In this section, we are going to find the decomposition matrix of GL2(q) and SL2(q) for p not dividing q. We split the procedure into several parts.
(Step 1) Elements of F×q and F×q2
Write q − 1 = pc1m1 and q2 − 1 = pc2m2, where c1, c2, m1, m2 are nonnegative integers, and p does not divide m1, m2. It is clear that c1 ≤ c2 and m1 | m2.
By Proposition 1.3, every element σ ∈ F×q2 decomposes into its p0-part σ0 and p-part υ. Write e ∈ F×q2 the identity, σ(d)0 ∈ F×q2 if it is p-regular of degree d, and υ(d) ∈ F×q2 if it is a p-element of degree d and is not identity. Then the following multiplication table gives the number of each kind of elements. Note that only {σ(1)0 } and {σ(1)0 υ(1)} are elements of F×q, the others are those of F×q2.
1 s1 := pc1 − 1 s2 := pc2 − pc1 r1 := m1 #{σ0(1)} #{σ0(1)υ(1)} #{σ(1)0 υ(2)} r2 := m2− m1 #{σ0(2)} #{σ0(2)υ(1)} #{σ(2)0 υ(2)}
Let τ ∈ F×q be an element of degree 1, τ0 for those p-regular. Similarly, let ω ∈ F×q2 be an element of degree 2, ω0 for those p-regular. The following table shows their corresponding p0-part. The entry − means the element is already p-regular.
element form p0-part form #
τ σ0(1) − − r1
σ0(1)υ(1) τ0 σ(1)0 r1s1
ω σ0(2) − − r2
σ0(2)υ(1) ω0 σ(2)0 r2s1 σ0(2)υ(2) ω0 σ(2)0 r2s2
σ0(1)υ(2) τ0 σ(1)0 r1s2
From now on, we write τ0, ω0, υ instead of σ0(1), σ(2)0 , υ(1), respectively.
(Step 2) Conjugacy classes of GL2
For simplicity, we write (σ) instead of (σ, (1)), and write e the identity of F×q2, in order to make difference from the partition (1) ` 1. The following table shows the conjugacy classes of GL2, where C2t:= t(t − 1)/2 for t ∈ N.
symbol (τ, (2)) (τ, (12)) (τ1◦ τ2) (ω)
(Step 3) Counting dimension of irreducible representations of GL2
We make use of the following dimension formulas.
Theorem A.8.
Proof. (1) follows from the definition of Harish-Chandra induction. (2) comes from [J, 6.5, 6.8], which gives dimFMF(σ, (k)). One may use (1) to extend the formula to (2).
For (3), note that Kostka numbers form a unitriangular matrix, hence we may find dimFSF(σ, (k)) one by one, or just use inverse matrix.
Hence we may calculate the dimension of irreducible FGL2-modules.
irr. rep’n LK(τ, (2)) LK(τ, (12)) LK(τ1◦ τ2) LK(ω)
# q − 1 q − 1 C2q−1 C2q
dimK 1 q q + 1 q − 1
The dimFLF(σ, λ) heavily depends on p, and have no general formula.
(Step 4) The p-regularization and James-Mathas theorem
The p-regularization is defined right before Definition 6.1. Roughly speaking, it consists the following 3 steps.
(1) Decompose each σi = σi0υi into its p0-part and p-part.
(2) Replace each (σi, λ(i)) by (σ0i, [fi]λ(i)), where fi = deg σi/ deg σi0 (3) If [σi] = [σj], then replace (σi, λ(i)) ◦ (σj, λ(j)) by (σi, λ(i) [+]λ(j)).
Theorem 6.4 tells that each LK(σ, λ) contains LF(σ∗, λ∗) of multiplicity 1, and every composition factor is of the form LF(σ∗, ν) with ν D λ∗.
There is another theorem useful to find the entries of a decomposition matrix, which classifies which LK(σ, λ) remains irreducible over F .
Theorem A.9 (James-Mathas theorem). Let V = LK(σ, λ) ∈ Irr0(GLn), and V be a reduction modulo p of V . Write (σ, λ) = (σ1, λ(1)) ◦ · · · ◦ (σa, λ(a)). Then V remains irreducible if and only if the following two conditions hold:
(1) If i 6= j, deg σi = deg σj, then σi, σj are not p-conjugate to each other.
