text:
\section{\large{Introduction}}
The essentially Speh representations have played an important role in the classification of the unitary dual of a general linear group over the $p$-adic field obtained in \cite{Tad9}.
The ladder representations are a generalization of the essentially Speh representations.
They were first studied in \cite{LapidMinguez,LapidMinguez1}. The importance of ladder representations is reflected in the way they simplify the proof of the unitary dual classification. Additionally, the class of ladder representations has an advantage over the class of essentially Speh representations on the level of Jacquet modules.
The results of \cite{KretLapid} show that the constituents of the Jacquet modules of ladder representations are tensor products of representations of the same type.
One can easily see that an analogue is generally not true for the Jacquet modules of the essentially Speh representations.
Since our approach in establishing the irreducible subquotients of an induced representation is based on the
Geometric Lemma, we consider the class of ladder representations as a more accessible one.

Motivated by the structure of representations in the unitary dual of a general linear group,
we study the composition series of representations induced from a ladder representation on the general linear part and a unitary representation on the classical part, which should provide insight into the structure of the unitary dual of other classical $p$-adic groups.
The reducibility of such induced representations is known if the representation on the classical part is cuspidal, see \cite{LT}. The goal of this paper is to obtain the composition series when the cuspidal support of the ladder representation contains only representations of the form $\nu^x\rho$ for a character $\nu(g)=|\text{det}(g)|_F$ of a general linear group and $x\ge\frac{1}{2}$.

The methods used in this paper rely on the structural formula for the Jacquet modules of induced representations obtained in \cite{Tad5}, which is an algebraization of the Geometric Lemma. Namely, together with the Langlands classification, it gives candidates for the subquotients of the considered induced representation.
All irreducible tempered representations which appear in our investigation happen to be strongly positive. We note that the strongly positive representations play an important role in determining which of the obtained candidates are indeed the subquotients of the considered induced representation.
M\oe glin and Tadić have shown in \cite{Moe2,MT1} that strongly positive representations are the fundamental objects in the construction of discrete series for classical $p$-adic groups. Their definition justifies their name since the exponents in the GL-cuspidal support of any induced representation into which they embed are positive.
Also, they are the only irreducible tempered representations which are completely determined by the cuspidal support.
In order to obtain subquotients of the considered induced representation we also use some sufficient conditions on the subquotients of certain Jacquet modules, mainly through the transitivity of the Jacquet modules and parabolic induction.

We briefly outline the content of the paper. In the following section we introduce the notation that will be used throughout the paper.
In the third section we determine the tempered irreducible subquotients of the representation induced from cuspidal and ladder representations such that the minimal exponent in the cuspidal support of the ladder representation is greater than or equal to $\frac{1}{2}$. In the fourth section we obtain a reducibility criterion based on the results in \cite{LT} and a description of the non-tempered irreducible subquotients of the considered representation.

For the reader's convenience we give a theorem that describes the length of a considered representation. It is an easy consequence of the main results of the paper, stated as Theorems \ref{tm1} and \ref{tm2}, which describe the irreducible subquotients of the considered representation.
In the following theorem, we consider the representation $\pi_L$, which is defined as the unique irreducible subrepresentation of
$$\delta([\nu^{a_1}\rho,\nu^{b_1}\rho])\times\ldots\times\delta([\nu^{a_t}\rho,\nu^{b_t}\rho])$$
for an irreducible cuspidal representation $\rho$ of the general linear group, a positive integer $t$ and real numbers $a_i, b_i$,  $i=1,\ldots,t$, such that $\frac{1}{2}\le a_1<\ldots<a_t$, $b_1<\ldots<b_t$ and $a_i\le b_i$ for $i=1,\ldots,t$.
Also, we let $\sigma_c$ denote an irreducible cuspidal representation of the special odd orthogonal or symplectic group over a  non-archimedean local field.
\begin{theorem}
    Let $m$ be a number of indices $i\in\{1,\ldots,t\}$ such that there is an $x\in \{a_i, a_i+1,\ldots, b_i\}$ for which $\nu^x\rho\rtimes\sigma_c$ reduces.
    The induced representation $\pi_L \rtimes \sigma_c$ is of the length $m+1$.
\end{theorem}
The author is thankful to Ivan Mati\'{c} for suggesting the problem and for many helpful comments and to Marcela Hanzer for several useful comments. Also, the author would like to thank Petar Baki\'{c} and Vanja Wagner for a number of suggestions and grammatical corrections.
\section{\large{Preliminaries}} \label{s2}
Throughout the paper, $F$ will denote a non-archimedean local field of characteristic different than two.

We will now describe the groups that we consider. We denote by $GL(n, F)$ the general linear group of rank $n$ over $F$. In what follows, we shall fix one of the series of classical groups $SO(2n+1, F)$ and $Sp(2n, F)$ and denote by $G_{n}$ a rank $n$ group belonging to this fixed series. The matrix realizations of these groups can be found in \cite[Sections 3,6]{Tad5}.

We fix a set of standard parabolic subgroups of $G_n$ by choosing a minimal $F$-parabolic subgroup in $G_{n}$ consisting of upper-triangular matrices.
Standard parabolic subgroups have the Levi decomposition $P=MN$, where $N$ is its unipotent radical and $M$ is the Levi factor of the form $GL(n_{1}, F) \times \ldots \times GL(n_{k}, F) \times G_{n'}$ where $\alpha=(n_1,\ldots,n_k)$ such that $n_1+\ldots+n_k\le n$. We use the notation $P=P_{\alpha}$, $M=M_{\alpha}$, $N=N_{\alpha}$.
For representations $\delta_{i}$ of $GL(n_{i}, F)$, $i = 1, 2, \ldots, k$ and a representation  $\tau$ of $G_{n'}$, we denote the normalized induction $\text{Ind}_{M}^{G_n}(\delta_{1} \otimes \ldots \otimes \delta_{k} \otimes \tau)$ by $\delta_{1} \times \ldots \times \delta_{k} \rtimes \tau$.

The set of all irreducible admissible representations of $GL(n,F)$ (resp.~$G_n$) will be denoted by $\Irr(GL(n,F))$ (resp.~$\Irr(G_n)$). We denote $\Irr(G)=\cup_{n\ge0} \Irr(G_n)$ (resp.~$\Irr(GL)=\cup_{n\ge0} \Irr(GL(n,F))$).
Let $R(G_n)$ denote the Grothendieck group of admissible, complex representations of finite length of $G_n$ and define $R(G)=\oplus_{n \geq 0} R(G_{n})$. Similarly, $R(GL) = \oplus_{n \geq 0} R(GL(n, F))$. For $\pi_1,\pi_2\in\Irr(G_n)$ the fact that $\pi_1$ is a subquotient of $\pi_2$ is denoted by $\pi_1\le\pi_2$. We also say that $\pi_1$ is contained in $\pi_2$.
The cuspidal support of an admissible representation $\pi$ of $GL(n,F)$ or $G_n$ is denoted by $[\pi]$.
The contragredient of a representation $\pi$ in $R(GL)$ or $R(G)$ is denoted by $\widetilde{\pi}$. We say $\pi$ is a selfcontragredient representation if $\widetilde{\pi}\simeq\pi$.
For $\pi,\pi_1,\pi_2\in \Irr(GL)$ and $\s\in\Irr(G)$ we have $\pi \rt \s = \widetilde{\pi} \rt \s$ and $\pi_1 \times \pi_2 \rt \s = \pi_2 \times \pi_1 \rt \s$ in $R(G)$.

For $\sigma \in \Irr(G_{n})$ and $\alpha=(n_1,\ldots,n_k)$ such that $n_1+\ldots+n_k\le n$ we denote by $r_{\alpha}(\sigma)$ the normalized Jacquet module of $\sigma$ with respect to the parabolic subgroup $P_{\alpha}$ with the Levi subgroup equal to $GL(n_1, F) \times \ldots \times GL(n_k,F)\times G_{n'}$. For $\alpha=(m)$ we identify $r_{(m)}(\sigma)$ with its semisimplification in $R(GL(m,F)) \otimes R(G_{n-m})$ and consider
\begin{equation*}
\mu^{\ast}(\sigma) = 1 \otimes \sigma + \sum_{m=1}^{n} r_{(m)}(\sigma) \in R(GL) \otimes R(G).
\end{equation*}
Similarly, for $\pi\in\Irr(GL)$, we denote by  $r_{(m)}(\pi)$ the normalized Jacquet module of $\pi$ with respect to the standard parabolic subgroup having the Levi factor equal to $GL(m, F) \times GL(n-m,F)$. We identify it with its semisimplification and define $m^*(\pi)=\sum_{m=0}^n r_{(m)}(\pi)$.
The mappings $\mu^*$ and $m^*$ extend to $R(G)$ and $R(GL)$ respectively. We define $M^*:R(GL)\to R(GL)\otimes R(GL)$ by
$$M^*=(m\otimes 1)\circ (\sim\otimes m^*)\circ s \circ m^*$$
where $m$ denotes $\times:R(GL)\otimes R(GL)\to R(GL)$ and $s: \sum x_i\otimes y_i \mapsto \sum y_i\otimes x_i$.
The following theorem is the main result of \cite{Tad5}.
\begin{theorem}
For $\pi\in R(GL)$ and $\sigma\in R(G)$
$$\mu^*(\pi\rtimes\sigma)=M^*(\pi)\rtimes\mu^*(\sigma) $$
where the right hand side is determined by $(\pi_1\otimes\pi_2)\rtimes(\pi'\otimes\sigma')=(\pi_1\times\pi')\otimes(\pi_2\rtimes\sigma')$.
\end{theorem}
The Geometric Lemma implies that $m^*$ is multiplicative, i.e. $m^*(\pi_1\times\pi_2)=m^*(\pi_1)\times m^*(\pi_2)$. Together with the definition of $M^*$, this implies $M^*$ is multiplicative.

We recall the Frobenius reciprocity:
$$ \text{Hom}_{G_n}(\pi,\delta_1\times\ldots\times\delta_k\rtimes\tau)\cong \text{Hom}_{M_{\alpha}}(r_{\alpha}(\pi),\delta_1\otimes\ldots\otimes\delta_k\otimes\tau). $$
Here $P_{\alpha} = M_{\alpha}N_{\alpha}$ is the standard parabolic subgroup of $G_n$ corresponding to the $k$-tuple $\alpha$.

The character $|det(g)|_F$ of $GL(n,F)$ is denoted by $\nu$, where $|\text{ }|_F$ is the normalized absolute value on $F$.
If $x,y$ are real numbers such that $y-x\in\mathbb{N}_0$, we denote the set $\{x,x+1,\ldots,y\}$ by $[x,y]$.
Let $\rho$ be an irreducible cuspidal representation of $GL(n, F)$.
Similarly, we refer to the set $\{ \rho, \nu \rho, \ldots, \nu^{m} \rho \}$ as a segment of cuspidal representations, and denote it by $[ \rho, \nu^{m} \rho]$.
Attached to each such segment is an irreducible essentially square-integrable representation $\delta ([\rho, \nu^{m} \rho ])$ of $GL(m \cdot n, F)$, which is the unique irreducible subrepresentation of $\nu^{m} \rho \times \ldots \times \nu \rho \times \rho$.
It is well-known that every irreducible essentially square-integrable representation $\delta \in \Irr(GL(n,F))$ is of the form $\delta([\nu^{x} \rho, \nu^{y} \rho])$, for an irreducible cuspidal representation $\rho \in \Irr(GL(n',F))$ and $x, y \in \mathbb{R}$ such that $y - x$ is a non-negative integer.
For an irreducible essentially square-integrable representation $\delta$, there is a unique $e(\delta) \in \mathbb{R}$ such that $\nu^{- e(\delta)} \delta$ is unitarizable. Note that $e(\delta([\nu^{x} \rho, \nu^{y} \rho])) = \frac{x+y}{2}$. We denote by $1$ the one dimensional representation of the trivial group and put $\delta([\nu^{x} \rho, \nu^{y} \rho]) = 1$ if $y=x-1$.

In the paper we favour the subrepresentation over the quotient version of the Langlands classification, which we now recall. Let $\delta_1, \ldots, \delta_l $ be the irreducible essentially square-integrable representations of general linear groups.
If $e(\delta_1) \leq \ldots \leq e(\delta_l)$, then $\delta_1\times\ldots\times\delta_l$ has a unique irreducible (Langlands) subrepresentation which we denote by $L(\delta_1,\ldots,\delta_l)$.
If $\tau \in \Irr(G_{n'})$ is a tempered representation and $e(\delta_1) \leq \ldots \leq e(\delta_l) < 0$, then $\delta_{1} \times \ldots \times \delta_{l} \rtimes \tau$ has a unique irreducible (Langlands) subrepresentation which we denote by $L(\delta_{1}, \ldots, \delta_{l} ; \tau)$.
Every irreducible non-tempered representation of $GL(n,F)$ (resp. $G_{n'}$) is equal to $L(\delta_1,\ldots,\delta_l)$ (resp. $L(\delta_{1}, \ldots, \delta_{l} ; \tau)$) for a unique choice of $\delta_1,\ldots,\delta_l$ (and $\tau$).

