# Some Combinatorics in the Cancellation of Poles of Eisenstein Series for
$GL(n,\mathbb{A}_{\mathbb{Q}})$

Zhuohui Zhang
Tel Aviv University<EMAIL_ADDRESS>
###### Abstract.

The cancellations of poles of degenerate Eisenstein series were studied by
Hanzer and Muić. This paper generalized the result to Eisenstein series
constructed from inducing two Speh representations
$\Delta(\tau,m)|\cdot|^{s_{1}}\otimes\Delta(\tau,n)|\cdot|^{s_{2}}$ for the
group $GL(m+n,\mathbb{A}_{\mathbb{Q}})$ for self-dual cuspidal automorphic
representation $\tau$ by describing the combinatorics of the relevant Weyl
group coset.

##  1\. Introduction

In [MW89], Mœglin and Waldspurger described the residual automorphic spectrum
of $GL(N)$. A residual automorphic representation can always be realized as a
_generalized Speh representation_ $\Delta(\tau,n)$, where $\tau$ is an
irreducible unitary cuspidal automorphic representation of the group $GL(a)$
with $a$ satisfying $N=an$. For convenience, we will assume the cuspidal
automorphic representation $\tau=\bigotimes_{v}\tau_{v}$ has a unitary central
character, and $\tau_{v}$ is tempered at each local place $v$. The automorphic
representation $\Delta(\tau,n)$ is a global Langlands quotient of a principal
series representation, and can be realized as the automorphic representation
generated by the residue of an Eisenstein series. For any partition
$\underline{N}=(N_{1},\ldots,N_{r})$ of $N$, denoting the standard parabolic
subgroup with Levi subgroup $M_{\underline{N}}$ isomorphic to
$GL(N_{1})\times\ldots GL(N_{r})$ by
$P_{\underline{N}}=M_{\underline{N}}U_{\underline{N}}$, we can choose cuspidal
automorphic representations $\tau_{1},\ldots,\tau_{r}$ of $GL(a_{i})$, where
$a_{i}n_{i}=N_{i}$, and construct the principal series representation

$I(\underline{\tau},\underline{s})=\mathrm{Ind}_{P_{\underline{N}}}^{G}\left(\Delta(\tau_{1},n_{1})|\cdot|^{s_{1}}\otimes\ldots\otimes\Delta(\tau_{r},n_{r})|\cdot|^{s_{r}}\right).$

This paper investigates the poles of the Eisenstein series constructed from a
section in $I(\underline{\tau},\underline{s})$ with isomorphic cuspidal data
$\tau_{1}=\tau_{2}=\tau$ in the case $r=2$.

######  Theorem A.

Fixing an irreducible unitary self-dual cuspidal automorphic representation
$\tau$, we construct the principal series

$I(\tau,\underline{s})=\mathrm{Ind}_{P_{\underline{N}}}^{G}\left(\Delta(\tau,m)|\cdot|^{s_{1}}\otimes\Delta(\tau,n)|\cdot|^{s_{2}}\right)$

where $\underline{N}=(m,n)$. In the region $s_{1}-s_{2}\geq 0$, the Eisenstein
series

$E(\cdot,\underline{s}):I(\tau,\underline{s})\longrightarrow\mathcal{A}_{G_{N}}$

has poles of maximal order 1 at
$s_{1}-s_{2}\in\frac{m+n}{2}-\\{0,1,\ldots,\min\\{m,n\\}-1\\}$.

The proof of this theorem is a generalization of the combinatorial method in
[HM15] to cuspidal automorphic induction data. The residues of this Eisenstein
series are described in the following corollary:

######  Corollary B.

The residue of Eisenstein series
$\mathrm{Res}_{s=\frac{m+n}{2}-\sigma}E(\cdot,\underline{s})$ defines an
intertwining operator sending $I(\tau,\underline{s})$ to the irreducible
representation
$\mathrm{Ind}_{P_{[m,n]}}^{G_{m+n}}\left(\Delta(\tau,m+n-\sigma)|\cdot|^{-\frac{m-n}{2}}\boxtimes\Delta(\tau,\sigma)|\cdot|^{\frac{m-n}{2}}\right)$.

##  2\. Principal Series and Eisenstein Series

Throughout this section, we denote the space of automorphic functions on
$G_{n}=GL(na)$ by $\mathcal{A}_{G_{n}}$. The standard parabolic subgroup with
Levi subgroup isomorphic to $GL(a)^{\times n}$ is denoted by
$P_{n}=M_{n}U_{n}$.

###  2.1. Induction from Cusp Forms

The cuspidal automorphic representation $\tau$ on the group $GL(a)$ can be
decomposed as a restricted tensor product $\tau=\bigotimes_{v}\tau_{v}$ of
irreducible unitary representations $\tau_{v}$ over local fields
$\mathbb{Q}_{v}$. One can choose a pure tensor vector $f=\otimes_{v}f_{v}$
which realizes the completed automorphic $L$-function as the normalization
factor of the intertwining operator (for the construction of such a vector
c.f. [Sha84]). Thus, if we consider the tensor product
$\bigotimes_{k=1}^{n}\tau|\cdot|^{\nu_{i}}$ as a representation on the group
$GL(a)^{\times n}$, we can still choose a pure tensor
$\bigotimes_{k=1}^{n}\left(\otimes_{v}f_{k,v}\right)\in\tau$ satisfying the
same property. This vector can be realized as an automorphic function
$f_{\underline{\nu}}\in\bigotimes_{k=1}^{n}\mathcal{A}_{G_{a}}$. By the
Iwasawa decomposition of $G_{n}=GL_{na}$, we can choose a section
$\tilde{f}_{\underline{\nu}}\in\mathrm{Ind}_{P_{n}}^{G}\left(\bigotimes_{k=1}^{n}\tau|\cdot|^{\nu_{i}}\right)$
of the principal series satisfying

$\tilde{f}_{\underline{\nu}}(umk)=f_{\underline{\nu}}(m)|m|^{\underline{\nu}}\tilde{f}_{\underline{\nu}}(k),$

where $|m|^{\nu}=|\det m_{1}|^{\nu_{1}}\ldots|\det m_{n}|^{\nu_{n}}$ for any
$m\in M_{[a^{n}]}$. The Eisenstein series

$E_{n}(f_{\nu},\underline{\nu})(g)=\sum_{\gamma\in P_{n}(k)\backslash
G(k)}\tilde{f}_{\underline{\nu}}(\gamma g)$

defines an intertwining operator between $G_{n}$ representations:

(1)
$E_{n}(\cdot,\underline{\nu}):I(\tau,\underline{\nu}):=\mathrm{Ind}_{P_{n}}^{G_{n}}\left(\bigotimes_{k=1}^{n}\tau|\cdot|^{\nu_{i}}\right)\longrightarrow\mathcal{A}_{G_{n}},$

and can be meromorphically continued to the whole $\mathbb{C}^{n}$ from the
domain of convergence $\mathrm{Re}(\nu_{i}-\nu_{j})>0$ for $i<j$.

###  2.2. The Rankin-Selberg $L$-Function

The Langlands-Shahidi method can be applied to define the Rankin-Selberg
$L$-function $L(s,\tau\times\hat{\tau})$ described in [JPSS83] for any
cuspidal automorphic representation $\tau$ as normalization factors of
intertwining operators. An exposition of such method can be found in [Sha10,
Chapter 5]. We will be mostly following the setup in [MW89, Appendice], but
will also point out other properties of the Rankin-Selberg $L$-function
$L(s,\tau\times\hat{\tau})$ which we will be using in this paper.

We consider two irreducible unitary cuspidal automorphic representations
$\tau_{1},\tau_{2}$ on $GL(n_{1},\mathbb{A})$ and $GL(n_{2},\mathbb{A})$,
respectively. The constant term of the Eisenstein series

$E_{n_{1},n_{2}}(\cdot,\underline{s}):\mathrm{Ind}_{P_{[n_{1},n_{2}]}}^{G_{n_{1}+n_{2}}}(\tau_{1}|\cdot|^{\frac{s}{2}}\otimes\tau_{2}|\cdot|^{-\frac{s}{2}})\longrightarrow\mathcal{A}_{G_{n_{1}+n_{2}}}$

will have only one summand unless if $n_{1}=n_{2}$. In any case, following
[Sha83], the normalization factor $r(s,w_{0})$ of the intertwining operator
$M(s,w_{0})$ can be expressed as

$r(s,w_{0})=\frac{L(s,\tau_{1}\times\hat{\tau}_{2})}{L(1+s,\tau_{1}\times\tau_{2})\epsilon(s,\psi,\tau_{1}\times\hat{\tau}_{2})}$

where $L(s,\tau_{1}\times\hat{\tau}_{2})$ is the completed Rankin-Selberg
$L$-function with its local factors described in [JPSS83]. In [MW89,
Appendice, Corollaire], it is shown that the function
$L(s,\tau_{1}\times{\hat{\tau}_{2}})$ is entire unless $n_{1}=n_{2}$ and there
exists a complex number $t$ such that $\tau_{1}=\tau_{2}|\cdot|^{t}$. The
poles of the function $L(s,\tau\times\hat{\tau})$ are simple and are located
at $s=0,1$. A zero-free region of $L(s,\tau\times\hat{\tau})$ is given in
[GL18] by

$\Re(s)>1-\frac{c}{(\log(|\Im(s)|+2))^{5}}.$

A consequence of the zero-free region given above is that there are no zeros
in the region $\Re(s)\geq 1$ and $\Re(s)\leq 0$. Therefore, one expects no
cancellations of poles between the numerators and the denominators apart from
the cancellations from the zeros on the critical strip.

###  2.3. Residues of Eisenstein Series and Speh Representation

In this section, we summarize the key facts concerning the poles and residues
of the Eisenstein series $E_{n}(\cdot,\nu)$ defined in (1) from the previous
section. By the Langlands constant term formula [MW95, II.1.7] and the
cuspidality of $\tau$, the constant term along the unipotent radical of
$U_{n}$ of the parabolic subgroup $P_{n}$ can be expressed as

$c_{U_{n}}E_{n}(f,\underline{\nu})(g)=\sum_{w\in W_{a}^{\times n}\backslash
W_{na}\cong\mathbb{S}_{n}}M(w,\underline{\nu})\tilde{f}(g).$

Each intertwining operator $M(w,\underline{\nu})$ is defined in the dominant
cone $\mathrm{Re}(\nu_{i}-\nu_{j})\gg 0$, where $i<j$, as the following formal
integral

$M(w,\underline{\nu})\tilde{f}(g)=\int_{U_{n}\cap w^{-1}U_{n}w\backslash
U_{n}}\tilde{f}(wng)dn.$

The operator $M(w,\underline{\nu})$ can be meromorphically continued to the
whole $\mathbb{C}^{n}$. By [MW89, Appendice], each summand
$M(w,\underline{\nu})\tilde{f}(g)$ has the same set of poles as the function
$r(w,\underline{\nu})$ known as the _normalization factor_ :

$r(w,\underline{\nu})=\prod_{\begin{subarray}{c}i<j\\\
w(i)>w(j)\end{subarray}}\frac{L(\nu_{i}-\nu_{j},\tau\times\hat{\tau})}{L(1+\nu_{i}-\nu_{j},\tau\times\hat{\tau})\epsilon(\nu_{i}-\nu_{j},\tau\times\hat{\tau},\psi)}.$

The normalized intertwining operator
$N(w,\underline{\nu})=r(w,\underline{\nu})^{-1}M(w,\underline{\nu})$ is
holomorphic in the region $\mathrm{Re}(\nu_{i}-\nu_{j})\geq 0$ with $i<j$.

The pole of the highest possible codimension $r(w,\underline{\nu})$ occurs
when $w(i)>w(j)$ for all $i<j$. In this case, the normalization factor
$r(w,\underline{\nu})$ has a simple pole at the intersection of the
hyperplanes $\nu_{i}-\nu_{i+1}=1$, i.e. at the point
$\underline{\nu}=-\underline{\lambda}_{n}=\left(\frac{n-1}{2},\frac{n-3}{2},\ldots,\frac{1-n}{2}\right)$
(pay attention to the minus sign). We can take the residue

$\mathrm{Res}_{n}E_{n}(f,\underline{\nu})=\lim_{\underline{\nu}\rightarrow-\underline{\lambda}_{n}}\prod_{i=1}^{n-1}(\nu_{i}-\nu_{i+1}-1)E_{n}(f,\underline{\nu}).$

of the Eisenstein series $E_{n}(f,\underline{\nu})$ at the point
$-\underline{\lambda}_{n}$. The irreducible automorphic representation
$\Delta(\tau,n)$ generated by $\mathrm{Res}_{n}E_{n}(f,\underline{\nu})$ is
usually referred to as the _Speh representation_ , which constitutes the whole
non-cuspidal discrete spectrum if we let $\tau$ run through all possible
choices of cuspidal automorphic representations for all block sizes $a$.

The composition $\mathrm{Res}_{n}E_{n}(\cdot,\underline{\nu})$ of the
Eisenstein series operator and the residue operator is a quotient operator
from the principal series
$\mathrm{Ind}_{P_{n}}^{G}\left(\bigotimes_{k=1}^{n}\tau|\cdot|^{\nu_{i}}\right)$
to $\mathcal{A}_{G_{n}}$. It factors through the embedding
$\Delta(\tau,n)\hookrightarrow\mathcal{A}_{G_{n}}$. By the Langlands
classification ([Lan89] and [Lan06]), on the pole $\underline{\nu}$ where a
residue exists, the representation $\Delta(\tau,n)$ is also the image of the
intertwining operator

$M(w_{0},\underline{\nu}):I(\tau,\underline{\nu})\longrightarrow
I(\tau,w_{0}\underline{\nu})$

for the longest element $w_{0}$ of the Weyl group $\mathbb{S}_{n}$. On each
finite place, the local component of the residual representation
$\Delta(\tau,n)$ is isomorphic to the Langlands quotient of the principal
series $I(\tau_{v},\underline{\nu})$.

###  2.4. Induction from Speh Representations

As in Theorem A, for $m\leq n$ and a irreducible cuspidal unitary automorphic
representation $\tau$, we construct the induced representation

$I(\tau,\underline{s}):=\mathrm{Ind}_{P_{[m,n]}}^{G_{m+n}}\left(\Delta(\tau,m)|\cdot|^{s_{1}}\boxtimes\Delta(\tau,n)|\cdot|^{s_{2}}\right)$

from two Speh representations $\Delta(\tau,m)$ and $\Delta(\tau,n)$ for the
groups $G_{m}$ and $G_{n}$, respectively. We would like to develop a
combinatorial method to understand the poles of the Eisenstein series

$E^{m,n}(\cdot,\underline{s}):I(\tau,\underline{s})\longrightarrow\mathcal{A}_{G_{n}}$

on a section $f$ of $I(\tau,\underline{s})$. Since the Speh representation
$\Delta(\tau,m)|\cdot|^{s_{1}}\boxtimes\Delta(\tau,n)|\cdot|^{s_{2}}$ can be
realized as a subrepresentation of
$I(m,\underline{\lambda_{m}})|\cdot|^{s_{1}}\boxtimes
I(n,\underline{\lambda_{n}})|\cdot|^{s_{2}}$, any vector of
$I(\tau,\underline{s})$ can be realized as a vector in the principal series

$I(m,n,\underline{s})=\mathrm{Ind}_{P_{[m,n]}}^{G_{m+n}}\left(I(m,\underline{\lambda_{m}})|\cdot|^{s_{1}}\boxtimes
I(n,\underline{\lambda_{n}})|\cdot|^{s_{2}}\right).$

For any such vector $f$, the constant term along the unipotent radical
$U_{m+n}$ can be expressed as the following sum of formal integrals over Weyl
group cosets:

$\displaystyle(c_{U_{m+n}}E^{m,n}(f,\underline{s}))(g)$
$\displaystyle=\sum_{\gamma\in P_{[m,n]}(k)\backslash
G_{m+n}(k)}\int_{U_{m+n}(k)\backslash U_{m+n}(\mathbb{A})}f(\gamma ng)dg$

By the Bruhat decompsition, $\gamma$ can be represented by product
$w\left(\begin{smallmatrix}1&\beta\\\ 0&1\end{smallmatrix}\right)$, and we can
decompose $n$ into the product $n=n_{1}n_{2}$, such that $n_{1}\in U_{[m,n]}$
and $n_{2}\in U_{m}\times U_{n}$. Writting
$n_{1}=\left(\begin{smallmatrix}1&a\\\ 0&1\end{smallmatrix}\right)$ and
$n_{2}=\left(\begin{smallmatrix}u_{1}&0\\\ 0&u_{2}\end{smallmatrix}\right)$,
we have

$\displaystyle\gamma n$ $\displaystyle=w\left(\begin{smallmatrix}1&a+\beta\\\
0&1\end{smallmatrix}\right)\left(\begin{smallmatrix}u_{1}&0\\\
0&u_{2}\end{smallmatrix}\right)$
$\displaystyle=w\left(\begin{smallmatrix}u_{1}&0\\\
0&u_{2}\end{smallmatrix}\right)\left(\begin{smallmatrix}u_{1}^{-1}&0\\\
0&u_{2}^{-1}\end{smallmatrix}\right)\left(\begin{smallmatrix}1&a+\beta\\\
0&1\end{smallmatrix}\right)\left(\begin{smallmatrix}u_{1}&0\\\
0&u_{2}\end{smallmatrix}\right)$
$\displaystyle=w\left(\begin{smallmatrix}u_{1}&0\\\
0&u_{2}\end{smallmatrix}\right)\left(\begin{smallmatrix}1&u_{1}^{-1}(a+\beta)u_{2}\\\
0&1\end{smallmatrix}\right).$

For any section $f\in I(m,n,\underline{s})$, since

$f(\gamma ng)=f\left(w\left(\begin{smallmatrix}u_{1}&0\\\
0&u_{2}\end{smallmatrix}\right)w^{-1}w\left(\begin{smallmatrix}1&u_{1}^{-1}(a+\beta)u_{2}\\\
0&1\end{smallmatrix}\right)g\right)$

and $f$ is invariant under the left regular action of $U_{m+n}$, we have

$f(\gamma ng)=f\left(w\left(\begin{smallmatrix}1&u_{1}^{-1}(a+\beta)u_{2}\\\
0&1\end{smallmatrix}\right)g\right).$

Therefore, for the section $f\in I(m,n,\underline{s})$, the constant term
$c_{U_{m+n}}E^{m,n}(f,\underline{s})$ can simply be written as the sum

$c_{U_{m+n}}E^{m,n}(f,\underline{s})=\sum_{w\in\mathbb{S}_{m}\times\mathbb{S}_{n}\backslash\mathbb{S}_{m+n}}\int_{U_{[m,n]}(\mathbb{A})\cap
w^{-1}U_{[m,n]}(\mathbb{A})w\backslash
U_{[m,n]}(\mathbb{A})}f(wng)dn=\sum_{w\in\mathbb{S}_{m}\times\mathbb{S}_{n}\backslash\mathbb{S}_{m+n}}M(w,\underline{s})f.$

We can summarize this result into the following lemma:

######  Lemma 1.

For any section $f\in I(m,n,\underline{s})\hookrightarrow
I(\tau,\underline{s})$, the Eisenstein series $E^{m,n}(f,\underline{s})$ has
the same set of poles with the constant term

$c_{U_{m+n}}E_{\underline{s}}^{m,n}(f)=\sum_{w\in\mathbb{S}_{m}\times\mathbb{S}_{n}\backslash\mathbb{S}_{m+n}}M(w,\underline{s})f.$

We can further organize the sum in Lemma 1 by grouping $M(w,\underline{s})f$
for different $w$’s based on which space $M(w,\underline{s})f$ lives in.
Denoting the representation

$\rho_{m,n}(\tau,\underline{s})=\left(\bigotimes_{i=1}^{m}\tau|\cdot|^{\frac{1-m}{2}+i-1+s_{1}}\right)\otimes\left(\bigotimes_{j=1}^{n}\tau|\cdot|^{\frac{1-n}{2}+j-1+s_{2}}\right)$

by $\rho_{m,n}(\tau,\underline{s})$, the intertwining operator
$M(w,\underline{s})f$ is a vector in the conjugation
$(\rho_{m,n}(\tau,\underline{s}))^{w}$. We will denote by
$\mathcal{O}_{\underline{s}}(w)$ the set of all Weyl group representatives
$w^{\prime}\in\mathbb{S}_{m}\times\mathbb{S}_{n}\backslash\mathbb{S}_{m+n}$
such that
$(\rho_{m,n}(\tau,\underline{s}))^{w}\cong(\rho_{m,n}(\tau,\underline{s}))^{w^{\prime}}$.
Thus the sum in Lemma 1 can be grouped into sums on orbits
$\mathcal{O}_{\underline{s}}(w)$:

(2) $\displaystyle
c_{U_{m+n}}E_{\underline{s}}^{m,n}(f)=\sum_{\mathcal{O}_{\underline{s}}(w)}\left(\sum_{w\in\mathcal{O}_{\underline{s}}(w)}M(w,\underline{s})f\right).$

##  3\. Segments and Weyl Group Elements

In this section, we classify the orbits of elements in
$\mathbb{S}_{m}\times\mathbb{S}_{n}\backslash\mathbb{S}_{m+n}$ and describe
the representatives of each orbit.

###  3.1. A Double Coset

We define $W_{m,n}$ as the collection of representatives of the double coset

$(W_{m}\times W_{n})\backslash
W_{m+n}/\\{e\\}=\mathbb{S}_{m}\times\mathbb{S}_{n}\backslash\mathbb{S}_{m+n},$

which has a one-to-one correspondence with the set of elements
$w\in\mathbb{S}_{m+n}$ preserving the order in the strings $\\{1,\ldots,m\\}$
and $\\{m+1,\ldots,m+n\\}$, respectively. Such an element satisifies the
property

(3) $\displaystyle w(1)<\ldots<w(m),\;w(m+1)<\ldots<w(m+n).$

The action of $w$ on the string $\\{1,\ldots,m+n\\}$ shuffles between the two
strings $\\{1,\ldots,m\\}$ and $\\{m+1,\ldots,m+n\\}$. We denote the
interlacing intervals originated from each one of these two strings by $v_{i}$
and $u_{i}$, respectively, as shown in the following diagram:

$v_{0}$$u_{1}$$v_{1}$$u_{2}$$\ldots$$u_{i}$$v_{i}$$\ldots$

.

The “light gray” $v_{i}$-intervals and the “dark gray” $u_{i}$-intervals
assemble back to the two original strings:

$12\ldots(m-1)m=$$v_{0}$$v_{1}$$v_{2}$$v_{3}$$\ldots$$(m+1)\ldots(m+n)=$$u_{1}$$u_{2}$$u_{3}$$u_{4}$$\ldots$

.

It is possible that $v_{0}$ is empty, in which case the sequence
$w(1\ldots(m+n))$ will start with the interval $u_{1}$:

$u_{1}$$v_{1}$$u_{2}$$\ldots$$u_{i}$$v_{i}$$\ldots$

.

For any interval $I$, we denote by $l(I)$ the index of its starting point on
the left and $r(I)$ the index of its ending point on the right. In the string
corresponding to $w$, if an interval $I_{1}$ appears to the left of $I_{2}$,
then we say $I_{1}$ _precedes_ $I_{2}$, and we denote this situation by
$I_{1}\prec I_{2}$.

###  3.2. Segments

The full principal series induced from the tensor product
$\rho_{m,n}(\tau,\underline{s})$ can be perceived as the automorphic analogue
of the induction from _Berstein-Zelevinsky segments_ $\Delta+s_{1}$ and
$\Delta^{\prime}+s_{2}$ (c.f. [Zel80]):

$\frac{1-m}{2}$$\frac{3-m}{2}$$\frac{5-m}{2}$$\ldots$$\frac{1-m}{2}+\sigma$$\ldots$$\frac{m-5}{2}$$\frac{m-3}{2}$$\frac{m-1}{2}$$+s_{1}$$\frac{1-n}{2}$$\frac{3-n}{2}$$\frac{5-n}{2}$$\ldots$$\frac{n-5}{2}$$\frac{n-3}{2}$$\frac{n-1}{2}$$+s_{2}$

The integer $\sigma\geq 0$ equals to the length of the overlapping part of the
two segments. The smallest possible $\sigma$ is equal to 0, in which case the
two segments are _juxtaposed_ , as shown in the following diagram:

$\frac{1-m}{2}$$\frac{3-m}{2}$$\frac{5-m}{2}$$\ldots$$\frac{m-5}{2}$$\frac{m-3}{2}$$\frac{m-1}{2}$$+s_{1}$$\frac{1-n}{2}$$\frac{3-n}{2}$$\frac{5-n}{2}$$\ldots$$\frac{n-5}{2}$$\frac{n-3}{2}$$\frac{n-1}{2}$$+s_{2}$

.

In general, if the length of the overlapping part of the two segments is
$\sigma$, we have:

$\frac{n-1}{2}+s_{2}=\frac{1-m}{2}+s_{1}-1+\sigma,$

and thus $\sigma$ determines $s=s_{1}-s_{2}=\frac{m+n}{2}-\sigma$, where
$\sigma$ is allowed to be any integer between $0$ and
$\lfloor\frac{m+n}{2}\rfloor$, so that the segments $\Delta+s_{1}$ and
$\Delta^{\prime}+s_{2}$ intersect in the following three ways:

  * •

Case I: $\sigma\leq\min\\{m,n\\}$:

$\Delta+s_{1}$$\Delta^{\prime}+s_{2}$

  * •

Case II: $m\leq\sigma\leq n$:

$\Delta+s_{1}$$\Delta^{\prime}+s_{2}$

  * •

Case III: $n\leq\sigma\leq m$:

$\Delta+s_{1}$$\Delta^{\prime}+s_{2}$

The case when $\sigma>\max\\{m,n\\}$ is called _$\Delta+s_{1}$ precedes
$\Delta^{\prime}+s_{2}$_, which will result in a negative $s=s_{1}-s_{2}$ and
will not be considered.

Since the representation $\rho_{m,n}(\tau,\underline{s})$ depends on the
choice of $\sigma$, the orbits $\mathcal{O}_{\underline{s}}(w)$ of Weyl group
cosets $\mathbb{S}_{m}\times\mathbb{S}_{n}\backslash\mathbb{S}_{m+n}$ also
depend on the choice of $\sigma$.

###  3.3. Orbits

This section generalizes the method developed in [HM15]. The condition for two
Weyl group representatives $w,w^{\prime}\in W_{m,n}$ belonging to the same
orbit is
$(\rho_{m,n}(\tau,\underline{s}))^{w}=(\rho_{m,n}(\tau,\underline{s}))^{w^{\prime}}$,
which is equivalent to requiring

$w^{\prime}(i)=w(i)\text{ or }w(m+n-\sigma+i).$

for $i\in\\{1,\ldots,m\\}$ and $m+n-\sigma+i\in\\{m+1,\ldots,m+n\\}$.
Therefore, there is a transitive action by a group
$G_{\sigma}=\mathbb{Z}_{2}^{r}$ (which is related to the $R$-group, and will
be described in the following sections) of _permissible moves_ for some
integer $r$ on each orbit $\mathcal{O}_{\underline{s}}$ permuting the pairs
$\\{w(i),w(m+n-\sigma+i)\\}$. For any choice of $\sigma$, we can choose the
_base point_ $w_{0}\in\mathcal{O}_{\underline{s}}$ to be the unique element in
the orbit satisfying

(4) $w_{0}(i)<w_{0}(m+n-\sigma+i)$

for all $i$ such that $i\in\\{1,\ldots,m\\}$ and
$m+n-\sigma+i\in\\{m+1,\ldots,m+n\\}$.

###  3.4. Permissible Moves on Basepoints

In this section, we describe the aforementioned group $G_{\sigma}$ of
_permissible_ moves.

$v_{i}$$u_{i+1}$$\ldots$$\ldots$

######  Definition 2.

Fixing an $s=s_{1}-s_{2}=\frac{m+n}{2}-\sigma$, a _permissible move_ $g\in
G_{\sigma}$ is a permutation of the string $\\{1,2,\ldots,(m+n)\\}$, such that

  1. (1)

For any element $w$ satisfying (3), $gw$ also satisfies (3).

  2. (2)

The two elements $w$ and $gw$ satisfy
$\rho_{m,n}(\tau,\underline{s})^{w}=\rho_{m,n}(\tau,\underline{s})^{gw}$, i.e.

$gw(i)\in\\{w(i),w(m+n-\sigma+i)\\}.$

Any element that can be affected by a permissible move is said to be _alive_ ,
otherwise it is said to be _dead_. If there is a continuous segment of living
elements, we refer to that segment as a _living segment_ , and vice versa for
_dead segments_.

For a basepoint $w_{0}$, we will color the _living_ elements of the string
$w_{0}(12\ldots(m+n))$ either green or red based on whether they originate
from the string $12\ldots m$ or $(m+1)\ldots(m+n)$, respectively. Recall that
we require the basepoint element $w_{0}$ to satisfy the property
$w_{0}(i)<w_{0}(m+n-\sigma+i)$, on the basepoint element, we will color
$w_{0}(i)$ green and $w_{0}(m+n-\sigma+i)$ red. We will call them _green
elements (intervals,resp.)_ and _red elements (intervals,resp.)_ for
convenience. Note that any green element lies on a “light gray”
$v_{i}$-interval while any red element lies on a “dark gray” $u_{i}$-interval.
We will use this diagram to prove properties of all permissible moves from a
basepoint $w_{0}$.

######  Lemma 3.

For a basepoint $w_{0}$, the following properties are true for green elements:

  1. (1)

All elements between two green elements on the same “light gray”
$v_{i}$-interval are green.

$v_{i}$$I_{i}$$u_{i+1}$$\ldots$$\ldots$

  2. (2)

The right endpoint of $v_{i}$ and the right endpoint of the corresponding
$I_{i}$ are aligned: $w(r(v_{i}))=w(r(I_{i}))$.

$v_{i}$$I_{i}$$u_{i+1}$$\ldots$$\ldots$

Similarly, for red elements, they form a gap-free interval on any $u_{i}$, and
their left endpoints are aligned with the left endpoint of $u_{i}$ they lie
in:

$v_{i}$$J_{i}$$u_{i+1}$$\ldots$$\ldots$

.

###### Proof.

These properties are all consequences of (1) in Definition 2. On any
$v$-interval, the green elements are not allowed to move beyond any dead
interval to its right:

$v_{i}$$\times$$u_{i+1}$$\ldots$$\ldots$

.

Therefore, on each $v$-interval, no gap is allowed on any of its green
subintervals. For the same reason, there are no dead elements to its right.
Therefore, the right end of $v_{i}$ and the right end of $I_{i}$ are aligned.
∎

######  Corollary A.

On the same $v_{i}$, the green elements form a gap-free interval, and on each
“light gray” interval $v_{i}$ there is only one interval of green elements. We
denote this interval by $I_{i}$. Similarly, on the same $u_{i}$, the red
elements form a unique gap-free interval, and we denote this interval by
$J_{i}$.

The following lemma describes the permissible moves near the dead elements:

######  Lemma 4.

On the string representing the basepoint $w_{0}$, the following statements are
true:

  1. (1)

If the element with index $w(r(I_{i}))+1$ lying to the right of a green
interval $I_{i}$ is dead, the permissible moves can only move $I_{i}$ into the
adjacent red interval in $u_{i+1}$, with their right endpoints aligned:

$v_{i}$$I_{i}$$J_{i}$$I_{i+1}$$J_{i+1}$$u_{i+1}$$v_{i+1}$$u_{i+2}$$\ldots$$\ldots$$\ldots$

.

Similarly, if the element with index $w(l(J_{i+1}))-1$ to the left of
$J_{i+1}\subset u_{i+2}$ is dead, the permissible moves can only move
$J_{i+1}$ leftwards to its adjacent green interval $I_{i+1}$ in $v_{i+1}$,
with their left endpoints aligned.

  2. (2)

If the intervals $I_{i}\subset v_{i}$ and $I_{i+1}\subset v_{i+1}$ are
connected (i.e. without any gap in between), then the whole $u_{i+1}$ is a red
interval $J_{i}$.

###### Proof.

The first part is a consequence of (1) in Definition 2 and Lemma 3 since there
is no green living interval adjacent to $I_{i}$ on the right. The second part
is a consequence of the first part and (2) in Definition 2. ∎

######  Remark 1.

There is no specification about how on any basepoint element $w_{0}$, the
elements are moved between continuous green/red intervals. One green interval
$I_{i}$ can be splitted across two $u$-intervals, and one red interval $J_{i}$
can be splitted across two $v$-intervals by a permissible move. The only fact
we know for sure is that, the left endpoint of a continuous interval satisfies
the same property as $I_{i+1}$ and $J_{i+1}$ in the Part (2) of the lemma
above, and the right endpoint of a continuous interval satisfies the same
property as $I_{i}$ and $J_{i}$ in the Part (1) of the lemma above.

The following lemma describes the gaps between two continuous intervals. In
fact, this lemma implies that there is actually no gap between two consecutive
living intervals.

######  Lemma 5.

  1. (1)

For the following configuration on four consecutive intervals
$v_{i},u_{i+1},v_{i+1},u_{i+2}$:

$v_{i}$$I_{i}$$J_{i}$$I_{i+1}$$J_{i+1}$$u_{i+1}$$v_{i+1}$$u_{i+2}$$\ldots$$\ldots$$J_{i}^{\prime}$$I_{i+1}^{\prime}$

Setting $J_{i}^{\prime}=u_{i+1}\backslash J_{i}$ and
$I_{i+1}^{\prime}=v_{i+1}\backslash I_{i+1}$, swapping $I^{\prime}_{i+1}$ and
$J_{i}^{\prime}$ is a permissible move.

  2. (2)

For the following configuration,

$v_{i}$$I_{i}$$J_{i}$$I_{i+k}$$J_{i+k}$$u_{i+1}$$v_{i+k}$$u_{i+k+1}$$\ldots$$\ldots$$J_{i}^{\prime}$$I_{i+k}^{\prime}$$\ldots$

one can construct a permissible move for the whole interval between
$r(I_{i})+1$ and $l(I_{i+k})-1$.

###### Proof.

  1. (1)

Since $J_{i}^{\prime}$ has the same length as $I_{i+1}^{\prime}$, swapping
$J_{i}^{\prime}$ with $I_{i+1}^{\prime}$ is in fact a permissible move.

  2. (2)

We can list the intervals between $J_{i}$ and $I_{i+k}$ as

$J_{i}^{\prime},v_{i+1},u_{i+2},\ldots,u_{i+k},I^{\prime}_{i+k}.$

Since the right endpoints of $I_{i}$ and $J_{i}$ correspond, and the left
endpoints of $I_{i+k}$ and $J_{i+k}$ correspond, denoting the lengths of the
“dark grey” intervals $J_{i}^{\prime},u_{i+2},\ldots,u_{i+k}$ by
$p_{1},\ldots,p_{k}$, and the lengths of “light grey” intervals
$v_{i+1},\ldots,I_{i+k}^{\prime}$ by $q_{1},\ldots,q_{k}$. We have
$\sum_{i=1}^{k}p_{i}=\sum_{i=1}^{k}q_{i}$. From the partitions
$(p_{1},\ldots,p_{k})$ and $(q_{1},\ldots,q_{k})$, we can obtain a refined
partition of the dark gray and light gray intervals. The move between
corresponding intervals is a permissible move.

∎

From the previous lemmas, we can find that on the segment $12\ldots m$, the
green elements only exist along
$(\max\\{\sigma-n+1,1\\},\ldots,\min\\{\sigma,m\\})$, and can be organized
into mutually connected intervals $K_{1},\ldots,K_{r}$, possibly straddling
across different $v$-intervals. Each $K_{t}$ moves as a whole under the action
of $G_{\sigma}$. The order of $G_{\sigma}$ is equal to $2^{r}$.

###  3.5. General Description of the Permissible Moves

Recall from Section 3.1 that the string

$w(12\ldots(m+n))$

can be expressed as the interlacing $v,u$ intervals:

$v_{0}$$u_{1}$$v_{1}$$u_{2}$$\ldots$$u_{i}$$v_{i}$$\ldots$

.

We would like to use this diagram, combined with the lemmas we stated in the
previous section to describe the green and red living intervals and the
corresponding permissible moves.

######  Proposition 6.

On the string representing a basepoint element $w_{0}$,

  1. (1)

The first green element is the first green element on the left such that there
is no dead element from $\\{m+1,\ldots,m+n\\}$ to its right.

  2. (2)

The last red element is the last red element on the right such that there is
no dead element from $\\{1,\ldots,m\\}$ to its left.

  3. (3)

An element $i\in\\{1,\ldots,\sigma\\}$ is dead if between $w(i)$ and
$w(m+n-\sigma+i)$ there is a dead element coming from
$\\{m+n-\sigma+1,\ldots,m+n\\}$.

###### Proof.

Part (1) and (2) follow from the definition of living elements and the proof
of Lemma 3. Part (3) is true simply because permissible moves do not allow any
move across dead elements. ∎

In the proof of Lemma 5, we have already seen that there is no dead element
between the first and the last living elements. Denoting the first living
green interval by $v_{k}^{\prime}\subset v_{k}$, and the last living red
interval by $u_{l}^{\prime}\subset u_{l}$. The positions of these living
elements are displayed as in the following diagram:

$v_{k}^{\prime}$$u_{k}$$v_{k+1}$$u_{k+1}$$\ldots$$u_{i}$$v_{i}$$\ldots$$u_{l}^{\prime}$

The permissible moves can be constructed with the following algorithm:

  1. (1)

Denoting the lengths of the intervals $v_{k}^{\prime},\ldots,v_{l}$ and
$u_{k},\ldots,u_{l}^{\prime}$ by $p_{k},\ldots,p_{l}$ and
$q_{k},\ldots,q_{l}$, for any $1\leq
r\leq\sum_{i=k}^{l}p_{i}=\sum_{i=k}^{l}q_{i}$, the collection of $S$ of “end
points of refined intervals” $\\{t_{1},\ldots,t_{N}\\}$ is defined as all the
$r$’s such that $r$ corresponds to an end point of either a green or a red
interval.

$v_{k}^{\prime}$$u_{k}$$v_{k+1}$$u_{k+1}$$\ldots$$u_{i}$$v_{i}$$\ldots$$u_{l}$

.

  2. (2)

Between these end points, we can obtain a refinement of the living green and
red intervals. We label each interval of each color from left to right with an
index:

$v_{k}^{\prime}$$u_{k}$$v_{k+1}$$u_{k+1}$$\ldots$$u_{i}$$v_{i}$$\ldots$$u_{l}$121323$N$

Following the convention of step 2, the labels of each color are
$\\{1,\ldots,N\\}$. Each labeled interval with index $i$ is denoted by
$\delta_{i}$.

  3. (3)

Now we can group the labeled intervals by collecting all the labels $k$ on the
red intervals which satisfy the condition

$\max\\{\text{labels of the green intervals to the left of }k\\}=k.$

We denote the set by $Q$ and the labels by ${k_{1},\ldots,k_{M}}$.

Thus we have proven the following proposition:

######  Proposition 7.

On an basepoint elemnt $w_{0}$, we can segment the green intervals into

$\displaystyle K_{1}$ $\displaystyle=(\Xi_{1},\ldots,\Xi_{k_{1}})$
$\displaystyle K_{2}$ $\displaystyle=(\Xi_{k_{1}+1},\ldots,\Xi_{k_{2}})$
$\displaystyle\ldots$ $\displaystyle K_{M}$
$\displaystyle=(\Xi_{k_{M-1}+1},\ldots,\Xi_{k_{M}}).$

The any permissible move moves $K_{i}$ as a whole to the corresponding red
interval $L_{i}$. Similar operations can be done for red intervals, and we can
define $L_{1},\ldots,L_{M}$ as the red intervals corresponding to each
$K_{i}$.

###  3.6. Dead Intervals

In this and the following sections, we will call the dead intervals to the
left of the living intervals the _head_ , and those dead intervals to the
right of the living intervals the _tail_. In this section, we describe the
head and tail intervals in all the three cases described in Section 3.2.

####  3.6.1. Head intervals for Cases I and II

In the cases I and II, we have $n\geq\sigma$, thus:

######  Proposition 8.

The _head_ can be constructed through the following procedure

$n-\sigma$$R_{1}$$R_{1}$$R_{2}$$R_{2}$$R_{3}$$R_{3}$$R_{4}=\emptyset$$\ldots$

Denote

  * •

the set of elements $i$ in $\\{1,\ldots,m\\}$ such that $w(i)\prec
w(m+n-\sigma)$ by $R_{1}$,

  * •

the set of elements $i$ in $\\{1,\ldots,m\\}$ such that $w(|R_{1}|)\prec
w(i)\prec w(m+n-\sigma+|R_{1}|)$ by $R_{2}$,

  * •

the set of elements $i$ in $\\{1,\ldots,m\\}$ such that
$w(|R_{1}|+|R_{2}|)\prec w(i)\prec w(m+n-\sigma+|R_{1}|+|R_{2}|)$ by $R_{3}$,

  * •

$\ldots$

  * •

the set of elements $i$ in $\\{1,\ldots,m\\}$ such that
$w(\sum_{j=1}^{k-1}|R_{j}|)\prec w(i)\prec
w(m+n-\sigma+\sum_{j=1}^{k-1}|R_{j}|)$ by $R_{k}$,

  * •

The head ends where $R_{k+1}=\emptyset$.

Then the set of all dead “light gray” elements in the head is the union of all
$R_{j}$’s.

###### Proof.

This theorem is a consequence of Proposition 6. There cannot be any dead
element on the right of a living element. Therefore, all $i$’s such that
$w(i)\prec w(m+n-\sigma)$ are dead, and so are all the elements $m+n-\sigma+i$
such that $i\in R_{1}$. As a consequence, all elements $i$ such that
$w(i)\prec w(m+n-\sigma+|R_{1}|)$ are also dead. We can continue with this
process, until any of the following three situations occur:

  1. (1)

All “light gray” elements from $\\{1,\ldots,m\\}$ are finished.

  2. (2)

All “dark gray” elements from $\\{m+1,\ldots,n\\}$ are finished.

  3. (3)

$R_{k+1}=\emptyset$ for some $k$.

If $(3)$ happens before $(1)$ or $(2)$ could happen, one will be able to
construct a permissible move on the living intervals to the right of the
element $\sum_{j=1}^{k}|R_{j}|+1$. ∎

######  Lemma 9.

If for any $1\leq j\leq k$ such that $R_{j}\neq\emptyset$, we have

  * •

$i\leq w(i)<n-\sigma+i$ for $i\in R_{1}$,

  * •

$n-\sigma+i\leq w(i)<n-\sigma+|R_{1}|+i$ for $i\in R_{2}$,

  * •

…

  * •

$n-\sigma+\sum_{j=1}^{k-2}|R_{j}|+i\leq
w(i)<n-\sigma+\sum_{j=1}^{k-1}|R_{j}|+i$ for $i\in R_{k}$.

###### Proof.

The inequalities follows simply from the enumeration of intervals. The strict
inequality is due to the nonemptyness of $R_{j}$. ∎

####  3.6.2. Head intervals for Case III

In Case III, we can construct the head following a similar process. In this
case, we have $\sigma>n$, and the dead intervals can be obtained via the
process described in the following proposition:

######  Proposition 10.

The _head_ can be constructed through the following procedure in Case III:

$\sigma-n$$R_{1}^{\prime}$$R_{1}^{\prime}$$R_{2}^{\prime}$$R_{2}^{\prime}$$R_{3}^{\prime}$$R_{3}^{\prime}$$R_{4}^{\prime}=\emptyset$$\ldots$

Denote

  * •

the set of elements $i$ in $\\{1,\ldots,n\\}$ such that $w(m+i)\prec
w(\sigma-n)$ by $R_{1}^{\prime}$,

  * •

the set of elements $i$ in $\\{1,\ldots,n\\}$ such that
$w(m+|R_{1}^{\prime}|)\prec w(m+i)\prec w(\sigma-n+|R_{1}^{\prime}|)$ by
$R_{2}^{\prime}$,

  * •

the set of elements $i$ in $\\{1,\ldots,n\\}$ such that
$w(m+|R_{1}^{\prime}|+|R_{2}^{\prime}|)\prec w(m+i)\prec
w(\sigma-n+|R_{1}^{\prime}|+|R_{2}^{\prime}|)$ by $R_{3}^{\prime}$,

  * •

$\ldots$

  * •

the set of elements $i$ in $\\{1,\ldots,n\\}$ such that
$w(\sum_{j=1}^{k-1}|R_{j}^{\prime}|)\prec w(m+i)\prec
w(\sigma-n+\sum_{j=1}^{k-1}|R_{j}^{\prime}|)$ by $R_{k}^{\prime}$,

  * •

The head ends where $R^{\prime}_{k+1}=\emptyset$.

Then the set of all dead “dark gray” elements in the head is the union of all
$R^{\prime}_{j}$’s.

###### Proof.

In Case III, since $\sigma>n$, all the elements between 1 and $\sigma-n$ are
dead, as well as the elements from $\\{m+1,\ldots,m+n\\}$ interlacing the
elements $1,\ldots,\sigma-n$. Therefore, we can imitate the process in
Proposition 8 to exhaust the dead elements in the head for Case III. The
process finishes when one of the following three situations occur:

  1. (1)

All “light gray” elements from $\\{1,\ldots,m\\}$ are finished.

  2. (2)

All “dark gray” elements from $\\{m+1,\ldots,n\\}$ are finished.

  3. (3)

$R_{k+1}^{\prime}=\emptyset$ for some $k$.

Similar to Proposition 8, if (3) occurs before (1) or (2), we can construct a
permissible move involving the element $\sum_{j=1}^{k}|R_{j}^{\prime}|+1$. ∎

The lengths of the $R_{j}^{\prime}$ intervals satisfy the following lemma:

######  Lemma 11.

If for any $1\leq j\leq k$, $R^{\prime}_{i}\neq\emptyset$, we have

  * •

$i\leq w(i)\leq i+|R_{1}^{\prime}|$ for $i\in\\{1,\ldots,\sigma-n\\}$

  * •

$i+|R_{1}^{\prime}|\leq w(i)\leq i+|R_{1}^{\prime}|+|R_{2}^{\prime}|$ for
$i\in\sigma-n+R_{1}^{\prime}$,

  * •

$i+|R_{1}^{\prime}|+|R_{2}^{\prime}|\leq w(i)\leq
i+|R_{1}^{\prime}|+|R_{2}^{\prime}|+|R_{3}^{\prime}|$ for
$i\in\sigma-n+R_{2}^{\prime}$,

  * •

…

  * •

$i+\sum_{j=1}^{k-1}|R_{j}^{\prime}|\leq w(i)\leq
i+\sum_{j=1}^{k}|R_{j}^{\prime}|$ for $i\in\sigma-n+R_{k-1}^{\prime}$,

  * •

$w(i)=i+\sum_{j=1}^{k}|R_{j}^{\prime}|$ for $i\in\sigma-n+R_{k}^{\prime}$,

At the right endpoint of each interval, the left part of the inequalities are
strict.

###### Proof.

The lemma follows easily from counting. The strict inequality at the right
endpoint of each interval follows from the nonzeroness of $|R_{j}^{\prime}|$
for $j\leq k$. ∎

####  3.6.3. Tail Intervals for Case I, III

We can construct the tail of the dead elements symmetric to the situation
described in Lemma 10:

$(\sigma+1,\ldots,m)$$S_{1}$$S_{1}$$S_{2}$$S_{3}$$S_{2}$$S_{3}$$\ldots$

######  Proposition 12.

The _tail_ can be constructed through the following algorithm: Denote

  * •

the elements $j$ in $\\{m+1,\ldots,m+n\\}$ such that $w(\sigma+1)\prec w(j)$
by $S_{1}$,

  * •

the elements $j$ in $\\{m+1,\ldots,m+n\\}$ such that $w(\sigma-|S_{1}|+1)\prec
w(j)\prec w(\sigma)$ by $S_{2}$,

  * •

the elements $j$ in $\\{m+1,\ldots,m+n\\}$ such that
$w(\sigma-|S_{1}|-|S_{2}|+1)\prec w(j)\prec w(\sigma-|S_{1}|)$ by $S_{3}$,

  * •

$\ldots$

  * •

the elements $j$ in $\\{m+1,\ldots,m+n\\}$ such that
$w(\sigma-\sum_{j=1}^{k-1}|S_{j}|+1)\prec w(j)\prec
w(\sigma-\sum_{j=1}^{k-2}|S_{j}|)$ by $S_{k}$

There exists a minimal $k$ such that $S_{k+1}=\emptyset$, and the set of all
dead light gray elements in the tail is the union of all nonempty $S_{j}$.

###### Proof.

In this situation, all the elements between $\sigma+1$ and $m$ are dead.
Therefore, the dead elements coming from $\\{m+1,\ldots,m+n\\}$ are those
which interlace with the elements $\sigma+1,\ldots,m$. Following the same
procedure as in the proof of Proposition 10 from right to left along the tail,
we can follow the process described in the statement of the proposition until
one of the following three situations occur

  1. (1)

All “light gray” elements from $\\{1,\ldots,m\\}$ are finished.

  2. (2)

All “dark gray” elements from $\\{m+1,\ldots,m+n\\}$ are finished.

  3. (3)

$S_{k+1}=\emptyset$ for some $k$.

If (3) happens before (1) and (2), the adjacent first element to the left of
$S_{k}$ is the last living element. ∎

Denoting the index of the right-most living element from $(m+1)\ldots(m+n)$ by
$q$, we have the following lemma

######  Lemma 13.

For any $1\leq j\leq k$ such that $S_{j}\neq\emptyset$, we have

  * •

$w(i)=q-m+i$ for $i\in S_{k}$,

  * •

$w(i)\geq q-m+i$ for $i\in S_{k-1}$,

  * •

$w(i)\geq q-m+|S_{k}|+i$ for $i\in S_{k-2}$,

  * •

…

  * •

$w(i)\geq q-m+\sum_{j=3}^{k}|S_{j}|+i$ for $i\in S_{1}$.

###### Proof.

The proof follows from the similar enumeration technique as in Lemma 11. ∎

####  3.6.4. Tail Intervals for Case II

We can construct the tail of the dead elements symmetric to the situation in
Proposition 8:

$(\sigma-m+1,\ldots,n)$$S_{1}^{\prime}$$S_{1}^{\prime}$$S_{2}^{\prime}$$S_{3}^{\prime}$$S_{2}^{\prime}$$S_{3}^{\prime}$$\ldots$

######  Proposition 14.

The _tail_ can be constructed through the following algorithm: Denote

  * •

the elements $j$ in $\\{1,\ldots,m\\}$ such that $w(\sigma-m+1)\prec w(j)$ by
$S_{1}^{\prime}$,

  * •

the elements $j$ in $\\{1,\ldots,m\\}$ such that
$w(\sigma-m-|S_{1}^{\prime}|+1)\prec w(j)\prec w(\sigma-m)$ by
$S_{2}^{\prime}$,

  * •

the elements $j$ in $\\{1,\ldots,m\\}$ such that
$w(\sigma-m-|S_{1}^{\prime}|-|S_{2}^{\prime}|+1)\prec w(j)\prec
w(\sigma-m-|S_{1}^{\prime}|)$ by $S_{3}^{\prime}$,

  * •

$\ldots$

  * •

the elements $j$ in $\\{1,\ldots,m\\}$ such that
$w(\sigma-m-\sum_{j=1}^{k-1}|S_{j}^{\prime}|+1)\prec w(j)\prec
w(\sigma-m-\sum_{j=1}^{k-2}|S_{j}^{\prime}|)$ by $S_{k}^{\prime}$

There exists a minimal $k$ such that $S_{k+1}^{\prime}=\emptyset$, and the set
of all dead light gray elements in the tail is the union of all nonempty
$S_{j}^{\prime}$.

###### Proof.

The proof follows the same procedure as in the proof of Proposition 8 from
right to left along the tail. Similarly, the algorithm terminates when one of
the following three situations occur:

  1. (1)

All “light gray” elements from $\\{1,\ldots,m\\}$ are finished.

  2. (2)

All “dark gray” elements from $\\{m+1,\ldots,m+n\\}$ are finished.

  3. (3)

$S^{\prime}_{k+1}=\emptyset$ for some $k$.

The adjacent element to the left of $S_{k}^{\prime}$ does not interlace with
any dead interval, and thus is an living element. ∎

Denoting the index of the right-most living element from $1\ldots m$ by
$q^{\prime}$, we have the following lemma

######  Lemma 15.

For any $1\leq i\leq n$ such that $S_{j}^{\prime}\neq\emptyset$, we have

  * •

$n-\sigma+q^{\prime}+|S_{n}^{\prime}|+i<w(i)\leq
n-\sigma+q^{\prime}+|S_{n}^{\prime}|+|S_{n-1}^{\prime}|+i$ for $i\in
S^{\prime}_{n}$,

  * •

$n-\sigma+q^{\prime}+|S_{n}^{\prime}|+|S_{n-1}^{\prime}|+i<w(i)\leq
n-\sigma+q^{\prime}+\sum_{j={n-2}}^{n}|S_{j}^{\prime}|+i$ for $i\in
S^{\prime}_{n-1}$,

  * •

…

  * •

$w(i)>n-\sigma+q^{\prime}+\sum_{j=1}^{n}|S_{j}^{\prime}|+i$ for $i\in
S_{1}^{\prime}$.

###### Proof.

The proof follows from the similar enumeration technique as in Lemma 9. ∎

##  4\. $L$-function Calculations

###  4.1. $L$-functions

For self-dual cuspidal automorphic representations $\tau$, the Rankin-Selberg
$L$-functions $L(s,\tau\times\hat{\tau})$ satisfies the following functional
equation

$L(s,\tau\times\hat{\tau})=\epsilon(s,\tau\times\hat{\tau})L(1-s,\tau\times\hat{\tau})$

with the epsilon factor $\epsilon=c^{s-\frac{1}{2}}$ for some rational number
$c$. Without incurring any confusion, we will simply denote
$L(s,\tau\times\hat{\tau})$ and $\epsilon(s,\tau\times\hat{\tau})$ by $L(s)$
and $\epsilon(s)$, respectively. By [MW89], the normalizing factor defined in
the previous section is equal to the product

$\displaystyle r(w,\underline{s})$
$\displaystyle=\prod_{\begin{subarray}{c}i<j\\\
w(i)>w(j)\end{subarray}}\frac{L(\nu_{i}-\nu_{j})}{L(1+\nu_{i}-\nu_{j})\epsilon(\nu_{i}-\nu_{j},\tau\times\hat{\tau})}.$

In the situation we are interested in, we specialize the parameters to the
following values:

$\displaystyle\nu_{i}$ $\displaystyle=\frac{1-m}{2}+i-1+s_{1},$
$\displaystyle\nu_{j}$ $\displaystyle=\frac{1-n}{2}+j-m-1+s_{2},$
$\displaystyle s$ $\displaystyle=s_{1}-s_{2}.$

For any element $w$ representable by the diagram

$v_{0}$$u_{1}$$v_{1}$$u_{2}$$\ldots$$u_{i}$$v_{i}$$\ldots$

,

the collection of inverted pairs $\mathrm{inv}(w)=\\{(i,j)\mid
i<j,w(i)>w(j)\\}$ is the set

$\mathrm{inv}(w)=(v_{1}\times u_{1})\bigcup(v_{2}\times(u_{1}\cup
u_{2}))\bigcup(v_{3}\times(u_{1}\cup u_{2}\cup u_{3}))\bigcup\ldots.$

Since $\nu_{i}-\nu_{j}=s+\frac{m+n}{2}+i-j$, For any
$w\in\mathbb{S}_{m}\times\mathbb{S}_{n}\backslash\mathbb{S}_{m+n}$, we denote
by $i_{w}$ the smallest $i$ such that there exists a $j>i$ with $w(j)<w(i)$.
For each fixed $i\geq i_{w}$, denoting the largest $j$ such that
$(i,j)\in\mathrm{inv}(w)$ by $j_{i}$, the product corresponding to all $j$’s
such that $(i,j)\in\mathrm{inv}(w)$ is

$\displaystyle\prod_{w(j)<w(i)}\frac{L(\nu_{i}-\nu_{j})}{L(1+\nu_{i}-\nu_{j})\epsilon(\nu_{i}-\nu_{j})}$
$\displaystyle=$
$\displaystyle\frac{L(s+\frac{m+n}{2}+i-(m+1))}{L(s+\frac{m+n}{2}+i-(m+1)+1)}\times\ldots\times\frac{L(s+\frac{m+n}{2}+i-j_{i})}{L(s+\frac{m+n}{2}+i-j_{i}+1)}\times\frac{1}{\text{products
of }\epsilon\text{'s}}$ $\displaystyle=$
$\displaystyle\frac{L(s+\frac{m+n}{2}+i-j_{i})}{L(s-\frac{m-n}{2}+i)}\times\frac{1}{\text{products
of }\epsilon\text{'s}}.$

Replacing $s$ by $s=\frac{m+n}{2}-\sigma+t$ and taking into account that

$w(i)=i+j_{i}-m,$

after plugging in all the parameters, the product of $\epsilon$-factors can be
expressed as

$\rho_{i}(w,t)=c^{\sum_{k=m+1}^{j_{i}}(s+\frac{m+n-1}{2}+i-k)}=c^{\left(w(i)-i\right)(n-\sigma+i-1+t)-\frac{1}{2}\left(w(i)-i\right){}^{2}},$

and the normalization factor of the intertwining operator is thus equal to

(5)
$r(w,t)=\prod_{i=i_{w}}^{m}\frac{L(n-\sigma+2i-w(i)+t)}{L(n-\sigma+i+t)}\rho_{i}(w,t)^{-1}.$

For any $\sigma$, we will be only interested in the analytic property of
$r(w,t)$ in the region $t\geq 0$, in which there will be no cancellation of
the poles of the denominator by the zeros of the denominator in the critical
strip.

The following lemma is a generalization of [HM15, Lemma 8.5]:

######  Lemma 16.

For any section $f\in I(\tau,\underline{s})$, the images of the normalized
intertwining operator $N(w,\underline{s})f$ are equal as long as $w$ lies in
the same orbit $\mathcal{O}_{\underline{s}}$.

###### Proof.

We would like to compare two $N(w_{1},\underline{s})$ and
$N(w_{2},\underline{s})$ for different $w_{1},w_{2}$ with isomorphic twists

$\rho_{m,n}(\tau,\underline{s})^{w_{1}}\cong\rho_{m,n}(\tau,\underline{s})^{w_{2}}.$

The normalized intertwining operators are holomorphic in the dominant chamber,
and each normalized intertwining operator

$N(w_{i},\underline{s}):\mathrm{Ind}_{P}^{G}\rho_{m,n}(\tau,\underline{s})\longrightarrow\mathrm{Ind}_{P}^{G}\rho_{m,n}(\tau,\underline{s})^{w_{i}},$

is invertible. Since the cuspidal representations $\tau$ are unitary, and by
[KS71, Proposition 38] the normalized intertwining operators satisfy

  1. (1)

$N(w_{1}w_{2},\underline{s})=N(w_{1},w_{2}\underline{s})N(w_{2},\underline{s})$

  2. (2)

$N(w,\underline{s})^{*}=N(w^{-1},-\overline{\underline{s}})$,

  3. (3)

$N(w,\underline{s})$ is unitary for $s=-\overline{s}$.

Since we can factorize the elements $w_{2}=uw_{1}$ such that $u$ is a
permissible move, which is known to be an involution. The intertwining
operator

$N(u,w_{2}\underline{s}):\mathrm{Ind}_{P}^{G}\rho_{m,n}(\tau,\underline{s})^{w_{2}}\longrightarrow\mathrm{Ind}_{P}^{G}\rho_{m,n}(\tau,\underline{s})^{w_{1}}$

is thus a self-adjoint, unitary involution. The operator
$N(u,w_{2}\underline{s})$ can be factorized further into:

$N(u,w_{2}\underline{s})=N(v^{-1},vw_{2}\underline{s})N(\iota,vw_{2}\underline{s})N(v,w_{2}\underline{s})$

where $v$ is the Weyl group element sending each pairs of corresponding blocks
in the permissible move to adjacent blocks, and the operator
$N(\iota,vw_{2}\underline{s})$ is the transposition of these adjacent blocks.
In fact, the intertwining operator $N(\iota,vw_{2}\underline{s})$ is a product
of self-intertwining operators $N(0)$ of
$\mathrm{Ind}_{P_{[a,a]}}^{G_{2a}}\left(\tau\boxtimes\tau\right)$
corresponding to the Weyl group element swapping the two Levi blocks. By
[KS88, Proposition 6.3], such an operator $N(0)=1$, and thus
$N(u,w_{2}\underline{s})=1$. ∎

As a result of this lemma, we can collect the normalization factors in the
constant term of $E^{m,n}(\otimes_{v}f_{v},\underline{s})$ into sums over
orbits, as in

(6)
$c_{U_{m+n}}E^{m,n}(\otimes_{v}f_{v},\underline{s})=\sum_{\mathcal{O}_{\underline{s}}(w)}\left(\sum_{w^{\prime}\in\mathcal{O}_{\underline{s}}(w)}r(w^{\prime},\underline{s})\right)\bigotimes_{v}N_{v}(w^{\prime},\underline{s})f_{v}.$

Each sum over an orbit is denoted by
$R(w,\underline{s})=\sum_{w^{\prime}\in\mathcal{O}_{\underline{s}}(w)}r(w^{\prime},\underline{s})$.
We will discuss the cancellations of poles in the sum $R(w,\underline{s})$.

###  4.2. Calculation on a Single Living Interval

Now we calculate the sums on a single living interval. In this section,
following Proposition 7, for each living interval $K_{i}$ with a starting
interval $\Xi_{k_{i-1}+1}$ and ending interval $\Xi_{k_{i}}$, we denote these
individual green intervals by

$\Xi_{k_{i-1}+1},\Xi_{k_{i-1}+2},\ldots,\Xi_{k_{i}}$

with $|\Xi_{k_{i-1}+1}|+\ldots+|\Xi_{k_{i}}|=|K_{i}|$. Their corresponding red
intervals are denoted by

$\Sigma_{k_{i-1}+1},\Sigma_{k_{i-1}+2},\ldots,\Sigma_{k_{i}}.$

Denoting by $A_{p}(t)$ the product of
$L(n-\sigma+2i-w(i)+t)\rho_{i}(w,t)^{-1}$ along each $K_{p}$, and $B_{p}(t)$
the product of $L(n-\sigma+2i-w(m+n-\sigma+i)+t)\rho_{i}(w,t)^{-1}$ along each
$L_{p}$. Since $i=s(\Xi_{k_{p-1}+1})+r$ and
$w(i)=n-\sigma+2s(\Xi_{k_{p-1}+1})+\sum_{\Sigma_{k_{p-1}+s}\prec
w(i)}|\Sigma_{k_{p-1}+s}|+r$, then

$n-\sigma+2i-w(i)=r-\sum_{\Sigma_{k_{p-1}+s}\prec w(i)}|\Sigma_{k_{p-1}+s}|.$

After swapping the corresponding green and red intervals, we obtain

$w(m+n-\sigma+i)=n-\sigma+2s(\Xi_{k_{p-1}+1})+\sum_{\Xi_{k_{p-1}+s}\prec
w(m+n-\sigma+i)}|\Xi_{k_{p-1}+s}|+r$

and thus

$n-\sigma+2i-w(m+n-\sigma+i)=r-\sum_{\Xi_{k_{p-1}+s}\prec
w(m+n-\sigma+i)}|\Xi_{k_{p-1}+s}|.$

###  4.3. Product of $L$ Factors

In this section we will calculate $A_{p}(t)$ and $B_{p}(t)$. We understand
that some of the $\Xi_{i},\Sigma_{i}$ intervals group together on the same
$v,u$ interval. If we denote the endpoints of these intervals by

$s(\Xi_{k_{p-1}+1})+\\{a_{1},\ldots,a_{r}\\},$

then the corresponding product of $L$-functions in $A_{p}(t)$ and $B_{p}(t)$
are given by

$A_{p}^{\prime}(t)=L(1+t)\ldots L(a_{1}+t)L(a_{1}-b_{1}+1+t)\ldots
L(a_{2}-b_{1}+t)\ldots L(a_{r-1}-b_{r-1}+1+t)\ldots L(a_{r}-b_{r-1}+t)$

and

$B_{p}^{\prime}(t)=L(-a_{1}+1+t)\ldots
L(b_{1}-a_{1}+t)L(b_{1}-a_{2}+1+t)\ldots L(b_{2}-a_{2}+t)\ldots
L(b_{r-1}-a_{r}+1+t)\ldots L(b_{r}-a_{r}+t).$

We can compare these two products, and it turns out that:

######  Lemma 17.

The two products above satisfy

$A^{\prime}_{p}(t)=c^{\sum_{r}\left(\frac{\sum_{\Xi_{k_{p-1}+s}\prec
w(m+n-\sigma+i)}|\Xi_{k_{p-1}+s}|-\sum_{\Sigma_{k_{p-1}+s}\prec
w(i)}|\Sigma_{k_{p-1}+s}|}{2}\right)}B^{\prime}_{p}(-t).$

###### Proof.

Since each $b_{i}\prec a_{i}$, we can reorder the factors in
$A^{\prime}_{p}(t)$ and $B^{\prime}_{p}(-t)$. If we represent these factors in
$A^{\prime}_{p}(t)$ as intervals, and they can be illustrated in the following
diagram:

$L(1+t)\ldots L(a_{1}+t)$$L(a_{1}-b_{1}+1+t)\ldots
L(a_{2}-b_{1}+t)$$\ldots$$\ldots$$L(a_{r-1}-b_{r-1}+1+t)\ldots
L(a_{r}-b_{r-1}+t)$

,

while the factors in $B^{\prime}_{p}(t)$ correspond to the intervals

$L(-a_{1}+1+t)\ldots L(b_{1}-a_{1}+t)$$L(b_{1}-a_{2}+1+t)\ldots
L(b_{2}-a_{2}+t)$$\ldots$$\ldots$$L(b_{r-1}-a_{r}+1+t)\ldots L(b_{r}-a_{r}+t)$

between which a bijection can be established after sending each index $p$ to
$1-p$. By the functional equation of the Rankin-Selberg $L$-function
$L(s,\tau\times\hat{\tau})$, we have

$A^{\prime}_{p}(t)=c^{\sum_{r}\left(r-\sum_{\Sigma_{k_{p-1}+s}\prec
w(i)}|\Sigma_{k_{p-1}+s}|-1/2+t\right)}B^{\prime}_{p}(-t).$

Applying the functional equation again, we can see that

$1=c^{\sum_{r}\left(r-\sum_{\Sigma_{k_{p-1}+s}\leq
w(i)}|\Sigma_{k_{p-1}+s}|-1/2+t\right)}c^{\sum_{r}\left(r-\sum_{\Xi_{k_{p-1}+s}\leq
w(m+n-\sigma+i)}|\Xi_{k_{p-1}+s}|-1/2-t\right)}.$

Therefore,

$\sum_{r}\left(r-\frac{\sum_{\Sigma_{k_{p-1}+s}\prec
w(i)}|\Sigma_{k_{p-1}+s}|+\sum_{\Xi_{k_{p-1}+s}\prec
w(m+n-\sigma+i)}|\Xi_{k_{p-1}+s}|}{2}-\frac{1}{2}\right)=0$

and the original factor becomes

$A^{\prime}_{p}(t)=c^{\sum_{r}\left(\frac{\sum_{\Xi_{k_{p-1}+s}\prec
w(m+n-\sigma+i)}|\Xi_{k_{p-1}+s}|-\sum_{\Sigma_{k_{p-1}+s}\prec
w(i)}|\Sigma_{k_{p-1}+s}|}{2}\right)}B^{\prime}_{p}(-t).$

∎

###  4.4. Product of $\rho$-factors

This subsection calculates the cancellations of poles in $R(w,\underline{s})$
from the living intervals.

######  Proposition 18.

Along each living interval, $A_{p}(t)+B_{p}(t)$ is holomorphic.

###### Proof.

Each individual $\rho$-factor can be calculated with the following procedure:

For any $i$, we have

$\displaystyle\rho_{i}(w,t)$
$\displaystyle=c^{(w(i)-i)(n-\sigma+i-1+t)-\frac{1}{2}(w(i)-i)^{2}}$
$\displaystyle=c^{(n-\sigma+s(\Sigma_{k_{p-1}+1})+\sum_{i}|\Sigma_{i}|)(n-\sigma+i-1+t)-\frac{1}{2}(n-\sigma+s(\Sigma_{k_{p-1}+1})+\sum_{i}|\Sigma_{i}|)^{2}}$

and

$\displaystyle\rho^{\prime}_{i}(w,t)$
$\displaystyle=c^{(w(m+n-\sigma+i)-i)(n-\sigma+i-1+t)-\frac{1}{2}(w(m+n-\sigma+i)-i)^{2}}$
$\displaystyle=c^{(n-\sigma+s(\Sigma_{k_{p-1}+1})+\sum_{i}|\Xi_{i}|)(n-\sigma+i-1+t)-\frac{1}{2}(n-\sigma+s(\Sigma_{k_{p-1}+1})+\sum_{i}|\Xi_{i}|)^{2}}.$

The differences of their products is

$\displaystyle
c^{\sum_{i}(\sum_{j}(|\Sigma_{j}|-|\Xi_{j}|))(n-\sigma+s+r-1+t)-(\sum_{j}(|\Sigma_{j}|-|\Xi_{j}|)))(n-\sigma+s+\frac{|\Sigma_{i}|+|\Xi_{i}|}{2})}=c^{\sum_{i}(\sum_{j}(|\Sigma_{j}|-|\Xi_{j}|))(r-1+t-\frac{|\Sigma_{i}|+|\Xi_{i}|}{2})}$

Therefore,

$A_{p}(t)=A^{\prime}_{p}(t)\rho_{i}(w,t)^{-1}=A^{\prime}_{p}(t)c^{-(n-\sigma+s(\Sigma_{k_{p-1}+1})+\sum_{i}|\Sigma_{i}|)(n-\sigma+i-1+t)+\frac{1}{2}(n-\sigma+s(\Sigma_{k_{p-1}+1})+\sum_{i}|\Sigma_{i}|)^{2}}$

and

$B_{p}(t)=B^{\prime}_{p}(t)\rho^{\prime}_{i}(w,t)^{-1}=B^{\prime}_{p}(t)c^{-(n-\sigma+s(\Sigma_{k_{p-1}+1})+\sum_{i}|\Xi_{i}|)(n-\sigma+i-1+t)+\frac{1}{2}(n-\sigma+s(\Sigma_{k_{p-1}+1})+\sum_{i}|\Xi_{i}|)^{2}}.$

The sum can be reduced to

$\displaystyle A_{p}(t)+B_{p}(t)$
$\displaystyle=\left(\prod_{i}\rho_{i}^{\prime}(w,t)^{-1}\right)\left(B_{p}^{\prime}(-t)c^{\sum_{r}\left(\frac{\sum_{j}\Xi_{j}-\sum_{j}\Sigma_{j}}{2}\right)}c^{\sum_{r}\left(\sum_{j}(\Xi_{j}-\Sigma_{j})(r-1+t-\frac{\Sigma_{j}+\Xi_{j}}{2})\right)}+B_{p}^{\prime}(t)\right)$
$\displaystyle=\left(\prod_{i}\rho_{i}^{\prime}(w,t)^{-1}\right)\left(B_{p}^{\prime}(-t)c^{\sum_{r}\left(\sum_{j}(\Xi_{j}-\Sigma_{j})(r-\frac{1}{2}+t-\frac{\Sigma_{j}+\Xi_{j}}{2})\right)}+B_{p}^{\prime}(t)\right).$

The exponent summing over all
$\sum_{j}(\Xi_{j}-\Sigma_{j})\left(r-\frac{1}{2}+t-\frac{\Sigma_{j}+\Xi_{j}}{2}\right)$
is in fact equal to

$\displaystyle\sum_{j}(\Xi_{j}-\Sigma_{j})\left(r-\frac{1}{2}+t-\frac{\Sigma_{j}+\Xi_{j}}{2}\right)$
$\displaystyle=\sum_{i\in\text{living elements}}(w(i)-w(m+n-\sigma+i))$
$\displaystyle\left(n-\sigma+2i-\frac{w(i)+w(m+n-\sigma+i)+1}{2}+t\right).$

Noting that the individual terms in the summation above can be rearranged into
the form:

$\displaystyle-\sum_{i}\frac{\left(n-\sigma+2i-w(i)\right)^{2}-\left(n-\sigma+2i-w(m+n-\sigma+i)\right)^{2}}{2}$
$\displaystyle+$
$\displaystyle\sum_{i}\frac{\left(n-\sigma+2i-w(i)\right)-\left(n-\sigma+2i-w(m+n-\sigma+i)\right)}{2}$
$\displaystyle-t$
$\displaystyle\sum_{i}\left(\left(n-\sigma+2i-w(i)\right)-\left(n-\sigma+2i-w(m+n-\sigma+i)\right)\right).$

The first sum can be further rearranged into the form of
$k^{2}-(1-k)^{2}=2k-1$, which cancels out the second sum. Therefore,

$A_{p}(t)+B_{p}(t)=\left(\prod_{i}\rho_{i}^{\prime}(w,t)^{-1}\right)\left(c^{-t\alpha}B_{p}^{\prime}(-t)+B_{p}^{\prime}(t)\right).$

Since both $A_{p}^{\prime}(t)$ and $B_{p}^{\prime}(t)$ has a simple pole at
$t=0$, and since $c^{-t\alpha}$ is holomorphic with zeroth order coefficient
1, we conclude that $A_{p}(t)+B_{p}(t)$ is holomorphic. ∎

###  4.5. Other Factors and Proof of the Theorem

Now we finish the calculation of the sum

$R(w,\underline{s})=\sum_{w^{\prime}\in\mathcal{O}_{s}(w)}r(w^{\prime},\underline{s})$

outside of the contributions of the living intervals. Recall that $i_{w}$ is
the first index $i$ such that there exist a $j>i$ such that $w(i)>w(j)$. The
location of the first living interval may affect $i_{w}$ when $1$ is a living
element. When that happens, we are forced to require $m+n-\sigma+1=m+1$, in
which case $n=\sigma$.

$v_{0}$$u_{1}$$v_{1}$$u_{2}$$\ldots$$u_{i}$$v_{i}$$\ldots$$u_{l}$

In this case, if $1$ is a living element for $w$, $R(w,\underline{s})$ can be
separated into two terms $R^{\prime}(t)$ and $R^{\prime\prime}(t)$ which
consist of the products of $L$-function factors with $i_{w}^{\prime}$ for
$w^{\prime}\in\mathcal{O}_{s}(w)$ the same as the $i_{w_{0}}$ of the basepoint
element $w_{0}$ or not, respectively.

