BLM realization for Frobenius--Lusztig Kernels of type A

0.24 MB·English

by  Qiang Fu

BLM REALIZATION FOR FROBENIUS–LUSZTIG KERNELS OF TYPE A
QIANGFU
3 Abstract. The infinitesimal quantum gl was realized in [1, §6]. We will realize Frobenius–
1 n
0 Lusztig Kernels of typeA in this paper.
2

n
a
J 1. Introduction

5
1 In1990, Ringel discovered theHall algebrarealization [19]of thepositivepartof thequantum

] enveloping algebras of finite type. Almost at the same time, the entire quantum gln was realized
A
by A. A. Beilinson, G. Lusztig and R. MacPherson in [1]. They first used q-Schur algebras to
Q
construct a Q(v)-algebra K (n), and then proved that the quantum enveloping algebra of gl
. Q n
h
t over Q(v) can be realized as a subalgebra of KQ(n).
a b
m Let U (n) be the the quantum enveloping algebra of gl over k with standard generators
k n
b
[ E(m), F(m), K±1 and Ki;0 , where k is a commutative ring containing a primitive l′th root ε
i i i t

1 of 1. Let p = chark. For h > 1, let u (n) be the k-subalgebra of U (n) generated by E(m),
v (cid:2) (cid:3) k h k i
3 F(m), K±1, Kj;0 for 1 6 i 6 n−1, 1 6 j 6 n and 0 6 m,t < lph−1, where l = l′ if l′ is odd,
i j t
1 and l = l′/2 otherwise. Then we haveeu (n) ⊆ u (n) ⊆ ··· ⊆ U (n). In the case where l′ is
4 (cid:2) (cid:3) k 1 k 2 k
3 an odd number, let u (n) = u (n) /hKl −1,··· ,Kl −1i. The algebra u (n) is called the
. k h k h 1 n k 1
1 infinitesimal quantum gl and the algebera u (n)eis called Frobenius–Lusztig Kernels of U (n)
0 n k h k
3 (cf. [7]). The algebra u (n) weas realized in [1, §6]. In this paper, we will realize the algebra
k 1
1
: uk(n)h forall h > 1. More precisely, we willfirstconstruct thek-algebra K ′(n)h in §4. Thenwe
v
Xi will prove in 5.5 that uk(n)h ∼= K ′(n)h in the case where l′ is odd, and that uk(n)h ∼= K ′(n)h
in the case where l′ is even and k is a field.
r
a Let S (n,r) be the q-Schur algebra over k. Certain subalgebra, denotedeby u (n,r) , of
k k h
S (n,r) was constructed in [12, §4]. It is proved in [13] that u (n,r) is isomorphic to the
k k 1
little q-Schur algebra introduced in [11, 14]. We will prove in 6.1 that the algebra ue(n,r) is a
k h
homomorphic image of u (n) . e
k h
e
Infinitesimal q-Schur algebras are certain important subalgebras of q-Schur algebras (cf. [6,
2, 3]). For h > 1 let s (en) be the k-subalgebra of U (n) generated by the algebra u (n) and
k h k k h
Kj;0 (1 6 j 6 n, t ∈ N). We will prove in 6.4 that the infinitesimal q-Schur algebra s (n,r)
t k h
is a homomorphic image of s (n) . e
(cid:2) (cid:3) k h
SupportedbytheNationalNaturalScienceFoundationofChina,theProgram NCET,FokYingTungEduca-
tion Foundation and theFundamentalResearch Fundsfor theCentral Universities.
1
2 QIANGFU
Throughout this paper, let Z = Z[v,v−1] where v is an indeterminate and let Q = Q(v) be
the fraction field of Z. For i ∈ Z let [i] = vi−v−i. For integers N,t with t > 0, let
v−v−1
N [N][N −1]···[N −t+1]
= ∈ Z
t [t]!
(cid:20) (cid:21)
where [t]! = [1][2]···[t]. For µ ∈Zn and λ ∈ Nn let µ = µ1 ··· µn .
λ λ1 λn
Let k be a commutative ring containing a primitive l′th root ε of 1 with l′ > 1. Let l > 1 be
(cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3)
defined by
l′ if l′ is odd,
l =
l′/2 if l′ is even.

