ASYMPTOTIC BEHAVIOUR OF THE LIE MODULE

KAY JIN LIM AND KAI MENG TAN

Abstract. We obtain a projective submodule of Lie(n) in prime characteristic p for n not a p-power, and show that the ratio of its dimension to that of Lie(n) approaches 1 as n tends to infinity.

1. Introduction

The Lie module Lie(n) of the symmetric group 𝔖_{n} appears in many contexts; in particular it is closely related to the free Lie algebra. It may be defined as the left ideal of the group algebra F 𝔖_{n} generated by the ‘Dynkin-Specht-Wever element’

υ_{n} := (1 − c_{2})(1 − c_{3}) ⋯ (1 − c_{n}),

where c_{k} is the descending k-cycle (k, k − 1, …, 1). (We use the convention of composing the elements of 𝔖_{n} from right to left.)

Our main motivation comes from the work of Selick and Wu [SW1]. Their problem is to find natural homotopy decompositions of the loop suspension of a p-torsion suspension where p is a prime. In [SW1] it is proved that this problem is equivalent to the algebraic problem of finding natural coalgebra decompositions of the primitively generated tensor algebras over the field with p elements. They determine the finest coalgebra decomposition of a tensor algebra (over arbitrary fields), which can be described as a functorial Poincaré-Birkhoff-Witt theorem [SW1, Theorem 6.5]. In order to compute the factors in this decomposition, one must know a maximal projective submodule, called Lie^{max}(n), of the Lie module Lie(n).

The projective modules for the symmetric groups over fields of positive characteristic are not known. Their structure depends on the decomposition matrices for symmetric groups, and the determination of the latter is a famous open problem, for which a complete solution does not seem forthcoming in the near future. However, according to [SW2], it would be interesting to know, even if the modules cannot be computed precisely, how quickly the dimensions grow, and whether or not the growth rate is exponential. Determination of Lie^{max}(6) and Lie^{max}(8) in characteristic 2 in [SW2] suggests that Lie^{max}(n) is relatively large compared with Lie(n). If this is true in general, it would have the desirable consequence that the factors in the functorial PBW theorem are relatively small.

Date: September 2010.

2000 Mathematics Subject Classification. 20C30, 20G05.

Supported by MOE Academic Research Fund R-146-000-135-112.

Some progress has been made in the understanding of Lie^{max}(n) for the case where n = pk with p ∤ k. Among other things, Erdmann and Schocker [ES] established a one-to-one correspondence between the indecomposable non-projective summands of Lie(pk) and the indecomposable summands of Lie(k). Erdmann and the second author [ET] provided an upper bound for the dimension of Lie^{max}(pk) and showed that the ratio of this upper bound to the dimension of Lie(pk) approaches 1 as k tends to infinity, and conjectured the same to hold for the ratio of the dimension of Lie^{max}(pk) to that of Lie(pk).

The main result of this paper proves this conjecture and more:

Theorem 1. Let A = ℤ^{+} \ {p^{m} | m ∈ ℤ_{≥0}}. Then

\frac{dim(Lie^{max}(n))}{dim(Lie(n))} → 1

as n → ∞ in A.

This theorem in particular shows that dim(Lie^{max}(n)) grows exponentially with n for n not a p-power.

Our approach is to exploit the decomposition theorem of Bryant and Schocker ([BS]) via the Schur functor. This provides a projective submodule of Lie(n) when n is not a p-power. We describe its dimension via a recurrence relation and show that its ratio to the dimension of Lie(n) approaches 1 as n tends to infinity. The second author thanks Roger Bryant for his suggestion of this approach.

Our paper is organised as follows: we introduce the notations and background theory, as well as obtain some simple preliminary results, in the next section, while in Section 3, we describe the dimension of a projective submodule of Lie(n) for n not a p-power via a recurrence relation, and show that its ratio to (n − 1)!, the dimension of Lie(n), approaches 1 as n tends to infinity.

