INNER PLETHYSM IN THE REPRESENTATION RING OF THE GENERAL LINEAR GROUP*

Essam A. Abotteen Emporia State University

The investigation of plethysms (inner and outer) in the representation theory of finite classical groups has been one of the important outstanding problems in the representation theory of the symmetric group [5], [7] and [8]. The fundamental theorem of the representation theory of the symmetric group has been more or less known since the origins of the subject with Frobenius at the turn of the century. This theorem states there is an isomorphism between the representation ring of the symmetric groups S_{n} and the ring of symmetric polynomials in an infinite number of variables. But for various reasons this isomorphism in its pure form seems not to have appeared until Atiyah [2] introduced the Steenrod power operations in K–Theory around 1966. In [2] Atiyah described how to use the complex representations of the symmetric group S_{k}, to

* This research was partially supported by Emporia State University summer fellowship. define and investigate operations in K–Theory. Atiyah’s main technical tool is the notation of λ–ring introduced by Grothendieck [3] in 1956 in an algebraic–geometric context. The notation of λ–ring has been used by Knutson [6], to study the fundamental theorem of the representation theory of the symmetric group which has been translated into that of plethysms by Hoffman [4], Uehara and myself [1] and [9]. The main purpose of this paper is to define and investigate the inner plethysm in the representation ring R(A_{n}), where A_{n} = GL(n, K) in the nth general linear group over a finite field K. In the case A_{n} = S_{n}, the symmetric group, Hoffman [4] investigated the inner and outer plethysms in the frame work of τ–rings. The authors of [1] and [9] studied the outer plethysms for A_{n} = S_{n}, A_{n} = GL(n, K) and A_{n} = [G]S_{n}, the wreath product of a finite group G by the symmetric group S_{n}.

Let A_{n} = GL(n, K) be the nth general linear group over a finite field K. The wreath product [A_{n}]S_{n} of A_{n} by the symmetric group S_{k} (the usual notation for [A_{n}]S_{k} is A_{n}≀S_{k}) is the set A_{n}^{k}×S_{k} with a multiplication defined by (a_{1}, …, a_{k}; σ)(a′_{1}, …, a′_{k}; σ′) = (a_{1}a′_{σ^{−1}(1)}, …, a_{k}a′_{σ^{−1}(k)}; σσ′) where a_{i}, a′_{i} ∈ A_{n} for k ≥ i ≥ 1 and σ, σ′ ∈ S_{k}. For a representation M of A_{n} and for k ≥ 1, the kth tensor product of M, M^{⊗k} = M ⊗ M ⊗ ⋯ ⊗ M (k factors) is a representation of [A_{n}]S_{k} with a group action given by (a_{1}, …, a_{k}; σ)(m_{1} ⊗ m_{2} ⊗ ⋯ ⊗ m_{k}) = a_{1}m_{σ^{−1}(1)} ⊗ ⋯ ⊗ a_{k}m_{σ^{−1}(k)} for any (a_{1}, …, a_{k}; σ) ∈ [A_{n}]S_{k} and m_{i} ∈ M. In what follows ⊗ means ⊗_{c} and we interpret M^{⊗0} as the A_{n}–representation C, on which A_{n} acts trivially.

For any finite group G let R(G) be the Grothendieck representation group of G, R(G) is the free abelian group generated by the isomorphism classes of irreducible complex representations of G. It is a ring with respect to the tensor product. For every integer k ≥ 1 we have a map

⊗k : R(A_{n}) → R([A_{n}]S_{k})

defined by ⊗k([M]) = [M^{⊗k}], where [M] ∈ R(A_{n}) is the class of M.

First we are going to show ⊗k is well defined (compare Atiyah [2], proposition 2.2). Let G be any finite group and consider the semiring M(G) = {(M, N)|M, N are G−modules } with addition and multiplication defined by

(M, N) + (M′, N′) = (M ⊕ M′, N ⊕ N′)

and

(M, N).(M′, N′) = (M ⊗ M′ ⊕ N ⊗ N′, M ⊗ N′ ⊕ M′ ⊗ N) .