Let p be the characteristic of k. We will regard k as a Z-module by specializing v to ε. When

v is specialized to ε, c specialize to the element c in k.
t t ε
(cid:2) (cid:3) (cid:2) (cid:3)
2. The BLM construction of quantum gl
n
Following [16] we define the quantum enveloping algebra U (n) of gl to be the Q(v)-algebra
Q n
with generators
E , F (1 6 i 6 n−1), K , K−1 (1 6 j 6 n)
i i j j
and relations
(a) K K = K K , K K−1 =1;
i j j i i i
(b) KiEj = vδi,j−δi,j+1EjKi;
(c) KiFj = vδi,j+1−δi,jFjKi;
(d) E E = E E , F F = F F when |i−j| > 1;
i j j i i j j i
(e) E F −F E = δ Kei−Kei−1, where K = K K−1;
i j j i i,j v−v−1 i i i+1
(f) E2E −(v+v−1)E E E +E E2 = 0 when |i−j| = 1;
i j i j i j i
e
(g) F2F −(v+v−1)F F F +F F2 = 0 when |i−j| = 1.
i j i j i j i
Following [17], let U (n) be the Z-subalgebra of U (n) generated by all E(m), F(m), K±1
Z Q i i i
and Ki;0 , where for m,t ∈ N,
t
(cid:2) (cid:3) Em Fm K ;0 t K v−s+1−K−1vs−1
E(m) = i , F(m) = i , and i = i i .
i [m]! i [m]! t vs−v−s
(cid:20) (cid:21) s=1
Y
Let Θ(n) be the set of all n×n matrices over N. Let Θ±(n) be the set of all A∈ Θ(n) whose
diagonal entries are zero. Let Θ+(n) (resp. Θ−(n)) be the subset of Θ(n) consisting of those
matrices (a ) with a = 0 for all i > j (resp. i 6 j). For A∈ Θ±(n), write A = A++A− with
i,j i,j
A+ ∈ Θ+(n) and A− ∈ Θ−(n). For A∈ Θ±(n) let
E(A+) = E(aij), F(A−) = F(ai,j)
s s
i6s<j j6s<i
16Yi,j6n 16Yi,j6n
BLM REALIZATION FOR FROBENIUS–LUSZTIG KERNELS OF TYPE A 3
where the ordering of the products is the same as in [1, 3.9]. According to [17, 4.5] and [18, 7.8]
we have the following result.
Proposition 2.1. The set
K ;0
{E(A+) Kδi i F(A−) | A∈ Θ±(n), δ,λ ∈ Nn, δ ∈ {0,1}, ∀i}
i λ i
i
16i6n (cid:20) (cid:21)
Y
forms a Z-basis of U (n).
Z
Usingthestabilizationpropertyofthemultiplicationofq-Schuralgebras,animportantalgebra
K (n) over Z (without 1), with basis {[A] | A ∈ Θ(n)} was constructed in [1, 4.5], where
Z
Θ(n)= {(a )∈ M (Z) |a > 0 ∀16 i 6= j 6 n}.
ij n ij
e
Following [1, 5.1], let K (n) be the vector space of all formal Q(v)-linear combinations
Q
e
β [A] satisfying the following property: for any x∈ Zn,
A∈Θ(n) A
e b
P
{A ∈ Θ(n) | β 6= 0, ro(A) = x}
A
(2.1.1) the sets are finite,
{A ∈ Θ(n) | β 6= 0, co(A) = x}
e A
where ro(A) = ( a ,··· , a ) and co(A) = ( a ,··· , a ) are the sequences of
j 1,j j n,ej i i,1 i i,n
row and column sums of A. The product of two elements β [A], γ [B] in
P P P A∈ΘP(n) A B∈Θ(n) B
e e
K (n) is defined to be β γ [A]·[B] where [A]·[B] is the product in K (n). Then K (n)
Q A,B A B P ZP Q
becomes an associative algebra over Q(v).
P
b b
For A∈ Θ±(n), δ ∈ Zn and λ ∈Nn let
µ
A(δ,λ) = vµ(cid:5)δ [A+diag(µ)] ∈ K (n);
Q
λ
µX∈Zn h i
A(δ) = vµ(cid:5)δ[A+diag(µ)] ∈ K (nb),
Q
µ∈Zn
X
b
where µ(cid:5)δ = µ δ .
16i6n i i
The next result is proved in [1, 5.5,5.7].
P
Theorem 2.2. There is an injective algebra homomorphism ϕ: U (n)→ K (n) satisfying
Q Q
E 7→ E (0), Kj1Kj2···Kjn 7→ 0(j), F 7→ E (0).
i i,i+1 1 2 n i i+1,i b
Furthermore the set {A(j) | A ∈ Θ±(n), j ∈ Zn} forms a Q(v)-basis for ϕ(U (n)).
