On Kronecker products of complex representations of the symmetric and alternating groups

C. Bessenrodt and A. Kleshchev

1 Introduction

Kronecker or inner tensor products of representations of symmetric groups (and many other groups) have been studied for a long time. But even for the symmetric groups no reasonable formula for decomposing Kronecker products of two irreducible complex representations into irreducible components is available (cf. [7, 5]). An equivalent problem is to decompose the inner product of the corresponding Schur functions into a linear combination of Schur functions.

In recent years, a number of partial results have been obtained. For example, the products of characters labelled by hook partitions or by two-row partitions [3, 8] have been computed, and special constituents, in particular of tensor squares, have been considered [10, 11, 12]. For general products, Dvir [2] and Clausen-Meier [1] determined the largest part and the maximal number of parts in a constituent of a product (this result is crucial in this paper).

In general, Kronecker products of irreducible representations have very many irreducible constituents (see e.g. [4, 2.9]). In this paper, we first consider the simple question: ‘when is the Kronecker product of two irreducible S_{n}-characters again irreducible?’ We prove that in fact such a product is always reducible, and even inhomogeneous, except for the obvious exception where one of the characters is of degree 1. Then we turn to the same question for the representations of the alternating group A_{n}. Here one can easily construct examples of non-trivial irreducible tensor products (actually, we observed this first using calculations with the MAPLE packages SF (by Stembridge) and ACE (by Veigneau et al.)). It turns out that the problem for A_{n} reduces to the classification of certain products of S_{n}-characters with 2 constituents. So we classify in general the Kronecker products of S_{n}-characters with 2 constituents, and even more generally, with two homogeneous components. We also obtain some partial results for products with 4 homogeneous components and conjecture a complete classification of the pairs (L_{1}, L_{2}) of irreducible complex S_{n}-representations such that L_{1} ⊗ L_{2} has at most 4 homogeneous components.

2 Preliminaries

We denote by ℕ the set {1, 2, …} of the natural numbers.

If G and H are two groups, L is a ℂG-module and M is a ℂH-module we write L ⊠ M for the outer tensor product of L and M (which is a module over G × H). If N is another ℂG-module we w rite L ⊗ N for the inner tensor (or Kronecker) product of L and N (which is a G-module).

A ℂG-module is called homogeneous if it is isomorphic to a direct sum of copies of one simple module. Every ℂG-module can be (uniquely) decomposed into a direct sum of its homogeneous components. Similarly we speak of the homogeneous characters and the homogeneous components of the characters.

We use the notions and notation of the representation theory of S_{n} and A_{n} and refer the reader to [4] for the most basic ones. In particular, we write λ = (λ_{1}, …, λ_{k}) ⊢ n if λ is a partition of n; in this case we also write |λ| for n. We often gather together equal parts of a partition and write, for example, (5^{2}, 3^{3}) for (5, 5, 3, 3, 3). The partition conjugate to λ is denoted by λ′. If λ = λ′ we say that λ is symmetric. We do not distinguish between a partition λ and its Young diagram λ = {(i, j) ∈ ℕ×ℕ | j ≤ λ_{i}}. Elements (i, j) ∈ ℕ×ℕ are called nodes. If λ = (λ_{1}, λ_{2}, …) and μ = (μ_{1}, μ_{2}, …) are two partitions we write λ ∩ μ for the partition (min(λ_{1}, μ_{1}), min(λ_{2}, μ_{2}), …) whose Young diagram is just the intersection of those for λ and μ. A node (i, λ_{i}) ∈ λ is called removable (for λ) if λ_{i} > λ_{i+1}. A node (i, λ_{i} + 1) is called addable (for λ) if i = 1 or i > 1 and λ_{i} < λ_{i−1}. We denote by

λ_{A} = λ \ {A} = (λ_{1}, …, λ_{i−1}, λ_{i} − 1, λ_{i+1}, …)

a partition of n − 1 obtained by removing a removable node A = (i, λ_{i}) from λ. Similarly

λ^{B} = λ ∪ {B} = (λ_{1}, …, λ_{i−1}, λ_{i} + 1, λ_{i+1}, …)

is a partition of n + 1 obtained by adding an addable node B = (i, λ_{i} + 1) to λ.

We denote by

h_{ij} = h_{ij}^{λ} = λ_{i} − j + λ′_{j} − i + 1

the (i, j)-hook length. If a partition λ has r nodes on the main diagonal and there are α_{i} (resp., β_{i}) nodes to the right of (resp., below) the node (i, i) then we may write λ in the Frobenius notation (cf. [6]):

F (λ) = ( \begin{matrix}α_{1} & ⋯ & α_{r} \\ β_{1} & ⋯ & β_{r} \end{matrix} ) .

If H_{λ} ≅ S_{λ_{1}} × S_{λ_{2}} × … < S_{n} is a Young subgroup we write M^{λ} for the permutation module ℂS_{n} ⊗_{ℂH_{λ}} 1_{H_{λ}}. The Specht module S^{λ} is explicitly defined as a submodule of M^{λ} (cf. [4]). The set {S^{λ} | λ ⊢ n} is a complete set of irreducible ℂS_{n}-modules (up to isomorphism). We write [λ] (or [λ_{1}, λ_{2}, …]) for the character of S^{λ}. Thus, {[λ] | λ ⊢ n} is a complete set of the irreducible characters of S_{n}. It is well known that S^{λ} is selfdual. Another fact (to be used without comment) is that S^{(1^{n})} is the 1-dimensional sign representation and S^{λ} ⊗ S^{(1^{n})} ≅ S^{λ′}. The standard inner product on the class functions on a group (symmetric or alternating, depending on the context) is denoted by ⟨·, ·⟩. If χ and ψ are two class functions we write χ·ψ for the function [g ↦ χ(g)ψ(g)]. The character of S^{λ} ⊗ S^{μ} is [λ] · [μ]. For λ, μ, ν ⊢ n we define the numbers d(μ, ν; λ) via

[μ] · [ν] = ∑_{λ} d(μ, ν; λ)[λ].

If α = (α_{1}, α_{2}, …) and β = (β_{1}, β_{2}, …) are two partitions then we write β ⊆ α if β_{i} ≤ α_{i} for all i. In this case we also consider the skew partition α/β. We do not distinguish between α/β and its Young diagram, which is the set of nodes α \ β.

If α/β is a skew Young diagram and A = (i, j) is some node we say A is connected with α/β if at least one of the nodes (i ± 1, j), (i, j ± 1) belongs to α/β. Otherwise A is disconnected from α/β.

If β ⊢ m, γ ⊢ n, α ⊢ m + n we write c_{βγ}^{α} for the corresponding Littlewood-Richardson coefficient, which may be defined as the multiplicity of S^{α} in the induced module

S^{β} ˆ⊗S^{γ} := (S^{β} ⊠ S^{γ}) ↑_{S_{m}×S_{n}}^{S_{m+n}} .

The character of this module will be denoted [β] ˆ⊗[γ]. The Littlewood–Richardson rule [4, 6] gives a combinatorial description of the coefficients c_{βγ}^{α} and will be repeatedly used in this paper. It says that c_{βγ}^{α} is the number of semistandard tableaux of skew shape α/β and content γ, which give a lattice permutation when the entries are read from right to left along the rows starting from the top row.

Let α and β be two partitions. Then the skew character [α/β] is defined to be the sum

[α/β] = ∑_{γ} c_{βγ}^{α}[γ].

Note that [α/β] = 0 unless β ⊆ α.

The following four results will be used repeatedly.

Theorem 2.1 [2, 1.6], [1, 1.1]. Let μ, ν be partitions of n. Then

max{λ_{1} | d(μ, ν; λ) ≠ 0 for some λ = (λ_{1}, λ_{2}, …)} = |μ ∩ ν|

and

max{m | d(μ, ν; λ) ≠ 0 for some λ = (λ_{1} ≥ … ≥ λ_{m} > 0)} = |μ ∩ ν′|

Since the skew characters can in principle be decomposed into the irreducible characters, the following theorem provides a recursive formula for the coefficients d(μ, ν; λ).

