April 5

Let V ◁ H ≤ G and let X be a ℂG-module.

Definition 1 Define

Inv_{V} X = {x ∈ X : vx = x for all v ∈ V } = X^{V} .

Observation 1 Inv_{V} X is a H-module since

vhx = h \underbrace{h^{−1}vh}_{∈V} x = hx

for v ∈ V, h ∈ H, and x ∈ Inv_{V} X. Now since V acts trivially on Inv_{V} X, we can regard Inv_{V} X as a H/V -module.

Definition 2 Let ξ be the character afforded by the G-module X. Write r_{H/V}^{G} ξ for the character afforded by Inv_{V} X. The operation r_{H/V}^{G} : Ch (G) → Ch (H/V) is called generalized restriction.

Proposition 1 Write Res_{H}^{G}ξ = ξ′ + ξ′′ where ξ′ is a sum of irreducible characters of H containing V in their kernels and ξ′′ is a sum of the remaining characters of H. Then regarded as characters of H/V, we have ξ′ = r_{H/V}^{G} ξ.

Proof. X as an H-module splits as X = Y ⊕ Z with Y the direct sum of irreducible H-modules on which V acts trivially and Z the direct sum of the other irreducible Hmodules. We have Y ≤ Inv_{V} X by definition. But if Y ≨ Inv_{V} X, then Y ⊕ Y′ = Inv_{V} X so that Y′ ≤ Z, a contradiction.

Proposition 2

r_{H/V}^{G} ξ (hV) = \frac{1}{|V |} ∑_{v∈V} ξ (hv) .

Proof. Define

e_{V} = \frac{1}{|V |} ∑_{v∈V} v ∈ ℂG.

Observe that Inv_{V} X = e_{V} X. Let \overline{h} ∈ H/V. Observe also that \overline{h}e_{V} = e_{V} \overline{h} since V ◁ H. Then

r_{H/V}^{G} ξ (\overline{h}) = Tr (\overline{h}) = Tr (\overline{h} · e_{V} X) = Tr (e_{V} \overline{h} · X) = \frac{1}{|V |} ∑_{v∈V} Tr (v\overline{h}X) = \frac{1}{|V |} ∑_{v∈V} ξ (vh) .

Definition 3 For λ ∈ Ch (H/V), write λ̃ for the lift of λ to a character of H. That is, λ̃ (h) = λ (hV).

Definition 4 Let λ ∈ Ch (H/V). Define i_{H/V}^{G} λ = Ind_{H}^{G} λ̃. The operation i_{H/V}^{G} : Ch (H/V) → Ch (G) is called generalized induction.

Proposition 3 ⟨ξ, i_{H/V}^{G} λ⟩_{G} = ⟨r_{H/V}^{G} ξ, λ⟩_{H/V}.

Proof. We have

⟨ξ, i_{H/V}^{G} λ⟩_{G} = ⟨ξ, Ind_{H}^{G} λ̃⟩_{G} = ⟨Res_{H}^{G}ξ, λ̃⟩_{H} = ⟨ξ′, λ̃⟩_{H} + ⟨ξ′′, λ̃⟩_{H} ,

but ⟨ξ′, λ̃⟩_{H} = ⟨ξ′, λ⟩_{H/V} and ⟨ξ′′, λ̃⟩_{H} = 0, since the constituents of ξ′′ are orthogonal

to the constituents of λ̃.

Proposition 4 Let ξ ∈ Irr (G) and suppose C_{H} (h) ∩ V = 1 for h ∈ H. Then ξ (h) = ξ′ (h).

Proof. Let \overline{·} : H → H/V be projection onto the quotient. We have that distinct elements of C_{H} (h) are in distinct cosets of V because if h_{1}h_{2}^{−1} ∈ V, we would have h_{1}h_{2}^{−1} = 1 since h_{1}h_{2}^{−1} ∈ C_{H} (x) ∩ V = 1. Therefore, we have |C_{H} (h)| = |\overline{C_{H} (h)}|. But \overline{C_{H} (h)} ≤ C\overline{H} (\overline{h}) so that

|C_{H} (h)| = |\overline{C_{H} (h)}| ≤ |C\overline{H} (\overline{h})|

Now by the second orthogonality relation, we have

∑_{\begin{matrix}χ∈IrrH \\ V ⊂ker χ \end{matrix}} |χ (h)|^{2} + ∑_{\begin{matrix}χ∈IrrH \\ V ⊄ker χ \end{matrix}} |χ (h)|^{2} = |C_{H} (h)| ≤ |C\overline{H} (\overline{h})| = ∑_{χ∈Irr\overline{H}} |χ (h)|^{2} = ∑_{\begin{matrix}χ∈IrrH \\ V ⊂ker χ \end{matrix}} |χ (h)|^{2}

Now let G be a finite group with split BN-pair and let P ≤ G be a parabolic subgroup.

