4
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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
    Nov 9, 2010 at 8:58

3 Answers 3

3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ The book was kind of unreadable. Anything easier on a noobie? $\endgroup$ 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? $\endgroup$ 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
    Oct 11, 2009 at 22:41
  • $\begingroup$ You might want to see Serge Lang's text: Sl(2,R). $\endgroup$ Nov 9, 2010 at 6:23
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.