Throughout this paper, we fix a field F of prime characteristic p. All our algebras and vector spaces are defined over F. All our tensor products are taken over F too.

2. Preliminaries

In this section, we give an account of the background theory and prove some preliminary results which we shall require.

2.1. Schur algebra. Let n, r ∈ ℤ^{+}. Let I(n, r) be the set of r-tuples of positive integers between 1 and n (both inclusive) and let Λ(n, r) be the set of compositions of r with n parts, i.e.

I(n, r) = {(i_{1}, …, i_{r}) ∈ (ℤ^{+})^{r} | 1 ≤ i_{1}, …, i_{r} ≤ n},
Λ(n, r) = {(λ_{1}, …, λ_{n}) ∈ (ℤ_{≥0})^{n} | ∑_{i=1}^{n} λ_{i} = r}.

To each (i_{1}, …, i_{r}) ∈ I(n, r), we associate a (λ_{1}, …, λ_{n}) ∈ Λ(n, r) by λ_{j} = |{k | i_{k} = j}|. The symmetric group 𝔖_{r} acts on the right of I(n, r) by place permutations, i.e. (i_{1}, …, i_{r})·σ = (i_{σ(1)}, …, i_{σ(r)}), and the orbits under this action are precisely indexed by Λ(n, r), i.e. two r-tuples in I(n, r) lie in the same 𝔖_{r}-orbit if and only if they are associated to the same composition of r. This induces a right action of 𝔖_{r} on I^{2} = (I(n, r))^{2} via (i, j) · σ = (i · σ, j · σ). We write I^{2}/ ∼ for a set of orbit representatives under the action.

We note also that 𝔖_{n} acts on the right of Λ(n, r) by place permutations. The Schur algebra S(n, r) has a distinguished basis {ξ_{i,j} | (i, j) ∈ I^{2}/ ∼}.

We write ξ_{α} for ξ_{i,i} where α ∈ Λ(n, r) is the composition of r associated to i. These ξ_{α}’s are mutually orthogonal idempotents of S(n, r), and 1_{S(n,r)} = ∑_{α∈Λ(n,r)} ξ_{α}. They thus provide a (vector space) decomposition of any left S(n, r)-module M = ⨁_{α∈Λ(n,r)} M^{α}, where M^{α} := ξ_{α}M is the weight space of M associated to α.

Let s ∈ ℤ^{+}. Given a left S(n, r)-module M and a left S(n, s)-module N, one may endow a natural left S(n, r + s)-module structure on M ⊗ N.

We collate together some results relating to weight spaces.

Lemma 2 ([G, (3.3a),(3.3c)]). Let n, r, s ∈ ℤ^{+}, and let M and N be finitedimensional left S(n, r)- and S(n, s)-modules respectively.

(1) If σ ∈ 𝔖_{n} and α ∈ Λ(n, r), then M^{α} ≅ M^{α·σ} as vector spaces.

(2) If γ ∈ Λ(n, r + s), then

(M ⊗ N)^{γ} = ⨁_{\begin{matrix}α∈Λ(n,r) \\ β∈Λ(n,s) \\ α+β=γ \end{matrix}} M^{α} ⊗ N^{β}.

(Here, and hereafter, if α = (α_{1}, …, α_{n}) ∈ Λ(n, r) and β = (β_{1}, …, β_{n}) ∈

Λ(n, s), then α + β = (α_{1} + β_{1}, …, α_{n} + β_{n}) ∈ Λ(n, r + s).)

In the case where n ≥ r, let ω_{r} = (\overbrace{1, …, 1}^{r times}, \overbrace{0, …, 0}^{n−r times}) ∈ Λ(n, r). It is wellknown that ξ_{ω_{r}}S(n, r)ξ_{ω_{r}} ≅ F 𝔖_{r} as F -algebras. This induces the Schur functor f_{r} : _{S(n,r)}mod → _{F 𝔖_{r}}mod which sends a left S(n, r)-module M to its weight space M^{ω_{r}}.

