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

Statement

Let $G$ be a finite group and let $V$ and $W$ be finite-dimensional complex representations of $G$ . Then, for every $g \in G$ :

$χ_{V \oplus W} (g) = χ_{V} (g) + χ_{W} (g)$ ;
$χ_{V \otimes W} (g) = χ_{V} (g) χ_{W} (g)$ ;
$χ_{V^{*}} (g) = \overline{χ_{V} (g)}$ .

Facts & Assumptions

Given: Finite-dimensional complex representations $V$ , $W$ of a finite group $G$ , and an element $g \in G$ .

[F1]

The character is $χ (g) = tr ρ (g)$ (The character $χ_{V} (g) = tr (ρ_{V} (g))$ of a finite-dimensional complex representation).

[F2]

The dual action is $(g \cdot f) (v) = f (g^{- 1} \cdot v)$ (The dual or contragredient complex representation).

[F3]

The tensor-product action is $g \cdot (v \otimes w) = (g \cdot v) \otimes (g \cdot w)$ (The tensor product of two complex representations).

[A1]

In a basis made of a basis of $V$ followed by a basis of $W$ , the matrix of the direct-sum action on $V \oplus W$ is block diagonal with the two action matrices on its diagonal, so its trace is the sum of the two block traces.

[A2]

Trace is linear, in particular additive, on the space of square matrices (Trace is a linear functional on $M_{n} (F)$ ).

[A3]

A basis of $V$ and a basis of $W$ give the basis $(v_{i} \otimes w_{j})_{i, j}$ of the tensor product (The elementary tensors of two bases form the product basis of the tensor product).

[A4]

For a complex character, $χ (g^{- 1}) = \overline{χ (g)}$ (For a complex character, $χ (1) = \dim V$ , $χ$ is a class function, and $∣ χ (g) ∣ \leq χ (1)$ with equality exactly at scalars).

Proof

technique · direct

1.1

Writing the direct-sum action in the concatenated basis of [A1], its trace is the sum of the traces of the two diagonal blocks, so $χ_{V \oplus W} (g) = tr (ρ_{V} (g) \oplus ρ_{W} (g)) = tr ρ_{V} (g) + tr ρ_{W} (g) = χ_{V} (g) + χ_{W} (g)$ by [F1]. This is claim 1.

F1A1A2given

1.2

By [F3] and [A3], the matrix of the tensor-product action in the basis $(v_{i} \otimes w_{j})$ has entry $ρ_{V} (g)_{i i^{'}} ρ_{W} (g)_{j j^{'}}$ at the position $(i, j) \mapsto (i^{'}, j^{'})$ , because $g \cdot (v_{i^{'}} \otimes w_{j^{'}}) = \sum_{i, j} ρ_{V} (g)_{i i^{'}} v_{i} \otimes ρ_{W} (g)_{j j^{'}} w_{j}$ .

F3A3given

1.3

With dual bases $(v_{i})$ of $V$ and $(v_{i}^{*})$ of $V^{*}$ , the matrix of the dual action of [F2] is the transpose of the matrix of $ρ_{V} (g^{- 1})$ : writing $g^{- 1} \cdot v_{j} = \sum_{i} ρ_{V} (g^{- 1})_{i j} v_{i}$ gives $(g \cdot v_{i}^{*}) (v_{j}) = v_{i}^{*} (g^{- 1} \cdot v_{j}) = ρ_{V} (g^{- 1})_{i j}$ .

F2given

2.1

Its trace is the sum of its diagonal entries: $\sum_{i, j} ρ_{V} (g)_{i i} ρ_{W} (g)_{j j} = (\sum_{i} ρ_{V} (g)_{i i}) (\sum_{j} ρ_{W} (g)_{j j})$ . By [F1] the two factors are $χ_{V} (g)$ and $χ_{W} (g)$ , so $χ_{V \otimes W} (g) = χ_{V} (g) χ_{W} (g)$ , which is claim 2.

F1step 1.2algebra

3.1

A matrix and its transpose have the same diagonal entries, hence the same trace, so $χ_{V^{*}} (g) = tr ρ_{V} (g^{- 1}) = χ_{V} (g^{- 1})$ . By [A4], $χ_{V} (g^{- 1}) = \overline{χ_{V} (g)}$ , which is claim 3.

F1A4step 1.3algebra∎

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Dependency tree · two levels

23 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

Peter Webb, A Course in Finite Group Representation Theory, Proposition 3.1.3 (standard reference, not scraped)
Pavel Etingof et al., Introduction to Representation Theory, Section 3.4 (standard reference, not scraped)