Generalized Schur Algebras

0.25 MB·

by  Robert D. May

Generalized Schur algebras
6
1 Robert May
0
Department of Mathematics and Computer Science
2
Longwood University
n
a 201 High Street, Farmville, Va. 23909
J [email protected]

7

05/10/2014
]
A
R
. 1 Introduction
h
t
a
In [3] and [1] “generalizedSchur algebras”,B(n,r), were defined corresponding
m
to a family of monoids S which includes the full transformation semigroup
r
[
τr =Tr, the partial transformation semigroup τ¯r =PTr, and the rook monoid

1 ℜ ofsizer. Inmanycasesaparameterizationoftheirreduciblerepresentations
r
v of the algebra B(n,r) was obtained. In [2] these algebras were redefined as
1
the “right generalized Schur algebras” , B , corresponding to certain “double
1 R
coset algebras”. Corresponding “left generalized Schur algebras” , B , were
7 L
1 also defined in [2]. In this paper we use the approach in [1] to study both
0 the right and left algebras. The algebras and their multiplication rules are
. described in sections 2 through 4. In section 5, we obtain filtrations of these
1
0 algebras. These lead, in most cases, to parameterizations of the irreducible
6 representations for both B and B for a field of characteristic 0 (in section 6)
R L
1 or positive characteristic p (in sections 7 and 8).
:
v
i
X 2 The semigroups S and the double coset alge-
r
r
a bra A(S ,G)
r
Letτ¯ bethesetofallmapsα:{0,1,...,r}→{0,1,...,r} such that α(0)=0.
r
τ¯ is a semigroup under composition and is isomorphic to the partial transfor-
r
mationsemigroupPT . Letτ bethefulltransformationsemigroupandS the
r r r
symmetric groupon {1,2,...,r}. Any α in τ or S canbe extended to α¯ ∈τ¯
r r r
by defining α¯(0) = 0, so we can regard S and τ as subsemigroups of τ¯ . We
r r r
will let S represent any subsemigroup of τ¯ which contains S . For example,
r r r
S couldbe the rook semigroup, ℜ = α∈τ¯ :∀i∈{1,2,...r}, α−1(i) 61 .
r r r
Our main examples for S will be S ,τ ,ℜ , and τ¯ . Note that S = τ ∩
r r(cid:8) r r r (cid:12) r (cid:12) r(cid:9)
ℜ and τ¯ =ℜ ·τ . (cid:12) (cid:12)
r r r r
1
EachS canbeidentifiedwithacertainsemigroupofmatrices. LetM (Z)
r r+1
be the set of all (r+1)×(r+1) matrices with entries in Z (considered as a
semigroup under matrix multiplication). For convenience, label the rows and
columns for each m ∈ M (Z) from 0 to r. To each α ∈ S assign a matrix
r+1 r
1 if i=α(j)
m(α) ∈ M (Z) by setting m(α) = . Note that column j
r+1 i,j 0 otherwise
(cid:26)
of m(α) contains exactly one non-zero entry, namely a 1 in row α(j). The 1
in column 0 is always in row 0. It is not hard to check that α 7→ m(α) gives
an injective semigroup homomorphism S → M (Z), so we will identify S
r r+1 r
with its image in M (Z). For α ∈ S we will usually write just α for the
r+1 r
corresponding matrix m(α) . If α ∈ R , then m(α) has at most one 1 in rows
r
1,2,...,r. If α ∈ τ , then m(α) has only one 1 in row 0 (in column 0). If
r
α∈S ,thenm(α)istheusualpermutationmatrixborderedwithanextrarow
r
0 and column 0.
