Representations of braid groups via conjugation actions on congruence subgroups

0.2 MB·English

by  Kevin P. Knudson

REPRESENTATIONS OF BRAID GROUPS VIA
CONJUGATION ACTIONS ON CONGRUENCE
4
0 SUBGROUPS
0
2
KEVINP. KNUDSON
n
a
J Abstract. We construct two families of representations of the braid

4 group Bn by considering conjugation actions on congruence subgroups
1 of GLn−1(Z[t±1,q±1]). Many of these representations are shown to be

faithful.
]
T
A
.
h 1. Introduction
t
a
Denote by B the braid group on n strings. The purpose of this note is
m n
to construct two families of representations
[
ρ (α) : B −→ SL (C)
2 n n n(n−2)
v
and
2 2
n
5 µ (α,β) :B −→ SL (C), N = −1,
n n N
0 (cid:18)2(cid:19)
1
where α and β are nonzero complex numbers. If α,β lie in some subfield F
0
4 of C, then the representations are defined over F (in fact, over Z[α±1,β±1]).
0 The starting point for ρ (α) is the reduced Burau representation β :
n n
h/ Bn → GLn−1(Z[t,t−1]). For each i ≥1, set
t
a Ki(α) = {A ∈ SL (C[t,t−1]) :A≡ I mod (t−α)i}.
n−1
m
Thesequence{Ki(α)} isacentralseriesinK(α) = K1(α)(i.e.,Ki+j(α) ⊇
: i≥1
v [Ki(α),Kj(α)]). Moreover, the graded quotients satisfy
i
X
Ki(α)/Ki+1(α) ∼= sl (C).
n−1
r
a The conjugation action of GL (C[t,t−1]) on K(α) induces a homomor-
n−1
phism
f (α) :GL (C[t,t−1]) −→ Aut(K(α)/K2(α)) ∼= GL (C)
n n−1 n(n−2)
(note that sl (C) is a vector space of dimension (n−1)2−1= n(n−2)).
n−1
We define
ρ (α) = f (α)◦β .
n n n
The kernel of ρ (1) is easily described: it is the subgroup P of pure
n n
braids. The map ρ (1) is thus a representation of the symmetric group Σ .
n n
1991 Mathematics Subject Classification. 20F36.
Partially supported by NSFgrant DMS-0242906, and by ORAU.
1
2 KEVINP.KNUDSON
Theorem2.2. Supposen ≥ 4. DenotebyV thestandard (n−1)-dimensional
representation of Σ and by W the representation corresponding to the par-
n
tition n−2,2. Then
2
ρ (1) ∼= V ⊕ V ⊕W.
n
^
Also, ρ (1) ∼= V ⊕ 2V.
3
V
Whenα 6= 1,however,theimageofρ (α)isinfinite. DenotebyΓ (α)
n n(n−2)
the subgroup of SL (Z[α±1]) consisting of matrices congruent to the
n(n−2)
identity modulo (α−1). Then we have the following result.
Proposition 2.1. The image of P under ρ (α) lies in Γ (α).
n n n(n−2)
For the representations µ (α,β), we begin with the Lawrence–Krammer–
n
Bigelow representation
κ :B −→ GL (Z[t±1,q±1]).
n n (n)
2
For each i ≥ 1, define a subgroup Li(α,β) by
Li(α,β) = {A ∈ SL (C[t±1,q±1]) :A ≡ I mod (t−α,q−β)i}.
(n)
2
The first graded quotient satisfies
L(α,β)/L2(α,β) ∼= sl(n)(C)×sl(n)(C).
2 2
The conjugation action of GL (C[t±1,q±1]) on L(α,β)/L2(α,β) is diago-
(n)
2
nal; that is, if (v,w) ∈L(α,β)/L2(α,β) then a matrix A acts as
A: (v,w) 7→ (AvA−1,AwA−1).
