From now on, we are going to establish lemmas used in Kleshchev-Tiep’s paper, which talk about the properties of σ ∈ F×q, and its p0-part σ0, under multiplication of some element τ ∈ F×q.
Lemma 4.16. Let σ ∈ F×q with p0-part σ0, and p-part υ, deg(σ) = kd and deg(σ0) = d.
If one of the following criterion holds, (1) p > 2 or p = 2, qd≡ 1 mod 4.
(2) p = 2, qd≡ 3 mod 4, k > 2.
(3) p = 2, qd≡ 3 mod 4, k = 2, |υ| = 4.
Then there exists some α ∈ Op(F×q), |α| = | gcd(k, q − 1)|p, such that σα is conjugate to σ over Fq.
Proof. If p does not divide q − 1, or k = 1, then just take α = 1, so assume p | q − 1 and k > 1. Hence p | qd− 1, and by Lemma 4.11 we have k = pb and the p-factor of gcd(k, q − 1) is pc for some integers b ≥ c ≥ 1.
For case (1), take α = σqdpb−c−1. Since (σ0)qd−1 = 1, α is a power of υ, thus a p-element. It suffices to prove that |α| = pc. By Lemma 4.9, the p-part of (qdpb − 1)/(qdpb−c − 1) is pc, thus σqdpb−1 = 1 implies αpc = 1. On the other hand, the p-part of (qdpb−1 − 1)/(qdpb−c − 1) is pc−1, so if αpc−1 = 1, then σqdpb−1−1 = 1, contradicts to deg(σ) = dpb.
For case (2) and (3), note that if 4 | q − 1 implies qd ≡ 1 mod 4, hence we only need to find |α| = 2. Take α = σq2b−1d−1. By similar argument in case (1), then α is a 2-element, and α 6= 1 or it will contradict to deg(σ) = 2ad. If k > 2, then b > 1, note that q2b−1d≡ 1 mod 4 is non-exceptional case of Lemma 4.9, so a similar argument in case (1) holds. If k = 2, then b = 1, since υ4 = 1, α2 = σ2(qd−1)− υ2(qd−1)= 1.
Lemma 4.17. Given σ ∈ F×q with p0-part σ0 and p-part υ, deg(σ) = kd, and deg(σ0) = d. Let I = {τ ∈ Op(F×q) | [σ] = [στ ]}. Then we have |I| divides | gcd(k, q − 1)|p. More precisely,
(1) |I| = | gcd(k, q − 1)|p, unless p = 2, qd≡ 3 mod 4, k = 2, and |υ| ≥ 8.
(2) In the exceptional case |I| = 1.
Proof. If p does not divide q − 1, then |I| = 1 = | gcd(k, q − 1)|p, so assume p | q − 1.
For any τ ∈ I, write στ = σqj for some integer j ≥ 0, then (σ0)1−qj = υqj−1τ−1 = 1.
Hence τ = υqj−1, and by deg(σ0) = d, we have (σ0)qj−1 = 1 if and only if d | j. Note that τq = τ ,
τk= τ1+qj+···+q(k−1)j = (υqj−1)1+qj+···+q(k−1)j = υqkj−1 = 1
since deg(υ) = kd by Corollary 4.12. Hence |I| | k, as well as | gcd(k, q − 1)|p.
Now (1) follows from an element α ∈ I of degree exactly | gcd(k, q − 1)|p found in Lemma 4.16. For (2), consider the case p = 2, qd ≡ 3 mod 4, k = 2, where
| gcd(k, q − 1)|p = 2 and |I| | 2. We show that if τ = −1 ∈ I, then |υ| = 4. Write again στ = σqj. From deg(σ) = 2d, j can be chosen from 0 ≤ j < 2j. Since q is odd, j = 0 is excluded. By (σ0)1−qj = υqj−1τ−1 = 1, we have d | j, thus d = j. Now τ = υqd−1 = −1 and qd ≡ 3 mod 4 gives |υ| = 4. Note that 1, −1 ∈ Fqd when qd ≡ 3 mod 4, so deg(υ) = 2d implies |υ| ≥ 4.
Lemma 4.18. Let d ∈ N and pc | (q − 1) for some integer c ≥ 0. Then there exists an p-element υ ∈ F×q, depending only on c, d, such that for any p0-element σ ∈ F×q of degree d, we have deg(συ) = dpc and |I| = pc, where I = {τ ∈ Op(F×q) | [συ] = [συτ ]}.
