text:
\chapter{Notation and Preliminary Results} \label{notation}

We fix a local non-archimedean field $F$ of characteristic zero.\footnote{We expect all the results to hold also
in positive characteristic. However, we would need to verify some facts such as the unitarizability of representations \eqref{intro-dist}.}
Let $G$ be the group of $F$-points of a reductive group defined over $F$.
In this paper, by a representation of $G$ we shall always mean a complex, smooth representation.
The Grothendieck group of the category $\Alg_{\fl}(G)$ of all finite length representations of
$G$ is denoted by $ \mathfrak R(G)$. It carries a natural ordering $\leq$.
We denote by $\ssm(\tau)$ the semi simplification of $\tau\in\Alg_{\fl}(G)$.
For brevity, if $\pi_1,\pi_2\in\Alg_{\fl}(G)$, the condition
$\ssm(\pi_1)\leq \ssm(\pi_2)$ will be written simply as $\pi_1\leq \pi_2$.

The contragredient representation of $\pi$ is denoted by $\tilde\pi$, while the complex conjugate representation is denoted by $\bar\pi$.
We call $\tilde{\bar \pi}$ the hermitian contragredient of $\pi$ and denote it by $\pi^+$.
Then, $\pi\mapsto \pi^+$ is an (exact) contravariant functor. It is well known that if $\pi$ is unitarizable, then $\pi^+\cong\pi$
(i.e., $\pi$ is hermitian).

\section{General linear groups}
Let $F'$ be either $F$ itself or a (separable) quadratic extension thereof.
(The second case is pertaining to unitary groups and the first case to all other classical groups considered below.)
If $F'\ne F$, then
$
\Theta
$
denotes the non-trivial $F$-automorphism of $F'$. Otherwise, $\Theta$ denotes the identity mapping on $F$. The representation
\begin{equation} \label{cont-var}
\check\pi=\tilde\pi\circ\Theta
\end{equation}
will be called the $F'/F$-contragredient of $\pi$.
The representation $\pi\circ\Theta$ will be denoted by $\pi^\Theta$.

We shall now recall notation for the general linear groups (mainly following \cite{MR584084}).
The modulus character of $F'$ is denoted by $\abs{\cdot}_{F'}$.
The character $\abs{\det}_{F'}$ of $\GL(n,F')$ will be denoted by $\nu$.

We fix the Borel subgroup of upper triangular matrices.
For $0\leq k\leq n$, let $P_{(k,n-k)}=M_{(k,n-k)}\ltimes N_{(k,n-k)}$ be the standard parabolic subgroup
of $\GL(n,F' )$ of type $(k,n-k)$ with Levi factor $M_{(k,n-k)}\cong\GL(k,F' )\times\GL(n-k,F' )$.
For $\pi_i\in\Alg_{\fl}(\GL(n_i,F' ))$, $i=1,2$ denote by $\pi_1\times
\pi_2\in\Alg_{\fl}(\GL(n_1+n_2,F' ))$ the representation parabolically inducted from
$\pi_1\otimes\pi_2$ on $P_{(n_1,n_2)}$ (normalized induction). Let
$R=\oplus_{n\geq 0} \mathfrak R(\GL(n,F' ))$ endowed with the structure of a commutative graded ring by $\times$.
The biadditive map $\times:R\times R\ra R$ gives rise to a map $m:R\otimes R\ra R$.

The normalized Jacquet module of $\pi \in \Alg_{\fl}(\GL(n,F' ))$ with respect
to $P_{(k,n-k)}$ is denoted by $r_{(k,n-k)}(\pi)$.
The comultiplication $m^*(\pi)$ of $\pi$ is defined by
$$
m^*(\pi)=\sum_{k=0}^n \ssm(r_{(k,n-k)}(\pi)) \ \in \ R\otimes R.
$$
One extends $m^*$ additively to a ring homomorphism $m^*:R\rightarrow R\otimes R$ in a natural way.
With $m$ and $m^*$, $R$ is a graded Hopf algebra.
The fact that $m^*$ is a ring homomorphism follows from the geometric lemma of Bernstein--Zelevinsky (see \cite{MR0579172}).

Denote by $\Cusp$ (resp., $\Irr$) the set of equivalence classes of all irreducible cuspidal
(resp., irreducible) representations of all $\GL(n,F' )$, $n\geq 1$ (resp., $n\ge0$).

By a \emph{$\Z$-segment} in $\R$ we shall mean a set of the form
$
\{x,x+1,\dots,x+n\},
$
where $x\in\R$ and $n\in\Z_{\geq 0}$. We shall denote the above set by
$
[x,x+n]_\Z.
$
For any $\Z$-segment $\D=[x,y]_\Z$ in $\R$ and $\rho\in\Cusp$, denote
$$
\D^{(\rho)}=[x,y]^{(\rho)}=[\nu^x\rho,\nu^y\rho]:=\{\nu^z\rho;z\in\D\}.
$$
The set $\D^{(\rho)}$ is called a segment in $\Cusp$.
The set of all segments in $\Cusp$ is denoted by $\SC$.
It is also convenient to set $\emptyset^{(\rho)}=\emptyset$.

We say that two segments $\D_1,\D_2\in\SC$ are \emph{linked} if
$\D_1\cup \D_2\in\SC$ and $\D_1\cup\D_2\not\in\{\D_1,\D_2\}$.
If the segments $\D_i=[x_i,y_i]^{(\rho)}$ are linked and $x_1<x_2$, then
we say that $\D_1$ precedes $\D_2$, and write
$$
\D_1\prec \D_2.
$$

For any set $X$, we denote by $M(X)$ the set of all finite multisets in $X$
(which we may view as functions $X\rightarrow \Z_{\geq0}$ with finite support;
note that finite subsets correspond to all functions $X\rightarrow \{0,1\}$ with finite support).
A typical elements of $M(X)$ will be denoted by $(x_1,\dots,x_n)$ (repetitions of elements can occur,
but the order of the $x_i$'s does not matter).
The set $M(X)$ has a natural structure of a commutative monoid whose zero element is the empty multiset.
The operation will be denoted additively: $(x_1,\dots,x_n)+(y_1,\dots,y_m)=(x_1,\dots,x_n,y_1,\dots,y_m)$.

For $\D\in\SC$ we define $\supp(\D)$ to be $\D$, but considered as an element of $M(\Cusp$).
For $a=(\D_1,\dots,\D_n)\in M(\SC)$ we define
$$
\supp(a)=\sum_{i=0}^n \supp(\D_i)\in M(\Cusp).
$$

\section{Classifications of admissible duals of general linear groups}

Fix a segment $\D=\{\rho,\nu\rho, \dots,\nu^n\rho\}\in\SC.$
Then, the representation
$$
\rho\times\nu\rho\times \dots\times\nu^n\rho
$$
has an irreducible socle, denoted by $\ms(\D)$, and an irreducible cosocle, denoted by $\delta(\D)$. We have
\begin{subequations}
\begin{gather} \label{m*d}
m^*(\delta([\rho,\nu^n\rho]))=\sum_{i=-1}^n \delta([\nu^{i+1}\rho,\nu^n\rho]) \otimes
\delta([\rho,\nu^{i}\rho]),\\
\label{m*s}
m^*(\ms([\rho,\nu^n\rho]))=\sum_{i=-1}^n \ms([\rho,\nu^{i}\rho])
\otimes\ms([\nu^{i+1}\rho,\nu^n\rho]).
\end{gather}
\end{subequations}

Let $a=(\D_1,\dots,\D_n)\in M(\SC)$. We can choose an enumeration satisfying
\begin{center}
if $\D_i\prec \D_j $ for some $1\leq i,j\leq n$, then $i>j$.
\end{center}
Then, up to isomorphism, the representations
\begin{align*}
&\zeta(a):= \ms(\D_1)\times \ms(\D_2)\times\dots\times \ms(\D_n),
\\
(\text{resp., }& \lambda(a):= \delta(\D_1)\times \delta(\D_2)\times\dots\times \delta(\D_n))
\end{align*}
are determined by $a$ and admit an irreducible socle (resp., cosocle) denoted by $Z(a)$ (resp., $L(a)$).
Now $Z$ (resp. $L$) is called the Zelevinsky (resp. Langlands) classification of irreducible representations of general linear groups
over $F'$. (We follow the presentation of these classifications by F. Rodier in \cite{MR689531}.)

Denote by $\Disc$ the set of equivalence classes of all irreducible essentially square-integrable representations
of $\GL(n,F')$, $n\geq1$, and by $\Disc_u$ the subset of all unitarizable classes in $\Disc$ (i.e. those having a unitary central character).
The mapping
\begin{equation} \label{esi}
(\rho,n)\mapsto \delta(\rho,n):=\delta([-\tfrac{n-1}2,\tfrac{n-1}2]^{(\rho)}), \quad \Cusp\times \Z_{>0}\ra \Disc
\end{equation}
is a bijection.

For $\delta\in \Disc$ define $\delta^u\in \Disc_u$ and $e(\delta)\in\R$ by the following requirement:
$$
\delta=\nu^{e(\delta)}\delta^u.
$$
Let $d=(\delta_1,\dots,\delta_n)\in M(\Disc)$, enumerated so that
$$
e(\delta_1)\geq e(\delta_2) \geq \dots \geq e(\delta_n).
$$
Let
$$
\lambda(d)=\delta_1\times\delta_2\times\dots\times\delta_n.
$$
Then, the representation $\lambda(d)$ has an irreducible cosocle, denoted by $L(d)$.
Again $d\mapsto L(d)$ is a version of Langlands classification for general linear groups (irreducible representations are
parameterized by elements of $M(\Disc)$).

For $d=(\delta_1,\dots,\delta_n)\in M(\Disc)$ denote $\tilde d=(\tilde\delta_1,\dots,\tilde\delta_n)\in M(\Disc)$,
$\bar d=(\bar\delta_1,\dots,\bar\delta_n)$, $d^+=(\delta_1^+,\dots,\delta_n^+)$ and
$d^\Theta=(\delta_1^\Theta,\dots,\delta_n^\Theta)$.
Then, $L(d)\tilde{\ }=L(\tilde d)$, $L(d)\bar{\ }=L(\bar d)$, $L(d)^+=L( d^+)$ and $L(d)^\Theta=L(d^\Theta)$.

Define a mapping $^t$ on $\Irr$ by $Z(a)^t=L(a), a \in M(\SC)$.
Extend $^t$ additively to $R$. Clearly, $^t$ is a positive mapping, i.e.
it satisfies: $r_1\leq r_2\implies r_1^t \leq r_2^t$. A non-trivial fact is that $^t$ is in fact a ring homomorphism
(see \cite{MR1285969} and \cite{MR1471867}). Furthermore, $^t$ is an involution, called Zelevinsky involution.

For $a\in M(\SC)$, define $a^t\in M(\SC)$ by the requirement
$$
L(a)^t=L(a^t).
$$

\section{Classical groups -- basic definitions}
We will mostly follow the notation of \cite{MR1896238} for classical $p$-adic groups.
The main difference is that the indexing of the classical groups will be slightly different.

Fix a Witt tower $\mathcal V=\{V_n\}_{n\ge0}$ of symplectic, quadratic or hermitian vector spaces over $F'$.
In the first two cases $F'=F$ and in the latter case $F'/F$ is a (separable) quadratic extension $F$
with Galois automorphism $\Theta$.
In all cases, a maximal isotropic subspace of $V_n$ has dimension $n$
(see sections III.1 and III.2 of \cite{KudlaNotes} for more details).\footnote{For some purposes a different indexing
of the groups $S_n$ may be more convenient -- see \cite{MR1896238}.}
In particular, $V_0$ is anisotropic.\footnote{In the symplectic case, $V_0=\{0\}$.}
Denote by $S_n$ the group of isometries of $V_n$.
For $0\leq k\leq n$, let $P_{(k)}$ be the stabilizer of a fixed $k$-dimensional isotropic subspace of $V_n$ --
see \cite[\S III.2]{KudlaNotes}.\footnote{One can find in \cite{MR1356358}
matrix realizations of the symplectic and split odd-orthogonal groups.
In a similar way one can make matrix realizations also for other classical groups.}
The Levi factor $M_{(k)}$ of $P_{(k)}$ is naturally isomorphic to $\GL(k,F')\times S_{n-k}$.
More generally, for any partition $\beta$ of $\ell\leq n$ we can in a natural way define a parabolic subgroup $P_\beta$
and its Levi subgroup $M_\beta$.
(For $M_\beta$ first consider $M_{(\ell)}$, and then apply the construction from the case of general linear groups.)

We remark that in the odd orthogonal case we may replace $S_n$ with the group of isometries of $V_n$ of determinant one.

We exclude in the paper the case of split even orthogonal groups, although we expect that all the results
hold also in this case, with the same proofs. (Split even orthogonal groups are not connected,
which requires some additional checks that we have not yet carried out.)

A minimal parabolic subgroup in $S_n$, which is the intersection of all $P_{(k)}$'s, will be fixed.
(Only standard parabolic subgroups with respect to the fixed minimal parabolic subgroup will be considered in this paper.)

For the rest of the paper we fix once and for all the series $\{S_n\}_{n\ge0}$ as above.
We denote by $\Cuspcl$, (resp., $\Discl$, $\Tempcl$, $\Irrcl$) the set of cuspidal (resp., square-integrable, tempered, all)
irreducible representations of $S_n$, $n\ge0$ (up to equivalence) and
by $\Cuspsd$ the subset of $\Cusp$ consisting of $F'/F$-selfcontragredient representations.
(We shall often apply Casselman's criteria from \cite{CassNotes} for representations to be square-integrable or tempered.)

