text:
\section{Introduction}
Nijenhuis operators on Lie algebras have been introduced in the theory of integrable systems in the work of Magri, Gelfand and Dorfman (see the book \cite{Dorfman}), and, under the name of hereditary operators, in that of Fuchssteiner and Fokas (\cite{FuchFoka}). In the sense of the theory of deformations of Lie algebras (\cite{Nijenhuis}), Nijenhuis operators generate trivial deformations naturally. More precisely,  A {\bf Nijenhuis operator} on a Lie algebra $(\g,[-,-]_\g)$ is a linear map $N:\g\longrightarrow\g$ satisfying
\begin{equation}
  [N(x),N(y)]_\g=N\big([N(x),y]_\g+[x,N(y)]_\g-N[x,y]_\g\big),\quad \forall~x,y\in\g.
\end{equation}
Furthermore, the deformed bracket $[-,-]_N:\wedge^2\g\longrightarrow\g$ given by
\begin{equation}\label{eq:deformbracket}
  [x,y]_N=[N(x),y]_\g+[x,N(y)]_\g-N([x,y]_\g),
\end{equation}
 is a Lie bracket and $N$ is a Lie algebra morphism from $(\g,[-,-]_N)$ to $(\g,[-,-]_\g)$.

 Poisson-Nijenhuis structures were defined by Magri and Morosi in 1984 (\cite{MaMo}) in their study of completely integrable systems. See \cite{Kosmann1,Kosmann2} for more details on Poisson-Nijenhuis structures. Recently, Ravanpak,   Rezaei-Aghdam and   Haghighatdoost introduced the notion of an $r$-$n$ structure on a Lie algebra $\g$, which is the infinitesimal of a right-invariant Poisson-Nijenhuis structure on the  Lie group $G$ integrating the Lie algebra $\g$ (\cite{rn}).  An   $r$-$n$ structure on a Lie algebra $\g$ is a pair $(\pi,N)$, where $\pi$ is an $r$-matrix, i.e. $\pi$ satisfies the classical Yang-Baxter equation $[\pi,\pi]_\g=0$,   and $N$ is a Nijenhuis operator on $\g$, such that some compatibility conditions are satisfied. In this paper, we will call such a structure an   $r$-matrix-Nijenhuis structure. An equivalent description of a bivector   $\pi\in\wedge^2\g$ being an  $r$-matrix is given by
\begin{equation}\label{eq:r-matrix}
  {[\pi^\sharp(\alpha),\pi^\sharp(\beta)]}_\g=\pi^\sharp(\ad^*_{\pi^\sharp(\alpha)}\beta-\ad^*_{\pi^\sharp(\beta)}\alpha),\quad\forall~\alpha,\beta\in\g^*,
\end{equation}
where   $\pi^\sharp:\g^*\longrightarrow\g$ is defined  by
$\langle\pi^\sharp(\alpha),\beta\rangle=\pi(\alpha,\beta)$ and
 $\ad^*$ is the coadjoint representation of the Lie algebra $\g$. See \cite{Drinfeld,Kosmann0} for more details.

\emptycomment{ Based on the importance of  Poisson-Nijenhuis structures, our work is motivated by the following questions:
 \begin{itemize}
  \item What is the equivalent description of the $r$-matrix-Nijenhuis structure in the language of the Rota-Baxter operator, or equivalently, what is the Rota-Baxter-Nijenhuis structure?

  \item We have seen that the Kupershmidt operators can unify the above two different operators of the $r$-matrix. What is the natural generalization of $r$-matrix-Nijenhuis structure or Rota-Baxter-Nijenhuis structure to the arbitrary Kupershmidt operators?
\end{itemize}
The Kupershmidt-(dual-)Nijenhuis structures that we introduce in this paper provides answers of above questions.
Let $(A,[-,-]_A,a_A)$ be a Lie algebroid on the manifold $M$. A bivector field $\pi\in\Gamma(\wedge^2 A)$ defines a Poisson structure on $A$ if and only if $\pi$ satisfies
\begin{equation}
\Courant{\pi,\pi}=0,
\end{equation}
where the bracket $\Courant{-,-}$ is a Schouten bracket on $A$. It is well-known that when the base manifold $M$ is a point, the Lie algebroid structure reduces to a Lie algebra structure. Now the Poisson structure $\pi$ is just the $r$-matrix, or a solution of classical Yang-Baxter equation for the Lie algebra. Let $(\g,[-,-]_\g$ be a Lie algebra and $\pi$ is an $r$-matrix. There are the following two main approaches to study $r$-matrix from a linear operator but the tensor form.
The other approach describes the $r$-matrix  $\pi$ as a linear map $\pi^\sharp:\g^*\longrightarrow\g$ given by
$$\langle\pi^\sharp(\alpha),\beta\rangle=\pi(\alpha,\beta)$$
satisfying
\begin{equation}\label{eq:r-matrix}
  {[\pi^\sharp(\alpha),\pi^\sharp(\beta)]}_\g=\pi^\sharp(\ad^*_{\pi^\sharp(\alpha)}\beta-\ad^*_{\pi^\sharp(\beta)}\alpha),\quad\forall~\alpha,\beta\in\g^*,
\end{equation}
where $\ad^*$ is the coadjoint representation of the Lie algebra $\g$. }

 Baxter introduced the concept of a Rota-Baxter algebra  for associative algebras   (\cite{Ba}) in his study of fluctuation theory in probability. It has been found many applications in recent years, including  the algebraic approach of Connes-Kreimer~(\cite{CK}) to renormalization of perturbative quantum field theory, tridendriform
algebras (\cite{BGN2013}), quantum analogue of Poisson geometry (\cite{Uchino1}), twisting on associative algebras (\cite{Uchino2}). In the Lie algebra context, a Rota-Baxter operator of weight zero was introduced independently in the 1980s as the operator form of the classical Yang-Baxter equation, named after the physicists C.-N. Yang and R. Baxter. See the book \cite{Gub} for more details. A linear operator $\huaR:\g\longrightarrow\g$ on a Lie algebra $\g$   is called a Rota-Baxter operator if the following condition is satisfied:
\begin{equation}\label{eq:Rota-Baxter}
  [\huaR(x),\huaR(y)]_\g=\huaR([\huaR(x),y]_\g+[x,\huaR(y)]_\g),\quad\forall~x,y\in\g.
\end{equation}
In \cite{Semonov-Tian-Shansky}, Semonov-Tian-Shansky studied the classical Yang-Baxter equation systematically.
He proved that if there is an $\ad$-invariant, non-degenerate, symmetric bilinear form on $\g$, then a Rota-Baxter operator and an $r$-matrix are equivalent.   Moreover, Kupershmidt restudied the classical Yang-Baxter equation (\cite{Kuper1}) and generalized the above representation to   an arbitrary representation $(V;\rho)$ of $\g$, and introduce the notion of a Kupershmidt operator (also called an $\huaO$-operator). By definition, a Kupershmidt operator on a Lie algebra $\g$ with respect to a representation $(V;\rho)$ is a
linear map $T:V\longrightarrow \g$ satisfying
\begin{equation}
  [T(u), T(v)]_\g=T\Big(\rho(T(u))(v)-\rho(T(v))(u)\Big),\quad \forall~u,v\in V.
\end{equation}
  Note that a Rota-Baxter operator given by  Semonov-Tian-Shansky is just a Kupershmidt operator on a Lie algebra with respect  to the adjoint representation $(\g;\ad)$ and an $r$-matrix is a Kupershmidt operator on a Lie algebra with respect to the coadjoint representation $(\g^*;\ad^*)$. Moreover, the notion of an extended $\huaO$-operator was introduced by Bai, Guo and Ni in \cite{BGN2010,BGN2011} and plays important role in the study of nonabelian generalized Lax pairs and the extended classical Yang-Baxter equation.

The purpose of this paper is to introduce the notion of a Kupershmidt-(dual-)Nijenhuis structure on a Lie algebra with respect to a representation that contains the aforementioned $r$-matrix-Nijenhuis structure as a special case, and give applications.

First we study   infinitesimal deformations of a Lie algebra with a representation and introduce the notion of a Nijenhuis pair. Similar to that a Nijenhuis operator gives a trivial infinitesimal deformation of a Lie algebra, a Nijenhuis pair  gives a trivial infinitesimal deformation of a Lie algebra with a representation. We also introduce the notion of a dual-Nijenhuis pair, which can be viewed as the dual of a Nijenhuis pair. Based on the Nijenhuis pair and dual-Nijenhuis pair, we introduce the notions of a Kupershmidt-Nijenhuis structure and a Kupershmidt-dual-Nijenhuis structure, respectively.
Just as that a Poisson-Nijenhuis structure gives rise to a hierarchy  of Poisson structures, a Kupershmidt-(dual-)Nijenhuis structure also gives rise to a hierarchy of Kupershmidt operators and these Kupershmidt operators are pairwise compatible. Moreover,  compatible Kupershmidt operators can also give rise to a Kupershmidt-dual-Nijenhuis structure. Our definition of a Kupershmidt-dual-Nijenhuis structure satisfies the criterion that a Kupershmidt-dual-Nijenhuis structure on a Lie algebra with respect to the coadjoint representation $(\g^*;\ad^*)$ is exactly an $r$-matrix-Nijenhuis structure.  Since a Rota-Baxter operator is a Kupershmidt operator with respect to the adjoint representation, we define a Rota-Baxter-Nijenhuis structure to be a Kupershmidt-Nijenhuis structure on a Lie algebra with respect to the adjoint representation $(\g;\ad)$. The relation between  Rota-Baxter-Nijenhuis structure and $r$-matrix-Nijenhuis structure is investigated (see Theorem \ref{thm:rmatrix-RBN}).

\emptycomment{ From the theory of the Kupershmidt-(dual-)Nijenhuis structures, the main results of $r$-matrix-Nijenhuis structures have been absorbed into this general framework.Therefore, once there is a structure satisfying the conditions of Kupershmidt-(dual-)Nijenhuis structures, an analogue of the results of $r$-matrix-Nijenhuis structures follow immediately. We give an example of Kupershmidt-dual-Nijenhuis structure through pseudo-Hessian-Nijenhuis structures on a pre-Lie algebra, which is introduced in \cite{WBLS} similar to the symplectic-Nijenhuis structures on a Lie algebra.
Furthermore, there is an important algebraic structure behind the Kupershmidt operators. It is the pre-Lie algebra (also called left-symmetric algebras, quasi-associative algebras, Vinberg algebras and so on). Pre-Lie algebras are a class of nonassociative algebras coming from the study of convex homogeneous cones, affine manifolds and affine structures on Lie groups, deformation of associative algebras and then  appeared in many fields in mathematics and mathematical physics, such as complex and symplectic structures on Lie groups and Lie algebras, integrable systems, Poisson brackets and infinite dimensional Lie algebras, vertex algebras, quantum field theory, operads and so on. See \cite{Andrada,Bakalov,ChaLiv,Lichnerowicz}, and the survey \cite{Pre-lie algebra in geometry} and the references therein for more details.
}

The paper is organized as follows. In Section 2, we study   infinitesimal deformations of a Lie algebra with a representation, and introduce the notion of a Nijenhuis pair on a Lie algebra and show that it generates a trivial deformation of a Lie algebra with a representation.  We also introduce the notion of a dual-Nijenhuis pair as the dual of a Nijenhuis pair. In Section 3, we introduce the notions of a Kupershmidt-Nijenhuis structure and a Kupershmidt-dual-Nijenhuis structure. Some properties of Kupershmidt-(dual-)Nijenhuis structures are studied. In Section 4, we first give the relations between Nijenhuis operators and Kupershmidt operators. Then, we prove that, on the one hand, Kupershmidt-(dual-)Nijenhuis structures give rise to hierarchies of Kupershmidt operators, which are pairwise compatible; on the other hand,  compatible Kupershmidt operators with a condition can give a Kupershmidt-dual-Nijenhuis structure. In Section 5, we introduce the notion of a Rota-Baxter-Nijenhuis structure on a Lie algebra and prove that it is equivalent to the $r$-matrix-Nijenhuis structure with some conditions.
\section{Infinitesimal deformations of a Lie algebra with a representation}

Let $(\g,[-,-]_\g)$ be a Lie algebra and $\rho:\g\longrightarrow\gl(V)$ a representation. Let $\omega:\wedge^2\g\longrightarrow\g$ and $\varpi:\g\longrightarrow\gl(V)$ be linear maps. Consider a $t$-parametrized family of bracket operations and linear maps:
\begin{eqnarray*}
  [x,y]_t&=&[x,y]_\g+t\omega(x,y),\\
  \rho_t(x)&=&\rho(x)+t\varpi(x).
\end{eqnarray*}
 If $(\g,[-,-]_t)$ are Lie algebras and $\rho_t$ are representations of $(\g,[-,-]_t)$ on $V$ for all $t$, we say that $(\omega,\varpi)$ generates a one-parameter infinitesimal deformation of the Lie algebra  $(\g,[-,-]_\g)$ with the representation $(V;\rho)$. We denote a one-parameter infinitesimal deformation of a Lie algebra  $(\g,[-,-]_\g)$ with a representation $(V;\rho)$ by $(\g,[-,-]_t,\rho_t)$.

