a(w

u) by P

. Therefore

deg λ

6 a(x) − a(w

x).

On the other hand, if deg λ

= a(x) − a(w

x), then there exists u ∈ W such that x 6

u

and deg h

= a(x). So, by P

, we get that x ∼

y. This shows (a) for n = 1.

Now, let ν : H → H denote the A-linear map such that ν(C

) = v

C

for all w ∈ W and let µ : H → H, h 7→ T

h. Then, if w ∈ W , we have

νµ(C

) = X

u6_Ly

v

λ

C

.

So, by the previous discussion, we have v

λ

∈ A

. Moreover, if u <

y, then v

λ

∈ A

. On the other hand, det µ = ±1 and det ν = 1. Therefore, if we write

µ

ν

(C

) = X

u6_Ly

β

C

, then β

∈ A

and, if u <

y, then β

∈ A

. Finally,

T

C

= µ

(C

)

= µ

ν

ν(C

)

= v

µ

ν

(C

)

= X

u6Ly

v

β

C

.

In other words, λ

= v

β

. This shows that (b) holds if n = −1. An elementary induction argument using P

shows that (a) and (b) hold in full generality.

Let us now prove (c). Let K be the field of fraction of A. Let C be a left cell of W and let c ∈ C. We set

H

= ⊕

w6Lc

AC

and H

= ⊕

w<Lc

AC

.

Then H

and H

are left ideals of H. The algebra KH = K ⊗

H being semisimple, there exists a left ideal I

of KH such that KH

= KH

⊕ I

.

We need to prove that, for all h ∈ I

,

T

h = v

h.

For this, we may, and we will, assume that n > 0. Let V

, V

,. . . , V

be irreducible sub-KH ⊗

K-modules of I

such that

I

= V

⊕ · · · ⊕ V

.

Let j ∈ {1, 2, . . . , n

}. Since T

is central and invertible in H, there exists ε ∈ {1, −1}

and i

∈ Γ such that

T

h = εv

h

for every h ∈ V

. By specializing v

7→ 1, we get that ε = 1. Moreover, by (a) and (b),

i

6 n(a(c) − a(w

c)). On the other hand, since det µ = ±1, we have det µ

= 1. But

det µ

= v

, where

r = X

C∈LC(W) nC

X

j=1

i

dim V

6 n X

C∈LC(W)

(a(C) − a(w

C))

n_C

X

j=1

dim V

= n X

C∈LC(W)

(a(C) − a(w

C))|C|

= n X

w∈W

(a(w) − a(w

w))

= 0.

Here, LC(W ) denotes the set of left cells in W and, if C ∈ LC(W ), a(C) denotes the value of a on C (according to P

). The fact that r = 0 forces the equality i

= n(a(C) −a(w

C)) for every left cell C and every j ∈ {1, 2, . . . , n

}.

Remark 1.5 - Assume here that w

is central in W and keep the notation of the proof of Proposition 1.4 (c). Let j ∈ {1, 2, . . . , n

}. Then there exists ε

(C) ∈ {1, −1} et e

(C) ∈ Γ such that T

h = ε

(C)v

h for every h ∈ V

.

Question: Let j, j

∈ {1, 2, . . . , n

}. Does ε

(C) = ε

(C) ?

A positive answer to this question would allow to generalize Proposition 1.4 (c) to the case where w

is central.

Corollary 1.6. Assume that W is finite and satisfies Lusztig’s conjectures P

, P

, P

, P

, P

and P

. Let w ∈ W and let n ∈ N . Then deg τ (T

C

) 6 −a(w) +n(a(w

w) − a(w)).

Moreover, deg τ (T

C

) = −a(w) + n(a(w

w) − a(w)) if and only if w

w

∈ D.

Proof - Assume first that n is even. In particular, w

= 1. By Proposition 1.4, we have τ (T

C

) = v

τ (C

) + X

x<Lw

λ

τ (C

).

But, if x <

w, then deg τ (C

) = −∆(x) 6 −a(x) 6 −a(w) by P

and P

. So, by Propo-

sition 1.4 (b), we have that deg λ

τ (C

) < −a(w) + n(a(w

w) −a(w)). Moreover, again

by P

, we have deg τ (C

) = −∆(w) 6 −a(w). This shows that deg τ (T

C

) 6 −a(w) +

n(a(w

w) − a(w)) and that equality holds if and only if ∆(w) = a(w), that is, if and only

if w ∈ D, as desired.