$R^{\prime}(t)=\sum_{\begin{subarray}{c}w^{\prime}\in\mathcal{O}_{s}(w)\\\
i_{w^{\prime}}=i_{w_{0}}\end{subarray}}r(w^{\prime},\underline{s})=\frac{\prod_{i\in\mathrm{head}\cup\mathrm{tail}}L(n-\sigma+2i-w_{0}(i)+t)\rho_{i}^{-1}(w_{0},t)}{\prod_{i=v_{0}+1}^{m}L(n-\sigma+i+t)}\prod_{i=1}^{r}(A_{i}(t)+B_{i}(t)).$

and the sum of the terms $r(w^{\prime},s)$ satisfying $i_{w^{\prime}}>1$ by

$R^{\prime\prime}(t)=\sum_{\begin{subarray}{c}w^{\prime}\in\mathcal{O}_{s}(w)\\\
i_{w^{\prime}}\neq
i_{w_{0}}\end{subarray}}r(w^{\prime},\underline{s})=\frac{\prod_{i\in\mathrm{head}\cup\mathrm{tail}}L(n-\sigma+2i-w_{0}(i)+t)\rho_{i}^{-1}(w_{0},t)}{\prod_{i=1}^{m}L(n-\sigma+i+t)}B_{0}(t)\prod_{i=1}^{r}(A_{i}(t)+B_{i}(t)),$

then

$R(w,\underline{s})=R^{\prime}(t)+R^{\prime\prime}(t).$

In this case, there are no indices to the left of all the movable indices. The
denominator is the product $L(1+t)\ldots L(m+t)$, and the tail does not
contribute to any poles since $w_{0}(i)<2i$ on tails.

On the other hand, if $1$ is not a movable element, then $R(w,\underline{s})$
simply takes the form

$R(w,\underline{s})=\sum_{\begin{subarray}{c}w^{\prime}\in\mathcal{O}_{s}(w)\end{subarray}}r(w^{\prime},\underline{s})=\frac{\prod_{i\in\mathrm{head}\cup\mathrm{tail}}L(n-\sigma+2i-w(i)+t)\rho_{i}^{-1}(w,t)}{\prod_{i=1}^{m}L(n-\sigma+i+t)}\prod_{i=1}^{r}(A_{i}(t)+B_{i}(t)).$

We will have to calculate the contribution from the head and the tails
separately.

####  4.5.1. Head Factors

Now we calculate the contribution to the poles from the head terms in the
numerator. In Case I and II, by Lemma 9, we have

  * •

$n-\sigma+2i-w(i)>i$ for $i\in R_{1}$,

  * •

$n-\sigma+2i-w(i)>i-|R_{1}|$ for $i\in R_{2}$,

  * •

$\ldots$

  * •

$n-\sigma+2i-w(i)>i-\sum_{j=1}^{k-1}|R_{j}|$ for $i\in R_{k}$.

Therefore, since the head terms can be expressed as

$\prod_{j=1}^{k}\prod_{i\in R_{j}}L(n-\sigma+2i-w(i)+t),$

the point $t=0$ is not a pole because all the $n-\sigma+2i-w(i)>1$ for
$i\in\cup_{j=1}^{k}R_{j}$. For Case III, by Lemma 11, we have

  * •

$n-\sigma+2i-w(i)\leq i-(\sigma-n)$ for $i\in\\{1,\ldots,\sigma-n\\}$, and
when $i=\sigma-n$ the inequality is strict,

  * •

$n-\sigma+2i-w(i)\leq i-(\sigma-n)-|R_{1}^{\prime}|$ for
$i\in\sigma-n+R_{1}^{\prime}$, and when $i=\sigma-n+r(R_{1}^{\prime})$ the
inequality is strict,

  * •

$\ldots$

  * •

$n-\sigma+2i-w(i)=i-(\sigma-n)-\sum_{j=1}^{k}|R_{j}^{\prime}|$ for
$i\in\sigma-n+R_{k}^{\prime}$, and when $i=\sigma-n+r(R_{k}^{\prime})$ we have
$n-\sigma+2i-w(i)=0$.

Therefore, in Case III when $\sigma>n$, the only possible contribution to the
pole on the numerator occurs when $i=\sigma-n+r(R_{k}^{\prime})$, and when
$\sigma=n$ there will be no head factors. However, this pole will be cancelled
by the denominator when $\sigma\geq n$ as in Case III since we allow
$i=\sigma-n$ and $i=\sigma-n+1$ in the factors of the denominator: the
denominator $\prod_{i=1}^{m}L(n-\sigma+i+t)$ has a pole of order 2 at $t=0$
when $\sigma>n$, and a pole of order 1 at $t=0$ when $\sigma=n$. In the case
when there are head factors, the contribution to the pole $t=0$ on the
numerator will be cancelled by the pole on the denominator.

####  4.5.2. Tail Factors

In the Cases I and III, the contribution to the poles from the tail factors
can be calculated with Lemma 13. Since by Lemma 13, on each interval $S_{k}$,
we have

  * •

$n-\sigma+2i-w(i)=i-(q-(m+n-\sigma))$ for $i\in S_{k}$,

  * •

$n-\sigma+2i-w(i)\geq i-(q-(m+n-\sigma))>1$ for $i\in S_{k-1}$,

  * •

$\ldots$

  * •

$n-\sigma+2i-w(i)>i-(q-(m+n-\sigma))-\sum_{j=3}^{k}|S_{j}|>1$ for $i\in
S_{1}$,

where the strict inequality is due to the nonvanishing of $|S_{j}|$, it turns
out that the only possible contribution to the pole at $t=0$ from the tail
factors occurs at $i=(q-(m+n-\sigma))+1$, which is the only contribution to
the poles of $R(w,\underline{s})$. However, as was discussed in the previous
subsection, in the cases when $\sigma>n$ or when $\sigma=n$ and $m>n$, this
pole will be cancelled by the denominator. When $m=n=\sigma$, there will be no
head or tail factors. On the other hand, in Case II, by Lemma 15, we have

  * •

$n-\sigma+2i-w(i)<i-(q^{\prime}+|S_{n}^{\prime}|)<0$ for $i\in
S_{n}^{\prime}$,

  * •

$n-\sigma+2i-w(i)<i-(q^{\prime}+|S_{n}^{\prime}|+|S_{n-1}^{\prime}|)<0$ for
$i\in S_{n-1}^{\prime}$,

  * •

$\ldots$

  * •

$n-\sigma+2i-w(i)<i-(q^{\prime}+\sum_{j=1}^{n}|S_{j}^{\prime}|)<0$ for $i\in
S_{1}^{\prime}$.

Therefore, there will be no contribution to the pole from the numerator of
$R(w,\underline{s})$. Thus, the tail factors will contribute to the poles of
$R(w,\underline{s})$ if and only if $0\leq\sigma<\min\\{m,n\\}$.

Thus, we have concluded the proof of Theorem A that the possible simple poles
occur only when $\sigma\in\\{0,1,\ldots,\min\\{m,n\\}-1\\}$.

###  4.6. The Residue of $E(\cdot,\underline{s})$

Fixing any $\sigma\in\\{0,1,\ldots,\min\\{m,n\\}-1\\}$, now we will prove
Corollary B of Theorem A which describes the residue of
$E(\cdot,\underline{s})$ at $s=s_{1}-s_{2}=\frac{m+n}{2}-\sigma$.

###### Proof.

For the character

$\underline{\Lambda}=(s_{1},\ldots,s_{m};t_{1},\ldots,t_{n}),$

denote the full Eisenstein series induced from $P_{[1,\ldots,1]}$ by

$I(\underline{\Lambda})=\mathrm{Ind}_{P_{m+n}}^{G_{m+n}}\left(\tau|\cdot|^{s_{1}}\boxtimes\ldots\boxtimes\tau|\cdot|^{s_{m}}\boxtimes\tau|\cdot|^{t_{1}}\boxtimes\ldots\boxtimes\tau|\cdot|^{t_{n}}\right).$

Assuming $\sigma\in\\{0,1\ldots,\min\\{n_{1},n_{2}\\}-1\\}$, we define the
following four Weyl group elements:

$\displaystyle w_{1}:$
$\displaystyle(1,2,\ldots,m;m+1,m+2,\ldots,m+n)\mapsto(m,m-1,\ldots,1;m+n,m+n-1,\ldots,m+1)$
$\displaystyle w_{\sigma}:$
$\displaystyle(1,2,\ldots,m;m+1,m+2,\ldots,m+n)\mapsto(m+1,\ldots,m+n-\sigma;1,\ldots,m;m+n-\sigma+1,\ldots,m+n)$
$\displaystyle w_{\sigma}^{\prime}:$
$\displaystyle(1,2,\ldots,m;m+1,m+2,\ldots,m+n)\mapsto(1,2,\ldots,m;m+\sigma+1,\ldots,m+n;m+1,\ldots,m+\sigma)$
$\displaystyle w_{1}^{\prime}:$
$\displaystyle(1,2,\ldots,m;m+1,m+2,\ldots,m+n)\mapsto(m+n-\sigma,\ldots,1;m+n-\sigma+1,\ldots,m+n).$

These four elements satisfy the relation
$w_{\sigma}w_{1}=w_{1}^{\prime}w_{\sigma}^{\prime}$. Consider following four
intertwining operators corresponding to these Weyl group elements:

By [MW89], for example, the image of the normalized intertwining operator
$N(w_{\sigma}w_{1},\underline{s})$ in $I(w_{\sigma}w_{1}\underline{\Lambda})$
is the representation

$I_{\sigma}=\mathrm{Ind}_{P_{[m,n]}}^{G_{m+n}}\left(\Delta(\tau,m+n-\sigma)|\cdot|^{-\frac{m-n}{2}}\boxtimes\Delta(\tau,\sigma)|\cdot|^{\frac{m-n}{2}}\right),$

which is irreducible if $\sigma\in\\{0,1,\ldots,\min\\{m,n\\}-1\\}$. Since
$w_{\sigma}w_{1}=w_{1}^{\prime}w_{\sigma}^{\prime}$, by the properties of the
normalized intertwining operators

$N(w_{1},\underline{\Lambda})=N(w_{\sigma}^{-1},w_{1}^{\prime}w_{\sigma}^{\prime}\underline{\Lambda})N(w_{1}^{\prime},w_{\sigma}^{\prime}\underline{\Lambda})N(w_{\sigma}^{\prime},\underline{\Lambda}),$

since $N(w_{1},\underline{\Lambda})\neq 0$, we have
$N(w_{\sigma}w_{1},\underline{\Lambda})\neq 0$. The image of the intertwining
operator $N(w_{1},\underline{\Lambda})$ is isomorphic to the image
$I(\tau,\underline{s})$ of the residue operator

$\mathrm{Res}_{(-\underline{\lambda_{m}}+s_{1},-\underline{\lambda_{n}}+s_{2})}=\prod_{1\leq
i\leq m-1}(s_{i}-s_{i+1}-1)\prod_{1\leq j\leq
n-1}(t_{j}-t_{j+1}-1)E(\cdot,\underline{\Lambda})\mid_{\underline{\Lambda}\to(-\underline{\lambda_{m}}+s_{1},-\underline{\lambda_{n}}+s_{2})}:I(\underline{\Lambda})\rightarrow
I(w_{1}\underline{\Lambda})$

as a submodule in $I(w_{1}\underline{\Lambda})$. By calculating the poles of
the normalized intertwining operator $N(w_{\sigma},w_{1}\underline{\Lambda})$,
the residue operator
$\mathrm{Res}_{s_{1}-s_{2}=\frac{m+n}{2}-\sigma}E(\cdot,\underline{s})$ kills
all the $M(w,\underline{\Lambda})f$ terms in the constant term of
$E(\cdot,\underline{\Lambda})$ except for those corresponding to the Weyl
group element $w=w_{2}w_{1}$ with $w_{2}$ satisfying

$w_{2}(m+1)<w_{2}(m+2)<\ldots<w_{2}(m+n-\sigma)<w_{2}(1)<w_{2}(2)<\ldots<w_{2}(m).$

as well as (3). For any such $w_{2}$,
$N(w_{2}w_{\sigma}^{-1},w_{\sigma}w_{1}\underline{\Lambda})$ maps $I_{\sigma}$
either isomorphically onto its image or to 0, and thus, the constant term of
$\mathrm{Res}_{s_{1}-s_{2}=\frac{m+n}{2}-\sigma}\mathrm{Res}_{(-\underline{\lambda_{m}}+s_{1},-\underline{\lambda_{n}}+s_{2})}E(f,\underline{\Lambda})$
is equal to the constant term of the Eisenstein series constructed from
$M(w_{\sigma}w_{1},w_{\sigma}w_{1}\underline{\Lambda})f$ as a vector in
$I_{\sigma}$ up to a nonzero constant scalar. Thus, the automorphic
representation generated by
$\mathrm{Res}_{s_{1}-s_{2}=\frac{m+n}{2}-\sigma}\mathrm{Res}_{(-\underline{\lambda_{m}}+s_{1},-\underline{\lambda_{n}}+s_{2})}E(\cdot,\underline{\Lambda})$
is isomorphic to $I_{\sigma}$. ∎