By direct calculation, we can deduce that $(\g,[-,-]_t,\rho_t)$ is a one-parameter infinitesimal deformation of the Lie algebra  $(\g,[-,-]_\g)$ with the representation $(V;\rho)$ if and only if
\begin{eqnarray}
 \label{eq:deform1}[\omega(x,y),z]_\g+ [\omega(z,x),y]_\g+ [\omega(y,z),x]_\g&=&\omega(x,[y,z]_\g)++\omega(z,[x,y]_\g)+\omega(y,[z,x]_\g),\\
 \label{eq:deform2}\omega(\omega(x,y),z)+\omega(\omega(z,x),y)+\omega(\omega(y,z),x)&=&0,\\
 \label{eq:deform3}\varpi(\omega(x,y))&=&[\varpi(x),\varpi(y)],\\
 \label{eq:deform4}\rho(\omega(x,y))+\varpi([x,y]_\g)&=&[\rho(x),\varpi(y)]+[\varpi(x),\rho(y)].
\end{eqnarray}
It is well-known that \eqref{eq:deform1} means that $\omega$ is a $2$-cocycle of the Lie algebra $\g$ with the coefficient in the adjoint representation and \eqref{eq:deform2} means that $(\g,\omega)$ is a Lie algebra. Furthermore, \eqref{eq:deform3} means that $\varpi$ is a representation of the Lie algebra $(\g,\omega)$ on $V$ and \eqref{eq:deform4} means that $\rho+\varpi$ is a representation of the Lie algebra $(\g,[-,-]_\g+\omega(-,-))$ on $V$.

 \begin{defi}Two one-parameter infinitesimal deformations, $(\g,[-,-]_t,\rho_t)$ and $(\g,[-,-]'_t,\rho'_t)$,  of a Lie algebra  $(\g,[-,-]_\g)$ with a representation $(V;\rho)$ are equivalent if there exists an isomorphism $({\Id}_\g+tN,{\Id}_V+tS)$ from $(\g,[-,-]'_t,\rho'_t)$ to $(\g,[-,-]_t,\rho_t)$, i.e.
 \begin{eqnarray*}
   ({\Id}_\g+tN)[x,y]'_t&=&[({\Id}_\g+tN)(x),({\Id}_\g+tN)(y)]_t,\\
   \quad ({\Id}_V+tS)\circ\rho'_t(x)&=&\rho_t(({\Id}_\g+tN)(x))\circ ({\Id}_V+tS).
 \end{eqnarray*}

   A one-parameter infinitesimal deformation  of a Lie algebra  $(\g,[-,-]_\g)$ with a representation $(V;\rho)$ is said to be {\bf trivial} if it is equivalent to $(\g,[-,-]_\g,\rho)$.
 \end{defi}
By straightforward computations, $(\g,[-,-]_t,\rho_t)$ is a trivial deformation if and only if
\begin{eqnarray}
\label{eq:trivialdeform1}\omega(x,y)&=&[N(x),y]_\g+[x,N(y)]_\g-N([x,y]_\g),\\
\label{eq:trivialdeform2}N\omega(x,y)&=&[N(x),N(y)]_\g,\\
\label{eq:trivialdeform3}\varpi(x)&=&\rho(N(x))+\rho(x)\circ S-S\circ\rho(x),\\
\label{eq:trivialdeform4}\rho(N(x))\circ S&=&S\circ \varpi(x).
\end{eqnarray}
It follows from \eqref{eq:trivialdeform1} and \eqref{eq:trivialdeform2} that $N$ must be a Nijenhuis operator on the Lie algebra $(\g,[-,-]_\g)$. It follows from \eqref{eq:trivialdeform3} and \eqref{eq:trivialdeform4} that $N$ and $S$ should satisfy the condition:
\begin{equation}\label{eq:Nijpair}
     \rho(N(x))(S(v))=S(\rho(Nx)(v))+S(\rho(x)S(v))-S^2(\rho(x)(v)).
   \end{equation}
 \begin{defi}
   A pair $(N,S)$, where $N\in\gl(\g)$ and $S\in\gl(V)$, is called a {\bf Nijenhuis pair} on a Lie algebra  $(\g,[-,-]_\g)$ with a representation $(V;\rho)$ if $N$ is a Nijenhuis operator on the Lie algebra $(\g,[-,-]_\g)$ and the condition \eqref{eq:Nijpair} holds.
 \end{defi}
We have seen that a trivial deformation of a Lie algebra with a representation could give rise to a Nijenhuis pair. In fact, the converse is also true.
\begin{thm}
  Let $(N,S)$ be a Nijenhuis pair on a Lie algebra  $(\g,[-,-]_\g)$ with a representation $(V;\rho)$. Then a deformation of $(\g,[-,-]_\g,\rho)$ can be obtained by putting
  \begin{eqnarray}
    \omega(x,y)&=&[N(x),y]_\g+[x,N(y)]_\g-N([x,y]_\g);\\
    \varpi(x)&=&\rho(N(x))+\rho(x)\circ S-S\circ\rho(x).
  \end{eqnarray}
  Furthermore, this deformation is trivial.
\end{thm}
\pf It is a straightforward computations. We omit the details.\qed
\vspace{3mm}

Similar to the definition of a Nijenhuis pair, we introduce the notion of a dual-Nijenhuis pair on a Lie algebra with a representation.
\begin{defi}
A pair $(N,S)$, where $N\in\gl(\g)$ and $S\in\gl(V)$, is called a {\bf dual-Nijenhuis pair} on a Lie algebra  $(\g,[-,-]_\g)$ with a representation $(V;\rho)$ if $N$ is a Nijenhuis operator on the Lie algebra $(\g,[-,-]_\g)$ and $S$ satisfies the following condition:
   \begin{equation}\label{eq:coNijpair}
    \rho({N(x)})(S(v))=S(\rho({N(x)})(v))+\rho({x})(S^2(v))-S(\rho({ x})(S(v))).
   \end{equation}
 \end{defi}
Let $\rho^*:\g\longrightarrow \gl(V^*)$ be the dual representation of $\rho$ defined by
\begin{eqnarray}
 \langle \rho^*(x)(\alpha),v\rangle=-\langle \alpha,\rho(x)(v)\rangle,\quad\forall~x,y\in\g,v\in V,\alpha\in V^*.
\end{eqnarray}
In fact, there is a close relationship between a Nijenhuis pair and a dual-Nijenhuis pair.
\begin{pro}\label{pro:Nij-coNij pair}
$(N,S)$ is a Nijenhuis pair on a Lie algebra  $(\g,[-,-]_\g)$ with a representation $(V;\rho)$ if and only if $(N,S^*)$ is a dual-Nijenhuis pair on the Lie algebra  $(\g,[-,-]_\g)$ with the representation $(V^*;\rho^*)$.
\end{pro}
\pf It follows from
\begin{eqnarray*}
   &&\langle\rho(N(x))S(v)-S\rho(N(x))(v)-S\rho(x)S(v)+S^2\rho(x)(v),\alpha\rangle\\
   &=&-\langle v,S^*\rho^*(N(x))(\alpha)-\rho^*(N(x))(S^*(\alpha))-S^*\rho^*(x)(S^*(\alpha))+\rho^*(x)((S^*)^2(\alpha))\rangle.\qed
\end{eqnarray*}

\begin{defi}
  A Nijenhuis pair $(N,S)$ on a Lie algebra  $(\g,[-,-]_\g)$ with a representation $(V;\rho)$ is called a {\bf perfect Nijenhuis pair}  if
  \begin{equation}\label{eq:pNijpair}
    S^2(\rho(x)(v))+\rho({x})(S^2(v))=2S(\rho({ x})(S(v))),\quad\forall~x\in\g,v\in V.
   \end{equation}
\end{defi}
It is obvious that a perfect Nijenhuis pair is not only a Nijenhuis pair but also a dual-Nijenhuis pair.

A Nijenhuis pair gives rise to a Nijenhuis operator on the semidirect product Lie algebra.
\begin{pro}\label{pro:bigN}
   Let $(N,S)$ be a Nijenhuis pair on a Lie algebra  $(\g,[-,-]_\g)$ with a representation $(V;\rho)$. Then $N+S$ is a Nijenhuis operator on the semidirect product Lie algebra $\g\ltimes_\rho V.$ Furthermore, if $(N,S)$ is a perfect  Nijenhuis pair, then $N+S^*$ is a also a Nijenhuis operator on the semidirect product Lie algebra $\g\ltimes_{\rho^*} V^*$.
 \end{pro}

 Define $\hat{\varrho}:\g\longrightarrow\gl(V)$ and $\tilde{\varrho}:\g\longrightarrow\gl(V)$, respectively, by
 \begin{eqnarray}
  \hat{ \varrho}(x)&=&\rho(Nx)+[\rho(x),S],\\
  \tilde{ \varrho}(x)&=&\rho(Nx)-[\rho(x),S],\quad\forall~x\in\g.
 \end{eqnarray}

 \begin{cor}\begin{itemize}
   \item[\rm(i)]If $(N,S)$ is a Nijenhuis pair on a Lie algebra  $(\g,[-,-]_\g)$ with a representation $\rho$, then $\hat{\varrho}$ is a representation of the Lie algebra $(\g,[-,-]_N)$ on $V$;
       \item[\rm(ii)]If $(N,S)$ is a dual-Nijenhuis pair on a Lie algebra  $(\g,[-,-]_\g)$ with a representation $\rho$, then $\tilde{\varrho}$ is a representation of the Lie algebra $(\g,[-,-]_N)$ on $V$;
 \end{itemize}

 \end{cor}
\pf (i) By Proposition \ref{pro:bigN}, $N+S$ is a Nijenhuis operator on the semidirect product Lie algebra  $\g\ltimes_\rho V$. The deformed bracket $[-,-]_{N+S}$ is given by
\begin{eqnarray*}
  [x+u,y+v]_{N+S}&=&[(N+S)(x+u),y+v]_\rho+[x+u,(N+S)(y+v)]_\rho-(N+S)[x+u,y+v]_\rho\\
  &=&[Nx,y]_\g+[x,Ny]_\g-N[x,y]_\g\\
  &&+\rho(Nx)(v)+\rho(x)S(v)-S\rho(x)(v)-\rho(y)S(u)-\rho(Ny)(u)+S\rho(y)(u)\\
  &=&[x,y]_N+\hat{\varrho}(x)(v)-\hat{\varrho}(y)(u),
\end{eqnarray*}
which implies that $\hat{\varrho}$ is a representation of the Lie algebra $(\g,[-,-]_N)$ on $V$.

(ii) By direct calculation, the dual map $\tilde{\varrho}^*$ of $\tilde{\varrho}$ is given by
\begin{equation}
\tilde{\varrho}^*(x)={\rho}^*(Nx)+[{\rho}^*(x),S^*],\quad\forall x\in\g.
\end{equation}
Since $(N,S)$ is a dual-Nijenhuis pair with a representation $\rho$, by Proposition \ref{pro:Nij-coNij pair}, $(N,S^*)$ is a Nijenhuis pair with a representation $\rho^*$. By (i), $\tilde{\varrho}^*$ is a representation of the Lie algebra $(\g,[-,-]_N)$ on $V^*$ and thus $\tilde{\varrho}$ is a representation of the Lie algebra $(\g,[-,-]_N)$ on $V$.\qed

\section{Kupershmidt-(dual-)Nijenhuis structures}

\emptycomment{Recall that a {\bf Kupershmidt operator} on a Lie algebra
$(\g,[-,-]_\g)$ associated to a representation $(V;\rho)$ is a
linear map $T:V\longrightarrow \g$ satisfying
\begin{equation}
  [T(u), T(v)]_\g=T\Big(\rho(T(u))(v)-\rho(T(v))(u)\Big),\quad \forall u,v\in V.
\end{equation}
If the representation is the adjoint representation $(\g;\ad)$, we obtain the usual Rota-Baxter  Lie algebra. More precisely, a Rota-Baxter  Lie algebra is a Lie algebra $(\g,[-,-]_\g)$ equipped with a linear operator $\huaR:\g\longrightarrow\g $ such that
\begin{equation}
  [\huaR(x),\huaR(y)]_\g=\huaR([\huaR(x), y]_\g+[x,\huaR(y)]_\g).
\end{equation}}

There is a close relationship between  Kupershmidt operators and pre-Lie algebras.
\begin{defi}  A {\bf pre-Lie algebra} is a pair $(V,\star)$, where $V$ is a vector space and  $\star:V\otimes V\longrightarrow V$ is a bilinear multiplication
satisfying that for all $x,y,z\in V$, the associator
$(x,y,z)=(x\star y)\star z-x\star(y\star z)$ is symmetric in $x,y$,
i.e.
$$(x,y,z)=(y,x,z),\;\;{\rm or}\;\;{\rm
equivalently,}\;\;(x\star y)\star z-x\star(y\star z)=(y\star x)\star
z-y\star(x\star z).$$
\end{defi}
Let $(V,\star)$ be a pre-Lie algebra. The commutator $
[x,y]^c=x\star y-y\star x$ defines a Lie algebra structure
on $V$, which is called the {\bf sub-adjacent Lie algebra} of
$(V,\star)$ and denoted by $V^c$. Furthermore,
$L:V\longrightarrow \gl(V)$ with $x\rightarrow L_x$, where
$L_xy=x\star y$, for all $x,y\in V$, gives a representation of
the Lie algebra $V^c$ on $V$. See \cite{Pre-lie algebra in geometry} for more details.

The following result establishes the connection between Kupershmidt operators and pre-Lie algebras.

\begin{thm}{\rm(\cite{Bai1})}
Let $T:V\to \g$ be a Kupershmidt operator  on a Lie algebra $(\g,[-,-]_\g)$ with respect to a representation $(V;\rho)$. Define a multiplication $\star^T$ on $V$ by
\begin{equation}
  u\star^T v=\rho(Tu)(v),\quad \forall u,v\in V.
\end{equation}
Then $(V,\star^T)$ is a pre-Lie algebra.
 \end{thm}

We denote by $(V,[-,-]^T)$ the sub-adjacent Lie algebra of the pre-Lie algebra $(V,\star^T)$. More precisely,
\begin{equation}\label{eq:bracketT}
 ~[u,v]^T=\rho(Tu)(v)-\rho(Tv)(u).
\end{equation}
Moreover,  $T$ is a Lie algebra homomorphism from $(V,[-,-]^T)$ to $(\g,[-,-]_\g)$.

