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The regular representation of $Z / 4 Z$ over $C$ splits into its four characters

Example

The regular representation of $Z / 4 Z$ over $C$ is the direct sum of its four one-dimensional irreducible summands, each occurring once.

Facts & Assumptions

Given: The cyclic group $G = Z / 4 Z$ .

[L1]

Over an algebraically closed field of characteristic prime to $∣ G ∣$ , the number of irreducible representations equals the number of conjugacy classes (If $k$ is algebraically closed and $char k ∤ ∣ G ∣$ , the number of irreducible representations of $G$ equals the number of conjugacy classes).

[L2]

Under the same hypotheses, the irreducible degrees satisfy the sum-of-squares formula (If $k$ is algebraically closed and $char k ∤ ∣ G ∣$ , then $\sum_{i} (\dim_{k} V_{i})^{2} = ∣ G ∣$ ).

[L3]

Under the same hypotheses, every irreducible representation occurs in the regular representation with multiplicity equal to its degree (If $k$ is algebraically closed and $char k ∤ ∣ G ∣$ , there are finitely many irreducible representations, and each occurs in the regular representation with multiplicity equal to its degree).

Verification

technique · direct

1.1

The group $G$ is abelian with four elements, so it has four conjugacy classes. Hence [L1] gives four irreducible complex representations, with degrees $d_{1}, d_{2}, d_{3}, d_{4}$ , and [L2] gives $d_{1}^{2} + d_{2}^{2} + d_{3}^{2} + d_{4}^{2} = 4.$

L1L2givenalgebra

2.1

Every $d_{i}$ is positive, so the only way four positive squares can sum to $4$ is $d_{1} = d_{2} = d_{3} = d_{4} = 1$ . Then [L3] says each irreducible occurs in the regular representation with multiplicity equal to $1$ , so the regular representation splits as the direct sum of those four one-dimensional summands.

L3step 1.1givenalgebra∎

Depends on

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Sources

Pavel Etingof et al., Introduction to Representation Theory, Section 3.3 (standard reference, not scraped)

The regular representation of Z/4ZZ/4Z over CC splits into its four characters