## Examples

###  .7. Case $m=n=2$

For $\sigma=0$, the two segments
$\Delta_{1}=\left[\frac{1}{2},\frac{3}{2}\right]$ and
$\Delta_{2}=\left[-\frac{3}{2},-\frac{1}{2}\right]$ are _juxtaposed_. The
orbits and the corresponding $R(w,\underline{s})$ are listed in the following
table:

Orbit | $\rho_{m,n}$ | $R(w,\frac{m+n}{2}-\sigma+t)$
---|---|---
${e}$ | $\left\\{\frac{1}{2},\frac{3}{2},-\frac{3}{2},-\frac{1}{2}\right\\}$ | $1$
${(1324)}$ | $\left\\{\frac{1}{2},-\frac{3}{2},\frac{3}{2},-\frac{1}{2}\right\\}$ | $\frac{c^{-t-\frac{5}{2}}L(t+3)}{L(t+4)}$
${(1342)}$ | $\left\\{\frac{1}{2},-\frac{3}{2},-\frac{1}{2},\frac{3}{2}\right\\}$ | $\frac{c^{-2t-4}L(t+2)}{L(t+4)}$
${(3124)}$ | $\left\\{-\frac{3}{2},\frac{1}{2},\frac{3}{2},-\frac{1}{2}\right\\}$ | $\frac{c^{-2t-4}L(t+2)}{L(t+4)}$
${(3142)}$ | $\left\\{-\frac{3}{2},\frac{1}{2},-\frac{1}{2},\frac{3}{2}\right\\}$ | $\frac{c^{-3t-\frac{11}{2}}L(t+2)^{2}}{L(t+3)L(t+4)}$
${(3412)}$ | $\left\\{-\frac{3}{2},-\frac{1}{2},\frac{1}{2},\frac{3}{2}\right\\}$ | $\frac{c^{-4t-6}L(t+1)L(t+2)}{L(t+3)L(t+4)}$

For $\sigma=1$, the poles of $r(w,\underline{s})$ in one of the orbits
$\\{(3142),(3412)\\}$ cancel:

Orbit | $\rho_{m,n}$ | $R(w,\frac{m+n}{2}-\sigma+t)$
---|---|---
${e}$ | $\\{0,1,-1,0\\}$ | $1$
${(1324)}$ | $\\{0,-1,1,0\\}$ | $\frac{c^{-t-\frac{3}{2}}L(t+2)}{L(t+3)}$
${(1342)}$ | $\\{0,-1,0,1\\}$ | $\frac{c^{-2t-2}L(t+1)}{L(t+3)}$
${(3124)}$ | $\\{-1,0,1,0\\}$ | $\frac{c^{-2t-2}L(t+1)}{L(t+3)}$
${(3142),(3412)}$ | $\\{-1,0,0,1\\}$ | $\frac{c^{-3t-\frac{5}{2}}L(t+1)\left(L(1-t)+L(1+t)\right)}{L(t+2)L(t+3)}$

For $\sigma=2$, more cancellations occur:

Orbit | $\rho_{m,n}$ | $R(w,\frac{m+n}{2}-\sigma+t)$
---|---|---
${e,(3412)}$ | $\left\\{-\frac{1}{2},\frac{1}{2},-\frac{1}{2},\frac{1}{2}\right\\}$ | $\frac{c^{-2t}\left(c^{2t}L(t+1)L(t+2)+L(1-t)L(2-t)\right)}{L(t+1)L(t+2)}$
${(1324),(1342),(3124),(3142)}$ | $\left\\{-\frac{1}{2},-\frac{1}{2},\frac{1}{2},\frac{1}{2}\right\\}$ | $\frac{c^{-t-\frac{1}{2}}(L(1-t)+L(t+1))^{2}}{L(t+1)L(t+2)}$

###  .8. Case $m=2,n=3$

For $\sigma=2$, the $R(w,\underline{s})$’s are displayed in the following
table:

Orbit | $\rho_{m,n}$ | $R(w,\frac{m+n}{2}-\sigma+t)$
---|---|---
${e}$ | $\left\\{-\frac{1}{4},\frac{3}{4},-\frac{5}{4},-\frac{1}{4},\frac{3}{4}\right\\}$ | $1$
${(13245)}$ | $\left\\{-\frac{1}{4},-\frac{5}{4},\frac{3}{4},-\frac{1}{4},\frac{3}{4}\right\\}$ | $\frac{c^{-t-\frac{3}{2}}L(t+2)}{L(t+3)}$
${(13425),(13452)}$ | $\left\\{-\frac{1}{4},-\frac{5}{4},-\frac{1}{4},\frac{3}{4},\frac{3}{4}\right\\}$ | $\frac{c^{-2t-2}(L(1-t)+L(t+1))}{L(t+3)}$
${(31245),(34512)}$ | $\left\\{-\frac{5}{4},-\frac{1}{4},\frac{3}{4},-\frac{1}{4},\frac{3}{4}\right\\}$ | $\frac{c^{-4t-2}\left(c^{2t}L(t+1)L(t+2)+L(1-t)L(2-t)\right)}{L(t+2)L(t+3)}$
${(31425),(31452),(34125),(34152)}$ | $\left\\{-\frac{5}{4},-\frac{1}{4},-\frac{1}{4},\frac{3}{4},\frac{3}{4}\right\\}$ |  $\frac{c^{-3t-\frac{5}{2}}(L(1-t)+L(t+1))^{2}}{L(t+2)L(t+3)}$.

###  .9. Case $m=3,n=2$

Similar to the previous subsection, for $\sigma=1$,

Orbit | $\rho_{m,n}$ | $R(w,\frac{m+n}{2}-\sigma+t)$
---|---|---
e | $\left\\{-\frac{1}{4},\frac{3}{4},\frac{7}{4},-\frac{5}{4},-\frac{1}{4}\right\\}$ | $1$
$(12435)$ | $\left\\{-\frac{1}{4},\frac{3}{4},-\frac{5}{4},\frac{7}{4},-\frac{1}{4}\right\\}$ | $\frac{c^{-t-\frac{5}{2}}L(t+3)}{L(t+4)}$
$(12453)$ | $\left\\{-\frac{1}{4},\frac{3}{4},-\frac{5}{4},-\frac{1}{4},\frac{7}{4}\right\\}$ | $\frac{c^{-2t-4}L(t+2)}{L(t+4)}$
$(14235)$ | $\left\\{-\frac{1}{4},-\frac{5}{4},\frac{3}{4},\frac{7}{4},-\frac{1}{4}\right\\}$ | $\frac{c^{-2t-4}L(t+2)}{L(t+4)}$
$(14253)$ | $\left\\{-\frac{1}{4},-\frac{5}{4},\frac{3}{4},-\frac{1}{4},\frac{7}{4}\right\\}$ | $\frac{c^{-3t-\frac{11}{2}}L(t+2)^{2}}{L(t+3)L(t+4)}$
$(14523)$ | $\left\\{-\frac{1}{4},-\frac{5}{4},-\frac{1}{4},\frac{3}{4},\frac{7}{4}\right\\}$ | $\frac{c^{-4t-6}L(t+1)L(t+2)}{L(t+3)L(t+4)}$
$(41235)$ | $\left\\{-\frac{5}{4},-\frac{1}{4},\frac{3}{4},\frac{7}{4},-\frac{1}{4}\right\\}$ | $\frac{c^{-3t-\frac{9}{2}}L(t+1)}{L(t+4)}$
$(41253)$ | $\left\\{-\frac{5}{4},-\frac{1}{4},\frac{3}{4},-\frac{1}{4},\frac{7}{4}\right\\}$ | $\frac{c^{-4t-6}L(t+1)L(t+2)}{L(t+3)L(t+4)}$
$(41523),(45123)$ | $\left\\{-\frac{5}{4},-\frac{1}{4},-\frac{1}{4},\frac{3}{4},\frac{7}{4}\right\\}$ | $\frac{c^{-5t-\frac{13}{2}}L(t+1)(L(1-t)+L(t+1))}{L(t+3)L(t+4)}$

###  .10. Case $m=3,n=3$

For $\sigma=3$, we have:

Orbit | $\rho_{m,n}$ | $R(w,\frac{m+n}{2}-\sigma+t)$
---|---|---
${e,(456123)}$ | $\\{-1,0,1,-1,0,1\\}$ | $\frac{c^{-6t}\left(c^{6t}L(t+1)L(t+2)L(t+3)+L(1-t)L(2-t)L(3-t)\right)}{L(t+1)L(t+2)L(t+3)}$
${(124356),(451623)}$ | $\\{-1,0,-1,1,0,1\\}$ | $\frac{c^{-5t-\frac{3}{2}}\left(c^{4t}L(t+1)L(t+2)^{2}+L(1-t)L(2-t)^{2}\right)}{L(t+1)L(t+2)L(t+3)}$
$\left\\{\begin{array}[]{c}(124536),(124563),\\\ (451236),(451263)\end{array}\right\\}$ | $\\{-1,0,-1,0,1,1\\}$ | $\frac{c^{-4t-2}(L(1-t)+L(t+1))\left(c^{2t}L(t+1)L(t+2)+L(1-t)L(2-t)\right)}{L(t+1)L(t+2)L(t+3)}$
$\left\\{\begin{array}[]{c}(142356),(145623),\\\ (412356),(415623)\end{array}\right\\}$ | $\\{-1,-1,0,1,0,1\\}$ | $\frac{c^{-4t-2}(L(1-t)+L(t+1))\left(c^{2t}L(t+1)L(t+2)+L(1-t)L(2-t)\right)}{L(t+1)L(t+2)L(t+3)}$
$\left\\{\begin{array}[]{c}(142536),(142563),\\\ (145236),(145263)\\\ (412536),(412563),\\\ (415236),(415263)\end{array}\right\\}$ | $\\{-1,-1,0,0,1,1\\}$ | $\frac{c^{-3t-\frac{5}{2}}(L(1-t)+L(t+1))^{3}}{L(t+1)L(t+2)L(t+3)}$

## References

  * [GL18] Dorian Goldfeld and Xiaoqing Li.  A standard zero free region for Rankin–Selberg $L$-functions.  International Mathematics Research Notices, 2018(22):7067–7136, 2018.
  * [HM15] Marcela Hanzer and Goran Muić.  On the images and poles of degenerate eisenstein series for $GL(n,\mathbb{A}_{\mathbb{Q}})$ and $GL(n,\mathbb{R})$.  American Journal of Mathematics, 137(4):907–951, 2015.
  * [JPSS83] Hervé Jacquet, Ilya I Piatetskii-Shapiro, and Joseph A Shalika.  Rankin-Selberg convolutions.  American journal of mathematics, 105(2):367–464, 1983.
  * [KS71] Anthony W Knapp and Elias M Stein.  Interwining operators for semisimple groups.  Annals of Mathematics, 93(3):489–578, 1971.
  * [KS88] C David Keys and Freydoon Shahidi.  Artin $L$-functions and normalization of intertwining operators.  In Annales scientifiques de l’École Normale Supérieure, volume 21, pages 67–89, 1988.
  * [Lan89] Robert P Langlands.  Irreducible representations of real algebraic groups.  Representation theory and harmonic analysis on semisimple Lie groups, 31:101–170, 1989.
  * [Lan06] Robert P Langlands.  On the functional equations satisfied by Eisenstein series, volume 544.  Springer, 2006.
  * [MW89] Colette Mœglin and J-L Waldspurger.  Le spectre résiduel de $GL(n)$.  In Annales scientifiques de l’École normale supérieure, volume 22, pages 605–674, 1989.
  * [MW95] Colette Mœglin and Jean-Loup Waldspurger.  Spectral Decomposition and Eisenstein Series, volume 113 of cambridge tracts in mathematics, 1995.
  * [Sha83] Freydoon Shahidi.  Local coefficients and normalization of intertwining operators for $gl(n)$.  Compositio Mathematica, 48(3):271–295, 1983.
  * [Sha84] Freydoon Shahidi.  Fourier transforms of intertwining operators and plancherel measures for $GL(n)$.  American Journal of Mathematics, 106(1):67–111, 1984.
  * [Sha10] Freydoon Shahidi.  Eisenstein series and automorphic $L$-functions, volume 58.  American Mathematical Soc., 2010.
  * [Zel80] Andrei V Zelevinsky.  Induced representations of reductive $\mathfrak{p}$-adic groups. ii. On irreducible representations of $GL(n)$.  In Annales scientifiques de l’École Normale Supérieure, volume 13, pages 165–210, 1980.