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The Frobenius characteristic dictionary for $S_{3}$

Example

For $n = 3$ , expand the three Schur functions $s_{(3)}, s_{(2, 1)}, s_{(1, 1, 1)}$ in the power-sum basis and recover the character table of $S_{3}$ . The expansion is

$s_{(3)} = h_{3} = \frac{p_{1}^{3} + 3 p_{1} p_{2} + 2 p_{3}}{6}, s_{(2, 1)} = h_{2} h_{1} - h_{3} = \frac{p_{1}^{3} - p_{3}}{3}, s_{(1, 1, 1)} = e_{3} = \frac{p_{1}^{3} - 3 p_{1} p_{2} + 2 p_{3}}{6},$

so the coefficients of $p_{ρ} / z_{ρ}$ give the table, on cycle types $(1^{3}), (2, 1), (3)$ respectively:

$χ^{(3)} = (1, 1, 1), χ^{(2, 1)} = (2, 0, - 1), χ^{(1, 1, 1)} = (1, - 1, 1) .$

The degrees are $f^{(3)} = f^{(1, 1, 1)} = 1$ and $f^{(2, 1)} = 2$ , matching the number of standard tableaux, and the columns are orthogonal with $\sum_{λ} f^{λ} χ^{λ} (1) = 1 + 4 + 1 = 6 = ∣ S_{3} ∣$ .

Facts & Assumptions

Given: The partitions $(3), (2, 1), (1, 1, 1)$ of $3$ and the cycle types $(1^{3}), (2, 1), (3)$ of $S_{3}$ .

[F1]

Jacobi–Trudi and dual Jacobi–Trudi: $s_{(3)} = h_{3}$ , $s_{(2, 1)} = \det (\begin{matrix} h_{2} & h_{3} \\ h_{0} & h_{1} \end{matrix}) = h_{2} h_{1} - h_{3}$ , and $s_{(1, 1, 1)} = e_{3}$ , with $h_{0} = e_{0} = 1$ (Jacobi–Trudi and dual Jacobi–Trudi identities).

[F2]

$h_{3} = \sum_{ρ ⊢ 3} p_{ρ} / z_{ρ} = \frac{p_{1}^{3} + 3 p_{1} p_{2} + 2 p_{3}}{6}$ , using $z_{(1^{3})} = 6$ , $z_{(2, 1)} = 2$ , $z_{(3)} = 3$ ; likewise $h_{2} = \frac{p_{1}^{2} + p_{2}}{2}$ and $h_{1} = p_{1}$ (Complete homogeneous functions expand in power sums with cycle-distribution coefficients).

[F3]

The involution $ω$ satisfies $ω (h_{r}) = e_{r}$ , $ω (p_{1}) = p_{1}$ , $ω (p_{2}) = - p_{2}$ , $ω (p_{3}) = p_{3}$ , and it is an algebra homomorphism, so $e_{3} = ω (h_{3}) = \frac{p_{1}^{3} - 3 p_{1} p_{2} + 2 p_{3}}{6}$ (The omega involution conjugates Schur functions).

[F4]

$ch (χ^{λ}) = s_{λ}$ for every $λ ⊢ 3$ , and $χ^{λ} (ρ)$ is the coefficient of $p_{ρ} / z_{ρ}$ in the power-sum expansion of $s_{λ}$ , i.e. $χ^{λ} (ρ) = ⟨ s_{λ}, p_{ρ} ⟩_{H}$ (The characteristic of a Specht character is a Schur function, Irreducible symmetric-group character values are power-sum coefficients).

[F5]

$f^{λ} = \dim_{C} S^{λ}$ is the number of standard $λ$ -tableaux, so $f^{(3)} = f^{(1, 1, 1)} = 1$ and $f^{(2, 1)} = 2$ (Standard polytabloids form a basis of a complex Specht module).

