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The decomposition of $L^{2}\left(S^{2}\right)$ under $SO\left(3,\mathbb{R}\right)$ is well-known.

Focus now on the hyperbolic plane $H$ presented as the quotient $SL\left(2,\mathbb{R}\right)/SO\left(2,\mathbb{R}\right)$. It is non-compact, therefore my understanding is that infinite-dimensional representations of $SL\left(2,\mathbb{R}\right)$ will appear in the decomposition of $L^{2}\left(H\right)$.

(a) Is there an algebraic part of the spectrum and does it have a description similar to the one in $L^{2}\left(S^{2}\right)$?

(b) How to classify the $SL\left(2,\mathbb{R}\right)$ representations and what is the whole spectrum?

(c) Consider $X_{0}\left(1\right):=SL\left(2,\mathbb{Z}\right)\setminus H$. How does $L^{2}\left(X_{0}\left(1\right)\right)$ decompose?

(d) The same for $X_{0}\left(N\right):=\Gamma_{0}\left(N\right)/H$. How does $L^{2}\left(X_{0}\left(N\right)\right)$ decompose?

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  • $\begingroup$ Richard Montgomery suggested below that you look at Lang's book $SL_2(\mathbb{R})$, and I'm reproducing the comment so you get notified. $\endgroup$
    – S. Carnahan
    Commented Nov 9, 2010 at 8:58

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a) Weyl's unitary trick implies there are no nontrivial irreducible finite dimensional unitary representations of $SL\left(2,\mathbb{R}\right)$. This is basically the opposite of $SO\left(3\right)$.

b) Wikipedia has a classification of all unitary irreps. An irreducible representation given as a space of functions on H can be viewed as a massive particle state in relativistic QM on $R^{\left(1,2\right)}$.

c) I think you get real-analytic Eisenstein series and discrete series. Eisenstein series form a continuous spectrum, while discrete series give modular forms. You can find more in Gelbart's book "Automorphic forms on adele groups"

d) Same thing, except the Eisenstein series involve a summation over a smaller range of cosets of translation, and the modular forms are invariant under a smaller group. I am told that the Maass forms and holomorphic forms for congruence groups that I mentioned only give a countable collection of unitary representations, while the principal series has a continuous parameter.

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  • $\begingroup$ The book was kind of unreadable. Anything easier on a noobie? $\endgroup$ Commented Oct 11, 2009 at 22:15
  • $\begingroup$ Scott, do you think I should post a similar question about \X_0(N) modular curve here or in separate post? $\endgroup$ Commented Oct 11, 2009 at 22:35
  • $\begingroup$ If it can be a reasonable part d, you might as well make it part d. $\endgroup$
    – S. Carnahan
    Commented Oct 11, 2009 at 22:41
  • $\begingroup$ You might want to see Serge Lang's text: Sl(2,R). $\endgroup$ Commented Nov 9, 2010 at 6:23
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DH's answer is not quite correct: The complementary series do not "appear in" $L^{2}\left(H\right)$, i.e., they do not appear in the support of a Plancherel measure.

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The infinite-dimensional unitary representations of $SL_{2}\left(\mathbb{R}\right)$ appearing in the right-regular representation on $L^{2}\left(H\right)$ are precisely the unitary representations of $SL_{2}\left(\mathbb{R}\right)$ possessing a $SO_{2}\left(\mathbb{R}\right)$-fixed vector. These are parametrized by $\mathbb{R}\cup\left[0,1\right]$, where $\mathbb{R}$ parametrizes unitary principal series representations and $\left[0,1\right]$ parametrizes the "complementary series" representations. This is implicit in Knapp's chapter in the Corvallis volume; see also Iwaniec's book on the spectral theory of automorphic forms for a classical treatment of this case.

Anyway, the point of this is that $L^{2}\left(H\right)$ has a "direct integral" decomposition into irreducible representations, so the proper analogy in this situation is not $L^{2}\left(S^{2}\right)$ but rather $L^{2}\left(\mathbb{R}\right)$. By contrast, the cofinite quotients $X_{0}\left(N\right)$ have a "mixed" spectral decomposition, that is $L^{2}\left(X_{0}\left(N\right)\right)$ breaks into a continuous part (Eisenstein series, parametrized by $\mathbb{R}$) and a discrete part, the so-called cusp forms. This theory is due to Selberg and is by no means straight-forward. Again, see Iwaniec's book for a nice classical treatment.

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