Now let $T:V\longrightarrow\g$ be a Kupershmidt operator and $(N,S)$ a (dual-)Nijenhuis pair on a Lie algebra  $(\g,[-,-]_\g)$ with a representation $(V;\rho)$.  We define the bracket $[-,-]^T_S:\wedge^2V\longrightarrow V$ to be the deformed bracket of $[-,-]^T$ by $S$, i.e.
\begin{eqnarray*}
 {[u,v]}^T_S&=&[S(u),v]^T+[u,S(v)]^T-S([u,v]^T),\quad\forall~u,v\in V.\label{eq:defieq33}
\end{eqnarray*}
Define the bracket $\{-,-\}_{\hat{\varrho}}^T:\wedge^2V\longrightarrow V$ and $\{-,-\}_{\tilde{\varrho}}^T:\wedge^2V\longrightarrow V$ similar as \eqref{eq:bracketT} using the representation $\hat{\varrho}$ and $\tilde{\varrho}$, respectively:
\begin{eqnarray}
 \label{eq:defieq22} \{u,v\}_{\hat{\varrho}}^T&=&\hat{\varrho}(Tu)(v)-\hat{\varrho}(Tv)(u),\\
  \label{eq:defieq23}\{u,v\}_{\tilde{\varrho}}^T&=&\tilde{\varrho}(Tu)(v)-\tilde{\varrho}(Tv)(u),\quad\forall~u,v\in V.
\end{eqnarray}
It is not true in general that the brackets $[-,-]^T_S$, $\{-,-\}_{\hat{\varrho}}^T$ and  $\{-,-\}_{\tilde{\varrho}}^T$ satisfy the Jacobi identity.

\begin{defi}
  A Kupershmidt operator $T:V\longrightarrow \g$   on a Lie algebra $(\g,[-,-]_\g)$ with respect to a representation $(V;\rho)$  and a (dual-)Nijenhuis pair $(N,S)$ are called {\bf compatible} if they satisfy the following conditions
 \begin{eqnarray}
 \label{eq:TN1}N\circ T&=&T\circ S,\\
 \label{eq:TN2} {[u,v]}^{N\circ T}&=&{[u,v]}^{T}_{S}.
  \end{eqnarray}
 The triple $(T,S,N)$ is called a {\bf Kupershmidt-(dual-)Nijenhuis structure} on the Lie algebra $(\g,[-,-]_\g)$ with respect to the representation $(V;\rho)$ if $T$   and $(N,S)$ are  compatible.
 \end{defi}
Note that if $(N,S)$ is a perfect Nijenhuis pair, then a Kupershmidt-Nijenhuis structure is also a Kupershmidt-dual-Nijenhuis structure.

\begin{lem}
\begin{itemize}
\item[$\rm(a)$] Let $(T,S,N)$ be a Kupershmidt-Nijenhuis structure. Then we have
 $$ {[u,v]}^T_S=\{u,v\}_{\hat{\varrho}}^T.$$
 \item[$\rm(b)$] Let $(T,S,N)$ be a Kupershmidt-dual-Nijenhuis structure. Then we have
 $$ {[u,v]}^T_S=\{u,v\}_{\tilde{\varrho}}^T.$$
\end{itemize}
\end{lem}
\pf(a) It follows  from \eqref{eq:TN1} directly.

(b) By \eqref{eq:TN1}, we have
$${[u,v]}^T_S+\{u,v\}_{\tilde{\varrho}}^T=2{[u,v]}^{N\circ T}.$$
Then by \eqref{eq:TN2}, we obtain ${[u,v]}^T_S=\{u,v\}_{\tilde{\varrho}}^T.$\qed\vspace{3mm}

Thus, if $(T,S,N)$ is a   Kupershmidt-(dual-)Nijenhuis structure, then the three brackets  ${[-,-]}^T_S,$ $\{-,-\}_{\hat{\varrho}}^T~(\{-,-\}_{\tilde{\varrho}}^T)$ and $[-,-]^{N\circ T}$ are the same. Moreover, we will see that they satisfy the Jacobi identity.

\begin{pro}\label{pro:S-Nijenhuis operator}
 Let $(T,S,N)$ be a Kupershmidt-(dual-)Nijenhuis structure. Then $S$ is a Nijenhuis operator on the sub-adjacent Lie algebra $(V,[-,-]^T)$. Thus, the brackets  ${[-,-]}^T_S,$ $\{-,-\}_{\hat{\varrho}}^T~(\{-,-\}_{\tilde{\varrho}}^T)$ and $[-,-]^{N\circ T }$ are all Lie brackets.
\end{pro}
\pf For the Kupershmidt-Nijenhuis structure $(T,S,N)$, by \eqref{eq:Nijpair} and substituting $x$ by $T(u)$, we get
    \begin{eqnarray*}
     0&=&\rho(NT(u))S(v)-S(\rho(NT(u))(v)+S\rho(T(u))S(v)-S\rho(T(u))(v))\\
     &=&\rho(TS(u))S(v)-S(\rho(TS(u))(v)+S\rho(Tu)S(v)-S\rho(Tu)(v))\\
     &=&S(u)\star^TS(v)-S(S(u)\star^Tv+u\star^TS(v)-S(u\star^Tv)),
     \end{eqnarray*}
     which implies that $S$ is a Nijenhuis operator on the pre-Lie algebra $(V,\star^T)$ (\cite{WBLS}). Thus $S$ is a Nijenhuis operator on the sub-adjacent Lie algebra $(V,[-,-]^T)$.

For the Kupershmidt-dual-Nijenhuis structure $(T,S,N)$, the proof is not direct. In fact, by the relation $[u,v]^{T\circ S}={[u,v]}^{T}_{S}$, one has
\begin{eqnarray}
 S(\rho({T\circ S(u)})(v))-S(\rho({T(v)})(S(u)))&=&\rho({T\circ S(u)})(S(v))-\rho({T(v)})(S^2(u));\label{eq:pn11}\\
 S^2(\rho({T(u)})(v))-S^2(\rho({T(v)})(u))&=& S(\rho({T(u)})(S(v)))-S(\rho({T(v)})(S(u))).
\end{eqnarray}
By the condition \eqref{eq:coNijpair}, we have
\begin{equation}\label{eq:pn22}
 \rho({ T(v)})(S^2(u))-\rho({T\circ S(v)})(S(u))=S(\rho({ T(v)})(S(u)))-S(\rho({T\circ S(v)})(u)).
\end{equation}
By \eqref{eq:pn11}-\eqref{eq:pn22}, we have
\begin{eqnarray*}
  &&[S(u),S(v)]^{T}- S([u,v]^{T}_{S})\\
  &=&\rho({T\circ S(u)})(S(v))-\rho({T\circ S(v)})(S(u))+S^2(\rho({T(u)})(v))-S^2(\rho({T(v)})(u))\\
  &&-S(\rho({T(u)})(S(v)))+S(\rho({T\circ S(v)})(u))-S(\rho({T\circ S(u)})(v))+S(\rho({T(v)})(S(u)))\\
&=&\rho({T\circ S(u)})(S(v))-\rho({T\circ S(v)})(S(u))-S^2(\rho({T(v)})(u))+S^2(\rho({T(u)})(v))\\
  &&-S(\rho({T(u)})(S(v)))+S(\rho({T\circ S(v)})(u))-\rho({T\circ S(u)})(S(v))+S(\rho({T(v)})(S^2(u)))\\
&=&\rho({T\circ S(u)})(S(v))-\rho({T\circ S(v)})(S(u))+ S(\rho({T(u)})(S(v)))-\rho({T(v)})(S(u))\\
  &&-S(\rho({T(u)})(S(v)))+S(\rho({T\circ S(v)})(u))-\rho({T\circ S(u)})(S(v))+\rho({T(v)})(S^2(u))\\
  &=&\rho({T(v)})(S^2(u))-\rho({T\circ S(v)})(S(u))-S(\rho({T(v)})(S(u)))+S(\rho({T\circ S(v)})(u))\\
  &=&0.
\end{eqnarray*}
Thus $S$ is a Nijenhuis operator on the Lie algebra $(V,[-,-]^{T})$.\qed
\emptycomment{By the proof of the above proposition, we can obtain a more strong conclusion for a Kupershmidt-Nijenhuis structure.
\begin{cor}
 Let $(T,S,N)$ be a Kupershmidt-Nijenhuis structure. Then $S$ is a Nijenhuis operator on the pre-Lie algebra $(V,\star^T)$.
\end{cor}}

\begin{thm}\label{pro:TS1}
 Let $(T,S,N)$ be a   Kupershmidt-(dual-)Nijenhuis structure. Then we have
  \begin{itemize}
\item[$\rm(i)$] $T$ is  a Kupershmidt operator on the deformed Lie algebra $(\g,[-,-]_N)$ with respect to the representation $(V; \hat{\varrho})$ $((V; \tilde{\varrho}))$;
 \item[$\rm(ii)$] $N\circ T$ is a Kupershmidt operator on the Lie algebra $(\g,[-,-]_{\g})$ with respect to the representation  $(V;\rho)$.
  \end{itemize}

\end{thm}
\pf
 We only prove the theorem for the Kupershmidt-Nijenhuis structure. The other one can be proved similarly.

(i)   Since $T$ is a Kupershmidt operator on the Lie algebra $(\g,[-,-]_\g)$ with respect to the representation $(V;\rho)$ and $T\circ S=N\circ T$, we have
\begin{eqnarray*}
T\{u,v\}^T_{\hat{\rho}}&=&T({[u,v]}^{T}_ {S})=T([S(u),v]^T+[u,S(v)]^T-S[u,v]^T)\\
&=&[T\circ S(u),T(v)]_\g+[T(u),T\circ S(v)]_\g-T\circ S[u,v]^T\\
&=&[N\circ T(u),T(v)]_\g+[T(u),N\circ T(v)]_\g-N\circ T[u,v]^T\\
&=&[T(u),T(v)]_N.
\end{eqnarray*}
Thus, $T$ is  a Kupershmidt operator on the deformed Lie algebra $(\g,[-,-]_N)$ with respect to the representation $(V; \hat{\varrho})$.

(ii) By \eqref{eq:TN2}, we have
\begin{eqnarray*}
 N\circ T([u,v]^{N\circ T})&=&N \circ T([u,v]^{T}_ {S})=N[T(u),T(v)]_N=[N\circ T(u),N\circ T(v)]_\g,
\end{eqnarray*}
which implies that $N\circ T$ is a Kupershmidt operator on the Lie algebra $(\g,[-,-]_{\g})$ with respect to the representation  $(V;\rho)$.
\qed\vspace{3mm}

The following theorem demonstrates that the Kupershmidt-Nijenhuis operator can give a Kupershmidt-dual-Nijenhuis operator with a
condition.
\begin{thm}\label{thm:dual-Nijenhuis and Nijenhuis}
 Let $(T,S,N)$ be a Kupershmidt-Nijenhuis structure. If $T$ is invertible, then $(T,S,N)$ is a Kupershmidt-dual-Nijenhuis structure.
\end{thm}
\pf We only need to prove that the Nijenhuis pair $(S,N)$ is also a dual-Nijenhuis pair. By \eqref{eq:TN2}, we have
\begin{eqnarray*}
{[u,v]}^{T}_{S}-{[u,v]}^{T\circ S}=\rho(T(u))(S(v))-\rho(T(v))(S(u))-S\Big(\rho(T(u))(v)-\rho(T(v))(u)\Big),
\end{eqnarray*}
which implies that
\begin{equation}\label{eq:TSN5}
\rho(T(u))(S(v))-\rho(T(v))(S(u))=S\Big(\rho(T(u))(v)-\rho(T(v))(u)\Big).
\end{equation}
Since $S$ is a Nijenhuis operator on the Lie algebra $(V,[-,-]^T)$ and ${[u,v]}^{T}_{S}={[u,v]}^{T\circ S}$, we have
$$S([u,v]^{T\circ S})=[S(u),S(v)]^T,$$
which means that
\begin{eqnarray}\label{eq:TSN6}
  S\Big(\rho(T\circ S(u))(v)-\rho(T\circ S(v))(u)\Big)=\rho(T\circ S(u))(S(v))-\rho(T\circ S(v))(S(u)).
\end{eqnarray}
By \eqref{eq:TSN5}, we have
\begin{eqnarray*}
 S\big(\rho(T\circ S(u))(v)\big)-\rho(T\circ S(u))(S(v))=S\big(\rho(T(v))(S(u))\big)-\rho(T(v))(S^2(u)).
\end{eqnarray*}
Thus \eqref{eq:TSN6} implies that
\begin{eqnarray*}
 0&=&S\big(\rho(T(v))(S(u))\big)-\rho(T(v))(S^2(u))- S\big(\rho(T\circ S(v))(u)\big)+\rho(T\circ S(v))(S(u))\\
 &=&S\big(\rho(T(v))(S(u))\big)-\rho(T(v))(S^2(u))- S\big(\rho(N\circ T(v))(u)\big)+\rho(N\circ T(v))(S(u)).
\end{eqnarray*}
Since $T$ is invertible and let $x=T(v)$, we have
$$S\big(\rho(x)(S(u))\big)-\rho(x)(S^2(u))- S\big(\rho(N(x))(u)\big)+\rho(N(x))(S(u))=0.$$
Thus the Nijenhuis pair $(S,N)$ is a dual-Nijenhuis pair. We finish the proof.\qed

\section{Hierarchy of Kupershmidt operators}

\subsection{Compatible Kupershmidt operators on Lie algebras}

\begin{defi}
  Let $T_1,T_2: V\longrightarrow \g$ be two
Kupershmidt operators on a Lie algebra $(\g,[-,-]_\g)$ with respect to a
representation $(V;\rho)$. If for all
$k_1,k_2$, $k_1T_1+k_2T_2$ is still a Kupershmidt operator, then $T_1$ and $T_2$ are called {\bf
compatible}.
\end{defi}

\begin{pro}\label{pro:compatible}
Let $T_1,T_2: V\longrightarrow \g$ be two Kupershmidt operators
on a Lie algebra $(\g,[-,-]_\g)$ with respect to a
representation $(V;\rho)$. Then $T_1$ and $T_2$ are compatible
if and only if the following equation holds:
\begin{eqnarray}
\nonumber [T_1(u), T_2(v)]_\g+ [T_2(u),
 T_1(v)]_\g&=&T_1\Big(\rho(T_2(u))(v)-\rho(T_2(v))(u)\Big)\\
 &&+T_2\Big(\rho(T_1(u))(v)-\rho(T_1(v))(u)\Big),\quad \forall u,v\in V.\label{eq:CN1}
\end{eqnarray}
\end{pro}

\pf It follows from a direct computation.\qed\vspace{3mm}

Using a Kupershmidt operator and a Nijenhuis operator, we can construct a pair of compatible Kupershmidt operators .

\begin{pro}\label{pro:NT}
Let $T: V\longrightarrow \g$ be a Kupershmidt operator on a
Lie algebra $(\g,[-,-]_\g)$ with respect to a representation
$(V;\rho)$ and $N$ a Nijenhuis operator on $(\g,[-,-]_\g)$.
Then $N\circ T$ is a Kupershmidt operator on the Lie algebra $(\g,[-,-]_\g)$ with respect to the representation $(V;\rho)$
if and only if for all $u,v\in V$, the following equation holds:
\begin{eqnarray}
\nonumber &&N\Big([NT(u), T(v)]_\g+[T(u), NT(v)]_\g\Big)\\
&=&N\Big(T\big(\rho(NT(u))(v)-\rho(NT(v))(u)\big)+NT\big(\rho(T(u))(v)-\rho(T(v))(u)\big)\Big).\label{eq:ON}
\end{eqnarray}
In this case, if in addition $N$ is invertible, then $T$ and $N\circ T$
are compatible. More explicitly, for any Kupershmidt operator $T$,
if there exists an invertible Nijenhuis operator $N$ such that $N\circ T$
is also a Kupershmidt operator, then $T$ and $N\circ T$ are compatible.
\end{pro}

\pf    Since $N$ is a Nijenhuis operator,
we have
$$[NT(u), NT(v)]_\g=N\Big([NT(u),T(v)]_\g+[T(u),
NT(v)]_\g\Big)-N^2([T(u), T(v)]_\g).$$ Note that
$$[T(u), T(v)]_\g=T\Big(\rho(T(u))(v)-\rho(T(v))(u)\Big).$$
Then
$$[NT(u), NT(v)]_\g=NT\Big(\rho(NT(u))(v)-\rho(NT(v))(u)\Big)$$
if and only if  (\ref{eq:ON}) holds.

