For completeness, we show that SLn have (U, p)-property for p not dividing q, which is a consequence of Kleshchev’s theorem.
Proposition 6.12. Given an p-regular admissible symbol [(τ , ν)], there is some n-admissible symbol [(σ, λ)], such that
(1) [(σ∗, λ∗)] = [(τ , ν)].
(2) κSLn(LK(σ, λ)) = κSLn(LF(τ , ν)).
Proof. (1) Write τ = (τ1, · · · , τa) and ν = (ν(1), · · · , ν(a)), the index set A = {1, · · · , a}.
Let κp(τ , ν) = pcfor some integer c ≥ 0, and κp0(τ , ν) = |J0|, where J0 = {β ∈ Op0(F×q) | β · [(τ , ν)] = [(τ , ν)]} = hρi. Since pc| (νi)0 for all i, choose λ(i) such that ν(i) = [pc]λ(i), and let λ = (λ(1), · · · , λ(a)).
Given some τi1, consider the orbit O ⊂ A such that [τi1], [ρτi1], · · · , [ρb−1τi1] are all distinct, [ρh−1τi1] = [τih] for some ih ∈ O, and the corresponding ν(ih)= ν(i1), 1 ≤ h ≤ b.
We may repick τih such that ρh−1τi1 = τih without changing [(τ , ν)]. By Proposition 4.6, all τih have same degree d, hence by Lemma 4.18, there is some p-element υ ∈ F×q, such that σih= τihυ have degree pcd.
Then σ = (σ1, · · · , σa) is constructed by considering all orbits of A. It is not hard to see that [(σ∗, λ∗)] = [(τ , ν)], so this proves (1).
(2) First we show that κp0(LK(σ, λ)) = κp0(τ , ν). Let I0 = {β ∈ Op0(F×q) | β ·[(σ, λ)] = [(σ, λ)]}. If β ∈ I0 send [σi1] to [σi2], then the corresponding λ(i1) = λ(i2), hence ν(i1) = ν(i2) and β also send [τi1] to [τi2], so I0 ⊂ J0.
Conversely, we want to prove ρ ∈ I0. If p not divide q − 1, then [(σ, λ)] = [(τ , ν)], so assume p | q − 1. From the orbit O above, we have τih = ρh−1τi1, thus ρ sends σih to σih+1 for 1 ≤ h ≤ b − 1. Thus it suffices to show that [ατi1] = [τi1] implies [ασi1] = [σi1] for α = ρb. There is some integer j ≥ 0 such that ατi1 = (τi1)qj, so α = (τi1)qj−1. Take
|υ| = pe for some integer e ≥ 0. Since p | q − 1, by Lemma 4.9 pe | qpe − 1. Then with αq = α we have
αpe = α1+qj+···+q(pe−1)j = (τi1)qjpe−1= (τi1υ)qjpe−1
hence [αpeσi1] = [σi1]. Now α is p-regular, so α also fixed [σi1].
Next we shall prove that κp(LK(σ, λ)) = κp(τ , ν), that is, |I| = pc, where I = {β ∈ Op(F×q) | β · [(σ, λ)] = [(σ, λ)]}. But since [(σ, λ)] is p-non-repeated by construction, this implies I =Ta
i=1Ii, Ii = {β ∈ Op(F×q) | [βσi] = [σi]} for each i. Then Lemma 4.18 claim that |Ii| = pc for each i, thus also for |I|.
Theorem 6.13. Let W be an irreducible F SLn-module. Then there is an irreducible KSLn-module WK, such that W is a composition factor of WK with multiplicity 1, and if U is a composition factor of WK, then U D W .
Proof. Write W = YF(τ , ν; j0). Then by Proposition 6.12, there is some [(σ, λ)] ∈
ΣK such that [(σ∗, λ∗)] = [(τ , ν)] and κSLn(LK(σ, λ)) = κSLn(LF(τ , ν)). A similar consideration to Theorem 6.8 shows that each YK(σ, λ; i) contains κ∗/κ = 1 composition factors of the form YF(σ∗, λ∗; j). Relabel i such that YK(σ, λ; i) ∼= YF(τ , ν; i) for i = 1, · · · , κ, then WK = YK(σ, λ; j0) is the required module.