We define an equivalence relation ∼ on M(G) by (M, N) ∼ (M′, N′) if and only if M ⊕ N′ ≃ M′ ⊕ N. We denote by [(M, N)] the equivalence class of (M, N) ∈ M(G). Then \overline{R}(G) = M(G)/ ∼ is a ring with 0 = [(D, D)] and −[(M, N)] = [(N, M)]. It is clear from the construction that the map h : \overline{R}(G) → R(G) defined by h([(M, n)]) = [M] − [N] is a ring isomorphism. For each integer k ≥ 1, we define a map △_{k} : M(A_{n}) → M([A_{n}]S_{k}) by △_{k}(M, N) = (M, N)^{k}.

\underline{Lemma 1}. The map △_{k} is compatible with the equivalence relation ∼ on M(A_{n}).

\underline{Proof}. We have to show that if D is any A_{n}–module and (M, N) ∈ M(A_{n}) then △_{k}(M, N) ∼ △_{k}(M ⊕ D, N ⊕ D). This can be proved by induction on k. If k = 1, △_{1}(M⊕D, N⊕D) = (M⊕D, N⊕D) = (M, N)+(D, D) = (M, N) . Assume the hypothesis is true for k − 1, then

△_{k}(M ⊕ D, N ⊕ D) = (M ⊕ D, N ⊕ D)^{k−1}.(M ⊕ D, N ⊕ D)
= (M, N)^{k−1}((M, N) + (D, D))
= (M, N)^{k−1}(M, N) = △_{k}(M, N) .

Now consider the diagram

M(A_{n}) ⟶^{△_{k}} M([A_{n}]S_{k})
↓^{P} ↓^{P}
\overline{R}(A_{n}) ⟶^{\overline{⊗}k} \overline{R}([A_{n}]S_{k})
↓^{h} ↓^{h}
R(A_{n}) ⟶^{⊗k} R([A_{n}]S_{k})

where \overline{⊗}k is the map induced by △_{k} and P is the projection map.

\begin{aligned}h ∘ p ∘ △_{k}(M, 0) & = h ∘ p(M, 0)^{k} = h ∘ p(M^{⊗k}, 0) \\ & = h([(M^{⊗k}, 0)] = [M^{⊗k}] = ⊗k[M] . \end{aligned}

Similarly ⊗k ∘ h ∘ p(M, 0) = ⊗k[M]. Thus it follows that the diagram commutes and ⊗k is also induced by △_{k}, and hence the map ⊗k is well defined.

Before we state the next result we recall the following.

\underline{Definition}. Let H be a subgroup of a finite group G and M is a complex representation of H the representation of G induced by M is given by Ind_{H}^{G}M = CG ⊗_{CH} M.

\underline{Lemma 2}. If (M, N) ∈ M(A_{n}), then for any k ≥ 1,

\begin{aligned}△_{k}(M, N) & = ( ∑_{{i=0 \atop i even}}^{k} Ind_{[A_{n}]S_{k−i}×[A_{n}]S_{i}}^{[A_{n}]S_{k}}(M^{⊗(k−i)} ⊗ N^{⊗i}), \\ & ∑_{{j=1 \atop j odd}}^{k} Ind_{[A_{n}]S_{k−j}×[A_{n}]S_{j}}^{[A_{n}]S_{k}}(M^{⊗(k−j)} ⊗ N^{⊗j})) . \end{aligned}

\underline{Proof}. The proof is by induction on k. If k = 1, this is evident. Assume that the hypothesis is true for all integers m ≤ k. Then we have △_{k+1} (M, N) = (M, N)^{k}(M, N)

= ( ∑_{{i=0 \atop i even}}^{k} Ind_{[A_{n}]S_{k−i}×[A_{n}]S_{i}}^{[A_{n}]S_{k}}(M^{⊗(k−i)} ⊗ N^{⊗i}),

∑_{{j=1 \atop j odd}}^{k} Ind_{[A_{n}]S_{k−j}×[A_{n}]S_{j}}^{[A_{n}]S_{k}}(M^{⊗(k−j)} ⊗ N^{⊗j}))(M, N))

= ( ∑_{{i=0 \atop i even}}^{k} Ind_{[A_{n}]S_{k−i}×[A_{n}]S_{i}}^{[A_{n}]S_{k}}(M^{⊗(k−i)} ⊗ N^{⊗i}) ⊗ M⊕