LetΛ(r,n)bethesetofallcompositionsofrwithnparts. Soforλ∈Λ(r,n)
n
we have λ= {λ ∈Z:i=1,2,...,n}, λ > 0, λ =r. For each λ∈Λ(r,n)
i i i
i=1
define “λ-blocks” of integers, bλ, and a “YoungPsubgroup”, S ⊆ S ⊆ S as
i λ r r
follows. First, let
bλ ={k ∈Z:0<k 6λ }and
1 1
bλi ={k ∈Z:λ1+λ2+···+λi−1 <k 6λ1+λ2+···+λi}for 1<i6n.
Thus bλ consists of λ consecutive integers and bλ = ∅ ⇔ λ = 0. Now let
i i i i
S bλ ⊆ S be the group of all permutations of bλ, so S bλ ∼= S . (Put
i r i i λi
S bλ = {identity element in S } when bλ = ∅.) Finally, define the Young
(cid:0) i(cid:1) r i (cid:0) (cid:1)
r r
sub(cid:0)gro(cid:1)up by Sλ = S bλi ∼= Sλi. Notice that each Sλ is generated by
i=1 i=1
{si :si ∈Sλ}whereQsi ∈(cid:0)Sr(cid:1)istheQelementarytranspositionwhichinterchanges
i and i+1. We often write G for the Young subgroup S .
λ λ
For λ,µ ∈ Λ(n,r), write M for the set of left S -cosets of the form
λ λ
S α, α ∈ S , M for the set of right S -cosets αS , α ∈ S , and M for
λ r µ µ µ r λ µ
the set of all double cosets S αS , α ∈ S . Write D(λ,α,µ) for the double
λ µ r
coset S αS . We may write just D for the double coset S αS when λ,µ
λ µ α λ µ
are clear.
Let G={G =S :λ∈Λ(r,n)}, a family of subgroups of the monoid S .
λ λ r
The double coset algebra A = A(S ,G) was defined in [2]. A is a complex
r
vectorspacewithbasis{X(λ,D,µ) : λ,µ∈Λ(n,r) , D ∈ M }. Theproduct
λ µ
in A is given by
0, if ν 6=ω;
X(λ,D ,ν)·X(ω,D ,µ)=
1 2  a(λ,µ,D1,D2,D)X(λ,D,µ), if ν =ω.
D∈λMµ
P
If D =D(λ,m,µ), then a(λ,µ,D ,D ,D) is the nonnegative integer
1 2
a(λ,µ,D ,D ,D)=#{(m ,m )∈D ×D :m m =m}.
1 2 1 2 1 2 1 2
2
As shown in [2], there is an alternative formula for the structure constants
a(λ,µ,D ,D ,D). ForanydoublecosetD ∈ M ,theproductgroupG ×G
1 2 λ µ λ µ
acts transitively on D by (σ ×π)(m) = σmπ−1 for m ∈ D,σ ∈ G ,π ∈ G .
λ µ
Thenfor any m∈D, let n(D)be the orderofthe subgroupG ofG ×G
λ,µ,m λ µ
which leaves m fixed: n(D) = # (σ×π):σ ∈G ,π ∈G ,σmπ−1 =m . It
λ µ
is easy to check that G , G are conjugate subgroups whenever m,n ∈
λ,µ,m λ,µ,n
(cid:8) (cid:9)
D, so the order n(D) is independent of the choice of m ∈ D. For D ∈
m
M ,D ∈ M ,D ∈ M , put N(D ,D ,D) = #{ρ∈G :mρn∈D}
λ ν n ν µ λ µ m n ν
(whichis independent ofthe choicesofm,n). Finally,let o(G ) be the orderof
ν
the group G . Then the structure constants are given by
ν
o(G )N(D ,D ,D)n(D)
ν m n
a(λ,µ,D ,D ,D)= .
m n
n(D )n(D )
m n
When S = S , the algebra A is isomorphic to the classical complex Schur
r r
algebra S(r,n), so for general S we call A the generalized Schur algebra asso-
r
ciated with S .
r
3 The left and right Z-forms for A
In [2], the left and right Z-forms for the algebra A = A(S ,G) were defined.