Corollary 6.14. SLn(Fq) has (U, p)-property for p not dividing q.
Proof. This is equivalent to the previous theorem.
7 Conclusion
The main result is Theorem 6.8, that SLn(q) has (C, p)-property of non-defining char-acteristic. In the following we list a table of related known result. Recall that (C) ⇒ (L00) ⇒ (L0) ⇒ (L) and (R) ⇒ (L), (U ).
(C) (L00) (L0) (L) (R) (U )
Sn ×
An ? ? × ?
p6 | q, GLn(q) ×
p6 | q, SLn(q) ×
abelian group
p-solvable group × × ×
Table 2: The property table for some families of groups
• Sn, symmetric group.
James’ Regularization Theorem [J1] is exactly (C), and (U ) is the consequence that the labels of irreducible Brauer characters are p-regular partitions. A coun-terexample of (R) is S6 for p = 3, which the decomposition matrix is listed in Table 1.
• An, alternating group.
Huang shows that (L0) holds [H]. The difficulty of alternating groups is that the Mullineax map distorts the dominance order of partitions, while (C) and (U ) needs some partial order. It is also not easy to prove that (L00) is true or not, either, again due to the Mullineax map, which combines entries when generating the decomposition matrix of An from Sn.
• GLn(q), p6 | q, finite general linear group of non-defining characteristic.
An analogue of James’ Regularization Theorem for GLn(q) gives property (C) [K, Theorem 5.4], which combines the result of [J] and the end-part of [D2].
Kleshchev-Tiep also proves (U ) for both GLn(q) and SLn(q), p6 | q [K, Theorem 6.3].
• SLn(q), p6 | q, finite special linear group of non-defining characteristic.
Kleshchev-Tiep only proves (U ), while he gives a powerful Kleshchev-Tiep’s theo-rem (Theotheo-rem 5.2), which can be use to prove (C), the main result of this thesis, implying (L00), (L0), (L).
• abelian group
Since every characters are of degree 1, they remains irreducible after reduction modulo p, hence (C) holds automatically. Since d is surjective, the decomposition matrix is of full rank, so (R) also holds.
• p-solvable group
The well-known Fong-Swan Theorem [S, Theorem 38] shows that (R) holds, while any non-abelian p-group is a counterexample of (L0).
Counterexamples of (R) for An, GLn(q) and SLn(q) are in Appendix A.4.
A Appendix
A.1 The Original Problem
Problem 1 in the introduction comes from exercise 16.6 in the textbook of Serre [S].
We start from the cde-triangle.
Fixed a finite group G and a prime p. Recall that given K ⊂ Q a field of characteristic 0, we may pick a valuation ring A, and obtain its residue field F = A/ m of characteristic p (see the first paragraph of section 1.6.) Here we need K and F to be sufficiently large for G.
Let RK(G) be the Grothendieck group of KG-modules (see the last paragraph of section 1.6), with the basis βK = {[Vi]K | Vi ∈ IrrK(G)}. Similarly, let RF(G) be the Grothendieck group of F G-modules, with the basis βF = {[Ej]F | Ej ∈ IrrF(G)}. Then the decomposition map is defined to be
d : RK(G) → RF(G), [V ]K 7→ [V ]F
where V is a reduction modulo p of V (see section 1.7). The decomposition matrix of G is the transpose of the matrix of d with respect to βK, βF.
Now we briefly recall the definitions and properties of projective modules.
Definition A.1. Let R be a ring, P be an left R-module.
(1) We say P is projective if any of the following equivalent condition holds:
(i) P is a direct summand of some free R-module.
(ii) Given any surjective R-module homomorphism f : E → E0, and any R-module homomorphism g0 : P → E0, there exists a homomorphism g : P → E such that g0 = f ◦ g.
(iii) For any exact sequence 0 → E1 → E → E2 → 0 of left R-modules, the sequence 0 → F (E1) → F (E) → F (E2) → 0 is also exact, where F is the functor E 7→ HomR(P, E).
