September 23, 2015 GOP primary, National, Politics.
Carly Fiorina, coming off a strong debate performance, this week rose to second place in the field of GOP presidential candidates, according to a CNN/ORC poll. Donald Trump with 24 percent still is the frontrunner. But Fiorina's 15 percent is five times the 3 percent support she showed in polls earlier this month. That sharp rise is causing many people and pundits to take a more serious look at Fiorina, the only woman in the Republican presidential field.
Former Texas Comptroller Susan Combs shares her opinion about why Fiorina is the best choice for Republicans, who need a candidate who's got game and gravitas to match the likely Democratic presidential nominee, Hillary Clinton. And New York Times columnist Charles Blow gives an opposing view of Fiorina, saying that she might be the flavor of the moment, but the more people know about her, the less likely they are to support her.
"Carly is not only the Republican Party's best rebuttal to Hillary Clinton, she also best reflects what conservatives are at heart: experienced, thoughtful and honest. That is why she presents such a compelling contrast to Donald Trump and I think he knows it.
This amount of drooling should come with a year's supply of bibs.
It is true that she found a way to effectively engage and combat the Republican front-runner, something her male counterparts had struggled to do. Good for her.
But before the fawning frenzy spins out of control, let's take a breath and an honest look at whom it is that is being cheered. . .
Blow provides different media descriptions of Fiorina, including one by Adele M. Stan, which Blow says captured Fiorina perfectly in a blistering assessment in The American Prospect.
Clearly a match-up between Clinton and Fiorina would be a terrific one as both are gifted campaigners. But there's a long way to go and still many GOP and several Democratic candidates to sift through before a Clinton-Fiorina contest is a reality. And that would not be reality TV, but historic.


text:
\section{Introduction}

 Let $(\mathbb{V},\omega)$ be a finite dimensional real symplectic  vector space and let
   $\vartheta: Mp(\mathbb{V},\omega)\to Sp(\mathbb{V},\omega)$ denote the $2:1$ covering of the symplectic group $G = Sp(\mathbb{V},\omega)$ by the metaplectic group $\tilde{G} = Mp(\mathbb{V},\omega).$ Let us denote the the maximal compact subgroup of $\tilde{G}$ by $K,$ and the complexification of
the Lie algebra of $G$  by $\mathfrak{g}.$  The complexification of the Lie algebra of the metaplectic group $\tilde{G}$ is isomorphic to $\mathfrak{g}$
 and thus, we shall denote it by the same letter.

 There exists a distinguished unitary  and faithful representation of the metaplectic group $\tilde{G}$ -- the so called
Segal-Shale-Weil representation.\footnote{Let us note that also the names oscillatory,  metaplectic or symplectic spinor representation are used in the literature.} We shall denote it by $\bf S$ in this text. Its elements  are usually called symplectic spinors. For a justification
 of this name, see  Kostant \cite{Kostant}.
Let $\mathbb{S}$ denote the underlying
Harish-Chandra $(\mathfrak{g},K)$-module of
$\bf S.$ When we   consider this $(\mathfrak{g},K)$-module with its $\mathfrak{g}$-module structure only, we call it the {\it oscillatory module}.
Notice that it is known that $\mathbb{S}$ splits into two irreducible submodules $\mathbb{S} \simeq \mathbb{S}_+\oplus \mathbb{S}_-.$

 Let us set $\mathbb{W}=\bigwedge^{\bullet}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}$ and consider the
appropriate tensor product representation of $\mathfrak{g}$ on $\mathbb{W}$  denoting it by $\rho$.
 In this paper, we first decompose the module $\mathbb{W}$ into irreducible
$\mathfrak{g}$-modules.

We shall also find generators of the
commutant algebra $$\mbox{End}_{\mathfrak{g}}(\mathbb{W})=\{T\in \mbox{End}(\mathbb{W}) \, | \,  T \rho(X)=\rho(X)T \mbox{ for all } X\in \mathfrak{g}\}$$
of the symplectic Lie algebra $\mathfrak{g}$ acting on $\mathbb{W}.$
 Let $p_{\pm}: \mathbb{S} \to \mathbb{S}_{\pm}$ be the unique $\mathfrak{g}$-equivariant projections.
These projections induce projection operators acting on the whole $\mathbb{W}$ in an obvious way. We keep denoting them by $p_{\pm}.$
We will introduce a representation $\sigma: \mathfrak{osp}(1|2)\to \mbox{End}(\mathbb{W})$
of the ortho-symplectic  super Lie  algebra $\mathfrak{osp}(1|2)$ on $\mathbb{W}$ and prove that
the image of $\sigma$ together with $p_+$ and $p_-$ generate the commutant.