r
These algebras, denoted here by A , A , are Z-algebras with unit such that
L R
there are isomorphisms of complex algebrasAL⊗ZC∼=A∼=AR⊗ZC . Eachof
A , A isafreeZ-modulewithabasis{f(λ,D,µ) : λ,µ∈Λ(n,r) , D ∈ M }
L R λ µ
corresponding to the collection of all double cosets. We now describe the (in-
teger) structure constants for these algebras as given in [2]. For D ∈ M ,
λ µ
note that G acts transitively on the set of left cosets C ∈ M,C ⊆ D. Let
µ λ
G (λ,C,µ) = {π ∈G :Cπ =C} be the subgroup of G which fixes a given
L µ µ
C. If C = G α, then G (λ,C,µ) = {π ∈G :∃σ ∈G s.t. απ =σα} and
λ L µ λ
we write G (λ,α,µ) = G (λ,C,µ) ⊆ G . For any two left cosets C ,C ∈
L L µ 1 2
M, C ,C ⊆D the subgroups G (C ),G (C ) are conjugate, so we can de-
λ 1 2 L 1 L 2
finen (D)tobetheorderofG (λ,C,µ)foranysuchC. IfD =D =G αG ,
L L α λ µ
wewriten (λ,α,µ)=n (D). Similarly,foranyrightcosetC =αG ⊆D ,let
L L µ
G (λ,C,µ) = {σ ∈G :σC =C} = {σ ∈G :∃π ∈G s.t. σα=απ} be the
R λ λ µ
subgroup of G which fixes C and let n (D) be the order of G (λ,C,µ) (in-
λ R R
dependent of the choice of C). We also write G (λ,α,µ)=G (λ,C,µ)⊆G .
R R λ
If D = D = G αG , we write n (λ,α,µ) = n (D). Let ∗ ,∗ be the prod-
α λ µ R R L R
ucts in A ,A respectively. Write f(λ,α,µ) for the basis element f(λ,D ,µ).
L R α
Then from [2] we have the following
Multiplication Rules:
f(λ,α,ν)∗ f(ω,β,µ)=f(λ,α,ν)∗ f(ω,β,µ)=0 if ν 6=ω
R L
n (D)N(D ,D ,D)
R α β
f(λ,α,ν)∗ f(ν,β,µ)= f(λ,D,µ)
R
n (λ,α,ν)n (ν,β,µ)
D∈XλMµ R R
3
n (D)N(D ,D ,D)
L α β
f(λ,α,ν)∗ f(ν,β,µ)= f(λ,D,µ)
L
n (λ,α,ν)n (ν,β,µ)
D∈XλMµ L L
Asshownin[2],thestructureconstantsappearinginthemultiplicationrules
are always nonnegative integers.
An alternative form of the multiplication rules is sometimes useful.
Lemma 3.1. (a) if ρ ,ρ are in the same right G (ν,β,µ)-coset of G , then
1 2 R ν
for any α∈S , αρ β and αρ β are in the same G −G double coset.
r 1 2 λ µ
(b) if ρ ,ρ are in the same left G (λ,α,ν)-coset of G , then for any β ∈ S ,
1 2 L ν r
αρ β and αρ β are in the same G −G double coset.
1 2 λ µ
Proof. a)Supposeρ =ρ κforsomeκ∈G (ν,β,µ). Thenthereexistsπ ∈G
2 1 R µ
such that κβ =βπ. Then αρ β =αρ κβ =αρ βπ ∈G (αρ β)G .
2 1 1 λ 1 µ
b) Suppose ρ = κρ for some κ ∈ G (λ,α,ν). Then there exists σ ∈ G
2 1 L λ
such that σα=ακ. Then αρ β =ακρ β =σαρ β ∈G (αρ β)G .
2 1 1 λ 1 µ
It follows that for any right G (ν,β,µ) coset, any D ∈ M , and any
R λ µ
α ∈ S , either αρβ ∈ D for every ρ in the coset or αρβ ∈/ D for any ρ in the
r
coset (and similarly for left G (λ,α,ν) cosets) .