If $N\circ T$ is a Kupershmidt operator and $N$ is invertible, then we have
$$[NT(u), T(v)]_\g+[T(u), NT(v)]_\g=
T\big(\rho(NT(u))(v)-\rho(NT(v))(u)\big)+NT\big(\rho(T(u))(v)-\rho(T(v))(u)\big),$$
which is exactly the condition that $N\circ T$ and $T$ are compatible. \qed\vspace{3mm}

A pair of compatible Kupershmidt operators can also give rise to a Nijenhuis operator under some conditions.

\begin{pro}\label{pro:TTN}
Let $T_1,T_2: V\longrightarrow \g$ be two Kupershmidt operators  on
a Lie algebra $(\g,[-,-]_\g)$ with respect to a representation
$(V;\rho)$. Suppose that $T_2$ is invertible. If $T_1$ and $T_2$
are compatible, then $N=T_1\circ T_2^{-1}$ is a Nijenhuis operator on the Lie algebra $(\g,[-,-]_\g)$.
\end{pro}

\pf For all $x,y\in \g$, there exist $u,v\in V$ such that
$T_2(u)=x, T_2(v)=y$. Hence $N=T_1\circ T_2^{-1}$ is a Nijenhuis operator
if and only if the following equation holds:
$$[NT_2(u), NT_2(v)]_\g=N([NT_2(u), T_2(v)]_\g+[T_2(u),
NT_2(v)]_\g)-N^2([T_2(u), T_2(v)]_\g).$$ Since $T_1=N\circ T_2$ is an
Kupershmidt operator, the left hand side of the above equation is
$$NT_2(\rho(NT_2(u))(v)-\rho(NT_2(v))(u)).$$
Since $T_2$ is a Kupershmidt operator which is compatible with
$T_1=N\circ T_2$, we have
\begin{eqnarray*}
 &&[NT_2(u),T_2(v)]_\g+[T_2(u), NT_2(v)]_\g\\
 &=&
T_2(\rho(NT_2(u))(v)-\rho(NT_2(v))(u))+NT_2(\rho(T_2(u))(v)-\rho(T_2(v))(u))\\
&=&T_2(\rho(NT_2(u))(v)-\rho(NT_2(v))(u))+N([T_2(u),T_2(v)]_\g).
\end{eqnarray*}
Let $N$ act on both sides, we get the conclusion. \qed\vspace{3mm}

By Proposition \ref{pro:NT} and \ref{pro:TTN}, we have
\begin{cor}\label{cor:NTN}
Let $T_1,T_2: V\longrightarrow \g$ be two Kupershmidt operators  on
a Lie algebra $(\g,[-,-]_\g)$ with respect to a representation
$(V;\rho)$. Suppose that $T_1$ and $T_2$ are invertible. Then
$T_1$ and $T_2$ are compatible if and only if $N=T_1\circ T_2^{-1}$ is a
Nijenhuis operator.
\end{cor}

In particular, as a direct application, we have the following conclusion.

\begin{cor} Let $(\g,[-,-]_\g)$ be a Lie algebra. Suppose that
$\huaR_1$ and $\huaR_2$ are two invertible Rota-Baxter operators. Then $\huaR_1$ and $\huaR_2$ are compatible in the sense that any
linear combination of $\huaR_1$ and $\huaR_2$ is still a Rota-Baxter
operator if and only if $N=\huaR_1\circ \huaR_2^{-1}$ is a
Nijenhuis operator.
\end{cor}

\subsection{Hierarchy of Kupershmidt  operators }
In the following, first we construct compatible Kupershmidt operators  from Kupershmidt-(dual-)Nijenhuis structures. Given a Kupershmidt-(dual-)Nijenhuis structure $(T,S,N)$, by Theorem \ref{pro:TS1}, $T$ and $T\circ S$ are Kupershmidt operators. In fact, they are compatible.
\begin{pro}\label{pro:TS2}
 Let $(T,S,N)$ be a Kupershmidt-(dual-)Nijenhuis structure. Then $T$ and $T\circ S$ are compatible Kupershmidt operators.
\end{pro}
\pf  We only prove the conclusion for the Kupershmidt-Nijenhuis structure. The other one can be proved similarly.
It is sufficient to prove that $T+T\circ S$ is a Kupershmidt operator.
It is obvious that
$$[u,v]^{T+T\circ S}=[u,v]^T+[u,v]^{T\circ S}=[u,v]^T+{[u,v]}^T_S.$$
Thus, we have
\begin{eqnarray*}
  &&(T+T\circ S)([u,v]^{T+T\circ S})\\
  &=&T([u,v]^T)+T\circ S([u,v]^T_S)+T\circ S([u,v]^T)+ T([u,v]^T_S)\\
  &=&T([u,v]^T)+T\circ S([u,v]^T_S)+T\circ S([u,v]^T)\\
  &&+T([S(u),v]^T+[u,S(v)]^T-S[u,v]^T)\\
  &=&T([u,v]^T)+T\circ S([u,v]^T_S)+T([S(u),v]^T+[u,S(v)]^T)\\
  &=&[T(u),T(v)]_\g+[T\circ S(u),T\circ S(v)]_\g+[T\circ S(u),T(v)]_\g+[T(u),T\circ S(v)]_\g\\
  &=&[(T+T\circ S)(u),(T+T\circ S)(v)]_\g,
\end{eqnarray*}
which means that $T+T\circ S$ is a Kupershmidt operator.\qed

\begin{lem}
 Let $(T,S,N)$ be a Kupershmidt-(dual-)Nijenhuis structure. Then for all $k,i\in\Nat$, we have
\begin{eqnarray}
   \label{eq:TSN3}T_k[u,v]^T_{S^{k+i}}=[ T_k(u), T_k(v)]_{N^i}.
\end{eqnarray}
\end{lem}
\pf
Since $T$ is a Kupershmidt operator and $T\circ S=N\circ T$, we have
\begin{eqnarray}
\nonumber T([u,v]^T_{S^i})&=&T\big([S^i(u), v]^T+[u, S^i(v)]^T-S^{i}([u, v]^T)\big)\\
\nonumber&=&[N^i(T (u)), T(v)]_\g+[T(u), N^i(T(v))]_\g-N^i([T(u), T(v)]_\g)\\
\label{eq:TSN2}&=& [T(u), T(v)]_{N^i}.
\end{eqnarray}
Since $S$ is a Nijenhuis operator on the associative algebra  $(V,\cdot^T)$, we have
\begin{equation}\label{eq:t1}
  S^k([u,v]^{T}_{S^{k+i}})=[S^k(u), S^k(v)]^{T}_{S^i}.
\end{equation}
Then by \eqref{eq:TSN2} and \eqref{eq:t1}, we have
\begin{eqnarray*}
 T_k([u,v]^{T}_{S^{k+i}}=T\circ S^k([u,v]^{T}_{S^{k+i}})=T([S^k(u),S^k(v)]^{T}_{S^i})= [T(S^k(u)),T(S^k(v))]_{N^i}.
\end{eqnarray*}
The proof is finished. \qed\vspace{3mm}

The proof of the following lemma is similar to the proof of Proposition 5.1 in \cite{Kosmann1}.
\begin{lem}
Let $(T,S,N)$ be a Kupershmidt-(dual-)Nijenhuis structure. Then for all $k,i\in\Nat$ such that $i\leq k$,
\begin{equation}\label{eq:TSN4}
  [u,v]^{T_k}= [u,v]^{T}_{S^k}=S^{k-i}[u,v]^{T_i},
\end{equation}
where $T_k=T\circ S^k=N^k\circ T$ and set $T_0=T$.
\end{lem}

\begin{pro}\label{pro:hierarchy}
Let $(T,S,N)$ be a Kupershmidt-(dual-)Nijenhuis structure with respect to the representation $(V;\rho)$. Then all $T_k=N^k\circ T$ are Kupershmidt operators with respect to the representation $(V;\rho)$ and for all $k,l\in\Nat$, $T_k$ and $T_l$ are compatible.
\end{pro}
\pf  We only prove the conclusion for the Kupershmidt-Nijenhuis structure. The other one can be proved similarly.

By  \eqref{eq:TSN3} and \eqref{eq:TSN4} with $i=0$, we have
\begin{eqnarray*}
 T_k[u,v]^{T_{k}}=[ T_k(u), T_k(v)]_\g,
\end{eqnarray*}
which implies that $T_k$ is a Kupershmidt operator associated the representation $(V;\rho)$.

For the second conclusion, we need  to prove that $T^k+T^{k+i}$ is a Kupershmidt operator for all $k,i\in\Nat$.
By  \eqref{eq:TSN4}, we have
$$[u,v]^{T_k+T_{k+i}}=[u,v]^{T_k}+[u,v]^{T_{k+i}}=[u,v]^{T_k}+[u,v]^{T_k}_{S^i}.$$
Thus, we have
\begin{eqnarray*}
  &&(T_k+T_{k+i})([u,v]^{T_k+T_{k+i}})\\
  &=&T_k([u,v]^{T_k})+T_k([u,v]^{T_k}_{S^i})+T_{k+i}([u,v]^{T_k})+T_{k+i}([u,v]^{T_k}_{S^i})\\
  &=&T_k([u,v]^{T_k})+T_{k+i}([u,v]^{T_k}_{S^i})+T_{k+i}([u,v]^{T_k}_{S^i})\\
  &&+T_k([S^i(u),v]^{T_k}+[u,S^i(v)]^{T_k}-S^i[u,v]^{T_k})\\
  &=&T_k([u,v]^{T_k})+T_{k+i}([u,v]^{T_k}_{S^i})+T^k([S^i(u),v]^{T_k})+T^k([u,S^i(v)]^{T_k})\\
  &=&[T_k(u),T_k(v)]_\g+[T_{k+i}(u),T_{k+i}(v)]_\g+[T_{k+i}(u),T_k(v)]_\g+[T_k(u),T_{k+i}(v)]_\g\\
  &=&[(T_k+T_{k+i})(u),(T_k+T_{k+i})(v)]_\g.
\end{eqnarray*}
Thus $T^k+T^{k+i}$ is a Kupershmidt operator. We finish the proof.\qed\vspace{3mm}

\emptycomment{

The Kupershmidt operator can be related to the solution of classical Yang-Baxter equation.
\begin{lem}{\rm\cite{Bai1}}
  $\pi^\sharp=T-T^*$ is an $r$-matrix in the Lie algebra $\g\ltimes_{\rho^*}V^*$ if and only if $T$ is a Kupershmidt operator associated to a representation $(V;\rho)$, where $T^*:\g^*\longrightarrow V^*$ is given by
  $$\langle T^*(\xi),u\rangle=\langle \xi,T(u)\rangle,\quad\forall\xi\in\g^*,u\in V.$$
\end{lem}

As a corollary of \ref{pro:hierarchy}, we have
\begin{cor}
 Let $(T,S,N)$ be a Kupershmidt-(dual-)Nijenhuis structure associated to the representation $(V;\rho)$. Then all $\pi^\sharp_k=N^k\circ T-T^*\circ (N^*)^k$ are $r$-matrix in the Lie algebra $\g\ltimes_{\rho^*}V^*$ and for all $k,l\in\Nat$, $\pi^\sharp_k$ and $\pi^\sharp_l$ are compatible.
\end{cor}
}

Compatible Kupershmidt operators  can give rise to Kupershmidt-dual-Nijenhuis structures.
\begin{pro}\label{pro:ComptoNS}
Let $T,T_1: V\longrightarrow \g$ be two Kupershmidt operators  on
a Lie algebra $(\g,[-,-]_\g)$ with respect to a representation
$(V;\rho)$. Suppose that $T$ is invertible. If
$T$ and $T_1$ are compatible, then
 \begin{itemize}
\item[$\rm(i)$]$(T,S=T^{-1}\circ T_1,N=T_1\circ T^{-1})$ is a Kupershmidt-dual-Nijenhuis structure;
\item[$\rm(ii)$]$(T_1,S=T^{-1}\circ T_1,N=T_1\circ T^{-1})$ is a Kupershmidt-dual-Nijenhuis structure.
\end{itemize}
\end{pro}
\pf (i) The proof of $(N,S)$ being a dual-Nijenhuis pair   is similar to the proof of Theorem \ref{thm:dual-Nijenhuis and Nijenhuis}. We omit the details.
It is obvious that $T\circ S=N\circ T$. Thus we only need to prove that the compatibility condition \eqref{eq:TN2} holds.
By the compatibility condition of $T$ and $T_1$ and Proposition \ref{pro:TTN}, $N=T_1\circ T^{-1}$ is a Nijenhuis operator on the Lie algebra $\g$. By Proposition \ref{pro:compatible}, we also have
\begin{eqnarray*}
 [T(u), T_1(v)]_\g+ [T_1(u),
 T(v)]_\g&=&T\Big(\rho(T_1(u))(v)-\rho(T_1(v))(u)\Big)\\
 &&+T_1\Big(\rho(T(u))(v)-\rho(T(v))(u)\Big),\quad \forall u,v\in V.
\end{eqnarray*}
Substituting $T_1$ with $T\circ S$, then we have
\begin{eqnarray}
 \nonumber[T(u), T\circ S(v)]_\g+ [T\circ S(u),
 T(v)]_\g&=&T\Big(\rho(T\circ S(u))(v)-\rho(T\circ S(v))(u)\Big)\\
 &&+T\circ S\Big(\rho(T(u))(v)-\rho(T(v))(u)\Big).\label{eq:Comp1}
\end{eqnarray}
Since $T$ is a Kupershmidt operator, we have
\begin{eqnarray*}
 [T(u), T\circ S(v)]_\g+ [T\circ S(u), T(v)]_\g&=&T\Big(\rho(T(u))(S(v))-\rho(T\circ S(v))(u)\\
 &&+\rho(T\circ S(u))(v)-\rho(T(v))(S(u))\Big).
\end{eqnarray*}
Since $T$ is invertible, \eqref{eq:Comp1} is equivalent to
\begin{eqnarray}\label{eq:ON3}
  S\Big(\rho(T(u))(v)-\rho(T(v))(u)\Big)=\rho(T(u))(S(v))-\rho(T(v))(S(u)).
\end{eqnarray}
On the other hand, we have
\begin{eqnarray*}
{[u,v]}^{T}_{S}-{[u,v]}^{T\circ S}=\rho(T(u))(S(v))-\rho(T(v))(S(u))-S\Big(\rho(T(u))(v)-\rho(T(v))(u)\Big).
\end{eqnarray*}
Thus, \eqref{eq:ON3} implies that $ {[u,v]}^{T}_{S}={[u,v]}^{T\circ S}$.
Therefore, $(T,S=T^{-1}\circ T_1,N=T_1\circ T^{-1})$ is a Kupershmidt-dual-Nijenhuis structure.