Q
We shall identify U (n) with ϕ(U (n)). According to [15, 4.2,4.3,4.4], we have the following
Q Q
result.
Proposition 2.3. The algebra U (n) is generated as a Z-module by the elements A(δ,λ) for
Z
A ∈ Θ±(n), δ ∈ Zn and λ ∈ Nn. Furthermore, each of the following set forms a Z-basis for
U (n) :
Z
4 QIANGFU
(1) {A(0)0(δ,λ) |A ∈Θ±(n), δ,λ ∈ Nn, δ ∈ {0,1},∀i};
i
(2) {A(δ,λ) | A∈ Θ±(n), δ,λ ∈ Nn, δ ∈ {0,1},∀i}.
i
We end this section by recalling an important triangular relation in K (n). For A= (a )∈
Z s,t
Θ(n) let
a if i< j
e σ (A) = s6i;t>j s,t
i,j

Ps>i;t6jas,t if i> j.
Following [1], for A,B ∈ Θ(n), define B 4PA if and only if σ (B) 6 σ (A) for all i 6= j. Put
 i,j i,j
B ≺ A if B 4 A and σ (B)< σ (A) for some i 6= j.
i,j i,j
e
According to [1, 5.5(c)], for A ∈ Θ±(n) and λ ∈ Zn the following triangular relation holds in
K (n):
Z
(2.3.1) E(A+)[diag(λ)]F(A−) = [A+diag(λ−σ(A))]+f
where σ(A) = (σ (A),··· ,σ (A)) with σ (A) = (a + a ) and f is a finite Z-linear
1 n i j<i i,j j,i
combination of [B] with B ∈ Θ(n) such that B ≺ A.
P
e 3. The algebra u (n)
k h
Let U (n) = U (n)⊗ k. We shall denote the images of E(m), F(m), A(δ,λ), etc. in U (n) by
k Z Z e i i k
the same letters. For h > 1 let u (n) be the k-subalgebra of U (n) generated by the elements
k h k
E(m), F(m), K±1, Kj;0 for 1 6 i 6 n−1, 1 6 j 6 n and 0 6 m,t < lph−1. If l′ is an odd
i i j t
e
number, we let
(cid:2) (cid:3)
(3.0.2) u (n) = u (n) /hKl −1,··· ,Kl −1i.
k h k h 1 n
The algebra u (n) is called Frobenius–Lusztig Kernels of U (n). We will construct several
k h e k
k-bases for u (n) in 3.7.
k h
We need some preparation before proving 3.7.
e
Lemma 3.1. Let m = m +lm , 06 m 6 l−1, m ∈ N. Then
0 1 0 1
m m m
= εl(t1l−t1m0−tm1) 0 1
t t t
(cid:20) (cid:21)ε (cid:20) 0 (cid:21)ε(cid:18) 1 (cid:19)
for 0 6 t 6 m, where t = t +lt with 0 6 t 6 l−1 and t ∈ N.
0 1 0 1
Lemma 3.2. The following identity hold in the field k : m+ph−1 = m for m ∈ Z and
s s
0 6 s <ph−1.
(cid:0) (cid:1) (cid:0) (cid:1)
Proof. We consider the polynomial ring k[x,y]. Since the characteristic of k is p we see that
ph−1
xjyph−1−j = (x+y)ph−1 = xph−1 +yph−1.
j
06jX6ph−1(cid:18) (cid:19)
BLM REALIZATION FOR FROBENIUS–LUSZTIG KERNELS OF TYPE A 5
It follows that ph−1 = 0 for 0< j < ph−1. This implies that
j
(cid:0) (cid:1)
m+ph−1 ph−1 m m
= =
s j s−j s
(cid:18) (cid:19) 06j6s(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
X
for m ∈ Z and 06 s < ph−1. (cid:3)
We now generalize 3.2 to the quantum case.