Let N = n 2−1 and let g (α,β) be the homomorphism
2 n
(cid:0) (cid:1) g (α,β) :GL (C[t±1,q±1]) −→ GL (C)
n (n) N
2
obtained by restricting the conjugation action to the first factor of sl (C).
(n)
2
Define µ (α,β) to be the composite
n
µ (α,β) = g (α,β)◦κ .
n n n
The maps µ (α,β) are more complicated than the ρ (α) mostly because
n n
the Burau matrices β (σ ) of the braid generators are block diagonal, while
n i
the matrices κ (σ ) are not. We content ourselves to analyze a few spe-
n i
cial cases. For example, the kernel of µ (1,1) is P and so µ (1,1) is a
n n n
representation of Σ . We give the decomposition of µ (1,1) for n= 3,4,5.
n n
In Section 4, we discuss the faithfulness of the maps ρ (α) and µ (α,β).
n n
Of course, none of them is faithful since the center of B lies in the kernel of
n
each. Moreover, since β is not faithful for n ≥ 5, there must be additional
n
elements in the kernel of ρ (α). However, the map κ is faithful for all n,
n n
and this allows us to deduce the following result.
Theorem 4.4. If α and β are algebraically independent, then the kernel of
the representation µ (α,β) is precisely the center of B .
n n
REPRESENTATIONS OF BRAID GROUPS 3
In the particular case of B , it is known [4] that β is faithful if and only
4 4
if the matrices β (σ σ−1) and β (σ σ σ−1σ−1) generate a freegroup of rank
4 3 1 4 2 3 1 2
2. We have the following result, which motivated the study of the ρ (α) in
n
the first place.
Theorem 4.1. There is a positive integer M such that for any m ≥ M, the
matrices β (σ σ−1)m and β (σ σ σ−1σ−1)m generate a free group of rank
4 3 1 4 2 3 1 2
2.
This was proved first by S. Moran [8], who showed also that one can take
M = 3. It was our hope that passing to the map ρ (α) would allow us to
4
showthat M = 1is possible. We show in Section 4that themethodof proof
fails for M = 1,2.
Acknowledgements. I thank Dan Cohen and Alex Suciu for inviting me to
participate in the Special Session on Arrangements in Topology and Alge-
braic Geometry at the AMS meeting in Baton Rouge, March, 2003. The
ensuing pressure to give a talk yielded these results.
2. The representations ρ (α)
n
Recall the reduced Burau representation β : B → GL (Z[t,t−1])
n n n−1
defined as follows. Let σ ,σ ,...,σ be the standard generators of B
1 2 n−1 n
and set
1
 ... 
−t 1 ··· 0
1 0 0
 0 1 ··· 0   
βn(σ1) =  ... ... , βn(σr)=  0t −0t 11 
   
 0 ··· 1   ... 
 
 1 
 
where the row containing t − t 1 in the matrix for σ is the rth row,
r
1 < r < n. The map β is not faithful for n ≥ 5 [2]. Let α ∈ C× and for
n
each i ≥ 1, define
Ki(α) = {A ∈ SL (C[t,t−1]) :A≡ I mod (t−α)i}.
n−1
OnecheckseasilythatK•(α)isadescendingcentralseriesinK(α) = K1(α).
If A ∈ Ki(α), we may write A = I +(t−α)iY mod (t−α)i+1, where Y is
a matrix with entries in C. Since detA = 1, we have trace(Y) = 0. Define
a map
π :Ki(α) −→ sl (C)
n−1
by π(A) = Y. This is easily seen to be a surjective homomorphism and the
kernel of π is the subgroup Ki+1(α). Thus, π induces an isomorphism
Ki(α)/Ki+1(α) ∼= sl (C).
