CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A complex character is irreducible if and only if its self-inner-product is $1$

Statement

Let $G$ be a finite group and let $χ$ be a complex character of $G$ . Then $χ$ is irreducible if and only if $⟨ χ, χ ⟩ = 1$ .

Facts & Assumptions

Given: A finite group $G$ and a complex character $χ = χ_{V}$ of a finite-dimensional representation $V$ , completely reduced as $V ≅ ⨁_{i} m_{i} V_{i}$ .

[F1]

The multiplicity of $V_{i}$ in $V$ is $m_{i} = ⟨ χ, χ_{i} ⟩$ (The multiplicity of an irreducible summand is a character inner product).

[F2]

Characters add on direct sums, so the decomposition $V ≅ ⨁_{i} m_{i} V_{i}$ gives $χ = \sum_{i} m_{i} χ_{i}$ (Characters add on direct sums, multiply on tensor products, and conjugate on duals).

[A1]

If $χ = \sum_{i} m_{i} χ_{i}$ and $m_{i} = ⟨ χ, χ_{i} ⟩$ , then conjugate-symmetry and linearity in the first argument give $⟨ χ, χ ⟩ = \sum_{i} m_{i} ⟨ χ_{i}, χ ⟩ = \sum_{i} m_{i} \overline{⟨ χ, χ_{i} ⟩} = \sum_{i} m_{i} \overline{m_{i}} = \sum_{i} m_{i}^{2}$ , because the multiplicities $m_{i}$ are nonnegative integers.

[A2]

A representation is irreducible exactly when it has one irreducible summand with multiplicity $1$ and no others.

Proof

technique · direct

1.1F1F2A1given

Completely reduce $V ≅ ⨁_{i} m_{i} V_{i}$ with $m_{i} \geq 0$ integers. By [F2] this decomposition gives $χ = \sum_{i} m_{i} χ_{i}$ , and by [F1] each $m_{i} = ⟨ χ, χ_{i} ⟩$ ; therefore [A1] gives $⟨ χ, χ ⟩ = \sum_{i} m_{i}^{2}$ .

2.1A2step 1.1

Assume $χ$ is irreducible. Then by [A2] exactly one multiplicity equals $1$ and the rest are $0$ , so the sum in step 1.1 is $1$ ; hence $⟨ χ, χ ⟩ = 1$ .

2.2A2step 1.1algebra

Conversely, assume $⟨ χ, χ ⟩ = 1$ . The sum $\sum_{i} m_{i}^{2}$ of nonnegative integers in step 1.1 can equal $1$ only when exactly one $m_{i}$ equals $1$ and all the others are $0$ . By [A2], $V ≅ V_{i}$ is irreducible, so $χ$ is irreducible.

3.1step 2.1step 2.2∎

Steps 2.1 and 2.2 prove the two implications, hence the biconditional.

Depends on

Used by

Dependency tree · two levels

10 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, Corollary 3.3.4 (standard reference, not scraped)
Pavel Etingof et al., Introduction to Representation Theory, Section 3.5 (standard reference, not scraped)

A complex character is irreducible if and only if its self-inner-product is 11