\emptycomment{
Furthermore, since $T$ and $T\circ S$ are Kupershmidt operators, thus
$$T\circ S([u,v]^{T\circ S})=[T\circ S(u),T\circ S(v)]_\g=T([S(u),S(v)]^T).$$
As $T$ is invertible, we have
$$S([u,v]^{T\circ S})=[S(u),S(v)]^T.$$
}

(ii) By direct calculation, we have
\begin{eqnarray*}
 &&{[u,v]}^{T_1}_{S}- {[u,v]}^{T_1\circ S}\\&=&\rho(T_1(u))(S(v))-\rho(T_1(v))(S(u))-S\Big(\rho(T_1(u))(v)-\rho(T_1(v))(u)\Big)\\
  &=&\rho(T\circ S(u))(S(v))-\rho(T\circ S(v))(S(u))-S\Big(\rho(T\circ S(u))(v)-\rho(T\circ S(v))(u)\Big)\\
  &=&[S(u),S(v)]^T-S[u,v]^{T\circ S}=0.
\end{eqnarray*}
Thus, $(T_1,S=T^{-1}\circ T_1,N=T_1\circ T^{-1})$ is also a Kupershmidt-dual-Nijenhuis structure.
 \qed

\section{Rota-Baxter-Nijenhuis structures and  $r$-matrix-Nijenhuis structures}
In the following, we first recall the definition of an $r$-$n$ structure on a Lie algebra $\g$, which is the infinitesimal of a right-invariant Poisson-Nijenhuis structure on the  Lie group $G$ integrating the Lie algebra $\g$ (\cite{rn}).  We call such a structure an $r$-matrix-Nijenhuis structure.
\begin{defi}\label{defi:rmnij}
Let $\pi$ be an $r$-matrix and $N:\g\longrightarrow \g$ a Nijenhuis operator on a Lie algebra $(\g,[-,-]_\g)$. A pair $(\pi,N)$ is a {\bf $r$-matrix-Nijenhuis structure} on the Lie algebra $(\g,[-,-]_\g)$ if for $x,y\in\g$ and $\alpha,\beta\in\g^*$, they satisfy
\begin{eqnarray}
 \label{eq:rmn1} N\circ \pi^\sharp&=&\pi^\sharp\circ N^*,\\
 \label{eq:rmn2} {[\alpha,\beta]}^{N\circ \pi^\sharp}&=&{[\alpha,\beta]}_{N^*}^{\pi^\sharp},
\end{eqnarray}
where
$\pi^\sharp:\g\longrightarrow \g$ is a linear operator induced by $\langle \pi^\sharp(\alpha),\beta\rangle=\pi(\alpha,\beta)$
, \eqref{eq:rmn2} is given by \eqref{eq:TN2} with $S=N^*$, $T=\pi^\sharp$ and the representation $\rho=\ad^*$.
\end{defi}
It is obvious that the triple $(\pi,S=N^*,N)$ is a Kupershmidt-dual-Nijenhuis structure on the Lie algebra $\g$ with respect to the representation $(\g^*;\ad^*)$.

Similar to the $r$-matrix-Nijenhuis structure, we give the definition of Rota-Baxter-Nijenhuis structure on a Lie algebra.
\begin{defi}
Let $(\g,[\cdot,\cdot]_\g)$ be a Lie algebra. Let $\huaR:\g\longrightarrow\g$ be a Rota-Baxter operator and $N:\g\longrightarrow \g$  a Nijenhuis operator on the Lie algebra $\g$. A pair $(\huaR,N)$ is a {\bf Rota-Baxter-Nijenhuis structure} on the Lie algebra $\g$  if for $x,y\in\g$, they satisfy
\begin{eqnarray}
 \label{eq:YBTS1}N\circ \huaR&=&\huaR\circ N,\\
 \label{eq:YBTS2}[x,y]^{N \circ\huaR}&=&[x,y]_{N}^{\huaR}.
\end{eqnarray}
where \eqref{eq:YBTS2} is given by \eqref{eq:TN2} with $T=\huaR$ and the representation $\rho=\ad$.
\end{defi}

It is obvious that if $(R,N)$ is Rota-Baxter-Nijenhuis structure, then the triple $(R,S=N,N)$ is a Kupershmidt-Nijenhuis structure on the Lie algebra $\g$ with respect to the representation $(\g;\ad)$.

In the following, we study the relation between Rota-Baxter-Nijenhuis structure and $r$-matrix-Nijenhuis structure. First we recall some notions which was given in the articles by Semonov-Tian-Shansky \cite{Semonov-Tian-Shansky} or Kosmann-Schwarzbach \cite{Kosmann1}.

Let $\g$ be a Lie algebra with an $\ad$-invariant, non-degenerate, symmetric bilinear form $B\in\g\otimes\g$. Then $B$ induces a bijective linear map $B^\sharp:\g^*\longrightarrow\g$ given by
\begin{equation}
  \langle B^\sharp(\alpha),\beta\rangle=B(\alpha,\beta),\quad\forall~\alpha,\beta\in\g^*.
\end{equation}
By the $\ad$-invariance of $B$, we have
\begin{equation}\label{eq:adinv}
  B^\sharp(\ad^*_x\alpha)=\ad_x(B^\sharp(\alpha)),\quad\forall~x\in\g,\alpha\in\g^*.
\end{equation}
A {\bf skew-symmetric endomorphism of $(\g,B)$} is a linear map $\huaR$ from $\g$ to $\g$ such that $\huaR\circ B^\sharp:\g^*\longrightarrow\g$ is skew-symmetric.

The following theorem demonstrates the relation between Rota-Baxter-Nijenhuis structure and $r$-matrix-Nijenhuis structure.
\begin{thm}\label{thm:rmatrix-RBN}
 Let $\huaR$ be a skew-symmetric endomorphism of $(\g,B)$, $N:\g\longrightarrow\g$ a Nijenhuis operator, and set $\pi^\sharp=\huaR\circ B^\sharp$. Assume that $B$ and $N$ are compatible, i.e.
 \begin{equation}\label{eq:ad-bilinear}
 B^\sharp\circ N^*=N\circ B^\sharp.
 \end{equation}
If $(\huaR,N)$ is a Rota-Baxter-Nijenhuis structure on the Lie algebra $\g$, then $(\pi,N)$ is an $r$-matrix-Nijenhuis structure on the Lie algebra $\g$. Conversely, let $(\pi,N)$ be a r-matrix-Nijenhuis structure on the Lie algebra $\g$ with an $\ad$-invariant, non-degenerate, symmetric bilinear form $B$. Then $(\huaR=\pi^\sharp\circ(B^\sharp)^{-1} ,N)$ is a Rota-Baxter-Nijenhuis structure on the Lie algebra $\g$.
\end{thm}
\pf By Semonov-Tian-Shansky's conclusion, $\pi$ is an $r$-matrix. By \eqref{eq:YBTS1} and \eqref{eq:ad-bilinear}, it is obvious that $N\circ \pi^\sharp=\pi^\sharp\circ N^*$.

Let $\alpha=(B^\sharp)^{-1}(x),\beta=(B^\sharp)^{-1}(y)$, by \eqref{eq:adinv}, we have
\begin{eqnarray*}
   {[\alpha,\beta]}^{\pi^\sharp}&=&\ad^*_{ \pi^\sharp((B^\sharp)^{-1}(x))}(B^\sharp)^{-1}(y)-\ad^*_{ \pi^\sharp((B^\sharp)^{-1}(y))}(B^\sharp)^{-1}(x)\\
   &=&\ad^*_{ \huaR(x)}(B^\sharp)^{-1}(y)-\ad^*_{ \huaR(y)}(B^\sharp)^{-1}(x)\\
   &=&(B^\sharp)^{-1}([\huaR(x),y]_\g-[\huaR(y),x]_\g)=(B^\sharp)^{-1}([x,y]^{\huaR}),
\end{eqnarray*}
which implies that
\begin{equation}\label{eq:rmn3}
 {[\alpha,\beta]}^{\pi^\sharp}=(B^\sharp)^{-1}([x,y]^{\huaR}).
\end{equation}
Thus by \eqref{eq:ad-bilinear} and \eqref{eq:rmn3}, one has
\begin{eqnarray*}
  &&{[\alpha,\beta]}^{N\circ \pi^\sharp}-{[\alpha,\beta]}_{N^*}^{\pi^\sharp}={[\alpha,\beta]}^{N\circ \pi^\sharp}-\big([N^*(\alpha),\beta]^{\pi^\sharp}+[\alpha,N^*(\beta)]^{\pi^\sharp}-N^*[\alpha,\beta]^{\pi^\sharp}\big)\\
  &=&(B^\sharp)^{-1}([x,y]^{N\circ \huaR})-\big([N^*((B^\sharp)^{-1}(x)),(B^\sharp)^{-1}(y)]^{\pi^\sharp}+[(B^\sharp)^{-1}(x),N^*((B^\sharp)^{-1}(y))]^{\pi^\sharp}\\
  &&-N^*[(B^\sharp)^{-1}(x),(B^\sharp)^{-1}(y)]^{\pi^\sharp}\big)\\
  &=&(B^\sharp)^{-1}([x,y]^{N\circ \huaR})-\big([(B^\sharp)^{-1}\circ N(x),(B^\sharp)^{-1}(y)]^{\pi^\sharp}+[(B^\sharp)^{-1}(x),(B^\sharp)^{-1}\circ N(y))]^{\pi^\sharp}\\
  &&-N^*[(B^\sharp)^{-1}(x),(B^\sharp)^{-1}(y)]^{\pi^\sharp}\big)\\
  &=&(B^\sharp)^{-1}([x,y]^{N\circ \huaR})-(B^\sharp)^{-1}([N(x),y]_\g)-(B^\sharp)^{-1}[x,N(y)]_\g+N^*\circ (B^\sharp)^{-1}([x,y]_\g)\\
  &&-N^*[(B^\sharp)^{-1}(x),(B^\sharp)^{-1}(y)]^{\pi^\sharp}\big)\\
  &=&(B^\sharp)^{-1}([x,y]^{N\circ \huaR})-(B^\sharp)^{-1}\big([N(x),y]_\g+[x,N(y)]_\g-N([x,y]_\g)\big)\\
  &=&(B^\sharp)^{-1}([x,y]^{N\circ \huaR}-[x,y]^{\huaR}_N)=0,
\end{eqnarray*}
which implies that ${[\alpha,\beta]}^{N\circ \pi^\sharp}={[\alpha,\beta]}_{N^*}^{\pi^\sharp}$.

By a similar proof, the converse follows immediately. We finish the proof.\qed\vspace{3mm}

\noindent {\bf Acknowledgement:} This work was supported by NSFC (11471139,  11425104), NSF of Jilin Province(20170101050JC) and Nanhu Scholars Program for Young Scholars of XYNU.

redpajama_set_name: RedPajamaArXiv


text:
EdinburghGuide » Edinburgh's Festivals » Festival 2018 Reviews
Forget Me Nots, Army@Summerhall, Review
By Imogen Rowe - Posted on 15 August 2018
Summerhall
Rokkur Friggjar
Anna íris Pétursdóttir (Writer & Director), Sophie May Nicholls (Producer), Viktor Ingi Guðmundsoon (Composer)
Fannar Arnarsson, David Barclay Fenne, Halla Sigriður Ragnarsdóttir
Making their home in the Army Base in conjunction with Army@TheFringe, British/Icelandic theatre company Rokkur Friggjar return to the fringe with a new play set in World War II.
After the British invaded Iceland in 1940, British and American troops were stationed in occupation. One such is Thomas (David Barclay Fenne), a young and untested British soldier who is stationed at Hvalfjordur, the home town of Siggi (Fannar Arnarsson) and Gréta (Halla Sigriður Ragnarsdóttir). The two friends have grown up together and spent their lives in each other's company and confidence, but now they are nearing adulthood things are changing between them. When they meet Thomas, everything is turned inside out.
Forget Me Nots is a beautiful tale about the three of them, about their love for each other and how that shapes them. It is told through a series overlapping and interactive monologues from Thomas and Siggi, set to physical theatre as first the characters meet, grow intimate and are forced apart. The progression between them is very moving and gracefully put forward, and their intimacy is elegantly realised through the physicality of their storytelling.
Although a mainly silent character, Greta plays a far more significant role in the relationship than either Thomas or Siggi would want, and yet by the end it is she who unites them. It is a sad story of hope and hopelessness but ultimately of love, and through Rokkur Friggjar's gentle narrative the lasting message of a forget-me-not overcomes the greatest obstacle of all.
A symbol of true love, fidelity and memory, Forget Me Nots is an intimate love letter to these characters.
Until 21st, not 20th

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Regex con números, sin incluir texto Quiero hacer un regex que recoja el número decimal, pero que no recoja cuando esta junto a un texto.
Ejemplo:

    123 -----> Sí
    12.34 ---> Sí
    as12 ----> No
    op12as --> No
    opa12.1 -> No

De momento tengo mi regex así /^((\d+\.\d+|\d)+)/g, y he estado intentado poner negadores, pero no me funciona correctamente.