At the end we decompose the $(\mathfrak{g}\times \mathfrak{osp}(1|2))$-module $\mathbb{W}$
into a direct sum $\bigoplus_{j=0}^l (\mathbb{E}_{jj}^-\oplus \mathbb{E}_{jj}^+) \otimes \mathbb{G}^{j},$
where $\mathbb{E}_{jj}^{\pm}$ is a certain irreducible infinite dimensional highest weight $\mathfrak{g}$-module
and $\mathbb{G}^j$ is an irreducible finite dimensional $\mathfrak{osp}(1|2)$-module. This establishes a Howe type duality in this case.

 The basic tool used to obtain the results was the decomposition of the $\mathfrak{g}$-module $\mathbb{W}$ into irreducible summands. To obtain this decomposition,
a theorem of Britten, Hooper, Lemire \cite{BHL} on the decomposition of the tensor product of an irreducible finite dimensional
$\mathfrak{sp}(\mathbb{V}^{\mathbb{C}},\omega)$-module and the oscillatory module $\mathbb{S}$ was crucial.

To give a 'pure' context of our research, let us remark that the Howe dualities are generalizations of the classical results of  Schur and Weyl. Whereas Schur studied the case of $GL(\mathbb{V})$ acting on the $k$-fold product $\bigotimes^k \mathbb{V},$ Weyl considered the $SO(\mathbb{V})$-module $\bigotimes^k \mathbb{V},$ $k\in \mathbb{N}.$
See Howe \cite{Howe} for a historical treatment of the cases studied by Schur and Weyl and for some of their generalizations
(examples of what one calls  Howe type dualities nowadays). In Howe \cite{Howe}, one can also find  several applications of
these dualities and a classical version of our $2:1$ duality. (The term $2:1$ will be explained below.)

  Let us also remark that a result related to the one presented here, is the result of Slupinski in \cite{Slupinski}. In his paper, Slupinski considers the case of (orthogonal) spinor valued exterior forms as a module over the appropriate spin group. Roughly speaking, he proves that $\mathfrak{sl}(2,\mathbb{C})$ is the
Howe dual partner to the spin group. One may rephrase this fact by saying that the situation studied in Slupinski is super symmetric to the one we are studying here.

 The motivation for our study of the Howe duality for exterior forms with values in the oscillatory module (i.e., for $\mathbb{W}$) comes from differential geometry and mathematical physics.  The duality we have proved here, makes it possible to do many computations in the so called symplectic spin geometry without using coordinates.  In this way, it sheds light into these structures. See, e.g., Habermann, Habermann \cite{HH} or Kr\'ysl \cite{SK2} for more details and examples.

For applications of symplectic spinors in mathematical physics we refer an interested reader to Shale \cite{Shale}
 where they were used in a quantization procedure for Klein-Gordon fields, and to Kostant \cite{Kostant}
 who used them in geometric quantization of quantum mechanics.

 In the second section of the paper, we introduce basic notation,  summarize  basic facts on the oscillatory module and
derive the mentioned decomposition of $\mathbb{W} = \bigwedge^{\bullet}(\mathbb{V}^{*})^{\mathbb{C}} \otimes \mathbb{S}$
into irreducible $\mathfrak{g}$-submodules (Theorem 3). The generators of $\mbox{End}_{\mathfrak{g}}(\mathbb{W})$  are given in the
third section (Theorem 8). In that section,
the representation $\sigma: \mathfrak{osp}(1|2)\to \mbox{End}(\mathbb{W})$ is introduced and the fact,
 that it is a representation is proved (Theorem 9).
 In the fourth section, the space $\mathbb{W}$ is  explicitly decomposed into irreducible
 submodules with respect to the joint action of $\mathfrak{g}$ and $\mathfrak{osp}(1|2)$  (Theorem 13).

\section{Decomposition of $\bigwedge^{\bullet}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}$}

Suppose that $\mathfrak{g}$ is a simple complex Lie algebra. Choose a Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{g}$ and a set of positive roots $\Phi^+.$ We denote the complex irreducible highest weight $\mathfrak{g}$-module with a highest weight $\mu \in \mathfrak{h}^*$ by $L(\mu).$ If $\mu$ happens to be dominant and integral with respect to the choice
$(\mathfrak{h}, \Phi^+),$ we denote $L(\mu)$ by $F(\mu)$ emphasizing the fact that the module $L(\mu)$ is finite dimensional.  For an integral dominant weight $\mu$ with respect to $(\mathfrak{h},\Phi^+),$ we denote the set of all weights of the irreducible representation $F(\mu)$ by $\Pi(\mu).$

