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I sometimes come across this notion called "unitary automorphic representation". But I have never seen the precise definition. When they say $(\pi, V)$ is a unitary automorphic representation of a group $G(\mathbb{A})$, does that mean that $\pi$ is unitary as an abstract representation of $G(\mathbb{A})$ (assuming $\pi$ is a smooth automorphic representation, so that the full group $G(\mathbb{A})$ acts) in the sense that the space $V$ of $\pi$ is a Hilbert space with $G(\mathbb{A})$-invariant inner product. Or should I assume that $\pi$ is only pre-unitary in the sense that $\pi$ has only a non-degenerate $G(\mathbb{A})$-invariant inner product but the space $V$ is possibly incomplete?

Also if $(\pi, V)$ is a unitary automorphic representation, can I assume that the space is indeed a subspace of the space of automorphic forms on $G(\mathbb{A})$ instead of subquotient?

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As you suspect, there are many implicit assumptions and abuses of language in this terminology.

First, it is not safe to assume that an "automorphic" repn of $G(\mathbb A)$ is literally a repn of that topological group. Sometimes, only the finite-prime groups are allowed to act, and at archimedean places one does not have a group representation, but only a $\mathfrak g,K$-module. A reason to want to be able to talk this way is so that the $K$-finite vectors at archimedean places be preserved, which would fail for a genuine group representation (at least if the archimedean groups were non-compact).

Second, "unitary" in these situations is used very casually, so, no, one should not presume that the action is on a Hilbert space. Indeed, the requisite completion(s) to a Hilbert space might wreck other assumptions about $K_v$-finiteness at both archimedean and finite places. That is, at finite places, the usual sense of "smoothness" would be ruined.

Nevertheless, sometimes there is reason to also be able to refer to the Hilbert-space completion of a pre-unitary repns of groups, as yet-another less-ambiguous isomorphism-class representative.

In any case, no, unitary automorphic repns need not be sub-representations of the space of automorphic forms. The easiest example is the Eisenstein series for $GL(2)$, with parameters making all the local representations unitary. Then these generate everywhere-locally-unitary (or pre-unitary, as you like) repns, but are not subrepresentations, only quotients. This is not a "pathology", any more than is the presence of continuous spectrum in the first place.

It is true that one can try to squirm out of this situation by trying to make "Hilbert integrals"'s integrands behave as well as do genuine sub-objections, but it really doesn't work out.

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  • $\begingroup$ Thanks for your quick reply, Prof. Garrett. Indeed, this is apparently related to my other question I asked before. Actually, here is another related question. Let $(\pi, V)$ be a (not necessarily irreducible) unitary automorphic representation in the sense that it has a non-degenerate G-invariant Hermitian structure but not assumed to be complete. Is it true that for a subspace $W\subseteq V$, we have $V=W\oplus W^\perp$? It seems your comment to my previous question implies this is true. $\endgroup$
    – Windi
    Nov 23, 2013 at 21:10
  • $\begingroup$ About orthogonal complements: if we further burden "unitary automorphic" with meaning "admissible, whether or not irreducible", then, yes, unlike general discussion of orthogonal complements, completeness is unnecessary, perhaps counter-intuitively. But it's about taking advantage of "admissibility", which algebraicizes the question, making it not so much about Hilbert spaces. But, again, there is some danger in making admissibility an only-implicit assumption about "automorphic repn". No guarantee that everyone agrees. (Irreducibles are admissible, by thm of Harish-Chandra, Bernstein.) $\endgroup$ Nov 23, 2013 at 21:46
  • $\begingroup$ Thanks for your quick reply, again. Could you let me know if there is any literature that proves that unitary + admissible implies complete reducibility? $\endgroup$
    – Windi
    Nov 23, 2013 at 21:54
  • $\begingroup$ Very likely you can prove as an exercise that an admissible $K$-space with inner product, in which $K$ acts unitarily, for compact topological group $K$, with every vector $K$-finite, has the complement property. Do you care about anything more general? $\endgroup$ Nov 23, 2013 at 21:59
  • $\begingroup$ Thanks, again and again. I greatly appreciate it. So does that mean that for a representation of an adele group to have the complement property, one needs the $K$-finiteness assumption only for a particular $K$, even like a compact $K_v=G(\mathcal{O}_v)$ at one finite $v$? $\endgroup$
    – Windi
    Nov 23, 2013 at 22:28

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