2.2. Lie module. Let n ∈ ℤ^{+}, and let 𝔖_{n} denote the symmetric group on n letters. We use the convention of composing the elements of 𝔖_{n} from right to left. The Lie module Lie(n) may be defined as the left ideal of F 𝔖_{n} generated by the Dynkin-Specht-Wever element

υ_{n} = (1 − c_{2})(1 − c_{3}) … (1 − c_{n}),

where c_{i} ∈ 𝔖_{n} is the descending i-cycle (i, i − 1, …, 1).

It is well-known that υ_{n}^{2} = nυ_{n}. Thus when p ∤ n, \frac{1}{n}υ_{n} is an idempotent, and hence Lie(n) is projective, while Lie(n) is non-projective when p | n. We also have the following identity.

Lemma 3. Let r, n ∈ ℤ^{+} with 2 ≤ r ≤ n. Then υ_{n} = −υ_{r−1}c_{r}υ_{n} (where υ_{1} = 1).

Proof. We note first that

υ_{r} = υ_{r−1}(1 − c_{r}) = υ_{r−1} − υ_{r−1}c_{r};
υ_{s}υ_{n} = υ_{s}υ_{s}(1 − c_{s+1}) ⋯ (1 − c_{n}) = sυ_{s}(1 − c_{s+1}) ⋯ (1 − c_{n}) = sυ_{n}

for all s = 1, …, n. Thus

(1 + υ_{r−1}c_{r})υ_{n} = (1 + υ_{r−1} − υ_{r})υ_{n} = υ_{n} + (r − 1)υ_{n} − rυ_{n} = 0.

□

For ρ ∈ 𝔖_{n} \ 𝔖_{{2,...,r}} with ρ(r) = 1, we note that ρυ_{r−1}c_{r} ∈ F 𝔖_{{2,...,n}}, where here, and hereafter, 𝔖_{{2,...,n}} = {σ ∈ 𝔖_{n} | σ(1) = 1}. Thus, using Lemma 3, we see that Lie(n) is spanned by {συ_{n} | σ ∈ 𝔖_{{2,...,n}}} as a vector space. It is well-known that this spanning set is in fact a basis, so that Lie(n) has dimension (n − 1)!.

2.3. Modular Lie powers. Let M be a left S(n, r)-module, and let s ∈ ℤ^{+}. The s-fold tensor product M^{⊗s} is then a left S(n, rs)-module, and it also admits another commuting right action of 𝔖_{s} by place permutations, i.e. (m_{1} ⊗ ⋯ ⊗ m_{s}) · σ = m_{σ(1)} ⊗ ⋯ ⊗ m_{σ(s)} (m_{1}, …, m_{s} ∈ M, σ ∈ 𝔖_{s}).

Let N_{1}, …, N_{s} be vector subspaces of M. Then (N_{1}⊗⋯⊗N_{s})σ is a vector subspace of M^{⊗s} for all σ ∈ 𝔖_{s}. We write [N_{1}, …, N_{s}] for (N_{1} ⊗⋯⊗N_{s})υ_{s}, where υ_{s} is the Dynkin-Specht-Wever element generating the Lie module Lie(s). This is a vector subspace of ∑_{σ∈𝔖_{s}}(N_{1} ⊗ ⋯ ⊗ N_{s})σ.

Lemma 4. Let N_{1}, …, N_{s} be vector subspaces of M whose sum is direct. For each i, let B_{i} be a basis for N_{i}. Then ∑_{σ∈𝔖_{s}}[N_{σ(1)}, …, N_{σ(s)}] has a basis

{(b_{1} ⊗ b_{2} ⊗ ⋯ ⊗ b_{s})συ_{s} | b_{i} ∈ B_{i}, σ ∈ 𝔖_{{2,...,s}}}.