A: en vez de usar |, se pueden usar ? y * para indicar que esas partes de la expresion regular no son obligatorias. ? nos indica 0 o 1 apariciones y * nos indica 0 o más apariciones

const regex = /^\s*\d+\.?\d*$/gm;

// Alternative syntax using RegExp constructor
// const regex = new RegExp('^\\d+\\.?\\d*$', 'gm')

const str = `123
 -----> Sí
          123
 -----> Sí
12.34
 ---> Sí
as12
 ----> No
op12as
 --> No
opa12.1
 -> No`;
let m;

while ((m = regex.exec(str)) !== null) {
    // This is necessary to avoid infinite loops with zero-width matches
    if (m.index === regex.lastIndex) {
        regex.lastIndex++;
    }

    // The result can be accessed through the `m`-variable.
    m.forEach((match, groupIndex) => {
        console.log(`Found match, group ${groupIndex}: ${match}`);
    });
}

A: Con esta expresión regular...
 /^[\d]([\d\.]+)\s/gm

puedes excluir si tiene un caracter diferente a digito al principio (letras) y que finalice un caracter espacio, adjunto imagen:

A: Número separado del texto
Te presento a \b. Es un límite de palabra, y no coincide con un caracter sino con una posición: una posición entre un caracter de texto ([a-zA-Z0-9_]) que no tiene al lado a otro caracter de texto.
\b coincide:

*

*entre el inicio del texto y un caracter de texto

*entre un caracter que no es de texto seguido por otro que sí

*entre un caracter de texto seguido de otro que no

*entre un caracter de texto y el final del texto

Si rodeamos al patrón para un número de \bs te podés asegurar de que no tenga caracteres alfabéticos ni _ alrededor:
/\b\d+(?:\.\d+)?\b/g

Tiene a favor que vas a poder hacer que coincida con algo como (23+12.3), pero la contra de que con ab67.8 coincidiría con 8 porque formalmente no está rodeado de caracteres de palabra.
Pero también con vocales acentuadas, ü, ñ y puntos
que no están incluidos en [a-zA-Z0-9_]... Para eso se nos complica apenas un poco porque tenemos que usar un lookbehind negativo (?<! ), que comprueba que no esté precedido por un patrón; y un lookahead negativo (?! ),  que verifica que no esté seguido de un patrón.
/(?<![\wáéíóúüñ.])\d+(?:\.\d+)?(?![\wáéíóúüñ.])/ig

redpajama_set_name: RedPajamaStackExchange


Commemorate when your New York Islanders started in the NHL when you get this 5″" x 7″" Bronze Coin mini mint from Highland Mint. It features a full color New York Islanders graphic with a bronze 39 mm minted coin with the team's name"," logo and when they started in the NHL.
Add this New York Islanders 12″" x 20″" Silhouette Bronze Coin Panoramic photo mint from Highland Mint to your already impressive fan cave. The squad's name as well as their home ice is prominently displayed so everyone will know who you support.
Show off your team devotion everywhere you go with this New York Islanders 2-pack vinyl car mat set. They feature vibrant New York Islanders graphics.
Let your favorite room in the house burn up with the neon light from this New York Islanders Neon clock and you'll be ready for the start of every game. Every minute of the New York Islanders game counts and this clock will help you keep track of all of them.
Keep your beverage cold while the New York Islanders are heating up the ice with this Neoprene Can Cooler. The vibrant New York Islanders graphics will make it clear where your loyalties lie.
Show off your team pride on game day in this stunning New York Islanders Lombard flex hat! The bright embroidered graphics will make you stand out in the arena while your New York Islanders are heating up the ice!
Show them you are the #1 fan of the New York Islanders with this 3 x 5 Acrylic Team Coin Display by Highland Mint! You can show off your New York Islanders spirit with this handsome etched acrylic display that features a 24kt gold flashed coin with New York Islanders and league graphics. It is a limited edition of 5″,"000. NHL Shop is your source for officially licensed New York Islanders gear.
Show them you are the #1 fan of the New York Islanders with this Highland Mint Stanley Cup Champs Ticket acrylic display! Relive your favorite memories of your New York Islanders lifting Lord Stanley's Cup every time you look at this attractive display that features a ticket depicting the Cup on one side and your New York Islanders's history raising it on the other. An acrylic display also holds a removable commemorative New York Islanders coin. NHL Shop is your source for officially licensed New York Islanders gear.
Show them you are the #1 fan of the New York Islanders with this set of five WinCraft Ultra Car decals! Driving to the arena or around town"," show off your passion for the New York Islanders with this set of five reusable cling decals on your vehicle. The printed graphics on these non-fading decals will let everyone know you are a fan of the New York Islanders and you can remove and attach them to figure out how to best display them. NHL Shop is your source for officially licensed New York Islanders gear.


Checked out Forever 21 and found this picture/style. Love everything about it. And even though I don't like tight tops on top personally I can see myself rocking a similar look because of the cardigan!
I also saw this cardigan on sale on Sheinside.com and as much as I liked it I decided not to get it becasue I have so many cardigans I don't need another one in my closet!
Me too, it's super cool!


The concept of Wuxing, the five elements or the five dynamic states of change, is central to all elements of Chinese thought, including science, philosophy, medicine, astrology, and Fengshui. The names Metal, Wood, Water, Fire and Earth of Wuxing are only substances whose properties resemble the respective Chi in the closest possible way. When referring to directions, these correspond to West, East, North, South and Centre respectively. When we refer to colors, these correspond to White, Green, Black, Red and Yellow.
These pillar candles have Chinese unique Fengshui significations and each named as Metal, Wood, Water, Fire and Earth respectively.


text:
Promising young fighter Victoria Lee dies at 18
The world of mixed martial arts is in mourning after the death of Victoria Lee, a member of ONE Championship, Asia's largest MMA promotion. Victoria, who was just 18 years old, was set to fight on January 14 at the first ONE show of the year.
However, it has now been confirmed by her sister and fellow martial arts star Angela Lee that Victoria passed away on December 26.
Announcment of Victoria's death
The Lee family has chosen to keep the news of Victoria's death private until now, and the cause of her death has not been disclosed.
In an Instagram post, Angela wrote, "On December 26, 2022 our family experienced something no family should ever have to go through," "It is incredibly difficult to say this … Our Victoria passed away.
She has gone too soon and our family has been completely devastated since then," "We miss her. More than anything in this world. Our family will never be the same. Life will never be the same". "Victoria was the most beautiful soul who ever lived," Angela Lee added.
"She was the best little sister in the world. The best daughter, the best granddaughter and the best auntie." "We miss you so much sis. More than you could ever realise. We're all broken. Because a piece of you was in each of us and when you left, those pieces were ripped out of us.
We will never be the same."
A post shared by Angela Lee Pucci (@angelaleemma)
Former two-division MMA champion Aung La N Sang commented, "I am so sorry to hear that, condolences to the Lee family".
"I am so sorry for your loss. Sending all my love to your family," former ONE fighter Bi Nguyen wrote in a reply. Victoria, who was a teammate of Angela's and was coached by Angela's husband, Bruno Pucci, was born and reared in Hawaii, where the Lee family had relocated from Canada.
Victoria was already well-known in the MMA community despite her early age. At the age of 16, she competed in her first professional match. She went on to win three matches with ONE Championship.


text:
The AllWays Lounge hosts Esoterotica: SeductionConnectionIntimacy Starts in the Mind
Event: Apr 19, 2017 7:00pm
Join Esoterotica for a reading of erotic writings by local writers at 7 p.m. on Wednesday, April 19, at The AllWays Lounge (2240 St. Claude Avenue). Admission is free and open to the public (…"but its always sexy to pay what…
The AllWays Lounge hosts Esoterotica Extended Universe: A Night of Nerd-rotica
Event: May 3, 2017 7:00pm
Join Esoterotica for original readings of erotic writings by local writers at 7 p.m. on Wednesday, May 3, at The AllWays Lounge (2240 St. Claude Avenue). Admission is free and open to the public (…"but its always sexy to pay what…
UNO presents an evening with Jennifer S. Davis and Mark Yakich on March 3
The University of New Orleans Creative Writing Workshop will host a reading by Jennifer S. Davis & Mark Yakich at 7 p.m. on Thursday, March 3, at the University of New Orleans Lakefront Campus, Liberal Arts Building, Room 140. The…
Whirlwind of poetry events to hit New Orleans next week!
I did my taxes last night, and the nice VITA fellow who helped me let me know that I had mis-checked a box on the prep form I filled out—under the question, "Do you live in an area affected by…
Get a Date This Saturday at the New Orleans Book Fair!
The tenth annual New Orleans Book Fair takes places this Saturday, Nov. 5, on Frenchmen Street from 11 a.m. – 6 p.m. There will be craploads of tables of books and other bound printed matter lining the sidewalks, and tons…
Room 220 Fall Reading Series Lineup Announced!!!


Share This: On Monday, 12th December, 2016, the wives of ambassadors visited the children at the Belgrade Shelter bringing them gifts that were received with a lot of joy.
Share This: The Children's Cultural Centre hosted an interactive children's New Year's programme. Fifty children participated in the programme, with a rich theatrical, ballet and music content for their peers.


The first-line treatment for anaphylaxis is epinephrine (adrenaline).
A food allergy reaction can cause symptoms that range from mild to life-threatening. In the United States, food allergy is the leading cause of severe, life-threatening allergic reactions (anaphylaxis) outside the hospital setting. People who have previously experienced only mild symptoms may suddenly experience a life-threatening reaction. Don't delay: Find an allergist today.
The first-line treatment for anaphylaxis is epinephrine (adrenaline). It is available by prescription in an auto-injector. A delay in using epinephrine is common in severe food allergic reaction deaths. Epinephrine works to reverse the life-threatening symptoms. Other medications, such as antihistamines, are not adequate substitutes and will not save you or your child's life, as they do not reverse swelling of the airway or raise low blood pressure. Allergists advise that all patients who have food allergies carry their epinephrine auto injector with them at all times. This is particularly important for those who have had a previous anaphylactic reaction; have both a food allergy and asthma; or have an allergy to peanuts, tree nuts, fish, or crustacean shellfish.
Be sure to have two doses available, as the severe reaction may reoccur. If you've had a history of severe reactions, take epinephrine as soon as you suspect you've eaten an allergy-causing food, or if you feel a reaction starting. Epinephrine should be used immediately if you experience severe symptoms such as shortness of breath, repetitive coughing, weak pulse, generalized hives, tightness in the throat, trouble breathing/swallowing, or a combination of symptoms from different body areas such as hives, rashes, or swelling on the skin coupled with vomiting, diarrhea, or abdominal pain. Repeated doses of epinephrine may be necessary.
If you are uncertain whether or not a reaction warrants epinephrine, use it right away, because the benefits of epinephrine far outweigh the risk that a dose may not have been necessary.
Common side effects of epinephrine may include anxiety, restlessness, dizziness, and shakiness. Rarely, the medication can lead to abnormal heart rate or rhythm, heart attack, sharp increase in blood pressure, and fluid build-up in the lungs, therefore patients with certain pre-existing conditions who may be at higher risk for adverse effects should speak to their allergist about when to use epinephrine.
Your allergist will provide you with a written emergency treatment plan that outlines which medications should be administered and when (note that between 10-20 percent of life-threatening severe allergic reactions have no skin symptoms.) Be sure you and those responsible for your children while not in your care understand how to properly and promptly use an epinephrine auto-injector.
This page was reviewed for accuracy 2/1/2018.
Epinephrine is always the first line of defense to treat severe allergic reactions.
Probably, but still best to avoid freezing.
No need for fear as kids head back to school.


text:
Is BATMAN Headed For Another DEATH IN THE FAMILY?
By George Marston, Newsarama Contributor September 1, 2016 08:26am ET
Credit: DC Comics
Batman has a bad track record with his sidekicks. Of the five Robins in current DC Comics continuity, all but one have died (or seemed to have died) and come back. Only Tim Drake has avoid the icy claws of death, but it appears possible that could soon change - or DC at least, wants you to think it will.
This week, DC has literally cut Tim Drake out of the Bat-family - ahead of a promised "change... you'd never expect" to the Bat-titles.
A newly-revised cover to September 14's Detective Comics #940 has removed Tim Drake from this cover illustrated by series artist Eddy Barrows. Now, covers change all the time for a variety of reasons, but removing one specific character is vastly different from re-soliciting brand new art or a variant. And those other Bat-folks sure do look mournful.
'Detective Comics #940' side by side covers
Adding to that curious revision is the solicitation for that issue, which reads "This epic concludes with a battle against the Colony that will change Batman's world forever-and in a way you'd never expect! Whether they win or lose, they've already lost…"
So "Batman's world" will be changed "forever" .. in an unexpected way ... with a major loss.
So is Tim really going to die? Not so fast...
The previously Detective Comics #939 hints at another different option, when Tim expresses his plans to leave Gotham and go to college at Ivy University – effectively ending his superheroing career while he's a full-time student.
But what's that old trope about being "days from retirement?" Lethal Weapon's Danny Glover would be proud, as the end of tha issue finds Tim pinned in by a veritable army of drones with mounted guns, right in that oft-used cliffhanger of a potentially fatal next issue.
But what if it's not that simple?
What if Tim has been removed from the cover – and left out of the already solicited covers for the next several issues – not because he's dead, but because he's retired and attending college? What if DC is teasing an absence - hoping fans will jump to the conclusion of a death - only to reassure us that Tim's still out there, even if he's not in his cape and pixie boots? Would DC be so bold as to outright fake his death in the story to give him a clean break?
If so, Tim would be no less than the fourth of Batman's sidekicks to be seemingly killed and brought back in secret. Hell, Stephanie Brown and Dick Grayson both underwent that exact same bait-and-switch. And as we said, killing Tim – fake or not – would complete the set of dead Robins.
But would DC really go back to that well again?
What if - and this is crazy – but what if Tim Drake is just going to college?
Ivy University isn't just some random school – it's the home of Professor Ray Palmer, the Atom. In DC Universe Rebirth #1, readers caught a brief glimpse of Palmer's successor, Ryan Choi, discovering a message from Palmer that he's trapped in the Microverse and needs Ryan's help. DC has promised this story would return. What if Drake is roped into that story thread originated by Geoff Johns?
What about that Detective Comics #940 solicitation, though? If the ending is something we'd truly "never expect," given that they laid out in #939 that Drake is planning on going to college wouldn't him going be rather... not unexpected?
While superhero deaths have become a arguably overused plot device, if there is a time to do it then it would be this month - 27 years to the month that Tim Drake debuted in comic books, August 1989.
How many promising careers have been cut short at 27? Kurt Cobain, Janis Joplin, Jimi Hendrix…
We'll find out if we're headed to Tim's funeral or his commencement when Detective Comics #940 hits stands on September 14.
DC Comics APRIL 2018 SOLICITATIONS
DC Comics MARCH 2018 SOLICITATIONS
DC Comics FEBRUARY 2018 SOLICITATIONS
DC Comics DECEMBER 2017 SOLICITATIONS


Pen and watercolour pet portrait. I have drawn many horse, dog, cat and other animals in Pen and watercolour and a selection of the 600+ portraits can be found below.
I have completed so many pieces to date I cannot post them all on my website but please visit www.facebook.com/louisejarvisart to see more of my work.
"YA HAFED" - The famous Race Horse "Nobby" made famous by Wocket Way and Batch.