Lemma 3.3. Assume 0 6 a < lph−1 and b ∈ Z. Then we have b+lph−1 = ε−alph−1 b . In
a ε a ε
particular, we have b+l′ph−1 = b
a ε a ε (cid:2) (cid:3) (cid:2) (cid:3)
(cid:2) (cid:3) (cid:2) (cid:3)
Proof. We write a = a +a l and b = b +b l with 0 6 a ,b < l, a ∈ N and b ∈ Z. If b ∈ N,
0 1 0 1 0 0 1 1
then by 3.1 and 3.2 we conclude that
b+lph−1 b b +ph−1
= ε−alph−1εl(a1l−a1b0−a1b1l−a0b1) 0 1
a a a
(cid:20) (cid:21)ε (cid:20) 0(cid:21)ε(cid:18) 1 (cid:19)
b b
= ε−alph−1εl(a1l−a1b0−a1b1l−a0b1) 0 1
a a
(cid:20) 0(cid:21)ε(cid:18) 1(cid:19)
b
= ε−alph−1 .
a
(cid:20) (cid:21)ε
Furthermore if b+lph−1 < 0, then −b+a−1−lph−1 > 0 and hence
b+lph−1 −b+a−1−lph−1 −b+a−1 b
= (−1)a = (−1)aεalph−1 = ε−alph−1 .
a a a a
(cid:20) (cid:21)ε (cid:20) (cid:21)ε (cid:20) (cid:21)ε (cid:20) (cid:21)ε
Now we assume −lph−1 6 b < 0. According to 3.1 we have
b+lph−1 b b
(3.3.1) = ε−alph−1εl(a1l−a1b0−ab1) 0 1 .
a a a
(cid:20) (cid:21)ε (cid:20) 0(cid:21)ε(cid:18) 1(cid:19)
If a −b −1 > 0 then b0 = (−1)a0 a0−b0−1 = 0 and hence, by 3.1 and (3.3.1), we have
0 0 a0 ε a0 ε
(cid:2) (cid:3) (cid:2) (cid:3)
b l(a −b )+(a −b −1)
= (−1)a 1 1 0 0
a a
(cid:20) (cid:21)ε (cid:20) (cid:21)ε
a −b −1 a −b
= (−1)aεl(a1l−a1(a0−b0−1)−a(a1−b1)) 0 0 1 1
a a
(cid:20) 0 (cid:21)ε(cid:18) 1 (cid:19)
= 0
b+lph−1
= εalph−1 .
a
(cid:20) (cid:21)ε
6 QIANGFU
Nowweassume−lph−1 6 b < 0anda −b −1< 0. Thena −b −1> 0and06 l+a −b −1< l.
0 0 1 1 0 0
According to 3.1 we have
b −b+a−1
= (−1)a
a a
(cid:20) (cid:21)ε (cid:20) (cid:21)ε
l(a −b −1)+(l+a −b −1)
= (−1)a 1 1 0 0
a
(cid:20) (cid:21)ε
l+a −b −1 a −b −1
= (−1)aεl(−a1(a0−b0−1)−a(a1−b1−1)) 0 0 1 1
a a
(cid:20) 0 (cid:21)ε(cid:18) 1 (cid:19)
b −l b
= (−1)a1l+a1εl(−a1(a0−b0−1)−a(a1−b1−1)) 0 1 .
a a
(cid:20) 0 (cid:21)ε(cid:18) 1(cid:19)
Since 0 6 a0 < l and [m+l]ε = ε−l[m]ε we see that b0a−0l = εa0l ab00 ε. This implies that
b (cid:2) (cid:3) (cid:2)b (cid:3) b
(3.3.2) = (−1)a1l+a1εl(a0−a1(a0−b0−1)−a(a1−b1−1)) 0 1 .
a a a
(cid:20) (cid:21)ε (cid:20) 0(cid:21)ε(cid:18) 1(cid:19)
Furthermore since ε2l = 1 and (a2l−a )−(a l+a )= −2a +la (a −1) is even, we see that
1 1 1 1 1 1 1
εl(a1l−a1b0−ab1) = εl(−2ab1−2a1b0−2a0+2a0a1)εl(a21l−a1)
εl(a0−a1(a0−b0−1)−a(a1−b1−1))
= εl(a21l−a1) = εl(a1l+a1) =(−1)a1(l+1).
Thus by (3.3.1) and (3.3.2) we conclude that b+lph−1 = ε−alph−1 b . The proof is completed.
a ε a ε
(cid:3)
(cid:2) (cid:3) (cid:2) (cid:3)
Corollary 3.4. Assume 0 6 a,b < lph−1 and a+b > lph−1. Then a+b = 0.
a ε
Proof. According to 3.3 we have a+b = ε−alph−1 a+b−lph−1 . Sinc(cid:2)e 0 6(cid:3) a+b−lph−1 < a, we
a ε a ε
see that a+b−lph−1 = 0. The assertion follows. (cid:3)
a ε (cid:2) (cid:3) (cid:2) (cid:3)
Let u0(cid:2)(n) be th(cid:3)e k-subalgebra of u (n) generated by K±1, Kj;0 for 1 6 j 6 n and
k h k h j t
0 6 t < lph−1. For h > 1 let
(cid:2) (cid:3)
e e
Nn = {λ ∈ Nn |0 6 λ < lph−1, ∀i}.