Proof. The existence of such υ follows from Lemma 4.11 and Corollary 4.12. For p = 2, q ≡ 3 mod 4, c = 1, we can pick |υ| = 4 by Corollary 4.12 (4). Therefore, we avoid the exceptional case in Lemma 4.17, taking k = pc = | gcd(k, q − 1)|p gives |I| = pc.
5 Kleshchev-Tiep’s Theorem
Here we are going to derive the main result of Kleshchev-Tiep.
Let F = K or F be sufficiently large for GLn. In [K], F is asked to be albegraically closed, but it seems that sufficiently large is enough. Recall that any irreducible KGLn -module is of the form LK(σ, λ) for some [(σ, λ)] ∈ ΣK, and any irreducible F GLn -module is of the form LF(σ, λ) for some [(σ, λ)] ∈ ΣF.
Definition 5.1. Let [(σ, λ)] be an n-admissible symbol, p a prime.
(1) The p0-branching number is defined by
κp0(σ, λ) = #{τ ∈ Op0(F×q) | τ · [(σ, λ)] = [(σ, λ)]}
(2) The p-branching number is defined by
κp(σ, λ) = | gcd(q − 1, ∆(λ0))|p (3) The ordinary branching number is defined by
κ(σ, λ) = #{τ ∈ F×q | τ · [(σ, λ)] = [(σ, λ)]}
When V = LF(σ, λ), we also denote κp0(V ), κp(V ) instead of κp0(σ, λ), κp(σ, λ).
The Kleshchev-Tiep’s Theorem is stated as following.
Theorem 5.2 (Kleshchev-Tiep (2008)).
(1) Let V = LF(σ, λ) be an irreducible F GLn-module corresponding to [(σ, λ)] ∈ ΣF. Then the branching number of V ↓SLn is
κSLn(V ) = κp0(σ, λ) · κp(σ, λ)
Note that κp0(σ, λ) and κp(σ, λ) are the p0-factor and p-factor of κSLn(V ).
(2) Let VK = LK(σ, λ) be an irreducible KGLn-module corresponding to [(σ, λ)] ∈ ΣK. Then the branching number of (VK)↓SLn is
κSLn(VK) = κ(σ, λ)
The proof of Theorem 5.2 splits into several steps. Recall that Tn and Rn satisfy SLn ≤ Tn, Rn≤ GLn, Tn/SLn = Op(GLn/SLn), Rn/SLn= Op0(GLn/SLn).
Using Lemma 3.6 to split κSLn(V ) into κTn(V ) · κRn(V ). The κTn(V ) part is rather simple, using Lemma 3.5 and Lemma 2.34. The rest is to find out κRn(V ). The strategy is to find a lower bound of κRn(V ), and use total counting to force κRn(V ) to match its for µ D λ, and exactly one of them is LF(σ, λ). Hence applying induction on dominance order on partition makes sense. t ≥ m. Since reduction modulo p and restriction commute,
SF(σ, λ)↓GLR n = LK(σ, λ)↓GLR n =
Now assume the contrary, LF(σ, λ)↓GLR n =Lm(λ)
i=1 Li, where Li ∈ IrrF(R) but m(λ) <
m. For convenience let L1 be a composition factor of V1. Since for any j, VK(j) ∼=
gVK(1) for some g ∈ GLn, thus gL1 is a composition factor of Vj. Hence there are at least t conjugates of L1, which cannot be all summands of LF(σ, λ)↓GLR n, and thus there are some Lµi ∼= gL1. But the indices of inertia group of Lµi and gL1 are m(µ), t, respectively, which leads to a contradiction as m(µ) ≥ m > t.
Lemma 5.4. Let λ ` k and 1 6= α ∈ F×q. Assume [σα] = [σ]. Then κR(LF(σ, λ)) ≥ |α|
for R = ker(LK(α, (n)))
Proof. By virtue of Lemma 2.34, LK(σ, λ)⊗LK(α, (n)) ∼= LK(σα, λ) ∼= LK(σ, λ). Hence the assumption of Lemma 3.5(2) is fulfilled, and thus κR(LK(σ, λ)) ≥ |α| as |α| = (GLn : R). Since λ is arbitrary, the result follows from Lemma 5.3.
Lemma 5.5. Let r ∈ N with r > 1, and |r − 1|p = pc with c ∈ N. Take N = rpa − 1 r − 1 for any a ∈ N. Then pa divides |N |p. More precisely,
(1) |N |p = pa, except the case p = 2, c = 1, r ≡ 3 mod 4.
(2) In the exceptional case, |N |p is a proper multiple of pa. Proof. Take k = pd in Lemma 4.9.
