By Emmanuel Kowalski
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Note that the assumption implies that the dimension of E is rG : Hs dim E1 , which is the dimension of IndG H pE1 q, as we have just seen, so the result is certainly plausible. We will construct an intertwining map Φ : IndG H pE1 q ÝÑ E, and show that it is injective. Since we also assume that dim E ă `8, this will be enough to finish the proof. The definition of Φ is easy: for f in the space F of the induced representation G IndH pE1 q, we define rG:Hs ÿ Φpf q “ pgi´1 qf pgi q. i“1 This defines a k-linear map to E, since f pgi q P E1 by definition of F .
We leave this to the reader: this is much helped by the fact that the action of H1 on this induced representation is also the regular representation. Next, we must check that T is an intertwining operator; but again, both F and F2 carry actions which are variants of the regular representation, and this should not be surprising – we therefore omit it... ) Taking g P G and h2 P H2 , denoting f “ T˜pψq, we again let the reader check that the following f ph2 gq “ ψph2 gqp1q “ ph2 ¨ ψpgqqp1q “ ψpgqph2 q “ ph2 qψpgqp1q “ ph2 qf pgq, makes sense, and means that T˜pψq is in F .
We now consider only the case k “ C. It is elementary that m is isomorphic to the m-th symmetric power of 1 for all m ě 0.