lph−1 i
Lemma 3.5. The set M0 = { Kδi Ki;0 | δ ∈ Nn, δ ∈ {0,1}, λ ∈ Nn } forms a
16i6n i λi i lph−1
k-basis for u0(n) .
k h Q (cid:2) (cid:3)
Proof. Let V = span M0. From 2.1, we see that the set M0 is linearly independent. Thus
e1 k
it is enough to prove that u0(n) = V . Let V be the k-submodule of u0(n) spanned by the
k h 1 2 k h
elements Kδi Ki;0 (δ ∈ Zn, λ ∈ Nn, 0 6 λ < lph−1, for all i). According to [17,
16i6n i λi i
2.3(g8)], for 06 t,t′ < lph−e1 we have e
Q (cid:2) (cid:3)
K ;0 K ;0 t+t′ K ;0 t+j −1 K ;0 K ;0
εt′t i i = i − (−1)jεt(t′−j) Kj i i .
t′ t t t+t′ j i t′−j t
(cid:20) (cid:21)(cid:20) (cid:21) (cid:20) (cid:21)ε(cid:20) (cid:21) 0<j6t′ (cid:20) (cid:21)ε (cid:20) (cid:21)(cid:20) (cid:21)
X
BLM REALIZATION FOR FROBENIUS–LUSZTIG KERNELS OF TYPE A 7
Note that by 3.4 we have t+t′ Ki;0 = 0 for 0 6 t,t′ < lph−1 with t+t′ > lph−1. Thus, by
t ε t+t′
induction on t′ we see that Ki;0 Ki;0 ∈ V for 0 6 t,t′ < lph−1. It follows that u0(n) = V .
(cid:2) t′(cid:3) (cid:2) t (cid:3) 2 k h 2
Furthermore, by the proof of [17, 2.14], for m > 0 and 0 6 t < lph−1 we have
(cid:2) (cid:3)(cid:2) (cid:3)
e
K ;0 K ;0 K ;0
Km+2 i = εt(εt+1−ε−t−1)Km+1 i +ε2tKm i ,
i t i t+1 i t
(cid:20) (cid:21) (cid:20) (cid:21) (cid:20) (cid:21)
K ;0 K ;0 K ;0
K−m−1 i = −ε−t(εt+1 −ε−t−1)K−m i +ε−2tK−m+1 i .
i t i t+1 i t
(cid:20) (cid:21) (cid:20) (cid:21) (cid:20) (cid:21)
If t+1= lph−1, thenεt(εt+1−ε−t−1)Km+1 Ki;0 = −ε−t(εt+1−ε−t−1)K−m Ki;0 = 0. Thusby
i t+1 i t+1
induction on m > 0 we see that K±m Ki;0 ∈ V for 0 6 t < lph−1. This implies that V = V .
i t (cid:2) (cid:3)1 (cid:2) (cid:3) 1 2
The assertion follows. (cid:3)
(cid:2) (cid:3)
We are now ready to prove 3.7. Let Θ±(n) = {A ∈Θ±(n) |0 6 a < lph−1, ∀s 6= t}.
h s,t
Lemma 3.6. The algebra u (n) is generated as a k-module by the elements A(δ,λ) for A ∈
k h
Θ±(n) , δ ∈ Zn and λ ∈ Nn .
h lph−1
e
Proof. Let V be the k-submodule of U (n) spanned by A(δ,λ) for A ∈ Θ±(n) , δ ∈ Zn and
h k h
λ ∈ Nn . According to [15, 3.5(1)] for A ∈Θ±(n) , 0 6 m < lph−1, 1 6 i6 n−1, δ ∈ Zn and
lph−1 h
λ ∈ Nn , we have
lph−1
(mE )(0)A(δ,λ)
i,i+1
= ft A+ t E − t E (δ+αt ,λ+βt ).
