Plancherel

L K_{H}

L : field G : finite group acting on L K := Fix_{G} L N : kernel of action H := G/N

Remark : L ≀ G semisimple ⟺ |N| is invertible in K.

Assume from now on |N| to be invertible in K. X_{i} : simple L ≀ G-modules, i ∈ [1, k] K_{i} := End_{L≀G} X_{i} x_{i} := dim_{K_{i}} X_{i} , d_{i} := [K_{i} : Z(K_{i})]^{1/2} , r_{i} := [K_{i} : K]

Remark : x_{i}d_{i}/|H| ∈ Z

Wedderburn isomorphism

\begin{matrix}L ≀ G & ⟶_{∼}^{ω} & ∏_{i∈[1,k]} End_{K_{i}} X_{i} \\ ξ & \overline{} & (ξω_{i})_{i} \end{matrix}

u : chosen element of L such that Tr_{L|K}(u) = 1 Bilinear form on L ≀ G :

(gx, g′x′) := ∂_{g,g^{′−1}} Tr_{L|K}(x^{g′}x′) = \frac{1}{|N|} Tr_{K} (u(−)gxg′x′) ,

where g, g′ ∈ G, x, x′ ∈ L.

Plancherel formula : For ξ, η ∈ L ≀ G, we have

(ξ, η) = ∑_{i∈[1,k]} \frac{x_{i}d_{i}}{|G|} \underbrace{tr_{K_{i}|K}}_{red. tr.} \underbrace{Tr_{K_{i}} ((ξη)ω_{i})}_{“matrix trace”}

⇝ explicit inverse to ω

Colength formula

[figure: L; K; H; T; s ∈ S]

S_{i} ⊆ K_{i} : maximal S-order V_{i} ⊆ X_{i} : suitable T ≀ G-lattice ℓ : Jordan-Hölder length of an S-module Discriminant valuations : S_{i}^{+} := {y ∈ K_{i} : tr_{K}(yS_{i}) ⊆ S} , δ_{S_{i}|S} := ℓ(S_{i}^{+}/S_{i}) T^{+} := {y ∈ L : Tr_{L|K}(yT) ⊆ S} , δ_{T |S} := ℓ(T^{+}/T)

Now Plancherel ⇝ Remark : The colength of the Wedderburn embedding

\begin{matrix}T ≀ G & ↪^{ω} & ∏_{i∈[1,k]} End_{S_{i}} V_{i} \\ ξ & \overline{} & (ξω_{i})_{i} \end{matrix}

is given by

ℓ(Cokern ω) = \frac{1}{2} (|G| (δ_{T |S} + v_{s}(|N|)|H|) − ∑_{i∈[1,k]} x_{i}^{2} (δ_{S_{i}|S} + v_{s}(\frac{x_{i}d_{i}}{|H|}) r_{i})) .

Open questions :

• Length of cokernel of T ≀ G ^{\overline{}ω_{i}} End_{S_{i}} V_{i} ?

(quasiblock colength, where quasiblock := Im(ω_{i}))

• Criterion for a quasiblock of an untwisted group ring to be a twisted group ring ? (examples in thesis of H. Weber)

Example

S := Z_{(3)} , s := 3 , t := (ζ_{9} − 1)(ζ_{9}^{−1} − 1) , T := S[t], a = 2 ρ : induced by ζ_{9} \overline{}ζ_{9}^{4} , restricted from Q(ζ_{9}) to T. G := ⟨ρ⟩ ≃ C_{3}

Then ρ̇ = (\begin{matrix}1 & 0 & 0 \\ 6 & −5 & 1 \\ 24 & −21 & 4 \end{matrix} ) (difficult to control), ṫ = (\begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 3 & −9 & 6 \end{matrix}), ẗ = (\begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 3 & 0 & 0 \end{matrix}).

So an isomorphic copy of T ≀ G is given by its Wedderburn image

\{(\begin{matrix}a_{0,0} & a_{0,1} & a_{0,2} \\ 3a_{1,2} & a_{1,0} & a_{1,1} \\ 3a_{2,1} & 3a_{2,2} & a_{2,0} \end{matrix}) ∈ Z_{(3)}^{3×3} : \underbrace{a_{0,1} ≡_{3} a_{1,1} ≡_{3} a_{2,1}}_{resulting from (t^{ρ} − t)^{1}} , \underbrace{a_{0,2} + a_{1,2} + a_{2,2} ≡_{3} 0}_{resulting from (t^{ρ} − t)^{2}}\} .

This can be used to calculate

H^{∗}(G, T; S) ≃ Z_{(3)}[h^{(1)}, h^{(2)}]/(3h^{(1)}, 3h^{(2)}, h^{(1)}^{2}) ,

where deg h^{(i)} = i.