L
Proposition 3.1. Let D ∈ M . Then
λ µ
N(D(λ,α,ν),D(ν,β,µ),D)=a n (ν,β,µ)=a n (λ,α,ν)
R,D R L,D L
where a = the number of distinct right cosets ρG (ν,β,µ) in G such that
R,D R ν
αρβ ∈ D, and a = the number of distinct left cosets G (λ,α,ν)ρ in G
L,D L ν
such that αρβ ∈D.
Proof. AllrightcosetsρG (ν,β,µ)haveo(G (ν,β,µ))=n (ν,β,µ)elements.
R R R
So
N(D(λ,α,ν),D(ν,β,µ),D)=#{ρ∈G :αρβ ∈D}=a n (ν,β,µ).
ν R,D R
Similarly, all left cosets G (λ,α,ν)ρ have n (λ,α,ν) elements, so
L L
N(D(λ,α,ν),D(ν,β,µ),D)=#{ρ∈G :αρβ ∈D}=a n (λ,α,ν).
ν L,D L
Substituting into the multiplication rules gives the following
Alternative forms of multiplication rules:
n (D) a
R R,D
f(λ,α,ν)∗ f(ν,β,µ)= f(λ,D,µ)
R
n (λ,α,ν)
D∈XλMµ R
4
n (D) a
L L,D
f(λ,α,ν)∗ f(ν,β,µ)= f(λ,D,µ).
L
n (ν,β,µ)
D∈XλMµ L
Some special cases of the multiplication rules will also be useful.
1, for 16i6r,
Case 1: ν =
i
(0, for i>r.
Then G = {id} and n (ν,β,µ) = 1 = n (λ,α,ν) for any λ,µ ∈ Λ(r,n) and
ν R L
α,β ∈S . Also
r
1, if D =G αβG ;
λ µ
N(D(λ,α,ν),D(ν,β,µ),D)=
(0, otherwise.
Then substituting in the multiplication rule gives
n (D(λ,αβ,µ))
R
f(λ,α,ν)∗ f(ν,β,µ)= f(λ,αβ,µ)
R
n (D(λ,α,ν))
R
For α=1 (the identity in S ), n (λ,1,ν)=1 giving
r R
f(λ,1,ν)∗ f(ν,β,µ)=n (D(λ,β,µ))f(λ,β,µ)
R R
If in addition λ=ν, then writing 1 =f(ν,1,ν) gives
ν
1 ∗ f(ν,β,µ)=f(ν,β,µ).
ν R
There are similar rules for ∗ :
L
n (D(λ,αβ,µ))
L
f(λ,α,ν)∗ f(ν,β,µ)= f(λ,αβ,µ)
L
n (D(λ,α,ν))
L
f(λ,α,ν)∗ f(ν,1,µ)=n (D(λ,α,µ))f(λ,α,µ)
L L
f(λ,α,ν)∗ 1 =f(λ,α,ν).
L ν
Case 2: α=1 (the identity in S ) and G ⊆G .
r ν λ
Then for any ρ∈G , we have αρβ =ρβ ∈G βG , so
ν λ µ
o(G ), if D =G βG ;
ν λ µ
N(D(λ,1,ν),D(ν,β,µ),D)=
(0, otherwise.
Also G (λ,1,ν)=G =G (λ,1,ν), so
R ν L
n (D(λ,1,ν))=o(G )=n (D(λ,1,ν)).
R ν L
5
Then substitution in the multiplication rule gives:
n (D(λ,β,µ))
R
f(λ,1,ν)∗ f(ν,β,µ)= f(λ,β,µ).
R
n (D(ν,β,µ))
R
In particular, if λ=ν and 1 =f(ν,1,ν) then
ν
1 ∗ f(ν,β,µ)=f(ν,β,µ).