Recycle ... Drive a Used Car!
All the cars we sell are meticulously reconditioned, mechanically and cosmetically, to the very highest of standards.
If we don't have it we can find it. We attend three auctions a week and have access to new car dealer inventory and rental agencies retired inventory.
Please email me and let me know what you are looking for.
Financing available through a local bank at very competitive rates. Please stop by and fill out a credit application and I will do the rest.
Trade-ins welcomed regardless of condition and money owed.
Want to sell your car? We are interested in buying nice low-mileage used cars with clean titles.


text:
Q: Copy images with EXIF(time) to new destination from several directories I have a problem, below is my root tree:
In my root I have that:
 ---dir1---sub1(images exif time 10:05:05 to 10:09:55)
        ---sub2(images exif time 10:11:15 to 10:15:42)
        ---sub3(images exif time 10:22:15 to 10:24:41)
        ---sub4(images exif time 10:28:15 to 10:35:40)

 ---dir2---sub1(images exif time 10:05:06 to 10:09:57)
        ---sub2(images exif time 10:11:15 to 10:15:40)
        ---sub3(images exif time 10:22:15 to 10:24:43)
        ---sub4(images exif time 10:28:15 to 10:35:40)
        ---sub5(images exif time 10:40:15 to 10:43:40)

 ---dir3---sub1(images exif time 10:05:05 to 10:09:54)
        ---sub2(images exif time 10:11:15 to 10:15:40)
        ---sub3(images exif time 10:22:15 to 10:24:41)
        ---sub4(images exif time 10:28:15 to 10:35:40)
        ---sub5(images exif time 10:40:15 to 10:43:42)

 ---dir4---sub1(images exif time 10:05:06 to 10:09:57)
        ---sub2(images exif time 10:11:15 to 10:15:40)
        ---sub3(images exif time 10:22:15 to 10:24:43)
        ---sub4(images exif time 10:28:15 to 10:35:40)
        ---sub5(images exif time 10:40:15 to 10:43:40)

 ---dir5---sub1(images exif time 10:05:05 to 10:09:54)
        ---sub2(images exif time 10:11:15 to 10:15:40)
        ---sub3(images exif time 10:22:15 to 10:24:41)
        ---sub4(images exif time 10:28:15 to 10:35:40)
        ---sub5(images exif time 10:40:15 to 10:43:42)

I have 5 dirs in my root and each contains  sub-folders(with images) number of sub-folders it's not a same all the time, What I want to do is  from first - dir1 get sub1 and put it to new destination folder after go to next dir (dir2) scan sub-folders to check exif(time) if its a same as sub1 from dir1 and copy it to same directory after go to next dir3 and do same for all others dir-s and subfolders, and after create newdir2 go and take sub2 from dir1 and do again same loop till end...
something like:
   ---newdir1---sub1(from dir1)
             ---sub1(from dir2)
             ---sub1(from dir3)
             ---sub1(from dir4)
             ---sub1(from dir5)

   ---newdir2---sub2(from dir1)
             ---sub2(from dir2)
             ---sub2(from dir3)
             ---sub2(from dir4)
             ---sub2(from dir5)

   ---newdir3---sub3(from dir1)
             ---sub3(from dir2)
             ---sub3(from dir3)
             ---sub3(from dir4)
             ---sub3(from dir5)

   ---newdir4---sub4(from dir1)
             ---sub4(from dir2)
             ---sub4(from dir3)
             ---sub4(from dir4)
             ---sub4(from dir5)

   ---newdir5---sub5(from dir2)
             ---sub5(from dir3)
             ---sub5(from dir4)
             ---sub5(from dir5)

I have a part of script with sort my images to dictionary by some time interval, how I can join it to my script?? to get my sub-s with same key to same dir ??:
import os
import exifread
from datetime import datetime, timedelta

TIME_RANGE = 2

src_root = 'F:\gopro_egouts\gopro_img_test\\2018-03-06'

dst_root = src_root + '-copie'

src_dirs_dict = {}

for cam_dir in os.listdir(src_root):
    laps_root = os.path.join(src_root, cam_dir)
    for lap_dir in os.listdir(laps_root):
        files_root = os.path.join(laps_root, lap_dir)
        min_time = None
        max_time = None
        for cam_file in os.listdir(files_root):
            with open(os.path.join(files_root, cam_file), 'rb') as f:
                tags = exifread.process_file(f, details=False, stop_tag="EXIF DateTimeOriginal")
                time_taken = tags.get("EXIF DateTimeOriginal")
                if time_taken:
                    file_time = datetime.strptime(str(time_taken), '%Y:%m:%d %H:%M:%S')
                    if min_time is not None:
                        if file_time < min_time:
                            min_time = file_time
                    else:
                        min_time = file_time

                    if max_time is not None:
                        if file_time > max_time:
                            max_time = file_time
                    else:
                        max_time = file_time

        is_key = None
        for key in src_dirs_dict.keys():
            if (min_time >= key[0] and min_time < key[1]) \
                    or (max_time >= key[0] and max_time < key[1]):
                is_key = key
                break
        min_time = min_time.replace(second=0)
        max_time = min_time + timedelta(minutes=TIME_RANGE)

        if is_key:

            key_min, key_max = is_key
            if min_time < key_min:
                key_min = min_time
            if max_time > key_max:
                key_max = max_time

            new_key = (key_min, key_max)

            if new_key == is_key:
                src_dirs_dict[new_key].append(files_root)
            else:

                src_dirs_dict[new_key] = src_dirs_dict.pop(is_key) + [files_root]
        else:
            new_key = (min_time, max_time)
            src_dirs_dict[new_key] = [files_root]

print(src_dirs_dict)

My print showing me that:
{(datetime.datetime(2018, 3, 6, 10, 31), datetime.datetime(2018, 3, 6, 10, 32)): ['F:\\gopro_egouts\\gopro_img_test\\2018-03-06\\CAM0101 1\\Time Lapse 3',...

I have a working script with working well but taking a sub-folders one by one , and when some Time-lapse is missing, there I have a problem, his mixing my sub-s(automatically taking next one from next dir with wrong time), where I have to add my exif script from above to here(how modify it)... how to join it together???
Any help will be appreciated.
from collections import defaultdict
import shutil
import os
import re
src_root = r'F:\gp\gp_test\\2018-03-06'

dst_root = src_root + '-copie'

#os.makedirs(dst_root, exist_ok=True)

src_dirname, src_folders, _ = next(os.walk(src_root))
src_folders = sorted(src_folders)

src_folders = [os.path.join(src_root, folder) for folder in src_folders]
print(src_folders)
job = defaultdict(list)

print('mes {} dossier cam'.format(len(src_folders)))

for folder in src_folders:
    print()
    dirname, src_sub_folders, _ = next(os.walk(os.path.join(src_dirname, folder)))
    src_sub_folders = sorted(src_sub_folders, key=lambda x: [re.search(r'(\D+)', x).group(1)] + list(map(int, re.findall(r'\d+', x))))
    print("mes 5 CAM avec {} time laps '{}'".format(len(src_sub_folders), folder))

    for index, sub_folder in enumerate(src_sub_folders, start=1):
        job['Time Lapse-{}'.format(index)].append(os.path.join(dirname, sub_folder))

#print()

for dst_folder, src_folders in sorted(job.items()):
    for index, src_folder in enumerate(src_folders, start=1):
        dst_new_folder = os.path.join(dst_root, dst_folder, 'CAM-{}'.format(index))
        print('{} -> {}'.format(src_folder, dst_new_folder))
        shutil.copytree(src_folder, dst_new_folder)
#shutil.rmtree(src_root)

for root, dirs, files in os.walk(dst_root):
    for f in files:
        prefix = os.path.basename(root)
        prefix1 = os.path.basename(src_root)
        os.rename(os.path.join(root, f), os.path.join(root, "{}-{}-{}".format(prefix1, prefix, f)))
        print("images rennomer ")

print("fini")
print("dossier supprimé")

I'm really sorry if that will be not to clear for users, but English it's not my strongest language ...

A: In a nutshell, you have images of the same set of events shot on a number of cameras.
Currently, they are grouped first by camera, then by event:
├── Camera1
│   ├── Event1
│   ├── Event2
│   ├── Event3
│   ├── Event4
│   └── Event5
├── Camera2
│   ├── Event1
│   ├── Event2
│   ├── Event3
│   ├── Event4
│   └── Event5
├── Camera3
│   ├── Event1
│   ├── Event2
│   ├── Event3
│   ├── Event4
│   └── Event5
├── Camera4
│   ├── Event1
│   ├── Event2
│   ├── Event3
│   ├── Event4
│   └── Event5
└── Camera5
    ├── Event1
    ├── Event2
    ├── Event3
    ├── Event4
    └── Event5

... where some events may be missing and the event numbering may not match because one or more events may not be recorded by all cameras.
And you want the same set of images grouped first by event, then by camera:
├── Event1
│   ├── Camera1
│   ├── Camera2
│   ├── Camera3
│   ├── Camera4
│   └── Camera5
├── Event2
│   ├── Camera1
│   ├── Camera2
│   ├── Camera3
│   ├── Camera4
│   └── Camera5
├── Event3
│   ├── Camera1
│   ├── Camera2
│   ├── Camera3
│   ├── Camera4
│   └── Camera5
├── Event4
│   ├── Camera1
│   ├── Camera2
│   ├── Camera3
│   ├── Camera4
│   └── Camera5
└── Event5
    ├── Camera1
    ├── Camera2
    ├── Camera3
    ├── Camera4
    └── Camera5

Here's an idea... I am kind of "thinking aloud" in pseudo-code:
Create the output directories {Event1..EventN}/Camera{1..N}
MissingDirectory=false
for each input directory Camera{1..N}
    if this directory has the full number of subdirectories
        copy all subdirectories to output area
    else
        MissingDirectory=true
    end if
end for

if MissingDirectory
    for each output Event directory
        get times of all files from all cameras for current event
        sort list and take median time of current event
    end for
    for each un-copied input directory
        get the mean time of all the files in it
        assign this directory's files to output directory with nearest median time
    end for
endif

You can convert your EXIF times to pure seconds since midnight (s) with:
s = (hours*3600) + (minutes*60) + seconds

Here's a way to get the time (in seconds since midnight) that an image was taken:
import exifread

def getImageTime(filename):
    "Read EXIF data of given file and return time in seconds since midnight"
    f=open(filename,'rb')
    tags=exifread.process_file(f)
    DateTime=tags["EXIF DateTimeOriginal"].printable
    # DateTime looks like: "2013:03:09 08:59:50"
    Time=DateTime.split()[-1]
    # Time looks like: "08:59:50"
    h,m,s=Time.split(":")
    # Return seconds since midnight: 32390
    return (int(h)*3600) + (int(m)*60) + int(s)

s=getImageTime("image.jpg")
print(s)

After some more thought, this won't work very well if one of the cameras is set, say, 20 minutes different from the others, since all its images from all the sequences will tend to get put into the first or last directory. Needs some more thought...


text:
Margie Bright Matthews (born February 10, 1963) is a Democratic member of the South Carolina Senate, representing the 45th District since 2015, when she won a special election to succeed Clementa Pinckney, who was killed in the Charleston church shooting in 2015.  She is an attorney.

Bright Matthews was among a number of African American women from around the United States who endorsed Hillary Rodham Clinton for President in 2016.

References

1963 births
Living people
African-American state legislators in South Carolina
Democratic Party South Carolina state senators
21st-century American politicians
People from Walterboro, South Carolina
21st-century African-American politicians
20th-century African-American people

Women state legislators in South Carolina
Women in the South Carolina State Senate

redpajama_set_name: RedPajamaWikipedia


text:
Argentina has competed at every edition of the Pan American Games since the first edition of the multi-sport event in 1951, in which it hosted.  Argentina competed in the first ever Pan American Winter Games in 1990, however it failed to get medals.

Medal count

To sort the tables by host city, total medal count, or any other column, click on the  icon next to the column title.

Summer

Notes
  Some sources appoint 47 silver medals and 39 bronze medals, instead of 44 and 38, respectively. This would result in a total of 154 medals earned during the 1951 Games, instead of 150.
  Some sources appoint 33 silver medals and 20 bronze medals, instead of 31 and 15, respectively. This would result in a total of 80 medals earned during the 1955 Games, instead of 73.
  Some sources appoint 22 silver medals and 12 bronze medals, instead of 19 and 11, respectively. This would result in a total of 43 medals earned during the 1959 Games, instead of 39.
  Some sources appoint 20 bronze medals, instead of 16. This would result in a total of 43 medals earned during the 1963 Games, instead of 39.
  Some sources appoint 13 silver medals and 11 bronze medals, instead of 14 and 12, respectively. This would result in a total of 32 medals earned during the 1967 Games, instead of 34.
  According to those sources, the historical medal table for Argentina counts 305 silver medals and 408 bronze medals, instead of 298 and 398, respectively. This would result in a total number of 992 Pan American medals.

Winter

Medals by summer sport
Argentines have won medals in most of the current Pan American sports.
The exceptions are , badminton and baseball.