Throughout the paper we fix cuspidal representations $\rho\in\Irr(GL(n,F))$ and $\s_c\in\Irr(G_{n'})$.
\begin{dfn}
A ladder representation is a representation of the form $$L(\delta([\nu^{x_1}\rho,\nu^{y_1}\rho]),\ldots,\delta([\nu^{x_k}\rho,\nu^{y_k}\rho]))$$
for a positive integer $k$, real numbers $x_1,\ldots,x_k$ and $y_1,\ldots,y_k$ such that $x_1<\ldots<x_k$, $y_1<\ldots<y_k$ and $x_i\le y_i$ for $i=1,\ldots,k$.
\end{dfn}
Now let $\pi=L(\delta([\nu^{x_1}\rho,\nu^{y_1}\rho]),\ldots,\delta([\nu^{x_k}\rho,\nu^{y_k}\rho]))$ be a ladder representation. We will frequently need to consider the Jacquet modules of $\pi \rtimes \sigma_c$ taken with respect to a standard maximal parabolic subgroup. From \cite[Theorem 2.1]{KretLapid} we see
\begin{gather}
    \label{lad}
    m^*(\pi)=
    \displaystyle \sum_{\Lad(\pi)} L(\delta([\nu^{c_1+1}\rho,\nu^{y_1}\rho]),\ldots,\delta([\nu^{c_k+1}\rho,\nu^{y_k}\rho]))\otimes  \\ \otimes L(\delta([\nu^{x_1}\rho,\nu^{c_1}\rho]),\ldots,\delta([\nu^{x_k}\rho,\nu^{c_k}\rho])) \nonumber
\end{gather}
where $\Lad(\pi)$ denotes the set of all $k$-tuples $(c_1,\ldots,c_k)$ of real numbers for which the following conditions hold:
\begin{itemize}
    \item $c_1<\ldots<c_k$,
    \item $x_i-1 \le c_i \le y_i$ for $i=1,\ldots,k$,
    \item $c_i-x_i\in\Z$ for $i=1,\ldots,k$.
\end{itemize}
We define $\Lad(\pi)'$ as the set of all pairs $\big((c_1,\ldots,c_k),(d_1,\ldots,d_k)\big)$ in $\Lad(\pi)\times\Lad(\pi)$ such that $c_i\le d_i$ for $i=1,2,\ldots,k$.
Combining (\ref{lad}) with \cite[Theorem 5.4]{Tad5} we have
\begin{gather}
    \label{struct}
    \mu^*(\pi\rtimes\sigma_c)=  \sum\limits_{\Lad(\pi)'} L(\delta([\nu^{-c_k}\widetilde{\rho},\nu^{-x_k}\widetilde{\rho}]),\ldots,\delta([\nu^{-c_1}\widetilde{\rho},\nu^{-x_1}\widetilde{\rho}]))\times \\
    \times L(\delta([\nu^{d_1+1}\rho,\nu^{y_1}\rho]),\ldots,\delta([\nu^{d_k+1}\rho,\nu^{y_k}\rho])) \otimes  \nonumber \\
    L(\delta([\nu^{c_1+1}\rho,\nu^{d_1}\rho]),\ldots,\delta([\nu^{c_k+1}\rho,\nu^{d_k}\rho])) \rtimes \s_c \nonumber.
\end{gather}
For a positive integer $t$ and real numbers $a_1,\ldots,a_t$ and $b_1,\ldots,b_t$ such that $ \frac{1}{2}\le a_1<\ldots<a_t$, $b_1<\ldots<b_t$ and $a_i\le b_i$ for $i=1,\ldots,t$, we define the ladder representation $$\pi_L=L(\delta([\nu^{a_1}\rho,\nu^{b_1}\rho]),\ldots,\delta([\nu^{a_t}\rho,\nu^{b_t}\rho])).$$
The goal of this article is to determine the composition series of $\pi_L\rtimes\s_c$.

For selfcontragredient $\rho$, there is a unique non-negative real number $x$ such that $\nu^x\rho\rtimes\sigma_c$ reduces. Otherwise, $\nu^x\rho\rtimes\sigma_c$ is irreducible for every real number $x$.

An irreducible representation $\sigma$ of $G_n$ is called
strongly positive if
\begin{equation*}
\sigma \hookrightarrow \nu^{x_{1}} \rho_{1} \times \ldots \times \nu^{x_{k}} \rho_{k} \rtimes \sigma_{cusp},
\end{equation*}
(where $\rho_i$, $\sigma_{cusp}$ are unitary cuspidal representations) implies $x_1,\ldots ,x_k$ are positive. By \cite{Arthur} and \cite[Th\'{e}or\`{e}me~3.1.1]{Moe2}, if $\nu^{x} \rho$ appears in the cuspidal support of $\sigma$, then $\rho$ is selfcontragredient and $2x \in \mathbb{Z}$.

Strongly positive representations play an important role in the classification of discrete series by M{\oe}glin-Tadi\'{c} \cite{Moe2,MT1}.
The classification of irreducible strongly positive square-integrable genuine representations of metaplectic groups over $F$ is obtained in \cite[Theorem 1.2]{Matic3} using a purely algebraic approach.
These results can be applied in the case of classical $p$-adic groups in a completely analogous manner. Every strongly positive representation whose cuspidal support is contained in $\{\nu^x\rho : x\in\R\}$ for some irreducible cuspidal selfcontragredient representation $\rho\in \Irr(GL(n,F))$ can be realized in the unique way as the unique irreducible subrepresentation of the induced representation
\begin{equation*}
 \delta([\nu^{\alpha -r +1 } \rho, \nu^{x_{1}} \rho])\times\ldots\times\delta([\nu^{\alpha } \rho, \nu^{x_{r}} \rho])  \rtimes \sigma_{cusp}
\end{equation*}
where $\sigma_{cusp}$ is an irreducible cuspidal representation of $G_{n'}$, $\alpha$ is the unique positive number such that $\nu^{\alpha} \rho \rtimes \sigma_{cusp}$ reduces,
$r = \lceil \alpha \rceil$ and $-\frac{1}{2} \le x_1 < x_2 < \ldots < x_r$, $x_i - \alpha \in \mathbb{Z}$ for $i = 1, \ldots,r$. Here $\lceil \alpha \rceil$ denotes the smallest integer which is not smaller than $\alpha$.

\section{\large{Tempered subquotients of $\pi_L\rtimes\s_c$}}
In this section we determine the tempered subquotients of $\pi_L\rtimes\s_c$. These results will be applied in the following section, with the goal of determining non-tempered subquotients of $\pi_L\rtimes\s_c$.
Namely, by the Langlands classification,  every non-tempered subquotient of $\pi_L\rtimes\sigma_c$ is of the form $L(\delta_1,\ldots,\delta_k;\tau)$.
If $L(\delta_1,\ldots,\delta_k;\tau)\le\pi_L\rtimes\s_c$, then the transitivity of Jacquet modules implies $L(\delta_1,\ldots,\delta_k)\otimes\tau\le\mu^*(\pi_L\rtimes\s_c)$. Now (\ref{struct}) implies that there exists $\big((c_1,\ldots,c_t),(d_1,\ldots,d_t)\big)\in \Lad(\pi_L)'$ such that $$\tau\le   L(\delta([\nu^{c_1+1}\rho,\nu^{d_1}\rho]),\ldots,\delta([\nu^{c_t+1}\rho,\nu^{d_t}\rho])) \rtimes \s_c.$$
Note that the exponents in the cuspidal support of a ladder representation on the general linear part are positive since $\frac{1}{2}\le a_1\le c_1+1$.

Note that any tempered subquotient of $\pi_L\rtimes\s_c$ is necessarily strongly positive.
To see this, suppose that $\pi_L \rtimes \sigma_c$ contains an irreducible tempered subquotient $\tau$ which is not strongly positive. Then
there exists an irreducible tempered representation $\tau'$ and $a,b\in\R$, $a\leq 0$, $a + b \geq 0$ such that $\tau$ is a subrepresentation of $\delta([\nu^a \rho,
\nu^b \rho]) \rtimes \tau'$.
If $a$ and $b$ are integers, then $\rho \in [\pi_L \rtimes \sigma_c]$, which is a contradiction.
If $a$ and $b$ are half-integers, then $[\pi_L \rtimes \sigma_c]$ contains the multiset $\{\nu^{\frac{1}{2}}\rho,\nu^{\frac{1}{2}}\rho\}$, which is again a contradiction.

The following theorem determines when $\pi_L\rtimes\sigma_c$ contains a tempered subquotient.
\begin{theorem}
\label{tm1}
The induced representation $\pi_L\rtimes\s_c$ has a tempered irreducible subquotient if and only if there exists $\alpha>0$ such that $\nu^{\alpha}\rho\rtimes\sigma_c$ reduces and $a_i=\alpha-t+i$ for $i=1,2,\ldots,t$.

In that case the tempered irreducible subquotient is unique; it is the unique irreducible subrepresentation of $\pi_L\rtimes\s_c$ and is isomorphic to the  strongly positive representation characterised as the unique irreducible subrepresentation of
\begin{gather}
    \delta([\nu^{\alpha -t +1 } \rho, \nu^{b_{1}} \rho])\times\ldots\times\delta([\nu^{\alpha } \rho, \nu^{b_{t}} \rho])  \rtimes \sigma_{c}. \label{zz}
\end{gather}
\end{theorem}
\begin{proof}
Assume that the induced representation $\pi_L\rtimes\s_c$ has a tempered irreducible subquotient. We denote it by $\tau$.
We have noted above that $\tau$ is necessarily a strongly positive representation. Then there exists a unique positive real number $\alpha$ such that $\nu^{\alpha}\rho\rtimes\sigma_c$ reduces, as seen in section \ref{s2}. Note that this implies $\rho\simeq \widetilde{\rho}$. We denote $r=\lceil\alpha\rceil$.
Thus there exists the unique increasing sequence of real numbers $x_1,\ldots,x_r$ such that $\alpha-r+i-1\le x_i$ and $x_i-\alpha\in\Z$ for $i=1,2,\ldots,r$ and $\tau$ is the unique irreducible subrepresentation of
\begin{equation}
\label{tau}
\delta([\nu^{\alpha -r +1 } \rho, \nu^{x_{1}} \rho])\times\ldots\times\delta([\nu^{\alpha } \rho, \nu^{x_{r}} \rho])  \rtimes \sigma_{c}.
\end{equation}
Therefore, it is also a subrepresentation of
$L(\delta([\nu^{\alpha -r +1 } \rho, \nu^{x_{1}} \rho]),\ldots,\delta([\nu^{\alpha } \rho, \nu^{x_{r}} \rho]) ) \rtimes \sigma_{c}.$
The Frobenius reciprocity and the transitivity of Jacquet modules imply that $$L(\delta([\nu^{\alpha -r +1 } \rho, \nu^{x_{1}} \rho]),\ldots,\delta([\nu^{\alpha } \rho, \nu^{x_{r}} \rho]) ) \otimes \sigma_{c}\le \mu^*(\pi_L\rtimes\s_c).$$
By the structural formula (\ref{struct}) there exists $\big((c_1,\ldots,c_t),(d_1,\ldots,d_t)\big)\in \Lad(\pi_L)'$ such that
\begin{gather}
\label{prva}
    L(\delta([\nu^{\alpha -r +1 } \rho, \nu^{x_{1}} \rho]),\ldots,\delta([\nu^{\alpha } \rho, \nu^{x_{r}} \rho]) ) \le  \\
    L(\delta([\nu^{-c_t}\rho,\nu^{-a_t}\rho]),\ldots,\delta([\nu^{-c_1}\rho,\nu^{-a_1}\rho]))
    \times \nonumber \\ \times L(\delta([\nu^{d_1+1}\rho,\nu^{b_1}\rho]),\ldots,\delta([\nu^{d_t+1}\rho,\nu^{b_t}\rho])) \nonumber
\end{gather}
and
\begin{gather}
\label{druga}
    \s_c\le  L(\delta([\nu^{c_1+1}\rho,\nu^{d_1}\rho]),\ldots,\delta([\nu^{c_t+1}\rho,\nu^{d_t}\rho])) \rtimes \s_c.
\end{gather}
Note that the exponents in a cuspidal support of a representation on the left hand side of (\ref{prva}) are positive and $-a_i,-c_i<0$ for indices $i\in\{1,2,\ldots,t\}$ for which $\delta([\nu^{-c_i}\rho,\nu^{-a_i}\rho])\not\simeq 1$. Since representations in (\ref{prva}) have the same cuspidal support, we conclude that $c_i=a_i-1$ for $i=1,2,\ldots,t$.
From (\ref{druga}) we get $c_i=d_i$ for $i=1,2,\ldots,t$, since $\sigma_c$ is a cuspidal representation. Thus (\ref{prva}) gives
$$ L(\delta([\nu^{\alpha -r +1 } \rho, \nu^{x_{1}} \rho]),\ldots,\delta([\nu^{\alpha } \rho, \nu^{x_{r}} \rho]) ) \le
 L(\delta([\nu^{a_1}\rho,\nu^{b_1}\rho]),\ldots,\delta([\nu^{a_t}\rho,\nu^{b_t}\rho])).$$
Since these representations are irreducible, we have the equality. Langlands classification now implies equality of the sets consisting of the essentially square-integrable representations defining them. Thus we have $a_i=\alpha-t+i$ and $b_i=x_{r-t+i}$ for $i=1,2,\ldots,t$.

