I'm trying to work out explicitly the spectral decomposition of $L^2(G(\mathbb{Q})\backslash G(\mathbb{A}))$ when $G$ is anisotropic -- it has no split tori defined over $\mathbb{Q}$. I know that this should mean that $L^2$ decomposes discretely into a direct sum of irreducible unitary representations. I'm looking at Arthur's papers "Eisenstein series and the trace formula" as well as "The trace formula for reductive groups I," (Duke 1978) and I'm having trouble seeing how at the end of the day, the decomposition one gets should end up being discrete, into a direct sum of cuspidal representations. I know that here $G=P=M$ and that $A$, the split component of the center of $M$, is trivial. How should the expansion simplify in this case?

In more detail, let's assume for simplicity that $G$ is adjoint (so we won't have to worry about the center) as well as anisotropic over $\mathbb{Q}$. Arthur takes $K$ to be a certain nice compact subgroup of $G(\mathbb{A})$, with the Haar measure of $K$ taken to be one. A Haar measure on $G(\mathbb{A})$ is also fixed.

Then it is known that $L^2(G(\mathbb{Q}) \backslash G(\mathbb{A}))=L^2_{cusp}(G(\mathbb{Q}) \backslash G(\mathbb{A}))=\bigoplus_\rho V_\rho$, where $\rho$ ranges over all irreducible unitary representations of $G(\mathbb{A})$, and each $V_\rho$ is a $G(\mathbb{A})$-invariant subspace of $L^2(G(\mathbb{Q}) \backslash G(\mathbb{A}))$ isomorphic under the action of $G(\mathbb{A})$ to a finite number of copies of $\rho$. (If $V_\rho \neq 0$ we call $\rho$ cuspidal.)

Arthur then goes further on page 925 of the paper. Let $\psi$ be a smooth function on $G(\mathbb{Q}) \backslash G(\mathbb{A})$ such that $\psi$ transforms under $K_\mathbb{R}$ according to an irreducible representation $W$. Then $\psi \in L^2(G(\mathbb{Q}) \backslash G(\mathbb{A}))$. Define $L^2(G(\mathbb{Q}) \backslash G(\mathbb{A}))_\rho$ to be the closed span of all functions of the form $\psi$ where $W$ varies over all irreducible representations of $K_\mathbb{R}$. Then $L^2(G(\mathbb{Q}) \backslash G(\mathbb{A}))=\bigoplus_\rho L^2(G(\mathbb{Q}) \backslash G(\mathbb{A}))_\rho$.

My first question is: This doesn't seem to depend on $\rho$. How does this definition make sense? (Arthur's notation is $\chi \in (M,\rho)$, but here $M=G$ so I used $\rho$ instead of $\chi$.)

Let $\Pi(G)$ denote the set of equivalence classes of irreducible unitary representations of $G(\mathbb{A})$. This is a discrete set. For $\pi \in \Pi(G)$, Arthur then defines $H_G^0(\pi)$ to be the space of functions $\phi:G(\mathbb{Q}) \backslash G(\mathbb{A}) \rightarrow \mathbb{C}$ which satisfy certain properties. Its completion, $H_G(\pi)$, is a Hilbert space. He then defines for each irreducible unitary representation $\rho$ of $G(\mathbb{A})$ a certain closed subspace $H_G(\pi)_\rho$ of $H_G(\pi)$ and says that $H_G(\pi)$ is the direct sum of the $H_G(\pi)_\rho$ -- he also says that he doesn't know if $H_G(\pi)_\rho$ can be nonzero for more than one $\rho$ - presumably this means for $\rho \neq \pi$!

Arthur then fixes an orthonormal basis of each $H_G(\pi)_\rho$ (thereby giving a basis for $H_G(\pi)$) and defines $\hat{L}_G$ to be the set of functions $F$ taking $\pi \in \Pi(G)$ to $H_G(\pi)$ such that $\Sigma_\pi ||F_\pi||^2 \ d\pi < \infty$. Then the map which sends $F$ to the function $x \mapsto \Sigma_\pi F_\pi(x)$, defined for $F$ is a dense subspace of $\hat{L}_G$, extends to a unitary map from $\hat{L}_G$ onto a closed $G(\mathbb{A})$-invariant subspace, whose completion is $L^2(G(\mathbb{Q}) \backslash G(\mathbb{A}))$.

Based on this, I think Arthur is saying that $L^2$ for $G$ anisotropic and adjoint is the completion of functions of the form $x \mapsto \Sigma_\pi \Sigma_\rho \Sigma_{\phi_{\pi,\rho,i} \in B_G(\pi)_\rho} \langle{F_\pi,\phi_{\pi,\rho,i} \rangle} \phi_{\pi,\rho,i}(x)$. Is that correct? I don't see how this is reconciled with the earlier description as a direct sum of cuspidal representations.