Equivalence of irreps on left ideals criterion

Let $𝐿 𝜇 𝛼$ and $𝐿 𝜈 𝛽$ be minimal left ideals transforming under the Regular group representation in irreps $Γ 𝜇$ and $Γ 𝜈$ respectively, and $𝑒 𝜇 𝛼$ and $𝑒 𝜈 𝛽$ be the generating primitive idempotents. Then $Γ 𝜇 ≅ Γ 𝜈$ iff $𝑒 𝜇 𝛼 * 𝑞 * 𝑒 𝜈 𝛽 \neq 0$ for some $𝑞 \in ℂ [𝐺]$ . #m/thm/rep

Proof

If $𝜇 = 𝜈$ then there exists an intertwiner $𝑆 : 𝐿 𝜇 𝛼 \to 𝐿 𝜇 𝛽$ with $𝑆 Γ 𝜇 𝛼 = Γ 𝜈 𝛽 𝑆$ and thus by lineärity

𝑆 P (𝑒 𝜇 𝛼) Λ (𝑟) 𝑝 = P (𝑒 𝜈 𝛼) Λ (𝑟) 𝑆 𝑝 𝑆 𝑟 * 𝑝 * 𝑒 𝜇 𝛼 = 𝑟 * 𝑆 𝑝 * 𝑒 𝜇 𝛼

for all $𝑝, 𝑟 \in ℂ [𝐺]$ . Then $𝑞 = 𝑆 𝑒 𝜇 𝛼 \in 𝐿 𝜈 𝛽$ has the required property, since

𝑒 𝜇 𝛼 * 𝑆 𝑒 𝜇 𝛼 * 𝑒 𝜈 𝛽 = 𝑆 𝑒 𝜇 𝛼 * 𝑒 𝜇 𝛼 * 𝑒 𝜇 𝛼 = 𝑆 𝑒 𝜇 𝛼

For the converse, let $𝑠 = 𝑒 𝜇 𝛼 * 𝑞 * 𝑒 𝜈 𝛽 \neq 0$ for some $𝑞 \in ℂ [𝐺]$ . Then

P (𝑠) Λ (𝑟) = Λ (𝑟) P (𝑠)

for all $𝑟 \in ℂ [𝐺]$ and in particular

P (𝑠) Γ 𝜇 𝛼 (𝑔) = Γ 𝜈 𝛽 (𝑔) P (𝑠) P (𝑠) P (𝑒 𝜇 𝛼) Λ (𝑔) = P (𝑒 𝜈 𝛼) Λ (𝑔) P (𝑠) P (𝑒 𝜇 𝛼 𝑒 𝜇 𝛼 𝑞 𝑒 𝜈 𝛽) Λ (𝑔) = Λ (𝑔) P (𝑒 𝜇 𝛼 𝑞 𝑒 𝜈 𝛽 𝑒 𝜈 𝛽) P (𝑠) Λ (𝑔) = Λ (𝑔) P (𝑠)

for all $𝑔 \in 𝐺$ , so by Schur's lemma the two irreps are equivalent.

Using lineärity arguments, it is sufficient to show $𝑒 𝜇 𝛼 * 𝛿 𝑔 * 𝑒 𝜈 𝛽 = 0$ for all $𝑔 \in 𝐺$ to prove the idempotents generate non-equivalent irreps.

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