If $a_i=\alpha-t+i$ for $i=1,2,\ldots,t$, then $$\pi_L\cong L(\delta([\nu^{\alpha -t +1 } \rho, \nu^{b_{1}} \rho]),\ldots,\delta([\nu^{\alpha } \rho, \nu^{b_{t}} \rho]) ). $$
Note that $\pi_L\rtimes\s_c$ is a subrepresentation of the representation (\ref{zz}). Since there exists $\alpha>0$ such that $\nu^{\alpha}\rho\rtimes\sigma_c$ reduces, we conclude that the representation (\ref{zz}) contains a unique irreducible tempered subquotient, denoted $\tau$.
It is a strongly positive representation and the unique irreducible subrepresentation of (\ref{zz}), hence it is also a subrepresentation of $\pi_L\rtimes\s_c$.
Every subquotient of $\pi_L\rtimes\sigma_c$ is a subquotient of representation (\ref{zz}), so $\tau$ appears in composition series of $\pi_L\rtimes\sigma_c$ with multiplicity one.
\end{proof}
\section{\large{Non-tempered subquotients of $\pi_L\rtimes\s_c$}}
In the last section we will determine non-tempered subquotients of $\pi_L\rtimes\s_c$ and thus obtain the complete composition series of a representation $\pi_L\rtimes\s_c$.
Firstly, we give a characterization of reducibility of the induced representation $\pi_L\rtimes\sigma_c$ based on the results in \cite{LT}.
\begin{prop}
\label{propi}
The induced representation $\pi_L\rtimes\sigma_c$ is reducible if and only if $\nu^{x}\rho\rtimes\sigma_c$ reduces for some $\nu^x\rho\in [\pi_L]$.
\end{prop}
\begin{proof}
  By Theorem 1.1 in \cite{LT}, if $\nu^{x}\rho\rtimes\sigma_c$ reduces for some $\nu^x\rho\in [\pi_L]$, then $\pi_L\rtimes\sigma_c$ is reducible.

  Conversely, assume that $\nu^{x}\rho\rtimes\sigma_c$ is irreducible for all $\nu^x\rho\in [\pi_L]$.
  Since $\pi_L$ is a ladder representation, Theorem 1.2 in \cite{LT} implies that the induced representation $\pi_L\rtimes\sigma_c$ is irreducible if and only if $(\pi_{L})_{+}\times(\widetilde{\pi_L})_+$ is irreducible.
  Here we use the notation from \cite[Definition 3.12.]{LT}: for a representation $\pi\simeq L(\delta(\Delta_1),\ldots,\delta(\Delta_k))$, let $(\pi)_+$ denote
  $L(\delta(\Delta_{i_1}),\ldots,\delta(\Delta_{i_l}))$, where $\{i_1,\ldots,i_l\}$ is a subset of $\{1,2,\ldots,k\}$ of all indices such that the sum of minimal and maximal exponent in $\Delta_i$ is positive.
  From our assumption $a_1\ge\frac{1}{2}$ it follows that $(\pi_L)_+\simeq\pi_L$ and from \cite[Lemma 4.2]{Matic3} we see $\widetilde{\pi_L}\simeq L(\delta([\nu^{-b_t}\widetilde{\rho},\nu^{-a_t}\widetilde{\rho}]),\ldots,\delta([\nu^{-b_1}\widetilde{\rho},\nu^{-a_1}\widetilde{\rho}]))$ so $(\widetilde{\pi_L})_+\simeq1$.
  Since $\pi_L$ is an irreducible representation, we get that the induced representation $\pi_L\rtimes\sigma_c$ is irreducible, thus finishing the proof.
\end{proof}
Since $\nu^{x}\rho\rtimes\sigma_c$ is irreducible for every $\nu^x\rho\in [\pi_L]$ for non-selfcontragredient $\rho$, the induced representation $\pi_L\rtimes\s_c$ is in that case irreducible and isomorphic to $$L(\delta([\nu^{-b_t}\rho,\nu^{-a_t}\rho]),\ldots,\delta([\nu^{-b_1}\rho,\nu^{-a_1}\rho]);\sigma_c).$$
For selfcontragredient $\rho$, there is a unique non-negative real number $x$ such that $\nu^x\rho\rtimes\sigma_c$ reduces and we denote it by $\alpha$.
From Proposition \ref{propi} we see that for $\alpha=0$ the induced representation $\pi_L\rtimes\s_c$ is irreducible.
Thus, in what follows we assume $\rho$ is selfcontragredient and $\alpha>0$.

Recall that the non-tempered subquotients of induced representations from $R(G)$ are, according to the Langlands classification, parametrized by $L(\delta_1,\ldots,\delta_l;\tau)$, where $\delta_1, \ldots, \delta_l $ are irreducible essentially square-integrable representations of general linear groups such that $e(\delta_1) \leq \ldots \leq e(\delta_l) < 0$ and  $\tau \in \Irr(G_{n'})$ is a tempered representation.
Since $L(\delta_1,\ldots,\delta_l;\tau)\le\pi_L\rtimes\s_c$ and  $L(\delta_1,\ldots,\delta_l)\otimes\tau\le \mu^*(L(\delta_1,\ldots,\delta_l;\tau))$, the transitivity of the Jacquet modules and the structural formula (\ref{struct}) imply that there exists a pair
$\big((c_1,\ldots,c_t),(d_1,\ldots,d_t)\big)\in \Lad(\pi_L)'$ such that
\begin{gather}
    \label{a}
    L(\delta_1,\ldots,\delta_l)\le
     L(\delta([\nu^{-c_t}\rho,\nu^{-a_t}\rho]),\ldots,\delta([\nu^{-c_1}\rho,\nu^{-a_1}\rho]))\times \\
    \times L(\delta([\nu^{d_1+1}\rho,\nu^{b_1}\rho]),\ldots,\delta([\nu^{d_t+1}\rho,\nu^{b_t}\rho])) \nonumber
\end{gather}
and
\begin{gather}
\label{b}
    \tau\le L(\delta([\nu^{c_1+1}\rho,\nu^{d_1}\rho]),\ldots,\delta([\nu^{c_t+1}\rho,\nu^{d_t}\rho])) \rtimes \s_c.
\end{gather}
Since $e(\delta_i)<0$ for $i=1,2,\ldots,l$, from (\ref{a}) we conclude $d_i=b_i$ for $i=1,2,\ldots,t$.
Namely, considering the cuspidal support of the representation on the right-hand side of (\ref{a}), we notice that there is no segment with both positive and negative exponents contained in it.
Recall that assumed restrictions on the exponents give $-a_1\le -\frac{1}{2}$ and $d_1+1\ge c_1+1\ge a_1 \ge\frac{1}{2}$. If $d_1+1\ge\frac{3}{2}$, then the difference between the maximal negative and minimal positive exponent is greater than or equal to two. If $d_1+1=\frac{1}{2}$, then $c_1+1=a_1=\frac{1}{2}$, so $\delta([\nu^{-c_1}\rho,\nu^{-a_1}\rho])\simeq 1$. Now the maximal negative exponent is less than or equal to $-a_2$, which is less than or equal to  $-\frac{3}{2}$, so we have the same conclusion as in the first case.

These conclusions reduce (\ref{a}) and (\ref{b}) to
\begin{gather*}
    L(\delta_1,\ldots,\delta_l)\le
     L(\delta([\nu^{-c_t}\rho,\nu^{-a_t}\rho]),\ldots,\delta([\nu^{-c_1}\rho,\nu^{-a_1}\rho]))
\end{gather*}
and
\begin{gather} \label{t1}
    \tau\le L(\delta([\nu^{c_1+1}\rho,\nu^{b_1}\rho]),\ldots,\delta([\nu^{c_t+1}\rho,\nu^{b_t}\rho])) \rtimes \s_c,
\end{gather}
respectively.
Denote by $i_M$ (resp.~$i_m$) the maximal (resp.~minimal) index $i\in\{1,2,\ldots,t\}$ such that $\alpha\in [a_i, b_i]$.
Note that if $i_M$ exists, then $\alpha\in [a_i, b_i]$ for every $i\in[i_m,i_M]$, since $\alpha\in[a_{i_M},b_{i_m}]\subseteq[a_i,b_i]$.
Further, denote by $k$ an arbitrary element of $[i_m,i_M]$.  We define $k_m$ to be a minimal index from $\{1,2,\ldots,t\}$ such that $\alpha-k+k_m\le b_{k_m}.$
Note that $\alpha-k+i\in[a_i,b_i],\forall i\in[k_m,k]$.
This follows by inductively applying the following inequalities which are the consequence of $a_1<\ldots<a_t$ and $b_1<\ldots<b_t$:
\begin{gather*}
 b_{k_m+1}\ge b_{k_m}+1\ge \alpha-k+(k_m+1) \text{ and } a_{k-1}\le a_k-1\le \alpha-1.
\end{gather*}
Consequently, $\alpha-k+i$ are positive numbers for $i\in [k_m,k]$.
We define $\sigma_k$
as a unique irreducible subrepresentation of
\begin{gather*}
    \delta([\nu^{\alpha-k+k_m}\rho,\nu^{b_{k_m}}\rho])\times\ldots\times\delta([\nu^{\alpha}\rho,\nu^{b_k}\rho]) \rtimes \s_c
\end{gather*}
and $\pi_k$, unless $k=t$ and $a_i=\alpha-t+i$ for $i=1,2,\ldots,t$, in which case we define it as
\begin{gather*}
    L(\delta([\nu^{-b_t}\rho,\nu^{-a_t}\rho]),\ldots,\delta([\nu^{-b_{k+1}}\rho,\nu^{-a_{k+1}}\rho]),\delta([\nu^{-\alpha+1}\rho,\nu^{-a_k}\rho]),\ldots, \\
    \delta([\nu^{-\alpha+k-k_m+1}\rho,\nu^{-a_{k_m}}\rho]),\delta([\nu^{-b_{k_m-1}}\rho,\nu^{-a_{k_m-1}}\rho]),\ldots,\delta([\nu^{-b_1}\rho,\nu^{-a_1}\rho]);\sigma_k).
\end{gather*}
Note that $\pi_k$ is well defined since $-b_{k+1}<-b_k<-\alpha+1$,  $b_{k_m-1}<\alpha-k+k_m-1$ because of the definition of index $k_m$ and $-(a_1+b_1)<0$.
\begin{theorem}
\label{tm2}
Assume that $\nu^{x}\rho\rtimes\sigma_c$ reduces for some $\nu^x\rho\in [\pi_L]$.
Then the induced representation $\pi_L\rtimes\s_c$ reduces. Furthermore, let $\alpha$ be the unique positive number such that $\nu^{\alpha}\rho\rtimes\sigma_c$ reduces. Then the semi-simplification of $\pi_L \rtimes \sigma_c$ is equal to
\begin{itemize}
    \item[(i)] $\s_t + L(\delta([\nu^{-b_t}\rho,\nu^{-a_t}\rho]),\ldots,\delta([\nu^{-b_1}\rho,\nu^{-a_1}\rho]);\s_c) + \displaystyle\sum\limits_{k\in[i_m,t-1]} \pi_k,$ \\ if $a_i=\alpha-t+i$ for $i=1,2,\ldots,t$,
    \item[(ii)] $L(\delta([\nu^{-b_t}\rho,\nu^{-a_t}\rho]),\ldots,\delta([\nu^{-b_1}\rho,\nu^{-a_1}\rho]);\s_c) + \displaystyle\sum\limits_{k\in[i_m,i_M]} \pi_k, $ otherwise.
\end{itemize}
\end{theorem}
\begin{proof}
As seen in Proposition \ref{propi}, if $\nu^{x}\rho\rtimes\sigma_c$ reduces for some $\nu^x\rho\in [\pi_L]$, then the induced representation $\pi_L\rtimes\s_c$ reduces.
The existence of a tempered subquotient $\s_t$ is established in Theorem \ref{tm1}, so we turn to the non-tempered case.
Since $\nu^{\alpha}\rho\in [\pi_L]$, there exists $i\in\{1,\ldots,t\}$ such that $\alpha\in [a_i,b_i]$ so $[i_m,i_M]\neq\emptyset$, while it is possible for $[i_m,t-1]$ to be empty.

