H, K \leq G

and a

K

-representation

W

Res_{H}^{G} Ind_{K}^{G} W ≅ ⨁_{x \in H \ G / K} Ind_{H \cap x K x^{- 1}}^{H} Res_{H \cap x K x^{- 1}}^{x K x^{- 1}} (^{x} W),

where

^{x} W (x k x^{- 1}) = W (k)

For

H = K = B

G = B ⊔ BwB, w = (01 - 1 0) .

Thus

B \ G / B

has representatives

1, w

. The identity double coset contributes

θ

, while

B \cap wB w^{- 1} = T

Res_{B}^{G} Ind_{B}^{G} θ ≅ θ \oplus Ind_{T}^{B} (θ^{w} ∣_{T}),

where

θ^{w} (t) = θ (w^{- 1} tw) .

For

t (a) = diag (a, a^{- 1})

, one has

w^{- 1} t (a) w = t (a^{- 1})

ii

Solution

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

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