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Let $G$ be a classical group defined over a number field $F$.

In his monumental book, (https://www.ams.org/books/coll/061/coll061-endmatter.pdf) Arthur described a spectral decomposition of $L_{disc}^2(G(F) \backslash G(\mathbb{A}))$, the discrete spectrum of $L^2(G)$, when $G$ is symplectic or special orthogonal groups.

Write $A(G)$, the space of automorphic forms on $G(F)$.

In some paper, it is written that since $A_2(G):=L_{disc}^2(G(F) \backslash G(\mathbb{A})) \cap A(G)$ is dense in $L_{disc}^2(G(F) \backslash G(\mathbb{A}))$, the study of spectral decomposition of $L_{disc}^2(G(F) \backslash G(\mathbb{A}))$ is equivalent to that of $A_2(G)$.

Does it mean that if $L_{disc}^2(G(F) \backslash G(\mathbb{A}))=\oplus m_{\widetilde{\pi}} \widetilde{\pi}$, then $A_2(G)=\oplus m_{\widetilde{\pi}} \pi$, where $\pi=\widetilde{\pi} \cap A_2(G)$ and vise versa?

If it is, I don't know why the denseness of $A_2(G)$ in $L_{disc}^2(G(F) \backslash G(\mathbb{A}))$ implies this.

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    $\begingroup$ Depends on the definition of automorphic form. If you mean K-finite and finite under the center of the universal enveloping algebra, then, well you need to make sense of the decomposition of $A_2$ first, as $G$ does not act. However, the Lie-algebra does and you consider the lot as $({\mathfrak g},K)$-module. In this sense, such a decomposition holds, and indeed, every summand lies dense in a unique summand on the other side. This follows by applying the decompostion with respect to the compact group, using admissibility and the fact that an irreducible has a central character. $\endgroup$
    – user473423
    Commented Jan 30, 2023 at 8:38

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