∑_{{j=1 \atop j odd}}^{k} Ind_{[A_{n}]S_{k−j}×[A_{n}]S_{j}}^{[A_{n}]S_{k}}(M^{⊗(k−j)} ⊗ N^{⊗j}) ⊗ N,

∑_{{i=0 \atop i even}}^{k} Ind_{[A_{n}]S_{k−i}×[A_{n}]S_{i}}^{[A_{n}]S_{k}}(M^{⊗(k−i)} ⊗ N^{⊗i}) ⊗ N ⊕ M⊗

∑_{{j=1 \atop j odd}}^{k} Ind_{[A_{n}]S_{k−j}×[A_{n}]S_{j}}^{[A_{n}]S_{k}}(M^{⊗(k−j)} ⊗ N^{⊗j}))

= ( ∑_{{i=0 \atop i even}}^{k+1} Ind_{[A_{n}]S_{k+1−i}×[A_{n}]S_{i}}^{[A_{n}]S_{k+1}}(M^{⊗(k+1−i)} ⊗ N^{⊗i}),

∑_{{j=1 \atop j odd}}^{k+1} Ind_{[A_{n}]S_{k+1−j}×[A_{n}]S_{j}}^{[A_{n}]S_{k+1}}(M^{⊗(k+1−j)} ⊗ N^{⊗j})) .

\underline{Corollary 3}. For any [M] − [N] ∈ R(A_{n}) and k ≥ 1,

⊗k([M] − [N]) = ∑_{i=0}^{k}(−1)^{i}[Ind_{[A_{n}]S_{k−i}×[A_{n}]S_{i}}^{[A_{n}]S_{k}}(M^{⊗(k−i)} ⊗ N^{⊗i})] .

\underline{Proof}. Since △_{k} induces ⊗k, apply h∘p to △_{k}(M, N) as given in the Lemma and we are done.

By construction [A_{n}]S_{k} is the semi–direct product A_{n}^{k} ×_{θ} S_{k}, where θ : S_{k} → Aut (A_{n}^{k}) is a group homomorphism given by

θ(σ)((a_{1}, …, a_{k})) = (a_{σ^{−1}(1)}, …, a_{σ^{−1}(k)})

for σ ∈ S_{k}, a_{i} ∈ A_{n}. In other words, the short exact sequence

1 → A_{n}^{k} → [A_{n}]S_{k} → S_{k} → 1

is split by the obvious maps α : S_{k} → [A_{n}]S_{k} where α(σ) = (1, 1, …, 1; σ). Also the image of A_{n} in [A_{n}]S_{k} as the diagonal subgroup

{(a, a, …, a)|a ∈ A_{n}}

of A_{n}^{k}, commutes elementwise with the image of S_{k} thus we get an embedding ψ : A_{n} ×S_{k} → [A_{n}]S_{k}. Hereafter, A_{n} ×S_{k} is considered as a subgroup of [A_{n}]S_{k}.

\underline{Definition}. By the inner plethysm T_{ϕ} associated with an element ϕ ∈ R_{c}^{∗}(S_{k}) = Hom _{c}(R(S_{k}), C), we mean the map

T_{ϕ} : R(A_{n}) ⟶^{R}(A_{n}) ⊗ C ≃ R(A_{n})
↓^{⊗k}
R(A_{n}) ⊗ R(S_{k})

defined by T_{ϕ} = (1 ⊗ ϕ) ∘ (⊗k). In the sequel for any [M] ∈ R(A_{n}) we denote T_{ϕ}([M]) by ϕ([M]), if no confusion arises. For a partition π = {1^{π1}, 2^{π2}, …, k^{πk}} of k (in notation π ⊢ k ), let S_{π} = S_{1}^{π1} × S_{2}^{π2} × ⋯ × S_{k}^{πk}. Then a trivial representation and a sign representation of S_{π} are denoted by 1_{S_{π}} and Alt S_{π} respectively. Let ρ_{π} = [Ind_{S_{π}}^{S_{k}}1_{S_{π}}] and η_{π} = [Ind_{S_{π}}^{S_{k}} Alt S_{π}].

