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The decomposition of L^2(S^2) under SO(3; R) is well-known.

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

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

(b) How to classify the SL(2; R) representations and what is the whole spectrum?

(c) Consider X_0(1) := SL(2; Z)\H. How does L^2(X_0(1)) decompose?

(d) The same for X_0(N) := \Gamma_0(N)/H. How does L^2(X_0(N)) decompose?

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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. –  S. Carnahan Nov 9 '10 at 8:58
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3 Answers

up vote 2 down vote accepted

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

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^(1,2).

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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The book was kind of unreadable. Anything easier on a noobie? –  Ilya Nikokoshev Oct 11 '09 at 22:15
    
Scott, do you think I should post a similar question about \X_0(N) modular curve here or in separate post? –  Ilya Nikokoshev Oct 11 '09 at 22:35
    
If it can be a reasonable part d, you might as well make it part d. –  S. Carnahan Oct 11 '09 at 22:41
    
You might want to see Serge Lang's text: Sl(2,R). –  Richard Montgomery Nov 9 '10 at 6:23
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The infinite-dimensional unitary representations of SL2(R) appearing in the right-regular representation on L2(H) are precisely the unitary representations of SL2(R) possessing a SO2(R)-fixed vector. These are parametrized by R \union [0,1], where R parametrizes unitary principal series representations and [0,1] parametrizes the "complimentary 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 L2(H) has a "direct integral" decomposition into irreducible representations, so the proper analogy in this situation is not L2(S^2) but rather L2(R). By contrast, the cofinite quotients X0(N) have a "mixed" spectral decomposition, that is L2(X0(N)) breaks into a continuous part (Eisenstein series, parametrized by 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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DH's answer is not quite correct: The complementary series do not "appear in" L^2(H), i.e., they do not appear in the support of a Plancherel measure.

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