n−1
4 KEVINP.KNUDSON
Now let i= 1. Consider the following matrices in K(α):
A = I +(t−α)e , i 6= j, 1 ≤ i,j ≤ n−1
ij ij
B = I +(t−α)e −(t−α)e −(t−α)e +(t−α)e ,
ii ii i+1,i+1 i,i+1 i+1,i
1 ≤ i≤ n−2
where e is the matrix having 1 in the i,j position and zeros elsewhere. As
ij
a basis of K(α)/K2(α), we choose the matrices A , 1 ≤ i,j ≤ n−1 and
ij
A = B A A−1 , 1 ≤ i ≤ n−2. Note that under the map π : K(α) →
ii ii i,i+1 i+1,i
sl (C), we have
n−1
π(A ) = e , and π(A ) = e −e .
ij ij ii ii i+1,i+1
The group GL (C[t,t−1]) acts on K(α) via conjugation and hence acts
n−1
on the quotient K(α)/K2(α). Denote by f (α) the map
n
f (α) :GL (C[t,t−1]) −→ Aut(K(α)/K2(α)) ∼= GL (C)
n n−1 n(n−2)
(note that dimsl (C) = (n−1)2−1 = n(n−2)). Restricting this action
n−1
to the β (σ ) gives a map
n i
ρ (α) : B −→ GL (C)
n n n(n−2)
(that is, ρ (α) = f (α)◦β ). The action of each σ on K(α)/K2(α) is given
n n n i
by the following formulæ:
σ : A 7→ A 3≤ i ≤ n−1
1 ij ij
2≤ j ≤ n−1
A 7→ A 3≤ i ≤ n−2
ii ii
A 7→ −αA 3≤ j ≤ n−1
1j 1j
A 7→ A +A 3≤ j ≤ n−1
2j 1j 2j
A 7→ −1A + 1A 3≤ i ≤ n−1
i1 α i1 α i2
A 7→ −αA
12 12
A 7→ 1A − 1A − 1A
21 α 12 α 21 α 11
A 7→ A −2A
11 11 12
A 7→ A +A
22 12 22
REPRESENTATIONS OF BRAID GROUPS 5
and for 2≤ k ≤ n−2 and i,j 6= k−1,k,k+1
σ : A 7→ A
k ij ij
A 7→ A
ii ii
A 7→ A
i,k−1 i,k−1
A 7→ A − 1A + 1A
ik i,k−1 α ik α i,k+1
A 7→ A
i,k+1 i,k+1
A 7→ A +αA
k−1,j k−1,j kj
A 7→ −αA
kj kj
A 7→ A +A
k+1,j kj k+1,j
A 7→ A +−1A + 1A +αA +A
k−1,k k−1,k−1 α k−1,k α k−1,k+1 k,k−1 k,k+1
A 7→ A +αA
k−1,k+1 k−1,k+1 k,k+1
A 7→ −αA
k,k−1 k,k−1
A 7→ −αA
k,k+1 k,k+1
A 7→ A +A
k+1,k−1 k,k−1 k+1,k−1
A 7→ −1A +A +A + 1A − 1A
k+1,k α kk k,k−1 k+1,k−1 α k,k+1 α k+1,k
A 7→ A −αA (k ≥ 3)
k−2,k−2 k−2,k−2 k,k−1
A 7→ A +2αA +A
k−1,k−1 k−1,k−1 k,k−1 k,k+1
A 7→ A −αA −2A
kk kk k,k−1 k,k+1
A 7→ A +A (k ≤ n−3)
k+1,k+1 k+1,k+1 k,k+1
and finally
σ : A 7→ A 1≤ i ≤ n−3
n−1 ij ij
1≤ j ≤ n−2
A 7→ A 1≤ i ≤ n−4
ii ii
A 7→ A +αA 1≤ j ≤ n−3
n−2,j n−2,j n−1,j
A 7→ −αA 1≤ j ≤ n−2
n−1,j n−1,j
A 7→ A − 1A 1≤ i ≤ n−3
i,n−1 i,n−2 α i,n−1
A 7→ −1A +αA +A
n−2,n−1 α n−2,n−1 n−1,n−2 n−2,n−2
A 7→ A −αA
n−3,n−3 n−3,n−3 n−1,n−2
A 7→ 2αA +A
n−2,n−2 n−1,n−2 n−2,n−2
Consider the particular case n = 3. We have dimρ (α) = 3. The 3-
3
dimensional representations of B were classified by Tuba and Wenzl [10]—
3
they are characterized uniquely be the eigenvalues of the matrices for σ
1
and σ . Using the basis e = −A , e = A , e = −αA , the matrices of
2 1 12 2 11 3 21
ρ (α)(σ ) and ρ (α)(σ ) are
3 1 3 2
−α 2 1 −1 0 0
α
ρ3(α)(σ1) =  0 1 1 , ρ3(α)(σ2)=  −1 1 0 .