Theorem 2.2 [2, 2.3]. Let μ, ν and λ = (λ_{1}, λ_{2}, …) be partitions of n, and set λ̂ = (λ_{2}, λ_{3}, …). Define

Y (λ) = {η | η ⊢ n, η_{i} ≥ λ_{i+1} ≥ η_{i+1} for all i ≥ 1}

Then

d(μ, ν; λ) = ∑_{{α⊢λ_{1} \atop α⊆μ∩ν}} ⟨[μ/α] · [ν/α], [λ̂]⟩ − ∑_{{η∈Y (λ) \atop η≠λ} ; η_{1}≤|μ∩ν|} d(μ, ν; η) .

Corollary 2.3 [2, 2.4], [1, 2.1(d)]. Let μ, ν and λ = (λ_{1}, λ_{2}, …) be partitions of n, and set λ̂ = (λ_{2}, λ_{3}, …), γ = μ ∩ ν. Assume that λ_{1} = |μ ∩ ν|. Then

d(μ, ν; λ) = ⟨[μ/γ] · [ν/γ], [λ̂]⟩ .

Corollary 2.4 [2, 2.4’]. Let μ and ν be partitions of n, and m = |μ ∩ ν′|. Let λ be a partition of n with m non-zero parts. Define λ̄ = (λ_{1} − 1, λ_{2} − 1, …, λ_{m} − 1). Then

d(μ, ν; λ) = ⟨[μ/(μ ∩ ν′)] · [ν/(μ′ ∩ ν)], [λ̄]⟩.

3 Homogenous S_{n}-products

Lemma 3.1 Let α, β, a, b be positive integers. Then

min(α + β + 1, a + b + 1) < min(α, a) + min(β, b) + min(α, b) + min(β, a).  (1)

Proof. We may assume that α ≤ β, a ≤ b and α + β ≤ a + b. So the left hand side in (1) is α + β + 1.

If β ≤ b, then the right hand side of (1) equals

min(α, a) + β + α + min(β, a)

which is greater than α + β + 1 since all numbers in this expression are positive integers.

If b < β, then the right hand side of (1) is

min(α, a) + b + min(α, b) + a ≥ min(α, a) + min(α, b) + α + β > α + β + 1, as claimed. ⋄ Lemma 3.2 Let μ, ν be partitions of n, both different from (n) and (1^{n}). Then

min(h_{11}^{μ}, h_{11}^{ν}) < |μ ∩ ν| + |μ ∩ ν′| − 2.

Proof. We write μ and ν in the Frobenius notation:

F (μ) = ( \begin{matrix}α_{1} & ⋯ & α_{r} \\ β_{1} & ⋯ & β_{r} \end{matrix} ) , F (ν) = ( \begin{matrix}a_{1} & ⋯ & a_{s} \\ b_{1} & ⋯ & b_{s} \end{matrix} ) .

We may assume that r ≤ s. Then

h_{11}^{μ} = α_{1} + β_{1} + 1 , h_{11}^{ν} = a_{1} + b_{1} + 1,

|μ ∩ ν| = r + ∑_{i=1}^{r}(min(α_{i}, a_{i}) + min(β_{i}, b_{i})),
|μ ∩ ν′| = r + ∑_{i=1}^{r}(min(α_{i}, b_{i}) + min(β_{i}, a_{i})).

Since r ≥ 1, it suffices to prove that

min(α_{1} + β_{1} + 1, a_{1} + b_{1} + 1) < min(α_{1}, a_{1}) + min(β_{1}, b_{1}) + min(α_{1}, b_{1}) + min(β_{1}, a_{1}). But this follows from Lemma 3.1 since our assumption on the partitions ensures that α_{1}, β_{1}, a_{1}, b_{1} > 0. ⋄ Proposition 3.3 Let μ, ν be partitions of n, both different from (n) and (1^{n}). If [λ] is a constituent of [μ] · [ν], then h_{11}^{λ} < |μ ∩ ν| + |μ ∩ ν′| − 1.

Proof. Put ℓ = |μ ∩ ν| + |μ ∩ ν′| − 1 . Take π to be an ℓ-cycle in S_{n}. By Lemma 3.2, either μ or ν does not have a hook of length ℓ. Hence, by the Murnaghan-Nakayama Rule [4, 2.4.7], either [μ](π) = 0 or [ν](π) = 0. So

([μ] · [ν])(π) = 0.  (2)

By Theorem 2.1, any constituent [λ] of [μ] · [ν] satisfies

λ_{1} ≤ |μ ∩ ν| and λ′_{1} ≤ |μ ∩ ν′|

where λ = (λ_{1}, …), λ′ = (λ′_{1}, …). So the maximal possible hook length in λ is ℓ. Moreover, λ contains a hook of length ℓ if and only if λ_{1} = |μ∩ν| and λ′_{1} = |μ∩ν′|, in which case this is the (1, 1)-hook whose leg length is |μ∩ν′|−1. In this case, using the Murnaghan-Nakayama Rule again, we get

[λ](π) = (−1)^{|μ∩ν′|−1}[λ \ H_{11}](1) ≠ 0

where λ \ H_{11} is the partition obtained from λ by removing the (1, 1)-hook H_{11}. Hence for every constituent [λ] of [μ] · [ν] containing an ℓ-hook we get a contribution on π of the same sign, and so no cancellation can occur. But this contradicts equation (2). ⋄

Theorem 3.4 Let μ, ν be partitions of n, both different from (n) and (1^{n}). Then [μ] · [ν] is not homogenous.

Proof. By Theorem 2.1, [μ] · [ν] has a constituent [λ] with λ_{1} = |μ ∩ ν| and a constituent [κ] with κ′_{1} = |μ ∩ ν′|. If λ = κ, then h_{11}^{λ} = |μ ∩ ν| + |μ ∩ ν′| − 1, which is impossible by Proposition 3.3. ⋄

Corollary 3.5 A product [μ] · [ν] is irreducible if and only if one of the two characters [μ], [ν] is of degree 1.

4 Kronecker products of S_{n}-representations with few components

The main result of this section is a description of the products of S_{n}-representations with two homogeneous components. First we need to know the product of any character with the character [n − 1, 1]:

Lemma 4.1 Let n ≥ 3 and μ be a partition of n. Then

[μ] · [n − 1, 1] = ∑_{A} ∑_{B}[(μ_{A})^{B}] − [μ]

where the first sum is over all removable nodes A for μ, and the second sum runs over all addable nodes B for μ_{A}.

Proof. This follows from the isomorphisms M^{(n−1,1)} ≅ S^{(n−1,1)}⊕S^{(n)} and S^{μ}⊗M^{(n−1,1)} ≅ (S^{μ} ↓_{S_{n−1}}) ↑^{S_{n}}. ⋄

Corollary 4.2 Let n ≥ 3 and μ be a partition of n. Then (i) [μ] · [n − 1, 1] has exactly one homogeneous component if and only if μ is (n) or (1^{n}).

(ii) [μ] · [n − 1, 1] has exactly two homogeneous components if and only if μ is a rectangle (a^{b}) for some a, b > 1. In this case we have

[a^{b}] · [n − 1, 1] = [a + 1, a^{b−2}, a − 1] + [a^{b−1}, a − 1, 1].

(iii) [μ] · [n − 1, 1] has exactly three homogeneous components if and only if n = 3 and μ = (2, 1). In this case we have

[2, 1] · [2, 1] = [3] + [2, 1] + [1^{3}].

(iv) [μ] · [n − 1, 1] has exactly four homogeneous components if and only if one of the following happens:

(a) n ≥ 4 and μ = (n − 1, 1) or (2, 1^{n−2});

(b) μ = (k + 1, k) or (2^{k}, 1) for k ≥ 2.