Firstly, assume that $\pi=L(\delta_1,\ldots,\delta_l;\tau)$ is a non-tempered subquotient of $\pi_L\rtimes\sigma_c$.
From (\ref{t1}) we can see that $\tau$, if not cuspidal, is necessarily equal to $\sigma_k$ for some $k\in[i_m,i_M]$.
Namely, Theorem \ref{tm1} gives the following conditions on real numbers $c_i$ defined in the discussion before Theorem \ref{tm2} since $c_1+1\ge a_1 \ge\frac{1}{2}$:
\begin{enumerate}
    \item For some fixed $k\in[i_m,i_M]$ such that $c_k+1=\alpha$, we get $c_i=b_i$ for $i=k+1,\ldots,t$. Similarly, since $\alpha-k+i>b_i$ for $i=1,2,\ldots,k_m-1$, we get $c_i=b_i$ for $i=1,2,\ldots,k_m-1$.
    \item There exists $k_0\in[k_m,k]$ such that $c_i=b_i$ for $i=k_m,\ldots,k_0-1$ and $c_i+1=\alpha-k+i$ for $i=k_0,\ldots,k$. We claim that $k_0=k_m$. Recall that $\alpha-k+i\in[a_i,b_i],\forall i\in[k_m,k]$.
    Thus if $k_0> k_m$, then $c_{k_0-1}=b_{k_0-1}$ and $c_{k_0}+1=\alpha-k+k_0$ imply
    $$\alpha-k+k_0-2=c_{k_0}-1 \ge c_{k_0-1}=b_{k_0-1}\ge \alpha-k+k_0-1, $$
    which is a contradiction.
\end{enumerate}
In this way, $\pi$ is either equal to $L(\delta([\nu^{-b_t}\rho,\nu^{-a_t}\rho]),\ldots,\delta([\nu^{-b_1}\rho,\nu^{-a_1}\rho]);\sigma_c)$ or to $\pi_k$ for some $k\in[i_m,i_M]$.

Conversely, we prove that $\pi_k$ is a subquotient of $\pi_L\rtimes\sigma_c$ for every $k\in [i_m,i_M]$. Let us fix some $k\in [i_m,i_M]$.
We define $\delta_i=\delta([\nu^{-b_i}\rho,\nu^{-a_i}\rho])$ for $i=1,\ldots,k_m-1,k+1,\ldots,t$, $\delta_i=\delta([\nu^{-\alpha+k-i+1}\rho,\nu^{-a_{i}}\rho])$ for $i=k_m,\ldots,k$,
\begin{gather*}
    \pi_k^{(1)} = \delta_t\times\ldots\times\delta_{k+1}\times\delta_k\times \ldots
    \times \delta_{k_m}\times\delta_{k_m-1}\times\ldots\times\delta_1, \\
    \Pi = \pi_k^{(1)}\times L(\delta([\nu^{\alpha-k+k_m}\rho,\nu^{b_{k_m}}\rho]),\ldots,\delta([\nu^{\alpha}\rho,\nu^{b_k}\rho])) \rtimes \s_c
\end{gather*}
and $\pi_k^{(2)} = L(\delta_t,\ldots,\delta_1)$, i.e. the Langlands subrepresentation of $\pi_k^{(1)}$.
\iffalse
Our goal is to obtain that both $\pi_k$ and $\pi_L\rtimes\sigma_c$ are the subquotients of $\Pi$, both contain $\pi_k^{(2)}\otimes\sigma_k$ in the Jacquet modules with respect to the appropriate parabolic subgroups and it appears in the Jacquet module with respect to the appropriate parabolic subgroup of $\Pi$ with multiplicity one. Since $\pi_k$ is irreducible, this would imply that $\pi_k\le\pi_L\rtimes\sigma_c$.
\fi
Note that $\sigma_k\hookrightarrow L(\delta([\nu^{\alpha-k+k_m}\rho,\nu^{b_{k_m}}\rho]),\ldots,\delta([\nu^{\alpha}\rho,\nu^{b_k}\rho])) \rtimes \s_c$
since $\sigma_k$ is the unique irreducible subrepresentation of $\delta([\nu^{\alpha-k+k_m}\rho,\nu^{b_{k_m}}\rho])\times\ldots\times\delta([\nu^{\alpha}\rho,\nu^{b_k}\rho]) \rtimes \s_c$.
Thus together with $\pi_k^{(2)}\hookrightarrow\pi_k^{(1)}$, we obtain $\pi_k\le \Pi$.
It is an easy consequence of Langlands classification that $$L(\delta([\nu^{a_{k_m}}\rho,\nu^{b_{k_m}}\rho]),\ldots,\delta([\nu^{a_k}\rho,\nu^{b_k}\rho]))$$ is a subquotient of
$$ \widetilde{\delta}_{k_m}\times\ldots\times\widetilde{\delta}_k\times L(\delta([\nu^{\alpha-k+k_m}\rho,\nu^{b_{k_m}}\rho]),\ldots,\delta([\nu^{\alpha}\rho,\nu^{b_k}\rho])). $$
This, together with the properties of $R(G)$,  clearly implies $\pi_L\rtimes\sigma_c \le \Pi$.

The Frobenius reciprocity implies that $\pi_k^{(2)}\otimes\sigma_k$ is a subquotient of $\mu^*(\pi_k)$ and $\mu^*(\Pi)$.
Also note that the above discussion shows that we can choose real numbers $c_i$ for $i=1,\ldots,t$ to obtain $\pi_k^{(2)}\otimes\sigma_k$ in $\mu^*(\pi_L\rtimes\sigma_c)$.
The transitivity of Jacquet modules now implies that the representation $\delta_t\otimes\ldots\otimes\delta_1\otimes\sigma_k$ is a subquotient of $r_{\beta}(\pi_k)$, $r_{\beta}(\pi_L\rtimes\sigma_c)$ and $r_{\beta}(\Pi)$ for the appropriate parabolic subgroup $P_{\beta}$.

To finish the proof, it remains to see that $\delta_t\otimes\ldots\otimes\delta_1\otimes\sigma_k$ appears in $r_{\beta}(\Pi)$ with multiplicity one.
Firstly, we want to determine all the constituents of $\mu^*(\Pi)$ of the form $\delta_t\otimes\pi$ for some $\pi\in \Irr(G)$.
According to the definition of $\pi_L$, $-b_t$ is the smallest exponent in the cuspidal support of $\Pi$. Also, $\nu^{-b_t}\rho$ appears in $[\Pi]$ with multiplicity one.
Using the multiplicativity of $M^{\ast}$, let $\pi_1 \otimes \pi_2$ denote an irreducible constituent of some factor of $\mu^{\ast}(\Pi)$ such that $\nu^{-b_t}\rho\in [\pi_1]$. We see that $\pi_1\otimes\pi_2$ is necessarily equal to $\delta_t\otimes 1\le  M^*(\delta_t)$.
Thus we get that
$$\delta_t\otimes \delta_{t-1}\times\ldots\times\delta_1\times L(\delta([\nu^{\alpha-k+k_m}\rho,\nu^{b_{k_m}}\rho]),\ldots,\delta([\nu^{\alpha}\rho,\nu^{b_k}\rho])) \rtimes \s_c $$
contains all constituents of $\mu^*(\Pi)$ of the form $\delta_t\otimes\pi$ for some $\pi\in \Irr(G)$.
Inductively, $\delta_t\otimes\ldots\otimes\delta_1\otimes\sigma_k$ appears in $r_{\beta}(\Pi)$ with the same multiplicity as $\delta_k\otimes\ldots\otimes\delta_1\otimes\sigma_k$ in the Jacquet module of
\begin{gather*}
    \Pi'=\delta_{k}\times\ldots\times\delta_1\times L(\delta([\nu^{\alpha-k+k_m}\rho,\nu^{b_{k_m}}\rho]),\ldots,\delta([\nu^{\alpha}\rho,\nu^{b_k}\rho])) \rtimes \s_c
\end{gather*}
with respect to the appropriate parabolic subgroup.

Let us now determine all the constituents of $\mu^*(\Pi')$ of the form $\delta_k\otimes\pi'$ for some $\pi'\in \Irr(G)$.
Note that $\nu^{-\alpha+1}\rho$ is not an element of the cuspidal support of $\delta_{k-1}\times\ldots\times\delta_{1}$ since minimal exponents in $[\delta_i]$ for $i=1,\ldots,k$ are in strictly increasing order.
According to the formula (\ref{struct}) we search for $\nu^{-\alpha+1}\rho$ in
\begin{gather*}
   L(\delta([\nu^{-x_{k}}\rho,\nu^{-\alpha}\rho]),\ldots,\delta([\nu^{-x_{k_m} }\rho,\nu^{-\alpha+k-k_m}\rho]))
\end{gather*}
where $x_{k_m}<\ldots<x_k$ and $\alpha-k+i-1 \le x_i\le b_i $ for $i=k_m,\ldots,k$.
Thus $-x_{i_0}=-\alpha+1$ for some $i_0\in\{k_m,\ldots,k-1\}$ and $x_i=\alpha-k+i-1$ for $i=i_0+1,\ldots,k$. This implies $-\alpha+1=-x_{i_0}>-x_k=-\alpha+1$, showing it is not possible.
Using the multiplicativity of $M^{\ast}$,
let $\pi_1'\otimes\pi_2'$ denote an irreducible constituent of some factor of $\mu^{\ast}(\Pi')$ such that $\nu^{-\alpha+1}\rho\in [\pi_1']$.
We see that $\pi_1'\otimes\pi_2'$ is necessarily equal to $\delta_k\otimes1\le M^*(\delta_k)$.
Inductively, $\delta_t\otimes\ldots\otimes\delta_1\otimes\sigma_k$ appears in $r_{\beta}(\Pi)$ with the same multiplicity as $\delta_{k_m-1}\otimes\ldots\otimes\delta_1\otimes\sigma_k$ in the Jacquet module of
\begin{gather*}
    \Pi''=\delta_{k_m-1}\times\ldots\times\delta_1\times L(\delta([\nu^{\alpha-k+k_m}\rho,\nu^{b_{k_m}}\rho]),\ldots,\delta([\nu^{\alpha}\rho,\nu^{b_k}\rho])) \rtimes \s_c
\end{gather*}
with respect to the appropriate parabolic subgroup.
Note that the greatest positive exponent in the cuspidal support of $\widetilde{\delta}_{k_m-1}\times\ldots\times\widetilde{\delta}_1$ is $b_{k_m-1}$, which is by definition of $k_m$ strictly less than $\alpha-k+k_m-1$.
Thus the cuspidal supports of representations  $\widetilde{\delta}_{k_m-1}\times\ldots\times\widetilde{\delta}_1$ and $L(\delta([\nu^{\alpha-k+k_m}\rho,\nu^{b_{k_m}}\rho]),\ldots,\delta([\nu^{\alpha}\rho,\nu^{b_k}\rho]))$ are disjoint multisets.
The constituents of $\mu^*(\Pi'')$ of the form $\pi''\otimes\sigma_k$ for $\pi''\in \Irr(GL)$ with negative exponents in $[\pi'']$ are consequently contained in
$$ \delta_{k_m-1}\times\ldots\times\delta_1\otimes L(\delta([\nu^{\alpha-k+k_m}\rho,\nu^{b_{k_m}}\rho]),\ldots,\delta([\nu^{\alpha}\rho,\nu^{b_k}\rho])) \rtimes \s_c .$$
Since $\sigma_k$ is a unique subrepresentation of
$$L(\delta([\nu^{\alpha-k+k_m}\rho,\nu^{b_{k_m}}\rho]),\ldots,\delta([\nu^{\alpha}\rho,\nu^{b_k}\rho])) \rtimes \s_c,$$
we finally conclude that $\delta_t\otimes\ldots\otimes\delta_1\otimes\sigma_k$ appears in $r_{\beta}(\Pi)$ with multiplicity one.
Hence it appears in $r_{\beta}(\pi_k)$ and $r_{\beta}(\pi_L\rtimes\s_c)$ with multiplicity one and we arrive to the conclusion that $\pi_k$ appears in $\pi_L\rtimes\sigma_c$ with multiplicity one.
\end{proof}
\bibliographystyle{siam}

redpajama_set_name: RedPajamaArXiv


Anchored in the Theravadan Buddhist Burmese lineage of Mahasi Sayadaw since 1974, Steven Smith was trained and sanctioned as a teacher by the late revered monk and meditation master Sayadaw U Pandita. Steven teaches Vipassana and the Divine Abodes (loving-kindness, compassion, joy, equanimity) in meditation retreats in Asia, Hawai'i, North America and Australia. He co-founded Vipassana Hawai'i in 1984 and in 1995 founded the MettaDana Project for educational and medical projects in Burma. Also in 1995 Steven helped establish the Kyaswa Valley Retreat Center in Burma, headed by the late Sayadaw U Lakkhana, Abbot of Kyaswa Monastery.
This retreat is suitable for beginners and experienced meditators.
Steven Smith is a founder of Vipassana Hawai'i and Kyaswa Valley Retreat Centre and MettaDana health and education project both in Myanmar. Steve has practised and studied vipassanā as a monk and layperson for more than 40 years and is anchored in the Theravāda tradition.