(2) Assume R is artinian. We say an R-module homomorphism f : E → E0 is essential, if f (E) = E0, and f (U ) 6= E0 for any proper submodule U of E. Note that if f is essential, then f must be surjective.
(3) We say P is a projective envelope of an R-module E, if P is projective, and there exists an essential homomorphism f : P → E.
In particular, the group ring F G is artinian.
Proposition A.2. Let E be an F G-module.
(1) There exists a projective envelope of E, unique up to isomorphism.
(2) Let Pj be the projective envelope of Ej ∈ IrrF(G) for each j. Then Pj is indecom-posible, and every projective F G-module is a direct sum of these Pj.
Proof. For a proof, see [S, Proposition 41] and its corollaries.
Therefore, we may define PF(G) to be the Grothendieck group of projective F G-modules with the basis βP = {[Pj]P}, where Pj is the projective envelope of Ej ∈ IrrF(G) in the same order as βF. Then we naturally have the Cartan homomorphism:
c : PF(G) → RF(G), [P ]P 7→ [P ]F
which roughly means the decomposition of a projective module into its composition factors.
To define the map e, we consider the projective AG-modules, where A is the valuation ring with F = A/ m. Given ˆE an AG-module, the quotient ˆE/m ˆE is an F G-module.
Denote this map πm. Proposition A.3.
(1) Every projective AG-module is a direct sum of indecomposible projective modules Qj, which is characterized (up to isomorphism) by πm(Qj) = Pj.
(2) Let PA(G) be the Grothendieck group of projective AG-modules. Then we may identify PA(G) and PF(G) via the map πm.
Proof. For a proof, see [S, Proposition 42] and its corollaries.
Now the map e is given by
e : PF(G) → RK(G), [P ]P 7→ [ ˆP ]K
where ˆP = KG ⊗AGπ−1m (P ). That is, first pass P to the AG-module via the inverse of πm, then tensor with K to obtain the KG-module ˆP . It roughly means that the there is an inverse of reduction modulo p on projective F G-modules.
Here we list some basic properties of the cde-triangle.
Proposition A.4.
(1) c = d ◦ e. Hence the following diagram commutes.
PF(G) RF(G)
RK(G)
c
e d
(2) With the basis βK, βF, βP, the matrix of d is transpose to the matrix of e. Hence the matrix of c (the Cartan matrix) is symmetric.
(3) d is surjective. Hence e is a split injection.
Proof. See section 15.4 and 16.1 of [S].
Note that in this thesis, the decomposition matrix of G is actually the matrix of e, the way as the decomposition matrix of symmetry groups list by James.
Despite of RK(G), we are often more interesting on R+K(G) = {[V ]K}, the case that there indeed exists a KG-module V , which can be characterized as {P
ici[Vi]K | ci ∈ Z, ci ≥ 0, Vi ∈ IrrK(G)}. Similarly we may define R+F(G) and P+F(G). Then in general, d no more sends RK+(G) onto R+F(G), while it sends RK(G) onto RF(G). We then have the condition (R).
(R) d(R+K(G)) = RF+(G). That is, d sends R+K(G) onto R+F(G).
For the map e, we may also consider a condition (E).
(E) e(P+F(G)) = e(PF(G)) ∩ R+K(G). That is, e sends R+K(G) onto the ”positive part”
of its original image.
The exercise 16.6 of Serre [S] asks the reader to prove that
The condition (E) is equivalent to the condition (R).
But Huang had difficulty to prove it. He then wrote a letter to the original author, J. P. Serre, asking for help. Serre sent back the modified exercise,
The condition (E) is equivalent to the condition (QR).
and gave a proof for it, with a slightly weaker condition.
(QR) There is some N ∈ N such that N · RF+(G) ⊂ d(R+K(G)).
In the last part of the letter, Serre left an open problem, that whether there exists any group G (and a prime p) such that (QR) is true but (R) is false, in order to ensure the original exercise is wrong. This is Problem 1, which remains unsolved in this thesis.
A proof for the modified exercise is in the Appendix A, B of [H]. Note that the condition (R), (QR) is equivalent to the property (R), (QR) listed in the introduction, respectively.