We then have:

\begin{aligned}[n − 1, 1] · [n − 1, 1] & = & [n] + [n − 1, 1] + [n − 2, 2] + [n − 2, 1^{2}], \\ [k + 1, k] · [2k, 1] & = & [k + 2, k − 1] + [k + 1, k] + [k + 1, k − 1, 1] + [k^{2}, 1], \end{aligned}

and the remaining products are obtained by conjugation.

Proof. The “if” parts and the decompositions of the products follow from Lemma 4.1.

We now prove the “only if” directions. We are going to use Lemma 4.1 again. First, observe that [μ] appears as a constituent in the product [μ] ⊗ [n − 1, 1] unless μ is a rectangle. Also note that (μ_{A})^{B} = (μ_{A′})^{B′} for two different pairs (A, B), (A′, B′) if and only if A = B and A′ = B′, in which case (μ_{A})^{B} = (μ_{A′})^{B′} = μ.

A partition with r removable nodes has exactly r + 1 addable nodes. So if μ has at least 2 removable nodes, say A_{1} and A_{2}, then μ_{A_{1}} and μ_{A_{2}} both have at least 2 addable nodes, which gives 4 composition factors in the product with the only common constituent [μ]. This proves the “only if” part of (i) and (ii). If μ has at least 3 removable nodes, then a similar argument shows that [μ] · [n − 1, 1] has at least 5 non-isomorphic constituents. So we may assume that μ has exactly two removable nodes: A_{1} and A_{2}. For [μ] · [n − 1, 1] to have exactly 3 components, both μ_{A_{1}} and μ_{A_{2}} should have only one removable node. This is only possible if n = 3 and μ = (2, 1). Finally, for [μ] · [n − 1, 1] to have exactly 4 components, one of μ_{A_{1}} and μ_{A_{2}} should have only one removable node and the other one should have two. This occurs exactly if μ or μ′ is (n − 1, 1), n ≥ 4, or (k + 1, k), k ≥ 2. ⋄

Lemma 4.3 Let λ be a partition of n. Then the square [λ]^{2} has at most 4 homogeneous components if and only if one of the following holds:

(i) λ = (n) or (1^{n}), when [λ]^{2} = [n];

(ii) n ≥ 4, λ = (n − 1, 1) or (2, 1^{n−2}), when [λ]^{2} = [n] + [n − 1, 1] + [n − 2, 2] + [n − 2, 1^{2}];

(iii) n = 3, λ = (2, 1), when [λ]^{2} = [3] + [2, 1] + [1^{3}]; (iv) n = 4, λ = (2^{2}), when [λ]^{2} = [4] + [2^{2}] + [1^{4}];

(v) n = 6, λ = (3^{2}) or (2^{3}), when [λ]^{2} = [6] + [4, 2] + [3, 1^{3}] + [2^{3}].

Proof. The “if” part follows from Corollary 4.2 and [4, Tables I.I].

In the other direction, let [λ]^{2} have at most 4 homogeneous components. We may assume that λ is not one of (n), (1^{n}), (n − 1, 1), (2, 1^{n−2}), and that n > 8 since for n ≤ 8 the results hold by [4, Tables I.I].

Clearly [λ]^{2} always contains [n]. Furthermore, by [10, Lemmas 1-3] and [12, 4.3] or by [11, 6.3], [λ]^{2} contains [n − 2, 2], and unless λ is a rectangle, it also contains [n − 1, 1], [n−2, 1^{2}] and [n−3, 3]. So we only have to deal with the case where λ = (a^{b}) is a rectangle. We already know that [λ]^{2} has the constituents [n] and [n − 2, 2]. If b > 2, then [λ]^{2} also has the constituent [n − 3, 3] by [10, Lemma 3] or [11, 6.3]. If n > 12, then also [n − 4, 4] occurs, see [10, Lemma 4]. Furthermore, by [11, 6.3], [n − 3, 1^{3}] appears as a constituent. Hence we can restrict ourselves to the cases λ = (k, k) or λ = (4^{3}).

Suppose λ = (k, k) (k ≥ 5). By Corollary 2.4, the components [μ] of [k, k]^{2} with μ′_{1} = 4 = |λ ∩ λ′| are of the form (ρ_{1} + 1, ρ_{2} + 1, ρ_{3} + 1, ρ_{4} + 1), where [(ρ_{1}, ρ_{2}, ρ_{3}, ρ_{4})], is a constituent of [k − 2, k − 2]^{2}. By what we have already proved, there are at least 3 such constituents. Thus [k, k]^{2} has at least 5 components.

Now, let λ = (4^{3}). We already know that [λ]^{2} contains [12], [10, 2], [9, 3] and [9, 1^{3}]. But it also contains some [μ] with μ′_{1} = 9 = |λ∩λ′|, thanks to Theorem 2.1. Alternatively, one may calculate [4^{3}]^{2} on a computer and find 52 (!) homogeneous components. ⋄

Lemma 4.4 Let μ, γ be partitions, γ ⊂ μ. Set I = {i | γ_{i} < μ_{i}}. Then the following assertions are equivalent:

(i) [μ/γ] is homogeneous;

(ii) [μ/γ] is irreducible;

(iii) I = {j, j + 1, …, k} for some j ≤ k, and one of the following holds:

(a) γ_{j} = γ_{j+1} = ⋯ = γ_{k};

(b) μ_{j} = μ_{j+1} = ⋯ = μ_{k};

Moreover, in this case [μ/γ] = [α], where α is the partition with the parts μ_{i} − γ_{i}, i ∈ I, sorted in the weakly decreasing order.

Proof. This follows from the Littlewood-Richardson Rule. ⋄

Remark. The situations described in (iii)(a) and (iii)(b) above correspond respectively to the pictures

γ  μ/γ  ^{γ  μ/γ}

Lemma 4.5 In the notation of Lemma 4.4 (and under the same assumptions), let A be a removable node of γ.

(1) If A is disconnected from μ/γ then

[μ/γ_{A}] = ∑_{B}[α^{B}]

where B runs over the addable nodes of α.

(2) Let A be connected with μ/γ.

In the case (iii)(a) we have

[μ/γ_{A}] = ∑_{B≠B_{0}}[α^{B}]

where B runs over the addable nodes of α, except for the bottom addable node B_{0}.

In the case (iii)(b) we have

[μ/γ_{A}] = [α^{B}]

where B is an addable node of α.

Proof. Again, this follows by the Littlewood-Richardson Rule. ⋄

The following two lemmas will be used in the proof of the main theorem of this section.

Lemma 4.6 Let μ ≠ ν be partitions of n, both different from (n), (1^{n}), (n − 1, 1) and (2, 1^{n−2}). Put γ = μ ∩ ν, m = |γ|. Assume that ν/γ is a row and that [μ/γ] is an irreducible character [α_{1}, α_{2}, …]. Then [m, α_{1}, α_{2}, …] appears in [μ] · [ν]. Moreover if an S_{n−m+1}-character [θ_{1}, θ_{2}, …] appears in

∑_{A removable for γ}[μ/γ_{A}] · [ν/γ_{A}] − ∑_{B addable for α}[α^{B}]  (3)

with a positive coefficient then [m − 1, θ_{1}, θ_{2}, …] appears in [μ] · [ν].

Proof. We have [ν/γ] = [n − m]. So Theorem 2.1 and Corollary 2.3 yield:

⟨[μ] · [ν], [m, α_{1}, α_{2}, …]⟩ = 1,  (4)

and

if λ ≠ (m, α_{1}, α_{2}, …) and ⟨[μ] · [ν], [λ]⟩ ≠ 0 then λ_{1} < m.  (5)

If λ is a partition of n with λ_{1} = m − 1, then in the notation of Theorem 2.2, we may write

{η ∈ Y (λ) | η ≠ λ, η_{1} ≤ m} = {(m, λ_{2}, …, λ_{i−1}, λ_{i} − 1, λ_{i+1}, …) | i ≥ 1, λ_{i} > λ_{i+1}}.