Proof. Let Y = {(b_{1} ⊗ b_{2} ⊗ ⋯ ⊗ b_{s})συ_{s} | b_{i} ∈ B_{i}, σ ∈ 𝔖_{{2,...,s}}}. Note first that [N_{σ(1)}, …, N_{σ(s)}] = (N_{1} ⊗ ⋯ ⊗ N_{s})συ_{s}. If σ ∉ 𝔖_{{2,...,s}}, we may choose t so that σ(t) = 1. Then συ_{s} = συ_{t−1}c_{t}υ_{s}, and συ_{t−1}c_{t} ∈ F 𝔖_{{2,...,s}}. Thus Y spans ∑_{σ∈𝔖_{s}}[N_{σ(1)}, …, N_{σ(s)}].

To show that Y is linearly independent, let B = {b_{1}⊗⋯⊗b_{s} | b_{i} ∈ B_{i} ∀i}, and suppose that ∑_{σ∈𝔖_{{2,...,s}}} ∑_{b∈B} λ_{b,σ}bσυ_{s} = 0. Since the sum of the N_{i}’s is direct, there exists a basis C for M such that B_{i} ⊆ C for all i. Then C = {c_{1} ⊗ ⋯ ⊗ c_{s} | c_{1}, …, c_{s} ∈ C} is a basis for M^{⊗s}, and C ⊇ B. Fix a b_{0} ∈ B; then b_{0}𝔖_{s} ⊆ C. Let ϕ be the projection of M^{⊗s} onto the vector subspace spanned by b_{0}𝔖_{s}, with ϕ(c) = 0 for all c ∈ C \ b_{0}𝔖_{s}. Then for b ∈ B and σ ∈ 𝔖_{{2,...,s}}, we have ϕ(bσυ_{s}) = δ_{b,b_{0}}b_{0}συ_{s}. Thus,

0 = ϕ ( ∑_{σ∈𝔖_{{2,...,s}}} ∑_{b∈B} λ_{b,σ}bσυ_{s}) = ∑_{σ∈𝔖_{{2,...,s}}} λ_{b_{0},σ}b_{0}συ_{s}.

This is a linear relation on {b_{0}συ_{s} | σ ∈ 𝔖_{{2,...,s}}}. The span of this set is isomorphic to Lie(s) via the correspondence b_{0}συ_{s} ↔ συ_{s}. Since {συ_{s} | σ ∈ 𝔖_{{2,...,s}}} is a well-known basis for Lie(s), the relation on {b_{0}συ_{s} | σ ∈ 𝔖_{{2,...,s}}} must be trivial. Hence Y is linearly independent. □

This immediately gives the following corollary.

Corollary 5. Let N_{1}, …, N_{s} be vector subspaces of M whose sum is direct. Then

∑_{σ∈𝔖_{s}}[N_{σ(1)}, …, N_{σ(s)}] = ⨁_{σ∈𝔖_{{2,...,n}}}[N_{σ(1)}, …, N_{σ(s)}].

Furthermore, if each N_{i} is finite-dimensional, then

dim ( ∑_{σ∈𝔖_{s}}[N_{σ(1)}, …, N_{σ(s)}]) = (s − 1)! ∏_{i=1}^{s} dim(N_{i}).

When N_{1}, …, N_{s} are S(n, r)-submodules of M, then (N_{1} ⊗ ⋯ ⊗ N_{s})σ (σ ∈ 𝔖_{s}) and [N_{1}, …, N_{s}] are all S(n, rs)-submodules of M^{⊗s}. We have the following lemma relating the weight spaces of [N_{1}, …, N_{s}] of those of N_{1}, N_{2}, …, N_{s}.

Lemma 6. Let M be a left S(n, r)-module, with submodules N_{1}, …, N_{s}, and let α ∈ Λ(n, rs). Then

[N_{1}, …, N_{s}]^{α} = ∑_{\begin{matrix}β^{(1)},...,β^{(s)}∈Λ(n,r) \\ β^{(1)}+···+β^{(s)}=α \end{matrix}}[N_{1}^{β^{(1)}}, …, N_{s}^{β^{(s)}}].