0 0 −1 1 −2 −α
 α   
These correspond to the matrices in [10] with λ = −α, λ = 1 and λ =
1 2 3
−1. We have thus found a naturally occurring one-parameter family of the
α
abstract representations defined in [10].
6 KEVINP.KNUDSON
Observe that the map ρ (α) is defined over the ring Z[α,α−1]. Denote
n
by Γ (α) the subgroup of GL (Z[α,α−1]) consisting of matrices
n(n−2) n(n−2)
congruent to I modulo (α−1). Let P be the subgroup of pure braids.
n
Proposition 2.1. The image of P under ρ (α) lies in Γ (α).
n n n(n−2)
Proof. The group P is generated by the σ2 along with some conjugates of
n i
these. Since the σ are all conjugate in B , so are the σ2. Thus it suffices
i n i
to show that ρ (α)(σ2) ∈ Γ (α). But this is easy:
n 1 n(n−2)
σ2 : A 7→ A 3 ≤i,j ≤ n−1
1 ij ij
A 7→ A 3 ≤i ≤ n−2
ii ii
A 7→ (1+2(α−1)+(α−1)2)A 2 ≤j ≤ n−1
1,j 1,j
A 7→ (1−α)A +A 3 ≤j ≤ n−1
2,j 1,j 2,j
A 7→ (1+ 1−α + 1−α)A + α−1A 3 ≤i ≤ n−1
i,1 α α2 i,1 α2 i,2
A 7→ A 3 ≤i ≤ n−1
i,2 i,2
A 7→ −(1−α)2A +(1+ 1−α + 1−α)A + 1−αA
21 α 12 α α2 21 α2 11
A 7→ A −2(1−α)A
11 11 12
A 7→ (1−α)A +A
22 12 22
(cid:3)
In particular, if α = 1, then P ⊆ kerρ (1). (This also follows from the
n n
factthatβ (P ) ⊂ K(1)andhencetheβ (σ2)acttrivially onK(1)/K2(1).)
n n n i
Thereverse inclusion also holds sincethe action of any ρ (1)(σ ) is the same
n i
as the action of β (σ ) evaluated at t = 1, and this collection of matrices
n i
is a faithful representation of the symmetric group Σ . For example, when
n
n= 3 we obtain (using the basis e = A , e = A , e = A −A +A )
1 12 2 21 3 11 12 21
−1 0 0 0 0 −1
ρ3(1)(σ1) =  0 0 −1  ρ3(1)(σ2)=  0 −1 0 .
0 −1 0 −1 0 0
   
The character of this is
(1) (12) (123)
χ 3 −1 0
ρ3(1)
If V denotes the standard 2-dimensional representation of Σ , then χ =
3 ρ3(1)
χ +χ . Thus, ρ (1) ∼= V ⊕ 2V.
V 2V 3
For nV= 4 we have V
(1) (12) (123) (1234) (12)(34)
χ 8 0 −1 0 0
ρ4(1)
Denote by W the representation corresponding to the partition 2,2. Then
an easy check shows that χ = χ +χ +χ .