(2) For each i and all nodes (r, t), (s, t) in the Young diagram of λ(i), write hrt, hst the corresponding hook length, di = deg σi. Then
|Nqdi(hrt)|p = |Nqdi(hst)|p
where Nr(m) = (rm− 1)/(r − 1), defined in §4.2.
Given a label [(σ, λ)] and its p-regularization [(σ∗, λ∗)], we say [(σ, λ)] is of type I, if in the construction of p-regularization above, all fi = |∆((λ∗)0)|p = pc in step (2), and step (3) never happens. Otherwise it is of type II.
The following table describes the number of p-regular and p-singular irreducible ordinary representations, their p-regularizations, and if they satisfy the criterion in James-Mathas theorem or not.
irr. rep’n form #p-reg #p-sing p-reg’n type* J M ?
LK(τ, (2)) τ0 r1 − I yes
τ0υ r1s1 (τ0, (2)) I yes
LK(τ, (12)) τ0 r1 − I/II **
τ0υ r1s1 (τ0, (12)) I/II **
LK(τ1◦ τ2) τ10 ◦ τ20 C2r1 − I yes τ10 ◦ τ20υ 2C2r1s1 (τ10 ◦ τ20) I yes τ10υ1◦ τ20υ2 C2r1s21 (τ10 ◦ τ20) I yes τ0◦ τ0υ r1s1 (τ0, (12)) II no τ0υ1◦ τ0υ2 r1C2s1 (τ0, (12)) II no
LK(ω) ω0 r2/2 − I yes
ω0υ r2s1/2 (ω0) I yes
ω0υ(2) r2s2/2 (ω0) I yes
τ0υ(2) r1s2/2 (τ0, (12)) II/I yes
* [type for p > 2] / [type for p = 2] ** yes, if p6 | q + 1; no, if p | q + 1.
(Step 5) The branching numbers
Given LF(σ, λ) ∈ IrrF(GLn), write κ, κ∗ the branching number from GLnto SLnover K, F , and write η, η∗ the branching number from SLnto GLn over K, F , respectively.
Write κ∗ = κp0κp for its p0-factor and p-factor.
Proposition A.10. Let G = GLn and S = SLn. Given the label [(σ, λ)].
(1) κ = #{ρ ∈ F×q | ρ · [(σ, λ)] = [(σ, λ)]}
(2) κp0 = #{ρ ∈ Op0(F×q) | ρ · [(σ, λ)] = [(σ, λ)]}
(3) κp = | gcd(n, q − 1, ∆(λ0))|p (4) ηκ = q − 1 and η∗κp0 = m1.
Proof. (1)(2)(3) is Theorem 5.2. By Proposition 3.11,
η = #{ orbit of [(σ, λ)] act by ρ ∈ F×q} η∗= #{ orbit of [(σ, λ)] act by ρ ∈ Op0(F×q)}
hence (4) follows by the orbit-stabilizer theorem.
Corollary A.11. Let G = GL2 and S = SL2.
(1) κ = 1, except for q odd, label (τ ◦ −τ ) or (ω) with ωq = −ω. In the exceptional case, κ = 2.
(2) κp0 = 1, except for q odd, p > 2, label (τ ◦ −τ ) or (ω) with ωq = −ω. In the exceptional case, κp0 = 2.
(3) In above two exceptional case, there are (q − 1)/2 labels of the form (τ ◦ −τ ), and another (q − 1)/2 labels of the form (ω) with ωq = −ω.
(4) κp = 1, except for p = 2, label (τ, (12)). In the exceptional case, κp = 2.
Proof. Assume ρ, τ ∈ F×q, ω ∈ F×q2, deg ω = 2. Then ρτ = τ implies ρ = e; ρτ1 = τ2, ρτ2 = τ1 implies ρ = e, −e, and it is clear that there are (q − 1)/2 labels of the form (τ ◦ −τ ); take ε2 a generator of F×q2, Then ρ = ε(q+1)j2 , and ω = εi2, 0 ≤ i < q2 − 1, q − 16 | i. Then ρω = ωq implies (q + 1)j = i(q − 1). Hence either q − 1 | j and ρ = e, or ρ 6= e, (q − 1)/2 | j, thus i = (2t + 1)(q + 1)/2, 0 ≤ t < q − 1. So there are (q − 1)/2 labels of the form (ω) with ωq = −ω. In order that −e ∈ F×q, we need q being odd.