Write P = L_{P} V_{P} for its Levi decomposition. Then V_{P} ◁ P ≤ G so that our results above hold in this situation. Observe also that L_{P} ≅ P/V_{P}.

Definition 5 The operation T_{L_{P}}^{G} : Ch (G) → Ch (L_{P}) defined T_{L_{P}}^{G} ξ = r_{P/V_{P}}^{G} ξ is called Truncation and the operation R_{L_{P}}^{G} : Ch (L_{P}) → Ch (G) defined R_{L_{P}}^{G} λ = i_{P/V_{P}}^{G} λ is called HaRish-Chandra induction.

Recall that if Q ≤ G is another parabolic subgroup with Q ≤ P and Levi decomposition Q = L_{Q}V_{Q}, then L_{P} ∩ Q is a parabolic subgroup of L_{P} with Levi decomposition L_{P} ∩ Q = L_{Q} (L_{P} ∩ V_{Q}) as in the following illustration.

P = (\begin{matrix}♣ & ♣ & ♣ & ♣ & ♣ \\ 0 & ♣ & ♣ & ♣ & ♣ \\ 0 & ♣ & ♣ & ♣ & ♣ \\ 0 & ♣ & ♣ & ♣ & ♣ \\ 0 & ♣ & ♣ & ♣ & ♣ \end{matrix})  L_{P} = (\begin{matrix}♣ & 0 & 0 & 0 & 0 \\ 0 & ♣ & ♣ & ♣ & ♣ \\ 0 & ♣ & ♣ & ♣ & ♣ \\ 0 & ♣ & ♣ & ♣ & ♣ \\ 0 & ♣ & ♣ & ♣ & ♣ \end{matrix})  V_{P} = (\begin{matrix}1 & ♣ & ♣ & ♣ & ♣ \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{matrix})

Q = (\begin{matrix}♣ & ♣ & ♣ & ♣ & ♣ \\ 0 & ♣ & ♣ & ♣ & ♣ \\ 0 & ♣ & ♣ & ♣ & ♣ \\ 0 & ♣ & ♣ & ♣ & ♣ \\ 0 & 0 & 0 & 0 & ♣ \end{matrix})  L_{Q} = (\begin{matrix}♣ & 0 & 0 & 0 & 0 \\ 0 & ♣ & ♣ & ♣ & 0 \\ 0 & ♣ & ♣ & ♣ & 0 \\ 0 & ♣ & ♣ & ♣ & 0 \\ 0 & 0 & 0 & 0 & ♣ \end{matrix})  V_{Q} = (\begin{matrix}1 & ♣ & ♣ & ♣ & ♣ \\ 0 & 1 & 0 & 0 & ♣ \\ 0 & 0 & 1 & 0 & ♣ \\ 0 & 0 & 0 & 1 & ♣ \\ 0 & 0 & 0 & 0 & 1 \end{matrix})

L_{P} ∩ Q = (\begin{matrix}♣ & 0 & 0 & 0 & 0 \\ 0 & ♣ & ♣ & ♣ & ♣ \\ 0 & ♣ & ♣ & ♣ & ♣ \\ 0 & ♣ & ♣ & ♣ & ♣ \\ 0 & 0 & 0 & 0 & ♣ \end{matrix})  L_{Q} = (\begin{matrix}♣ & 0 & 0 & 0 & 0 \\ 0 & ♣ & ♣ & ♣ & 0 \\ 0 & ♣ & ♣ & ♣ & 0 \\ 0 & ♣ & ♣ & ♣ & 0 \\ 0 & 0 & 0 & 0 & ♣ \end{matrix})  L_{P} ∩ V_{Q} = (\begin{matrix}1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & ♣ \\ 0 & 0 & 1 & 0 & ♣ \\ 0 & 0 & 0 & 1 & ♣ \\ 0 & 0 & 0 & 0 & 1 \end{matrix})

Proposition 5 (Transitivity of R and T) We have

T_{L_{Q}}^{G} = T_{L_{Q}}^{L_{P}} ∘ T_{L_{P}}^{G}  and  R_{L_{Q}}^{G} = R_{L_{P}}^{G} ∘ R_{L_{Q}}^{L_{P}}.

Proof. The proof is linguistic. We have

ξ afforded by X a G-module
T_{L_{P}}^{G} ξ afforded by Inv_{V_{P}} X an L_{P} -module
T_{L_{Q}}^{L_{P}}T_{L_{P}}^{G} ξ afforded by Inv_{L_{P} ∩V_{Q}}Inv_{V_{P}} X an L_{Q}-module.

But Inv_{L_{P} ∩V_{Q}}Inv_{V_{P}} X = Inv_{(L_{P} ∩V_{Q})V_{P}} X = Inv_{V_{Q}}X, the module affording T_{L_{Q}}^{G}ξ.