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

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.

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.