This gives (1) and (3). For (2), in order that −e ∈ Op0(F×q), we need further p > 2.
Finally, (4) follows by checking every labels of GL2 for the criterion in James-Mathas theorem.
(Step 6) The decomposition matrix We split into several cases. Note that p6 | q.
• p > 2, p6 | q2 − 1.
That is, p6 | |GL2| and p6 | |SL2|. By Proposition 43 of [S], both the decomposition matrix of GL2 and SL2 are identity. We notice that this result is consistent to James-Mathas theorem, that is, every label [(σ, λ)] satisfies the irreducibility criterion.
• GL2, p > 2, p | q − 1
In this case, p6 | q + 1. Thus m2 = (q + 1)m1 and r2 = m1q. Also c1 = c2 gives s2 = 0.
dimK IrrK(GL2) # p-reg’n type JM
1 LK(τ, (2)) q − 1 (τ0, (2)) I yes q LK(τ, (12)) q − 1 (τ0, (12)) I yes q + 1 LK(τ1◦ τ2) (q − 1)(q − 2 − s1)/2 (τ10 ◦ τ20) I yes (q − 1)s1/2 (τ10, (12)) II no
q − 1 LK(ω) (q − 1)q/2 (ω0) I yes
q2− 1 dimF IrrF(GL2) #
1 LF(τ0, (2)) r1 q LF(τ0, (12)) r1
q + 1 LF(τ10 ◦ τ20) r1(r1− 1)/2 q − 1 LF(ω0) r1q/2
r1(r1+ q + 3)/2
dimF 1 q q + 1 q − 1
dimK type (τ0, (2)) (τ0, (12)) (τ10 ◦ τ20) (ω0)
1 (τ, (2)) I 1
q (τ, (12)) I 1
q + 1 (τ1 ◦ τ2) I 1
II 1 1
q − 1 (ω) I 1
Table 3: Decomposition matrix of GL2, p > 2, p | q − 1
The bold 1 means we find it by Theorem 6.4. The underlined 1 means we find it by counting the dimension.
• SL2, p > 2, p | q − 1, q odd
• SL2, p > 2, p | q − 1, q even
We have κ = κp0 = κp = 1, η = q − 1, η∗ = r1 for all labels. Hence the decomposition matrix of SL2 is similar to that of GL2.
dimK IrrK(SL2) # type
1 YK(e, (2)) 1 I
q YK(e, (12)) 1 I
q + 1 YK(e ◦ τ ) (q − 2 − s1)/2 I
s1/2 II
q − 1 YK(ω) q/2 I
(q − 1) + 2 dimF IrrF(SL2) #
1 YF(e, (2)) 1 q YF(e, (12)) 1
q + 1 YF(e ◦ τ0) (r1− 1)/2 q − 1 YF(ω0) q/2
(r1+ q − 1)/2 + 2
dimF 1 q q + 1 q − 1
dimK type (e, (2)) (e, (12)) (e ◦ τ0) (ω0)
1 (e, (2)) I 1
q (e, (12)) I 1
q + 1 (e ◦ τ ) I 1
II 1 1
q − 1 (ω) I 1
Table 5: Decomposition matrix of SL2, p > 2, p | q − 1, q even
• GL2, p > 2, p | q + 1.
