By Huntington E.V.
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Extra info for An Improved Equal-Frequency Map of the Normal Correlation Surface, Using Circles Instead of Ellipses
Sample text
We suppose of course that π (b)ϕ = ξψ (b)ϕ if b is in BF and ϕ is in V . Let A be an isomorphism of V with V such that Aπ(g) = π (g)A for all g . Let L be the linear functional L(ϕ) = Aϕ(1) on V . Then a 0 0 1 L π ϕ = Aϕ(a) so that A is determined by L. If we could prove the existence of a scalar λ such that L(ϕ) = λϕ(1) it would follow that Aϕ(a) = λϕ(a) for all a such that Aϕ = λϕ. This equality of course implies the theorem. Observe that L π 1 x 0 1 ϕ =π 1 x 0 1 Aϕ(1) = ψ(x)L(ϕ). 1) Thus we need the following lemma.
It follows from the previous lemma that a 0 0 1 WΦ is zero for almost all a if fΦ is 0. Since WΦ a 0 0 1 is a locally constant function on F × it must vanish everywhere. We have incidentally proved the following lemma. 3 Suppose |µ1 ( )µ−1 2 ( )| = | | with s > −1 and W belongs to W (µ1 , µ2 ; ψ). If W a 0 0 1 =0 for all a in F × then W is 0. 3 Let µ1 and µ2 be two quasi-characters of F × . −1 (i) If neither µ1 µ−1 2 nor µ1 µ2 is αF the representations ρ(µ1 , µ2 ) and ρ(µ2 , µ2 ) are equivalent and irreducible.
Because of (i) we need only show that every linear transformation of X which commutes with all the operators Cn (ν) is a scalar. Suppose T is such an operator. If ϕ belongs to V let Tϕ be the function from F × to X defined by T ϕ(a) = T ϕ(a) . Observe that T ϕ is still in V . This is clear if ϕ belongs to V0 and if ϕ = π(w)ϕ0 we see on examining the Mellin transforms of both sides that T ϕ = π(w)T ϕ0 . Since V = V0 + π(w)V0 the observation follows. T therefore defines a linear transformation of V which clearly commutes with the action of any g in PF .