ν R
Similarly,
n (D(λ,β,µ))
L
f(λ,1,ν)∗ f(ν,β,µ)= f(λ,β,µ)
L
n (D(ν,β,µ))
L
1 ∗ f(ν,β,µ)=f(ν,β,µ).
ν L
Case 3: β =1 and G ⊆G .
ν µ
Then for any ρ∈G , we have αρβ =αρ∈G αG , so
ν λ µ
o(G ), if D =G αG ;
ν λ µ
N(D(λ,α,ν),D(ν,1,µ),D)=
(0 otherwise.
Also G (ν,1,µ)=G =G (ν,1,µ), so
R ν L
n (D(ν,1,µ))=o(G )=n (D(ν,1,µ)).
R ν L
Then substitution in the multiplication rule gives:
n (D(λ,α,µ))
R
f(λ,α,ν)∗ f(ν,1,µ)= f(λ,α,µ)
R
n (D(λ,α,ν))
R
In particular, if µ=ν and 1 =f(ν,1,ν) then
ν
f(λ,α,ν)∗ 1 =f(λ,α,ν).
R ν
Similarly,
n (D(λ,α,µ))
L
f(λ,α,ν)∗ f(ν,1,µ)= f(λ,α,µ)
L
n (D(λ,α,ν))
L
f(λ,α,ν)∗ 1 =f(λ,α,ν).
L ν
Case 4: α=β =1, λ=µ, G ⊆G =G .
ν λ µ
As a special case of either case 2 or case 3,
o(G )
λ
f(λ,1,ν)∗ f(ν,1,λ)= f(λ,1,λ)=f(λ,1,ν)∗ f(ν,1,λ).
R L
o(G )
ν
6
In particular, when ν =λ=µ,
1 ∗ 1 =1 =1 ∗ 1 .
ν R ν ν ν L ν
So in both A and A , each1 =f(ν,1,ν) is anidempotent. In fact for either
R L ν
algebra, cases 2 and 3 can be used to check that {1 :ν ∈Λ(r,n)} is a family
ν
of orthogonal idempotents with ⊕ 1 =1 (the identity).
ν
ν∈Λ(r,n)
4 The left and right generalized Schur algebras
Let R be a commutative domain with unit 1 and let ψ :Z→R be the natural
ring homomophism such that ψ(1) = 1. Regard R as a right Z-module via
ψ and form the tensor products LGSR = R⊗ZAL and RGSR = R⊗ZAR
which we call the left and right generalized Schur algebras over R (for the
monoid S ). (As in [2], for S =S , LGS is isomorphic to the classical Schur
r r r R
algebra over R, S (r,n), while RGS is isomorphic to the opposite algebra
R R
S (r,n)op.) The R algebras LGS , RGS are both free as R modules with
R R R
bases {1⊗f(λ,D,µ):λ,µ∈Λ(r,n),D ∈ M }. We will often denote a basis
λ µ
element 1⊗ f(λ,D,µ) by just f(λ,D,µ). Then all the multiplication rules
given in section 3 apply if we identify each coefficient a ∈ Z occurring with
its image ψ(a) ∈ R. As shown in [2], the isomorphism A ∼= C⊗AL ≡ LGSC
is given by matching basis elements X(λ,D,µ) ↔ 1⊗f(λ,D,µ), where n (D)
nL(D) L
is the number of elements in any left G coset C ⊆ D ∈ M . Similarly,
λ λ µ
the isomorphism A ∼= C⊗AR ≡ RGSC is given by matching basis elements
X(λ,D,µ) ↔1⊗f(λ,D,µ),wheren (D)isthe numberofelementsinanyright
nR(D) R
G coset C ⊆D ∈ M .
