Calculating the residue of Eisenstein series from the residue of the intertwining operator

Question

I've been reading the article Forms of $\operatorname{GL}(2)$ from the analytic point of view by Gelbart and Jacquet in Corvallis and am confused on a particular claim (equation 5.17 on page 232).

The claim essentially boils down to a assertion about the residue of an integral of an Eisenstein series against a pseudo-Eisenstein series (called "P-series" in the article):

$$\operatorname{Res}_{s = 1/2} \langle E(\hat{f_1}(s),g), F_2(g) \rangle = c\sum\limits_{\chi^2 = \omega} \langle F_1, \chi \circ \det \rangle \overline{\langle F_2, \chi \circ \det \rangle}.\tag{$\ast$}$$

Here the pairing $\langle-,-\rangle$ is the inner product in $L^2(G(\mathbb Q)Z(\mathbb A) \backslash G(\mathbb A, \omega)$. Another residue calculation is sketched in the article earlier, namely

$$\operatorname{Res}_{s=1/2}( M(s)\hat{f}_1(s), \hat{f}_2(\overline{s})) = c\sum\limits_{\chi^2 = \omega} \langle F_1, \chi \circ \det \rangle \overline{\langle F_2, \chi \circ \det \rangle} \tag{$\ast \ast$}$$ (Theorem 4.19 and equation 4.22 on page 227-228), where $M(s)$ is an intertwining operator. So equation $(\ast)$ actually asserts that these residues are equal. I haven't been able to see why this is, or why the residue calculation with the Eisenstein series in ($\ast$) follows from that of the intertwining operator in ($\ast \ast$). I wonder if there is a nontrivial calculation that needs to be done which the article does not mention.

Notation:

Here $c$ is a known constant, $F_i(g) = \sum\limits_{\gamma \in P(\mathbb Q) \backslash G(\mathbb A)} f_i(\gamma g)$, $E(\hat{f}_i(s),g)) = \sum\limits_{\gamma \in P(\mathbb Q) \backslash G(\mathbb Q)} \hat{f}_i(g,s)$ for $\operatorname{Re}(s) > 1/2$ (but has a meromorphic continuation), $f_i$ is compactly supported mod $N(\mathbb A)P(\mathbb Q) Z(\mathbb A)$ and satisfies $f(n\gamma zg) = \omega(z)f(g)$, and

$$\hat{f}(g,s) = \int_{0}^{\infty} f(\begin{pmatrix} t \\ & 1 \end{pmatrix}g)|t|^{-s-1/2}d^{\ast}t$$ is the Mellin transform lying in the space $\mathbf H(s)$ induced to $G(\mathbb A)$ from the character $n \gamma z\begin{pmatrix} u \\ & v \end{pmatrix} \mapsto \omega(z)|u/v|^s$ of $N(\mathbb A)Z(\mathbb A)T(\mathbb Q) T(\mathbb R)^0$. There is a natural pairing on $\mathbf H(s) \times \mathbf H(-\overline{s})$. The intertwining operator $M(s): \mathbf H(s) \rightarrow \mathbf H(-s)$ is defined for $\operatorname{Re}(s) > 1/2$ by

$$M(s)\phi(g) = \int\limits_{N(\mathbb A)} f(wng) \space dn$$

but has a meromorphic continuation.

Answer 1

I seem to have solved it, modulo some convergence issues I'm not yet confident about. We have

$$\langle E(\hat{f}_1(s),g), F_2(g) \rangle = \int\limits_{G(\mathbb Q)Z(\mathbb A) \backslash G(\mathbb A)} E(\hat{f}_1(s),g)\overline{F_2(g)}dg$$ where a well known calculation expresses this as as an integral over the constant term of $E(\hat{f}_1(s),g)$ paired with $f_2$:

$$\int\limits \hat{f}_1(g,s) \overline{f_2(g)} dg + \int\limits M(s)\hat{f}_1(g,s) \overline{f_2(g)} dg$$ where both integrals are over $P(\mathbb Q) N(\mathbb A)Z(\mathbb A) \backslash G(\mathbb A)$. As far as residues go, we only have to worry about the second term, which can also be expressed as (analogous calculation on page 222)

$$\int\limits_K \int\limits_{\mathbb A^{\ast 1}/\mathbb Q} \int_0^{\infty} M(s)\hat{f}_1\Bigg[\begin{pmatrix} ta & \\& 1 \end{pmatrix}k,s\Bigg] \overline{\Bigg[f_2(\begin{pmatrix} ta & \\ 1 \end{pmatrix}k\Bigg]} |t|^{-1}d^{\ast}t d^{\ast}a dk$$

Now since $M(s)\hat{f}_1(s) \in \mathbf H(-s)$, the inner integral is

$$\int_0^{\infty} |t|^{-s+1/2}M(s)\hat{f}_1\Bigg[ \begin{pmatrix} a & \\& 1 \end{pmatrix}k,s\Bigg] \overline{\Bigg[f_2(\begin{pmatrix} ta & \\& 1 \end{pmatrix}k\Bigg]}|t|^{-1}d^{\ast}t$$

$$= M(s)\hat{f}_1\Bigg[ \begin{pmatrix} a & \\& 1 \end{pmatrix}k,s\Bigg] \int_0^{\infty}\overline{\Bigg[f_2(\begin{pmatrix} ta & \\ &1 \end{pmatrix}k\Bigg]} |t|^{-s-1/2} d^{\ast}t$$

$$ = M(s)\hat{f}_1\Bigg[ \begin{pmatrix} a & \\& 1 \end{pmatrix}k,s\Bigg] \overline{\hat{f}_2\Bigg[ \begin{pmatrix} a & \\& 1 \end{pmatrix}k, \overline{s}\Bigg]}.$$

Therefore,

$$\int\limits M(s)\hat{f}_1(g,s) \overline{f_2(g)} dg = \int\limits_K \int\limits_{\mathbb A^{\ast 1}/\mathbb Q}M(s)\hat{f}_1\Bigg[ \begin{pmatrix} a & \\& 1 \end{pmatrix}k,s\Bigg] \overline{\hat{f}_2\Bigg[ \begin{pmatrix} a & \\& 1 \end{pmatrix}k, \overline{s}\Bigg]} d^{\ast}adk$$

$$ = (M(s)\hat{f}_1(s), \hat{f}_2(\overline{s})).$$

So, the point is that the difference between $\langle E(\hat{f_1}(s),g), F_2(g) \rangle$ and $(M(s)\hat{f}_1(s), \hat{f}_2(\overline{s}))$ is holomorphic at $s=1/2$, so they have the same residue.