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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29
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The averaging operator projects onto the fixed subspace

Statement

Let ρ:G→GL⁡(V) be a finite-dimensional representation of a finite group G over C. The averaging operator

P:=1∣G∣∑g∈Gρ(g)

satisfies P2=P and has image exactly VG; consequently tr⁡P=dim⁡VG.

Facts & Assumptions

Given: A finite group G and a finite-dimensional complex representation ρ:G→GL⁡(V).

[F1]

The fixed subspace is VG={v∈V:g⋅v=v for every g∈G} (The fixed subspace VG of a representation).

[F2]

An idempotent endomorphism has trace equal to the dimension of its image as a field scalar (The trace of an idempotent is its rank as a field scalar).

Proof

technique · direct
1.1givenalgebra

For h∈G, ρ(h)P=1∣G∣∑gρ(hg) =1∣G∣∑gρ(g)=P, because left translation by h permutes G, so the sums run over the same index set.

2.1F1step 1.1given

Hence for every v∈V, the vector Pv satisfies h⋅(Pv)=ρ(h)Pv=Pv for every h∈G, so Pv∈VG by [F1]; thus im⁡P⊆VG.

3.1F1step 2.1algebra

If v∈VG, then ρ(g)v=v for every g by [F1], so Pv=1∣G∣∑gv=∣G∣∣G∣v=v. Thus VG⊆im⁡P, and with step 2.1, im⁡P=VG exactly.

4.1step 2.1step 3.1algebra

For v∈VG, step 3.1 gives P(Pv)=Pv=v=Pv; for general v, Pv∈VG by step 2.1, so P2v=P(Pv)=Pv. Hence P2=P.

5.1F2step 4.1step 3.1∎

Apply [F2] to the idempotent P from step 4.1. Over C this gives tr⁡P=dim⁡(im⁡P), and step 3.1 identifies im⁡P with VG.

Depends on

Used by

Dependency tree · two levels

7 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