\section{Twisted Hopf algebra structure}
For $\pi\in\Alg_{\fl}(\GL(k,F'))$ and $\sigma\in\Alg_{\fl}(S_{n-k})$, the representation parabolically
induced from $\pi\otimes\sigma$ is denoted by
$$
\pi\rtimes\sigma.
$$
We shall often use that
\begin{equation} \label{asso}
\pi_1\rtimes (\pi_2\rtimes\sigma )\cong (\pi_1\times\pi_2)\rtimes\sigma.
\end{equation}

For $\pi$ as above holds
\begin{equation} \label{check}
\ssm(\pi\rtimes\sigma)=\ssm(\check\pi\rtimes\sigma).
\end{equation}
Therefore, if $\pi \rtimes \sigma$ is irreducible, then
$\pi\rtimes\sigma\cong\check\pi\rtimes\sigma$.
We say that a representation $\pi$ of a general linear group
over $F'$ is $F'/F$-selfcontragredient if
$
\pi\cong\check\pi.
$

The normalized Jacquet module of $\tau\in\Alg_{\fl}(S_n)$ with respect to $P_{(k)}$ is denoted by $s_{(k)}(\tau)$.
Let $\tau$ and $\omega$ be irreducible representations of $\GL(p,F)$ and $S_q$, respectively, and let $\pi$ be an admissible
representation of $S_{p+q}$. Then, by Frobenius reciprocity
$$
\Hom_{_{S_{p+q}}}(\pi,\tau \rtimes \omega)\cong
\Hom_{_{\GL(p,F)\times S_q}}(s_{(p)}(\pi),\tau \otimes \omega),
$$
while the second adjointness implies
$$
\Hom_{_{S_{p+q}}}(\tau \rtimes \omega, \pi)\cong
\Hom_{_{\GL(p,F)\times S_q}}(\check \tau \otimes \omega,s_{(p)}(\pi)).
$$

Denote
$$
R(S)=\underset {n\geq 0} \oplus \mathfrak R(S_n).
$$

Now $\rtimes$ induces in a natural way a mapping $R\times R(S)\ra R(S)$, which is denoted again by $\rtimes$.
For $\tau\in \Alg_{\fl}(S_n)$, denote
$$
\mu^*(\tau)=\sum_{k=0}^n \ssm\left(s_{(k)}(\tau)\right).
$$
We extend $\mu^*$ additively to $\mu^*:R(S)\ra R\otimes R(S)$. Denote
\begin{equation} \label{M*}
 M^*= (m \otimes 1) \circ (\, \check{\ }\, \otimes m^\ast) \circ
\kappa \circ m^\ast : R\ra R\otimes R,
\end{equation}
where $\check{\ }\,:R\ra R$ is a
group homomorphism determined by the requirement that
$\pi\mapsto\check\pi$ for all $\pi\in\Irr$, and $\kappa:R\times R\ra R\times R$ maps $\sum x_i\otimes y_i$ to $\sum y_i\otimes x_i.$
The action $\rtimes$ of $R\otimes R$ on $R\otimes R(S)$ is defined in a natural way. Then,
\begin{equation}
\label{mu*}
\mu^*(\pi \rtimes \sigma)= M^*(\pi) \rtimes \mu^*(\sigma)
\end{equation}
holds for $\pi\in R$ and $\sigma \in R(S)$.

For any finite length representation $\pi$ of $\GL(k,F')$, the component of $M^*(\pi)$ which is in
$
\mathfrak R(\GL(k,F'))\otimes \mathfrak R(\GL(0,F'))
$, will be denoted by
$$
M^*_{\GL}(\pi)\otimes 1.
$$

Let $\pi$ be a representation of $\GL(k,F')$ of finite length, and let $\sigma\in\Cuspcl$. Suppose
that $\tau$ is a subquotient of $\pi\rtimes \sigma$. Then, we shall denote $s_{(k)}(\tau)$ also by
\begin{equation*}
s_{\GL}(\tau).
\end{equation*}
If in addition, $\tau$ is irreducible, then we shall say that
\begin{equation} \label{pcs}
\s
\end{equation}
is the partial cuspidal support of $\tau$.
We say that $\theta\in\Cusp$ is a factor of $\tau$ if there exists an irreducible subquotient
$\beta\otimes\s$ of $s_{\GL}(\tau)$ such that $\theta$ is in the support of $\beta$.

Let $\pi$ be a finite length representation of a general linear group, and let $\tau$ be a representation of
$S_n$ as above. Then, \eqref{mu*} implies
\begin{equation} \label{s-GL}
\ssm(s_{\GL}(\pi\rtimes\tau))=M^*_{\GL}(\pi)\times
\ssm(s_{\GL}(\tau))
\end{equation}
($\times$ in the above formula denotes multiplication in $R$ of
$M^*(\pi)$ with the factors on the left-hand side of $\otimes$ in $\ssm(s_{\GL}(\tau))$).

Let $\tau$ be a representation of some $\GL(m,F)$ and let
$
m^{\ast}(\tau)=\sum x\otimes
y.
$
Then, the formula \eqref{M*} implies directly
\begin{equation} \label{M-GL}
M^*_{\GL}(\tau)=\sum x\times \check{y}.
\end{equation}
Furthermore, the sum of the irreducible subquotients of the form $1\otimes \ast$ in $M^{\ast}(\tau)$ is
\begin{equation} \label{1o}
1\otimes \tau.
\end{equation}

Now assume that $\pi$ is a representation of $\GL(d,F)$ and $\s$ is a representation of a classical group.
Let $\pi_1\otimes\pi_2\otimes\pi_3$ be an irreducible subquotient of some $r_{(n_1,n_2,n_3)}(\pi)$ ($n_1+n_2+n_3=n$)
and let $\pi_4\otimes\s_0$ be an irreducible subquotient of some $s_{(m_1)}(\s)$ ($m_1\leq m$). Then,
$$
\pi_1 \times\pi_4\times \tilde\pi_3 \otimes \pi_2\rtimes\s_0
$$
is a subquotient of the corresponding Jacquet module (see \cite[Lemma 5.1]{MR1356358} and the discussion preceding it).

\section{Some formulas for \texorpdfstring{$M^*$}{M*}} \label{some form.}
Let $\rho\in\Cuspsd$. Suppose that $x,y\in \R$ satisfy $y-x\in \Z_{\geq0}$.
Then, one directly gets from \eqref{m*d} and \eqref{M*}
\begin{equation} \label{M-seg}
M^*\big(\delta([x,y]^{(\rho)})\big) =
\sum_{i= x-1}^{ y}\sum_{j=i}^{ y}
\delta([-i,-x]^{(\rho)})\times\delta([j+1,y]^{(\rho)}) \otimes\delta([i+1,j]^{(\rho)}),
\end{equation}
where $y-i,y-j\in \Z_{\geq 0}$ in the above sums. In particular
\begin{equation}\label{M-seg-GL}
M_{\GL}^*\big(\delta([x,y]^{(\rho)})\big) =
\sum_{i= x-1}^{ y}\delta([-i,-x]^{(\rho)})\times\delta([i+1,y]^{(\rho)}).
\end{equation}

In a similar way, one gets for Zelevinsky segment representations
$$
M^*(\ms([x,y]^{(\rho)}))=\sum_{x -1\leq i \leq y}\sum_{ x -1\leq j \leq i}
\ms ([-y,-i -1]^{(\rho)})\times\ms ([x ,j]^{(\rho)})\otimes \ms ([j +1,i]^{(\rho)}).
$$

More generally, let $\pi=L(\D_1,\dots,\D_k)$ be a ladder representations, i.e., we can write $\D_i=[a_i,b_i]^{(\rho)}$ where
$a_k<\dots<a_1$ and $b_k<\dots<b_1$ (we continue to assume below
$
\rho\cong\check\rho).
$
Then, using \cite{MR2996769} we get

\begin{equation}\label{lad-GL}
M^*_{\GL}(\pi)=\sum_{\substack{a_i-1\leq x_i\leq b_i,
\\x_k<\dots<x_1}}L(\ ([-x_i,-a_i]^{(\rho)})_{1\leq i\leq k} \ )
\times L(\ ([x_i+1,b_i]^{(\rho)})_{1\leq i\leq k}\ ).
\end{equation}

\section{Langlands classification for classical groups
\texorpdfstring{(\cite{MR0507262}, \cite{MR1721403}, \cite{MR2050093}, \cite{MR2567785}, \cite{MR584084})}{}} \label{LCCG}
Denote
$$
\Disc_+=\{\delta\in \Disc: e(\delta)>0\}.
$$
For
$
t=((\delta_1,\delta_2,\dots,\delta_k),\tau)\in
M(\Disc_+)\times \Tempcl
$
take a permutation $p$ of $\{1,\dots,k\}$ such that
\begin{equation}\label{dec}
e(\delta_{p(1)})\geq e(\delta_{p(2)})\geq\dots\geq e(\delta_{p(k)}).
\end{equation}
Then, the representation
$$
\lambda(t):=\delta_{p(1)}\times\delta_{p(2)}\times\dots\times\delta_{p(k)}\rtimes \tau
$$
has an irreducible cosocle, denoted by
$$
L(t).
$$
The mapping
$$
t\mapsto L(t)
$$
defines a bijection between $M(\Disc_+)\times \Tempcl$ and $\Irrcl$. This is the Langlands classification for classical groups.
The multiplicity of $L(t)$ in $\lambda(t) $ is one.

Write $t=(d;\tau)$. Then, $L(d;\tau)\bar{\ }\cong L(\bar d;\bar\tau)$ and $L(d;\tau)\tilde{\ }\cong L(d^\Theta;\tilde\tau)$.

Let $t=((\delta_1,\delta_2,\dots,\delta_k),\tau) \in M(\Disc_+)\times \Tempcl$ and suppose that a permutation $p$ satisfies \eqref{dec}.
Suppose that $\delta_{p(i)}$ is a representation of $\GL(n_i,F)$ and $L(t) $ a representation of $S_n$. Define
$$
e_*(t)=(\underbrace{e(\delta_{p(1)}),\dots,e(\delta_{p(1)})}_{n_1\text{ times}},\dots,
\underbrace{e(\delta_{p(k)}),\dots,e(\delta_{p(k)})}_{n_k\text{ times}},\underbrace{0,\dots,0}_{n'\text{ times}}),
$$
where $n'=n-(n_1+\dots+n_k)$. Consider the partial ordering on $\R^n$ given by
$$
(x_1,\dots,x_n)\leq (y_1,\dots,y_n)\iff\sum_{i=1}^j x_i\leq \sum_{i=1}^j y_i, \quad 1\leq j\leq n.
$$
Suppose $t,t'\in M(\Disc_+)\times \Tempcl$ and $L(t')$ is a subquotient of $\lambda(t)$. Then,
\begin{equation} \label{BPLC}
e_*(t')\leq e_*(t), \text{ with an equality }\iff t'=t.
\end{equation}
(See \cite[\S6]{MR1266251} for the symplectic groups -- the same proof works for all classical groups
other than the split even orthogonal groups.)

For $\Delta\in \mathcal S$ define $\mathfrak c(\Delta)$ to be $e(\delta(\Delta))$. Let
$$
\SC_+=\{\Delta\in\SC; \mathfrak c (\Delta)>0\}.
$$
In this way we can define in a natural way the Langlands classification $(a,\tau)\mapsto L(a;\tau)$ using $M(\SC_+)\times \Tempcl$
as the parameters.

\section{Irreducible subquotients of induced representations of classical groups} \label{irr-sq}
We will recall a very useful fact from \cite{MR2504024}.

For $d=(\delta_1,\dots,\delta_k)\in M(\Disc)$ denote by
$$
d^\uparrow
$$
the element of $M(\Disc_+)$ obtained from $d$ by removing every $\delta_i$ such that $e(\delta_i)=0$
and replacing every $\delta_i$ for which $e(\delta_i)<0$ by $\check\delta_i$. Also, denote by
$$
d_u
$$ the multiset in $M(\Disc)$ obtained from $d$ by retaining only the $\delta_i$'s such that $e(\delta_i)=0$.

\begin{proposition}[\cite{MR2504024}] \label{prop: addparms}
Let $d\in M(\Disc)$ and $t=(d',\tau)\in M(\Disc_+)\times\Tempcl$. The tempered representation
$
\lambda(d_u)\rtimes\tau
$
is unitarizable and multiplicity free.
For every irreducible constituent $\tau'$ of $\lambda(d_u)\rtimes\tau$,
the representation
$$
L(d^\uparrow+d';\tau')
$$
occurs with multiplicity one in the Jordan--H\"older sequence of the induced representation
$$
L(d)\rtimes L(d';\tau).
$$
\end{proposition}

\section{Involution}
The Zelevinsky involution is a special case of an involution $D_G$ which exists on the Grothendieck group of the representations
of any connected reductive $p$-adic group. This involution is constructed in \cite{MR1285969} and \cite{MR1471867}.
It takes any irreducible representation to an irreducible representation up to a sign.
For any irreducible representation $\pi$, let $\pi^t$ be the irreducible representation such that $D_G(\pi)=\pm\pi^t$.
We call $\pi^t$ the \ASS involution of $\pi$, or \ASS dual of $\pi$.

This involution is compatible with parabolic induction in the sense that
$$
(\pi\rtimes\tau)^t=\pi^t\rtimes\tau^t
$$
(on the level of Grothendieck groups).

Furthermore, for Jacquet modules, the mapping
$$
\pi_1\otimes\dots\pi_l\otimes\mu\mapsto \check\pi_1^t\otimes\dots\check\pi_l^t\otimes\mu^t,
$$
is a bijection from the semi simplification of $s_\beta(\pi)$ onto the semisimplification of $s_\beta(\pi^t)$
($\beta$ is the partition which parametrizes the corresponding parabolic subgroup).

\section{Reducibility point and generalized Steinberg representations}

Let $\rho\in\Cusp$ with a unitary central character and $\sigma\in\Cuspcl$.
If $\rho\notin\Cuspsd$ then $\nu^x\rho\rtimes\s$ is irreducible for all $x\in\R$.
Otherwise,
\begin{equation}\label{alpha}
\nu^{\alpha_{\rho,\sigma}}\rho\rtimes\s
\end{equation}
is reducible for a unique $\alpha_{\rho,\sigma}\geq0$ (\cite{MR577138}). C. M\oe glin has proved that
$
\alpha_{\rho,\sigma}\in \tfrac12\Z.
$

Given $\rho\in\Cuspsd$, either $\alpha_{\rho,\sigma}\in\Z$ for all $\sigma\in\Cuspcl$ or
$\alpha_{\rho,\sigma}\in\frac12+\Z$ for all $\sigma\in\Cuspcl$.
We say that $\rho$ is of odd type (or parity) in the former case and of even type (or parity) in the latter.
The parity of $\rho$ depends only on $\rho$ and the Witt tower and can be detected by the existence of a pole for a suitable $L$-function.
For instance, for odd orthogonal groups, $\rho$ is of odd (resp., even) type if and only if
the exterior (resp., symmetric) square $L$-function of $\rho$ has a pole at $s=0$.
These conditions are reversed for symplectic groups.

From now on we fix $\rho\in\Cuspsd$ and $\sigma\in\Cuspcl$
and denote the reducibility point $ \alpha_{\rho,\sigma}$ simply by
$$
\alpha.
$$

The representation
$
\nu^{\alpha+ n} \rho \times \nu^{\alpha+ n-1} \rho \times \cdots \times
\nu^{\alpha+ 1} \rho \times \nu^\alpha\rho \rtimes \sigma
$
admits an irreducible socle, which is denoted by
$
\delta([\nu^\alpha\rho, \nu^{\alpha+ n} \rho]; \sigma) \ \ (n\geq 0).
$
It is square-integrable and called a generalized Steinberg representation. We have
\begin{equation} \label{eq: muforgenstn}
\mu^*\left(\delta([\nu^\alpha\rho, \nu^{\alpha+ n}\rho];\sigma)\right) =
\sum^n_{k=-1} \delta([\nu^{\alpha+ k + 1}\rho, \nu^{\alpha+n} \rho]) \otimes \delta([\nu^\alpha\rho,
\nu^{\alpha+ k}\rho]; \sigma),
\end{equation}
$$
\delta([\nu^\alpha\rho, \nu^{\alpha+ n} \rho];\sigma)\tilde{\ }
\cong
\delta([\nu^\alpha\rho,\nu^{\alpha+ n} \rho]; \tilde{\sigma}).
$$
Applying the \ASS involution, we get
\begin{multline} \label{eq: muforgenstnt}
\mu^*\left(L(\nu^{\alpha+n} \rho. \dots , \nu^{\alpha+ 1}\rho, \nu^{\alpha}\rho; \sigma)\right)=
\\\sum^n_{k=-1}L(\nu^{-(\alpha+ n)}\rho,\dots , \nu^{-(\alpha+ k+2)}\rho,\nu^{-(\alpha+k+1)} \rho)
\otimes L(\nu^{\alpha+k} \rho. \dots , \nu^{\alpha+ 1}\rho, \nu^{\alpha}\rho; \sigma).
\end{multline}

\label{sec: Castrick}

The generalized Steinberg representation and its \ASS dual are the only unitarizable irreducible subquotients of $
\nu^{\alpha+ n} \rho \times \nu^{\alpha+ n-1} \rho \times \cdots \times
\nu^{\alpha+ 1} \rho \times \nu^\alpha\rho \rtimes \sigma$ (\cite{MR2652536}, \cite{MR3052686};
see also \cite[\S 13]{MR3969882}).