\underline{Lemma 3}.

{ρ_{π}|π ⊢ k} and {η_{π}|π ⊢ k} are bases for R(S_{k}) .

This fact is known, for example, see Knutson [6]. Let us consider the elements λ_{π} and σ_{π} in R^{∗}(S_{k}) defined by

λ_{π}([M]) = \{ \begin{matrix}1, & if M = Alt S_{π}; \\ 0, & otherwise. \end{matrix}

and

σ_{π}([M]) = \{ \begin{matrix}1, & if M = 1_{S_{π}}; \\ 0, & otherwise. \end{matrix}

Define a map μ : R(S_{k}) → R^{∗}(S_{k}) by μ([M])([N]) =< M, N > for [M], [N] ∈ R(S_{k}) where <, > is the Schur inner product in R(S_{k}). Then it is known that μ is a ring isomorphism and μ(ρ_{π}) = σ_{π} and μ(η_{π}) = λ_{π}, where π ⊢ k. If E is an A_{n}–representation and V is an S_{k}–representation, then Hom_{S_{k}}(V, E^{⊗k}) can be considered as an A_{n}–representation when a group action is defined by a•f = a^{⊗k} ∘f for all f ∈ Hom_{S_{k}}(V, E^{⊗k}) and a ∈ A_{n}. It is well known that if {V_{π}|π ⊢ k} is a complete set of irreducible S_{k}–representations then there exists an A_{n} × S_{k}–isomorphism.

θ : ∑_{π⊢k} Hom_{S_{k}}(V_{π}, E ⊗ k) ⊗ V_{π} → E^{⊗k}

defined by θ(f ⊗ x) = f(x) for f ∈ Hom_{S_{k}}(V_{π}, E^{⊗k}) and x ∈ V_{π}.

\underline{Theorem 5}. For any λ_{τ} ∈ R^{∗}(S_{k}) with τ ⊢ k and for any A_{n}– module M, we have

λ_{τ} ([M])[Hom_{S_{k}}(Ind_{S_{τ}}^{S_{k}}Alt S_{τ} , M^{⊗k})]

\underline{Proof}. First consider the A_{n} × S_{k} decomposition

M^{⊗k} ≃ ∑_{π⊢k} Hom_{S_{k}}(V_{π}, M^{⊗k}) ⊗ V_{π} .

Then by definition

λ_{τ} ([M]) = (1 ⊗ λ_{τ}).(⊗k)([M])

= (1 ⊗ λ_{τ})([M^{⊗k}])

= ∑_{π⊢k} Hom_{S_{k}}(V_{π}, M^{⊗k})λ_{τ} ([V_{π}])

= [Hom_{S_{k}}(∑ λ_{τ} ([V_{π}])V_{π}, M^{⊗k})]

However,

∑_{π⊢k} λ_{τ} ([V_{π}])V_{π} = ∑_{π⊢k} μ_{τ} (η_{τ})([V_{π}])V_{π}

= ∑_{π⊢k} < Ind_{S_{τ}}^{S_{k}}Alt S_{τ} , V_{π} > V_{π}

= Ind_{S_{τ}}^{S_{k}}Alt S_{τ} .

Hence we obtain λ_{τ} ([M]) = [Hom_{S_{k}}(Ind_{S_{τ}}^{S_{k}}Alt S_{τ} , M^{⊗k})]. \underline{Theorem 6}. For any partition τ = \{1^{τ1}, 2^{τ2}, …, k^{τk}\} ⊢ k and for any A_{n}–representation M, we have

λ_{τ} ([M]) = λ_{1}([M])^{τ1}λ_{2}([M])^{τ2} ⋯ λ_{k}([M])^{τk} .