So (4) and (5) imply ∑_{{η∈Y (λ) \atop η≠λ} ; η_{1}≤m} d(μ, ν; η) = ε, where

ε = \{ \begin{matrix}1 & if λ̂ = α^{B} for some addable node B of α \\ 0 & otherwise. \end{matrix}  (6)

Now, by Theorem 2.2, for a partition λ of n with λ_{1} = m − 1 we have

⟨[μ] · [ν], [λ]⟩ = ∑_{A}⟨[μ/γ_{A}] · [ν/γ_{A}], [λ̂]⟩ − ε.  (7)

where the sum is over all removable nodes A of γ.

Let [θ] be a constituent of [μ/γ_{A}] · [ν/γ_{A}]. Then [θ] is a constituent of [β] · [δ] with [β] a constituent of [μ/γ_{A}] and [δ] a constituent of [ν/γ_{A}]. It follows from the definition of skew characters that β ⊆ μ, δ ⊆ ν. Hence β ∩ δ ⊆ μ ∩ ν = γ. In view of Theorem 2.1, this implies

θ_{1} ≤ |β ∩ δ| ≤ |μ ∩ ν| = m .

If θ_{1} = m, then β ∩ δ = γ, therefore β ⊇ γ and δ ⊇ γ. However, ν/γ_{A} is a union of a row and a node, so either δ = (n − m + 1) or δ = (n − m, 1). If δ = (n − m + 1), then μ ∩ ν ⊆ δ implies μ ∩ ν = (m). But then either μ or ν is (n), which contradicts the assumptions of the lemma. If δ = (n − m, 1), then we conclude similarly that μ ∩ ν = (m − 1, 1). Since neither μ nor ν is equal to (n − 1, 1) or its conjugate and μ/γ should be connected by Lemma 4.4, then the only possibilities are: ν = (m−1, n−m+1), μ = (m−1, 1^{n−m+1}) or ν = (n−2, 1^{2}), μ = (n−2, 2) (in the latter case n−m = 1). In both cases m−1 ≥ n−m+1, so θ_{1} ≤ m − 1 since θ is a partition of n − m + 1. This contradiction shows that we may assume that θ_{1} ≤ m − 1 for any [θ] appearing in [μ/γ_{A}] · [ν/γ_{A}].

This, together with (7), shows that any S_{n−m+1}-character [θ_{1}, θ_{2}, …] appearing in (3) gives rise to the character [m − 1, θ_{1}, θ_{2}, …] appearing in [μ] · [ν]. ⋄

Lemma 4.7 Let μ ≠ ν be partitions of n, both different from (n), (1^{n}), (n − 1, 1), and (2, 1^{n−2}). Put γ = μ ∩ ν. Assume that ν/γ is a row, [μ/γ] is irreducible, and [μ] · [ν] has 2 homogeneous components.

If there exists a removable node A_{0} of γ, disconnected from ν/γ, then the following condition holds:

(*) [μ/γ_{A_{0}}] is 1-dimensional, μ/γ is connected with all removable nodes of γ, ν/γ is connected with all removable nodes of γ except A_{0}.

Proof. Let A_{0} be a removable node of γ disconnected from ν/γ, and put m = |γ|. Since μ ≠ ν, we have n − m > 0. Let α be the partition of n − m defined by [μ/γ] = [α]. Note that [ν/γ] = [n − m].

By Lemma 4.6, it suffices to show that the expression (3) contains at least two distinct irreducible characters unless the conditions (*) hold.

Since A_{0} is disconnected from ν/γ, we have by Lemma 4.5(1):

[ν/γ_{A_{0}}] = [n − m + 1] + [n − m, 1].  (8)

In view of Lemmas 4.4 and 4.5, we have three cases to consider: (a) when A_{0} is disconnected from μ/γ; (b) when A_{0} is connected with μ/γ and we are in the case (iii)(a) of Lemma 4.4; (c) when A_{0} is connected with μ/γ and we are in the case (iii)(b) of Lemma 4.4 (the cases (b) and (c) overlap when μ/γ is a rectangle).

(a) In this case A_{0} is disconnected from μ/γ. Then, by Lemma 4.5(1), we get

[μ/γ_{A_{0}}] = ∑_{B}[α^{B}]

where the sum runs over all addable nodes B of α. So (3) contains

([n − m + 1] + [n − m, 1]) · ( ∑_{B}[α^{B}]) − ∑_{B}[α^{B}] = [n − m, 1] · ( ∑_{B}[α^{B}]).

If there is a non-linear character among the [α^{B}], we are done by Corollary 4.2(i). Otherwise α = (1), but even in this case the expression above contains two different characters:

[2] and [1^{2}]. This completes the case (a).

In particular, we now may assume that every removable node A of γ disconnected from ν/γ is connected with μ/γ.

Note that [μ/γ_{A_{0}}] contains [α^{B_{1}}] for some addable node B_{1}, see Lemma 4.5. So, in view of (8) and Lemma 4.1 , [μ/γ_{A_{0}}] · [ν/γ_{A_{0}}] contains ∑_{B}[α^{B}]. Hence any removable node A_{1} ≠ A_{0} of γ yields a positive contribution of [μ/γ_{A_{1}}] · [ν/γ_{A_{1}}] to the expression (3). If A_{1} is disconnected from ν/γ then [ν/γ_{A_{1}}] = [n − m, 1] + [n − m + 1], and the product [μ/γ_{A_{1}}] · [ν/γ_{A_{1}}] is not homogeneous. If A_{1} is connected with ν/γ but disconnected from μ/γ then, by Lemma 4.5, [μ/γ_{A_{1}}] is not irreducible and [ν/γ_{A_{1}}] is [n − m, 1] or [n − m + 1]. So the product [μ/γ_{A_{1}}] · [ν/γ_{A_{1}}] is not homogeneous again, thanks to Lemmas 4.1 and 4.5. Thus we may always assume that

(**) μ/γ is connected with all removable nodes of γ, and ν/γ is connected with all removable nodes of γ different from A_{0}.

(b) In this case Lemma 4.5 yields

[μ/γ_{A_{0}}] = ∑_{B≠B_{0}}[α^{B}]

where the sum runs over all addable nodes B of α except for the bottom one B_{0}. Consider the constituent [μ/γ_{A_{0}}] · [ν/γ_{A_{0}}] − ∑_{B}[α^{B}] of (3). By (8), it is equal to

([n − m + 1] + [n − m, 1]) · ( ∑_{B≠B_{0}}[α^{B}]) − ∑_{B}[α^{B}] = [n − m, 1] · ( ∑_{B≠B_{0}}[α^{B}]) − [α^{B_{0}}]. (9)

Since α ≠ ∅, it has at least 2 addable nodes. Let B_{1} be an addable node of α, different from B_{0}, and let r be the number of removable nodes of α^{B_{1}}. Then, using Lemma 4.1, we can rewrite (9) as follows:

[n − m, 1] · ([α^{B_{1}}] + ∑_{B≠B_{0},B_{1}}[α^{B}]) − [α^{B_{0}}]

= (r − 1)[α^{B_{1}}] + [α^{B_{0}}] + ∑_{B≠B_{0},B_{1}}[α^{B}] + ∑_{C,D}[(α^{B_{1}})_{C}^{D}]

+[n − m, 1] · ( ∑_{B≠B_{0},B_{1}}[α^{B}]) − [α^{B_{0}}]

= (r − 1)[α^{B_{1}}] + ([n − m + 1] + [n − m, 1]) · ( ∑_{B≠B_{0},B_{1}}[α^{B}]) + ∑_{C,D}[(α^{B_{1}})_{C}^{D}]

where the sum ∑_{C,D} is over all removable nodes C of α^{B_{1}}, different from B_{1}, and over all

addable nodes D of (α^{B_{1}})_{C}, different from C.