\section{Representations of segment type} \label{segment}
We shall recall the formulas for Jacquet modules obtained in \cite{MR3360752}.\footnote{The results of \cite{MR3360752}
are only stated for symplectic and split odd-orthogonal groups but the proof works for all classical groups.
Note that the proof does not use the classification of $\Discl$ in terms of $\Cuspcl$.}
We fix $\rho\in\Cuspsd$ and $\sigma\in\Cuspcl$ and
consider irreducible subquotients of $\delta([\nu^{-c}\rho,\nu^{d}\rho])\rtimes\s$, where
$c+d\in \Z_{\geq 0}$.
As above, $\alpha\in\tfrac12\Z_{\geq0}$ denotes the reducibility exponent \eqref{alpha}.

The representation $\delta([\nu^{-c}\rho,\nu^{d}\rho])\rtimes\s$ is multiplicity free of length at most three.
It is reducible (resp., of length three) if and only if $[-c,d]_\Z\cap\{-\alpha,\alpha\}\ne \emptyset$ (resp.,
$\{-\alpha,\alpha\}\subseteq[-c,d]_\Z$ and $c\ne d$).

Assume that $d\ge c$ and $d-\alpha\in\Z$.
We define terms $\delta([\nu^{-c}\rho,\nu^{d}\rho]_+;\s)$, $\delta([\nu^{-c}\rho,\nu^{d}\rho]_-;\s)$ and $L_\alpha([\nu^{-c}\rho,\nu^{d}\rho];\s)$.\footnote{In \cite{MR3360752}
we denoted the last term by $L_\alpha(\delta([\nu^{-c}\rho,\nu^{d}\rho]);\s)$.}
Each of these terms is either an irreducible representation or the trivial (zero-dimensional) representation.
They satisfy
\begin{equation} \label{uuvodu}
\delta([\nu^{-c}\rho,\nu^{d}\rho])\rtimes\s=\delta([\nu^{-c}\rho,\nu^{d}\rho]_+;\s)+\delta([\nu^{-c}\rho,\nu^{d}\rho]_-;\s)+
L_\alpha([\nu^{-c}\rho,\nu^{d}\rho];\s)
\end{equation}
in the corresponding Grothendieck group.

Suppose first that $\delta([\nu^{-c}\rho,\nu^{d}\rho])\rtimes\s$ is irreducible.
Then, we define that $\delta([\nu^{-c}\rho,\nu^{d}\rho]_-;\s)=0$ and require that
$\delta([\nu^{-c}\rho,\nu^{d}\rho]_+;\s)\ne0$ if and only if $[-c,d]\subseteq[-\alpha+1,\alpha-1]$.
By \eqref{uuvodu} this determines $L_\alpha([\nu^{-c}\rho,\nu^{d}\rho];\s)$.

Suppose now that $\delta([\nu^{-c}\rho,\nu^{d}\rho])\rtimes\s$ is reducible.
If $c=d$, we define that $L_\alpha([\nu^{-c}\rho,\nu^{d}\rho];\s)=0$. Otherwise, $L_\alpha([\nu^{-c}\rho,\nu^{d}\rho];\s)=L([\nu^{-c}\rho,\nu^{d}\rho];\s)$.

If $\alpha> 0$, then there is a unique irreducible subquotient $\g$ of $\delta([\nu^{-c}\rho,\nu^{d}\rho])\rtimes\s$
which has in $s_{\GL}(\g)$ an irreducible subquotient $\tau\otimes\s$ such that $\tau$ is generic,
and $e(\theta)\geq 0$ for all $\theta$ in $\supp(\tau)$.
We denote this $\g$ by $\delta([\nu^{-c}\rho,\nu^{d}\rho]_+;\s)$.

If $\alpha= 0$, we write $\rho \rtimes \sigma$ as a sum of irreducible subrepresentations
$\tau_+ \oplus \tau_{-}$.
We denote also $\tau_\pm$ by $\delta([\rho]_\pm;\s)$.
Then, there exists a unique irreducible subquotient of $\delta([\nu^{-c}\rho,\nu^{d}\rho]) \rtimes \s$ that
contains an irreducible representation
of the form $\tau \otimes \tau_\pm$ in its Jacquet module with respect to an appropriate standard parabolic subgroup,
such that $\tau$ is generic, and $e(\theta)\geq 0$ for all $\theta$ in $\supp(\tau)$.
We denote this subquotient by $\delta([\nu^{-c}\rho,\nu^{d}\rho]_\pm;\s)$.

If $c = d$ or the length of $\delta([\nu^{-c}\rho,\nu^{d}\rho])\rtimes\s$ is three, then
$\delta([\nu^{-c}\rho,\nu^{d}\rho])\rtimes\s$ contains a unique irreducible subrepresentation different from
$\delta([\nu^{-c}\rho,\nu^{d}\rho]_+;\s)$ and we denote it by $\delta([\nu^{-c}\rho,\nu^{d}\rho]_-;\s)$.
Otherwise, we take $\delta([\nu^{-c}\rho,\nu^{d}\rho]_-;\s)=0$.

The representations $\delta([\nu^{-c}\rho,\nu^{d}\rho]_\pm;\s)$ are called representations of segment type.

The representation $\delta([\nu^{-c}\rho,\nu^{d}\rho]_+;\s)$ is square-integrable if and only if $c\ne d$
and $\{-\alpha,\alpha\}\subseteq [-c,d]$ or $\alpha=-c$.
If $\delta([\nu^{-c}\rho,\nu^{d}\rho]_+;\s)$ is square-integrable, then $\delta([\nu^{-c}\rho,\nu^{d}\rho]_-;\s)$
is also square-integrable if it is non-zero.
Conversely, if $\delta([\nu^{-c}\rho,\nu^{d}\rho]_-;\s)$ is square-integrable (and non-zero),
then $\delta([\nu^{-c}\rho,\nu^{d}\rho]_+;\s)$ is square-integrable (and non-zero).

In the two formulas below, we symmetrize notation in the following way. We define
$$
\delta([\nu^{-d}\rho,\nu^{c}\rho]_+;\s),\
\delta([\nu^{-d}\rho,\nu^{c}\rho]_-;\s)\text{ and
}L_\alpha([\nu^{-d}\rho,\nu^{c}\rho];\s)
$$
to denote $\delta([\nu^{-c}\rho,\nu^{d}\rho]_+;\s)$, $ \delta([\nu^{-c}\rho,\nu^{d}\rho]_-;\s)$ and
$L_\alpha([\nu^{-c}\rho,\nu^{d}\rho];\s)$ respectively (assumptions on $c$ and $d$ are as above).

\begin{remark}
Now we recall the formulas for the Jacquet modules of segment representations and associated Langlands quotient from \cite{MR3360752}.
We take this opportunity to correct several typographical errors in \cite{MR3360752}.
The upper limit in the first sum of the second row of \eqref{jm-seg-ds} below is $d-1$
(instead of $c$).\footnote{This correction refers to the formulas on page 441 and Corollaries 4.3, 5.4 and 6.4 of \cite{MR3360752}.}
The limits of the first sum in the third row of \eqref{jm-seg-ds} is $-c-1\leq i\leq c-1$ (instead of $-c-1\leq i\leq d$)
in fact, the indices between $c$ and $d$ do not give any contribution.
Further, the limits in the first sum in the second row of \eqref{jm-seg-q} are $-c-1\leq i\leq d-1$ (instead of $-c-1\leq i\leq d$)
since the index $d$ does not contribute show up in the formula.\footnote{Each of these two not so essential modifications
also refers to the formulas on page 441 and Corollaries 4.3, 5.4 and 6.4 of \cite{MR3360752}.}
\end{remark}

If $\delta([-c,d]^{(\rho)})\rtimes\s$ is reducible (with notation as above), then
we have the following equality
\begin{equation} \label{jm-seg-ds}
\begin{aligned}
\mu^*\big(&\delta([-c,d]^{(\rho)}_\pm;\s)\big)=
\sum_{i=-c-1}^{\pm \alpha- 1}
\delta([-i,c]^{(\rho)}) \times \delta([i+1,d]^{(\rho)})\otimes\s
 \\
+&\sum_{i= -c -1}^{ d-1}\sum_{j=i+1}^{d} \delta([-i,c]^{(\rho)}) \times \delta([j+1,d]^{(\rho)}) \otimes
\delta([i+1,j]^{(\rho)}_\pm;\s)
\\
+&\underset {\underset{ i+j<-1} {-c-1\le i\le c-1 \quad i+1\le j\le c}}{\sum \quad \sum}
\delta([-i,c]^{(\rho)}) \times
\delta([j+1,d]^{(\rho)}) \otimes
L_\alpha([i+1,j]^{(\rho)};\s).
\end{aligned}
\end{equation}
If additionally $c\ne d$, and either $c < \alpha$ or $\alpha\leq c<d$, then we have
\begin{equation}
\begin{gathered} \label{jm-seg-q}
\mu^*\big(L([-c,d]^{(\rho)};\sigma)\big)=
\mu^*\big(L_\alpha([-c,d]^{(\rho)};\sigma)\big)=
\sum_{i=\alpha}^dL([-i,c]^{(\rho)}, [i+1,d]^{(\rho)}) \otimes\sigma+
\\\sum_{-c-1\le i\le d-1}\ \sum_{\substack{i+1\le j\le d\\0\le i+j}}
L([-i,c]^{(\rho)},[j+1,d]^{(\rho)}\big) \otimes L_\alpha([i+1,j]^{(\rho)};\sigma).
\end{gathered}
\end{equation}

\section{Jordan blocks}
Now we shall recall the definition of the Jordan blocks $\Jord(\pi)$ of
an irreducible square-integrable representation $\pi$ of $S_n$.

\begin{definition}
For any $\pi\in\Discl$ denote by $\Jord(\pi)$ the set of all square-integrable representations $\delta(\rho,a)\in\Disc$ where
$\rho\in\Cuspsd$ and $a\in\Z_{>0}$ is of the same parity of $\rho$,
such that $\delta(\rho,a)\rtimes\pi$ is irreducible.
For any $\rho\in\Cuspsd$ we denote
\[
\Jord_\rho(\pi)=\{a:(\rho,a)\in \Jord(\pi)\}\subset\Z_{>0}.
\]
\end{definition}

The set $\Discl$ is classified by admissible triples (see \cite{MR1896238} for details).
Any $\pi\in\Discl$ is parameterized by a triple $(\Jord(\pi),\e_\pi,\pi_{cusp})$, where $\e_\pi$ is a
function\footnote{It is called partially defined function attached to $\pi$.} defined on a subset of $\Jord(\pi)\cup \Jord(\pi)\times \Jord(\pi)$ which takes values in $\{\pm1\}$, and $\pi_{cusp}$ is the partial cuspidal support
(which was defined earlier).

The construction of irreducible square-integrable representations in \cite{MR1896238} starts with strongly positive representations ($\epsilon_\pi$ is alternating function on  $\Jord_\rho(\pi)$ in this case).
The simplest example of such representations is the generalized Steinberg representations.
We shall give one more example of strongly positive representations.

Assume that the reducibility point $\alpha=\alpha_{\rho,\s}$ is strictly positive.
Take $k\in\Z_{\geq0}$ such that $k<\alpha$. Then, the representation
$
\nu^{\alpha-k}\rho\times \nu^{\alpha-k+1}\rho\times\dots\times \nu^{\alpha}\rho\rtimes\s
$
admits an irreducible socle, which we denote by
$$
\delta([\nu^{\alpha-k}\rho],[ \nu^{\alpha-k+1}\rho],\dots,[ \nu^{\alpha}\rho];\s).
$$
This is an example of strongly positive (square-integrable) representation.

Sometimes when we deal with strongly positive representations, to stress this we shall add subscript $\spsi$
(we shall not do this for the generalized Steinberg representations).
Therefore, the above representations we shall also denote by
$\delta_{\spsi}([\nu^{\alpha-k}\rho],[ \nu^{\alpha-k+1}\rho],\dots,[ \nu^{\alpha}\rho];\s).$

Now we recall Proposition 6.1 from \cite{MR2504024}, which we use several
times in the paper. (Note that in (vii) of Proposition 6.1 in
\cite{MR2504024}, the condition on parity was forgotten by mistake.)

\begin{proposition} \label{Pr-red-si}
Let $\pi\in\Discl$, $\rho\in\Cuspsd$ and $a>0$. Then,
\begin{enumerate}
\item $\nu^a\rho\rtimes\pi$ is reducible if and only if $\nu^{-a}\rho\rtimes\pi$ is reducible.
\item If $a\notin\tfrac12\Z$, then $\nu^a\rho\rtimes\pi$ is irreducible.
\item $\rho\rtimes \pi$ is reducible if and only if $\rho$ has odd parity and $1\not\in\Jord_\rho(\pi)$.
\item
\label{Pr-red-si-item: a-a+1-irr}
If $a\not\in\Jord_\rho(\pi)$, then $\nu^{(a+1)/2}\rho\rtimes\pi$ is irreducible.
\item
\label{Pr-red-si-item: a-not-a+2-red}
If $a\in\jrp$ and $a+2\not\in\jrp$, then $\nu^{(a+1)/2}\rho\rtimes\pi$ is reducible.
\item
\label{Pr-red-si-item: a-a+1-epsilon}
Suppose that $a$ and $a+2$ are in $\jrp$. Then,
$\nu^{(a+1)/2}\rho\rtimes\pi$ is reducible if and only if
$\e((\rho,a))=\e((\rho,a+2))$.
\item $\nu^{1/2}\rho\rtimes\pi$ is reducible if and only if $\rho$ is of even parity and either $2\not\in \jrp$,
or $2\in\jrp$ and $\e((\rho,2))=1$.

\noindent
In other words, $\nu^{1/2}\rho\rtimes\pi$ is irreducible if and only if
either $\rho$ is of odd parity or $2\in \jrp$ and $\e((\rho,2))=-1$.
\end{enumerate}
\end{proposition}

\subsection{On computing of Jordan blocks}

Next, we recall how to compute the Jordan blocks of any $\pi'\in\Discl$.

Suppose that $\pi'$ is a subquotient of
$$
\rho_1\times\dots\times\rho_r\rtimes\sigma,
$$
where $\rho_i\in\Cusp$ and $\sigma\in\Cuspcl$.
Fix $\rho\in\Cuspsd$ and let $\alpha=\alpha_{\rho,\sigma}$.
Let $\eta=0$ if $\alpha\in\Z$ and $\eta=1$ otherwise.

