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I recently learned about automorphic spectral decomposition from the book "Spectral decomposition and Eisenstein series" by Moeglin and Waldspurger. (Let me call it M-W)

I have a question about the characterization of discrete spectra.

Let me explain the basic notation as in M-W.

Let $G$ be a connected reductive group over algebraic field $k$ and $\xi$ be a unitary character of $Z_G(A)$.

Let $L^2(G(k) \setminus G(A))_\xi$ be $L^2$-functions on $G(k)\setminus G(A)$ with central character $\xi$.

Then, $L^2(G(k) \setminus G(A))_\xi$ decomposes into the space generated by iterated residues of Eisenstein series and its complement, that is described by direct integrals of Eisenstein series.(M-W, IV 2.1)

Let me call the first space $L^2_d$.

(I think that $L^2_d$ is the closure of span of $L^2$ automorphic forms in $L^2(G(k) \setminus G(A))_\xi$.)

Let me call the semi-simple part i.e. Hilbert direct sum of topologically irreducible subrepresentations of $L^2(G(k) \setminus G(A))_\xi$, by a name $L^2_{ss}$.

Definition of discrete spectrum and continuous and basic properties

In the article above, it is called discrete spectrum.

My questions are

  1. Are $L^2_d$ and $L^2_{ss}$ the same?
  2. If so, how to prove it? Can we prove it by means of elementary functional analysis (e.g. the knowledge of the book "Functional Analysis" by Walter Rudin) like the proof of the theorem of Gelfand-Graev-Patetski-Shapiro i.e. like in cuspidal case?

I think that it is obvious that $L^2_d$ contains $L^2_{ss}$, but I wonder whether the converse is true. I would appreciate any clues to resolve this question. Thanks!

Edited: I added one more question and definition of $L^2_{ss}$ in line with the comments. Thanks for the comments!

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    $\begingroup$ What do you mean by $L^{2}_{ss}$? $\endgroup$
    – Asaf
    Commented Aug 19, 2020 at 17:31
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    $\begingroup$ @Asaf Thank you for your comment. I should have crarified the definition. By $L^2_{ss}$, I mean the Hilbert space direct sum of all the topologically irreducible subrepresentations of right regular representation of $G(A)$ on $L^2(G(k) \backslash G(A))_{ξ}$. $\endgroup$
    – user163485
    Commented Aug 19, 2020 at 19:33
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    $\begingroup$ You don't have such a direct sum in general, but only a direct integral decomposition, this already appears in simpler cases i.e. $L^{2}(\mathbb{R})$. $\endgroup$
    – Asaf
    Commented Aug 19, 2020 at 19:43
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    $\begingroup$ @Asaf As in "Automorphic forms, representations, and L-functions", p196-197, I think that by the theorem of Gelfand-Piatetski-Shapiro, cuspidal representation decomposes into the topologically irreducible subrepresentations in the $L^2(G(k) \backslash G(A))_{ξ}$, so at least these representations appear in $L^2_{ss}$. In your $L^2(R)$ case, I think that $L^2(R)_{ss}$ =0 (because any nonzero closed sub-representation has infinite length) and $L^2(R)_{ss}$ exists. Please tell me if there is anything wrong or anything I may have misunderstood. Thanks! $\endgroup$
    – user163485
    Commented Aug 19, 2020 at 20:11
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    $\begingroup$ (By the way, "compliment" should be "complement".) It seems to me that your clarified definition of $L^2_{ss}$ is exactly your $L^2_d$, if you have in hand the theorem of the decomposition of $L^2$. Namely, if all continuous parts are excluded, what is left is cuspforms together with some $L^2$ iterated residues of Eisenstein series. But your question seems to grant that at the beginning. Or is the truth of that assertion your question? $\endgroup$
    – paul garrett
    Commented Aug 19, 2020 at 20:44

1 Answer 1

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It is true by Harish-Chandra's admissibility theorem.(c.f. 1.7 of the article by Borel-Jacquet in the Corvallis book.

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