j,c,k u i,u u i+1,u j,c,k j,c,k
t∈Λ(n,mX),06j6λi (cid:18) Xu6=i uX6=i+1 (cid:19)
tu6ai+1,u,∀u6=i+1
06k6λi+1,06c6min{ti,j}
where αt = t +λ −j−c e + λ −k− t e , βt = (t +j−c−λ )e +
j,c,k i>u u i i i+1 i+1>u u i+1 j,c,k i i i
(k−λ )e with e = (0,··· ,0,1,0··· ,0) ∈ Nn, and
i+1 i+1(cid:0)P i (cid:1) (cid:0) P (cid:1)
i
ft = εgjt,k ai,u+tu −ti ti+j −c ti ti+1
j,c,k t λ −j t c λ −k
u6=i(cid:20) u (cid:21)ε(cid:20) i (cid:21)ε(cid:20) i (cid:21)ε(cid:20) (cid:21)ε(cid:20) i+1 (cid:21)ε
Y
with gt = a t − a t + t t −t δ +t δ +2jt −
j,k j>u,j6=i i,j u j>u,j6=i+1 i+1,j u u′6=i,i+1,u<u′ u u′ i i i+1 i+1 i
kt . If A+ t E − t E 6∈ Θ±(n) then a +t > lph−1 for some u 6= i.
i+1 P u6=i u i,u Pu6=i+1 u i+1,u P h i,u u
From 3.4 we see that ai,u+tu = 0 and hence ft = 0. Furthermore, if λ+βt 6∈ Nn then
P tu Pε j,c,k j,c,k lph−1
(λ+βt ) = t +j−c > lph−1. From 3.4 we see that ti+j−c = 0 and hence ft = 0. Thus
j,c,k i i (cid:2) (cid:3) ti ε j,c,k
we conclude that
(cid:2) (cid:3)
(3.6.1) (mE )(0)V ⊆ V ,
i,i+1 h h
for 0 6 m < lph−1 and 16 i 6 n−1. Similarly, using [15, 3.4,3.5(2)] we see that
(3.6.2) (mE )(0)V ⊆ V and 0(γ,µ)V ⊆ V
i+1,i h h h h
8 QIANGFU
for 06 m < lph−1, 1 6 i6 n−1, γ ∈ Zn and µ ∈ Nn . Combining (3.6.1) with (3.6.2) implies
lph−1
that
(3.6.3) u (n) ⊆ u (n) V ⊆ V .
k h k h h h
On the other hand, from [15, 3.4e] we see theat for A ∈Θ±(n)h, δ ∈ Zn and λ ∈ Nnlph−1,
co(A)
(3.6.4) A(0)0(δ,λ) = εco(A)(cid:5)(δ+λ)A(δ,λ)+ εco(A)(cid:5)(δ+λ−j) A(δ−j,λ−j).
j
j∈Nn,0<j6λ (cid:20) (cid:21)
X
This implies that
(3.6.5) V = span {A(0)0(δ,λ) |A ∈ Θ±(n) , δ ∈ Zn, λ ∈ Nn }.
h k h lph−1
Furthermore, combining (2.3.1) with 2.3 shows that for A ∈ Θ±(n) , δ ∈ Zn and λ ∈ Nn ,
h lph−1
K ;0
E(A+)F(A−) Kδi i = E(A+)F(A−)0(δ,λ) = A(0)0(δ,λ)+f
i λ
i
16i6n (cid:20) (cid:21)
Y
where f is a k-linear combination of B(0)0(γ,µ) with B ∈ Θ±(n), B ≺ A, γ ∈ Zn and µ ∈ Nn.
From (3.6.3) and (3.6.5) we see that f must be a k-linear combination of B(0)0(γ,µ) with
B ∈ Θ±(n) , B ≺ A, γ ∈ Zn and µ ∈ Nn . Thus we conclude that
h lph−1
(3.6.6)
K ;0
V = span E(A+)F(A−) Kδi i |A ∈ Θ±(n) , δ ∈ Zn, λ ∈ Nn ⊆ u (n) .
h k i λ h lph−1 k h
i
16i6n (cid:20) (cid:21)
(cid:8) Y (cid:9)
e
The assertion follows. (cid:3)
Proposition 3.7. Each of the following set forms a k-basis for u (n) :
k h
(1) M:= {E(A+) Kδi Ki;0 F(A−) | A∈ Θ±(n) , δ ∈ Nn, δ ∈ {0,1}, ∀i, λ ∈ Nn };
16i6n i λi h e i lph−1
(2) B:= {A(δ,λ) | A∈ Θ±(n) , δ ∈ Nn, δ ∈{0,1}, ∀i, λ ∈ Nn };
Q (cid:2)h (cid:3) i lph−1
(3) B′ := {A(0)0(δ,λ) | A∈ Θ±(n) , δ ∈ Nn, δ ∈ {0,1}, ∀i, λ ∈Nn }.
