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By Garrett P.
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Proof: Suppose the functional equation holds. The functions eKgK certainly span H(G, K) over k. We have (eKgK ϕ)(h) = ϕ(hx) dx / meas (KgK) = KgK ϕ(hθ1 gθ2 ) dθ1 dθ2 / meas (K)2 = = K K ϕ(hθg) dθ / meas (K) = K since ϕ is right K-invariant. Then this is = ϕ(h)ϕ(g) = Λ(eKgK ) ϕ(h) by definition of ϕ and by the functional equation. Thus, Rη ϕ = Λ(η)ϕ Then Λ(η1 ∗ η2 )ϕ = Rη1 ∗η2 ϕ = = Rη1 Rη2 ϕ = Λ(η1 )Λ(η2 )ϕ Thus, ϕ is an eigenfunction and Λ is a k-algebra homomorphism, and then also ϕ(1) = 1. On the other hand, suppose that Λ is a k-algebra homomorphism and ϕ a normalized eigenvector.
Admissibility A smooth representation (π, K) of G is admissible if, for every compact open subgroup K of G the K-fixed vector space V K is finite-dimensional. This following assertion, and all further assertions in this section, follow directly from the ‘complete reducibility’ of smooth representations of compact totally disconnected groups. Proposition: An equivalent characterization of admissibility is that, for a fixed compact open subgroup K of G, every irreducible smooth representation δ of K has finite multiplicity in Res G K π.
A K-spherical representation of G is an irreducible (smooth) representation π of G with a (non-zero) k-spherical vector. Lemma: Let v = 0 be a K-spherical vector in a K-spherical representation π of G. Then πK = k · v Proof: As usual, let eK = chK / meas (K) The irreducibility of π implies that H(G) · v = π where H(G) is the full Hecke algebra. Therefore, for any w ∈ π K there is η ∈ H(G) so that ηv = w Then w = eK w = eK ∗ ηv = eK ∗ η ∗ eK v = Λ(eK ∗ η ∗ eK )v ∈ k · v ♣ This is the desired result.