Verification

technique · direct

1.1F2given

The cycle types of $S_{3}$ are $(1^{3}), (2, 1), (3)$ , with $z_{(1^{3})} = 1^{3} \cdot 3! = 6$ , $z_{(2, 1)} = 2 \cdot 1 = 2$ and $z_{(3)} = 3$ , so [F2] gives $h_{1} = p_{1}$ , $h_{2} = \frac{p_{1}^{2} + p_{2}}{2}$ and $h_{3} = \frac{p_{1}^{3} + 3 p_{1} p_{2} + 2 p_{3}}{6}$ .

1.2F1F3

The three relevant Schur functions are $s_{(3)} = h_{3}$ , $s_{(2, 1)} = h_{2} h_{1} - h_{3}$ and $s_{(1, 1, 1)} = e_{3} = ω (h_{3})$ by [F1] and [F3].

1.3F5given

The numbers of standard tableaux are $f^{(3)} = 1$ , $f^{(2, 1)} = 2$ and $f^{(1, 1, 1)} = 1$ by [F5].

2.1F3step 1.1step 1.2algebra

Substituting step 1.1 into step 1.2: $s_{(3)} = \frac{p_{1}^{3} + 3 p_{1} p_{2} + 2 p_{3}}{6}$ ; $s_{(2, 1)} = \frac{p_{1}^{2} + p_{2}}{2} p_{1} - \frac{p_{1}^{3} + 3 p_{1} p_{2} + 2 p_{3}}{6} = \frac{3 p_{1}^{3} + 3 p_{1} p_{2} - p_{1}^{3} - 3 p_{1} p_{2} - 2 p_{3}}{6} = \frac{p_{1}^{3} - p_{3}}{3}$ ; and $s_{(1, 1, 1)} = ω (h_{3}) = \frac{p_{1}^{3} - 3 p_{1} p_{2} + 2 p_{3}}{6}$ .

3.1F4step 2.1algebra

By [F4], $χ^{λ} (ρ)$ is the coefficient of $p_{ρ} / z_{ρ}$ in the expansion of step 2.1; since $z_{(1^{3})} = 6$ , $z_{(2, 1)} = 2$ , $z_{(3)} = 3$ , the coefficients of $p_{(1^{3})} / 6, p_{(2, 1)} / 2, p_{(3)} / 3$ in $s_{(3)}, s_{(2, 1)}, s_{(1, 1, 1)}$ are respectively $(1, 1, 1)$ , $(2, 0, - 1)$ and $(1, - 1, 1)$ .

4.1step 1.3step 3.1algebra∎

The identity-column sum is $1 \cdot 1 + 2 \cdot 2 + 1 \cdot 1 = 6 = ∣ S_{3} ∣$ . The weighted squared row norms are $\frac{1}{6} + \frac{1}{2} + \frac{1}{3} = 1$ , $\frac{4}{6} + 0 + \frac{1}{3} = 1$ , and $\frac{1}{6} + \frac{1}{2} + \frac{1}{3} = 1$ in the displayed row order. The weighted products of the three distinct row pairs are $\frac{2}{6} - \frac{1}{3} = 0$ , $\frac{1}{6} - \frac{1}{2} + \frac{1}{3} = 0$ , and $\frac{2}{6} - \frac{1}{3} = 0$ . The column squared norms are $1 + 4 + 1 = 6$ , $1 + 0 + 1 = 2$ , and $1 + 1 + 1 = 3$ , and the distinct column products are $1 - 1 = 0$ , $1 - 2 + 1 = 0$ , and $1 - 1 = 0$ . Thus the rows are orthonormal with weights $1 / z_{ρ}$ , and the columns are orthogonal with squared norms $z_{ρ}$ .

Depends on

Used by

The Young permutation characteristic for shape (2,1) Example

Dependency tree · two levels

37 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

I. G. Macdonald, Symmetric Functions and Hall Polynomials, 2nd ed., Chapter I §7 (standard reference, not scraped)
G. D. James, The Representation Theory of the Symmetric Groups, §6 (standard reference, not scraped)

The Frobenius characteristic dictionary for S3S3​