h i lph−1
Proof. According to 2.1 and 2.3, it is enough to prove that u (n) = span M = span B =
k h k k
span B′. From 3.5, 3.6, (3.6.5) and (3.6.6) we see that u (n) = span M = span B′. For
k k h k k
A∈ Θ±(n) , δ ∈ Zn and λ ∈Nn we have e
h lph−1
e
A(δ,λ) = ελi(ελi+1−v−λi−1)A(δ−e ,λ+e )+ε2λiA(δ−2e ,λ)
i i i
= −ε−λi(ελi+1−ε−λi−1)A(δ+e ,λ+e )+ε−2λiA(δ+2e ,λ)
i i i
Notethatifλi+1 = lph−1 thenελi(ελi+1−v−λi−1)A(δ−ei,λ+ei)= −ε−λi(ελi+1−ε−λi−1)A(δ+
e ,λ+e )= 0. This together with 3.6 shows that u (n) = span B. (cid:3)
i i k h k
e
BLM REALIZATION FOR FROBENIUS–LUSZTIG KERNELS OF TYPE A 9
4. The algebra K ′(n)
h
We willconstructthealgebra K ′(n) in thissection. We willprove in5.5 thealgebra K ′(n)
h h
is the realization of u (n) .
k h
Let K (n) = K (n)⊗ k, where k is regarded as a Z-module by specializing v to ε. For
k Z Z
e
A∈ Θ(n) let
[A] = [A]⊗1 ∈ K (n).
ε k
e
Let Θ(n) be the set of all A= (a )∈ Θ(n) such that a < lph−1 for all i 6= j. We will denote
h i,j i,j
by K (n) the k-submodule of K (n) spanned by the elements [A] with A ∈ Θ(n) .
h k ε h
e e
To construct the algebra K ′(n) we need the following lemma (cf. [1, 6.2] and [14, 5.1]).
h
e
Lemma 4.1. (1) K (n) is a subalgebra of K (n). It is generated by [mE +diag(λ)] and
h k h,h+1 ε
[mE +diag(λ)] for 0 6 m < lph−1, 16 h 6 n−1 and λ ∈ Zn.
h+1,h ε
(2) Let D be any diagonal matrix in Θ(n). The map τ : K (n) → K (n) given by [A] →
D h h ε
[A+l′ph−1D] is an algebra homomorphism.
ε
e
Proof. Let A = (a ) ∈Θ(n) and 0 6 m < lph−1. Assume that B = (b ) ∈Θ(n) is such that
s,t h s,t h
B −mE is a diagonal matrix such that co(B) = ro(A). According to [1, 4.6(a)] we have
i,i+1
e e
a +t
[B] ·[A] = εβ(t,A) i,u u A+ t (E −E )
ε ε u i,u i+1,u
t
t∈Λ(n,m) 16u6n(cid:20) u (cid:21)ε(cid:20) 16u6n (cid:21)ε
X Y X
∀u6=i+1,tu6ai+1,u
where β(t,A) = a t − a t + t t . Assume that A + t (E −
j>u i,j u j>u i+1,j u u<u′ u u′ u u i,u
E ) 6∈ Θ(n) for some t; then a +t > lph−1 for some u 6= i. Since 0 6 a ,t < lph−1,
i+1,u h P P i,u u P i,uPu
by 3.4, we conclude that ai,u+tu = 0 and hence [B] · [A] ∈ K (n) . Similarly, we have
e tu ε ε ε h
[C] ·[A] ∈ K (n) , where C is such that C −mE is a diagonal matrix such that co(C) =
ε ε h (cid:2) (cid:3) i+1,i
ro(A). Now using [1, 4.6(c)], (1) can be proved in a way similar to the proof of [1, 6.2].
According to [1, 4.6(a),(b)] and 3.3 we see that τ ([A′] [A] ) = τ ([A′] )τ ([A] ) for any A′
D ε ε D ε D ε
of the form B,C as above. Since K (n) is generated by elements like [B] , [C] above, we
h ε ε
conclude that τ is an algebra homomorphism. (cid:3)
D
Let Θ′(n) be the set of all n×n matrices A= (a ) with a ∈ N, a < lph−1 for all i 6= j
h i,j i,j i,j
and a ∈ Z/l′ph−1Zfor all i. We have an obvious map pr : Θ(n) → Θ′(n) definedby reducing
i,i h h
e
the diagonal entries modulo l′ph−1Z.