Proof. We have

[N_{1}, …, N_{s}] = [ ⨁_{β^{(1)}∈Λ(n,r)} N_{1}^{β^{(1)}}, …, ⨁_{β^{(s)}∈Λ(n,r)} N_{s}^{β^{(s)}} ]

= ∑_{β^{(1)},...,β^{(s)}∈Λ(n,r)}[N_{1}^{β^{(1)}}, …, N_{s}^{β^{(s)}}].

Since (N_{1}^{β^{(1)}} ⊗ ⋯ ⊗ N_{s}^{β^{(s)}})σ ⊆ (M^{β^{(1)}} ⊗ ⋯ ⊗ M^{β^{(s)}})σ ⊆ (M^{⊗s})^{∑_{i=1}^{s} β^{(i)}} for all σ ∈ 𝔖_{s} by Lemma 2(2), we have

[N_{1}^{β^{(1)}}, …, N_{s}^{β^{(s)}}] = (N_{1}^{β^{(1)}} ⊗ ⋯ ⊗ N_{s}^{β^{(s)}})υ_{s} ⊆ (M^{⊗s})^{∑_{i=1}^{s} β^{(i)}}.

Thus, ξ_{γ}[N_{1}^{β^{(1)}}, …, N_{s}^{β^{(s)}}] = 0 for all γ ∈ Λ(n, rs) \ {∑_{i=1}^{s} β^{(i)}}, so that

[N_{1}, …, N_{s}]^{α} = ξ_{α}[N_{1}, …, N_{s}]

= ξ_{α} ( ∑_{β^{(1)},...,β^{(s)}∈Λ(n,r)}[N_{1}^{β^{(1)}}, …, N_{s}^{β^{(s)}}])

= ∑_{\begin{matrix}β^{(1)},...,β^{(s)}∈Λ(n,r) \\ β^{(1)}+···+β^{(s)}=α \end{matrix}}[N_{1}^{β^{(1)}}, …, N_{s}^{β^{(s)}}].

□

An n-dimensional vector space V has a natural left S(n, 1)-module struc-

ture. Thus V^{⊗r} and L^{r}(V) := [\overbrace{V, …, V}^{r times}] are naturally left S(n, r)-modules. When n ≥ r, it is well-known that the Schur functor f_{r}, sends V^{⊗r} and L_{r}(V) to F 𝔖_{r} and Lie(r) respectively.

3. Main results

In this section, we describe the dimension of a projective submodule of Lie(n) when n is not a p-power via a recurrence relation, and show that its ratio to dim(Lie(n)) = (n − 1)! approaches 1 as n tends to infinity.

We begin with the decomposition theorem of Bryant and Schocker.

Theorem 7 ([BS, Theorem 4.4]). Let V be an n-dimensional vector space, and let k be a positive integer not divisible by p. Then for each r ∈ ℤ_{≥0}, there exists a direct summand B_{p^{r}k} of V^{⊗p^{r}k} (as an S(n, p^{r}k)-submodule) such that

L^{p^{m}k}(V) = L^{p^{m}}(B_{k}) ⊕ L^{p^{m−1}}(B_{pk}) ⊕ ⋯ ⊕ L^{1}(B_{p^{m}k})

for all m ∈ ℤ_{≥0}.

We may assume that n is ‘large’ for all our purposes; then under the direct-sum-preserving Schur functor f_{p^{m}k}, we have a decomposition of

(∗) Lie(p^{m}k) = f_{p^{m}k}L^{p^{m}}(B_{k}) ⊕ f_{p^{m}k}L^{p^{m−1}}(B_{pk}) ⊕ ⋯ ⊕ f_{p^{m}k}L^{1}(B_{p^{m}k}).