ρ4 V 2V W
The character of ρ5(1) is V
(1) (12) (123) (1234) (12345) (12)(34) (12)(345)
χ 15 3 0 −1 0 −1 0
ρ5(1)
REPRESENTATIONS OF BRAID GROUPS 7
Theorem2.2. Suppose n≥ 4. DenotebyV thestandard (n−1)-dimensional
representation of Σ and by W the representation corresponding to the par-
n
tition n−2,2. Then
2
ρ (1) ∼= V ⊕ V ⊕W.
n
^
Proof. First note that dimV = n−1, dim 2V = (n−1)(n−2) and dimW =
2
n(n−3) ([5], p. 50). Thus, V
2
(n−1)(n−2) n(n−3)
(n−1)+ + = n(n−2)
2 2
so that dimρ (1) = dim(V ⊕ 2V ⊕ W). Now if C , i = (i ,i ,...,i )
n i 1 2 n
denotes the conjugacy class inVΣn consisting of cycles with i1 1-cycles, i2
2-cycles, etc., then
(χ +χ +χ )(C ) = i (i −2)
V 2V W i 1 1
(this follows from [5], 4.15, pV. 51). Direct calculation shows that
χ ((12)) = (n−2)(n−4) = i (i −2)
ρn(1) 1 1
χ ((123)) = (n−3)(n−5) = i (i −2)
ρn(1) 1 1
χ ((12)(n−1,n)) = (n−4)(n−6) = i (i −2),
ρn(1) 1 1
etc. Thus, χ = χ +χ +χ and since V, 2V, and W are distinct
ρn(1) V 2V W
irreducible representations,Vthe result follows. V (cid:3)
3. The representations µ (α,β)
n
Let R = Z[t±1,q±1] be the ring of Laurent polynomials in t,q and let A
be the free R-module
A = Rx .
ij
1≤Mi<j≤n
The Lawrence–Krammer–Bigelow representation
κ :B −→ GL (Z[t±1,q±1])
n n (n)
2
is defined by
κ (σ ) : x 7→ tq2x
n k k,k+1 k,k+1
x 7→ (1−q)x +qx i < k
ik ik i,k+1
x 7→ x +tqk−i+1(q−1)x i < k
i,k+1 ik k,k+1
x 7→ tq(q−1)x +qx k+1 < j
kj k,k+1 k+1,j
x 7→ x +(1−q)x k+1 < j
k+1,j kj k+1,j
x 7→ x i < j < k or k+1 < i < j
ij ij
x 7→ x +tqk−i(q−1)2x i < k < k+1 < j.
ij ij k,k+1
The map κ is faithful for all n [3],[7]. This is the only known faithful
n
representation of B , n ≥ 4.
n
For each i≥ 1 define, for α,β ∈C×,
Li(α,β) = {A ∈ SL (C[t±1,q±1]): A≡ I mod (t−α,q−β)i}
(n)
2
8 KEVINP.KNUDSON
(here, (t−α,q−β) denotes the ideal generated by t−α and q−β). This is
a descending central series in L(α,β) = L1(α,β). The graded quotients
are more complicated here, but we are interested only in the first one:
L(α,β)/L2(α,β). If A ∈ L(α,β), write
A = I +(t−α)A +(q−β)A +X,
t q
where A ,A are n × n matrices over C and X ≡ 0 mod (t−α,q−β)2.
t q 2 2
Again, the condit(cid:0)ion(cid:1) de(cid:0)tA(cid:1) = 1 forces tr(At) = 0= tr(Aq). Define a map
π : L(α,β) −→ sl (C)×sl (C)
(n) (n)
2 2
by π(A) = (A ,A ). This is a surjective group homomorphism with kernel
t q
L2(α,β).
