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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-29
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The class-function inner product χV,χW equals dimHomG(W,V)

Statement

Let G be a finite group and let V and W be finite-dimensional complex representations of G with characters χV and χW. Then

χV,χW=dimHomG(W,V).

Facts & Assumptions

Given: Finite-dimensional complex representations V and W of a finite group G, with characters χV and χW.

[F1]

The inner product of class functions is φ,ψ=1GgGφ(g)ψ(g) (The standard inner product on cf(G)).

[F2]

For a finite-dimensional complex representation U, the averaging operator PU=1GgρU(g) has image UG and trace dimUG (The averaging operator projects onto the fixed subspace).

[F3]

The fixed points of the diagonal representation on WV are exactly the intertwiners WV, so (WV)GHomG(W,V) (For finite-dimensional complex V, the intertwiners VW are exactly the fixed points of VW).

[F4]

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

[A1]

The character of a representation is by definition the trace of its action operator, χU(g)=trρU(g).

Proof

technique · direct
1.1

By [F3], dimHomG(W,V)=dim(WV)G. By [F2] applied to U=WV, this equals trPU for the averaging operator PU=1GgρWV(g).

F2F3given
1.2

Trace is linear, so trPU=1GgtrρWV(g)=1GgχWV(g), the second equality by [A1].

A1algebragiven
1.3

By [F4], the tensor-product and dual clauses give χWV(g)=χW(g)χV(g)=χW(g)χV(g).

F4given
2.1

Combining steps 1.1 through 1.3, dimHomG(W,V)=1GgχW(g)χV(g)=1GgχV(g)χW(g), reordering the product of two complex numbers in each summand. This is χV,χW by [F1].

F1step 1.1step 1.2step 1.3algebra

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Dependency tree · two levels

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