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The square of the two-dimensional $S_{3}$ character decomposes as $1 + sgn + χ_{2}$

Example

The tensor square of the two-dimensional irreducible representation of $S_{3}$ has character $χ_{2}^{2}$ , and

$χ_{2}^{2} = 1 + sgn + χ_{2} .$

Facts & Assumptions

Given: The irreducible characters $1$ , $sgn$ , $χ_{2} = χ_{s t d}$ of $S_{3}$ .

[F1]

The character table of $S_{3}$ gives the values $χ_{2} = (2, 0, - 1)$ on the classes of $1$ , $(12)$ , $(123)$ (The character table of $S_{3}$ ).

[F2]

Characters multiply on tensor products (Characters add on direct sums, multiply on tensor products, and conjugate on duals).

[F3]

The multiplicity of an irreducible character in a given character is the inner product with it (The multiplicity of an irreducible summand is a character inner product).

[A1]

The inner product is $⟨ f, h ⟩ = \frac{1}{6} \sum_{σ} f (σ) \overline{h (σ)}$ .

Verification

technique · direct

1.1

By [F2], the tensor square of the two-dimensional representation has character $χ_{2}^{2}$ ; from [F1] its values are $(2, 0, - 1)^{2} = (4, 0, 1)$ on the three classes.

F1F2given

2.1

Using [A1] and the values of step 1.1, the multiplicities are $⟨ χ_{2}^{2}, 1 ⟩ = \frac{1}{6} (4 + 0 + 2) = 1$ , $⟨ χ_{2}^{2}, sgn ⟩ = \frac{1}{6} (4 + 0 + 2) = 1$ , and $⟨ χ_{2}^{2}, χ_{2} ⟩ = \frac{1}{6} (8 + 0 - 2) = 1$ . By [F3] these are the coefficients of $1$ , $sgn$ , and $χ_{2}$ .

A1F3step 1.1algebra

3.1

The value at $1$ checks: $χ_{2}^{2} (1) = 4 = 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 2$ , the sum of the degrees with the multiplicities of step 2.1; hence $χ_{2}^{2} = 1 + sgn + χ_{2}$ .

F1step 2.1algebra∎

Depends on

Used by

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Sources

Pavel Etingof et al., Introduction to Representation Theory, Example 3.16 (standard reference, not scraped)

The square of the two-dimensional S3S3​ character decomposes as 1+sgn⁡+χ21+sgn+χ2​