Award-winning freelance art director Nancy Roy feeds her wanderlust as a self-proclaimed digital gypsy, splitting her time among Denver, New York, Chicago and Washington, DC. Her work has been recognized by PRINT Magazine, SPD, Folio:, ASBPE, ASHPE, Smashing Magazine, Custom Content Council and is featured in the Big Book of Layouts. Her designs have appeared in Entrepreneur, WebMD the Magazine, WebMD's Medscape, Diabetes Forecast, Denver Magazine, HemAware, AOA Focus, National Underwriter and the Consumer Electronics Association's Vision magazine.


The following article is entirely the opinion of Hope Collier and does not reflect the views of the Inquisitr.
Tennis living legend Serena Williams shocked the world when she posted and quickly deleted a Snapchat post announcing that she is 20 weeks pregnant. While most people responded with jubilation at the announcement, there were some who took her pregnancy as a sign that she was on her way to retirement at the apparently ripe old age of 35.
As someone who has watched Serena Williams evolve from the sister of Venus to the greatest tennis player ever (yeah, I said it), I am here to tell anyone who thinks that pregnancy is going to stop her from achieving more success on the court that they are beyond delusional.
Let's just take a look at all of the adversity that Serena has faced throughout her 20-plus year career.
Back in 1997, Serena was a mere footnote in the story of Venus Williams, who was in the midst of going from a child prodigy to a full-fledged adult phenom on the WTA circuit. It was Venus who demanded the headlines, who everyone saw as the transformational figure that emerged from the legacy laid by the likes of Althea Gibson, Zina Garrison, and, of course, Arthur Ashe.
Two decades ago, Serena had a dismal world ranking of 304. Nobody was thinking about her and then BOOM! – she knocks out two top-10 players in a tournament (Mary Pierce and Monica Seles). Suddenly, the world notices that maybe, just maybe, Venus's little sister might be someone to watch.
Cut to just two years later and Serena begins racking up singles titles left and right, culminating in a stunning victory over then No. 1 Martina Hingis at the U.S. Open.
After her 1999 U.S. Open win, Serena became a solid top-10 player, but injuries prevented her from being a consistent force on the women's circuit. Then, in 2001, she began to face a series of professional and personal setbacks that would have broken most athletes. However, Serena emerged stronger and on a destined path to greatness.
In 2001, Venus and Serena were set to go head to head in a semifinal at a tournament at Indian Wells, California. Elena Dementieva, fresh off of a loss against Venus in a quarterfinal, decided to be a hater and speak about the long-rumored, yet unsubstantiated claim that the sister's father/coach Richard Williams fixed their matches.
When Venus withdrew from the semifinal match, Serena faced tremendous scrutiny that culminated in her being booed and subjected to overt racist remarks from the crowd when she defeated Kim Clijsters in the final match.
That experience prompted both Venus and Serena to boycott Indian Wells for years.
Serena's dominance on the court continued. She earned her first "Serena Slam." With each tournament win, she seemed unstoppable and then came the shocking murder of her sister Yetunde Price in 2003.
The death of her sister coupled with a series of devastating injuries left Serena with a reputation for being wildly inconsistent on the court. Although she had talent and an awe-inspiring work ethic, she was consistently labeled as someone that didn't have the drive to compete. That reputation stuck with Serena until fairly recently when she joined forces with coach Patrick Mouratoglou and began her drive towards the most Grand Slams in the Open era.
One critique that Serena consistently faced was that she relied on power while other players used intelligence and strategy. This criticism has always struck me as the most racially motivated of all. To some, Serena will always just be some muscular force who simply overpowers opponents.
To this day, Serena continues to be maligned and called masculine simply because she doesn't fit some outdated vision of what a woman athlete should like. The staggering amount of discipline it takes to achieve a body like Serena's is rarely mentioned in discussions on her game, and in that silence is the implicit notion that she somehow has an advantage due to African ancestry.
In an interview with Common, Serena spoke at length about the frustrations she experienced when not getting the credit due for the intelligence behind her dominating play, as well as her confidence.
From the 1936 Olympics to today, there is an overwhelming narrative that exists about black athletes in the media. Credit is always given for physical prowess and little else. Serena Williams is someone who not only challenges that point of view, but she turns that racist notion on its and head and exposes the idiocy of it.
You don't get to where Serena has gotten in her career without being the smartest person at all times. She anticipates her opponent's every move with a precision the game has never seen and quite possibly will never see again.
Serena's level of dominance doesn't just happen because she lucked out genetically.
In recent years, analysts have finally begun to acknowledge that Serena possesses a finesse and skill on the court, but a disproportionate level of credit is given to Patrick Mouratoglou. Yes, he certainly has helped her to refine her level of play and mindset, but at the end of the day, she is the one on the court playing.
When you look back on Serena's still-vital career and all of the challenges she has faced, how could anyone actually believe that her pregnancy news means that she is done with tennis? How could anyone think that this superwoman from Compton, California, is incapable of simultaneously being a mother and a world-class athlete?
I will admit that Serena's pregnancy news is a shock and that it will most certainly delay her overtaking Margaret Court as the all-time Women's Grand-Slam recipient, but it won't stop her from doing it.
Until very recently, Roger Federer faced a career slump that lasted years, and I don't recall anyone saying that he should just quit now and go be a father. No, people cheered him on and encouraged him to keep fighting through mediocre play.
Just like her sister Venus, who is on track to play well into her 40s, Serena defies what it means to be an aging athlete every day. To think that pregnancy will cause her to retire right as she is on the precipice of breaking what was thought to be an unbreakable record is stupid, to say the least.
If ever there was an athlete that knows how to multitask, it is Serena Williams. I can only imagine that motherhood will kick her ability to juggle multiple things at once into complete overdrive.
With news from Reuters confirming that Serena will indeed return to tennis after sitting out this year to become a mother, I believe that she will transform how we view athletes who are parents. So for all of the folks out there ready to write Serena's career postmortem, please remember this: whether you like it or not, she ain't going anywhere.
I look forward to seeing how Serena's next incredible chapter with a new baby will play out.


text:
Q: Counting Number of Paired Un-set bits At Even Boundaries Given a 64-bit number what's the best way to find out the number of paired un-set bits at even boundaries. The extra zero padding after the MSB should be ignored.
For example:
For the two numbers 25223 and 10578
25223 -- 01 10 00 10 10 00 01 11
          7  6  5  4  3  2  1  0
Count = 2, (at positions 2 and 5)

10578 -- 00 10 10 01 01 01 00 10
          7  6  5  4  3  2  1  0
Count = 1, (at position 1. Ignore position 7)

I could do a mask, shift-by-2 and compare, but I'm looking for something better. Is there anything faster than this:
def PairedCount(n):
    c=0
    while(n!=0):
        if((n & 3) == 0):
            c+=1
        n >>= 2;
    return c

What if I want to count the number of paired non-zero bits at even boundaries? What's the best method for that?

A: It's a simple question, but the way you put it scares me :)
Let's first try doing it to pairs of 1 s (you'll see why) for 32 bits:
unsigned count_pairs_1(unsigned n){
    n = n & ( n >> 1);  // bit N will be set if bits N and N+1 were set
    n &= 0x55555555;    // we need just those on even position, so ANDing with 0b01..0101
    return count_set_bits(n);  // now we need the number of 1 bits in the result
};

All we need now it count_set_bits(unsigned) , that is very known function: http://www-graphics.stanford.edu/~seander/bithacks.html#CountBitsSetTable
To count zero bits use count_pairs(~n) or
unsigned count_pairs_0(unsigned n){
    n = n | ( n >> 1); // bit N will be zero iff bits N and N+1 were zero
    n |= 0xAAAAAAAA; // all odd bits are set
    return 32 - count_set_bits(n);  // every remaining zero bit corresponds to zero pair in the input
};

EDIT: just observed the remark Given the 64 bit number... The extra zero padding after the MSB should be ignored. After what MSB? Do you mean the input is a byte? or word?

A: unsigned count_pairs_0_n(unsigned n){
  unsigned int i=n;
  unsigned int l=0;
  while(i){l=i;i&=(i-1);}
  n=((l<<1) -1) &(~n);
  return count_pairs_1(n);
}

based on @rusliks answer, I tried making my answer a bit short.

A: This is for 32bits .. 0x55555555 is a dependency .. is order of number of set bit
   int countpairs(int n){
      int o=n;
      int c=0;

      unsigned int i=n;
      unsigned int l=0;
      while(i){l=i;i&=(i-1);}

      n=((l<<1) -1) &(~n);

      while(n){
        unsigned int k= n&(n-1);
        unsigned int k2=k&(k-1);
        unsigned int k3=(n^k) + (k^k2);
        if((n^k) && k^k2 && (n^k)*2 == (k^k2) && ((n^k) & 0x55555555)) {
            c++;
        }
        n=k;
      }
     return c;
    }

A: This is not much cheaper (one loop per pair of zeroes + overhead) but it's just to expose a few bit tricks.
size_t count_pairs_of_zeros( your_uint_type x );
{
   // create a mask with even bits set like 0x55555555
   // but independent of bit length
   your_uint_type mask = 0;
   mask -= 1;
   mask /= 3;

   // replace 01 and 10 pairs by 11
   x |= x>>1;
   x &= mask;
   x |= x<<1;

   // count the pairs of zeros up to most significant bit
   size_t count = 0;
   while( x & (x+1) )
   {
      count++;
      // remove next pair of zeros
      x |= x+1;
      x |= x+1;
   }
   return count;
}

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On 2nd July 2015, a somewhat cloudy and showery day, a trip to Benfield Nature Reserve (see Brighton & Hove nature reserves) to examine the grassland habitats and see why perhaps the glowworms had declined, provided a wonderful scene of flowering plants and insects. The site is cared for by the Benfield Wildlife Group and they have noticed that the glowworms have declined. Why this should be is unclear and may be a result of habitat change, light levels, pesticides, parasites or other reasons (see the Glow Worm Survey site for more information on these insects). Much of the area visited (details to follow) appeared ideal for glowworms.
Although wet and warm, the drizzle stopped and the sun partly came out and with it some insects including butterflies: Meadow Brown, Ringlet, Marbled White, Small Skipper, and Small Tortoiseshell. No blues were seen. A male Meadow Browns found a female, the pair danced and eventually dropped to the ground and mated.
More pictures from this trip to be added.


1920s Living Room Ideas have several images associated one another. Obtain the most update pictures of 1920s Living Room Ideas on this page, and you may bring the images from this blog for personal usage. 1920s Living Room Ideas photos are posted by our team on July 20, 2018 at 11:26 pm. The image collection of 1920s Living Room Ideas presented on this page is some of the best pictures around the internet and has been seen by 94 viewers.
We get some stunning images which you can utilize as design ideas and/or for informative usage of 1920s Living Room Ideas. The team have faith that you are all enjoying our compilation of the most outstanding photo of 1920s Living Room Ideas on this site and utilize them for personal purpose. We also provide the images in High Quality (HQ) Resolution that can be loaded to your laptop. What you need is only click on the compilation of 1920s Living Room Ideas underneath.


- We are representing the German company, SAP, which in the near future is planning to open an office in Azerbaijan. SAP is a global leader in software manufacturing and associated services. Our company is a major integrator of ICT solutions, with some 150 offices around the world. We offer IT solutions for large, medium-sized and small businesses, as well as the local implementation of solutions and development of applications.
Currently SAP is poorly represented in Azerbaijan; therefore our participation at Bakutel is important for telling others about our brand, which is necessary in order to attract customers to our soon-to-be-opened office. The exhibition, of course, has been very fruitful for us, and yesterday there was a lot of activity on our stand. Mainly, we met with representatives of other businesses – we held many meetings and made useful discoveries. From our participation in Bakutel we expect to develop our business in Azerbaijan, which will allow us to achieve the important task of reaching a new place in the market here.


Who will play in this years 103rd Grey Cup Eastern Finals? Stay in touch with Quality Plus Tickets to find out! This year, the Grey Cup will be held at the Investors Group Field in Winnipeg on November 29th at 5pm.
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text:
What is The Owners Club?
Why was The Owners Club created?
What is the concept of member ownership?
Why is The Owners Club a more sensible option than buying a conventional vacation home?
Why is The Owners Club a better option than renting or staying at a hotel on your vacation?
What does a typical home look like?
What kinds of amenities and services are offered?
Where are The Owners Club locations?
What is ClubCorp?
The Owners Club is a vacation home membership concept that provides a unique, cost-effective alternative to conventional second home ownership. The "membership home" concept offers fractional ownership of private luxury vacation homes with club membership privileges in some of the most desirable vacation destinations. Members also enjoy reciprocal privileges at affiliate Owners Clubs, present and future. And because The Owners Club is affiliated with ClubCorp, members also receive associate club membership privileges at nearly 170 ClubCorp country clubs, business and sports clubs and golf resorts around the world.
The Owners Club was established by ClubCorp in 1994 as a membership club that provided a vacation "home away from home." ClubCorp asked members what they wanted when it came to second home ownership, vacations and active retirement. They answered that they wanted the services and amenities of the private club experience while they were on vacation. They wanted the privacy of a single-family, detached home. They wanted the investment value of deeded ownership. They also wanted the flexibility of having more frequent (often briefer) vacations in private, single-family accommodations in a variety of exquisite destinations. Hence the creation of The Owners Club, the "vacation country club."
As a member of The Owners Club, you own part of a private vacation home – an investment with real value. Most are single-family homes with only 13 members, and each development has a small number of homes, which creates an exclusive community environment among residents. Members can share their vacation experience with family, friends and business associates for 27 or more days each year. Owners have priority at their home club, but they also enjoy reciprocal privileges at affiliate Owners Clubs, present and future. As an added benefit, they also receive associate membership at nearly 170 ClubCorp country clubs, athletic and city clubs around the world.
With your Owners Club membership, you buy a fraction of a well-appointed private home that includes the benefits of belonging to an exclusive club community with amenities and personal concierge service not available to conventional homeowners. This is a smart investment because most people today use a vacation home on only holidays or long weekends. As a member you have the luxury of deciding the length of each stay in your "home away from home," be it a long weekend, a few days or a week or two at a time. The Owners Club takes care of maintaining the properties to ClubCorp standards of excellence, with none of the expenses and headaches of owning and maintaining a conventional second home. Because members enjoy reciprocal privileges at affiliate Owners Club properties, it's actually like having several vacation homes plus private club memberships for a fraction of what a single home would cost.
You stay in a private, single family, well-appointed home. When you become a member, you are investing in an asset that can be passed down to your children. The Owners Club is not a lodging experience, it's a resort/club experience in a private vacation home. There is also tremendous flexibility in the length of the stay; you choose when you want to go and how long you want to stay. As a member, another advantage is that you can get to know your neighbors as they return year after year because The Owners Club is a community of members.
The Owners Club homes are detached, single family, two-bedroom, three-bath homes ranging from 2,000 to 2,700 square feet in size. Homes are fully furnished with high-end furniture and finishes, and they include amenities such as two master bedrooms, big-screen TVs, washers and dryers. Every home has a balcony or patio to enjoy the exquisite views.
The Owners Club does it the way you would do it if you could, which enables you to enjoy the leisure time without having to do errands or home maintenance. For instance: you may call ahead to the concierge with a shopping list so your refrigerator is stocked when you arrive; the Club can arrange private valet airport transportation; and planned activities such as golf, kids' clubs and more can be reserved in advance so that you can really make the most of your vacation time.
The Owners Club at Barton Creek in Austin, Texas
The Owners Club at Hilton Head Island, South Carolina
The Owners Club at The Homestead in Hot Springs, Virginia
The Owners Club at Marina Vallarta in Puerto Vallarta, Mexico
Members enjoy reciprocal privileges at affiliate Owners Clubs, present and future.
Founded in 1957, Dallas-based ClubCorp is the world leader in delivering golf, private club and resort experiences. Internationally, the privately-held company owns or operates nearly 170 country clubs, business and sports clubs and golf resorts.

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text:
Home / news / COPPIN STATE COMMEMORATES 60TH ANNIVERSARY OF A RAISIN IN THE SUN
COPPIN STATE COMMEMORATES 60TH ANNIVERSARY OF A RAISIN IN THE SUN
By bmorenews on February 15, 2019
(BALTIMORE—February 15, 2019) – Coppin Repertory Theatre at Coppin State University continues its 2018-2019 Commemorative Theatre Season with a special production in celebration of the 60th anniversary of Lorraine Hansberry's A Raisin in the Sun. Performances will run March 8-11, 2019 in the James Weldon Johnson Auditorium, 2500 West North Avenue.
A Raisin in the Sun Show Times
Friday, March 8th ……………………………….………………………7:30 p.m.
Saturday, March 9th……….……………………………..2:00 p.m. & 7:30 p.m.
Sunday, March 10th …………………………….………………………3:00 p.m.
Monday, March 11th ………………………………..……………………..7:30 p.m.
Based on the Langston Hughes poem Dream Deferred, A Raisin in the Sun tells the story of an African American family living on Chicago's South Side during the late 1950's, struggling for their piece of the American Dream. Alternating between humor and drama, Hansberry's play chronicles the struggles of members of the Younger family as they attempt to break out of the vicious cycle of poverty. Each has a dream, at times at odds with each other's, but always deferred.
A Raisin in the Sun made its historic Broadway debut on March 11, 1959 at the Ethel Barrymore Theatre, where it ran for 530 performances. The production, directed by Lloyd Richards, featured a stellar cast including Sidney Poitier, Ruby Dee, Claudia McNeil and Diana Sands.
The play won the 1959 New York Drama Critics Circle Award and a Pulitzer Prize nomination. The 1961 movie adaptation, with the original cast, won the Gary Cooper Award at the 1962 Cannes International Film Festival. A subsequent musical version of the play won Tony and Grammy Awards in 1974 for Best Musical.
Hansberry's play was revived on Broadway in April 2004, in celebration of its 45th anniversary.
Directed by Kenny Leon, the cast included Sean Combs, Phylicia Rashad, Audra McDonald and Sanaa Lathan. Rashad and McDonald received Tony and Drama Desk Awards for Best Actress in a Play and Best Featured Actress in a Play, respectively. Rashad also made history, becoming
the first African American actress to win the Tony for Leading Actress in a Play. During its run, the revival of A Raisin in the Sun was the highest grossing non-musical on Broadway.
Coppin's Professor Willie O. Jordan is directing the campus production.
The cost of admission is $25.00 for the general public, $20.00 for seniors and non-Coppin students.
Coppin students with ID are admitted for $10.00. Group discounts are also available.
To purchase tickets online, go to www.raisinatcoppin.brownpapertickets.com. For additional
information, call Dr. Garey Hyatt at (410) 951-3369.
Related ItemsCOPPIN STATE COMMEMORATES 60TH ANNIVERSARY OF A RAISIN IN THE SUNhomepage
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text:
Fish Conservation
What are greenhorns on a crab vessel -- and why are they called that?
by Debra Ronca
Browse the article What are greenhorns on a crab vessel -- and why are they called that?
Crab fishermen who haul in pots face furious and frigid waves.
Erik Hill/Anchorage Daily News/Getty Images
Wanted: Crew members for dangerous deep-sea crabbing. Location: The middle of an icy ocean. Description: Endure merciless hazing by fellow crewmen; work 15 hours or more a day, seven days a week; risk your life; get paid only half of what everyone else gets paid on the boat.
Believe it or not, each year, people fight tooth and nail to land this position. Why? Because as brutal as the work is, the pay can be handsome. Crew members on a crab fishing boat can earn up to $15,000 in one month [source: Alaska Fishing Jobs]. Unfortunately, as with any career, before you earn top dollars, you have to start at the bottom. Entry-level crabbers are called "greenhorns," and you'll find at least one on every crab boat.
To understand what a greenhorn is, let's first take a look at the meaning of the word. Originally, a greenhorn referred to an animal with green (or young) horns. By the mid-1600s, a greenhorn referred to a newly enlisted soldier. Then, in the 1830s, the word came to describe a very inexperienced person or a novice in a particular trade [source: Safire]. That last definition aptly describes the crab fishing greenhorn.
On a crab boat, a greenhorn is a newbie who may or may not have previous fishing experience. A greenhorn could be completely new to the fishing industry, or he or she could be several seasons in -- still learning the ropes on a particular boat. Commercial crab fishing is punishing and dangerous work, and many greenhorns quit after the first trip. To learn more about the work, read Why was Alaskan fishing named the most dangerous job in the world?.
We're not exaggerating about the difficulty of a greenhorn's job. In season two of the TV show "Deadliest Catch," the conditions on the Aleutian Ballad pushed the boat's greenhorn to his breaking point. Desperate to escape the vessel, Lysander Smith threatened to jump into the frigid water -- which would have meant certain death -- claiming he'd wait for the Coast Guard to rescue him. The ship had to make an unscheduled stop to return Smith to the harbor and pick up a fresh greenhorn. Smith's meltdown caused the Aleutian Ballad to lose hours of valuable fishing time. On dry land, Smith admitted that the job was simply more than he could handle.
Crab fishing is not for the weak of spirit.
How to Become a Greenhorn
You can see why tasks like this one might lead to a man overboard situation.
Donald Bland/Discovery Communications LLC
A greenhorn is the lowest man on the totem pole on a crab boat. He or she works for a season or two on the deck, helping the deckhands and performing the grunt work. We'll cover those grunt work tasks in more detail in the next section.
What's the payoff? A successful crab boat can bring in nearly a million dollars in one trip. Crew members might go home with more than $50K for only a few weeks of work [source: Miller]. It's that promise of quick cash that lures many greenhorns into the crabbing lifestyle. But it takes a while to earn those spoils. Greenhorns typically receive only half of what the regular crew members make, what the industry dubs a half-share. Crew members earn a high-share or full-share. Also, many greenhorns and deckhands are expected to pay up front for their equipment and food, or the cost is taken out of their cut of the haul [source: Hutton]. Greenhorns and deckhands must also pay for their own fishing licenses. It may seem like a thankless job, but the point is this: Put in your time, and eventually you'll get promoted to full-share deckhand or crew member.
Besides the money, many greenhorns join the crab fishing industry for the challenge and the adventure. TV coverage, like the show "Deadliest Catch," has attracted more and more people to the industry. So how does one become a greenhorn? As with most jobs -- it's often who you know more than what you know. Most boats are family-run, and captains will only hire people they trust [source: ADF&G]. Knowing the new hire is important -- captains don't want to wind up with a greenhorn who has a nervous breakdown mid-trip. The Alaskan Bering Sea is extremely dangerous, and a captain usually won't take out anyone who doesn't have deck experience and general know-how.
If you're looking for a greenhorn position, job centers in Alaska will point you in the right direction. Some of them suggest that you simply hanging around at the docks and nearby bars to let skippers and captains know you're available for work. It's possible to get hired as a greenhorn deckhand with no previous experience if you can convince the captain that you'll work hard. There's no formal crab fishing training -- greenhorns learn on the job. However, fishing experience is always a plus. Many deckhands start with salmon fishing in the summer months in order to get a feel for the industry. Crab processing jobs are also a good way to get a foot in the door, even though pay is usually low.
With the recent crab rationalization measures put in place, fewer vessels go out each year, reducing the amount of available deckhand jobs. (Find out more about crab rationalization in How is a crab boat like a floating city?.) About 250 Alaskan crab boats go out each season and each crew typically has between three and nine deckhands [source: Deadliest Catch Facts]. Because of the pay, deckhand jobs don't turn over often.
Other Greenhorns
Greenhorns aren't just found on fishing boats. Many industries dub their newbies "greenhorns." Here are some examples:
Greenhorns are a new breed of young farmers and agriculturalists.
Seasoned cowboys call novice ranchers greenhorns or tenderfoots.
A greenhorn pro wrestler is someone new and inexperienced in the ring.
Greenhorn Fishing Jobs: Dirty Work
A greenhorn embrace -- deckhand Jake Harris and greenhorn Joshua Harris, onboard an Alaskan king crab boat
Zac McFarlane/Discovery Communications LLC
Once a greenhorn gets a job on a boat, he has to prove himself. Many captains don't trust greenhorns until they've proven that they work hard and they work safely. Greenhorns are a mixed blessing on a boat -- they're eager and ready to work, but also inexperienced.
Greenhorns work alongside deckhands, learning the labor and performing some of the least desirable tasks. For example, a typical greenhorn duty is to prepare bait for the crab pots. That means chopping up thousands of pounds of dead fish. Later, the greenhorn will have to crawl into the crab pot to hook the 12 pounds of bait to the trap -- a backbreaking chore. Greenhorns also are put to the task of chopping ice off boat rails with a sledgehammer. Crews do this to ensure the boat doesn't get weighed down by hundreds of pounds of ice that can quickly form in the freezing temperatures. Falling ice can also cause serious injury or death. Now imagine doing these tasks -- with sledge hammer or knife in hand -- on a boat that's listing from side to side, is slippery with ice, in 60 mph (80 kph) winds. And you haven't slept in days.
Because there's so much money at stake, crab fishing is very competitive. The northern latitudes only get about five full hours of sunlight, which makes the work even more difficult. Once a boat finds a honey hole -- a secret spot with a lot of crab -- the crew wants to haul in as much crab as possible for as long as the weather allows. Conditions on the Bering Sea are unpredictable and dangerous -- one rogue wave or mechanical breakdown can destroy an entire crabbing expedition.
Greenhorns also need to stay mentally alert without much sleep. Physical and mental exhaustion can lead to mistakes. In an industry as dangerous as crab fishing, mistakes can be very harmful. A deckhand unlucky enough to get hit with a swinging crab pot could get his or her head smashed in or legs crushed. One errant step into a moving coil of rope could fling him overboard. If he doesn't pay attention at the bait-cutting machine, he could lose an entire arm.
As if the physical and mental demands aren't enough, greenhorns also endure good-natured hazing from their fellow fishermen. Every crew has its own hazing rituals. Sometimes the captain will take the greenhorn's sharp fishing knife and replace it with a butter knife until he feels the greenhorn is experienced enough to handle a real knife. Some crews make the newest greenhorn bite the head off a dead herring for good luck and a good season. Crew members also will periodically throw fish at an unsuspecting greenhorn's head -- or maybe make him wear a Viking helmet with green horns [source: Stanley].
What happens if a greenhorn can't handle the job? When you're on a boat in the middle of the Bering Sea, you can't just get up and leave. Once you're out there, you're out there. If the situation is dire, though, a boat may return to dock. Seasickness can become very critical very quickly. People with severe seasickness become dehydrated and then delirious, necessitating professional medical attention and intravenous fluids [source: Hillstrand].
For more information on crab fishing, haul up some links to related articles on the next page.
Ay Ay, Captain!
Crab boat captains offer words of advice to greenhorns:
Do what you are told and ask questions after.
Don't act like you know what you are doing.
Pay attention to what people are trying to teach you.
Know your role.
And shut your hole.
Work your butt off and don't complain.
No cry babies!
Leave your ego at the dock.
Our boat works our way.
Never let up!
[source: Deadliest Catch Wiki]
Crab Fishing Quiz
How Rogue Waves Work
How to Avoid Hypothermia
The Bering Sea
10 Most Dangerous Jobs in America
Why was Alaskan fishing named the most dangerous job in the world?
How is a crab boat like a floating city?
How to Survive a Sinking Ship
What do I do if I fall overboard?
Inside "Deadliest Catch"
How to Survive a Shipwreck
How Deep-sea Rescue Works
Alaska Job Finder
Crab Fishing 101
Alaska Crabbing Jobs
ADF&G Division of Commercial Fisheries. "Frequently Asked Questions." 2008. (Aug. 5, 2008) http://www.cf.adfg.state.ak.us/geninfo/about/faq/cf_faq.php
Alaska Fishing Jobs Center. "Crab Fishing in Alaska." 2008. (Aug. 4, 2008) http://www.fishingjobs.com/index.php?option=com_content&view=article&id=14&Itemid=15
Bryant, Charles W. "Inside 'Deadliest Catch'." HowStuffWorks.com. July 29, 2008. (Aug. 13, 2008) https://entertainment.howstuffworks.com/deadliest-catch.htm/printable
Deadliest Catch. Season 1, Episode 1. "Greenhorns." Apr. 12, 2005. (Aug. 5, 2008)
Deadliest Catch. Season 2, Episode 4. "The Finish Line." Apr. 18, 2006. (Aug. 5, 2008)
Deadliest Catch Wiki. "Words of Advice for a Greenhorn." June 17, 2008. (Aug. 5, 2007) http://www.deadliest-catch-wiki.discovery.com/page/Words+of+Advice+for+a+Greenhorn
Division of Employment Security. "Seafood and Fishing Jobs in Alaska." 2008. (Aug. 13, 2008) http://labor.state.ak.us/esd_alaska_jobs/seafood.htm
Gwinn, Mary Ann. "Mining Crab in a Merciless Sea." The Seattle Times. Jan. 12, 1997. (Aug. 13, 2008) http://community.seattletimes.nwsource.com/archive/?date=19970112&slug=2518396
Hillstrand, Johnathan. "Live Chat." The Patriot Resource. May 1, 2007. (Aug. 6, 2008) http://www.patriotresource.com/deadliest/articles/0705chatjohn.html
Hutton, Ryan. "Former Local Man in Deep Water." North Adams Transcript. June 20, 2007. (Aug. 4, 2008) http://deadliestreports.wordpress.com/2007/06/20/fv-wizard-greenhorn-guy-says-he-wont-do-it-again/
Mapes, Lynda V. "Low Tech -- Eight Hours at a Desk Isn't You? There's Still a Paycheck in the Old Standbys." The Seattle Times. Nov. 24, 1997. (Aug. 13, 2008) http://community.seattletimes.nwsource.com/archive/?date=19971124&slug=2574374
Miller, Kerry. "Worst Jobs with the Best Pay." Business Week. Sept. 14, 2006. (Aug. 7, 2008) http://www.businessweek.com/careers/content/sep2006/ca20060914_736742.htm?chan=search
Mills, Pete. "Bonerack: Crab Therapy." WOAI.com: San Antonio News. Jul. 29, 2008. (Aug. 13, 2008) http://www.woai.com/content/blogs/bonerack/story.aspx?content_id=644636b2-70fe-476b-a52c-190001a91a81
Moriarty, Colin. "Deadliest Catch: 'Trials of the Greenhorns' Review." June 6, 2007. (Aug. 6, 2008) http://tv.ign.com/articles/794/794620p1.html
Safire, William. "On Language; Greenhorn." New York Times. June 23, 2001. (Aug. 7, 2008)http://query.nytimes.com/gst/fullpage.html?res=9D0CE7D9143CF930A15755C0A967958260
Stanley, Doug. "Flying Fish." Production Diary. June 10, 2008. (Aug. 6, 2008) http://blogs.discovery.com/deadliest_catch_diaries/2008/06/flying-fish.html
Weatherby, Cooper. "Live Chat." Deadliest Catch Wiki. June 17, 2008. (Aug. 4, 2008) http://www.deadliest-catch-wiki.discovery.com/page/Cooper+Weatherby%2C+Page+1


text:
Луи́с Альфо́нсо Бурбонский (; ; родился 25 апреля 1974 года, Мадрид) — член испанской королевской семьи, легитимистский претендент на трон Франции. Правнук короля Испании Альфонсо XIII.

Биография
Луис Альфонсо родился в Мадриде в 1974 году в семье принца Альфонсо и доньи Марии дель Кармен Мартинес-Бордью и Франко, внучки испанского диктатора Франсиско Франко. На тот момент отца принца — Альфонсо Хайме, герцога Анжу и Сеговии, — часть французских легитимистов считали законным претендентом на трон Франции. После его смерти в 1975 году Альфонсо стал главой Французского Королевского Дома и в этом качестве пожаловал своему второму сыну титул герцога Туреньского.

7 февраля 1984 старший брат Луиса Альфонсо Франсиско погиб в результате автокатастрофы. С этого времени Луис Альфонсо стал наследником своего отца и носил титул герцога Бурбонского. В 1989 году, после гибели отца, Луис Альфонсо стал главой Дома Бурбонов и претендентом на французский престол; он принял титул герцога Анжуйского.

Семья
О подготовке Луиса Альфонсо к женитьбе на наследнице венесуэльского знатного рода Марии Маргарите Варгас Сантаэлле было объявлено в ноябре 2003. Гражданский брак был заключён в Каракасе 5 ноября 2004, церковный — 6 ноября 2004 в Ла-Романе, Доминиканская Республика. Никто из членов испанской королевской семьи не присутствовал на свадьбе. С 2005 года супруги живут в Венесуэле; Луис Альфонсо работает в Banco Occidental de Descuento.

В этом браке родились четверо детей — дочь и трое сыновей:
 Эухения (для французских легитимистов — принцесса Эжени де Бурбон) (род. 5 марта 2007 в Синайском медицинском центре в Майами);
 Луис, герцог Бургундский (род. 28 мая 2010 года с братом-близнецом);
 Альфонсо,  герцог Беррийский (род. 28 мая 2010 года с братом-близнецом);
 Энрике, герцог Туреньский (род. 1 февраля 2019 года).

Титулы и звания
Луис Альфонсо поддерживается частью легитимистов как претендент на трон Франции; королевская семья Испании не поддерживает его в этом качестве. В настоящее время Луи Альфонсо претендует на следующие титулы и статусы:
 Титулярный герцог Анжуйский
 Титулярный король Франции
 Титулярный король Наварры
 Глава Дома Бурбонов
 Бальи Большого Креста Почета и Преданности Мальтийского ордена

Литература

 Thierry Ardisson. Louis XX. Contre-enquête sur la monarchie. — Olivier Orban, 1986. — ISBN 2-85565-334-7
 Apezarena, José. Luis Alfonso de Borbón: Un príncipe a la espera. — Forthcoming.
 Cassani Pironti, Fabio. Bref crayon généalogique de S.A.R. la Princesse Marie-Marguerite, Duchesse d'Anjou, née Vargas Santaella. — Le Lien Légitimiste, n. 16, 2007.
 Opfell, Olga S. H.R.H. Louis-Alphonse, Prince of Bourbon, Duke of Anjou: Royal House of France. // House of Bourbon, Royalty Who Wait: The 21 Heads of Formerly Regnant Houses of Europe. — Jefferson: McFarland & Company, Inc., 2001. — P. 11—32.