Assume now that n = 2k+1 for some natural number k. Recall that, by [17, Proposition 11.4], T

= P

u∈W

(−1)

p

C

. Therefore T

C

= X

u∈W

(−1)

p

T

C

C

= X

u,x∈W x6Lw and x6Ru

(−1)

p

h

T

C

.

This implies that

τ (T

C

) = X

u,x∈W x6_Lw andx6_Ru

(−1)

p

h

τ (T

C

),

so deg τ (T

C

) 6 max

u,x∈W x6Lw and x6Ru

deg(p

h

τ (T

C

)).

Let u and x be two elements of W such that x 6

w and x 6

u. Since n + 1 is even and by the previous discussion, we have

deg τ (T

C

) 6 −a(x) + (n + 1)(a(w

x) − a(x)).

By P

and P

, deg p

6 −a(w

u) 6 −a(w

x). Moreover, deg h

6 a(x). Conse- quently,

deg(p

h

τ (T

C

)) 6 −a(x) + n(a(w

x) − a(x)) 6 −a(w) + n(a(w

w) − a(w)).

Moreover, equality holds if and only if w

u ∈ D, deg h

= a(x) = a(w) and x ∈ D.

We first deduce that

deg τ (T

C

) 6 −a(w) + n(a(w

w) − a(w)) which is the first assertion of the proposition.

Assume now that deg τ (T

C

) = −a(w) + n(a(w

w) − a(w)). Then there exists u and x in W such that x 6

w, x 6

u, w

u ∈ D, deg h

= a(x) = a(w) and x ∈ D.

Since deg h

= a(x) and x ∈ D, we deduce from P

that w = u

, which shows that w

w

∈ D.

Conversely, assume that w

w

belongs to D. To show that deg τ (T

C

) = −a(w) +

n(a(w

w) −a(w)), it is sufficient to show that there is a unique pair (u, x) of elements of W

such that x 6

w, x 6

u, w

u ∈ D, deg h

= a(x) = a(w) and x ∈ D. The existence

follows from P

(take u = w

and x be the unique el´ement of D belonging to the left cell

containing w). Let us now show unicity. Let (u, x) be such a pair. Since deg h

= a(x)

and x ∈ D, we deduce from P

that u = w

. Moreover, since a(x) = a(w) and x 6

w,

we have x ∼

w by P

. But, by P

, x is the unique element of D belonging to the left

cell containing w.

2. Preliminaries on type B (asymptotic case) From now on, we are working under the following hypothesis.

Hypothesis and notation: We assume now that W = W

is of type B

, n > 1. We write S = S

= {t, s

, . . . , s

} as in [3, §2.1]: the Dynkin diagram of W

is given by

i i i

· · ·

t s

s

s

We also assume that Γ = Z

and that Γ is ordered lexicographically:

(a, b) 6(a

, b

) ⇐⇒ a < a

or (a = a

and b 6 b

).

We set V = v

and v = v

so that A = Z [V, V

, v, v

] is the Laurent polynomial ring in two algebraically independent indeterminates V and v.

If w ∈ W

, we denote by ℓ

(w) the number of occurences of t in a reduced expression of w. We set ℓ

(w) = ℓ(w) − ℓ

(w). Then ℓ

and ℓ

are weight functions and we assume that L = L

: W

→ Γ, w 7→ (ℓ

(w), ℓ

(w)). So H = H

= H(W

, S

, L

). We denote by S

the subgroup of W generated by {s

, . . . , s

}: it is isomorphic to the symmetric group of degree n.

We now recall some notation from [3, §2.1 and 4.1]. Let r

= t

= t and, if 1 6 i 6 n −1, let r

= s

r

and t

= s

t

s

. If 0 6 l 6 n, let a

= r

r

. . . r

. We denote by S

, W

, S

, W

the standard parabolic subgroups of W

generated by {s

, s

, . . . , s

}, {t, s

, s

, . . . , s

}, S

\ {t, s

} and S

\ {s

} respectively. The longest element of S

is denoted by σ

. Let

Y

= {a ∈ S

| ∀ σ ∈ S

, ℓ(aσ) > ℓ(σ)}.