Lemma 5.6. Assume that σ is an p0-element, pc | gcd(n, q − 1) with c ≥ 1. Then for any λ ` k with pc | λ0, we have κRn(LF(σ, λ)) ≥ pc.
Proof. Let d = deg(σ). Take a = dpc in Lemma 4.11, there exists some υ such that deg(συ) = dpc. If p = 2, q ≡ 3 mod 4, c = 1, by Corollary 4.12 we can take |υ| = 4.
Therefore, by Lemma 4.16 there is some α ∈ F×q, |α| = | gcd(pc, q − 1)|p = pc, such that [συ] = [συα]. In particular, α 6= 1, which matches the assumption of Lemma 5.4. Thus κR(LF(συ, ν)) ≥ |α| for R = ker(LC(α, (n))) and all ν ` n/ deg(στ ) = n/(dpc) = k/pc. By (GLn : R) = |α| = pc≤ (GLn: Rn), we have R ≥ Rn and κRn(LF(συ, ν)) ≥ pc for any ν ` k/pc. Now since pc| λ0, there exists some µ ` k/pc such that λ = [pc]µ. Then LF(στ, µ) ∼= LF(σ, λ) by Corollary 4.15 and the proof is complete.
Lemma 5.7. Assume pc| gcd(n, q − 1)) with c ≥ 1. Let V = LF(σ, λ) be an irreducible F GLn-module, [(σ, λ)] ∈ ΣF, and pc| ∆(λ0). Then κRn(V ) ≥ pc.
Proof. Write σ = (σ1, · · · , σa), λ = (λ(1), · · · , λ(a)). For i = 1, · · · , a, deg(σi) = di, λ(i) ` ki, and n = Pa
i=1kidi. Apply induction on a. The case a = 1 is Lemma 5.6.
Assume a ≥ 2, set
r = k1d1, A = GLr, A1 = Rr, WA= LF(σ1, λ(1)),
s = n − r, B = GLs, B1 = Rs, WB = LF(σ2, λ(2)) ◦ · · · ◦ LF(σa, λ(a)).
then WA ∈ IrrF(A), WB ∈ IrrF(B) and V = WA◦ WB. Since pc | (λ(i))0 for any i, we have pc| gcd(r, q − 1) and pc| gcd(s, q − 1). By induction hypothesis,
κA1(WA) = κA1×B(WA⊗ WB) = pα ≥ pc κB1(WB) = κA×B1(WA⊗ WB) = pβ ≥ pc
Now choose suitable x ∈ A, y ∈ B satisfied det(x) = τ , hτ i = Op(F×q), det(y) = τ−1. Then ¯x = xA1 generates A/A1 and ¯y = yB1 generates B/B1. Since GLn/SLn ∼= F×q is independent of n, we have (A : A1) = (B : B1) = (GLn: Rn).
Consider the standard parabolic subgroup P = QL < GLn with upper unitriangular subgroup Q and Levi subgroup L = GLr× GLs = A × B. Let H = hA1, B1, xyi, then H/(A1× B1) = h¯x¯yi and applying Lemma 3.7,
κH(WA⊗ WB) = gcd(pα, pβ) ≥ pc
Note that H = L ∩ Rn. For this, observe that hA1, B1, xyi ≤ L ∩ Rn, and check (L : H) = (L/(A1× B1) : K/(A1× B1)) = (GLn: Rn) = (L : L ∩ Rn). Multiply Q on the left gives QH = P ∩ Rn. The following lattice may help.
GLn P = QL L
Rn QH H
From P · Rn= GLn we have [P \G/Rn] = {1}. Then
V ↓GLRnn = (inflPL(WA⊗ WB))↑GLP n ↓GLRnn
∼= (inflPL(WA⊗ WB))↓PQH ↑RQHn
∼=
inflQHH (WA⊗ WB)↓LH
↑RQHn Hence κGLRnn(V ) ≥ κLH(WA⊗ WB) ≥ pc.
Lemma 5.8. Let V = LF(σ, λ) be an irreducible F GLn-module, [(σ, λ)] ∈ ΣF. Then κRn(V ) = gcd ((GLn: Rn), ∆(λ0)).
Proof. Write σ = (σ1, · · · , σa), λ = (λ(1), · · · , λ(a)). For i = 1, · · · , a, deg(σi) = di, λ(i) ` ki, and n =Pa
i=1kidi.