If α is not a rectangle, then ∑_{B≠B_{0},B_{1}} is non-empty, so our expression involves at least two different irreducible characters. Let α be a rectangle. If [α] is not of degree 1, then α^{B_{1}} is not a rectangle, so r > 1, and thus our expression involves [α^{B_{1}}]. Moreover, α^{B_{1}} has a removable node C ≠ B_{1}, so for an addable node D ≠ C of (α^{B_{1}})_{C} we get the contribution [(α^{B_{1}})_{C}^{D}] ≠ [α^{B_{1}}]. Finally, let [α] = [μ/γ] be of degree 1. If [α^{B_{1}}] is not of degree 1, then it is [2, 1^{(n−m−1)}]. So for n − m ≥ 2, we have r = 2, and so [2, 1^{n−2}] and [3, 1^{n−3}] appear in our expression. However, if n − m = 1, then [μ/γ_{A_{0}}] is of degree 1. So, in view of (**), all the conditions in (*) hold.

(c) In this case by Lemma 4.5 we have

[μ/γ_{A_{0}}] = [α^{B_{1}}]

for some addable node B_{1} of α. Then the constituent [μ/γ_{A_{0}}] · [ν/γ_{A_{0}}] − ∑_{B}[α^{B}] of (3) is

([n − m + 1] + [n − m, 1]) · [α^{B_{1}}] − ∑_{B}[α^{B}]

= ∑_{C,D}[(α^{B_{1}})_{C}^{D}] − ∑_{B}[α^{B}]
= ∑_{B}[α^{B}] + ∑_{C,D; C≠B_{1}}[(α^{B_{1}})_{C}^{D}] − ∑_{B}[α^{B}]
= ∑_{C,D; C≠B_{1}}[(α^{B_{1}})_{C}^{D}].  (10)

In the last sum C runs through the removable nodes of α^{B_{1}}, different from B_{1}, and D runs through the addable nodes of (α^{B_{1}})_{C}. So (10) has at least two different irreducible constituents, unless it is empty. Hence we may assume that α^{B_{1}} is a rectangle. If [α] is of degree 1 then [α^{B_{1}}] = [μ/γ_{A_{0}}] is also of degree 1, and, in view of (**), we are in the exceptional case (*). So we may assume that α = (a^{b−1}, a − 1) for some a > 1, b > 1, and α^{B_{1}} = (a^{b}). This together with Lemma 4.4 implies that γ has a removable node A_{1}, different from A_{0}. We know that it must be connected with ν/γ and μ/γ, thanks to (**). If there was a third removable node of γ , A_{2} say, then again by (**), both A_{1} and A_{2} would be connected with both ν/γ and μ/γ. But this is impossible since γ = μ ∩ ν. So we may assume that γ has exactly two removable nodes. Now, by Lemma 4.5(2), we have

[μ/γ_{A_{1}}] = [α^{B_{2}}] with B_{2} the top or the bottom, but not the middle, addable node of α, and [ν/γ_{A_{1}}] is either [n − m, 1] or [n − m + 1]. The corresponding pictures are:

ν/γ  μ/γ A_{1}A_{0}  ^{ν/γ μ/γ A_{1} A_{0}}

In the first case, [α^{B_{2}}] · [n − m, 1] contributes at least two constituents by Theorem 3.4. In the second case ν = (n − 1, 1). ⋄ Theorem 4.8 Let μ, ν be partitions of n. Then [μ] · [ν] has exactly two homogenous components if and only if one of the partitions μ, ν is a rectangle (a^{b}) with a, b > 1, and the other is (n − 1, 1) or (2, 1^{n−2}). In these cases we have :

[n − 1, 1] · [a^{b}] = [a + 1, a^{b−2}, a − 1] + [a^{b−1}, a − 1, 1],
[2, 1^{n−2}] · [a^{b}] = [b + 1, b^{a−2}, b − 1] + [b^{a−1}, b − 1, 1].

Proof. The “if” part is proved in Corollary 4.2 (note that S^{(2,1^{n−2})} ≅ S^{(n−1,1)} ⊗ sign). To prove the “only if” part, assume that

[μ] · [ν] = x[κ] + y[λ] for some x, y ∈ ℕ ,

with κ > λ in the lexicographic order. Clearly, μ, ν ∉ {(n), (1^{n})}. If μ or ν is (n − 1, 1) or (2, 1^{n−2}) the result follows from Corollary 4.2. Assume μ, ν ∉ {(n − 1, 1), (2, 1^{n−2})}. By Theorem 2.1 and Proposition 3.3, we have

κ_{1} = |μ ∩ ν|, λ′_{1} = |μ ∩ ν′| and λ_{1} < |μ ∩ ν| = κ_{1}.

By Lemma 4.3, μ ≠ ν, and hence κ_{1} < n. Put γ = μ ∩ ν, m = |γ|. By Corollary 2.3, we must have

[μ/γ] · [ν/γ] = x[κ̂]

where κ̂ = (κ_{2}, κ_{3}, …). So, in view of Theorem 3.4, one of the following happens:

(i) x = 1 and one of the characters [μ/γ], [ν/γ] is of degree 1, while the other is irreducible;

(ii) one of the characters [μ/γ], [ν/γ] is equal to [κ̂], the other is of the form z[n − m] + w[1^{n−m}] with some z, w ∈ ℕ , and κ̂ = κ̂′. By the Littlewood-Richardson rule, a skew character contains both [n − m] and [1^{n−m}] only if its diagram is a set of disconnected nodes. So we must have n − m = 2, since otherwise such a skew character has more than 2 constituents. But there is no symmetric partition of 2, i.e. κ̂ ≠ κ̂′. This contradiction allows us to assume that we are in the case (i).

Without loss of generality, suppose that [ν/γ] is of degree 1 and [μ/γ] = [α] is irreducible. Then the shape of ν/γ is a row or a column. Passing, if necessary, from μ, ν to μ′, ν′, we may assume that ν/γ is a row. Now, by Lemma 4.7 we may assume that one of the following holds:

(a) ν/γ is connected with every removable node of γ.

(b) There exists a removable node A_{0} of γ disconnected from ν/γ, [μ/γ] and [μ/γ_{A_{0}}] are of degree 1, μ/γ is connected with every removable node of γ, and ν/γ is connected with every removable node of γ different from A_{0}.

Case (a). In this case ν must be a rectangle, and γ must have a removable node A_{0} such that [ν/γ_{A_{0}}] = (n − m, 1) for otherwise ν = (n).

ν/γ γ  A_{0}

Let us first assume that μ/γ is disconnected from A_{0}. Then, in view of Lemmas 4.5(1) and 4.1, the expression (3) contains

[ν/γ_{A_{0}}] · [μ/γ_{A_{0}}] − ∑_{B}[α^{B}] = [n − m, 1] · ( ∑_{B}[α^{B}]) − ∑_{B}[α^{B}]

= ∑_{B} ∑_{C} ∑_{D}[(α^{B})_{C}^{D}] − 2 ∑_{B}[α^{B}]

= ∑_{B}(r_{B} − 2)[α^{B}] + ∑_{B} ∑_{C} ∑_{D≠C}[(α^{B})_{C}^{D}]  (11)

where B runs over the addable nodes of α, C runs over the removable nodes of α^{B} (for the respective node B), D runs over the addable nodes of (α^{B})_{C} and r_{B} denotes the number of removable nodes of α^{B}.