If w ∈ W

is such that ℓ

(w) = l, then [3, §4.6] there exist unique a

, b

∈ Y

, σ

∈ S

such that w = a

a

σ

b

. Recall that ℓ(w) = ℓ(a

) + ℓ(a

) + ℓ(σ

) + ℓ(b

).

2.A. Some submodules of H . If l is a natural number such that 0 6 l 6 n, we set T

= ⊕

w∈Wn

ℓt(w)=l

AT

, T

= ⊕

w∈Wn

ℓt(w)6l

AT

, T

= ⊕

w∈Wn

ℓt(w)>l

AT

,

C

= ⊕

w∈Wn

ℓt(w)=l

AC

, C

= ⊕

w∈Wn

ℓt(w)6l

AC

and C

= ⊕

w∈Wn

ℓt(w)>l

AC

.

Let Π

: H

→ T

and Π

: H

→ C

be the natural projections (for ? ∈ {l, 6 l, > l}).

Proposition 2.1. Let l be a natural number such that 0 6 l 6 n. Then:

(a) T

and C

are sub-H( S

)-modules-H( S

) of H

. The maps Π

and Π

are mor- phisms of H( S

)-modules-H( S

).

(b) T

= C

.

(c) C

is a two-sided ideal of H

.

Proof - (a) follows from [3, Theorem 6.3 (b)]. (b) is clear. (c) follows from [3, Corollary 6.7].

The next proposition is a useful characterization of the elements of the two-sided ideal C

.

Proposition 2.2. Let h ∈ H

. The following are equivalent:

(1) h ∈ C

.

(2) ∀x ∈ H

, (T

− V )xh = 0.

(3) ∀x ∈ H( S

), (T

− V )xh = 0.

Proof - If ℓ

(w) = n, then tw < w so (T

− V )C

= 0 by [17, Theorem 6.6 (b)]. Since C

is a two-sided ideal of H

(see Proposition 2.1 (c)), we get that (1) implies (2). It is also obvious that (2) implies (3). It remains to show that (3) implies (1).

Let I = {h ∈ H

| ∀x ∈ H( S

), (T

− V )xh = 0}. Then I is clearly a sub-H( S

)- module-H

of H

. We need to show that I ⊂ C

. In other words, since C

⊂ I , we need to show that I ∩ C

= 0. Let I

= I ∩ C

= I ∩ T

(see Proposition 2.1 (b)).

Let X = {w ∈ W

| τ (I

T

) 6= 0}. Showing that I

= 0 is equivalent to showing that X = ∅ .

Assume X 6= ∅ . Let w be an element of X of maximal length and let h be an element of I

such that τ (hT

) 6= 0. Since h ∈ T

, we have ℓ

(w) 6 n − 1. Moreover, T

h = V h, so τ (T

hT

) = τ (hT

T

) 6= 0. By the maximality of ℓ(w), we get that tw < w. So, there exists s ∈ S

such that sw > w and s 6= t. Then

τ (T

hT

) = τ (hT

T

)

= τ (hT

) + (v − v

)τ (hT

) 6= 0,

the last inequality following from the maximality of ℓ(w) (which implies that τ (hT

) = 0). But T

h ∈ I

and so sw ∈ X. This contradicts the maximality of ℓ(w).

2.B. Some results on the Kazhdan-Lusztig basis. In this subsection, we study the elements of the Kazhdan-Lusztig basis of the form C

where 0 6 l 6 n and σ ∈ S

. Proposition 2.3. Let σ ∈ S

and let 0 6 l 6 n. Then C

C

= C

and C

C

= C

. Proof - Let C = C

C

. Then C = C and

C − T

= X

w<alσ

λ

T

with λ

∈ A for w < a

σ. To show that C = C

, it is sufficient to show that λ

∈ A

.

But, C

= T

+ P

x<al

V

β

T

with β

∈ Z [v, v

] (see [3, Theorem 6.3 (a)]).

Hence,

C = T

+ X

τ <σ

p

T

+ X

x<al

V

β

T

C

.