Note that (GLn: Rn) is a power of p, thus gcd ((GLn: Rn), ∆(λ0)) = pc(V ) for some non-negative integer c(V ). Since pc(V )divides Pa
i=1|(λ(i))0| = n and (GLn : Rn) divides
| F×q | = q − 1, applying Lemma 5.7 gives κRn(V ) ≥ pc(V ) if c(V ) ≥ 1. The case c(V ) = 0 holds trivially.
To force κRn(V ) to match their lower bound, we gives a counting argument. Take S = Rn and G = A = GLn in Lemma 3.10, the set X containing all irreducible summand of restriction of any non-isomorphic irreducible F GLn-module, is exactly the set Y containing all non-isomorphic irreducible F Rn-module.
Now |X| equals P κRn(V ), sum over V = LF(σ, λ) for [(σ, λ)] ∈ ΣF. On the other hand, by Proposition 1.11, |Y | equals to P gcd ((GLn : Rn), ∆(λ)), sum over all [(σ, λ)] ∈ ΣF. Since λ only depends on k, [(σ, λ)] ∈ ΣF means [(σ, µ)] ∈ ΣF for any µ ` λ. In particular, we may write |Y | = P gcd ((GLn : Rn), ∆(λ0)) = P pc(V ), sum over V = LF(σ, λ) for [(σ, λ)] ∈ ΣF. This forces the inequality of each κRn(V ) ≥ pc(V ) is actually equality, hence proves the lemma.
Now we can prove Theorem 5.2.
Proof. By Lemma 3.6 we have κSLn(V ) = κTn(V ) · κRn(V ). From Lemma 3.5, κTn(V ) =
#{L ∈ IrrF(GLn/Tn) | V ∼= V ⊗ L}. The tensor product is describe in Lemma 2.34,
which proves exactly κTn(V ) = κp0(σ, λ). Lemma 5.8 gives κRn(V ) = κp(σ, λ), which proves (1).
In the case F = K, any finite group G is a p0-group, and Op0(G) is G itself. Hence again by Lemma 3.5 and Lemma 2.34 we have κSLn(VK) = κ(σ, λ).
6 Main Theorem
6.1 The Canonical Composition Factor
Given VK = LK(σ, λ) ∈ IrrK(GLn), where [(σ, λ)] is an n-admissable symbol, we con-sider VK, its reduction modulo p. Then (σ, λ) may not be p-regular, even not p-non-repeated. Nevertheless, we can find a corresponding p-regular n-admissable symbol [(σ∗, λ∗)], called the p-regularization of [(σ, λ)], such that V = LF(σ∗, λ∗) ∈ IrrF(GLn) has some good properties related to VK.
Let [(σ, λ)] be an n-admissable symbol, where σ = (σ1, · · · , σa), λ = (λ(1), · · · , λ(a)) with λ(i) ` ki. For each σi, let σi0 = (σ)p0, υi = (σ)p, di = deg(σi) and fi = deg(σi)/ deg(σi0). Consider the index set A = {1, · · · , a} with the equivalence rela-tion i1, i2 ∈ A, i1 ∼ i2 if [σi01] = [σ0i2]. Under this equivalence relation, A is split into parts A1, · · · , Ab. For each j = 1, · · · , b, pick some ih ∈ Aj and take σ∗j = σi0h as a repre-sentative. Let λ(j)∗ =[+]i
h∈Aj[fih]λ(ih), and set σ∗ = (σ∗1, · · · , σb∗), λ∗ = (λ(1)∗, · · · , λ(b)∗).
It is clear that (σ∗, λ∗) is p-regular and p-non-repeated.
Definition 6.1. The p-regular n-admissable symbol [(σ∗, λ∗)] defined above is called the p-regularization of [(σ, λ)].
By Proposition 4.6, [σ1] = [σ2] implies [σ01] = [σ20]. Hence the p-regularization is well-defined.
Lemma 6.2. Let VK = LK(σ, α(1)) ◦ · · · ◦ LK(σ, α(m)) and α =[+]mi=1α(i). Then VK has a composition factor of LK(σ, α) of multiplicity one, and all other factor is of the form LK(σ, β) with β D α.
Proof. By Theorem 2.24, in the Grothendieck group of KGLn-modules we have LK(σ, α) ◦ LK(σ, β) = X
γ`|α|+|β|
cγαβLK(σ, γ)
where cγαβ are Littlewood-Richardson coefficients. By Proposition 2.25 if δ = α [+] β, then cδαβ = 1 and cγαβ > 0 implies γ D δ. Now the result follows by induction on m.