If α has at least 3 addable nodes, say B_{0}, B_{1}, B_{2}, then we have the following contribution to the expression above:

(r_{B_{0}} − 2)[α^{B_{0}}] + [α^{B_{1}}] + [α^{B_{2}}] + (r_{B_{1}} − 2)[α^{B_{1}}] + [α^{B_{0}}] + [α^{B_{2}}]
+(r_{B_{2}} − 2)[α^{B_{2}}] + [α^{B_{0}}] + [α^{B_{1}}]
= r_{B_{0}}[α^{B_{0}}] + r_{B_{1}}[α^{B_{1}}] + r_{B_{2}}[α^{B_{2}}].

By Lemma 4.6, this yields 3 irreducible components in [μ] · [ν].

So α has exactly two addable nodes, say B_{0}, B_{1}, i.e. α is a rectangle. Then we have the following contribution to the expression (11):

(r_{B_{0}} − 2)[α^{B_{0}}] + [α^{B_{1}}] + (r_{B_{1}} − 2)[α^{B_{1}}] + [α^{B_{0}}]

If α is not a row or a column then both r_{B_{0}}, r_{B_{1}} are at least 2, and in view of Lemma 4.6, we get two irreducible constituents for [μ] · [ν], both different from [κ]. Let α be a row or a column. Assume that α is a row, the column case being similar. Then (11) equals [n − m, 1] + [n − m − 1, 2] + [n − m − 1, 1^{2}] if n − m > 2, and [2, 1] + [1^{3}] if n − m = 2. By Lemma 4.6, this yields at least two constituents in [μ] · [ν] different from [κ]. Finally, let n − m = 1. Then (11) equals 0. Note that γ must have a removable node A_{1} ≠ A_{0}, since otherwise ν = (1^{n}). If μ/γ is disconnected from A_{1}, then

[μ/γ_{A_{1}}] · [ν/γ_{A_{1}}] = [2] + [1^{2}],

and we are done by Lemma 4.6. If μ/γ is connected with A_{1}, then μ = (2^{k}, 1^{2}), ν = (2^{k+1}) (and k > 1 since μ is not of the form (2, 1^{n−2})). Then the expression (3) equals [1^{2}]. So, by Lemma 4.6, [n − 1, 1] and [n − 2, 1^{2}] are constituents of [μ] · [ν]. But |μ ∩ ν′| = 4, so there also must be a constituent with 4 non-zero rows, thanks to Theorem 2.1.

This completes the consideration of the case where [μ/γ] is disconnected from A_{0}. Let μ/γ be connected with A_{0}. Then, in view of Lemmas 4.4 and 4.5(2), we have

[μ/γ_{A_{0}}] = ∑_{B≠B_{0}}[α^{B}]

where B_{0} is the bottom addable node of α. Let B_{1} be the top addable node of α. Then we get a contribution to (3) from the following expression:

[ν/γ_{A_{0}}] · [μ/γ_{A_{0}}] − ∑_{B}[α^{B}] = [n − m, 1] · ∑_{B≠B_{0}}[α^{B}] − ∑_{B}[α^{B}]

= ∑_{B≠B_{0}} ∑_{C} ∑_{D}[(α^{B})_{C}^{D}] − ∑_{B≠B_{0}}[α^{B}] − ∑_{B}[α^{B}]
= ∑_{C≠B_{1}} ∑_{D}[(α^{B_{1}})_{C}^{D}] + ∑_{B≠B_{0},B_{1}} ∑_{C} ∑_{D}[(α^{B})_{C}^{D}] − ∑_{B≠B_{0}}[α^{B}]  (12)

where B runs through the addable nodes of α, C runs through the removable nodes of α^{B} (for the respective node B) and D runs through the addable nodes of (α^{B})_{C}.

If α has a third addable node, say B_{2}, then α^{B_{1}} is not a rectangle, and hence there exists a node C_{1} ≠ B_{1} which is removable from α^{B_{1}}. This shows that the first sum in (12) contains [α^{B_{1}}]. Moreover, the second sum in (12) contains ∑_{D}[α^{D}], and so both [α^{B_{0}}] and [α^{B_{1}}] are constituents of (12). Now we can apply Lemma 4.6.

If B_{0} and B_{1} are the only addable nodes of α, then α is a rectangle. Let C_{1} be the corner node of α.

If α is not a row, then α^{B_{1}} also has the removable node C_{1}. In this case, (12) is

∑_{D}[(α^{B_{1}})_{C_{1}}^{D}] − [α^{B_{1}}]

which gives at least two contributions, except in the case where α = (1^{2}) when (12) equals [3]. If γ has a further removable node A_{1}, then this leads to a further contribution [2, 1] to (3). But if γ is a rectangle, then μ = (2^{3}) and ν = (3^{2}), and we can apply Lemma 4.3.

If α is a row then γ must have a removable node A_{1} ≠ A_{0}, since otherwise μ = (n). Note that (12) equals −[n − m + 1]. Also [ν/γ_{A_{1}}] · [μ/γ_{A_{1}}] = [n − m + 1] + [n − m, 1]. By Lemma 4.6, the product [μ] · [ν] contains [m, n − m] and [m − 1, n − m, 1]. Note that our assumptions yield μ = (k + n − m, k − n + m), ν = (k, k) with k − n + m ≥ 2. But in this case |μ ∩ ν′| ≥ 4. So Theorem 2.1 implies that [μ] · [ν] has a constituent with 4 rows.

Case (b). Since [μ] is not of degree 1, the assumption [μ/γ] and [μ/γ_{A_{0}}] are of degree 1 implies that γ must have a removable node A_{1} ≠ A_{0}. By assumption, A_{1} is connected with both μ/γ and ν/γ, and since [μ/γ] is of degree 1, A_{1} and A_{0} are the only removable nodes of γ.

Since [μ/γ_{A_{0}}] is 1-dimensional, we conclude from Lemmas 4.4 and 4.5 that [μ/γ_{A_{1}}] = [n − m, 1] or the conjugate. So if n − m > 1 and [ν/γ_{A_{1}}] = [n − m, 1] then (3) equals [n − m, 1] · [n − m, 1] or the conjugate. Now we apply Corollary 4.2 and Lemma 4.6. Otherwise μ = (k, k), ν = (k + n − m, k − n + m) or μ = (2^{k}), ν = (2^{k−1}, 1^{2}). But these cases have already been considered. ⋄

Thus we have classified all pairs μ, ν such that [μ] · [ν] has at most 2 homogeneous components. The “if-parts” of the following conjecture are proved in Corollary 4.2 and Lemma 4.3.

Conjecture

(i) [μ] · [ν] has 3 homogeneous components if and only if n = 3 and μ = ν = (2, 1) or n = 4 and μ = ν = (2, 2).

(ii) [μ] · [ν] has 4 homogeneous components if and only if one of the following happens

(a) n ≥ 4 and μ, ν ∈ {(n − 1, 1), (2, 1^{n−2})};

(b) n = 2k + 1 for some k ≥ 2, and one of μ, ν is in {(2k, 1), (2, 1^{2k−1})} while the

other one is in {(k + 1, k), (2^{k}, 1)}

(c) n = 6 and μ, ν ∈ {(2^{3}), (3^{2})}.

The following theorem proves the conjecture in the special case when both μ and ν are symmetric.

Theorem 4.9 Let μ and ν be symmetric partitions of n. Then [μ] · [ν] has at most 4 homogeneous components if and only if one of the following holds:

(i) n = 1.

(ii) n = 3, μ = ν = (2, 1), when [μ]^{2} = [3] + [2, 1] + [1^{3}];

(iii) n = 4, μ = ν = (2^{2}), when [μ]^{2} = [4] + [2^{2}] + [1^{4}].

Proof. Let γ = μ ∩ ν, m = |γ|. Then γ is a symmetric partition, and at least one of the skew diagrams μ/γ, ν/γ has no box on the main diagonal. Say it is μ/γ. Because of the symmetry, we can then write μ/γ as a disjoint union α∪α′, where α and α′ are some skew shapes which are conjugate to each other. In particular, n − m is even. By [6, (5.7)],

[μ/γ] = [α] ˆ⊗[α′].