Now, let us restrict our attention to the studied symplectic case.
Consider a $2l$ dimensional real symplectic vector space
$(\mathbb{V},\omega).$ Let $\mathbb{V} = \mathbb{L}\oplus \mathbb{L}'$ be a direct sum decomposition of the vector space $\mathbb{V}$ into two Lagrangian subspaces $\mathbb{L}$ and $\mathbb{L}'$ of $(\mathbb{V},\omega).$ Let $\{e_i\}_{i=1}^{2l}$ be an adapted  symplectic basis of $(\mathbb{V},\omega),$ i.e., $\{e_i\}_{i=1}^{2l}$ is a symplectic basis of $(\mathbb{V},\omega)$ and $\{e_i\}_{i=1}^{l}\subseteq \mathbb{L}$ and $\{e_{i}\}_{i=l+1}^{2l}\subseteq \mathbb{L}'.$ Because the notion of a symplectic basis is not unique, let us fix one for a later use.
We call a basis $\{e_i\}_{i=1}^{2l}$ of $\mathbb{V}$ symplectic basis of $(\mathbb{V},\omega)$ if for $\omega_{ij}=\omega(e_i,e_j),$ we have
\begin{itemize}
\item[] $\omega_{ij}=1$ if an only if $i\leq l$ and $j=i+l,$
\item[] $\omega_{ij}=-1$ if and only if $i>l$ and $j=i-l$ and
\item[] $\omega_{ij}=0$ in other cases.
\end{itemize}
For $i,j =1,\ldots, 2l,$ let us define numbers $\omega^{ij}$ by $\sum_{k=1}^{2l}\omega_{ik}\omega^{jk}=\delta_i^j.$
The basis of $\mathbb{V}^*$ dual to the basis $\{e_i\}_{i=1}^{2l}$
will be denoted by $\{\epsilon^i\}_{i=1}^{2l}.$

Let us denote the symplectic group $Sp(\mathbb{V},\omega)$ by $G.$ In this text, the metaplectic group $Mp(\mathbb{V},\omega)$ is denoted by $\tilde{G}$ and the
appropriate two-fold covering map $\tilde{G} \to G$ by $\vartheta.$
For some technical reasons, let us suppose that $l\geq 2,$  and consider this choice holds throughout the rest of this article.  We shall denote the complexification of the Lie algebra of $(\mathbb{V},\omega),$ i.e., the Lie algebra  $\mathfrak{sp}(\mathbb{V}^{\mathbb{C}},\omega)$ by $\mathfrak{g}.$
(The complexified symplectic form on $\mathbb{V}^{\mathbb{C}}$
will still be denoted by $\omega$.) If we choose a Cartan subalgebra
$\mathfrak{h}\subseteq \mathfrak{g}$ and a
system of positive roots $\Phi^+ \subseteq
\mathfrak{h}^*,$ then the set of fundamental weights
$ \{ \varpi_i \}_{i=1}^l$ is uniquely determined. Further this set
determines a basis $\{\epsilon_i\}_{i=1}^l$ of
$\mathfrak{h}^*$ given by
$\varpi_i=\sum_{j=1}^i\epsilon_j$ for $i=1,\ldots, l.$\footnote{Let
us mention that we are distinguishing the sets
$\{\epsilon^i\}_{i=1}^{l} \subseteq \mathbb{V}^*$ and
$\{\epsilon_i\}_{i=1}^l \subseteq \mathfrak{h}^{*}$ from each other,
although we could have identified them by a choice of the first
inclusion from the chain of inclusions $\mathfrak{h}\subseteq \mathfrak{g} \subseteq (\mathbb{V}^{*}\otimes \mathbb{V})^{\mathbb{C}}$ and
a set of positive roots.}

For $\mu=\sum_{i=1}^l\mu_i\epsilon_i,$ we shall often denote $L(\mu)$ by $L(\mu_1,\ldots,\mu_l)$ simply.

The Segal-Shale-Weil representation is a faithful unitary representation of the metaplectic group
$Mp(\mathbb{V},\omega)$ on the complex vector  space $L^2(\mathbb{L})$ of complex valued square Lebesgue integrable functions defined on $\mathbb{L}.$ Because we would like to omit problems caused by dealing with unbounded operators, we shall consider the underlying Harish-Chandra $(\mathfrak{g},K)$-module of the Segal-Shale-Weil representation.
This module considered as a $\mathfrak{g}$-module only will be called the {\it oscillatory module} and the appropriate representation
$$L: \mathfrak{g} \to \mbox{End}(\mathbb{S})$$ will be called the {\it oscillatory representation} (see Howe \cite{Howe}).
One can prove that  $\mathbb{S}_+ \simeq
L(\mu^0)$  and $\mathbb{S}_-\simeq L(\mu^1)$ where $\mu^0= -\frac{1}{2}\varpi_l$ and $\mu^1=\varpi_{l-1}-\frac{3}{2}\varpi_l.$  Let us notice that as a vector space
$\mathbb{S}_+$ is isomorphic to the space of even polynomials in $\mathbb{C}[z^1,\ldots, z^l]$ and $\mathbb{S}_-$ to the space of the odd ones.
For more information on the Segal-Shale-Weil representation, see
Weil \cite{Weil} and Kashiwara, Vergne \cite{KV}. For information on the oscillatory module, see Britten, Hooper, Lemire \cite{BHL}.