e e
Let K ′(n) be the free k-module with basis {[A] | A ∈ Θ′(n) }. We shall define an algebra
h ε h
structure on K ′(n) as follows. If the column sums of A are not equal to the row sums of A′
h
e
(as integers modulo l′ph−1), then the product [A] ·[A′] for A,A′ ∈ Θ′(n) is zero. Assume now
ε ε h
that the column sums of A are equal to the row sums of A′ (as integers modulo l′ph−1). We can
e
then represent A, A′ by elements A, A′ ∈ Θ(n) such that the column sums of A are equal to
h
the row sums of A′ (as integers). According to 4.1(1), we can write[A] ·[A′] = ρ [A′′]
ε ε A′′∈I A′′ ε
e e e ee e
P
e e e e
10 QIANGFU
(product in K (n) ) where I = {A′′ ∈ Θ(n) | ro(A′′) = ro(A), co(A′′) = co(A′)} (a finite set)
h h
and ρ ∈ k. Then the product [A] ·[A′] is defined to be ρ [pr(A′′)] . From 4.1(2) we
A′′ ε ε A′′∈I A′′ ε
e e e e ee ee e
see that the product is well defined and K ′(n) becomes an associative algebra over k.
h P
e
In the case where l′ is odd, the algebra K ′(n) is the algebra K ′ constructed in [1, 6.3].
1
Furthermore, it was remarked at the end of [1] that K ′ is “essentially” the algebra defined in
[17, §5] for type A. We will prove in 5.5 that K ′(n) is isomorphic to the algebra u (n) in the
h k h
case where l′ is odd.
Mimicking the construction of K (n), we define K (n) to be the k-module of all formal k-
Q k
linearcombinations β [A] satisfyingtheproperty(2.1.1). Theproductoftwoelements
A∈Θ(n) A ε
e b b
β [A] , γ [B] inK (n)is definedtobe β γ [A] ·[B] where[A] ·[B]
A∈Θ(n) A ε BP∈Θ(n) B ε k A,B A B ε ε ε ε
e e
is the product in K (n). Then K (n) becomes an associative algebra over k.
P Pk k P
b
WeendthissectionbyinterpretingK ′(n) asak-subalgebraofK (n). Forh > 1letZ =
h k l′ph−1
b
Z/l′ph−1Z and let ¯: Zn → (Z )n be the map defined by (j ,j ,··· ,j ) = (j ,j ,··· ,j ).
l′ph−1 1 2 n 1 2 n
b
For A ∈Θ±(n) and µ¯ ∈ (Z )n let
h l′ph−1
(4.1.1) [[A+diag(µ¯)]] = [A+diag(ν)] .
h ε
ν∈Zn
X
µ¯=ν¯
Let W (n) bethek-submoduleof K (n) spannedby theset {[[A+diag(λ¯)]] |A ∈ Θ±(n) , λ¯ ∈
k h k h h
(Z )n}. From 4.1 we see that W (n) is a k-subalgebra of K (n). Furthermore, it is easy
l′ph−1 k h k
b
to prove that there is an algebra isomorphism
b
(4.1.2) W (n) →∼ K ′(n)
k h h
defined by sending [[A]] to [A] for A ∈Θ′(n) .
h ε h
5. Reaelization of u (n)
k h
For A∈ Θ±(n), δ ∈ Zn and λ ∈Nn let
µ
A(δ,λ) = εµ(cid:5)δ [A+diag(µ)] ∈ K (n).
ε ε k
λ ε
µX∈Zn h i
b
LetV (n)bethek-submoduleof K (n)spannedby theelements A(δ,λ) forA ∈Θ±(n), δ ∈Zn
k k ε
andλ ∈ Nn. Forh >1letV (n) bethek-submoduleofK (n)spannedbytheelements A(δ,λ)
k h k ε
b
for A ∈ Θ±(n)h, δ ∈ Zn and λ ∈ Nnlph−1. We will prove in 5.5 that uk(n)h ∼= Vk(n)h ∼= K ′(n)h
b
in the case where l′ is odd, and that uk(n)h ∼= Vk(n)h ∼= K ′(n)h in the case where l′ is even and
k is a field.
e
Let K (n) be the Z-submodule of K (n) consisting of the elements β [A] with
Z Q A∈Θ(n) A
e
β ∈ Z. Then K (n) is a Z-subalgebra of K (n). There is a natural algebra homomorphism
A Z Q P
b b
b θ :KZ(nb)⊗Z k → Kk(n)
b b