If every constituent of [α] ˆ⊗[α′] belongs to M = {[n−m], [1^{n−m}], [n−m−1, 1], [2, 1^{n−m−2}]} then by the Littlewood-Richardson Rule, every constituent of [α] and [α′] would have to belong to {[(n − m)/2], [1^{(n−m)/2}], [(n − m)/2 − 1, 1], [2, 1^{(n−m)/2−2}]} . But even then, if n − m ≥ 6, the Littlewood-Richardson Rule implies that there are components of [α] ˆ⊗[α′] not in M.

Assume first that n − m ≥ 6. Then, by the Littlewood-Richardson Rule again, [ν/γ] contains a constituent not in M. Now Theorems 3.4 and 4.8 imply that [μ/γ] · [ν/γ] contains at least three different irreducible constituents, say [ρ̂_{1}], [ρ̂_{2}], [ρ̂_{3}]. Then [μ] · [ν] contains the corresponding constituents [ρ_{1}], [ρ_{2}], [ρ_{3}], thanks to Corollary 2.3. Since μ and ν are symmetric, [μ] · [ν] also contains the conjugate constituents [ρ′_{1}], [ρ′_{2}], [ρ′_{3}]. Now, by Proposition 3.3 no constituent can have at the same time the maximal length and width among all the constituents. Hence [ρ_{i}] ≠ [ρ′_{j}] for all i, j. Thus we have found 6 different irreducible constituents.

The case n−m = 0 follows from Lemma 4.3. So we may now assume that n−m = 2 or 4. Note that in the first case n > 7 since for n ≤ 7 there is only one symmetric partition, and in the second case n > 8, since the intersection of the two different symmetric partitions for n = 8 is a partition of 6. Then by the Littlewood-Richardson Rule and Corollary 2.3, we know that [μ] · [ν] has the constituents [n − 2, 2], [n − 2, 1^{2}] and their conjugates if n − m = 2, and it has the constituents [n − 4, 3, 1] and [n − 4, 2, 1^{2}] and their conjugates if n − m = 4. By the remark above, n is sufficiently large in both cases so that the four constituents are all different.

Assume that these are all the constituents of [μ] · [ν]. Consider the case n − m = 4. We compute the character values on (n − 1)-cycles and (n − 2)-cycles. Since |γ| = n − 4, we know that min(h_{11}^{μ}, h_{11}^{ν}) < n − 2. Hence on an (n − 1)-cycle z_{n−1} and an (n − 2)-cycle z_{n−2} in S_{n} we have by the Murnaghan-Nakayama rule:

[μ](z_{n−1}) · [ν](z_{n−1}) = 0 = [μ](z_{n−2}) · [ν](z_{n−2}).

On the other hand, if n is even, then

[n − 4, 2, 1^{2}](z_{n−1}) = −1 = [4, 2, 1^{n−6}](z_{n−1})

and

[n − 4, 3, 1](z_{n−1}) = 0 = [3, 2^{2}, 1^{n−7}](z_{n−1})

gives a contradiction. If n is odd, then similarly

[n − 4, 2, 1^{2}](z_{n−2}) = 0 = [4, 2, 1^{n−6}](z_{n−2})

and

[n − 4, 3, 1](z_{n−2}) = 1 = [3, 2^{2}, 1^{n−7}](z_{n−2})

gives a contradiction. The case n − m = 2 is considered similarly using z_{n} and z_{n−1}. ⋄

5 Homogeneous Kronecker products of A_{n}-representations

We first recall the classification of the complex irreducible A_{n}-representations (cf. [4, 2.5]). If μ is a non-symmetric partition of n then the restrictions S^{μ} ↓_{A_{n}} and S^{μ′} ↓_{A_{n}} are irreducible and isomorphic to each other. We denote the corresponding irreducible A_{n}-module by T^{μ} or T^{μ′}. Thus T^{μ} ≅ T^{μ′} for μ ≠ μ′. On the other hand, if μ = μ′ then S^{μ} ↓_{A_{n}} splits into a direct sum of two non-isomorphic A_{n}-modules, say T_{+}^{μ} and T_{−}^{μ}. Moreover, the modules T_{+}^{μ} and T_{−}^{μ}, as μ runs over all symmetric partitions of n, together with the modules T^{μ}, as μ runs over a system of representatives of the pairs {μ, μ′} for the non-symmetric partitions μ of n, form a complete system of the non-isomorphic irreducible A_{n}-modules. It is well known that T_{±}^{μ} is obtained from T_{∓}^{μ} by twisting with an automorphism of A_{n}, which comes from a conjugation by an element g ∈ S_{n} \ A_{n}. The character of T_{(±)}^{μ} will be denoted by {μ}_{(±)}.

Lemma 5.1 Let μ, ν be non-symmetric partitions of n, both different from (n) and (1^{n}). Then T^{μ} ⊗ T^{ν} is homogeneous if and only if S^{μ} ⊗ S^{ν} ≅ x S^{λ} ⊕ y S^{λ′} for some λ ≠ λ′, x, y ∈ IN.

Proof. This follows from the definition of the modules T^{μ} and Theorem 3.4. ⋄

Lemma 5.2 Let μ, ν be partitions of n, both different from (n), (1^{n}). Assume that μ ≠ μ′, ν = ν′. Then T^{μ} ⊗ T_{±}^{ν} is homogeneous if and only if S^{μ} ⊗ S^{ν} ≅ x S^{λ} ⊕ y S^{λ′} for some λ ≠ λ′, x, y ∈ IN.

Proof. The “if-part” is clear.

If T^{μ} ⊗ T_{+}^{ν} ≅ x T_{±}^{λ} for some λ = λ′, then, conjugating by g ∈ S_{n} \ A_{n}, we get T^{μ} ⊗ T_{−}^{ν} ≅ x T_{∓}^{λ}. So

T^{μ} ⊗ (T_{+}^{ν} ⊕ T_{−}^{ν}) ≅ x(T_{+}^{λ} ⊕ T_{−}^{λ}) .

The lift to S_{n} gives S^{μ} ⊗ S^{ν} ≅ x S^{λ}, which is impossible by Theorem 3.4.

If T^{μ} ⊗ T_{+}^{ν} ≅ x T^{λ} for some λ ≠ λ′, then as above we have T^{μ} ⊗ T_{−}^{ν} ≅ x T^{λ}, so the lift to S_{n} gives S^{μ} ⊗ S^{ν} ≅ y S^{λ} ⊕ z S^{λ′} (with y + z = x). ⋄

Lemma 5.3 Let ν be a symmetric partition of n, and let ϕ, ψ be irreducible A_{n}-characters both different from {ν}_{+} and {ν}_{−}. Then

⟨ψ · {ν}_{+}, ϕ⟩ = ⟨ψ · {ν}_{−}, ϕ⟩ .

Proof. By [4, 2.5.13], we have

⟨ψ · \{ν\}_{±}, ϕ⟩ = \frac{1}{|A_{n}|} ∑_{g∈A_{n}} ψ(g)\{ν\}_{±}(g)\overline{ϕ(g)}

= \frac{1}{|A_{n}|}( ∑_{g∈A_{n}\(C_{ν}^{+}∪C_{ν}^{−})} ψ(g)\{ν\}_{±}(g)\overline{ϕ(g)} + ∑_{g∈C_{ν}^{+}} ψ(g)\frac{1}{2}(ε_{ν} ± \sqrt{ε_{ν} ∏_{i} h_{ii}^{ν}})\overline{ϕ(g)}

+ ∑_{g∈C_{ν}^{−}} ψ(g)\frac{1}{2}(ε_{ν} ∓ \sqrt{ε_{ν} ∏_{i} h_{ii}^{ν}})\overline{ϕ(g)})

where ε_{ν} = (−1)^{(n−k)/2} and C_{ν}^{±} denote the two conjugacy classes in A_{n} which consist of elements of cycle type (h_{11}^{ν}, …, h_{kk}^{ν}). Since ψ, ϕ correspond to partitions different from ν, each of them takes the same value on C_{ν}^{+} and C_{ν}^{−}, so the last expression is the same for {ν}_{+} and {ν}_{−}. ⋄

Lemma 5.4 Let ν be a symmetric partition of n and let ψ be an irreducible A_{n}-character different from {ν}_{+} and {ν}_{−}. Then

⟨ψ · {ν}_{+}, {ν}_{+}⟩ = ⟨ψ · {ν}_{−}, {ν}_{−}⟩ and ⟨ψ · {ν}_{+}, {ν}_{−}⟩ = ⟨ψ · {ν}_{−}, {ν}_{+}⟩ .