\underline{Proof}. By the Frobenius reciprocity law we have

Hom_{S_{k}}(Ind Alt S_{τ} , M^{⊗k}) ≃ Hom(Alt S_{τ} , Res_{S_{τ}}^{S_{k}}M^{⊗k}) since

Alt S_{τ} ≃ (Alt S_{1})^{⊗τ1} ⊗ ⋯ ⊗ (Alt S_{k})^{⊗τk}

and

Res_{S_{π}}^{S^{k}}(M^{⊗k})M^{⊗τ1} ⊗ (M^{⊗2})^{⊗τ2} ⊗ ⋯ ⊗ (M^{⊗k})⊗τk ,

we obtain

Hom_{s_{τ}} (Alt _{S_{τ}} , Res_{s_{τ}}^{S_{k}}M^{⊗k}) ≃ ⊗_{i=1}^{k}(Hom_{S_{i}}(Alt S_{i}, M^{⊗i}))^{⊗τi} By Theorem 5 we have

λ_{τ} ([M]) = Hom_{S_{k}}(Ind_{S_{τ}}^{S_{k}}Alt S_{τ} , M^{⊗k}))]

= ∏_{i=1}^{k}[Hom_{S_{i}}((Alt S_{i}, M^{⊗i}))]^{⊗τi}

= λ_{1}([M])^{τ1}λ_{2}([M])^{τ2} ⋯ λ_{k}([M])^{τk} .

This completes the proof. \underline{Theorem 7}. For any σ_{τ} ∈ R^{∗}(S_{k}) where τ = \{1^{τ1}, …, k^{τk}\} and any A_{n}–representation M, we have

\begin{aligned}σ_{τ} ([M]) & = [Hom_{S_{k}}(ρ_{τ} , M^{⊗k})] \\ & = σ_{1}([M])^{τ1}σ_{2}([M])^{τ2} ⋯ σ_{k}([M])^{τk} \end{aligned}

The proof is similar to that of Theorem 5 and Theorem 6. To compute the inner plethysm associated with λ_{k} and σ_{k} for a general element [M] − [N] ∈ R(A_{n}), we need the following lemmas:

\underline{Lemma 8}. Let H ⊆ G ⊆ S_{n} be groups and let N be a representation of H. Then Hom_{G}(Alt G, Ind_{H}^{G}N) and Hom_{H}(Alt H, N) are isomorphic.

\underline{Proof}. We construct a linear map

p : Hom_{G}(Alt G, Ind_{H}^{G}N) → Hom_{H}(Alt H, N)

and its inverse σ. Let {e = r_{0}, r_{1}, … r_{t}} be a complete set of coset representatives for G/H. Then

Ind_{H}^{G}N ≃ N ⊕ r_{1}N ⊕ ⋯ ⊕ r_{t}N .

If U ∈ Hom_{G}(Alt G, Ind_{H}^{G}N) then there are n_{i} ∈ N_{i} such that

U(1) = n_{0} + r_{1}n_{1} + ⋯ + r_{t}n_{t} .

We let p be the linear map from C to N defined by p(U)(1) = n_{0}. p is an H–homomorphism because if h ∈ H, then

hp(U)(1) = hn_{0} = sgn(h)n_{0} = p(u)(sgn(h)) = p(f)(h1) . We now construct σ. If ω ∈ Hom_{H}(Alt H, N) and ω(1) = n_{0}, let σ be the linear map from C to N ⊕ r_{1}N ⊕ ⋯ ⊕ r_{t}N defined by

σ(ω)(1) = ∑_{i=0}^{t} sgn(r_{i})r_{i}n_{0} .

σ is a G–homomorphism because if g ∈ G, then

gσ(ω)(1) = ∑_{i=0}^{t} sgn(r_{i})r_{i}n_{0} .

Furthermore, since {gr_{0}, gr_{1}, … gr_{t}} is a set of coset representatives for G/H, there exist elements h_{0}, …, h_{t} ∈ H and there is a permutation τ of {0, …, t} such that gr_{i} = r_{τ} (i)h_{i}. Hence,

∑_{i=0}^{t} sgn(r_{i})gr_{i}n_{0} = ∑_{i=0}^{t} sgn(r_{i})r_{τ} (i)h_{i}n_{0}

= ∑_{i=0}^{t} sgn(r_{i})sgn(h_{i})r_{τ} (i)n_{0}

= ∑_{i=0}^{t} sgn(g)sgn(r_{τ} (i))r_{τ} (i)n_{o}

= ∑_{i=0}^{t} sgn(g)sgn(r_{i})r_{i}n_{o}

= sgn(g)σ(ω)(1) = σ(ω)(sgn(g)) = σ(ω)(g • 1) .

