I'm reading James Arthur's notes on the trace formula and am confused on a point on pages 65 and 66. For $G$ a reductive group over $\mathbb Q$ we are going over the decomposition of the space $L^2(G(\mathbb Q) \backslash G(\mathbb A))$. First, I'm a bit surprised we are not going modulo center or modulo $A_G(\mathbb R)^{\circ}$ and wonder if Arthur means this space to be left $G(\mathbb Q$)-invariant functions which are square integrable on $G(\mathbb Q) A_G(\mathbb R)^{\circ} \backslash G(\mathbb A)$, rather than on $G(\mathbb Q) \backslash G(\mathbb A)$. If so, then $L^2(G(\mathbb Q) \backslash G(\mathbb A)^1)$ would then be isomorphic to the subspace of functions in $L^2(G(\mathbb Q) \backslash G(\mathbb A))$ which are moreover $A_G(\mathbb R)^{\circ}$-invariant.

Anyway, whatever is meant by $L^2(G(\mathbb Q) \backslash G(\mathbb A))$, Arthur says it decomposes as a direct sum of subspaces $L^2_{\chi}(G(\mathbb Q) \backslash G(\mathbb A))$, as $\chi$ runs through the cuspidal automorphic data. Here $\chi$ is an equivalence class of pairs $(P,\sigma)$, where $P = MN$ is a standard parabolic subgroup of $G$, and $\sigma$ is a cuspidal automorphic representation of $M(\mathbb A)^1$. Another pairs $(P',\sigma')$ is equivalent to $(P,\sigma)$ if there is a $w$ in the Weyl group such that $wMw^{-1} = M'$ and $w(\sigma) \cong \sigma'$ as representations of $M'(\mathbb A)^1$.

The space $L^2_{\chi}(G(\mathbb Q) \backslash G(\mathbb A))$ is the closed subspace of $L^2(G(\mathbb Q) \backslash G(\mathbb A))$ generated by pseudo-Eisenstein series

$$E \psi(g) = \sum\limits_{\delta \in P(\mathbb Q) \backslash G(\mathbb Q)}\psi(\delta g)$$ where $\psi: N(\mathbb A)M(\mathbb Q)\backslash G(\mathbb A) \rightarrow \mathbb C$ is a function obtained by Mellin transform

$$\psi(x) = \int\limits_{\mathfrak a_P^{\ast}} e^{\langle \lambda + \rho_P, H_P(x) \rangle} \Psi(\lambda ,x) \space d \lambda$$

of a Paley-Weiner section of an induced space $\mathcal H_{P,\sigma}$ (definition on page 65). So, from

$$L^2(G(\mathbb Q) \backslash G(\mathbb A)) = \bigoplus\limits_{\chi} L^2_{\chi}(G(\mathbb Q) \backslash G(\mathbb A))$$

Arthur says we should use the obvious analogous decomposition

$$L^2(G(\mathbb Q) \backslash G(\mathbb A)^1) = \bigoplus\limits_{\chi} L^2_{\chi}(G(\mathbb Q) \backslash G(\mathbb A)^1).$$

My questions are:

1 . What is meant by $L^2_{\chi}(G(\mathbb Q) \backslash G(\mathbb A)^1)$? Does this mean the closed subspace of $L^2(G(\mathbb Q) \backslash G(\mathbb A)^1)$ generated by Pseudo-Eisenstein series $E\psi(g)|G(\mathbb A)^1$ which are restricted to $G(\mathbb A)^1$?

2 . For a much more basic question, what is meant by $L^2(G(\mathbb Q) \backslash G(\mathbb A))$? Is it the plain meaning, of measurable $G(\mathbb Q)$-invariant functions on $G(\mathbb A)$ which are square-integrable on $G(\mathbb Q) \backslash G(\mathbb A)$? Or is it the space of measurable $G(\mathbb Q)$-invariant functions which are square-integrable on $G(\mathbb Q) A_G(\mathbb R)^{\circ} \backslash G(\mathbb A)$?


1 Answer 1


Arthur doesn't work mod centre or $A_G(\mathbb R)^0$ because he works with $G(\mathbb A)^1$ instead. For a nice explanation of the subtle difference between these choices, take a look at Section 6 of Knapp's Theoretical Aspects of the Trace Formula for GL(2). Essentially, it comes down to the nonuniqueness of extending characters from $Z(\mathbb A)^1 G(F)$ to $Z(\mathbb A)G(F)$, both trivial on $G(F)$. With this in mind, then the $L^2$-spaces under consideration are really as the notation suggests, and hopefully this should clarify both your questions.

  • $\begingroup$ So $L^2(G(\mathbb Q) \backslash G(\mathbb A))$ is really as the notation suggests, and consists of all measureable functions on $G(\mathbb A)$ which are $G(\mathbb Q)$-invariant and satisfy $\int\limits_{G(\mathbb Q) \backslash G(\mathbb A)} |f(g)|^2 dg < \infty$? This seems strange, since $G(\mathbb Q) \backslash G(\mathbb A)$ doesn't usually have finite volume. $\endgroup$
    – D_S
    Sep 8, 2022 at 21:37
  • $\begingroup$ I meant that one should always work with $L^2(G(\mathbb Q)\backslash G(\mathbb A)^1)$ or something like $L^2(Z(\mathbb A)G(\mathbb Q)\backslash G(\mathbb A))$ for the reasons you mentioned. $\endgroup$
    – Tian An
    Sep 9, 2022 at 16:04

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