6
$\begingroup$

I have a short question concerning the spectral theory of automorphic forms. What is the main property of the unipotent group $N$, which consist of matrices in the form \begin{pmatrix} 1 & t \\ 0 &1 \end{pmatrix} in $GL(2)$, which provides that the cuspidal spectrum decomposes discretely?

The background: Consider $G = GL_2(\mathbb{A})$, $\Gamma =GL_2(\mathbb{Q})$ and $Z$ the centrum of $G$, then we decompose $$Ind_{ \Gamma Z}^G 1 = \pi \oplus \pi^\bot,$$ where $\pi = Ind_{N\Gamma Z}^G$. The projection onto $\pi$ is given in terms of the integral $$ P : \phi(g) \mapsto \int\limits_{N(\mathbb{A})} \phi(ng) d g.$$ Now given a bounded function $f$ on $\Gamma \backslash G / \Gamma$, with suitable decay properties, we can define the operator

$$Tf : \phi \mapsto f * \phi.$$

Why is $(1-P)Tf(1-P)$ a Hilbert Schmidt operator?

My guess is that an answer should include the Iwasawa decomposition and an according decomposition of the integral operator.

Improve this question
$\endgroup$
3
  • 3
    $\begingroup$ I'll offer a vague answer to what seems to me to be a vague question. N represents a loop around a cusp. If you're perpendicular to $\pi$ then you satisfy some boundedness criterion at the cusps, and adding the cusps to a modular curve makes it compact, so now you're looking at functions on a compact space and then spectral theory works very well and you get discrete decompositions. Is this any help or are you looking for something totally different? $\endgroup$
    – Kevin Buzzard
    Commented Apr 10, 2011 at 13:31
  • $\begingroup$ So is it true, that Maass forms live on the compactified surface? I will try to make the question more specific. $\endgroup$
    – Marc Palm
    Commented Apr 10, 2011 at 14:52
  • 1
    $\begingroup$ @pm: Maass forms don't live on the compactified surface. $\endgroup$
    – GH from MO
    Commented Apr 10, 2011 at 18:27

1 Answer 1

Reset to default
6
$\begingroup$

This result is due to Gelfand, Graev, Piatetski-Shapiro and has a short proof. I suggest you read Bump: Automorphic forms and representations, Prop. 3.2.3, pp. 285-289.

Let me switch to $G=\mathrm{PGL}_2(\mathbb{R})$ and $\Gamma=\mathrm{PGL}_2(\mathbb{Z})$ for simplicity. The idea is to consider right convolutions $\rho(\phi)$ by smooth and compactly supported functions $\phi:G\to\mathbb{C}$, which act on the space of automorphic functions $L^2(\Gamma\backslash G)$. This family of operators is sufficiently rich to distinguish automorphic functions: if $f\neq g$ then there is $\phi$ such that $\rho(\phi)f\neq\rho(\phi)g$. Thinking of $\rho(\phi)$ as an operator on $(\Gamma\cap N)\backslash G$, its kernel is given by

$$K(g,h)=\sum_{\gamma\in\Gamma\cap N}\phi(g^{-1}\gamma h).$$

As $N\cong\mathbb{R}$ and $\Gamma\cap N\cong\mathbb{Z}$, the sum can be analyzed by Poisson summation, using the characters of $N$. It turns out that by subtracting the term corresponding to the trivial character, i.e.

$$K_0(g,h)=\int_{N}\phi(g^{-1} n h) \ dn,$$

the rest of the kernel decays rapidly at infinity, hence defines a compact operator. However, on the cuspidal space the operator with kernel $K_0(g,h)$ acts by zero, hence the operator $\rho(\phi)$ is compact on the cuspidal space. In particular, $\rho(\phi)$ has an eigenvalue with finite multiplicity on any right $G$-invariant subspace of cuspidal functions, and from here the result follows by a standard linear algebra argument.

EDIT: I fixed some inaccuracies.

Improve this answer
$\endgroup$
3
  • $\begingroup$ I accept this as the answer, because it answers the question as stated above. But I was hoping something more conceptual. If you'd no glur how to achieve the spectral decomposition for $GL_2$, how do you come with integrating over $N$? Why is it reasonable for $GL_n$ to guess that all the mess comes from the induction from a maximal unipotent subgroup, i.e. from the smaller $GL_m, m<n$? $\endgroup$
    – Marc Palm
    Commented Apr 11, 2011 at 6:06
  • 2
    $\begingroup$ I think it is not true that all the mess comes from a maximal unipotent subgroup. One needs to consider all parabolic subgroups containing a fixed minimal parabolic subgroup, define a correction kernel for each of them, and modify the original kernel with the sum of these. Then one needs to show that the modified operator is compact which boils down to a detailed understanding of the group and its parabolic subgroups. Admittedly, I am no expert in the general theory, but I don't know about a more conceptual proof. $\endgroup$
    – GH from MO
    Commented Apr 11, 2011 at 8:36
  • 2
    $\begingroup$ The argument for discreteness of cuspforms in Lax-Phillips' "Scattering theory for automorphic forms", pp 204-6, is essentially independent of the rest of the book, and for $SL_2$ may give a more intuitive argument than the integral operators argument, athough not clearly suggesting the general case. $\endgroup$
    – paul garrett
    Commented Jul 16, 2011 at 16:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .