Let $(\pi, V_\pi)$ and $(\pi^\prime, V_{\pi^\prime})$ be cuspidal, unitary, irreducible automorphic representations of $GL_n(\mathbb{A})$ and $GL_m(\mathbb{A})$ respectively. We assume $m < n$. Let $\varphi \in V_\pi$ and $\varphi \in V_{\pi^\prime}$. To pair $\varphi$ and $\varphi^\prime$ suitably together, we first need to project $\varphi$ correspondingly.

Let $\psi$ be a additive continuous automorphic character of $\mathbb{A}$. We can extend it to a character of $N_n(\mathbb{A})$, the standard Borel subgroup of $GL_n(\mathbb{A})$, in the standard way: $\psi(u) = \psi\left(\sum_{i=1}^{n-1}u_{i,i+1}\right),$ for $u = (u_{i,j}) \in N_n(\mathbb{A})$. Let $Y=Y_{n, m}$ be the standard unipotent radical associated to the partition $(m+1, 1, \dots, 1)$ of $GL_n$, i.e., $Y=\left\{\left(\begin{array}{cc}I_{m+1}&*\\ 0&u\end{array}\right): u \in N_{n-m-1}\right\} \subset N_n.$ Let $P_{m+1}$ be the mirabolic subgroup of $GL_{m+1}$, then for $p \in P_{m+1}(\mathbb{A})$ define $\mathbb{P}^n_m \varphi(p) = |\det p|^{-\frac{n-m-1}{2}}\int_{Y(k)\backslash Y(\mathbb{A})} \varphi\left(y\left(\begin{array}{cc} p& \\ & I_{n-m-1} \end{array}\right)\right)\psi^{-1}(y) dy.$

Now we can pair $\varphi$ and $\varphi^{\prime}$ in the following way: $I(s; \varphi, \varphi^\prime) = \int_{GL_m(k)\backslash GL_m(\mathbb{A})} \mathbb{P}^n_m \varphi\left(\begin{array}{cc} h& \\ &1\end{array}\right)\varphi^\prime(h)|\det h|^{s-\frac{1}{2}}dh.$ The above integral converges absolutely for $s \in \C$. For any matrix $x$, let $x^\iota$ be the transpose inverse of $x$. If we do the change of variables $h \mapsto h^\iota$, then we can get the global functional equation $$$I(s; \varphi, \varphi^\prime) = \widetilde{I}(1-s; \tilde{\varphi}, \tilde{\varphi^\prime}), \label{20150712-1}$$$

where $\tilde{\varphi}(g) = \varphi(g^\iota)$ and $\widetilde{I}(s; \varphi, \varphi^\prime) = \int_{GL_m(k)\backslash GL_m(\mathbb{A})} \widetilde{\mathbb{P}}^n_m \varphi\left(\begin{array}{cc} h& \\ &1\end{array}\right)\varphi^\prime(h)|\det h|^{s-\frac{1}{2}}dh,$ with $\tilde{\mathbb{P}}^n_m = \iota \circ \mathbb{P}^n_m \circ \iota$.

The unfolding of the left hand side of $\eqref{20150712-1}$ can be found in James Cogdell’s note on $L$-functions and Converse Theorem for $GL_n$, with additional details to what have been told above. This post will do the unfolding of the right hand side of $\eqref{20150712-1}$. I learned the unfolding from James Cogdell.

Notations: I will write $G_n$ for $GL_n$ and denote $G(k) \backslash G(\mathbb{A})$ as $\underline{G}$ for various $G$. Inside a matrix, I will use “$[\;]$” to group blocks together to form one single block. Let $M=M_{n-m-1,m}$ denote the group of all $(n-m-1) \times m$ matrices. Dimensions of zero vectors will be omitted. Also, let $\begin{equation*} w_n = \left(\begin{array}{ccc} &&1 \\ &\unicode{x22F0}& \\ 1&&\end{array}\right),\;\; w_{n, m} = \left(\begin{array}{cc} I_m& \\ &w_{n-m}\end{array}\right) \end{equation*}$ Let $\rho$ be the regular action. Let $W_{\varphi} \in \mathcal{W}(\pi, \psi^{-1})$ be the Fourier expansion of $\varphi$ with respect to $\psi^{-1}$, i.e., $W_\varphi(g) = \int_{\underline{N_n}} \varphi(ug)\psi(u)du.$ Similarly, one can define $W_{\varphi^\prime} \in \mathcal{W}(\pi^\prime, \psi)$.

Using the above notations, the main theorem of the post is the following unfolding.

Theorem 1.$\displaystyle \widetilde{I}(s; \varphi, \varphi^\prime) = \widetilde{\Psi}(s; \rho(w_{n, m})W_{\varphi}, W_{\varphi^\prime})$, where $\widetilde{\Psi}(s; W, W^\prime) = \int_{N_m(\mathbb{A})\backslash G_m(\mathbb{A})} \int_{M(\mathbb{A})}W\left(\begin{array}{cc} h & \\ \left[\begin{array}{cc}x\\0\end{array}\right]&I_{n-m}\end{array}\right) dx W^\prime(h) |\det h|^{s-\frac{n-m}{2}}dh.$

In order to obtain the above unfolding, we need to understand $\widetilde{\mathbb{P}}^n_m \varphi\left(\begin{array}{cc} h& \\ &1\end{array}\right)$ first. To this end, we have

Theorem 2.$\widetilde{\mathbb{P}}^n_m \varphi\left(\begin{array}{cc} h& \\ &1\end{array}\right) = |\det h|^{-\frac{n-m-1}{2}}\sum_{\gamma\in N_m(k)\backslash G_m(k)}\int_{M(\mathbb{A})}\rho(w_{n,m})W_\varphi\left(\begin{array}{cc} \gamma h & \\ \left[\begin{array}{cc}x\\0\end{array}\right]&I_{n-m}\end{array}\right) dx.$

Proof. By definition of $\widetilde{\mathbb{P}}^n_m$, $\widetilde{\mathbb{P}}^n_m \varphi\left(\begin{array}{cc} h& \\ &1\end{array}\right) = |\det h|^{\frac{n-m-1}{2}}\int_{\underline{Y}}\varphi\left(y^\iota \left(\begin{array}{cc} h& \\ &I_{n-m} \end{array}\right) \right)\psi^{-1}(y)dy.$ For $y \in \underline{Y}$, we have $y=\left(\begin{array}{cc} I_m&[0 \; x] \\ &y_1\end{array}\right),$ where $x \in {^t\underline{M}}$ and $y_1 \in \underline{N_{n-m}}$. Then $y^\iota=\left(\begin{array}{cc} I_m& \\ -y_1^\iota\left[\begin{array}{cc}0\\{^tx}\end{array}\right] &y_1^\iota \end{array}\right).$ If we replace the integral over $\underline{Y}$ by the double integrals over $\underline{N_{n-m}}$ and $\underline{^tM}$, then with the change of variables $x \mapsto -{y_1^\prime} \cdot {^tx}, \text{where } {^ty_1} = \left(\begin{array}{c} [1 \; 0] \\ y_1^\prime\end{array}\right),$ one can get $$$\begin{split} \widetilde{\mathbb{P}}^n_m \varphi & \left(\begin{array}{cc} h& \\ &1\end{array}\right) = |\det h|^{\frac{n-m-1}{2}} \cdot\\ &\int_{\underline{N_{n-m}}}\int_{\underline{M}} \varphi\left(\left(\begin{array}{cc} I_m& \\ \left[\begin{array}{cc}0\\x\end{array}\right]&y^\iota\end{array}\right)\left(\begin{array}{cc} h& \\ &I_{n-m}\end{array}\right)\right)\psi^{-1}(y)dxdy. \end{split} \label{20150712-2}$$$

Since $\varphi$ is automorphic, the argument inside $\varphi$ in $\eqref{20150712-2}$ can be replaced by $\begin{equation*} \begin{split} &w_{n,m}\left(\begin{array}{cc} I_m& \\ \left[\begin{array}{cc}0\\x\end{array}\right]&y^\iota\end{array}\right)\left(\begin{array}{cc} h& \\ &I_{n-m}\end{array}\right) \\ &= \left(\begin{array}{cc} I_m& \\ \left[\begin{array}{cc}w_{n-m-1}x\\0\end{array}\right]&w_{n-m}y^\iota w_{n-m}\end{array}\right)\left(\begin{array}{cc} h& \\ &I_{n-m}\end{array}\right)w_{n,m} \end{split} \end{equation*}$ Therefore, from $\eqref{20150712-2}$ together with the change of variables $x \mapsto w_{n-m-1}x, \;\; y \mapsto w_{n-m}y^\iota w_{n-m},$ and noticing that $\psi^{-1}(w_{n-m}y^\iota w_{n-m}) = \psi(y)$, it can be deduced that $\begin{equation*} \begin{split} \widetilde{\mathbb{P}}^n_m \varphi & \left(\begin{array}{cc} h& \\ &1\end{array}\right) = |\det h|^{\frac{n-m-1}{2}} \cdot\\ &\int_{\underline{N_{n-m}}}\int_{\underline{M}} \varphi\left(\left(\begin{array}{cc} I_m& \\ \left[\begin{array}{cc}x\\0\end{array}\right]&y\end{array}\right)\left(\begin{array}{cc} h& \\ &I_{n-m}\end{array}\right)w_{n,m}\right)\psi(y)dxdy. \end{split} \end{equation*}$ Replacing $\varphi$ by its Fourier expansion $\varphi(g) = \sum_{\alpha \in N_{n-1}(k) \backslash G_{n-1}(k)}W_{\varphi}\left(\left(\begin{array}{cc} \alpha& \\ &1\end{array}\right)g\right),$ and switching orders, we can get $\begin{equation*} \begin{split} \widetilde{\mathbb{P}}^n_m \varphi & \left(\begin{array}{cc} h& \\ &1\end{array}\right) = |\det h|^{\frac{n-m-1}{2}} \int_{\underline{N_{n-m}}}\sum_{\alpha \in N_{n-1}(k) \backslash G_{n-1}(k)} \int_{\underline{M}}\\ &W_{\varphi}\left(\left(\begin{array}{cc} \alpha& \\ &1\end{array}\right)\left(\begin{array}{cc} I_m& \\ \left[\begin{array}{cc}x\\0\end{array}\right]&y\end{array}\right)\left(\begin{array}{cc} h& \\ &I_{n-m}\end{array}\right)w_{n,m}\right)\psi(y)dxdy. \end{split} \end{equation*}$ Now, $\left(\begin{array}{cc} I_m& \\ \left[\begin{array}{cc}x\\0\end{array}\right]&y\end{array}\right) = \left(\begin{array}{cc} I_m& \\ &y\end{array}\right)\left(\begin{array}{cc} I_m& \\ y^{-1}\left[\begin{array}{cc}x\\0\end{array}\right]&I_{n-m}\end{array}\right),$ so if we do the change of variables $x \mapsto y^\prime x, \text{ where } y = \left(\begin{array}{cc} y^\prime & * \\ & 1 \end{array}\right),$ and change orders, we can see $$$\widetilde{\mathbb{P}}^n_m \varphi \left(\begin{array}{cc} h& \\ &1\end{array}\right) = |\det h|^{\frac{n-m-1}{2}} \int_{\underline{M}} \sum_{\alpha \in N_{n-1}(k) \backslash G_{n-1}(k)} A(\alpha; h, x)dx, \label{20150712-3}$$$

where $\begin{equation*} \begin{split} A(\alpha; h, x) &= \int_{\underline{N_{n-m}}} W_{\varphi}\left(\left(\begin{array}{cc} \alpha& \\ &1\end{array}\right)\left(\begin{array}{cc} I_m& \\ &y\end{array}\right) \right.\cdot \\ &\left.\left(\begin{array}{cc} I_m& \\ \left[\begin{array}{cc}x\\0\end{array}\right]&I_{n-m}\end{array}\right)\left(\begin{array}{cc} h& \\ &I_{n-m}\end{array}\right)w_{n,m}\right)\psi(y)dy. \end{split} \end{equation*}$ For $y \in \underline{N_{n-m}}$, let’s write $y=[y_1 \, \dots \, y_{n-m}],$ with each $y_i$ being column vector such that the $i$-th entry is $1$ and $j$-th entry is $0$ if $j > i$. With this notation, $\left(\begin{array}{cc} I_m& \\ &y\end{array}\right)=\left(\begin{array}{cc} I_m& \\ &[e_1 \, \dots \, e_{n-m-1} \, y_{n-m}]\end{array}\right) \cdots \left(\begin{array}{cc} I_m& \\ &[y_1 \, e_2 \, \dots \, e_{n-m}]\end{array}\right),$ where entries of the column vector $e_i$ are all zeros except being $1$ at $i$-th entry. Therefore, we can rewrite the integral over $y$ in $A(h; \alpha, x)$ into multiple integrals over $y_{n-m}, \dots, y_2$ (Note that $y_1$ is just constant $e_1$): $\begin{equation*} \begin{split} A(\alpha; &h, x) = \int_{\underline{N_{n-m}}} W_{\varphi}\left(\left(\begin{array}{cc} \alpha& \\ &1\end{array}\right)\left(\begin{array}{cc} I_m& \\ &[e_1 \, \dots \, e_{n-m-1} \, y_{n-m}]\end{array}\right) \cdots \right. \\ &\left(\begin{array}{cc} I_m& \\ &[y_1 \, y_2\, e_3 \, \dots \, e_{n-m}]\end{array}\right) \left. \left(\begin{array}{cc} I_m& \\ \left[\begin{array}{cc}x\\0\end{array}\right]&I_{n-m}\end{array}\right)\left(\begin{array}{cc} h& \\ &I_{n-m}\end{array}\right)w_{n,m}\right)\\ &\prod_{i=2}^{n-m}\psi(y_{i,i-1})dy_{n-m} \cdots dy_{2}. \end{split} \end{equation*}$ If we write $\alpha = (\alpha_{i, j})$ and $y_{n-m} = (y_{n-m, i}) = \left(\begin{array}{cc}y^\prime_{n-m} \\ 1\end{array}\right)$, then $\begin{equation*} \begin{split} & \left(\begin{array}{cc} \alpha& \\ &1\end{array}\right)\left(\begin{array}{cc} I_m& \\ &[e_1 \, \dots \, e_{n-m-1} \, y_{n-m}]\end{array}\right) \\ &= \left(\begin{array}{cc} I_{n-m-1}&\alpha \left[\begin{array}{cc}0\\y^\prime_{n-m}\end{array}\right] \\ &1\end{array}\right)\left(\begin{array}{cc} \alpha& \\ &1\end{array}\right). \end{split} \end{equation*}$ For convenience, let $g$ be the product of matrices in the arguments of $W_\varphi$ not relating to $\alpha$ and $y_{n-m}$. Then if we focus on the integral over $y_{n-m}$, and use the fact that $W_\varphi(ug) = \psi^{-1}(u)W_\varphi(g)$ for $u \in N_n(\mathbb{A})$, we can get $\begin{equation*} \begin{split} A(\alpha; & h, x) = \int \int W_\varphi\left(\left(\begin{array}{cc} \alpha& \\ &1\end{array}\right)g\right) \\ & \psi(-\alpha_{n-1,m+1}y_{n-m,1}-\cdots-(\alpha_{n-1, n-1}-1)y_{n-m,n-m-1}) dy_{n-m} \\ & \prod_{i=2}^{n-m-1}\psi(y_{i, i-1}) dy_{n-m-1} \dots dy_2 \end{split} \end{equation*}$ The inner integral is zero, expect when $\alpha_{n-1, m+1} = \cdots = \alpha_{n-1, n-2} = 0$ and $\alpha_{n-1, n-1} = 1$, in which case the inner integral equals $W_{\varphi}\left(\left(\begin{array}{cc} \alpha& \\ &1\end{array}\right)g\right).$ In this case, $\alpha$ can be written as $\left(\begin{array}{cc} *&* \\ {[* \;0]}&1 \end{array}\right) = \left(\begin{array}{cc} I_{n-2}&* \\ &1 \end{array}\right) \left(\begin{array}{cc} *&0 \\ {[* \;0]}&1 \end{array}\right).$ Since $\alpha \in N_{n-1}(k) \backslash G_{n-1}(k)$, we can choose $\alpha$ to be in the form $\left(\begin{array}{cc} *&0 \\ {[* \;0]}&1 \end{array}\right).$ Now repeating the above process again for $y_{n-m-1}, \dots, y_2$ in order, we can conclude that $A(\alpha; h, x) = 0$ unless $\alpha$ has the form $\alpha = \left(\begin{array}{cc} \gamma&0 \\ \beta&I_{n-m-1} \end{array}\right),$ where $\gamma \in N_m(k) \backslash G_m(k)$ and $\beta \in M(k)$. When $\alpha$ is in this form, $A(\alpha; h, x)$ equals $W_{\varphi}\left(\left(\begin{array}{cc} \gamma& \\ \left[\begin{array}{cc}\beta\\0\end{array}\right]&I_{n-m}\end{array}\right)\left(\begin{array}{cc} I_m& \\ \left[\begin{array}{cc}x\\0\end{array}\right]&I_{n-m}\end{array}\right)\left(\begin{array}{cc} h& \\ &I_{n-m}\end{array}\right)w_{n,m}\right),$ which is the same as $W_\varphi\left(\left(\begin{array}{cc} \gamma h & \\ \left[\begin{array}{cc}(\beta+x)\\0\end{array}\right]h&I_{n-m}\end{array}\right)w_{n,m}\right).$ Gathering the analysis of $A(\alpha; h, x)$ into $\eqref{20150712-3}$, $\begin{equation*} \begin{split} \widetilde{\mathbb{P}}^n_m \varphi & \left(\begin{array}{cc} h& \\ &1\end{array}\right) = |\det h|^{\frac{n-m-1}{2}} \int_{\underline{M}} \sum_{\beta \in M(k)}\sum_{\gamma \in N_{m}(k) \backslash G_{m}(k)} \\ & W_\varphi\left(\left(\begin{array}{cc} \gamma h & \\ \left[\begin{array}{cc}(\beta+x)\\0\end{array}\right]h&I_{n-m}\end{array}\right)w_{n,m}\right)dx. \end{split} \end{equation*}$ Combining $\int_{\underline{M}}$ and $\sum_{\beta \in M(k)}$ and then switching orders, $\begin{equation*} \begin{split} \widetilde{\mathbb{P}}^n_m \varphi & \left(\begin{array}{cc} h& \\ &1\end{array}\right) = |\det h|^{\frac{n-m-1}{2}} \sum_{\gamma \in N_{m}(k) \backslash G_{m}(k)} \int_{M(\mathbb{A})}\\ & W_\varphi\left(\left(\begin{array}{cc} \gamma h & \\ \left[\begin{array}{cc}xh\\0\end{array}\right]&I_{n-m}\end{array}\right)w_{n,m}\right)dx. \end{split} \end{equation*}$ Let’s now do the change of variable $x \mapsto xh^{-1}$ with $d(xh^{-1}) = |\det h|^{-(n-m-1)}dx,$ then $\begin{equation*} \begin{split} \widetilde{\mathbb{P}}^n_m \varphi & \left(\begin{array}{cc} h& \\ &1\end{array}\right) = |\det h|^{-\frac{n-m-1}{2}} \sum_{\gamma \in N_{m}(k) \backslash G_{m}(k)} \int_{M(\mathbb{A})}\\ & W_\varphi\left(\left(\begin{array}{cc} \gamma h & \\ \left[\begin{array}{cc}x\\0\end{array}\right]&I_{n-m}\end{array}\right)w_{n,m}\right)dx. \end{split} \end{equation*}$ This finishes the proof of Theorem 2. $\square$