Since B_{p^{m}k} is a direct summand of V^{⊗p^{m}k}, we see that f_{p^{m}k}L^{1}(B_{p^{m}k}) = f_{p^{m}k}B_{p^{m}k} is a projective submodule of Lie(p^{m}k). Its dimension is then a lower bound for dim(Lie^{max}(p^{m}k)). Unfortunately, the abstract description of B_{p^{m}k} given in [BS] makes its image under the Schur functor difficult to describe. We get around this by using (∗) to provide a recurrence relation on dim(f_{p^{m}k}B_{p^{m}k}).

We have the following description of f_{rs}(L^{s}(M)), in which we use the following notations. Let n, r, s ∈ ℤ^{+} with n ≥ rs. We write [rs] for {1, 2, …, rs}. Let A denote the set of r-element subsets of [rs], and let ≥ be any total ordering on A. If A ∈ A, we write ω_{A} ∈ Λ(n, rs) for the composition (I_{A}(1), I_{A}(2), …, I_{A}(n)), where I_{A}(x) = 1 if x ∈ A, and = 0 otherwise.

Proposition 8. Let n, r, s ∈ ℤ^{+} with n ≥ rs, and let M be an S(n, r)- module. Then

f_{rs}L^{s}(M) = ⨁_{\begin{matrix}A_{1},...,A_{s}∈A \\ A_{1}>···>A_{s} \\ ⋃_{i=1}^{s} A_{i}=[rs] \end{matrix}} ⨁_{σ∈𝔖_{\{2,...,s\}}}[M^{ω_{A_{σ(1)}}} , …, M^{ω_{A_{σ(s)}}} ].

In particular, if M is finite-dimensional, then

dim(f_{rs}L^{s}(M)) = \frac{(rs)!}{s} (\frac{dim(f_{r}M)}{r!})^{s} .

Proof. We have

f_{rs}L^{s}(M) = (L^{s}(M))^{ω_{rs}} = ∑_{\begin{matrix}α^{(1)},...,α^{(s)}∈Λ(n,r) \\ α^{(1)}+···+α^{(s)}=ω_{rs} \end{matrix}}[M^{α^{(1)}}, …, M^{α^{(s)}}]

by Lemma 6. Note that if α^{(1)}, …, α^{(s)} ∈ Λ(n, r) such that ∑_{i=1}^{s} α^{(i)} = ω_{rs}, then for each i, α^{(i)} = ω_{A_{i}} for some A_{i} ∈ A, and ⋃_{i=1}^{s} A_{i} = [rs]. Thus,

f_{rs}L^{s}(M) = ∑_{\begin{matrix}A_{1},...,A_{s}∈A \\ A_{1}>···>A_{s} \\ ⋃_{i=1}^{s} A_{i}=[rs] \end{matrix}} ∑_{σ∈𝔖_{s}}[M^{ω_{A_{σ(1)}}} , …, M^{ω_{A_{σ(s)}}} ].

The outer sum is direct by Lemma 2(2), while the inner sum, by Corollary 5, equals

⨁_{σ∈𝔖_{{2,...,s}}}[M^{ω_{A_{σ(1)}}} , …, M^{ω_{A_{σ(s)}}} ],

and has dimension (s − 1)! ∏_{i=1}^{s} dim(M^{ω_{A_{σ(i)}}} ) = (s − 1)!(dim(M^{ω_{r}}))^{s} by Lemma 2(1) when M is finite-dimensional. Thus

dim(f_{rs}L^{s}(M)) = \frac{(rs)!}{(r!)^{s}s!}(s − 1)! dim(f_{r}M)^{s},

and the proposition follows. □

Definition 9. For each m ∈ ℤ_{≥0} and each k ≥ 2 be an integer not divisible by p, let z_{m,k} = dim(f_{p^{m}k}B_{p^{m}k})/(p^{m}k − 1)!.

