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Outer induction is not the Kronecker product

Statement refuted

The claim refuted is: the outer induction product of two symmetric-group characters is the same operation as the Kronecker (tensor) product of two representations of one symmetric group, so that for $f, g \in R (S_{n})$ the outer product $f \circ g \in R (S_{2 n})$ coincides with the tensor product $f \cdot g \in R (S_{n})$ . Take $n = 1$ and let $1_{1}$ be the trivial character of $S_{1}$ . The outer product $1_{1} \circ 1_{1} = {Ind}_{S_{1} \times S_{1}}^{S_{2}} (1_{1} ⊠ 1_{1})$ is the permutation character of $S_{2}$ on the two cosets of $S_{1} \times S_{1}$ , with values $(2, 0)$ on the cycle types $(1^{2}), (2)$ and decomposition $χ^{(2)} + χ^{(1, 1)}$ ; its characteristic is $p_{1}^{2} = s_{(2)} + s_{(1, 1)}$ . By contrast, the tensor product of the two one-dimensional trivial representations of $S_{1}$ has dimension $1 \cdot 1 = 1$ , lies in degree $1$ and remains a representation of $S_{1}$ . Hence the outer induced product and the same-rank tensor product are different operations: an outer coefficient such as $c_{(1), (1)}^{(2)} = 1$ records a multiplicity for representations of $S_{1} \times S_{1}$ inducing to $S_{2}$ , not a tensor-product multiplicity for two representations of one symmetric group.

Facts & Assumptions

Given: The trivial characters $1_{1}$ of $S_{1}$ and $1_{2}$ of $S_{2}$ , the transposition $τ \in S_{2}$ , and the subgroup $H : = S_{1} \times S_{1} = {1} \leq S_{2}$ .

[F1]

The outer product is $f \circ g = \sum_{i, j} a_{i} b_{j} {Ind}_{S_{m} \times S_{n}}^{S_{m + n}} (χ_{i} ⊠ ψ_{j})$ for $f = \sum_{i} a_{i} χ_{i} \in R (S_{m})$ , $g = \sum_{j} b_{j} ψ_{j} \in R (S_{n})$ , with $(χ_{i} ⊠ ψ_{j}) (σ, τ) = χ_{i} (σ) ψ_{j} (τ)$ ; for $m = n = 1$ the subgroup $S_{1} \times S_{1}$ is the trivial subgroup of $S_{2}$ (The outer induction product of symmetric-group characters).

[F2]

Frobenius' formula: ${Ind}_{H}^{G} θ (g) = \frac{1}{∣ H ∣} \sum_{x \in G : x^{- 1} g x \in H} θ (x^{- 1} g x)$ for a character $θ$ of a subgroup $H \leq G$ (Frobenius' formula for the character of an induced representation).

[F3]

$ch (f) = \sum_{ρ ⊢ n} f (ρ) p_{ρ} / z_{ρ}$ is linear, and $ch (f \circ g) = ch (f) ch (g)$ for $f \in R (S_{m})$ , $g \in R (S_{n})$ (The Frobenius characteristic map, The Frobenius characteristic preserves outer products).

[F4]

$ch (χ^{λ}) = s_{λ}$ for the Specht characters of $S_{n}$ , and $χ^{(2)}, χ^{(1, 1)}$ are the two pairwise inequivalent irreducible characters of $S_{2}$ (The characteristic of a Specht character is a Schur function, Specht modules classify the complex irreducibles of $S_{n}$ , Distinct complex Specht modules are inequivalent, Column antisymmetrizers, polytabloids, and Specht modules).

[F5]

Jacobi–Trudi and dual Jacobi–Trudi: $s_{(2)} = \det (h_{2}) = h_{2}$ and $s_{(1, 1)} = \det (e_{2}) = e_{2}$ (Jacobi–Trudi and dual Jacobi–Trudi identities).

[F6]

The involution $ω$ satisfies $ω (h_{2}) = e_{2}$ and $ω (p_{r}) = (- 1)^{r - 1} p_{r}$ (The omega involution conjugates Schur functions).

[F7]

$h_{2} = \frac{p_{1}^{2} + p_{2}}{2}$ in $Λ_{Q}^{2}$ , since $z_{(1, 1)} = 2 = z_{(2)}$ (Complete homogeneous functions expand in power sums with cycle-distribution coefficients).

[F8]

The tensor product of two finite-dimensional complex representations of a group $G$ is a representation of $G$ on the same group, with $χ_{V \otimes W} (g) = χ_{V} (g) χ_{W} (g)$ and $\dim (V \otimes W) = (\dim V) (\dim W)$ (The tensor product of two complex representations, Characters add on direct sums, multiply on tensor products, and conjugate on duals).

