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Let $G = \operatorname{GL}_2$, $f \in C_c^{\infty}(G(\mathbb A)/Z(\mathbb A))$, and $V = L^2( G(\mathbb Q)Z(\mathbb A)\backslash G(\mathbb A))$ (trivial central character). Then the operator $R(f)$ on $V$ is integral with kernel $K(x,y) = \sum\limits_{\gamma \in Z(G) \backslash G(F)} f(x^{-1}\gamma y)$. I'm trying to understand this kernel as it appears on $V_{\operatorname{cont}} := V_{\operatorname{cusp}}^{\perp}$, in particular as an integral over Eisenstein series.

Let $\mu$ be a character of $\mathbb A^{\ast 1}/\mathbb Q^{\ast}$, extended to a character of $\mathbb A_k^{\ast}$ by making it trivial on the archimedean connected component, and then to a character of the maximal torus $T(\mathbb A)/Z(\mathbb A)$ by $\mu \begin{pmatrix} a_1 & \\ &a_2 \end{pmatrix} = \mu(a_1/a_2)$. Let

$$I(\mu,s) = \operatorname{Ind}_{B(\mathbb A)}^{G(\mathbb A)} \mu e^{\langle s \alpha , H_B(-) \rangle}$$

(normalized Mackey induction), and for $\phi \in I(\mu,0)$, set $\phi_s(x) = \phi(x) e^{\langle s\alpha, H_B(-) \rangle} \in I(\mu,s)$. To this we can associate the Eisenstein series

$$E(x,\phi,s) = \sum\limits_{\gamma \in B(\mathbb Q) \backslash G(\mathbb Q)} \phi_s(\gamma x)$$

which for fixed $\phi$ and $s$, lies in $L^2(G(\mathbb Q) Z(\mathbb A) \backslash G(\mathbb A))_{\operatorname{cont}}$ as a function of $x$. The series converges for $\operatorname{Re}(s) > 1$ but admits a meromorphic continuation. Then as I understand it,

$$x \mapsto \int_{-\infty}^{\infty} E(x,\phi, it)dt \tag{$\ast$}$$

also lies in $V_{\operatorname{cont}}$, and as $\phi$ runs through an orthonormal basis of $I(\mu,0)$, these integrals $\ast$ run through an orthonormal basis of direct summand $V_{\mu}$ of $V$. The kernel $K(x,y)$ is supposed to decompose as a sum $\sum\limits_{\chi} K_{\chi}(x,y)$ over the cuspidal automorphic data $\chi$ of $V$. For $\chi = \mu$, I want to understand why

$$K_{\chi}(x,y) = \sum\limits_{\phi} \int_{-\infty}^{\infty} E(x, I(\mu,it)f(\phi), it) \overline{E(y,\phi,it)} dt$$

where the sum $\phi$ is over an orthonormal basis of $I(\mu,0)$. This is stated, but not proved, in equation (1.2), pg. 18 of Stephan Gelbart's book Lectures on the Arthur-Selberg trace formula. James Arthur's notes An Introduction to the Trace Formula gives a hint for this on pg. 37 by saying this is related to the Fourier inversion formula

$$f(-x+y) = \frac{1}{2\pi i} \int_{i\mathbb R}\int\limits_{\mathbb R} f(u) e^{\lambda u}e^{\lambda x} \overline{e^{\lambda y}} d\lambda du$$

but I don't understand the connection.

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  • $\begingroup$ Did you try Gelbart-Jacquet's Corvallis article? Or Knapp's notes from a later PSYMP? $\endgroup$
    – Kimball
    Feb 13, 2021 at 13:07
  • $\begingroup$ I had considered about that gap, and as far as I remember, it was resolved by Lemma 2.7 in the Arthur's "The Selberg trace formula for groups of F-rank one". $\endgroup$
    – Aut
    Mar 6, 2021 at 14:48
  • $\begingroup$ I consulted the lemma, but I don't see what it has to do with what I'm asking. It never mentions an orthonormal basis $\endgroup$
    – D_S
    Mar 7, 2021 at 0:07
  • $\begingroup$ Sorry for my having misunderstood about your question. In general principle, to compute kernel of some operators on hilbert spaces, we need to take a orthogonal basis of the space. In this case, the spectral decomposition gives the basis, and that's why normalized induction appears. The reference is, Arthur's p35 of "An Introduction..", or §3 of "A trace formula for reductive groups I" or Moeglin-Waldspurger VI 2.1.The direct computation is given in 2.6 of Konno Takuya's "A Survey on the Arthur-Selberg Trace Formula" (but "the Fourier inversion" step uses the argument like the "Lemma 2.7") $\endgroup$
    – Aut
    Mar 7, 2021 at 7:18

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