But, if x < a

, then ℓ

(x) < l. This shows that λ

∈ A

for every w < a

σ. This shows the first equality. The second one is obtained by a symmetric argument.

Proposition 2.3 shows that it can be useful to compute in different ways the elements C

to be able to relate the Kazhdan-Lusztig basis of H

to the Kazhdan-Lusztig basis of H( S

). Following the work of Dipper-James-Murphy [4], Ariki-Koike [2] and Graham- Lehrer [11, §5], we set

P

= (T

+ V

)(T

+ V

) . . . (T

+ V

)

= X

06k6l

V

X

16i1<···<ik6l

T

.

Lemma 2.4. P

is central in H

.

Proof - First, P

commutes with T

(indeed, tt

= t

t > t

for 1 6 i 6 n). By [2, Lemma 3.3], P

commutes with T

for 1 6 i 6 n − 1. Since the notation and conventions are somewhat different, we recall here a brief proof. First, if j 6∈ {i, i + 1}, s

t

= t

s

> t

so T

commutes with T

. Therefore, it is sufficient to show that T

commutes with (T

+ V

)(T

+ V

). This follows from a straightforward computation using the fact that s

t

> t

, that t

s

< t

and that s

t

= t

s

.

Proposition 2.5. If 0 6 l 6 n, then C

= P

T

l

= T

l

P

.

Proof - The computation may be performed in the subalgebra of H

generated by {T

, T

, . . . , T

} so we may, and we will, assume that l = n. First, we have T

P

= V P

. Since P

is central in H

, it follows from the characterization of C

given by Proposition 2.2 that P

∈ C

.

Now, let h = C

− P

T

. Then, by Proposition 2.1 (a), we have h ∈ C

. Moreover, it is easily checked that h ∈ T

= C

. So h = 0.

Corollary 2.6. If 0 6 l 6 n and σ ∈ S

, then Π

(C

) = V

T

l

C

. In particular, τ (C

) = V

τ (T

C

).

Proof - Since Π

is a morphism of right H( S

)-modules (see Proposition 2.1 (a)) and

since C

= P

T

C

(see Propositions 2.3 and 2.5), we have Π

(C

) = Π

(P

)T

C

.

But, Π

(P

) = V

. This completes the proof of the corollary.

3. Two-sided cells

The aim of this section is to show that, if x and y are two elements of W such that ℓ

(x) = ℓ

(y) = l, then x 6

y if and only if σ

6

σ

(see Theorem 3.5). Here, 6

is the preorder 6

defined inside the parabolic subgroup S

. For this, we adapt an argument of Geck [7] who was considering the preorder 6

.

We start by defining an order relation 4 on W . Let x and y be two elements of W . Then x ≺ y if the following conditions are fulfilled:

(1) ℓ

(x) = ℓ

(y), (2) x 6 y,

(3) a

< a

or b

< b

, (4) σ

6

σ

.

We write x 4 y if x ≺ y or x = y. If y ∈ W

, we set Γ

= T

C

C

T

y

.

Lemma 3.1. Let y ∈ W . Then Γ

∈ T

+ ⊕

A

T

.

Proof - First, recall that Γ

is a linear combination of elements of the form T

T

T

y

with z 6 a

σ

, so it is a linear combination of elements of the form T

with x 6 y.

Let l = ℓ

(y) and σ = σ

. We have Γ

= T

T

T

T

y

+ X

τ <σ

p

T

T

T

T

+ X

a<al, τ6σ

p

p

T

T

T

T

.

If τ < σ, then T

T

T

T

y

= T

ya_lτ b⁻_y¹

by [3, §4.6]. On the other hand, if a < a

, then ℓ

(a) < l so T

T

T

T

y

is a linear combination, with coefficients in Z [v, v

] of elements T

with ℓ

(w) = ℓ

(a) < l (because a

, τ and b

are elements of S

). Since V

p

∈ Z [v, v

] by [3, Theorem 6.3 (a)], this proves the lemma.

Lemma 3.2. If y ∈ W

, then

Γ

= Γ

+ X

x≺y

ρ

Γ

where the ρ

’s belong to Z [v, v

].

Proof - Let l = ℓ

(y). Then T

a⁻¹y

= T

+ X

a∈Yl,n−l

x∈S_l,n−l ax<ay

R

T

T

.