µ λ µ
5 Filtrations of generalized Schur algebras and
irreducible representations
Let Λ=Λ(r,n) be the set of all compositions of r into n parts; Λ+ =Λ+(r)⊆
Λ, the set of all partitions of r. For λ ∈ Λ and 0 6 k 6 r, put L(λ,k) =
#{i:λ =k} (= number of rows of length k in λ ). Define an equivalence
i
relation on Λ by λ ∼ λ ⇔ ∀i,L(λ ,i) = L(λ ,i). Notice that for each
1 2 1 2
λ ∈ Λ there is exactly one partition λ+ ∈ Λ+ with λ ∼ λ+. Let < be the
“lexicographic” order on Λ+: For λ,µ ∈ Λ+, λ < µ ⇔ ∃k s.t. λ = µ for i <
i i
k while λ < µ . Then < gives a total ordering of the partitions or of the
k k
equivalence classes of compositions. It extends to a partial order on Λ: for
λ,µ ∈ Λ, λ 6 µ ⇔ λ ∼ µ or λ+ < µ+ where λ+,µ+ ∈ Λ+ are the partitions
corresponding to λ,µ. (This partial order differs slightly from that used in [2]
.) The smallest partition, ν¯ = 1(r), has ν¯ = 1 for all i. The largest partition,
i
λ=(r), has λ =r,λ =0,i>1.
1 i
7
Write B for either of the R-algebras LGS or RGS . As mentioned above,
R R
{1 =f(λ,1,λ):λ∈Λ}isasetoforthogonalidempotentsinBwith ⊕ 1 =1
λ λ
λ∈Λ
(the identity in B). For each partition λ ∈ Λ+ define e = ⊕ 1 and
λ µ
µ∈Λ,µ<λ
e¯ = e ⊕ ⊕ 1 . (For the smallest partition ν¯ = 1(r) define e = 0.)
λ λ µ ν¯
µ∈Λ,µ∼λ
(cid:18) (cid:19)
Note the following results: each e and e¯ is an idempotent; for the largest
λ λ
partition, (r), we have e¯ = 1 (the identity in B); λ ,λ ∈ Λ+ ⇒ e¯ e¯ =
(r) 1 2 λ1 λ2
e¯ and e e =e ; e¯ e =e e¯ =e .
min(λ1,λ2) λ1 λ2 min(λ1,λ2) λ λ λ λ λ
Obtain a filtration of the algebra B by defining Bλ =e¯ Be¯ ,λ∈Λ+. Then
λ λ
B(r) = B and λ < λ ⇒ Bλ1 ⊆ Bλ2. Each Bλ is an R-algebra with unit e¯ .
1 2 λ
Next define Kλ =(Be B)∩Bλ =Bλe Bλ. Then Kλ is the two-sided ideal in
λ λ
Bλ generated by e . Finally, define Qλ = Bλ/Kλ. Then Qλ is an R-algebra
λ
with unit e¯ mod Kλ .
λ
By standard idempotent theory, if I is an irreducible (left) B-module, then
either e¯ I = 0 or e¯ I is an irreducible Bλ(=e¯ Be¯ )-module. Define an irre-
λ λ λ λ
ducible B-module I to be at level λ if e¯ I 6=0 and e I =0. Then isomorphism
λ λ
classesofirreducibleB-modulesatlevels6λcorrespondtothosewithe¯ I 6=0,
λ
which in turn correspond to isomorphism classes of irreducible Bλ-modules.
AnirreducibleQλ-modulecorrespondstoanirreducibleBλ-moduleonwhich
Kλ acts trivially, and thus to an irreducible B-module I such that e¯ I 6=
λ
0 but Kλe¯ I =0. But then Kλe¯ I =0⇒e e¯ I =0⇒e I =0⇒I is at level
λ λ λ λ λ
λ. Since any irreducible B-module lies at exactly one level, it corresponds to
an irreducible Qλ-module for exactly one λ ∈ Λ+. Thus a classification of the
irreducible Qλ-modules for all λ∈Λ+ will give a classificationof all irreducible
B-modules.
We will make one further reduction. Let B¯λ =(1 +e )B(1 +e )⊆Bλ.
