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I was reading about the classification of unitary representations of $G=GL_2({\mathbb Q}_p)$ in Automorphic representations and ... by Goldfeld-Hundley yesterday and could not understand a very basic thing. I can see three sets of irreducible representations:

$S_1$ = {unitarizable smooth irreducible representations of $G$ over $\mathbb C$ up to an equivalence of representations},

$S_2$ = {unitary smooth irreducible representations of $G$ over $\mathbb C$ up to a unitary equivalence of representations},

$S_3$ = {Hilbert-irreducible representations of $G$ on Hilbert spaces over $\mathbb C$ up to a bounded equivalence}.

The difference is that in $S_2$ a Hermitian form is fixed, and in $S_3$ the space must be complete without closed invariant subspaces. They explain the classification for $S_1$: it falls into special, principal series and supercuspidal... classical stuff

There are natural functions forgetting the form $F: S_2 \rightarrow S_1$ and completion $C: S_2 \rightarrow S_3$.

Are $F$ and $C$ bijections? Where is it explained?

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  • $\begingroup$ The inverse to $F$ should "choosing a form" but I cannot understand why any two choices must be equivalent. Similarly, the inverse to $C$ is "taking smooth vectors" but here I do not understand why smooth vectors should form a non-zero irreducible representation. Cheers! $\endgroup$
    – Bugs Bunny
    Jul 4, 2012 at 11:34

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If an irreducible representation is unitarizable, it has at most one unitarization (up to equivalence). Going back from $S_2$ to $S_1$, you only have to consider the invariant subspace of smooth vectors, which has a naturally topology associated to it. The smooth vectors are always a dense subset, so cannot be zero.

For the second question, I am not quite sure what bounded equivalence means, but it seems that you do not get surjectivity from $S_2$ to $S_3$. I assume it means isomorphism, but not necessarily unitary. There are smooth, admissible representations, which are not unitarizable. A necessary condition for unitarizability is that the central character is unitary. This is also a sufficient criteria for the Steinberg and the supercuspidal representations. These are square integrable. For the principal series, it is not. Only the continuous series and complementary series representation are unitarizable. So $S_2$ is a proper subset of $S_3$.

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  • $\begingroup$ Steinberg=special. $\endgroup$
    – Marc Palm
    Jul 8, 2012 at 8:46
  • $\begingroup$ Btw, either unitarizable smooth or unitary. Unitary is the completed stuff, which does not preserve smoothness, and the smooth vectors are almost never a Hilbert space, at most a pre-Hilbert space. $\endgroup$
    – Marc Palm
    Jul 8, 2012 at 8:51
  • $\begingroup$ Thanks! Unfortunately, you have neither explained, nor given any references. Say, the standard proof of uniqueness of the unitary structure in the finite dimensional case goes via a morphism $\phi$ defined by $<\phi (x),y> = (x,y)$ and the endomorphism property $End (V)={\mathbb C}$. Here one has the endomorphism property, indeed, but I do not understand why $\phi$ is well-defined!! Essentially, so defined $\phi (x)$ is an element of the complection, and not necessarily of the original representation. How do you go around it? $\endgroup$
    – Bugs Bunny
    Jul 8, 2012 at 15:36
  • $\begingroup$ I am less certain what you are saying about $C$. All admissible irreducible reps are smooth and some are not unitarizable, as sure as devine carrots. How does it give a rep in $S_3$, not in $C(S_2)$? I am kind of thinking that $C$ is not surjective, but this should come from a representation on a Hilbert space without any nonzero smooth vectors!! I think $C$ is injective but again one needs to argue why two different reps cannot have isomorphic completions. This should correspond to "too many" smooth vectors forming a reducible representation... $\endgroup$
    – Bugs Bunny
    Jul 8, 2012 at 15:50
  • $\begingroup$ For $1$: Schur's lemma holds for densely-defined, closed intertwiner. Use the graph norm for extending Schur's lemma. For $2$: There is no nice correspondance between the categories $S_1$ and $S_3$. If you change the equivalence on the Hilbert-reps. to Naimark equivalence, then $S_1$ is contained in $S_3$ and $S_1$ is equal to $S_3$. I am not sure what the correct category terms are here, but I guess you understand what I want to say. Note that there exists unbounded intertwiner between non-unitarizable principal series representations. $\endgroup$
    – Marc Palm
    Jul 8, 2012 at 17:16

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