[F9]

Finite-dimensional complex representations of a finite group are determined up to isomorphism by their characters (Finite-dimensional complex representations of a finite group are determined up to isomorphism by their characters).

Counterexample

technique · direct

1.1F1F2

The subgroup $H = S_{1} \times S_{1}$ is the trivial subgroup of $S_{2}$ , and $1_{1} ⊠ 1_{1}$ is the trivial character of $H$ . By [F2], ${Ind}_{H}^{S_{2}} (1_{1} ⊠ 1_{1}) (1) = \frac{1}{1} \sum_{x \in S_{2}} 1 = 2$ , while for the transposition $τ$ the condition $x^{- 1} τ x \in H = {1}$ fails for every $x$ , so $Ind (τ) = 0$ .

1.2F3given

Since $1_{1}$ is the trivial character of $S_{1}$ , whose only cycle type is $(1)$ with $z_{(1)} = 1$ , [F3] gives $ch (1_{1}) = p_{1}$ ; hence $ch (1_{1} \circ 1_{1}) = ch (1_{1})^{2} = p_{1}^{2}$ .

1.3F6F7algebra

By [F7], $h_{2} = \frac{p_{1}^{2} + p_{2}}{2}$ , so $e_{2} = ω (h_{2}) = \frac{p_{1}^{2} - p_{2}}{2}$ by [F6], and therefore $h_{2} + e_{2} = p_{1}^{2}$ .

1.4F8given

The tensor product of two $1$ -dimensional complex representations of $S_{1}$ is a $1$ -dimensional complex representation of $S_{1}$ with character the product of the two characters, so it lies in $c f (S_{1})$ and has degree $1$ ; the value of the product of two trivial characters at the identity is $1$ .

2.1F5step 1.3algebra

By [F5] and step 1.3 applied to the identities $s_{(2)} = h_{2}, s_{(1, 1)} = e_{2}$ , one has $s_{(2)} + s_{(1, 1)} = h_{2} + e_{2} = p_{1}^{2}$ .

2.2F3F4step 1.3

By [F4] and [F3], $ch (χ^{(2)}) = s_{(2)} = h_{2} = \frac{p_{1}^{2} + p_{2}}{2}$ and $ch (χ^{(1, 1)}) = s_{(1, 1)} = e_{2} = \frac{p_{1}^{2} - p_{2}}{2}$ ; reading the coefficients of $p_{ρ} / z_{ρ}$ in these expansions with $z_{(1, 1)} = z_{(2)} = 2$ gives $χ^{(2)} = (1, 1)$ and $χ^{(1, 1)} = (1, - 1)$ on the cycle types $(1^{2}), (2)$ .

3.1step 1.2step 2.1

Steps 1.2 and 2.1 give $ch (1_{1} \circ 1_{1}) = p_{1}^{2} = s_{(2)} + s_{(1, 1)}$ in $Λ^{2}$ .

4.1F4F9step 1.1step 3.1step 2.2

Step 3.1 and step 2.2 show that the character of the induced module $1_{1} \circ 1_{1}$ equals $χ^{(2)} + χ^{(1, 1)}$ , namely $(2, 0)$ as computed in step 1.1; by [F9] the induced module is isomorphic to $S^{(2)} \oplus S^{(1, 1)}$ , so the outer coefficient $c_{(1), (1)}^{(2)} = 1$ is the multiplicity of $S^{(2)}$ in a module induced from $S_{1} \times S_{1}$ to $S_{2}$ .

5.1step 1.4step 4.1∎

The two sides are therefore objects attached to different symmetric groups: the outer product $1_{1} \circ 1_{1}$ is an element of $R (S_{2})$ , namely the degree- $2$ character $(2, 0)$ with $ch (1_{1} \circ 1_{1}) = p_{1}^{2}$ , while the tensor product of the two one-dimensional trivial representations of $S_{1}$ is an element of $R (S_{1})$ of degree $1$ with value $1$ at the identity [step 1.4]; a class function on $S_{2}$ with value $2$ at $1$ and a class function on the one-element group $S_{1}$ cannot be the same function, and an outer product coefficient records a multiplicity for representations of $S_{m} \times S_{n}$ inducing to $S_{m + n}$ , not a tensor-product multiplicity inside one $R (S_{n})$ . This refutes the identified claim.

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Sources

I. G. Macdonald, Symmetric Functions and Hall Polynomials, 2nd ed., Chapter I §7 (standard reference, not scraped)