λ λ λ λ
This is an R-algebra with unit 1 +e . Let K¯λ = Kλ ∩B¯λ = B¯λe B¯λ, the
λ λ λ
two-sided ideal in B¯λ generated by e , and define Cλ = B¯λ/K¯λ. Cλ is an
λ
R-algebrawith unit 1 modK¯λ and can be identified with a subalgebra of Qλ:
λ
Cλ = 1 modKλ Qλ 1 modKλ . We will see that irreducible Qλ-modules
λ λ
correspond to irreducible Cλ-modules.
No(cid:0)tice that if µ(cid:1),λ∈(cid:0)Λ and µ ∼(cid:1)λ, then there exists α∈S ⊆S such that
r r
G =αG α−1.
µ λ
Lemma 5.1. If µ,λ ∈ Λ, µ ∼ λ, and G = αG α−1, then for X = L or R,
µ λ
1 =f(µ,α,λ)∗ 1 ∗ f λ,α−1,µ .
µ X λ X
Proof. For any ρ∈G we h(cid:0)ave αρα−(cid:1)1 ∈G =G 1G , so
λ µ µ µ
o(G ), if D =D(µ,1,µ),
N D(µ,α,λ),D λ,α−1,µ ,D = λ
(0, otherwise.
(cid:0) (cid:0) (cid:1) (cid:1)
For any σ ∈G , let π =α−1σα∈G . Then σα=απ. Then G (µ,α,λ)=G
µ λ R µ
and n (µ,α,λ) = o(G ). Reversing the roles of λ,µ gives n λ,α−1,µ =
R µ R
o(G ). Also n (µ,1,µ) = o(G ). Substituting into the multiplication rule
λ R µ
(cid:0) (cid:1)
8
then gives
f(µ,α,λ)∗ 1 ∗ f λ,α−1,µ =f(µ,α,λ)∗ f λ,α−1,µ
R λ R R
(cid:0) (cid:1) (cid:0) (cid:1)
o(G )o(G )
µ λ
= f(µ,1,µ)=1 .
µ
o(G )o(G )
µ λ
By a similar argument, n (µ,α,λ) = o(G ), n λ,α−1,µ = o(G ), and
L λ L µ
n (µ,1,µ)=o(G ). The multiplication rule again gives
L µ
(cid:0) (cid:1)
f(µ,α,λ)∗ 1 ∗ f λ,α−1,µ =f(µ,α,λ)∗ f λ,α−1,µ
L λ L L
(cid:0) (cid:1) (cid:0) (cid:1)
o(G )o(G )
µ λ
= f(µ,1,µ)=1 .
µ
o(G )o(G )
µ λ
Corollary 5.1. 1 modKλ is a “full idempotent” in Qλ, that is, the two sided
λ
ideal generated by 1 modKλ in Qλ is all of Qλ.
λ
Proof. Modulo Kλ , e¯ = 1 . By the lemma, the two-sided ideal gen-
λ µ
µ∈Λ,µ∼λ
erated by 1 contains each 1P, µ ∼ λ . So the two sided ideal generated by
λ µ
1 modKλ contains the identity e¯ modKλ in Qλ and hence all of Qλ.
λ λ
Proposition 5.1. Isomorphism classes of irreducible Cλ-modules correspond
one to one with isomorphism classes of irreducible Qλ-modules, and hence to
isomorphism classes of irreducible B-modules at level λ.
Proof. By generalidempotent theory, irreducible Cλ-modules correspondto ir-
reducible Qλ-modules I such that 1 modKλ I 6= 0. But by the corollary
λ
above, we have a full idempotent, and for a full idempotent 1 modKλ I =
(cid:0) (cid:1) λ
0 ⇒ I = 0. So every irreducible Qλ-module corresponds to a Cλ-module as
(cid:0) (cid:1)
claimed.
Remark: Suppose there are d compositions µ in Λ(r) with µ ∼ λ ∈ Λ+(r).