With Theorem 2 at hand, now it’s easy to prove Theorem 1.

Proof of Theorem 1. By Theorem 2, $\begin{equation*} \begin{split} \widetilde{I}(s; &\varphi, \varphi^\prime) = \int_{\underline{G_m}} \widetilde{\mathbb{P}}^n_m \varphi\left(\begin{array}{cc} h& \\ &1\end{array}\right)\varphi^\prime(h)|\det h|^{s-\frac{1}{2}}dh \\ &=\int_{\underline{G_m}}|\det h|^{s-\frac{n-m}{2}} \sum_{\gamma \in N_{m}(k) \backslash G_{m}(k)} \int_{M(\mathbb{A})} \\ & \;\;\;\;\;\;\;\; \rho(w_{n,m})W_\varphi\left(\begin{array}{cc} \gamma h & \\ \left[\begin{array}{cc}x\\0\end{array}\right]&I_{n-m}\end{array}\right)dx\varphi^\prime(h)dh. \end{split} \end{equation*}$ Since $\varphi^\prime$ is automorphic and $|\det \gamma^{-1}|=1$, if we do the change of variables $h \mapsto \gamma^{-1}h$, then $\begin{equation*} \begin{split} \widetilde{I}(s; \varphi, \varphi^\prime) = \int_{N_m(k)\backslash G_m(\mathbb{A})} & |\det h|^{s-\frac{n-m}{2}} \int_{M(\mathbb{A})} \\ & \rho(w_{n,m})W_\varphi\left(\begin{array}{cc} h & \\ \left[\begin{array}{cc}x\\0\end{array}\right]&I_{n-m}\end{array}\right)dx\varphi^\prime(h)dh. \end{split} \end{equation*}$ Noticing that $|\det u| = 1$ for $u \in \underline{N_m}$, we can rewrite the integral over $N_m(k)\backslash G_m(\mathbb{A})$ into double integrals over $N_m(\mathbb{A})\backslash G_m(\mathbb{A})$ and $\underline{N_m}$ as follow. $$$\begin{split} \widetilde{I}(s; \varphi, \varphi^\prime) = \int_{N_m(\mathbb{A})\backslash G_m(\mathbb{A})} & |\det h|^{s-\frac{n-m}{2}} \int_{\underline{N_m}} \int_{M(\mathbb{A})} \\ & \rho(w_{n,m})W_\varphi\left(\begin{array}{cc} uh & \\ \left[\begin{array}{cc}x\\0\end{array}\right]&I_{n-m}\end{array}\right)dx\varphi^\prime(uh)dh. \end{split} \label{20150712-4}$$$

Since $\left(\begin{array}{cc} uh & \\ \left[\begin{array}{cc}x\\0\end{array}\right]&I_{n-m}\end{array}\right) = \left(\begin{array}{cc} u & \\ &I_{n-m}\end{array}\right)\left(\begin{array}{cc} h & \\ \left[\begin{array}{cc}x\\0\end{array}\right]&I_{n-m}\end{array}\right),$ so $W_\varphi\left(\begin{array}{cc} uh & \\ \left[\begin{array}{cc}x\\0\end{array}\right]&I_{n-m}\end{array}\right) = \psi^{-1}(u)W_\varphi\left(\begin{array}{cc} h & \\ \left[\begin{array}{cc}x\\0\end{array}\right]&I_{n-m}\end{array}\right).$ Now if we separate the integral over $u$ in $\eqref{20150712-4}$, we get $\int_{\underline{N_m}}\varphi^\prime(uh)\psi^{-1}(u)du = W_{\varphi^\prime}(h).$ Hence, from $\eqref{20150712-4}$, we obtain $\widetilde{I}(s; \varphi, \varphi^\prime) = \widetilde{\Psi}(s; \rho(w_{n, m})W_{\varphi}, W_{\varphi^\prime}),$ as desired. $\square$