Why is the cuspidal spectrum discrete? 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. 
 A: 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.