It can be shown that Qλ is isomorphic as an R-algebra to the algebra of d
by d matrices with entries in Cλ. This leads to an alternative verification of
Proposition 5.1.
Inthe remainderofthis paper wewill classifythe irreducible B-modules (in
many cases) by classifying the irreducible Cλ-modules for all partitions λ. It
will be useful to rewrite Cλ slightly: since
B¯λ =(1 +e )B(1 +e )=1 B1 +e B1 +1 Be +e Be
λ λ λ λ λ λ λ λ λ λ λ λ
and
e B1 +1 Be +e Be ⊆Be B∩B¯λ =K¯λ
λ λ λ λ λ λ λ
we have
Cλ =B¯λ/K¯λ =1 B1 /(1 B1 ∩Be B)
λ λ λ λ λ
(where e = 1 ).
λ µ
µ∈Λ,µ<λ
P
9
6 Characteristic zero
In this section we will assume R is a field k of characteristic zero. We continue
to write B for either of the k-algebras LGS or RGS and let Bλ,Cλ be as in
k k
section 5. Let ν =1(r) be the “smallest” partition.
Theorem 6.1. For a field k of characteristic zero, Cν ∼= k[S ] (the monoid
r
algebra for S over k) while Cλ =0, λ6=ν. The irreducible left B-modules are
r
all at level ν and correspond to irreducible left k[S ]-modules. (The irreducible
r
leftk[S ]-modulesinturncorrespondtoirreducibleleftk[S ]-modulesforvarious
r i
i6r.)
Proof. Let X be either L or R. For the smallest partition ν we have e =
ν
0, Cν =1 B1 . ThenCν isak-vectorspacewithbasis{f(ν,D,ν):D ∈ M }.
ν ν ν ν
Since G = {1}, the ν −ν double cosets D = G αG = α consist of single
ν α ν ν
elements of S . Then M = S and f(ν,D ,ν) 7→ α gives a vector space
r ν ν r α
isomorphism φ:Cν →k[S ]. By special case 1 of the multiplication law,
r
n (D(ν,αβ,ν))
X
f(ν,α,ν)∗ f(ν,β,ν)= f(ν,αβ,ν)=f(ν,αβ,ν)
X
n (D(ν,α,ν))
X
(where we used the fact that n (D(ν,γ,ν)) = 1 for any γ ∈ S ). So φ is an
X r
isomorphism of k-algebras. So irreducible left B-modules at level ν correspond
to irreducible left Cν ∼=k[S ]-modules as stated.
r
Now suppose λ ∈ Λ+, λ 6= ν. To show Cλ = 0 it suffices to prove that
1 B1 ⊆Be B,whichwillbetrueif1 ∈Be B. Wewillshowthat1 ∈B1 B.
λ λ λ λ λ λ ν
Then since 1 = e¯ = e¯ e , we will have 1 ∈ B1 B = Be¯ e B ⊆ Be B as
ν ν ν λ λ ν ν λ λ
desired. By special case 4 of the multiplication rule,
f(λ,1,ν)∗ 1 ∗ f(ν,1,λ)=f(λ,1,ν)∗ f(ν,1,λ)
X ν X X
o(G )
λ
= f(λ,1,λ)=o(G )1 .
λ λ
o(G )
ν
Since k has characteristic zero, o(S )6=0 in the field k and we have
λ
1
1 = f(λ,1,ν)∗ 1 ∗ f(ν,1,λ)∈B1 B
λ o(S ) X ν X ν
λ
as claimed, completing the proof that Cλ =0.
7 Characteristic p
We now consider the case where R is a field k with positive characteristic p.
Definition 7.1. A partition λ∈Λ+(r) is a p-partition if for each i, 16i6n,
either λi =0 or λi =pki for some integer power ki >0 of p.
Theorem 7.1. If λ is not a p-partition, then Cλ =0.
10