Претенденты на трон Франции
Герцоги Анжуйские
Титулярные герцоги де Бурбон
Члены королевского дома Испании
Испанские Бурбоны
Герцоги Турени
Принцы Франции (Бурбоны)

redpajama_set_name: RedPajamaWikipedia


text:
One on One: Jim Crow is still alive in North Carolina
Published 2:38 pm Sunday, December 13, 2020
By Guest Columnist
D.G. Martin
By D.G. Martin
Unbelievable when you think about it, how North Carolina maintained its segregated and discriminatory racial system for such a long time. The remnants of what is commonly called "Jim Crow" are still with us, a daily reminder of the horrors of the past.
A new book, "Jim Crow in North Carolina: The Legislative Program from 1865 to 1920," by Richard A. Paschal helps us understand the impact of Jim Crow on North Carolina. It catalogues and summarizes race-based laws passed by the General Assembly from the end of the Civil War until 1920.
Paschal, a Raleigh lawyer who holds a graduate degree in history, argues that it was not so much the laws on the books that brought about and maintained the segregated and oppressive system sometimes branded as "Jim Crow." More important, he says, were the longstanding community standards and customs and the allocation of community resources by the dominating white power structure.
Paschal challenges the views of respected historians such as C. Vann Woodward, whose classic book, "The Strange Career of Jim Crow," argued that, even under slavery, the two races had not been as divided as they were under the Jim Crow laws of the late 1890s and afterwards. Further, Woodward said that during Reconstruction, there was significant racial mixing in economic and political matters. The segregating of the races, he wrote, was a relative newcomer to the region.
The core of Paschal's book is his lists of laws passed by the N. C. General Assembly from 1865 through 1920. The lists include numerous laws passed before the 1890s that restricted blacks or discriminated against them.
While having separate schools for blacks and whites was not controversial in the time after the Civil War, unequal funding was a continuing blight. For instance, local communities had the power and responsibility for establishing and funding schools, supposedly on an equal basis. But one method approved by the legislature provided that the white schools would be funded by taxes on white people and black schools would be funded by taxes on "colored persons." In 1885 the N. C. Supreme Court recognized the discrimination and struck down the plan.
Paschal asserts that many of the Jim Crow restrictions and much of the subjugation of blacks were not solely "de jure," that is, mandated by specific laws. Much of Jim Crow came about without specific legislation.
Even the most blatant result of the 1898 and 1900 white supremacy campaigns, the disenfranchisement of blacks, was accomplished by implementation of a literacy requirement for voting. Paschal points out that a significant number of blacks were literate. The law permitted literate blacks to vote. But in its implementation, even the most literate blacks failed the tests imposed by community voting officials who were, of course, all white.
With respect to jury participation by blacks, Paschal found no racial restrictions in the law, but blacks rarely were allowed to serve. "The lack of African Americans in the jury box worked to the detriment of the black community and black criminal defendants for decades. The racial incitement of the white supremacy campaigns created a chasm between legal text and the application and operation of the law, as local officials tilted the playing field decisively against the state's African American population."
In his conclusion Paschal writes, "In order to fully understand the shadow that Jim Crow continues to cast over us today, it is necessary to know how ostensibly democratic government at all levels and in all places used law to advance white interests while disadvantaging the interests of African Americans and other minorities."
He might have added that in too many places, it is still happening.
D.G. Martin hosts "North Carolina Bookwatch," Sundays at 3:30 p.m. and Tuesdays at 5 p.m. on UNC-TV. The program also airs on the North Carolina Channel Tuesdays at 8 p.m. and other times.
FOR MORE COLUMNS AND LETTERS TO THE EDITOR, CHECK OUT OUR OPINION SECTION HERE.
Memories and Musings: Getting lost is a special skill
One on One: Civil war – our people versus my people?
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The Nature Corner: Worth a thousand words
Letter to the Editor: Good deed appreciated


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The purpose of this course is to develop competence in: physical fitness, locomotor and non-locomotor expertise, participation expertise, social habits, and techniques for bodily activities. Requiring individuals to take part in bodily schooling actions, resembling dodge ball , flag soccer , and other aggressive sports activities stays a controversial subject due to the social affect these have instances physical education schemes have been cut. Malina RM. Adherence to physical activity from childhood to adulthood: A perspective from monitoring studies.
There are a wide range of degree options that an aspiring PE Teacher can select from including BS in Heath Science, BS in Health and Wellness, Master of Public Heath, Master of Healthcare Administration, and more. Janelle Farley is a certified sport administration professional who is keen about everyone main a healthy and lively life-style. Recent scientific studies have confirmed that students, who give equal importance to games along with research, are achievers in professional life as in comparison with those who neglect sports.
In Indonesia , college students starting from Kindergarten to High School have PE built-in with their curriculum. When viewed from the actual fact, this model is extra more likely to be utilized to the lower lessons, especially from the beginning tamn childhood to grade three elementary school. This is one reason why physical education is an integral part in a faculty's curriculum, especially for grade college to high school students. Thus, Physical Education will not be solely geared toward bodily development but also contains the event of the individual as a whole. Students are also taught physical exercises to enhance and maintain good well being.
UNESCO will work with governments and other stakeholders to develop a methodology for the worldwide follow-up to the Sport Charter. Adopted in 1978, the unique Charter was perceived as progressive on the time – because it was the primary rights-based document to state that the practice of physical education and sport is a basic proper for all". Meanwhile actions in physical education is much more extensively, which could be a sport or other physical activities comparable to recreation, adventure, social activity, or vary of motion base.
In Malaysia , pupils from primary colleges to secondary schools are expected to do 2 durations or 1 hour of PE all year long except a week earlier than examination In most secondary faculties , video games like badminton , sepak takraw , soccer , netball , basketball and tennis are available. Various kinds of video games, sports activities, and physical workout routines are taught in faculties as part of bodily schooling program.


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Japan and other countries warn citizens about traveling to the US due to mass shootings
Photo: Nicole S Glass/Shutterstock
After two mass shootings this weekend left 31 people dead in the US, Venezuela, Japan, and Uruguay are urging their citizens to reconsider their vacations. In a statement issued by the Venezuelan government, people are advised to stay away from cities that are considered the most violent, including Cleveland, Detroit, and Baltimore. A similar statement was issued by Uruguay, encouraging citizens to take extra precautions while traveling to the US due to its "indiscriminate violence" and "racism and discrimination that cost the lives of more than 250 people in the first seven months of this year."
Uruguay's government specifically advises people to avoid places where there are high concentrations of people, like theme parks, malls, festivals, sporting events, mass protests, and religious gatherings, which might be targets for potential shooters.
The warnings out of Latin America come on the heels of the revelation that the El Paso shooting suspect had posted a manifesto admitting that the massacre was racially motivated.
Similarly, Japan's Ministry of Foreign Affairs issued an alert cautioning Japanese citizens to be "aware of potential for gunfire incidents everywhere in the United States," and goes on to describe the US as a "gun society."
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And when Adam Curry calls the iPod a platform, he's overlooking one important thing that users and developers like to see in a platform: choice of vendor. It's laughable to think that iTunes will ever work with another vendor's hardware, it's not very likely to happen, in our lifetimes, even if we live to a ripe old age. I literally did laugh when I considered the idea. That's not the way Apple works.
iTunes for Windows can transfer a variety of audio and music files to iPod, including MP3 and AAC encoded files. Other MP3 players do not work with iTunes for Windows.
So I wonder, are those players listed on the Mac OS X Compatible Players page mostly legacy things, either deals that existed before the iPod took off, or what? I mean, the list is not long, and there's only a few companies represented on it. I know I get just a little nervous at each iTunes release that my Rio 500 will no longer be supported.
Of course the Apple-fan can explain this as such: Apple wants you to buy an iPod, and a Mac, but if you won't buy a Mac, at least buy an iPod, and we'll give you the nice iPod software called iTunes, and it'll work with the iPod, and a few other devices, if you have a Mac, but not if you have a Windows PC. Or something like that… Just buy an iPod… and a Mac.
Posted on Thursday, September 2nd, 2004 at 11:30 am. Subscribe to the comments feed. Both comments and pings are currently closed. Post on your own blog.


In October 2013, Rupen Pandya was appointed President and Chief Executive Officer of SaskBuilds, a Treasury Board Crown Corporation with the Government of Saskatchewan. Rupen leads a team responsible for advising the Saskatchewan Government on infrastructure priorities and alternative financing solutions for large and complex capital projects. SaskBuilds is currently managing a portfolio of projects totaling approximately $2 billion.
Previously, Rupen was the Assistant Deputy Minister responsible for the Labour Market Development Division with the Ministry of the Economy, and the Assistant Deputy Minister for Immigration with the Ministry of Advanced Education, Employment and Immigration. In these senior leadership roles, Rupen was responsible for provincial labour market agreement negotiations as well as developing the province's immigration and labour market strategies. Immigration, which accounts for more than 65 per cent of Saskatchewan's population growth, is central to the province's long-term growth planning.
With almost 20 years of public service in Saskatchewan, Rupen has extensive professional experience in infrastructure procurement, advanced policy development and implementation encompassing: labour market policy, post-secondary education, housing, social services and intergovernmental affairs.
He holds Bachelor and Master of Arts degrees from the University of Regina. Rupen was raised in Regina where he resides with his wife Terri and two young daughters.


The battle for on-court supremacy between the North Providence Police and Fire Departments will play out next Thursday, March 28, in the North Providence High School gym, 1828 Mineral Spring Ave. Police Chief David Tikoian got the idea for this charity game after attending a charity football game between the North Providence and Johnston Fire Departments in November of 2017. "I considered an event that would benefit a nonprofit in our town," he said. He floated the idea in January, and it was immediately supported by Mayor Charles Lombardi and Fire Chief John Silva. All proceeds from the event will benefit Tri-Town Community Action, a local agency providing needed social services to town residents. "I'm confident this will be an exciting event for police and fire professionals, their families and the community to support," Tikoian said.


Where Lost Briton road, Bridgnorth.
Other Info Nervous rescue dog who escaped from her new home. Please ring if sighted.
I am truly sorry to read such a tragic update for Persia.
Our lovely little Piti (Persia), who went missing from her new home on 7th June, has been found killed by a car. She had a significant head injury. The vet said she would have died instantly, our only relief is to know that she didn't suffer.
Piti had been in the shelter for a long time, living most of it with one other dog, Negrut. Dora always felt she could be rehomed, even though she was shy and nervous, she had a lovely soul. She was a very sweet girl and we all hoped she would find a lovely family and have a happy life in the UK.
When she was found today near Market Drayton, she was in good condition considering she'd been wandering alone for over 3 months. So she knew how to look after herself, darling girl, but she was seen so many times near busy roads it was always a risk that she would get run over one day. We tried so hard to find her, so many people helped, hours & hours of searching and putting up posters, leaflets through doors, sharing on Facebook, sending emails. As always it was a team effort and it's impossible to thank everyone individually, you all know who you are, thank you so much for helping.
Piti's life was tragically cut short by this accident, but in her life she did know love from Dora and the whole Amicii team her room mate Negrut and the team at Forest Dog Rescue, in particular Sam who worked with her since April and loved her so much.
This is not what we bring dogs from Romania for, to lose their lives on the UK roads. Please let her legacy be that we all take more care, especially when new dogs arrive, and we give them time to settle and bond with their human family so that they don't run away and get lost like she did.
Sighting during the period 16-20 June on the car park at The Royal Oak in Alveley. Visited there yesterday, landlady confirmed definitely was Persia, she was trying to drink the water from the hanging baskets, lady took her a bowl of water and then tried to catch her, she ran off. Posters put up all over Alveley and Arley village yesterday and leaflet drop in farms locally. No further sightings.
Oh :( Gutted to see that this little sweetheart still hasn't been found. Is there any news? Have there been any sighting? What is being done to get this little girl found?
Tweeted again, hope she is found soon, never give up! We've been out for dog walks in the Benthall area, no sign of Persia so far.
Shared face book Twitter and local FB groups.
Lost notices put up around Benthall which is a few miles away from Persia's home.
Shared to Dog Lost West Midlands FB page and asked if sighted to ring immediately and not to chase. J.


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Q: Parse.com Facebook Login stays on "Loading" iOS I'm using Parse, with the LogInViewController. I added Facebook on the logIn View.
When someone wants to connect with Facebook, it opens the browser, the user accepts the application and then he returns to the app. But when he returns on the app, he stays on "loading" and nothing happens.
However if the user restart the app, it works.
Only when it's the first connexion for the user I have this problem.
- (void)viewDidAppear:(BOOL)animated {
    [super viewDidAppear:animated];

    if (![PFUser currentUser]) {
        // Customize the Log In View Controller
        PFLogInViewController *logInViewController = [[PFLogInViewController alloc] init];
        [logInViewController setDelegate:self];
        [logInViewController setFacebookPermissions:[NSArray arrayWithObjects:@"friends_about_me", nil]];
        [logInViewController setFields:  PFLogInFieldsFacebook | PFLogInFieldsDismissButton];

        // Present Log In View Controller
        [self presentViewController:logInViewController animated:YES completion:NULL];
    }
}

I think the problem isn't in my code because I use the default code of Parse but I'm not sure.

A: - (BOOL)application:(UIApplication *)application openURL:(NSURL *)url sourceApplication:(NSString *)sourceApplication annotation:(id)annotation {
    return [FBAppCall handleOpenURL:url
                  sourceApplication:sourceApplication
                        withSession:[PFFacebookUtils session]];
}

- (void)applicationDidBecomeActive:(UIApplication *)application {
    /*
     Restart any tasks that were paused (or not yet started) while the application was inactive. If the application was previously in the background, optionally refresh the user interface.
     */
    [FBAppCall handleDidBecomeActiveWithSession:[PFFacebookUtils session]];
}

- (void)applicationWillTerminate:(UIApplication *)application {
    /*
     Called when the application is about to terminate.
     Save data if appropriate.
     See also applicationDidEnterBackground:.
     */
    [[PFFacebookUtils session] close];
}


Camou-coloured line (three colours) with high durability. Ideal for method feeder fishing, whether in competition or when specimen hunting. In fact, whenever you need a reliable, inconspicuous line. Very fast sinking like a plummet!