If no $\nu^x\rho$, $x\in \tfrac12\Z$ is a factor of $\pi'$, then
$$
\Jord_\rho(\pi')=\{\eta+1,\eta+3,\dots,2\alpha-3,2\alpha-1\}.
$$
(Note that for $\alpha=0$, $\Jord_\rho(\pi')=\emptyset.)$

In the general case, we recall Proposition 2.1 of \cite{MR1896238},
from which we can compute Jordan blocks in general.

\begin{proposition} \label{JBcomputation}
Let $x,y \in\frac12\Z$ be such that $x-y\in \Z_{\geq 0}$ and $x-\alpha\in\Z$.
Suppose that $\pi\in\Discl$ embeds in the induced representation
$$
\pi \hookrightarrow \nu^x \rho\times \cdots \times
\nu^{x-i+1}\rho
\times \cdots \times \nu^y \rho\rtimes \pi'.
$$
Then,
\begin{enumerate}
\item If $y>0$, then $2y-1\in\Jord_\rho(\pi')$ and
$$
\Jord_\rho(\pi)=\left(\Jord_\rho(\pi')\setminus\{2y-1\}\right)\cup\{2x+1\}.
$$
\item If $y\leq 0$ , then $2x+1$ and $-2y+1$ are not in $\Jord_\rho(\pi')$ and
$$
\Jord_\rho(\pi)=\Jord_\rho(\pi')\cup \{2x+1,-2y+1\}.
$$
\end{enumerate}
\end{proposition}

\begin{remark} \label{general-basic}
The Local Langlands correspondence for the general linear group (\cite{MR1876802}, \cite{MR1738446}) gives a bijection $\Phi$
between the set $\Disc$ (resp., $\Cusp$) and the irreducible representations of the Weil--Deligne (resp., Weil) group.\footnote{An alternative
classification of $\Cusp$ is given in \cite{MR1204652}.}
(Recall that $(\rho,n)\mapsto \delta(\rho,n)$ gives a parametrization of $\Disc$ by $\Cusp\times \Z_{>0}$.)

J. Arthur has obtained in \cite[Theorem 1.5.1]{MR3135650} a classification of irreducible tempered representations of classical groups
attaching to them pairs of admissible homomorphisms and characters of the component groups of the admissible homomorphisms.
C. M\oe glin has proved in \cite[Theorem 1.3.1]{MR2767522} that the admissible homomorphism attached to $\pi\in\Discl$ is
\begin{equation*} \label{eq-sum-L}
\underset{\s\in \Jord(\pi)}\oplus\Phi(\s).
\end{equation*}
In this way one gets a classification of $\Discl$ in terms of (certain) finite sets of $\Cuspsd$ and functions on these sets with values in $\{\pm1\}$.
In addition, \cite[Theorem 1.5.1]{MR2767522} $\pi\in\Cuspcl$ if and only if
for any $\rho\in\Cuspsd$, $\e_\pi(\delta(k,2))=-1$ whenever $2\in\Jord_\rho(\pi)$ and if $k\in\Jord_\rho(\pi)$ with $k>2$ then
$k-2\in\Jord_\rho(\pi)$ and $\e_\pi((\delta,k))\ne\e_\pi((\delta,k-2))$.
Finally, if $\sigma\in\Cuspcl$ and $\Jord_\rho(\sigma)\ne\emptyset$ then
$$
\alpha_{\rho,\sigma}=\tfrac{1+\max\Jord_\rho(\pi)}2
$$
while if $\Jord_\rho(\sigma)=\emptyset$ then $\alpha_{\rho,\sigma}$ is $0$ or $\frac12$ according to the parity of $\rho$
(see \cite{MR1913095} or \cite{MR1896238}).
One can find in \cite{MR3156864} more details and precise statements related to the above discussion.
\end{remark}

\section{Induction of \texorpdfstring{$\GL$}{GL}-type} \label{GL-LT}
Next we shall recall the results of \cite{1703.09475}, except that we shall formulate them in terms of the Langlands classification.
As above, $\alpha=\alpha_{\rho,\s}$ denotes the reducibility point (then $\rho\cong\rho\check{\ }$).
Let $\pi\in\Irr$.

If $\supp(\pi)$ contains $\nu^\alpha\rho$ or $\nu^{-\alpha}\rho$, then $\pi\rtimes\s$ is reducible (\cite{1703.09475}).

Suppose now that $\supp(\pi)$ does not contain $\nu^\alpha\rho$ or $\nu^{-\alpha}\rho$.
Assume that all members of $\supp(\pi)$ are contained in $\{\nu^{k+x}\rho:k\in\Z\}$, for some fixed $x\in\tfrac12\Z$.
Write $\pi=L(d)$, for some $d\in M(\Disc)$. Denote by $d_{>0}$ (resp. $d_{<0}$) the multiset consisting of all $\delta$ in $d$
such that $e(\delta)>0$ (resp $e(\delta)<0$), counted with multiplicities.
Then, if $\pi$ is a ladder representation or if $\alpha\leq 1$ and all members of $\supp(\pi)$ are contained in $\{\nu^{k+\alpha}\rho;k\in\Z\}$, then holds
\begin{equation} \label{eq: ladcrit}
L(d)\rtimes\s \text{ is reducible } \iff L(d_{>0})\times L(d_{<0})\check{\ } \text{ is reducible }.
\end{equation}

A very special case is the following very useful result proved already in \cite{MR1658535}. For $\D\in \SC$ holds:
$$
\delta(\D)\rtimes\s \text{ is reducible }\iff \theta\rtimes\s \text{ is reducible for some }\theta \in \D.
$$

\begin{remark} \label{rem-irr}
\begin{enumerate}
\item
We shall often use the following simple consequence of Proposition 3.2 of \cite{MR2504024}.
Let $\rho\in\Cuspsd$ and assume that $\pi\in\Irr$ is supported on
$$
\{\nu^{x+z}\rho:z\in \Z\} \quad \text{ for some fixed } \quad x\in \R\, \setminus \,\tfrac12 \Z.
$$
Then, $\pi\rtimes\s$ is irreducible.

\item
One can combine the above fact with the Jantzen decomposition (see section 8 of \cite{MR3969882}) to get further irreducibilities.
We shall do it later in the paper. One can get these irreducibilities also directly from Proposition 3.2 of \cite{MR2504024}.

\end{enumerate}
\end{remark}

\section{Technical lemma on irreducibility}
\begin{lemma} \label{lemma-irr}
Let $d_1,d_2,d_3\in M(\Disc_+)$ and $\tau\in \Tempcl$. Write $d_i=(\delta_1^{(i)},\dots, \delta_{k_i}^{(i)})$, $i=1,2,3$. Suppose
\begin{enumerate}
\item
\label{lemma-irr-item: irr1}
 $L(d_1)\times L(d_2)$ is irreducible;

\item
\label{lemma-irr-item: irr2}
 $L(d_1)\times L(d_2)\check{\ }$ is irreducible;

\item
\label{lemma-irr-item: irr3}
 $L(d_1)\times L(d_3;\tau)$ is irreducible;

\item
\label{lemma-irr-item: bigger}
$e(\delta_j^{(i)})\geq e(\delta_l^{(3)})$ for all \ $i=1,2$, \ $1\leq j\leq k_i$, \ $1\leq l \leq k_3$;

\item
\label{lemma-irr-item: theta}
$\bar d_i\cong d_i^\Theta$, $i=1,2,3.$
\end{enumerate}
Then,
$$
L(d_1)\rtimes L(d_2+d_3;\tau)
$$
is irreducible.
\end{lemma}

\begin{proof}
First, $L(d_2+d_3;\tau)$ is the unique irreducible quotient of $\lambda(d_2+d_3;\tau)$.
Condition (\ref{lemma-irr-item: bigger}) implies that $L(d_2)\rtimes L(d_3;\tau) $ is also quotient of $\lambda(d_2+d_3;\tau) $.
Therefore, $L(d_2+d_3;\tau)$ is (the unique irreducible) quotient of $L(d_2)\rtimes L(d_3;\tau) $.
This implies that $L(d_1)\rtimes L(d_2+d_3;\tau)$ is a quotient of $L(d_1)\times L(d_2)\rtimes L(d_3;\tau) $.
Since $L(d_1)\times L(d_2)\rtimes L(d_3;\tau) = L(d_1+d_2)\rtimes L(d_3;\tau) $, by condition (\ref{lemma-irr-item: irr1})
the last representation is a quotient of $\lambda(d_1+d_2+d_3;\tau)$.
This implies that $L(d_1)\rtimes L(d_2+d_3;\tau)$ has a unique irreducible quotient, which is $L(d_1+d_2+d_3;\tau)$,
and that this quotient occurs with multiplicity one.
Observe that (\ref{lemma-irr-item: irr1}) -- (\ref{lemma-irr-item: irr3}) imply $L(d_1)\times L(d_2)\rtimes L(d_3;\tau)\cong L(d_1)\check{\ }\times L(d_2)\rtimes L(d_3;\tau)$.
Therefore, $L(d_1)\rtimes L(d_2+d_3;\tau)$ is a quotient of $L(d_1)\check{\ }\times L(d_2)\rtimes L(d_3;\tau)$.

Obviously, $L(d_1)\check{\ }\rtimes L(d_2+d_3;\tau)$ is a quotient of $L(d_1)\check{\ }\times L(d_2)\rtimes L(d_3;\tau)$,
which implies that $L(d_1+d_2+d_3;\tau)$ is a quotient of $L(d_1)\check{\ }\rtimes L(d_2+d_3;\tau)$.
Now observe that (\ref{lemma-irr-item: theta}) implies $L(d_1+d_2+d_3;\tau)^+\cong L(d_1+d_2+d_3;\tau)$ and $L(d_2+d_3;\tau)^+\cong L(d_2+d_3;\tau)$.
Therefore,
$$
L(d_1+d_2+d_3;\tau)\h (L(d_1)\check{\ })^+\rtimes L(d_2+d_3;\tau)\cong L(d_1)\rtimes L(d_2+d_3;\tau).
$$
This implies the irreducibility of $L(d_1)\rtimes L(d_2+d_3;\tau)$ since $L(d_1+d_2+d_3;\tau)$ is a unique irreducible quotient
of $L(d_1)\rtimes L(d_2+d_3;\tau)$, and it has multiplicity one in $L(d_1)\rtimes L(d_2+d_3;\tau)$.
\end{proof}

In the paper, we shall use the following special case of the above lemma:

\begin{corollary} \label{irr-simple}
Let $\rho\in\Cuspsd$ and $\tau\in \Tempcl$.
Suppose that $d_1,d_2 \in M(\Disc_+)$ are such that all elements in their supports are contained
in either $\{\nu^{k+\frac12}\rho:k\in\Z\}$ or $\{\nu^{k}\rho:k\in\Z\}$.
If the following three representations
$$
L(d_1)\times L(d_2), \quad L(d_1)\times L(d_2\check{\ }), \quad L(d_1)\rtimes\tau
$$
are irreducible, then
$$
L(d_1)\rtimes L(d_2;\tau)
$$
is irreducible.\qed

\end{corollary}

\section{Distinguished irreducible subquotient in induced representation} \label{dist}
Fix $\rho\in\Cuspsd$ and $\sigma\in\Cuspcl$.
Let $c$ be a multiset of elements of representations in
$$
\{\nu^{k+\frac12}\rho:k\in\Z\}\ (\subseteq M(\Cusp)\subseteq M(\Disc)).
$$
Then, $\lambda(c^\uparrow)$ has a unique generic irreducible subquotient (which has multiplicity one in $\lambda(c^\uparrow)$).
Denote it by $\lambda(c^\uparrow)_{gen}$.
Now the formula \eqref{mu*} directly implies that the multiplicity of $\lambda(c^\uparrow)_{gen}\otimes\s$ in
$s_{\GL}(\lambda(c)\rtimes\s)$ is one.
This implies that $\lambda(c)\rtimes\s$ has a unique irreducible subquotient $\pi$ which contains
$\lambda(c^\uparrow)_{gen}\otimes\s$ in $s_{\GL}(\pi)$ as a subquotient. We denote this $\pi$ by
$$
\lambda(c;\rho)_+.
$$
Clearly, it has multiplicity one in $\lambda(c)\rtimes\s$.

Let now $c$ be a (finite) multiset of elements of
$$
\{\nu^{k}\rho:k\in\Z\}.
$$
Then, $\lambda(c^\uparrow+c_u)$ has a unique generic irreducible subquotient (which has multiplicity one in $\lambda(c^\uparrow+c_u)$).
Denote it by $\lambda(c^\uparrow+c_u)_{gen}$.
Again the formula \eqref{mu*} directly implies that the multiplicity of
$\lambda(c^\uparrow+c_u)_{gen}\otimes\s$ in $s_{\GL}(\lambda(c)\rtimes\s)$ is $2^{m(\rho,c)}$,
where $m(\rho,c)$ is the multiplicity of $\rho$ in $c$.
In this case we have the following

\begin{lemma} \label{lem: crho+}
Suppose that $c$ does not contain the reducibility point $\nu^\alpha\rho$, or that $\alpha>0$.
Then, $\lambda(c)\rtimes\s$ has a unique irreducible subquotient $\pi$ which contains
$\lambda(c^\uparrow+c_u)_{gen}\otimes\s$ in $s_{\GL}(\pi)$ as a subquotient. We denote this $\pi$ by
$$
\lambda(c;\rho)_+.
$$
It has multiplicity one in $\lambda(c)\rtimes\s$ and its Jacquet module contains $\lambda(c^\uparrow+c_u)_{gen}\otimes\s$ with
multiplicity $2^{m(\rho,c)}$.
\end{lemma}

\begin{proof}
It suffices to show that the Jacquet module of $
\lambda(c;\rho)_+.
$ contains $\lambda(c^\uparrow+c_u)_{gen}\otimes\s$ with multiplicity $2^{m(\rho,c)}$. We now prove this claim.

First consider the case when $c=\sum_{i=1}^n \D_i$, for some $\D_i\in\SC$ such that $\D_i\check{\ }=\D_i$ for all $i$.
We shall see by induction that in this case the lemma holds, and we shall show that
$\lambda(c;\rho)_+$ is a subrepresentation of $ (\prod _{i=1}^n \delta(\D_i))\rtimes\s$.
By the theory of $R$-groups it is enough to prove the claim when all $\D_i$ are different, and all $\D_i$ contain $\nu^\alpha\rho$.

Let $i=1$. Denote $\D_{1, \alpha\leq }:=\{\nu^\beta\rho\in\D_i;\beta\geq \alpha\}$.
Consider $\delta(\D_1\setminus \D_{1, \alpha\leq })\rtimes\delta(\D_{1, \alpha\leq };\s)$ and $\delta(\D_1)\rtimes\s$.
The last representation has length two.
In the Jacquet module of both representations, $\lambda(c^\uparrow+c_u)_{gen}\otimes\s$ has multiplicity 2 (as well as of $\lambda(c)\rtimes\s$).
From Jacquet module easily follows that
$\delta(\D_1)\rtimes\s\not\leq \delta(\D_1\setminus \D_{1, \alpha\leq })\rtimes\delta(\D_{1, \alpha\leq };\s)$.
This, together with the above multiplicities of $\lambda(c^\uparrow+c_u)_{gen}\otimes\s$, imply the claim.