In other words, z_{m,k} is the ratio of the dimension of the projective submodule f_{p^{m}k}B_{p^{m}k} to that of the Lie module Lie(p^{m}k).

Theorem 10. Let k ≥ 2 be an integer not divisible by p. Then z_{m,k} satisfies the following recurrence relation:

z_{0,k} = 1,  and  z_{m,k} = 1 − ∑_{i=0}^{m−1} \frac{z_{i,k}^{p^{m−i}}}{(p^{i}k)^{p^{m−i}−1}} (m > 0).

Proof. From (∗) with m = 0, we see that Lie(k) = B_{k}, so that z_{0,k} = 1. By Proposition 8, we have

dim(f_{p^{m}k}L^{p^{m−i}}(B_{p^{i}k})) = \frac{(p^{m}k)!}{p^{m−i}} (\frac{dim(f_{p^{i}k}B_{p^{i}k})}{(p^{i}k)!})^{p^{m−i}}

= \frac{(p^{m}k − 1)!}{(p^{i}k)^{p^{m−i}−1}} z_{i,k}^{p^{m−i}}.

The theorem now follows from (∗). □

Theorem 11. Let A = ℤ^{+} \ {p^{m} | m ∈ ℤ_{≥0}}. For n ∈ A, let z_{n} = z_{m,k} where n = p^{m}k with p ∤ k. Then z_{n} → 1 as n → ∞ in A.

Proof. Let a_{i} = (p^{i}k)^{p^{m−i}−1}. For 1 ≤ i ≤ m − 2, we have

\frac{a_{i}}{a_{i+1}} = \frac{(p^{i}k)^{p^{m−i}−1}}{(p^{i+1}k)^{p^{m−i−1}−1}} = \frac{(p^{i}k)^{p^{m−i}−p^{m−i−1}}}{p^{p^{m−i−1}−1}} = \frac{(p^{i}k)^{p^{m−i−1}(p−1)}}{p^{p^{m−i−1}−1}}

≥ (\frac{p^{i}k}{p})^{p^{m−i−1}} = (p^{i−1}k)^{p^{m−i−1}} ≥ 1.

Thus

1 ≥ z_{m,k} = 1 − ∑_{i=0}^{m−1} \frac{z_{i,k}^{p^{m−i}}}{a_{i}} ≥ 1 − \frac{1}{a_{0}} − \frac{m − 1}{a_{m−1}} = 1 − \frac{1}{k^{p^{m}−1}} − \frac{m − 1}{(p^{m−1}k)^{p−1}} .

The theorem now follows. □

We end the paper by noting that Theorem 1 follows immediately from Theorem 11.

References

[BS] R. M. Bryant and M. Schocker, ‘The decomposition of Lie powers’, Proc. London Math. Soc. (3) 93 (175–196), 2006.

[ES] K. Erdmann, M. Schocker, ‘Modular Lie powers and the Solomon descent algebra’, Math. Z. 253 (295–313), 2006.

[ET] K. Erdmann, K. M. Tan, ‘The Lie module of the symmetric group’, Int. Math. Res. Not. 2010, article ID rnp244, 23 pages, doi:10.1093/imrn/rnp244, 2010.

[G] J. A. Green, Polynomial Representations of GL_{n}, Lecture Notes in Mathematics 830, Springer, 2007.

[SW1] P. Selick, J. Wu, ‘Natural coalgebra decomposition of tensor algebras and loop suspensions’, Mem. Amer. Math. Soc. 148, 2000.

[SW2] P. Selick, J. Wu, ‘Some calculations for Lie(n)^{max} for low n’, J. Pure Appl. Algebra 212 (2570–2580), 2008.

Department of Mathematics, National University of Singapore, Block S17, 10 Lower Kent Ridge Road, Singapore 119076.

E-mail address, K. J. Lim: matlkj@nus.edu.sg

E-mail address, K. M. Tan: tankm@nus.edu.sg