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The complete internal product table for S_3

Example

For $S_{3}$ the three irreducible complex representations are $S^{(3)}$ (the trivial representation, dimension $1$ ), $S^{(2, 1)}$ (the standard representation, dimension $2$ ) and $S^{(1, 1, 1)}$ (the sign representation, dimension $1$ ). Their characters on the three conjugacy classes, indexed by cycle type in the order $(1^{3}), (2, 1), (3)$ , are

representation	$(1^{3})$	$(2, 1)$	$(3)$
$S^{(3)}$	$1$	$1$	$1$
$S^{(2, 1)}$	$2$	$0$	$- 1$
$S^{(1, 1, 1)}$	$1$	$- 1$	$1$

The complete internal (Kronecker) product table is therefore $S^{(3)} \otimes X ≅ X for every irreducible X,$ $S^{(2, 1)} \otimes S^{(2, 1)} ≅ S^{(3)} \oplus S^{(2, 1)} \oplus S^{(1, 1, 1)}, S^{(2, 1)} \otimes S^{(1, 1, 1)} ≅ S^{(2, 1)}, S^{(1, 1, 1)} \otimes S^{(1, 1, 1)} ≅ S^{(3)} .$ Equivalently, all Kronecker coefficients $g (λ, μ, ν)$ with $λ, μ, ν ⊢ 3$ are $0$ or $1$ , and the table is invariant under all permutations of the three arguments. The character table is recomputed here from the Frobenius dictionary rather than imported from an examples page.

Facts & Assumptions

Given: The partitions of $3$ : $(3)$ , $(2, 1)$ , $(1^{3})$ (Partitions, English diagrams, and conjugation); the class-function inner product $⟨ φ, ψ ⟩ = \frac{1}{6} \sum_{σ \in S_{3}} φ (σ) \overline{ψ (σ)}$ ; and the cycle-type data $z_{(1^{3})} = 6$ , $z_{(2, 1)} = 2$ , $z_{(3)} = 3$ , so the three conjugacy classes have sizes $6 / z = 1, 3, 2$ respectively (The conjugacy classes of $S_{n}$ are indexed by the tuples $(c_{1}, \dots, c_{n})$ with $\sum k c_{k} = n$ , If $σ \in S_{n}$ has $c_{k}$ cycles of length $k$ , then $∣ C_{S_{n}} (σ) ∣ = \prod_{k = 1}^{n} k^{c_{k}} c_{k}!$ ).

[F1]

$ch (χ^{λ}) = s_{λ}$ for every $λ ⊢ n$ , and $ch (f) = \sum_{ρ ⊢ n} f (ρ) p_{ρ} / z_{ρ}$ (The characteristic of a Specht character is a Schur function, The Frobenius characteristic map).

[F2]

For partitions $ρ$ and $λ$ of the same integer, $h_{λ} = \sum_{ρ} N (λ, ρ) p_{ρ} / z_{ρ}$ in $Λ_{Q}$ , in particular $h_{1} = p_{1}$ , $h_{2} = \frac{1}{2} p_{1}^{2} + \frac{1}{2} p_{2}$ and $h_{3} = \frac{1}{6} p_{1}^{3} + \frac{1}{2} p_{1} p_{2} + \frac{1}{3} p_{3}$ (Complete homogeneous functions expand in power sums with cycle-distribution coefficients).

[F3]

The family ${p_{ρ} / z_{ρ} : ρ ⊢ n}$ is a $Q$ -basis of $Λ_{Q}^{n}$ , so coefficients in that basis are unique (Power sums form a rational but not integral stable basis).

[F4]

$g (λ, μ, ν) = \frac{1}{6} \sum_{σ \in S_{3}} χ^{λ} (σ) χ^{μ} (σ) χ^{ν} (σ)$ (The Kronecker coefficient is a three-character inner product), and $g$ is symmetric in the three arguments (Kronecker coefficients are nonnegative integers with full S_3 and conjugation symmetries).

[F5]

The tabloid module $M^{μ}$ is the permutation representation on the $μ$ -tabloids (Young subgroups, tabloids, and permutation modules), and Young's rule gives $M^{μ} ≅ ⨁_{λ ⊢ 3} (S^{λ})^{\oplus K_{λ μ}}$ , so the Young permutation character is $φ^{μ} = \sum_{λ} K_{λ μ} χ^{λ}$ by additivity of characters (Young's rule for complex permutation modules, Characters add on direct sums, multiply on tensor products, and conjugate on duals); in particular $φ^{(3)} = χ^{(3)}$ and $φ^{(2, 1)} = χ^{(3)} + χ^{(2, 1)}$ because content $(2, 1)$ consists of two $1$ 's and one $2$ : its only semistandard shapes are $(3)$ and $(2, 1)$ , with a unique filling for each. A three-cell column is impossible (Semistandard tableaux and Kostka numbers).

[F6]

$ch (φ^{μ}) = h_{μ}$ for every $μ ⊢ 3$ (The characteristic of a Young permutation character is complete homogeneous).

[F7]

$s_{λ} = \det (h_{λ_{i} - i + j})$ for every partition $λ$ , so $s_{(3)} = h_{3}$ (Jacobi–Trudi and dual Jacobi–Trudi identities).