Lemma 6.3. Given σ an p0-element, let V = LF(σ, λ(1)) ◦ · · · ◦ LF(σ, λ(m)) and λ =
[+]mi=1λ(i). Then V has a composition factor of LF(σ, λ) of multiplicity one, and all other factor is of the form LF(σ, µ) with µ D λ.
Proof. By Theorem 2.28, if deg(σ) = d, β ` k, then in the Grothendieck group of F GLkd-modules we may write
LK(σ, β) = LF(σ, β) + X
αDβ,α6=β
wβαLF(σ, α)
This gives an integer lower unitriangular matrix (wβα) with index as partition ordered by dominance order. Its inverse (xβα) is also an integer lower unitriangular matrix. Note that the set ΦD = {(α, β) | α Dβ} is a closed subset (see §2.2) of Φ≥= {(α, β) | α ≥ β}.
Hence if all (α, β)-entries of (wβα) with (α, β) ∈ Φ≥\ΦD are zero, then so do such entries of (xβα). Therefore we may write
LF(σ, β) = LK(σ, β) + X
αDβ,α6=β
xβαLK(σ, α)
Now replace β by λ(i) and apply Harish-Chandra induction to get
V = VK+X
xα(1),··· ,α(m)LK(σ, α(1)) ◦ · · · ◦ LK(σ, α(m))
which sum over α(i)D λ(i) for all i and α(j) 6= λ(j) for at least one j, VK = LK(σ, λ(1)) ◦
· · · ◦ LK(σ, λ(m)), and xα(1),··· ,α(m) ∈ Z.
Now if LF(σ, µ) is a composition factor of V , then either it’s a composition factor of VK, which µ = λ with multiplicity one or µ D λ and µ 6= λ; or it’s a composition factor of some LK(σ, α(1)) ◦ · · · ◦ LK(σ, α(m)), which µ D[+]mi=1α(i)D[+]mi=1λ(i) = λ, and µ 6= λ by the strict inequality of the second D. This proves the lemma.
Theorem 6.4. Let VK = LK(σ, λ) be an irreducible KGLn-module. Then VK has LF(σ∗, λ∗) as a composition factor with multiplicity one, and all other factor is of the form LF(σ∗, ν) with ν D λ∗.
Proof. Adopt the notation before Definition 6.1, such as σi, σi0, ki, di, fi, partition set A1, · · · , Ab, and the corresponding σ∗i, λ∗i.
Now consider some i ∈ Aj for some j. Then by Corollary 4.15,
LK(σi, λ(i)) = LF(σi, λ(i)) + X
βDλ(i),β6=λ(i)
x(i)β LF(σi, β)
= LF(σi0, [fi]λ(i)) + X
βDλ(i),β6=λ(i)
x(i)β LF(σi0, [fi]β)
in the Grothendieck group of F GLkidi-modules, x(i)β ∈ Z.
For i1, · · · , im ∈ Aj, let VK,j = LK(σi1, λ(i1)) ◦ · · · ◦ LK(σim, λ(im)) and let Vj = LF(σj∗, [fi1]λ(i1)) ◦ · · · ◦ LF(σ∗j, [fim]λ(im)), then
VK,j = Vj+X
xβ(i1),··· ,β(im)LF(σ∗j, fi1β(i1)) ◦ · · · ◦ LF(σj∗, [fim]β(im))
which sum over β(ih)D λ(ih) for all ih ∈ Aj, and β(ih) 6= λ(ih) for at least one ih ∈ Aj, xβ(i1),··· ,β(im) ∈ Z.
By Lemma 6.3, any composition factor of RHS is of the form LF(σ∗j, µ), either it is a composition factor of Vj, which has µ = λ(j)∗ with multiplicity one or µ D λ(j)∗ and µ 6= λ(j)∗; or it is a composition factor of some LF(σj∗, [fi1]β(i1)) ◦ · · · ◦ LF(σ∗j, [fim]β(im)), which has
µ D[+]
ih∈Aj
[fih]β(ih)D[+]
ih∈Aj
[fih]λ(ih) = λ(j)∗
with strict inequality from the second D. Hence in particular LF(σj∗, λ(j)∗) is a compo-sition factor of VK,j with multiplicity one.
Finally, taking Harish-Chandra induction over all VK,j gives VK. Hence LF(σ∗, λ∗) is a composition factor of VK with multiplicity one.
Corollary 6.5. GLn(q) has (C, p)-property, hence (L00, p)-property, (L0, p)-property, and (L, p)-property for p not dividing q.
Proof. This is the direct result of the previous theorem.