Proof. We compute the scalar products using [4, 2.5.13] as in the previous proof, and use the facts that {ν}_{+}(g) = {ν}_{−}(g) for any g ∈ A_{n} \ (C_{ν}^{+} ∪ C_{ν}^{−}) and ψ(g) = ψ(h) for any g, h ∈ C_{ν}^{+} ∪ C_{ν}^{−}. ⋄

From the previous two results we deduce:

Proposition 5.5 Let μ, ν be symmetric partitions of n, μ ≠ ν. Then {μ}_{+} · {ν}_{+} is homogeneous if and only if {μ}_{+} · {ν}_{−} is homogeneous.

Now we can classify the homogeneous Kronecker products of irreducible A_{n}-characters. Note that if n > 4 then the only 1-dimensional character is the trivial one. For n = 3 and 4 there are two more 1-dimensional characters in each case: {2, 1}_{±} and {2^{2}}_{±}.

Theorem 5.6 Let ϕ, ψ be irreducible A_{n}-characters both of degrees greater than 1. Then ϕ · ψ is homogeneous if and only if n = a^{2} for some a > 2 and one of the characters is {n − 1, 1}, while the other is {a^{a}}_{+} or {a^{a}}_{−}. In the exceptional case:

{n − 1, 1} · {a^{a}}_{±} = {a + 1, a^{a−2}, a − 1} .

Proof. The “if-part” follows from Corollary 4.2(ii).

Let ϕ and ψ correspond to partitions μ and ν, respectively. If μ and ν are both non-symmetric, then by Lemma 5.1 and Theorem 4.8 the tensor product T^{μ} ⊗ T^{ν} is not homogeneous. If one of the partitions μ, ν is symmetric and the other is not, use Lemma 5.2 and Theorem 4.8. So we may assume that μ and ν are both symmetric. If μ ≠ ν, then by Lemmas 5.3, 5.4 and 5.5, if one of the four products {μ}_{±} · {ν}_{±} is homogeneous then the product [μ] · [ν] has at most two homogeneous components, contradicting Theorems 3.4 and 4.8. Indeed, consider for example the case where {μ}_{−} · {ν}_{−} is homogeneous. Since {λ}_{±} is obtained from {λ}_{∓} by conjugating with an element g ∈ S_{n} \A_{n}, we conclude that {μ}_{+}·{ν}_{+} is also homogeneous. Moreover, if {μ}_{−}·{ν}_{−} = x{λ} then {μ}_{+}·{ν}_{+} = x{λ}, and if {μ}_{−} · {ν}_{−} = x{κ}_{±} then {μ}_{+} · {ν}_{+} = x{κ}_{∓}. By Proposition 5.5, we also have that {μ}_{±} · {ν}_{∓} are homogeneous. Moreover, Lemmas 5.3, 5.4 imply {μ}_{±} · {ν}_{∓} = {λ} or {κ}_{± or ∓}. Thus [μ] · [ν] is x[λ] + y[λ′] or x[κ].

Now let μ = ν be symmetric. We have to consider three cases: {μ}_{±} · {μ}_{±} and {μ}_{+} · {μ}_{−}. Using conjugation with g ∈ S_{n} \ A_{n}, we can eliminate one of them, and work only with {μ}_{+} · {μ}_{+} and {μ}_{+} · {μ}_{−}. Let us consider the first case (the second one is similar). So let {μ}_{+} · {μ}_{+} = xψ for some irreducible A_{n}-character ψ.

If the dual character {μ}_{+}^{∗} is equal to {μ}_{+}, then

⟨{n}, {μ}_{+} · {μ}_{+}⟩ = ⟨{μ}_{+}, {μ}_{+}⟩ = 1 ,

so we deduce {μ}_{+} · {μ}_{+} = {n}, which is impossible as {μ} is not of degree 1. If {μ}_{+}^{∗} = {μ}_{−}, then

⟨{n}, {μ}_{+} · {μ}_{+}⟩ = ⟨{μ}_{−}, {μ}_{+}⟩ = 0

and

⟨{n − 1, 1}, {μ}_{+} · {μ}_{+}⟩ = ⟨{n} + {n − 1, 1}, {μ}_{+} · {μ}_{+}⟩
= ⟨{n − 1} ↑^{A_{n}}, {μ}_{+} · {μ}_{+}⟩
= ⟨{μ}_{−} ↓_{A_{n−1}}, {μ}_{+} ↓_{A_{n−1}}⟩

Consider the case where μ is not a square. Then, by the Branching Rule, both restrictions in the last expression contain some {λ} where λ is a non-symmetric partition of n − 1. So the scalar product above is non-zero, whence {μ}_{+} · {μ}_{+} = x{n − 1, 1}. Take z ∈ A_{n} of cycle type (n − 2, 2), if n is even and of cycle type (n − 2, 1, 1), if n is odd. As μ is symmetric it does not have a hook of length n − 2. Hence by [4, 2.5.13] and the Murnaghan-Nakayama Rule we have

{μ}_{+}(z){μ}_{+}(z) = 0 .

On the other hand, x{n − 1, 1}(z) = ±x ≠ 0, when n is odd or even, respectively. This is a contradiction.

It remains to deal with the case where {μ}_{+}^{∗} = {μ}_{−} and μ is a square. Consider

⟨{n − 2} ↑^{A_{n}}, {μ}_{+} · {μ}_{+}⟩ = ⟨{μ}_{−} ↓_{A_{n−2}}, {μ}_{+} ↓_{A_{n−2}}⟩

By the Branching Rule, the last scalar product is non-zero. But

{n − 2} ↑^{A_{n}}= {n} + 2{n − 1, 1} + {n − 2, 2} + {n − 2, 1^{2}} ,

and the product {μ}_{+} · {μ}_{+} can not be of the form x{n} or x{n − 1, 1} by the same arguments as before. So we may assume that

{μ}_{+} · {μ}_{+} = x{n − 2, 2} or {μ}_{+} · {μ}_{+} = x{n − 2, 1^{2}} .

In the first case, we evaluate both sides on an element of cycle type (n − 2, 1^{2}) if n is odd, and on an element of cycle type (n − 1, 1) if n is even. Then the left hand side gives zero whereas the right hand side is ±x, giving a contradiction.

In the second case, we evaluate both sides on an element of cycle type (n) if n is odd, and on an element of cycle type (n − 3, 1^{3}) if n is even. This gives zero on the left hand side and ±x on the right hand side. ⋄

Acknowledgement. We are grateful to the Mathematical Sciences Research Institute, University of Oregon and the Isaac Newton Institute where the research for this paper was done. The authors were also supported by the NSF (grants # DMS 9022140 and # DMS-9600124) and the Deutsche Forschungsgemeinschaft (grant Be 923/6-1).

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Christine Bessenrodt, Fakultät für Mathematik, Otto-von-Guericke-Universität Magdeburg, D-39016 Magdeburg, Germany

Email address: bessen@mathematik.uni-magdeburg.de

Alexander Kleshchev, Department of Mathematics, University of Oregon, Eugene, Oregon, 97403, USA

Email address: klesh@darkwing.uoregon.edu