In this case, p6 | q − 1. Thus c1 = s1 = 0 and r1 = m1 = q − 1, so every element of F×q is p-regular. Also we have q − 1 | m2, q − 1 | r2, hence we may write r0 := r2/(q − 1).
dimK IrrK(GL2) # p-reg’n type JM
1 LK(τ, (2)) q − 1 (τ0, (2)) I yes q LK(τ, (12)) q − 1 (τ0, (12)) I no q + 1 LK(τ1◦ τ2) (q − 1)(q − 2)/2 (τ10 ◦ τ20) I yes q − 1 LK(ω) (q − 1)(q − s2)/2 (ω0) I yes (q − 1)s2/2 (τ0, (12)) II yes q2− 1
dimF IrrF(GL2) # 1 LF(τ0, (2)) q − 1 q − 1 LF(τ0, (12)) q − 1
q + 1 LF(τ10 ◦ τ20) (q − 1)(q − 2)/2 q − 1 LF(ω0) r2/2
(r2+ (q − 1)(q + 2))/2
dimF 1 q − 1 q + 1 q − 1
dimK type (τ0, (2)) (τ0, (12)) (τ10 ◦ τ20) (ω0)
1 (τ, (2)) I 1
q (τ, (12)) I 1 1
q + 1 (τ1 ◦ τ2) I 1
q − 1 (ω) I 1
II 1
Table 6: Decomposition matrix of GL2, p > 2, p | q + 1
• SL2, p > 2, p | q + 1, q odd
When q is odd with label (τ ◦ −τ ) or (ω) with ωq = −ω, we have κ = κp0 = 2, otherwise 1. We have κp = 1, η = (q − 1)/κ, η∗ = r1/κp0.
Pick ε2 a generator of F×q2. If ωq = −ω, then ω = ε(2t+1)(q+1)/2
2 for some t ∈ N.
Then the order of ω is 2(q−1), which is prime to p, hence every such ω is p-regular.
Take ω0 = ε(q+1)/22 . In particular, ωq0 6= ω0, hence deg ω0 = 2.
• SL2, p > 2, p | q + 1, q even
We have κ = κp0 = κp = 1, η = η∗ = (q − 1) for all labels. Hence again the decomposition matrix of SL2 is similar to that of GL2.
dimK IrrK(SL2) # type
1 YK(e, (2)) 1 I
q YK(e, (12)) 1 I q + 1 YK(e ◦ τ ) (q − 2)/2 I q − 1 YK(ω) (q − s2)/2 I
s2/2 II
(q − 1) + 2 dimF IrrF(SL2) #
1 YF(e, (2)) 1 q − 1 YF(e, (12)) 1
q + 1 YF(e ◦ τ0) (q − 2)/2 q − 1 YF(ω0) r0/2
(r0+ q − 2)/2 + 2
dimF 1 q − 1 q + 1 q − 1
dimK type (e, (2)) (e, (12)) (e ◦ τ0) (ω0)
1 (e, (2)) I 1
q (e, (12)) I 1 1
q + 1 (e ◦ τ ) I 1
q − 1 (ω) I 1
II 1
Table 8: Decomposition matrix of SL2, p > 2, p | q + 1, q even
• GL2, p = 2, q odd.
In this case, p | q − 1 and p | q + 1. Let r0 = r2/r1 and s0 = s2/(s1+ 1). For type I, if all fi = pc for some c, write Ic for distinction.
dimK IrrK(GL2) # p-reg’n type JM
1 LK(τ, (2)) q − 1 (τ0, (2)) I yes q LK(τ, (12)) q − 1 (τ0, (12)) II no q + 1 LK(τ1◦ τ2) (q − 1)(q − 2 − s1)/2 (τ10 ◦ τ20) I yes
(q − 1)s1/2 (τ10, (12)) II no q − 1 LK(ω) (q − 1)(q − s0)/2 (ω0) I0 yes
(q − 1)s0/2 = r1s2/2 (τ0, (12)) I1 yes q2− 1
dimF IrrF(GL2) # 1 LF(τ0, (2)) r1 q − 1 LF(τ0, (12)) r1
q + 1 LF(τ10 ◦ τ20) r1(r1− 1)/2 q − 1 LF(ω0) r2/2
r1(r1+ r0+ 3)/2
dimF 1 q − 1 q + 1 q − 1
dimK type (τ0, (2)) (τ0, (12)) (τ10 ◦ τ20) (ω0)
1 (τ, (2)) I 1
q (τ, (12)) II 1 1
q + 1 (τ1 ◦ τ2) I 1
II 2 1
q − 1 (ω) I0 1
I1 1
Table 9: Decomposition matrix of GL2, p = 2, q odd
• SL2, p = 2, q odd