IfZ ∈ GL (C[t±1,q±1]), wemaywriteZ = Z Z ,whereZ ∈GL (C)
(n) 0 1 0 (n)
2 2
and Z ∈L(α,β). Then Z acts trivially on L(α,β)/L2(α,β) and so Z acts
1 1
on L(α,β)/L2(α,β) by
Z :(A ,A ) 7→ (Z A Z−1,Z A Z−1);
t q 0 t 0 0 q 0
that is, the action is diagonal. Consider the action on the first factor (call
it L):
gn(α,β) : GL(n)(C[t±1,q±1]) −→ Aut(L)∼= GLN(C)
2
where N = n 2−1. Restricting this to the image of B under κ yields a
2 n n
map (cid:0) (cid:1)
µ (α,β) : B −→ GL (C)
n n N
(i.e., µ (α,β) = g (α,β)◦κ ).
n n n
We shall not write down a formula for the µ (α,β) in general. Indeed,
n
the following formula for µ (α,β) shows that the general case is hopelessly
3
complicated. Using the basis described in (1) and (2) below, the matrix of
µ (α,β)(σ ) is
3 1
αβ(β−1) αβ2 −αβ(β−1)3 αβ(β−1) 0 0 −2α(β−1)2 α(β−1)2
 αβ 0 −αβ(β−1)2 0 0 0 −2α(β−1) α(β−1) 
 0 0 0 0 βα2 0 0 0 
 00 00 α01β 00 −−(ββα−−ββ211) β01 00 00 
 00 00 −1(β−−β1β1)2 β0 (β−β021)3 −(β0−β1)2 1−1β 2(β0−1) 
 0 0 1−β1 0 −(β−β21)2 1−β1 1 −1 
and that of µ (α,β)(σ ) is
3 2
−(β−β1)2 (β−β1)3 β1 −(β−β1)2 0 0 −2(ββ−1) β−β1
 0 −βα−β21 0 αβ12 0 0 0 0 
 β −β(β−1) 0 0 0 0 0 0 
 00 α01β 00 −αβ2(0β−1)2 00 α0β2 −αβ2(0β−1) 2αβ2(0β−1) .
 0 0 αβ(β−1) −αβ(β−1)3 αβ αβ(β−1) −αβ(β−1)2 2αβ(β−1)2 
 1−β (β−1)2 0 1−β 0 0 −1 1 
 0 0 0 1−β 0 0 0 1 
REPRESENTATIONS OF BRAID GROUPS 9
Note however that µ (α,β) is defined over Z[α±1,β±1] and we have the
n
following result. Denote by Γ (α,β) the subgroup of GL (Z[α±1,β±1])
N N
consisting of those matrices that are congruent to I modulo (α−1,β −1).
Proposition 3.1. The image of P under µ (α,β) lies in Γ (α,β).
n n N
Proof. It suffices to check this for the µ (α,β)(σ2). First note the following
n k
formula for κ (σ2):
n k
x 7→ t2q4x
k,k+1 k,k+1
x 7→ (1+(q−1)+(q−1)2)x +(1−q)qx +tqk−i+1(q−1)x
ik ik i,k+1 k,k+1
x 7→ (1−q)x +(1+(q−1))x +t2qk−i+3(q−1)x
i,k+1 ik i,k+1 k,k+1
x 7→ t2q3(q−1)x +(1+(q−1))x +q(1−q)x
kj k,k+1 kj k+1,j
x 7→ tq(q−1)x +(1+(q−1)+(q−1)2)x +(1−q)x
k+1,j k,k+1 k+1,j kj
x 7→ x +tqk−i(q−1)2(1+tq2)x
ij ij k,k+1
where the ranges on i,j are (i < k), (i < k), (k + 1 < j), (k + 1 < j),
(i < j < k or k+1 < i < j), and (i < k < k +1 < j), respectively. Note
that t2q4 ≡ 1+2(t−1)+4(q−1) mod (t−1,q−1)2. Thus,after evaluating
κ (σ2) at t = α, q = β, we may write
n k
κ (σ2) = X Y ,
n k k k
where X ∈ L(α,β) and Y ≡ I mod (α−1,β −1). The action of κ (σ2)
k k n k
on L is via conjugation by Y . Write
k
Y = I +(α−1)Y +(β −1)Y +Z
k α β
where Z ≡ 0 mod (α−1,β −1)2. Then if A = I +(t−α)A +X (X ≡ 0
t
mod (t−α,q−β)2), we have
YkAYk−1 = (1+(α−1)Yα+(β−1)Yβ+Z)(I+(t−α)At+X)(I−(α−1)Yα−(β−1)Yβ+U)
≡ I+(t−α)At+(α−1)(t−α)[Yα,At]+(β−1)(t−α)[Yβ,At] mod(t−α,q−β)2
≡ A mod(α−1,β−1).