We now show that σ ∘ p is the identity. Consider

U(1) = ∑_{i=1}^{t} r_{i}n_{i}

and

σ ∘ p(U)(1) = ∑_{i=0}^{t} sgn(r_{i})r_{i}n_{0} .

It suffices to show that sgn(r_{k})n_{0} = n_{k} for all k. Since U is a G–homomorphism,

r_{k}U(1) = U(r_{k}1) = U(sgn(r_{k})) = sgn(r_{k}) ∑_{i=0}^{t} r_{i}n_{i} .

On the other hand,

r_{k}U(1) = ∑_{i=0}^{t} r_{k}r_{i}n_{i} .

Hence sgn(r_{k})r_{k}n_{k} = r_{k}n_{0} and sgn(r_{k})n_{k} = 0. The proof is complete, since it is obvious that P ∘ σ is the identity.

\underline{Lemma 9}. Let H ⊆ G ⊆ S_{n} be groups and let N be a representation of H. Then Hom_{G}(1_{G}, Ind_{H}^{G}N) and Hom_{H}(1_{H}, N) are isomorphic.

\underline{Proof}. Similar to the proof of Lemma 8.

\underline{Theorem 10}. For any [M] − [N] ∈ R(A_{n}), where M is assumed to have even grading and N to have odd grading, we have:

(i) λ_{k}([M] − [N]) = ∑_{i=0}^{k}(−1)^{i}λ_{k−i}([M])σ_{i}([N])

(ii) σ_{k}([M]) − [N]) = ∑_{i=0}^{k}(−1)^{i}σ_{k−i}([M])λ_{i}([N]) .

\underline{Proof}. (i) By definition and since N have odd grading, λ_{k}([M] − [N]) = (1 ⊗ λ_{k}) ∘ (⊗k)([M] − [N])

= (1 ⊗ λ_{k})(∑_{i=0}^{k}(−1)^{i}[Ind_{[A_{n}]S_{k−i}×[A_{n}]S_{i}}^{[A_{n}]S_{k}}(M^{⊗(k−i)} ⊗ N^{⊗i})])

= ∑_{i=0}^{k}(−1)^{i}(1 ⊗ λ_{k})[Ind_{[A_{n}]S_{k−i}×[A_{n}]S_{i}}^{[A_{n}]S_{k}}(M^{⊗(k−i)} ⊗ N^{⊗i})]

= ∑_{i=0}^{k}(−1)^{i}[Hom_{S_{k}}(Alt S_{k}, Ind_{[A_{n}]S_{k−i}×[A_{n}]S_{i}}^{[A_{n}]S_{k}}(M^{⊗(k−i)} ⊗ N^{⊗i}))]

= ∑_{i=0}^{k}(−1)^{i}[Hom_{S_{k−i}×S_{i}}(Alt (S_{k−i} × S_{i}), M^{⊗(k−i)} ⊗ N^{⊗i})]

= ∑_{i=0}^{k}(−1)^{i}[Hom_{S_{k−i}×S_{i}}(Alt S_{k−i} ⊗ S_{i}, M^{⊗(k−i)} ⊗ N^{⊗i})]

= ∑_{i=0}^{k}(−1)^{i}[Hom_{S_{k−i}}(Alt S_{k−i}M^{⊗(k−i)}) ⊗ Hom_{S_{i}}(Alt S_{i}, N^{⊗i})]

= ∑_{i=0}^{k}(−1)^{i}[Hom_{S_{k−i}}(Alt S_{k−i}M^{⊗(k−i)})][Hom_{s_{i}}(Alt S_{i}, N^{⊗i})]

= ∑_{i=0}^{k}(−1)^{i}[Hom_{S_{k−i}}(Alt S_{k−i}M^{⊗(k−i)})][Hom_{S_{i}}(1_{S_{i}}, N^{⊗i})]

= ∑_{i=0}^{k}(−1)^{i}λ_{k−i}(([M]σ_{i}([N]) .

The proof of (ii) is similar to (i). Hence, the proof is complete.

Acknowledgement. The author wishes to thank Professor Hiroshi Uehara for suggestions which improved this paper, and the Emporia State University Research and Grants Center for the financial support.

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