For $i=2$, we consider $\delta(\D_1)\rtimes \lambda(\D_2;\rho)_+$ and $\delta(\D_2)\rtimes \lambda(\D_1;\rho)_+$.
We conclude in a similar way. Multiplicity of $\lambda(c^\uparrow+c_u)_{gen}\otimes\s$ is now 4 in both Jacquet modules (and in Jacquet module of $\lambda(c)\rtimes\s$).

For the general case, we consider $\delta(\D_1)\rtimes \lambda(\D_2+\dots+\D_n;\rho)_+$ and
$\delta(\D_n)\rtimes \lambda(\D_1+\dots+\D_{n-1};\rho)_+$ (see also Proposition 5.1 of \cite{MR3123571}, and its proof).

Now we go to the proof of the general case.
The first observation is that one can easily show that there exists $c'\in M(\Cusp)$ such that
\begin{enumerate}
\item $\ssm(\lambda(c)\rtimes\s)=\ssm(\lambda(c')\rtimes\s)$
\item
\label{6conditions}
 there exist $\D_1,\dots,\D_k,\G_1,\dots,\G_l\in \SC$ such that

\begin{enumerate}
\item $c'=\D_1+\dots+\D_k+\G_1+\dots+\G_l$;

\item $\mathfrak c(\D_i)\geq 0$ and $\rho\in\D_i$, $i=1,\dots,k$;

\item $\D_{i+1}\cup\D_{i+1}\check{\ }\subseteq \D_{i}\cap\D_{i}\check{\ }$, $i=1,\dots, k-1$;

\item $\mathfrak c(\G_j)> 0$ and $\rho\not\in \G_j$, $j=1,\dots,l$;

\item $\G_j$ is not linked to any other $\G_{j'}$, or any $\D_i$, $j=1,\dots,l$

\item $\G_j\check{\ }$ is not linked to any $\D_i$, $j=1,\dots,l$\footnote{To get these segments,
one consider $\nu^x\rho\in c$ with maximal $|x|$.
Then, $\nu^{|x|}\rho$ is the right end of $\D_1$ or $\G_1$ (it depends on the fact if by the process that follows one
will reach $\rho$ or not).
Then, one looks if $\nu^{|x|-1}\rho\in c$ or $\nu^{-(|x|-1)}\rho\in c$
(if there is no such a member, then the first segment consists of $\nu^{|x|}\rho$ and we repeat above
search with $c-(\nu^x\rho)$).
If yes, one has the next point of the segment of cuspidal representations and we continue the above procedure
(looking for an exponent which is smaller for one then the previous exponent) as long as we can,
in forming the first segment of cuspidal representations.
After we cannot continue the above procedure, we have got the first segment (which is $\D_1$ or $\G_1$,
depending if $\rho$ is in it, or not).
Now we repeat the above procedure with $c$ from which we have removed terms used in the above process.
We repeat these steps as long as there are remaining members of $c$. In this way one gets segments in (\ref{6conditions}).}.

\end{enumerate}

\end{enumerate}

Observe that $k=m(\rho,c)$.

Suppose that the claim does not hold for this $c$ (and $\s$).
This implies that in $\lambda (c)\rtimes\s$ there exists an irreducible subquotient $\pi$ such that
the multiplicity $m$ of $\lambda(c^\uparrow+c_u)_{gen}\otimes\s$ in $s_{\GL}(\pi)$ satisfies $0<m<2^k$.
We know
\begin{equation}\label{leq1}
\pi\leq ( \prod_{i=1}^k\delta(\D_i))\times ( \prod_{j=1}^l\delta(\G_j))\rtimes\s,
\end{equation}
since the multiplicity of $\lambda(c^\uparrow+c_u)_{gen}\otimes\s$ in the Jacquet module of the right-hand side is $2^k$,
which is the same as it is in $\lambda(c')\rtimes\s$ (we shall use this argument also below, without repeating this explanation).
The above inequality implies
$$
( \prod_{i=1}^k\delta(\D_i\check{\ }\setminus \D_i))\rtimes\pi\leq
( \prod_{i=1}^k\delta(\D_i\check{\ }\setminus \D_i))\times
( \prod_{i=1}^k\delta(\D_i))\times ( \prod_{j=1}^l\delta(\G_j))\rtimes\s.
$$
Denote $\D_i''=\D_i\cup \D_i\check{\ }$ and $c''=\D_1''+\dots+\D_k''+\G_1+\dots+\G_l$.
Considering how on the right-hand side we get $\lambda((c'')^\uparrow+c''_u)_{gen}\otimes\s$
in the Jacquet module (all of them we get from terms of $\lambda(c^\uparrow+c_u)_{gen}\otimes\s$ multiplying with
$\delta(\D_i\setminus\D_i\check{\ })$'s and taking appropriate subquotient),
we conclude that its multiplicity in the left-hand side is $m$.
Therefore, there is an irreducible subquotient $\pi''$ of the left-hand side which has
$\lambda((c'')^\uparrow+c''_u)_{gen}\otimes\s$ in its Jacquet module with multiplicity $m$.
Now in the same way as in the case of \eqref{leq1}, we conclude
\begin{equation} \label{leq2}
\pi''\leq ( \prod_{j=1}^l\delta(\G_j)) \times ( \prod_{i=1}^k\delta(\D_i''))\rtimes\s.
\end{equation}
Write $\G_j=[\nu^{g_{j,b}}\rho,\nu^{g_{j,e}}\rho]$. Denote
$\G_j'=[\nu^{-g_{j,e}}\rho, \nu^{g_{j,b}-1}\rho]$,
$\G_j''=\G_j\cup \G_j'$, $c'''=\D_1''+\dots+\D_k''+\G_1''+\dots+\G_l''$. Then,
$$
( \prod_{j=1}^l\delta(\G_j'))\rtimes\pi''\leq ( \prod_{j=1}^l\delta(\G_j'))\times ( \prod_{j=1}^l\delta(\G_j)) \times ( \prod_{i=1}^k\delta(\D_i''))\rtimes\s.
$$
Considering how on the right-hand side we get $\lambda((c''')^\uparrow+c'''_u)_{gen}\otimes\s$ in the Jacquet module (all of them we get from terms of
$\lambda((c'')^\uparrow+c_u'')_{gen}\otimes\s$ multiplying with the following two $\delta([\rho,\nu^{g_{j,e}}\rho])\times\delta([\nu\rho,\nu^{g_{j,b}-1}\rho])$, $\delta([\nu\rho,\nu^{g_{j,e}}\rho])\times\delta([\rho,\nu^{g_{j,b}-1}\rho])$ subquotients of $M^*(\delta(G_j'))$'s, and taking appropriate irreducible subquotient),
we conclude that its multiplicity is $2^lm$ in the left-hand side, which is strictly smaller then $2^{k+l}$. Directly follows that this multiplicity is positive.
This is a contradiction with the first part of the proof. The proof is now complete.
\end{proof}

\begin{remark}
\begin{enumerate}
\item If $
\s$ is generic, then $\lambda(c;\rho)_+$ is generic $($then $\alpha\in\{\frac12,1\}$; we do not consider here reducibility at $0$).

\item Since $\Cusp\subseteq \Disc$, then $M(\Cusp)\subseteq M(\Disc)$.
We say that $\pi\in\Irr$ is cogeneric if $\pi=L(d)$ for some $d\in M(\Cusp)$.

Let $c$ be a multiset of elements of $\{\nu^{k+\frac12}\rho:k\in\Z\}$.
Then, $\lambda(c^\uparrow)\check{\ }$ has a unique cogeneric irreducible subquotient
(which has multiplicity one in $\lambda(c^\uparrow)\check{\ }$).
Denote it by $\lambda(c_\downarrow)_{cogen}$.
Now the formula \eqref{mu*} directly implies that the multiplicity of $\lambda(c_\downarrow)_{cogen}\otimes\s$
in $s_{\GL}(\lambda(c)\rtimes\s)$ is one.
This implies that $\lambda(c)\rtimes\s$ has a unique irreducible subquotient $\pi$ which contains
$\lambda(c_\downarrow)_{cogen}\otimes\s$ in $s_{\GL}(\pi)$ as a subquotient. We denote this $\pi$ by
$$
\lambda(c;\rho)_-.
$$
We define representation $\lambda(c;\rho)_-$ analogously for a multiset $c$ of representations in $\{\nu^{k+\frac12}\rho:k\in\Z\}$
(in this case we consider $\lambda(c^\uparrow\check{\ }+c_u)_{cogen}\otimes\s$).
Then, the analogue of Lemma \ref{lem: crho+} holds for $\lambda(c;\rho)_-$.
Furthermore,
\begin{equation} \label{dist-formula}
\lambda(c;\rho)_+^t=\lambda(c;\rho)_-.
\end{equation}
\end{enumerate}
\end{remark}

Suppose that $\pi=\lambda(c;\rho)_+$ is square-integrable, and $\D$ is a segment contained in
$\{\nu^{k+\frac12}\rho:k\in\Z\}$ or $\{\nu^{k}\rho:k\in\Z\}$, such that $\D=\D\check{\ }$.
Then, we denote $\pi=\lambda(c+\D;\rho)_+$ by
$$
\tau(\D_+;\pi).
$$
One directly sees that $\tau(\D_+;\pi)\leq \delta(\D)\rtimes\pi$.
Furthermore, if $\delta(\D)\rtimes\pi$ is reducible, then it decomposes into a direct sum of two nonequivalent irreducible (tempered) representations.
The other one we denote by
$$
\tau(\D_-;\pi).\footnote{Note that $\tau(\D_-;\pi)$ is not related to the representations of type $\lambda(c;\rho)_-$.
Furthermore, note that if $\pi$ is a cuspidal representation $\s$, then $\tau(\D_\pm;\pi)=\delta(\D_\pm;\pi)$.}
$$

\section{Some well-known ways of obtaining unitarizability} \label{ways of getting unitary}
The standard way of obtaining new unitarizable representations from old ones is by parabolic induction,
which preserves unitarity (unitary parabolic induction).

One can also deduce unitarizability in the opposite direction.
Namely, if $\theta$ is an irreducible hermitian representation of a Levi subgroup $M$
of a parabolic subgroup $P$ of a reductive group $G$, and if Ind$_P^G(\theta)$ is irreducible and unitarizable, then $\theta$ is unitarizable.
This method of proving unitarizability will be called \emph{unitary parabolic reduction}.

A third way of proving unitarizability is by considering limits.
If $\pi_n$ is a sequence of irreducible unitarizable representations of a reductive group $G$,
$\tau_i$ irreducible representations of $G$ and $m_i\in\Z_{>0}$ such that distribution characters
$\Theta_{\pi_n}$ of $\pi_n$ converge pointwise to $\sum_im_i\Theta_{\tau_i}$, then all $\tau_i$ are unitarizable (\cite{MR0324429}).

A fourth way of proving unitarizability is by considering families.
If a continuous family of irreducible hermitian representations of a reductive groups $G$
contains at least one unitarizable representation, then all representations in the family are unitarizable.
(For a definition of continuous family of representation see \cite[\S 3 (b)]{MR1181278}.)

Furthermore, if a continuous family of irreducible hermitian representations is parameterized by unbounded set of unramified parameters,
then all the representations in the family are non-unitarizable (see \cite{MR733166} for more details).

The above methods of proving unitarizability can be easily modified for proving non-unitarizability.

\subsection{}
We will use the following elementary, but powerful, consequence of unitarizability.
\begin{lemma} \label{lem: nonunit}
Let $P$ be a parabolic subgroup of a reductive group $G$ over a non-archimedean local field.
Let $\sigma$ be an irreducible representation of the Levi part $M$ of $P$ and let $\Pi=\Ind_P^G\sigma$.
Assume that $\sigma$ or $\sigma^t$ is unitarizable.
Then, the length $\ell$ of $\Pi$ is at most the multiplicity $m$ of $\sigma$ in the Jacquet module $J_P(\Pi)$.
Moreover, if $\ell=m$, then
there exists a direct summand of $J_P(\Pi)$ isomorphic to $m\cdot\sigma=\overbrace{\sigma\oplus\dots\oplus\sigma}^m$.
\end{lemma}

\begin{proof}
Using duality, it is enough to consider the case where $\sigma$ unitarizable.
In this case $\Pi$ is unitarizable, hence semisimple and therefore
$\ell(\Pi)\le\dim\Hom_G(\Pi,\Pi)$ which by Frobenius reciprocity is equal $\dim\Hom_M(J_P(\Pi),\sigma)\le m$.

Moreover, if $\ell=m$ then necessarily $m=\dim\Hom_M(J_P(\Pi),\sigma)$.
Upon replacing $\sigma$ by its contragredient and $P$ by an opposite parabolic, we also get that
$m=\dim\Hom_M(\sigma,J_P(\Pi))$.
Therefore,
\[
\omega:=\sum_{\varphi\in\Hom_M(\sigma,J_P(\Pi))}\Img(\varphi)\cong m\cdot\sigma\text{ and }
J_P(\Pi)=\omega\oplus\bigcap_{\varphi\in\Hom_M(J_P(\Pi),\sigma)}\Ker\varphi
\]
as claimed.
\end{proof}

\section{Reduction of unitarizability to the weakly real case} \label{reduction-u-w}
$\pi\in\Irrcl$ is called weakly real if it is a subquotient of a representation of the form
$$
\nu^{r_1}\rho_1\times\dots \times\nu^{r_k}\rho_k\rtimes\s,
$$
where $\rho_i\in \Cusp$ satisfy $\rho_i\cong\rho_i\check{\ }$,
$r_i\in \R$ and $\s\in\Cuspcl$.
Now we recall \cite[Theorem 4.2 (i)]{MR2504024}:

\begin{theorem} \label{unitary-red-1}
For any unitarizable $\pi\in\Irrcl$ is unitarizable there exist a unitarizable $\theta\in\Irr$ and a weakly real unitarizable
$\pi'\in\Irrcl$ such that
$$
\pi\cong\theta\rtimes\pi'.
$$
\end{theorem}

Note that \cite[Theorem 4.2 (ii)]{MR2504024} gives a more precise reduction then the above theorem.
Since \cite[Theorm 7.5]{MR870688} (which we recall below) gives a classification of unitary duals of general linear groups,
the above theorem reduces the unitarizability problem for classical $p$-adic groups to the weakly real case.

For $\delta\in \Disc_u$ and $m\geq 1$ denote by $u(\delta,m)$ the unique irreducible quotient of
$\nu^{\frac{m-1}2}\delta\times\nu^{\frac{m-1}2-1}\delta\times\dots \times\nu^{-\frac{m-1}2}\delta,$
which is called a Speh representation. Let
$
B_{rigid}
$
be the set of all Speh representations, and
\[
B=B(F)=B_{rigid}\cup \{\nu^\alpha\s\times\nu^{-a}\s;\s\in B_{rigid}, 0<\alpha<1/2\}.
\]
Now the following simple theorem solves the unitarizability for archimedean and non-archimedean general linear groups in the uniform way:

\begin{theorem} \label{ud-gl}
The mapping $(\s_1,\dots,\s_k)\mapsto \s_1\times\dots\times\s_k$
defines a bijection between $M(B)$ and $\cup_{n\geq0}\GL(n,F)\widehat{\ }$.
\end{theorem}

\section{Computing irreducible subquotients}
The following simple lemma will be often used for computing irreducible subquotients.
\begin{lemma} \label{lem: simpac} \ \
\begin{enumerate}
\item Suppose that we have a diagram
\[
\xymatrix{
0\ar[r] & \Sigma_2 \ar[r] & \Sigma_1 \ar@{->>}[d] \ar[r] & \Pi \ar[r]&0\\
& & \Sigma
}
\]
where the top row is a short exact sequence.
Assume that $\Sigma$ is finitely generated and $\Sigma\not\le\Sigma_2$.
Then, the cosocle of $\Sigma$ and the cosocle of $\Pi$ have a common irreducible constituent.
In particular, if the cosocle of $\Pi$ is irreducible, then it occurs in the cosocle of $\Sigma$.
\item Suppose that we have a diagram
\[
\xymatrix{
0\ar[r] & \Sigma_2 \ar[r] & \Sigma_1 \ar@{->>}[d] \ar[r] & \Pi \ar[r]&0\\ & & \Sigma
}
\]
where the top row is a short exact sequence.
Assume that $\Sigma_2$ is finitely generated and $\Sigma\not\le\Pi$.
Then, $\Sigma$ and the cosocle of $\Sigma_2$
have a common irreducible subquotient.
In particular, if the cosocle of $\Sigma_2$ is irreducible, then it occurs in $\Sigma$.
\end{enumerate}
\qed
\end{lemma}

Indeed, in the first part we get a map $\Pi\twoheadrightarrow\overline{\Sigma}$ where
$\overline{\Sigma}$ is the quotient of $\Sigma$ by the image of the composition
$\Sigma_2\rightarrow\Sigma_1\rightarrow\Sigma$.
In the second part, the composition $\Sigma_2\rightarrow\Sigma_1\rightarrow\Sigma$ is non-zero.

\begin{remark} \label{rem: mdtp}
Let $\omega=\delta([-a,a]^{(\rho)})$ be a square-integrable representation and
let $\pi$ be an irreducible subquotient of $\Pi$. Then, the multiplicity of $\omega\otimes\pi$
in $\mu^*(\omega\rtimes\Pi)=M^*(\omega)\rtimes\mu^*(\Pi)$ is equal the multiplicity of $\omega\otimes\pi$ in
\[
\Big(2\sum_{i=-a}^a\delta([-i,a]^{(\rho)})\otimes\delta([i+1,a]^{(\rho)})+1\otimes\omega\Big)\rtimes\mu^*(\Pi)
\]
\end{remark}

\endinput


TITLE: Proximal Operator of Summation of $ {L}_{1} $ Norm and $ {L}_{2, 1} $ Norm
QUESTION [3 upvotes]: I would like the proximal operator of the following function:
\begin{align}
  f(x) &= \lambda_1 \|x\|_1 + \lambda_2 \|x\|_{2,1} \\
       &= \lambda_1 \|x\|_1 + \lambda_2 \sum_g \|x_g\|_2
\end{align}
where $x$ is the concatenation of all $x_g$, $\|x\|_1=\sum_i|x_i|$, and $\|y\|_2=\sqrt{\sum_i y_i^2}$.
I appreciate any help that is offered.

REPLY [3 votes]: This is a special case of the more general class of summative proximable functions, i.e. functions of the form
$$
x \mapsto f(x) + g(x)
$$
which satisfy $\mathrm{prox}_{f + g}(x) = \mathrm{prox}_{f} \left( \mathrm{prox}_g(x)
\right)$. See Appendix D of A Primer on Coordinate Descent Algorithms
for more details and a proof.
For your specific instance, notice that since $\| x \|_1 = \sum_{g \in G} \| x_g \|_1$ from your definition of $x$, you can decompose your proximity operator into
$G$ separate instances of the form
$$
f_g(x) = \lambda_1 \| x_g \|_1 + \lambda_2 \| x_g \|_2.
$$
Each of these is then solvable using the rule for summative functions, i.e.
$$
\mathrm{prox}_{f_g}(x) = \mathrm{prox}_{\lambda_1 \| \cdot \|_1} \big(
\mathrm{prox}_{\lambda_2 \| \cdot \|_2}(x) \big).
$$
Both of these proximal operators are known in closed form.
See Chapter 6 in First Order Methods in Optimization.


TITLE: What is the largest number of squares that can be cut by the sides of the triangle in this picture?
QUESTION [1 upvotes]: I came across a question in the Mathematics Olympiad exercise booklet for primary school students in Australia. The questions is presented in the following picture. I can't quite understand what the question is about. Would anyone please help make the question clearer and suggest a solution? Thank you very much in advance.

REPLY [1 votes]: If you keep vertex $A$ at the lower left corner of the grid, and you allow one vertex of the triangle to be anywhere on the right side of the rectangle enclosing the grid, and the other vertex to be anywhere on the top of the rectangle enclosing the grid, what is the maximum possible number of squares that the edges of the triangle will pass through?  You can see in the example figure that the triangle's bottom edge passes through $6$ squares; the top edge passes through another $6$ squares; and the left edge passes through $6$ squares.  But for the bottom and top vertices, the squares are counted twice, so this triangle passes through $6 + 6 + 6 - 2 = 16$ different squares.  By changing the position of the top and right vertices of the triangle, can you make it pass through more squares?


TITLE: Could you please help me understand the logic behind the 'New Years Chaos' question?
QUESTION [1 upvotes]: This is a coding challenge, but first I need to understand the problem and the solution from a purely thought/logical point of view.

I kind of worked out how to find the people in the list who have bribed their way forward, but I can't quite workout the logic behind finding out how many times they have bribed their way forward.
This is how much I know so far:

For a queue of people that looks like this: 1, 2, 5, 3, 7, 8, 6, 4 I have figured out that by checking all the people in front of each person, the people with larger ID (number) would be those who have bribed their way forward.

So at the end, I get the following list of bribers: 5, 7, 8, 6
But how do I find how much times there has been a bribing. Basically, I need to know how much the briber has moved forward. That is the bit that I fail to grasp.
I can't simply compare their initial position to their final position. In the thought experiments I ran, some of the bribers ended up in their original queue slot because of other briber's movements.
This might help, an algorithm that I know works, but is not my solution so the problem still stands, however, this might help someone to understand the complete solution.
def minimumBribes(q):
    # variables
    # last index is 7
    lastIndex = len(q) - 1
    moves = 0
    swap = False

    # Check if too chaotic?
    for i, v in enumerate(q):
        if (v - 1) - i > 2:
            print('Too chaotic')
            return

    # figure out number of moves
    # 'i' is a range from 0 to 7
    for i in range(len(q)):
        # 'j' is in range from 0 to 7
        for j in range(lastIndex):
            if q[j] > q[j+1]:
                q[j], q[j+1] = q[j+1], q[j]
                moves += 1
                swap = True

        # check if any swap have occurred - if not it means the queue is sorted
        if swap:
            swap = False
            lastIndex -= 1
        else:
            print(moves)
            return
    print(moves)

REPLY [1 votes]: It's important to note here that we're interested in the minimum number of bribes.
Firstly, note that a person can only move position in the queue if they either bribe someone, or accept a bribe. Other people's actions that don't involve them have no effect on their position. So since a person can only bribe a maximum of $2$ times, they can only be in $3$ possible positions. Their initial position ($k$), the position in front of their initial position ($k-1$) or the position two in front of their initial position ($k-2$). This is obviously considering the case where the queue length is $3$ or greater.
So now let's break the problem down into cases, by considering a queue of length $k$, and doing analysis on the position of person $k$ and using this position to break the problem into subproblems.
If person $k$ is at the back of the queue we can assume they never moved, as we're looking for a minimum, so just remove them and consider the ($k-1$)-person queue that's made of the first $k-1$ people in the queue we're looking at, as we can assume that person $k$ was never involved in any bribes.
If person $k$ is at position $k-1$, assume the sequence of bribes ended with them moving one step forward (this is fine as this doesn't affect the other bribes, so the order here doesn't matter). So the queue went from $(n_1,...,n_{k-1},k)$ to $(n_1,...,n_{k-2},k,n_{k-1})$ with this final step. So before this, person $k$ didn't move. So consider the $(k-1)$ person queue $(n_1,...,n_{k-1})$ and then add $1$ for the final bribe to the total that the subproblem gives us.
Finally, if person $k$ is at position $k-2$, assume the sequence ended with 2 bribes from them (this is fine again for the same reasons as above). So before these two bribes the queue looks like: $(n_1,...,n_{k-2},n_{k-1},k)$, and ends up looking like: $(n_1,...,n_{k-3},k,n_{k-2},n_{k-1})$. So consider the subproblem on the $k-1$ length queue $(n_1,...,n_{k-2},n_{k-1})$. Then add $2$ to the total that this subproblem gives us.
These three cases cover all possible legal queues. If person $k$ is at any other position, we know they must have bribed more than twice, which isn't possible. So the algorithm would return "too chaotic" in these cases.
Of course they could accept more bribes, but they'd only be able to move back as far as the final position in the queue. There's no way for them to be outside of these three positions, and we're only interested in the minimum possible, so we don't need to consider sequences of bribes that put us back into a position we've already been in, as these aren't minimal.

REPLY [0 votes]: In your example of $1, 2, 5, 3, 7, 8, 6, 4$ you identified the bribers by the fact they are ahead of a smaller number
The number of bribes paid by a briber is at least (and optimally is equal to) the number of smaller numbers they are ahead of since they must have bribed them to get ahead of them, so you can conclude:

$5$ is ahead of $3$ and $4$ so paid two bribes
$6$ is ahead of $4$ so paid one bribe
$7$ is ahead of $6$ and $4$ so paid two bribes
$8$ is ahead of $6$ and $4$ so paid two bribes

so $7$ bribes were paid in total.  A possible example is
Briber  Bribee  Pattern
                1   2   3   4   5   6   7   8
5       4       1   2   3   5   4   6   7   8
5       3       1   2   5   3   4   6   7   8
6       4       1   2   5   3   6   4   7   8
7       4       1   2   5   3   6   7   4   8
7       6       1   2   5   3   7   6   4   8
8       4       1   2   5   3   7   6   8   4
8       6       1   2   5   3   7   8   6   4


TITLE: Wouldn't real set be indeed smaller rather than bigger than the integer set?
QUESTION [1 upvotes]: This is linked to Natural and Real sets of numbers, which one is bigger than another? but clearly my question is of a different nature.
Hello there,
I would like to ask the community a question : Why is it that one say that "real" are "bigger" than "integer" ??

Now i'm well aware of the diagonal argument, but it seems to me that there is no logical connection to bigger, and even that the real set is indeed smaller than the integer set. I'll try to explain, using only logical arguments, because i don't know much about mathematics, but i hope that by using logic i'll be able to communicate with the mathematical community and be able to grasp what's wrong in my understanding.

I would like to try to define a few words i'll be using first :
Let's assume that there exists a turing complete language L, that you can write programs with, such that the program can process a finite string, and produce another finite string, and has a terminal instruction (ie programs can terminate).
Then let's call Program, something written in L, that take as input an integer (let's say as a finite string of 0/1), and output either 0 or 1 (but has no guarantee to terminate).
Let's call "List" a program that take as input an integer, and output a finite string, and is guaranteed to terminate for any integer input.
Let's call "Subpart of the Integer set" (shorten as SI) a Program that somehow is guaranteed to terminate for any integer input. (So any SI is both a List and and a program)
Let's call maximal List (of a criterion C), a list (if there exists one) that given the criterion C list all output string that meets C.

So now we can say that there is indeed a maximal list of Programs (trivial to explicit), and that there is no maximal List of SI (because if there was, then you'd be able to have a program that answer the termination problem for any input).
Now, there is another way of noticing that there is no maximal List of SI, and that is to apply the "diagonal strategy".
Let's presume that there exist an explicit List of all SI called List_SI, then you can build a new SI :
new_SI (n) :
- let a = List_SI (n) (n). If a = 0 return 1, else return 0;
new_SI is in 0/1 and terminate, so it is definitely a SI.
Now, this look to me as exactly the diagonal argument used by Cantor, wich enable to conclude that "Real Set" is "Bigger" than "Integer Set".
Only we replaced the real set with the list of programs that represents a subpart of N, and the integer set with a generalization to programs that represents a subpart of N but might leave some integer neither into or out of the the subpart it represents.
Thus, we have to conclude that the Program that terminate Set (SI) is Bigger than the set of all programs (Program). Wich seems a bit counter intuitive to me. Indeed i would tend to assume that the set of program that terminate is indeed smaller.

And there i am, asking the mathematical community : wouldn't be the "real set" indeed be "smaller" than the "integer set" ?

REPLY [2 votes]: Sorry: I will use the word "program" in its correct meaning, i.e. something having as an input a number and as an output a number, or it does not terminate.
I recall you the notion of recursively enumerable set: a subset $A \subseteq \Bbb{N}$ is recursively enumerable if there exists a program $L(n)$ such that
$$\{ L(n): n \in \Bbb{N} \} = A$$
It is true that you can list all programs: this gives you a universal list $U(n)$ listing all programs. In particular
$$\{ n \in \Bbb{N} : U(n) \mbox{ is a program} \} = \Bbb{N}$$
is recursively enumerable.
Now, what you have proved is that
$$\{ n \in \Bbb{N} : U(n) \mbox{ is a SI (i.e. it always terminates with $0$ or $1$)} \}$$
is not recursively enumerable. This is an important result in computability theory, due to Turing himself.
Despite being recursively enumerable, $\Bbb{N}$ has some non recursively enumerable subsets. But this does not mean that the collection of SI is bigger than the collection of programs. Not at all.
This result means that you can list all programs, but you have totally no way of deciding which one of them is a SI and which one isn't. So, in particular, there is no way to ensure that a program is a list (this is referred to your sentence "is guaranteed to terminate for any integer input": you cannot have a guarantee).


TITLE: projective geometry and relationship of cross-ratios
QUESTION [1 upvotes]: Define for pairwise different points $P_i=[v_i]$ the cross-ratio $\operatorname{CR}(P_1,P_2,P_3,P_4) = \frac{\det(v_1,v_2)}{\det(v_2,v_3)}\cdot\frac{\det(v_3,v_4)}{\det(v_4,v_1)}$ on $\mathbb{KP^1}$ where $\det$ is a determinant function on the underlying vector space. $\operatorname{CR}$ is independent of the choice of $\det$ and scaling of the $v_i$.
Show
$$\operatorname{CR}(P_1,P_2,P_3,P_4) = \operatorname{CR}(P_3,P_2,P_1,P_4) \implies \operatorname{CR}(P_1,P_2,P_3,P_4) = -1$$

Using that $\det$ is alternating, I showed that $\operatorname{CR}(P_3,P_2,P_1,P_4) = (\operatorname{CR}(P_1,P_2,P_3,P_4))^{-1}$ and hence from $\operatorname{CR}(P_1,P_2,P_3,P_4) = \operatorname{CR}(P_3,P_2,P_1,P_4) = (\operatorname{CR}(P_1,P_2,P_3,P_4))^{-1}$ concluded that $\operatorname{CR}(P_1,P_2,P_3,P_4) \in \{-1,1\}$.
But how to eliminate now the option $1$?
Based on that I am asked tho show for pairwise points $P_1,P_2,P_3,P_4 \in \mathbb{R}$ (how should I interpret this here?), that the following are equivalent:
a. $\operatorname{CR}(P_1,P_2,P_3,P_4) = -1$ (what is CR here? we actually defined it for $\mathbb{KP^1}$)
b. either $P_3$ or $P_4$ lies between $P_1$ and $P_2$ and $\frac{|P_1-P_3|}{|P_3-P_2|} = \frac{|P_1-P_4|}{|P_4-P_2|}$
I do not even get the formulation of this second part..

REPLY [0 votes]: Funny, I did the proof which excludes $1$ in a lecture today. The basic idea is that you can show that if the cross ratio is $1$, then two points must coincide. Without loss of generality you may assume
$$
v_1=\begin{pmatrix}a\\1\end{pmatrix} \qquad
v_2=\begin{pmatrix}b\\1\end{pmatrix} \qquad
v_3=\begin{pmatrix}c\\1\end{pmatrix} \qquad
v_4=\begin{pmatrix}d\\1\end{pmatrix}
$$
You can do this because the cross ratio is independent from your choice of basis, so you will always find a basis such that none of your four points is $(1,0)^T$, and then you can choose representants as above.
With this choice you get
\begin{align*}
\operatorname{CR}(P_1,P_2,P_3,P_4) &= 1\\
\frac{(a-b)(c-d)}{(b-c)(d-a)} &= 1\\
(a-b)(c-d)&=(b-c)(d-a) \\
ac-ad-bc+bd&=bd-ba-cd+ca \\
-ad-bc&=-ba-cd \\
ab-ad&=bc-cd \\
a(b-d)&=c(b-d)
\end{align*}
From this you can conclude that the cross ratio will be $1$ if and only if $P_1=P_3$ or $P_2=P_4$. So as your statement guarantees four distinct points, you know that the cross ratio cannot be $1$.
Note that these $a,b,c,d\in\mathbb R$ above are the inhomogenous desciption your second part talks about. Adding a one in the last coordinate is one possible way of homogenizing them, thus turning them into coordinates in $\mathbb{RP}^1$. So this difference computation is almost like mine, except that it uses absolute values so it has to check for order separately.


text:
import for_mathlib.homology
import for_mathlib.exact_lift_desc

noncomputable theory

open category_theory category_theory.limits opposite

variables {𝓐 : Type*} [category 𝓐] [abelian 𝓐]
variables {A B C : 𝓐} (f : A ⟶ B) (g : B ⟶ C) (w : f ≫ g = 0)

lemma homology.lift_desc (X Y Z : 𝓐) (f : X ⟶ Y) (g : Y ⟶ Z) (w)
  (U : 𝓐) (e : _ ⟶ U) (he : f ≫ e = 0) (V : 𝓐) (t : V ⟶ _) (ht : t ≫ g = 0) :
  homology.lift f g w (t ≫ cokernel.π _) (by simp [ht]) ≫
  homology.desc' _ _ _ (kernel.ι _ ≫ e) (by simp [he]) =
  t ≫ e :=
begin
  let s := _, change s ≫ _ = _,
  have hs : s = kernel.lift _ t ht ≫ homology.π' _ _ _,
  { apply homology.hom_to_ext,
    simp only [homology.lift_ι, category.assoc, homology.π'_ι, kernel.lift_ι_assoc] },
  simp [hs],
end

lemma homology.lift_desc' (X Y Z : 𝓐) (f : X ⟶ Y) (g : Y ⟶ Z) (w)
  (U : 𝓐) (e : Y ⟶ U) (he : f ≫ e = 0) (V : 𝓐) (t : V ⟶ Y) (ht : t ≫ g = 0)
  (u v) (hu : u = t ≫ cokernel.π _) (hv : v = kernel.ι _ ≫ e) :
  homology.lift f g w u (by simpa [hu]) ≫ homology.desc' _ _ _ v (by simpa [hv]) = t ≫ e :=
begin
  subst hu,
  subst hv,
  apply homology.lift_desc,
  assumption'
end

subset_name: curated


TITLE: Using Square area finding quadrilateral area
QUESTION [1 upvotes]: Area of square ABCD is 169 and that of square EFGD is 49. Find area of quadrilateral FBCG

I am stuck and just thinking which way can be helpful for me finding this area of quadrilateral FBCG. Instead of hint can please suggest me perfect answer , so using this idea answer, i will solve other questions which are near about similar.

REPLY [1 votes]: By symmetry, the quadrilaterals $FBCG$ and $FBAE$ are congruent, so they have the same area. Their combined area is the area of the region which is in the big square but outside the little square. So their combined area is $169-49$, that is, $120$. It follows that quadrilateral $FBCG$ has area $60$.
Remark: The above is the quickest way to find the answer.
But we can do it in various other ways, that may be useful to you in other problems.   For example, the little square has area $49$, so it has sides $7$ each. The big square has area $169$, so has sides $13$.
It follows that the height $CG$ of the trapezoid $FBCG$ is $13-7=6$. The two bases are $7$ and $13$. So by a standard formula for the area of a trapezoid, the area is $(6)\left(\frac{7+13}{2}  \right)$.
Or else you can think of the region $FBCG$ as made up of a $6\times 7$ rectangle, with half of a $6\times 6$ square on top. That gives area $(6)(7)+\left(\frac{1}{2}\right)(6)(6)$.
Another computational approach is by using triangles. Because of your picture, we calculate the area of $ABFE$, which is the same as the area of our target region. This is made up, as your picture shows, of two triangles. Each has height $13-7=6$. Their bases are $7$ and $13$, Now huse the standard formula for the area of a triangle.


TITLE: Where does the norm come from in the definition of differentiability?
QUESTION [0 upvotes]: Let $f: \mathbb{R}^n \to \mathbb{R}$. We define $\triangledown f(x) \in M_{n \times 1}(\mathbb{R})$ , the instantaneous rate of change of $f$ at $x \in \mathbb{R}^n$, as the linear transformation matrix such that $\triangledown f(x) x = f(x)$ and $\triangledown f(x) (x+h) = f(x+h)$ as $h \to 0$. Or, by linearity, as $$\triangledown f(x) \in M_{n \times 1}{\mathbb{(R)}}:\lim_{h \to 0} \triangledown f(x) (x + h - x) = \lim_{h \to 0}f(x+h) - f(x)$$
$$\triangledown f(x) \in M_{n \times 1}{\mathbb{(R)}}:\lim_{h \to 0} \triangledown f(x)h  = \lim_{h \to 0}f(x+h) - f(x)$$
$$\triangledown f(x) \in M_{n \times 1}{\mathbb{(R)}}:\lim_{h \to 0} f(x+h) - f(x) - \triangledown f(x)h = 0$$
In this case we say $f$ is differentiable at $x$, because it has an instantaneous rate of change at $x$.
However, in the definition of differentiability I am presented with, the norm is unaccountably present. I.e. $f: \mathbb{R}^n \to \mathbb{R}$ is differentiable at $x$  if
$$\lim_{h \to 0} \cfrac{f(x+h)-f(x) - \triangledown f(x)h}{|h|} = 0 $$
Where am I going wrong?
Edit
A lot of answers are emphasizing the importance of the norm in taking multivariate limits. That I understand. Let me try to recast the question.
We define the (average) rate of change between $x, y \in \mathbb{R}^n$ with respect to $f:\mathbb{R}^n \to \mathbb{R}$ as the linear transformation matrix $T$ such that $T(y-x) = f(y) - f(x)$; the instantenous rate of change, it would seem to follow, as the linear transformation matrix $T$ such that
$\lim_{y \to x} T(y-x) - f(y) - f(x) = 0$
Where does the norm show up in this reasoning (besides implicity, in the $\epsilon-\delta$ definition)?

REPLY [1 votes]: Note that as peek-a-boo noted, the first definition (assuming $h$ tends to the zero vector) isn't even correct, as this limit tends to zero even if $\nabla f$ is not the correct derivative. Let $\mathbf{h} = (p, q)$. For example, if $f$ is smooth, the definition you gave will assert that $\nabla f = [4, 2]$ at the origin if $f = x + y$, because
$$\lim_{\mathbf{h} \to \mathbf{0}} f(\mathbf{x} + \mathbf{h}) - f(\mathbf{x}) - \nabla f (\mathbf{x}) \cdot \mathbf{h} = \lim_{p, q \to 0} p + q - 4p - 2q = \lim_{p, q \to 0} -3p - q = 0$$
What we want is for this difference to not only tend to zero, but to tend to zero faster than a linear function of $\mathbf{h}$. If this is true, then as we take $\mathbf{h}$ closer and closer to $\mathbf{0}$, then $\nabla f \cdot h$ dominates $f(\mathbf{x} + \mathbf{h}) - f(\mathbf{x})$, so $f(\mathbf{x + h})$ is approximated to leading order by $f(\mathbf{x}) + \nabla f (\mathbf{x}) \cdot \mathbf{h}$. Note that above, the difference $f(\mathbf{x} + \mathbf{h}) - f(\mathbf{x}) - \nabla f (\mathbf{x}) \cdot \mathbf{h}$ is a linear function of $\mathbf{h}$, so it is not what we are looking for.
Up till this point, I haven't completely been thorough about what it means to "decay faster than linearly." For a smooth-enough function $g : \mathbb{R}^n \to \mathbb{R}$ to decay faster than linearly around the origin is to have for any linear function $A : \mathbb{R}^n \to \mathbb{R}$,
$$\lim_{\mathbf{h} \to \mathbf{0}} \frac{| g(\mathbf{h}) |}{|A \mathbf{h}|} = 0$$
barring the cases where $A \mathbf{h} = 0$. In other words, for any $A$, as long as we are close enough to the origin, $g(\mathbf{h})$ is significantly closer to zero than $A \mathbf{h}$.
Interestingly, a way to enforce that a given function $g(\mathbf{h})$ decays faster than linearly around the origin is to mandate that $|g(\mathbf{h})| / \|\mathbf{h}\|$ tends to zero, because for any linear transformation $A$,
$$\frac{\|A \mathbf{h}\|}{\|\mathbf{h}\|} \geq \lambda$$
for a fixed, finite, positive value of $\lambda$ (it is actually the smallest positive singular value of $A$). So it follows that
$$\lim_{\mathbf{h} \to \mathbf{0}} \frac{| g(\mathbf{h}) |}{|A \mathbf{h}|} \leq \lim_{\mathbf{h} \to \mathbf{0}} \frac{| g(\mathbf{h}) |}{\lambda \|\mathbf{h}\|}$$
and hence if $|g(\mathbf{h})| / \|\mathbf{h}\|$ tends to zero, then so does the left-hand side of the above equation. Thus, by choosing $g = f(\mathbf{x} + \mathbf{h}) - f(\mathbf{x}) - \nabla f (\mathbf{x}) \cdot \mathbf{h}$ for any $\mathbf{x}$, it follows from what we've discussed that if
$$\lim_{\mathbf{h} \to \mathbf{0}} \frac{|f(\mathbf{x} + \mathbf{h}) - f(\mathbf{x}) - \nabla f (\mathbf{x}) \cdot \mathbf{h}|}{\|\mathbf{h}\|} = 0$$
then the numerator decays faster than linearly, which is exactly what we want in defining the derivative. This above, is usually given as the definition of $f$ being differentiable. (Note that you may see it without the absolute value in the numerator, but the two are equivalent.)
EDIT: What is the reason for wanting $\nabla f$ to be such that $f(\mathbf{x} + \mathbf{h}) - f(\mathbf{x}) - \nabla f (\mathbf{x}) \cdot \mathbf{h}$ to decay faster than linearly?
Answer: This is how we generalize the definition of the derivative. In one dimension, one might say the motivation for calculating a derivative of some smooth function $f$ at some point $x_0$ is to linearize $f$ around $x_0$, i.e. best approximate the function as a line going thru $(x_0, f(x_0))$. Linearizing the function is equivalent to finding a linear transformation $L$ such that $f(x_0) + L(x - x_0)$ best approximates $f(x)$ around $x_0$ to linear order. Note that in one dimension, the linear function $L$ will simply be multiplication by a constant.
You probably have two questions in mind. First, why would we want to do this? And second, what does best approximate even mean?
To answer the first question, we are often interested in the leading approximation to $f(x)$ with $x$ close to $x_0$. If you are familiar with the Taylor series, you will know that many smooth functions can be written as
$$f(x) = f(x_0) + f'(x_0) (x - x_0) + \frac{f''(x_0)}{2} (x - x_0)^2 + \cdots $$
You can observe in the above expression that when $x$ is very close to $x_0$ (i.e. $|x - x_0|$ is small), first two terms dominate. For example, is $x - x_0$ was say, $0.01$, then the first correction (the second term) would be a correction on the order of $1 \%$, while the second correction is a on the order of $0.01 \%$, which is much smaller. (The following corrections are even smaller!)
Answering the second question gets us to the crux of your question. A best linear approximation is intuitively one that captures fully the effect to linear order of deviating from $x_0$. More formally, we want $L$ to be the linear function such that
$$|f(x) - f(x_0) - L(x - x_0)| < |f(x) - f(x_0) - L'(x - x_0)|$$
for any other linear function $L'$, so long as $|x - x_0|$ is small enough. From this it follows that $f(x) - f(x_0) - L(x - x_0)$ should decay faster than linearly in $x - x_0$. If instead it decayed to linear order, say
$$f(x) - f(x_0) - L(x - x_0) = L'(x - x_0) + o\left(x - x_0\right)$$
for some linear function $L'$, then $f(x_0) + (L + L')(x - x_0)$ would be a better linear approximation, because
$$|f(x) - f(x_0) - (L + L')(x - x_0)| = o\left(x - x_0\right)$$
would eventually be smaller than
$$|f(x) - f(x_0) - L (x - x_0)| = \Theta(x - x_0)$$
as $x - x_0$ becomes smaller and smaller.
In higher dimensions, the same motivation holds: we want a mathematical construct that will help us "linearize" a function around some point $\mathbf{x_0}$. The same definition of "best" holds: we want some linear transformation $L = \nabla f$ such that for all other linear transformations $L'$,
$$|f(\mathbf{x}) - f(\mathbf{x_0}) - L(\mathbf{x} - \mathbf{x_0})| < |f(\mathbf{x}) - f(\mathbf{x_0}) - L' (\mathbf{x} - \mathbf{x_0})|$$
and using similar logic, you can deduce that
$$f(\mathbf{x}) - f(\mathbf{x_0}) - L(\mathbf{x} - \mathbf{x_0})$$
must decay faster than linearly.


TITLE: Fourier Coefficients Defined
QUESTION [0 upvotes]: Consider the following function $$ \Pi(t) = \begin{cases} 1 \ \ \text{if} \ |t| < 1/2 \\ 0 \ \ \text{if} \ |t| \geq 1/2 \end{cases} $$
This is not a periodic function. Why can we define the fourier coefficients of $\Pi(t)$ as $$c_n = \frac{1}{T} \int_{-T/2}^{T/2} e^{-2 \pi int/T} \Pi(t) \ dt $$
$$ = \frac{1}{T} \int_{-1/2}^{1/2} e^{-2 \pi int/T} \Pi(t) \ dt $$
if the function is not periodic? I thought fourier coefficients were only defined for periodic functions?

REPLY [0 votes]: Fourier coefficients are initially explained and taught as having applying to periodic functions, since they involve $\sin()$ and $\cos()$ functions, and since they find great utility among periodic functions.  But the process of taking the integral
$\displaystyle \int \exp(ikx) f(x) dx $
has no inherent limit in period length.  It can be taken for periods as long as you like, i.e. periods approaching infinity, and can therefore be applied to non-periodic functions.  It's just a matter of whether 1) the integral converges and 2) whether the numbers that pop out are useful somehow, and they often times are.


TITLE: Hardness of approximation - additive error
QUESTION [25 upvotes]: There is a rich literature and at least one very good book setting out the known hardness of approximation results for NP-hard problems in the context of multiplicative error (e.g. 2-approximation for vertex cover is optimal assuming UGC).  This also includes well understood approximation complexity classes such as APX, PTAS and so on.
What is known when additive error is to be considered?  A literature search shows a few upper bound type results, most notably for bin packing (see for example http://www.cs.princeton.edu/courses/archive/spr03/cs594/dpw/lecture2.ps), but is there a more comprehensive complexity class classification or is there a reason why it is not so interesting or relevant?
As a further comment, for bin packing, for example, there is as far as I know no theoretical reason why a poly time algorithm which is always within an additive distance from optimal of 1 couldn't be found (although I stand to be corrected).  Would such an algorithm collapse any complexity classes or have any other significant theoretical knock-on effect?
EDIT: The key phrase I didn't use is "asymptotic approximation class" (thanks Oleksandr). It seems that there is some work in this area but it hasn't got to the same stage of maturity yet as the theory of classic approximation classes.

REPLY [6 votes]: There is a recent work on asymptotic approximation classes and their comparison with classical counterparts.
Erik Jan van Leeuwen and Jan van Leeuwen. Structure of Polynomial-Time Approximation. Technical Report UU-CS-2009-034. December 2009.


TITLE: why do we treat independent and dependent variables in a differential equations differently?
QUESTION [1 upvotes]: let $g'(x)=\frac{1}{x^2};x \neq 0$-------(1)
since, $\int g'(x)dx= \left\{\begin{array}{ll}\frac{1}{x}+c_1;x>0\\ \mathbf{and} \\ \frac{1}{x}+c_2;x<0\end{array}\right.$-----(1.1)
because, g'(x) is a piece-wise function and the pieces of the integral of g'(x) can differ by a constant and still have the same derivative.
then why do we have,
$\int dy=\int ydx$
$\int \frac{dy}{y}=\int dx$-------(2)
$=>\left\{\begin{array}{ll}ln|y|=x+c_1;y>0\\ \mathbf{or}  \\ln|y|=x+c_2;y<0\end{array}\right.$
$=>\left\{\begin{array}{ll}y=c_1e^x;y>0\\ \mathbf{or}  \\y =c_2e^x;y<0\end{array}\right.$-------(3)
i know, that having or  in (3) makes sense (because if we had and instead then, the resulting curve would not be a function and thus derivatives would not be defined at all.)
but if we follow the same method we used for integrating (1), to integrate (2), then we should have and in (3) instead of or.
so, for what reason do we end up with and in (1.1) but or in (3)?
to make clear my point , let's consider
$\int \frac{dy}{y}=\int \frac{dx}{x}$-----(4)
$=>\left\{\begin{array}{ll} \left\{\begin{array}{ll}ln|y|=ln|x|+c_1;y>0,x>0\\ \mathbf{or}  \\  ln|y|=ln|x|+c_2;y<0,x>0  \end{array}\right. \\ \mathbf{and} \\  \left\{\begin{array}{ll}ln|y|=ln|x|+c_3;y>0,x<0\\ \mathbf{or}  \\ln|y|=ln|x|+c_4;y<0,x<0\end{array}\right. \end{array}\right.$
see! , when it is the independent variable we use and(we say it's the same function) , but when its the dependent variable we use or( we say there are actually two different solutions, one of which will be eliminated by the initial condition). why do we do this ?

REPLY [0 votes]: Whenever you encounter a problem in your equation, the typical fix is to find a solution for a specific portion without the problem. When you do that for two different portions of the dependent variable, you wind up with two different solutions that suggest different outputs for the same input, so they can’t both be the answer. When you do that to the input, then you’re defining solutions for two different sections so the different solutions can both be part of the same function.


TITLE: "Thin film evolution" (Reference request)
QUESTION [7 upvotes]: Ok this is my first$^*$ question on overflow, my apologies if this is not the right place to ask what follows!
I observed the following phenomenon: I put a  (vitamin) tablet into water, then after a while a thin film forms (I have attached a picture of it). That made me wonder - is there a particular pde that models such a process (or particular stages of it)? I realize that this phenomenon is quite complex, but perhaps there are particular stages that can be modelled and studied generically (for example I think soap bubbles are studied mathematically). What interests me in particular is the transition from the stage where there is only the surface of the water to the stage where a second film has formed.
Here is the picture:

Thanks very much for pointers to sources where I can learn more.
$^*$(Edit: I just realised it is in fact my second question here)

REPLY [7 votes]: The phenomenon you are seeing when you dissolve a tablet in water is called effervescence: the escape of gas, carbon dioxide in this case, from an aqueous solution, when it becomes supersaturated.
The expansion of bubbles in the liquid and the collapse when they reach the surface has been extensively studied, for an intro (with some equations...) see this story in Europhysics News.
The film which is produced in your glass happens by the same process that children use to make soap bubbles by blowing soap through a metal ring. The escaping gas plays the role of the blowing child and the rim of the glass plays the role of the ring. For this to work, the liquid in your glass should have sufficient surface tension, like soap. I would imagine that if you take care to use a clean glass with pure water, the film will not develop.


TITLE: Minimizing $\mathbb E((X-m)^2)$
QUESTION [1 upvotes]: Let $X$ be a real random variable such that $X^2$ is integrable. I have to find $m$ minimizing $\mathbb E((X-m)^2)$.
How I want to do this is by saying:
$$
\begin{align}
\mathbb E((X-m)^2) &= \int_\Omega (X-m)^2d\mathbb P \\
                   &= \int_\Omega X^2-2Xm+m^2d\mathbb P \\
                   &= \int_\Omega X^2d\mathbb P - 2m\int_\Omega Xd\mathbb P + \int_\Omega m^2d\mathbb P \\
                   &= \mathbb E(X^2) - 2m\mathbb E(X) + m^2
\end{align}
$$
Which is easy to minimize, yielding $m=\mathbb E(X)$. But aren't I assuming that $X$ is integrable? How do I justify this calculation given that I only know $X^2$ is integrable?

REPLY [3 votes]: More simply without invoking the Cauchy-Schwarz inequality,
$$|x| \leq 1+x^2\ \ \forall\, x \in \mathbb R \Rightarrow E[|X|] \leq 1 + E[X^2] < \infty.$$


TITLE: Involution on the set of isomorphism classes of 2n-dimensional vector bundles induced by an involution of $O(2n)$
QUESTION [3 upvotes]: Denote by $\varphi$ the automorphism of $O(2n)$ given by $\varphi (A)=det(A)\cdot A$.This induces a self-map $B\varphi$ of $BO(n)$, so it induces a self-map (actually an involution)
$B\varphi ^*$ on $[X,BO(2n)]\cong Vect_{2n}(X)$ where $Vect_{2n}(X)$ is the set of isomorphic classes of $2n$-dimensional real vector bundles over the topological
space $X$.  Or in terms of charts and transition functions, given a bundle $E$ with trivializing neighborhoods  $U_i$ and transition functions $g_{i,j}$, one can construct a bundle $B\varphi ^*(E)$ with same trivializing neighborhoods  $U_i$ and transition functions $\varphi \circ g_{i,j}$.
Question: is there any geometric interpretation of this involution?  For example, can we describe geometrically $B\varphi ^*(T(RP^4))$ where $T(RP^4)$
denotes the tangent bundle over $RP^4$?

REPLY [4 votes]: The involution described could also be called "twisting by the determinant line bundle". I think that the following is a geometric description: let $\mathcal{E}$ be an orthogonal bundle on $X$, let $\mathcal{L}=\det \mathcal{E}$ be the determinant line bundle (whose transition functions are given by the determinant as in the question). Take $f:Y\to X$ to be the associated double covering. Then the projection formula implies
$$
f_\ast f^\ast \mathcal{E}\cong \mathcal{E}\oplus(\mathcal{E}\otimes\mathcal{L}^{-1}).
$$
But for the real bundle, the line bundles $\mathcal{L}$ and $\mathcal{L}^{-1}$ are isomorphic. So the result of the involution can be described as $\operatorname{coker}(\mathcal{E}\to f_\ast f^\ast\mathcal{E})$. For the tangent bundle on $\mathbb{RP}^4$, the covering associated to the determinant line bundle is the orientation double cover $S^4\to \mathbb{RP}^4$.


TITLE: Reference for spherical distance
QUESTION [1 upvotes]: Here we can find a formula for a distance on the sphere $\Bbb S^n=\{(x_1,\dots,x_{n+1})\in\Bbb R^{n+1}\;:\;x_1^2+\cdots+x_{n+1}^2=1\}\subset\Bbb R^{n+1}$.
The one point compactification of $\Bbb R^n$ is homeomorphic to $\Bbb S^n$ via stereographic projection:
\begin{align*}
\varphi_n&\colon\overline{\Bbb R^n}\to\Bbb S^n\\
\varphi_n(x)&=\frac2{\|x\|^2+1}\left(x,\frac{\|x\|^2-1}{2}\right),\;\;x\neq\infty\\
\varphi_n(\infty)&=(0,\dots,0,1)
\end{align*}
This allows us to define a notion of distance on any such $\overline{\Bbb R^n}$ as
$$
d_n(x,y):=\arccos(\varphi_n(x)\cdot\varphi_n(y)),\;\;\;x,y\in\overline{\Bbb R^n}\;\;,
$$
where $\cdot$ denotes the scalar product in $\Bbb R^{n+1}$.
I need to quote this result, so can somebody tell me a book in which I can find it?

REPLY [0 votes]: Brannan, D. A, Matthew F Esplen, and Jeremy Gray. 1999. Geometry. Cambridge: Cambridge University Press.  and the proof and explanations are on p.144


TITLE: Suspicion about simple modules
QUESTION [3 upvotes]: It's stated here that for a commutative ring $R$, every simple module over $R$ is isomorphic as an $R$-module to a quotient ring of $R$ by a maximal ideal.
Intuitively this seems likely, since a quotient ring is a field if any only if we quotient by a maximal ideal, and a field has no nonzero proper ideals. This makes me suspect that an $R$-module $M$ is simple if and only if $M\cong R/I$ (as $R$-modules) for a maximal ideal $I$ of $R$.
Is there a proof of whether this is true or not?

REPLY [2 votes]: Let $M$ be a simple module and let $m \in M$ be nonzero. By simplicity, $\{ rm : r \in R \} = R$, so the map $r \mapsto rm$ defines a natural surjective homomorphism $R \to M$ of $R$-modules giving an isomorphism $M \cong R/I$ where $I$ is the kernel. If $I$ is not maximal, then letting $m$ be any maximal ideal containing it we have another surjection $R/I \to R/m$ with nontrivial kernel, and this contradicts simplicity.

REPLY [1 votes]: If $M$ is a simple module for each $x\in M$ with $x\neq 0$, $Rx=M$, so take a $x\in M$ with $x\neq 0$, and consider the next morfism $f:R\longrightarrow M$ given by $f(a)=ax$, by the last observation is an epimorphism. Now apply the first isomorphism theorem and you get $M/\ker f\cong M$. Now the $\ker f=I$ is an ideal of $R$. To see that $I$ is maximal use the theorem of the bijective correspondence. Also the same theorem gives you a proof of the return.


TITLE: How many numbers less than $500$ are divisible by $2$ or $3$ or $4$ or $5$ or $7$ or $21$?
QUESTION [0 upvotes]: I tried this ,
No of prime no's below $500$ is $95$
Excluding $2,3,5,7$, it is $91$
No's not divisible $2,3,4,5,7,21$, are
$11 \cdot (11,13,17,19,23,29,31,37,41,43)$ i.e $10$
$13 \cdot (13,17,19,23,29,31,37)$ i.e $7$
$17 \cdot (17,19,23,29)$ i.e $4$
$19 \cdot (19,23)$ i.e $2$
initially $1$ is also not divisible
total = $91+10+7+4+2+1= 115$
answer $500-115 = 385$
Is this correct or  incorrect??

REPLY [4 votes]: It's seems that you've made some small mistakes, as the number of non-divisible numbers is 115, but there are $499$ integers less than $500$. Nonetheless your method might fail because you are prone to missing some numbers, like product of three primes greater than $7$, which isn't the case here, as $11^3 > 500$
Anyway probably the best way to go here is the Inclusion - Exclusion Principle. Namely let $N(c_1,c_2,...,c_n)$ is the number of elements that satisfy the condition $c_1, c_2,..., c_n$. Now if $c_1$ is the condition "divisible by 2", $c_2$ is "divisible by 3" and so on we have:
$N(\overline{c_1},\overline{c_2},\overline{c_3},\overline{c_4}) = N - \sum_{i} N(c_i) + \sum_{i,j} N(c_i,c_j) - \sum_{i,j,k} N(c_i,c_j,c_k) + N(c_1,c_2,c_3,c_4)$
The result will be:
$$499 - \left[\frac{499}{2}\right] - \left[\frac{499}{3}\right] - \left[\frac{499}{5}\right] -\left[\frac{499}{7}\right] + \left[\frac{499}{6}\right] + \left[\frac{499}{10}\right] + \left[\frac{499}{14}\right] + \left[\frac{499}{15}\right] +\left[\frac{499}{21}\right] + \left[\frac{499}{35}\right] - \left[\frac{499}{30}\right] - \left[\frac{499}{42}\right] - \left[\frac{499}{70}\right] - \left[\frac{499}{105}\right] + \left[\frac{499}{210}\right] = 115$$


TITLE: Bounds on Riemann Zeta type Function?
QUESTION [0 upvotes]: Given a sequence $a_n$ such that $\sum_{n\ge1} \dfrac{|a_n|}{n^s}$ is convergent for $s>0$. Given that $\sum_{n\ge1} \frac{|a_n|}{n} < 1$, is it be possible to impose some sort of an upperbound for $\sum_{n\ge1} \frac{|a_n|}{n^s}$ for $1>s>k>0$ for some fixed $k$?

REPLY [1 votes]: Given an integer $m$, define a sequence $a_n$ to be $m$ at the $m$-th place and $0$ otherwise.
Then
$$\sum_{n\ge 1} {a_n\over n^s}=m^{1-s}$$
As any upper bound must work for all $m$, we see that no upper bound is possible for $s<1$.