Thus, κ (σ2) acts as the identity on L modulo (α − 1,β − 1); that is,
n k
µ (α,β)(σ2) ∈ Γ (α,β). (cid:3)
n k N
Corollary 3.2. The kernel of µ (1,1) is the subgroup P and so µ (1,1) is
n n n
a representation of Σ . (cid:3)
n
As a basis of L we take the matrices
n
(1) A = I +(t−α)e 1≤ i,j ≤ ,i 6= j,
ij ij
(cid:18)2(cid:19)
and
n
(2) A ≡ I +(t−α)e −(t−α)e 1 ≤ i ≤ −1,
ii ii i+1,i+1
(cid:18)2(cid:19)
ordered as A ,A ,...,A ,A ,...,A ,A ,A ,.... It is possible to
12 13 1,n 21 n,n−1 11 22
give a simple formula for µ (1,1). Note that upon evaluating κ (σ ) at
n n k
10 KEVINP.KNUDSON
t = 1,q = 1, one obtains the permutation matrix
τ : x 7→ x
k k,k+1 k,k+1
x 7→ x i < k
ik i,k+1
x 7→ x i < k
i,k+1 ik
x 7→ x k+1 < j
kj k+1,j
x 7→ x k+1 < j
k+1,j kj
x 7→ x i < j < k,k+1 < i <j,i < k < k+1< j.
ij ij
Order the basis of A = Rx as
ij
1≤Mi<j≤n
e = x ,e = x ,...,e = x ,e = x ,...,e = x
1 12 2 13 n−1 1,n n 23 (n) n−1,n
2
and consider τ as a permutation of the set {1,2,..., n }. The action of
k 2
µn(1,1) may then be described as (cid:0) (cid:1)
A 7→ A i 6= j
ij τk(i),τk(j)


 τk(i+1)−1

µn(1,1)(σk): Aii 7→  ℓ=Xττkk((ii))−1Aℓℓ τk(i) < τk(i+1)

 − Aℓℓ τk(i+1) < τk(i).
The characters of µ (1,1) forℓ=nτXk=(i+31,)4,5 are as follows:
n
(1) (12) (123)
χ 8 0 −1
µ3(1,1)
(1) (12) (123) (1234) (12)(34)
χ 35 3 −1 3 −1
µ4(1,1)
(1) (12) (123) (1234) (12345) (12)(34) (12)(345)
χ 99 15 0 −1 −1 3 0
µ5(1,1)
A straightforward calculation then shows the following.
µ (1,1) ∼= (alt)⊕(triv)⊕V⊕3
3
2
µ (1,1) ∼= (alt)⊕(triv)⊕2⊕( V)⊕3⊕W⊕4⊕V⊕5
4
^
2
µ (1,1) ∼= (V ⊗(alt))⊕(triv)⊕2⊕(W ⊗(alt))⊕3⊕( V)⊕4⊕W⊕6⊕V⊕6
5
^
(recall that V is the standard (n−1)-dimensional representation and W is
the representation corresponding to the partition n−2,2).
Note that the A , 1 ≤ i ≤ n −1 form a Σ -submodule of L.
ii 2 n
(cid:0) (cid:1)
Proposition 3.3. The submodule U spanned by the A , 1≤ i ≤ n −1, is
ii 2
isomorphic to V ⊕W. (cid:0) (cid:1)