''As of the conclusion of the 2019 Pan American Games

Best results in non-medaling sports:

References

External links
COA - Comité Olimpico Argentino Official site.


text:
Smith named women's basketball head coach
Jackie Smith named head women's basketball coach
Jackie Smith, who most recently served as an assistant coach at George Washington University, has been named the new head women's basketball coach at Connecticut College.
"We are very excited to welcome Jackie Smith into our community," said Fran Shields, the Katherine Wenk Christoffers '45 Director of Athletics. "We are confident that she can take a healthy existing program even higher given her great mentors and her valuable coaching and recruiting experience. Jackie has made it clear that she will also place a strong emphasis on the student-athlete experience."
Smith will be serving as a head coach for the first time in her professional career after being the top assistant under her longtime mentor Jennifer Rizzotti, a UConn alumna and 2013 Women's Basketball Hall of Fame inductee, for the past five seasons at Hartford and George Washington. The Colonials most notably advanced to the 2018 NCAA Tournament by winning the Atlantic 10 Conference crown and also qualified for the 2017 WNIT Tournament in her first season on the sidelines at the institution.
At George Washington, Smith served as the program's recruiting coordinator and was heavily involved with the team's scouting, analysis and game strategy for the past two seasons. She worked directly with the guards and successfully developed two all-conference players, including the MVP of the 2018 Atlantic 10 Tournament. Furthermore, the assistant coach was the program's admissions, academic and alumni relations liaison.
Smith came to George Washington with Rizzotti after spending three years with her in the same role at Hartford, her alma mater. She began her professional career as an assistant coach for two seasons at Stonehill College in Massachusetts.
"I am thrilled about this opportunity," Smith said. "The athletic department's pursuit of excellence, both academically and athletically, will match perfectly with my values. I am also thankful to be taking over a program which has dedicated student-athletes that have had success in the classroom and on the court, and I look forward to building upon what former head coach Brian Wilson was able to accomplish."
A native of Westfield, Massachusetts, Smith earned a bachelor's degree in elementary and special education with a minor in sociology from Hartford in 2011. She will begin her duties at Connecticut College on Aug. 4.
Join the herd. Learn more
Women's ice hockey looks to extend 7-game win streak


Today we have a trip to a recovery facility and a prison. New Brunswick this time of year is spectacular. All the trees are filled out, flowers in bloom and the fields lush and green. As we travel, we comment upon the spring season and the new life it brings.
As we arrive to our first destination we are excited. No two trips are ever the same, as God is infinite, so are His ways. Today we have Marg, Becka, Bill, Doc(Steve) and Tony. As the room begins to fill up we pray and then commence our ministry.
Bill opens with the introduction of Prison Christian Ministries. We are one of many concerned groups who have answered the call of God according to Matthew 25: 35-36. Prison Christian Ministries conducts 30 functions in 30 days, putting close to 6000 Km on our van. We visit all New Brunswick Federal and Provincial institutions including Halfway houses and recovery centers. We also have an African outreach to 11 more prisons in Kenya supplying food and necessities to men and women and children within the confines of their prison system. Our agent there does roughly 30 visits per month too. Locally we also have an after care facility which will open in the near future to help house inmates who have no where to go upon release.
Today we are honored with the attendance of people in recovery. Doc speaks first, and brings a well thought out teaching regarding the love of our Heavenly Father, and how that love is so necessary for recovery to be successful. Many times the use of drugs and alcohol, or other life controlling issues can be traced to a deep need to cover pain within the heart of the individual. Frequently, unforgiveness and bitterness are at the root of a lot of pain. Until we realize that unconditional love, and hate are opposed to each other, we will have a problem letting go of our pain.
Becka shared a marvelous contribution to the over all continuity of the teaching. She added that rest is also a necessity to recovery, we must rest. She is a musician, and coming from a musician point of view, she explained that in music many compositions have "rests" in the written piece which helps to make the composition complete. So it is with us in recovery, we must also learn to rest, to enable us to have complete recovery. The people responded to that description, God was touching hearts even more profoundly.
Next Bill spoke about the necessity of relationship. When Jesus said, "it is finished", he was proclaiming that the relationship between God and man had been restored. The cross has a vertical member and a horizontal member. The vertical member is our relationship with Christ and the horizontal member is our relationship with others. Recovery is not a solo process, it not only requires restored relationship with the Father but also with others. Recovery never is complete and restorative without relationship with others.
Next Tony came forward. His evangelistic gifting is so evident as he proclaims the power, love, and completeness of Christ. He expounds upon the necessity and need for allowing God to gain access to our souls and closes with an invitation for prayer. Later, after we meet and greet we understand that souls have been touched, and we proceed to have a meal together. This time is so refreshing as we meet others who were not able to be in the meeting but arrive in time for a community meal. We meet old acquaintances and have good fellowship. The time goes fast, now we must say our goodbyes and head on to our next stop, a Federal Institution.


New or added value services to current customers.
A special offer to encourage lapsed customers to come back.
An incentive or trial to good prospects that didn't convert to business.
A package for introducers and partners who could refer business to you.
Contact with quality contacts on Outlook that you haven't spoken to in a while.
It isn't just about 'cold calling'. Telemarketing or telesales as it is also known is one of a number of direct marketing routes to market. It is the most direct approach. It works best when it is integrated into other activities such as direct mail, email marketing, events and so on. However, it is also effective when used on its own.
Aside from online ecommerce, most transactions happen face to face. Without qualified sales meetings and new business appointments this cannot happen. Therefore, telemarketing lead generation still has an important role to play in sales campaigns.
Telemarketing is a proactive medium. No waiting for the phone to ring, a response to a mailer or advertisement or hoping someone will walk into your office.
Q. Why does cold calling have a bad name?
Cold calling sometimes gets a bad name and it is assumed this means haranguing people like has happened in the past with double-glazing. Actually, nothing could be further from the truth especially when it comes to b2b telemarketing. If you are a reputable company promoting appropriate services to a marketplace that uses those services, it is highly unlikely that the phone will be slammed down on you. Telemarketing can provide an effective way to build rapport as part of your new business marketing activities.
A clearly defined target audience. Target marketing is extremely important. Select a market for your goods and services that understands what you offer and could have use for them.
Develop a strong reason for your prospects to engage with you. Calls that have no basis or value leave the recipient wondering why you're wasting their time.
Make sure you have good people making the calls who are mature, confident, assertive and who have an excellent telephone manner.
Q: Who are the clients of GSA Business Development?
At GSA, we undertake senior level telemarketing campaigns. We are not a call centre. Our team is made up of senior people used to calling decision-makers.
Our work includes appointment generation for PR and advertising agencies as well as for associations, federations, financial services and many more. Please view examples of our work and view our testimonials.
Q: Do you make face to face appointments for us?
Q: What type of companies and decision makers can you get me in front of?
We make appointments with Managing Directors, Finance Directors, Marketing Directors, Compliance directors as well as owners and more. Whoever your decision-maker is, we can help you generate new business.
Our work has included telemarketing appointments with many blue-chip companies including Virgin, Tesco, Body Shop, BP and many others. Visit our testimonials page to see what our clients are saying.
Q: Who does my telemarketing?
Our team are senior people who have had excellent professional careers. They are used to making sales calls and will represent your business and protect and enhance your reputation.
Q: What makes a compelling proposition?
A good sales proposition is all important.
Why should I listen to this person?
Why should I see them?
Why should I see them now as opposed to some time in the future?
These are important questions and you should try to identify a strong reason to see you not something that everyone else can claim.
Q: Can I get a trial period?
Yes. GSA can undertake a trial period at low cost. Contact us for more details.
Following a trial period, we work either on a fixed fee basis or a hybrid (lower daily rate and bonus for appointments) or on an appointment only basis.
GSA provides in depth training for its telemarketing team and also offers telesales training for clients' in-house telemarketers. Ask us about our telemarketing workshops and top telemarketing tips.
Q: I want to recruit a telemarketer, can you help?
We are able to help undertake telemarketer recruitment and telemarketing training if you feel that you would like to recruit and develop your outbound calling internally.
We undertake a new business development strategy workshop with all clients before commencement. We also run these and marketing workshops for clients outside of a telemarketing campaign.
Q: How do you monitor the success of my campaign?
We track every call made and provide reports so that clients can see details of every call.
This no problem as GSA regularly conducts list sourcing for clients. We can identify a target market and secure a good list. We also build lists for clients using the web, phone and other measures.
We use FIVE CRM and some clients wish to use this as their CRM database. For a low monthly fee, clients can house all of their customer and prospect data in one place to make marketing to them much easier and more effective.
Q: Why should I outsource my telemarketing?
Often new business is everyone's responsibility and no one takes responsibility. Managing in-house staff is often costly and time-consuming. Regularly, companies fail to provide the tools, the data, the systems, the briefing and the environment to make telemarketing work. Training isn't provided and sales staff can be transient thereby incurring companies extra recruitment cost on top of the salary and NI.
By using an outsourced telemarketing agency, you are passing the responsibility to a company whose job it is to do telemarketing day in and day out. You reduce risks and often enhance success. If you don't believe us, take a look at our testimonials and give us a call.
Find out how GSA Business Development can help Generate Growth for your Business, by contacting us now on 0845 658 8192 or by sending us an email using the form below.


text:
Q: Date ranges from start date and duration New here, and to perl as well.
I'm trying to write a perl script to do some data archiving at work. Our directory structure is separated at one level by year, then has every day/month inside that.
So, for example:
\2016\1-1
\2016\1-2
....
\2016\12-20
etc...

I'm trying to make this script usable for every test (some of which are run in the future), so I have a prompt for the user to enter the year, which I use to populate the directory structure in the code. I'm having trouble with the month-date portion.
Usually tests are 3-4 days long. Right now I'm asking the user for the test length, then subtracting one. I do that because a 4 day test starting on the 8th, would be the 8 ,9 ,10, and 11th. If I don't subtract one, and just add the duration to the start date, it would go to the 12th.
I can't figure out how to take a start date and duration, and get that range of dates into an array, that I can then call when I need to find/create directories for the archiving.
Ideally, I'd like to have the user enter the year, start date, and length of the test, then create an array that will hold the month-day for all dates of the test. Then I could call those array dates when I create/find the needed directories.
Right now I'm still working on getting the inputs from the user, so that is all that I have.
Couple notes: I don't have every Perl module available to me. I have been using DateTime, and I'm not 100% sure what other modules I could use. Also, the directory structure does not have leading zeros on the month/date (i.e., 2-5, NOT 02-05).
Here is the input code that I have so far. It gives me a start date of today (which I need to change), and gives me the correct end date based on the test length. As you can see, I have not been able to get very far in this:
use warnings;

use DateTime;

my @test_dates;

print "What NAS?";
my $NAS = <STDIN>;
chomp($NAS);

print "What is the Multi-Mission mode (ops/nonops)?";
my $MM_Mode = <STDIN>;
chomp($MM_Mode);

print "What is the data mode (ops/sim/tst)?";
my $Data_Mode = <STDIN>;
chomp($Data_Mode);

print "Which spacecraft (NPP/J01)?";
my $sc = <STDIN>;
chomp($sc);

print "What LOM is being cleaned?";
my $LOM = <STDIN>;
chomp($LOM);

print "What MDMZ is being cleaned?";
my $MDMZ = <STDIN>;
chomp($MDMZ);

print "How many days is the test?";
my $Length   = ( <STDIN> - 1 );
my $date     = DateTime->from_epoch( epoch => time );
my $Duration = DateTime::Duration->new( days => $Length );

print "NAS is $NAS\n";
print "Multi-Mission mode is $MM_Mode\n";
print "Data Mode is $Data_Mode\n";
print "LOM is $LOM\n";
print "MDMZ is $MDMZ\n";

printf $date->ymd('/');

print @test_dates;

A: I don't think you have it clear in your mind exactly what you need, but this should help
use strict;
use warnings 'all';

use Time::Piece;
use Time::Seconds 'ONE_DAY';

print 'Enter start date (YYYY-MM-DD): ';
chomp(my $start = <>);
$start = Time::Piece->strptime($start, '%Y-%m-%d');

print 'Enter number of days: ';
chomp(my $duration = <>);

{
    my $end = $start + ONE_DAY * ($duration-1);
    die "Period crosses end of year" if $start->year != $end->year;
}

my $date = $start;

while ( --$duration ) {
    my $dir = $date->strftime('\%Y\%m-%d');
    $dir =~ s/\D\K0+//g;
    print $dir, "\n";
    $date += ONE_DAY;
}

output
E:\Perl\source>perl date_duration.pl
Enter start date (YYYY-MM-DD): 2016-7-18
Enter number of days: 22
\2016\7-18
\2016\7-19
\2016\7-20
\2016\7-21
\2016\7-22
\2016\7-23
\2016\7-24
\2016\7-25
\2016\7-26
\2016\7-27
\2016\7-28
\2016\7-29
\2016\7-30
\2016\7-31
\2016\8-1
\2016\8-2
\2016\8-3
\2016\8-4
\2016\8-5
\2016\8-6
\2016\8-7

A: Here is an approach that uses DateTime. You don't need to create a DateTime::Duration object, DateTime does that implicitly for you whenever you use date math.
use strict;
use warnings 'all';
use DateTime;
use feature 'say';

my $duration = 4;
my ($year, $month, $day) = (2016, 4, 9);
my $start = DateTime->new( year => $year, month => $month, day => $day);

my @dates;
foreach my $i ( 0 .. ( $duration - 1 )) {
    my $date = $start->clone->add( days => $i )->strftime('\%Y\%m-%d');
    $date =~ s/(?<=\D)0+//g;
    push @dates, $date;
}

say for @dates;

I left out the user input stuff and just set some values. Basically it builds each day, creates the folder names and puts them into an array.
Output
\2016\4-9
\2016\4-10
\2016\4-11
\2016\4-12


text:
Showing results by author "Mick Mars" in All Categories
Confessions of the World's Most Notorious Rock Band
By: Tommy Lee, Vince Neil, Nikki Sixx, and others
Narrated by: Sebastian York, Roger Wayne, Fred Berman, and others
Celebrate over 30 years of the world's most notorious rock band with the audiobook edition of The Dirt - the outrageous, legendary, no-holds-barred autobiography of Mötley Crüe. Fans have gotten glimpses into the band's crazy world of backstage scandals, celebrity love affairs, roller-coaster drug addictions, and immortal music in Mötley Crüe books like Tommyland and The Heroin Diaries, but now the full spectrum of sin and success by Tommy Lee, Nikki Sixx, Vince Neil, and Mick Mars is an open book in The Dirt.
1000x more detail than the movie
By C. K. Lyons on 06-27-19
By: Tommy Lee, Vince Neil, Nikki Sixx, Mick Mars
Narrated by: Sebastian York, Roger Wayne, Fred Berman, MacLeod Andrews, Hillary Huber
5 out of 5 stars 2,865 ratings