 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/19j/a/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Apply Frobenius reciprocity and then the decomposition from part (i):

⟨ Ind_{B}^{G} θ, Ind_{B}^{G} φ ⟩_{G} = ⟨ Res_{B}^{G} Ind_{B}^{G} θ, φ ⟩_{B} = ⟨ θ, φ ⟩_{B} + ⟨ Ind_{T}^{B} (θ^{w} ∣_{T}), φ ⟩_{B} = ⟨ θ, φ ⟩_{B} + ⟨ θ^{w} ∣_{T}, φ ∣_{T} ⟩_{T} .

Since all the representations in the last line are one-dimensional,

⟨ Ind_{B}^{G} θ, Ind_{B}^{G} φ ⟩_{G} = 1_{θ = φ} + 1_{θ^{w} ∣_{T} = φ ∣_{T}} .

It is therefore

2

when both equalities hold,

1

when exactly one holds, and

0

 New to topics? Read the docs here!