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Sign twist corresponds to the omega involution

Statement

Let $sgn$ be the sign representation of $S_{n}$ and let $V$ be a finite-dimensional complex representation of $S_{n}$ with character $χ_{V}$ and characteristic $F = ch (χ_{V})$ . Then

$ch (χ_{V \otimes sgn}) = ω (F),$

where $ω$ is the involutive algebra endomorphism of $Λ$ with $ω (e_{r}) = h_{r}$ , extended to $Λ_{C}$ by complex linearity. In particular, for every $λ ⊢ n$ ,

$ch (χ_{S^{λ} \otimes sgn}) = ω (s_{λ}) = s_{λ^{'}},$

so $S^{λ} \otimes sgn ≅ S^{λ^{'}}$ and $χ^{λ^{'}} (w) = (- 1)^{n - ℓ (ρ)} χ^{λ} (w)$ on elements of cycle type $ρ$ .

Facts & Assumptions

Given: An integer $n \geq 0$ , a finite-dimensional complex representation $V$ of $S_{n}$ with character $χ_{V}$ , the sign representation $sgn$ , and $w \in S_{n}$ of cycle type $ρ ⊢ n$ with $ℓ (ρ)$ parts.

[F1]

$ch (χ) = \sum_{ρ ⊢ n} χ (ρ) p_{ρ} / z_{ρ}$ for a character $χ$ , where $χ (ρ)$ is its value on cycle type $ρ$ and $z_{ρ} = \prod_{i} i^{m_{i} (ρ)} m_{i} (ρ)!$ (The Frobenius characteristic map).

[F2]

For finite-dimensional complex representations $V, W$ of $S_{n}$ one has $χ_{V \otimes W} (g) = χ_{V} (g) χ_{W} (g)$ for every $g$ , where $V \otimes W$ is the tensor product representation with $g \cdot (v \otimes w) = g v \otimes g w$ (Characters add on direct sums, multiply on tensor products, and conjugate on duals, The tensor product of two complex representations).

[F3]

The sign representation of $S_{n}$ is one-dimensional with $σ \cdot a = sgn (σ) a$ ; a $k$ -cycle has sign $(- 1)^{k - 1}$ , and $sgn (w) = (- 1)^{n - c (w)}$ , where $c (w) = ℓ (ρ)$ is the number of cycles of $w$ counted with fixed points (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).

[F4]

The $Z$ -algebra endomorphism $ω : Λ \to Λ$ with $ω (e_{r}) = h_{r}$ is an involution and satisfies $ω (p_{r}) = (- 1)^{r - 1} p_{r}$ in $Λ_{Q}$ and $ω (s_{λ}) = s_{λ^{'}}$ for every partition $λ$ ; it extends to a $C$ -linear algebra endomorphism of $Λ_{C}$ (The omega involution conjugates Schur functions).

[F5]

$ch (χ^{λ}) = s_{λ}$ for every $λ ⊢ n$ , where $χ^{λ}$ is the character of the Specht module $S^{λ}$ (The characteristic of a Specht character is a Schur function).

[F6]

Two finite-dimensional complex representations of a finite group are isomorphic if and only if their characters are equal (Finite-dimensional complex representations of a finite group are determined up to isomorphism by their characters).

[F7]

The characteristic map is injective on $c f (S_{n})$ (The Frobenius characteristic is an isometry).

Proof

technique · direct

1.1F2F3

By [F2] and [F3], for $w$ of cycle type $ρ$ the tensor-product character is $χ_{V \otimes sgn} (w) = χ_{V} (w) sgn (w) = χ_{V} (ρ) (- 1)^{n - ℓ (ρ)}$ .

1.2F4given

Since $ω$ is a $C$ -algebra endomorphism of $Λ_{C}$ [F4] and $p_{ρ} = \prod_{i} p_{ρ_{i}}$ with $ℓ (ρ)$ factors, $ω (p_{ρ}) = \prod_{i} ω (p_{ρ_{i}}) = \prod_{i} (- 1)^{ρ_{i} - 1} p_{ρ_{i}} = (- 1)^{\sum_{i} (ρ_{i} - 1)} p_{ρ} = (- 1)^{n - ℓ (ρ)} p_{ρ}$ .

2.1F1step 1.1

From the definition of the characteristic [F1] and step 1.1, $ch (χ_{V \otimes sgn}) = \sum_{ρ ⊢ n} χ_{V \otimes sgn} (ρ) p_{ρ} / z_{ρ} = \sum_{ρ ⊢ n} χ_{V} (ρ) (- 1)^{n - ℓ (ρ)} p_{ρ} / z_{ρ}$ .

2.2F1F4step 1.2

By $C$ -linearity of $ω$ [F4], [F1] and step 1.2, $ω (F) = ω (\sum_{ρ ⊢ n} χ_{V} (ρ) p_{ρ} / z_{ρ}) = \sum_{ρ ⊢ n} χ_{V} (ρ) ω (p_{ρ}) / z_{ρ} = \sum_{ρ ⊢ n} χ_{V} (ρ) (- 1)^{n - ℓ (ρ)} p_{ρ} / z_{ρ}$ .

3.1step 2.1step 2.2

Steps 2.1 and 2.2 exhibit two equal expressions, so $ch (χ_{V \otimes sgn}) = ω (F)$ for every finite-dimensional complex representation $V$ of $S_{n}$ .

4.1F4F5step 3.1

Taking $V = S^{λ}$ in step 3.1 and using [F5], $ch (χ_{S^{λ} \otimes sgn}) = ω (s_{λ}) = s_{λ^{'}}$ by [F4].

5.1F5F6F7step 1.1step 4.1∎

Since $ch$ is injective on $c f (S_{n})$ [F7] and $ch (χ^{λ^{'}}) = s_{λ^{'}}$ [F5], step 4.1 gives $χ_{S^{λ} \otimes sgn} = χ^{λ^{'}}$ ; by [F6] this equality of characters is equivalent to $S^{λ} \otimes sgn ≅ S^{λ^{'}}$ . Evaluating the character identity from step 1.1 for $V = S^{λ}$ gives $χ^{λ^{'}} (w) = (- 1)^{n - ℓ (ρ)} χ^{λ} (w)$ for $w$ of cycle type $ρ$ .

Depends on

Used by

Sign twist conjugates the (3,1) character of S₄ Example

Dependency tree · two levels

49 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 and §16 (standard reference, not scraped)