[F8]

The sign representation is one-dimensional with $sgn (σ) = (- 1)^{3 - c (σ)}$ , where $c (σ)$ is the number of cycles of $σ$ counted with fixed points; the sign twist of $S^{(3)}$ is $S^{(1, 1, 1)}$ (Sign twist corresponds to the omega involution); hence, once step 1.1 identifies $S^{(3)}$ as trivial, $χ^{(1, 1, 1)}$ has values $1, - 1, 1$ on $(1^{3}), (2, 1), (3)$ (The sign representation of $S_{n}$ and the restriction ${Res}_{H}^{G} (V)$ of a representation to a subgroup, A $k$ -cycle has sign $(- 1)^{k - 1}$ , and $sgn (σ) = (- 1)^{n - c (σ)}$ when fixed points are counted as cycles).

Verification

technique · direct

1.1F1F2F3F7

Character of $S^{(3)}$ : by [F7] $s_{(3)} = h_{3}$ , and [F2] expands $h_{3} = \frac{1}{6} p_{1}^{3} + \frac{1}{2} p_{1} p_{2} + \frac{1}{3} p_{3}$ in the basis ${p_{ρ} / z_{ρ}}$ . By [F1] the coefficient of $p_{ρ} / z_{ρ}$ in $s_{(3)} = ch (χ^{(3)})$ is $χ^{(3)} (ρ)$ , so uniqueness [F3] gives the values $χ^{(3)} (1^{3}) = 1$ , $χ^{(3)} (2, 1) = 1$ , $χ^{(3)} (3) = 1$ .

1.2F1F2F3F5F6algebra

Character of $S^{(2, 1)}$ : the Young permutation character of the tabloid module $M^{(2, 1)}$ has characteristic $ch (φ^{(2, 1)}) = h_{2} h_{1}$ by [F6], and $φ^{(2, 1)} = χ^{(3)} + χ^{(2, 1)}$ by Young's rule [F5]. Applying [F1] therefore gives $ch (χ^{(2, 1)}) = h_{2} h_{1} - h_{3}$ ; by the expansions [F2], $h_{2} h_{1} - h_{3} = (\frac{1}{2} p_{1}^{2} + \frac{1}{2} p_{2}) p_{1} - (\frac{1}{6} p_{1}^{3} + \frac{1}{2} p_{1} p_{2} + \frac{1}{3} p_{3}) = \frac{1}{3} p_{1}^{3} - \frac{1}{3} p_{3}$ . By [F1] and uniqueness [F3], the values are $χ^{(2, 1)} (1^{3}) = 6 \cdot \frac{1}{3} = 2$ , $χ^{(2, 1)} (2, 1) = 0$ and $χ^{(2, 1)} (3) = 3 \cdot (- \frac{1}{3}) = - 1$ .

2.1F1F8step 1.1

Character of $S^{(1, 1, 1)}$ : step 1.1 gives the trivial character of $S^{(3)}$ , so by [F8] the sign character has values $1, - 1, 1$ on $(1^{3}), (2, 1), (3)$ , so its characteristic is $ch (χ^{(1, 1, 1)}) = \frac{1}{6} p_{1}^{3} - \frac{1}{2} p_{1} p_{2} + \frac{1}{3} p_{3}$ by [F1]; this is consistent with $ch (χ^{(1, 1, 1)}) = s_{(1, 1, 1)} = e_{3}$ from [F1].

3.1F4step 1.1step 2.1step 1.2algebra

Triple products. Write $T = (3)$ , $V = (2, 1)$ and $E = (1^{3})$ . Their character rows from steps 1.1, 1.2 and 2.1 are $(1, 1, 1)$ , $(2, 0, - 1)$ and $(1, - 1, 1)$ on classes of sizes $1, 3, 2$ . Formula [F4] gives $g (T, T, T) = (1 + 3 + 2) / 6 = 1$ , $g (T, E, E) = (1 + 3 + 2) / 6 = 1$ , $g (T, V, V) = g (E, V, V) = (4 + 0 + 2) / 6 = 1$ and $g (V, V, V) = (8 + 0 - 2) / 6 = 1$ . The other unordered triples give $g (T, T, E) = g (E, E, E) = (1 - 3 + 2) / 6 = 0$ and $g (T, T, V) = g (T, E, V) = g (E, E, V) = (2 + 0 - 2) / 6 = 0$ . These are all ten unordered triples; [F4] supplies every ordering.

4.1F4step 3.1∎

Since the multiplicity of $S^{ν}$ in $S^{λ} \otimes S^{μ}$ is $g (λ, μ, ν)$ , step 3.1 gives exactly the displayed table: $S^{(2, 1)} \otimes S^{(2, 1)} ≅ S^{(3)} \oplus S^{(2, 1)} \oplus S^{(1, 1, 1)}$ , $S^{(2, 1)} \otimes S^{(1, 1, 1)} ≅ S^{(2, 1)}$ , $S^{(1, 1, 1)} \otimes S^{(1, 1, 1)} ≅ S^{(3)}$ and $S^{(3)} \otimes X ≅ X$ for every irreducible $X$ . All coefficients are $0$ or $1$ , and the table equals each of its permutations by [F4].

Remarks

The example gives the complete internal product table for $S_{3}$ and exhibits the multiplicities of the tensor square of the standard representation, which decomposes as the direct sum of all three irreducibles.
The three characters were recomputed from the Frobenius dictionary rather than cited from a character table on another page, and the count of elements in each class used $1, 3, 2 = 6 / z_{ρ}$ .

Depends on

Used by

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Sources

Igor Pak and Greta Panova, Bounds on the Kronecker coefficients, arXiv:1406.2988v2 (2014) (standard reference, not scraped)
I. G. Macdonald, Symmetric Functions and Hall Polynomials, 2nd ed., Oxford University Press 1995 (standard reference, not scraped)