In order to derive the studied Howe type duality, we shall need the symplectic Clifford multiplication $\mathbb{V}^{\mathbb{C}}\times \mathbb{S} \to \mathbb{S}$, which enables us to multiply symplectic spinors by  element from $\mathbb{V}^{\mathbb{C}}.$ It is given by the following prescription
\begin{eqnarray}
(e_i.f)(x)=\frac{\partial f}{\partial x^i}(x), \, \mbox{ } (e_{i+l}.f)(x)=\imath x^i f(x),\, i=1,\ldots, l \label{sc}
\end{eqnarray} where $x = \sum_{i=1}^lx^ie_i \in \mathbb{L}, f\in \mathbb{S}$ and it is extended linearly to the whole $\mathbb{V}^{\mathbb{C}}.$
(The symplectic Clifford multiplication is basically the canonical  quantization prescription.)

Now, for $i = 0, 1$ and a dominant integral weight $\mu=\sum_{j=1}^l\mu_j \varpi_j \in \mathfrak{h}^{*},$ let us introduce two sets $T^i_{\mu}$.  A weight
$\mu' \in \mathfrak{h}^*$ is an element of $T^i_{\mu}$ if and only if the numbers $d_j,$ $j=0,\ldots, l,$ defined by
$\mu'-\mu = \sum_{j=1}^ld_j\epsilon_j,$ satisfy the following  conditions
\begin{itemize}
\item[1)] $d_j + \delta_{l,j}\delta_{1,i} \in \mathbb{N}_0$ for $j=0,\ldots,l$
\item[2)] $0\leq d_j \leq \mu_j$ for $j=0,\ldots, l - 1$ and  $0 \leq d_l +\delta_{1,i} \leq 2\mu_l +1$
\item[3)] $\sum_{j=1}^l d_j$ is even.
\end{itemize}

In what follows, we shall  need a result on the decomposition of the tensor product of a finite dimensional $\mathfrak{g}$-module with one of the modules $L(\mu^i),$ $i=0,1,$ into irreducible submodules.  This result was published in Britten, Hooper, Lemire \cite{BHL}. Because there is a misprint in the cited article, let us  write the corrected result explicitly here.

{\bf Theorem 1:}
For $i=0,1$ and an integral dominant weight $\mu$ with respect to $(\mathfrak{h},\Phi^+),$ we have
$$F(\mu) \otimes L(\mu^i) \, \simeq  \bigoplus_{\kappa \in T_{\mu}^i \cap \Pi(\mu)}L(\mu_i + \kappa).$$

{\it Proof.} See Britten, Hooper, Lemire \cite{BHL}. The mentioned misprint occurs in the definition of $T_{\mu}^1$ in that paper (private communication with F. W. Lemire). $\Box$

For convenience let us introduce a function $\mbox{sgn}:\{+,-\}\to \{0,1\}$ given by the prescription
$\mbox{sgn}(+)=0$ and $\mbox{sgn}(-)=1.$  Following modules will be used in the decomposition theorem
$$\mathbb{E}_{ij}^{\pm}:=L(\underbrace{\frac{1}{2},\cdots,\frac{1}{2}}_{j},\underbrace{-\frac{1}{2},\cdots,
-\frac{1}{2}}_{l-j-1},-1+\frac{1}{2}(-1)^{i+j+\mbox{sgn}(\pm)})$$ where $i=0,\ldots,l-1,$   $j=0,\ldots, i$ and $i=l,$   $j=0,\ldots, l-1.$ For $i=j=l,$
we set $\mathbb{E}^+_{ll}:=L(\frac{1}{2}, \cdots, \frac{1}{2})$ and $\mathbb{E}^-_{ll}:=L(\frac{1}{2}, \cdots, \frac{1}{2}, -\frac{5}{2}).$
For $i=l+1,\ldots,2l$ and $j=0,\ldots, 2l-i,$ we set $\mathbb{E}_{ij}^{\pm}:=\mathbb{E}_{(2l-i)j}^{\pm}.$
In order to write the results  as short as possible, let us introduce the following notation.
For $i=0,\ldots, l,$ let us define $m_i=i$ and for $i=l+1,\ldots, 2l,$   $m_i=2l-i.$ With these conventions, let  us set
$$\Xi=\{(i,j)|i=0,\ldots, 2l; j= 0,\ldots, m_i\} \, \mbox{  and further  }
 \mathbb{E}_{ij}^{\pm} = 0 \mbox{ for } (i,j)\in \mathbb{Z}^2 \setminus \Xi.$$

Now, let us derive the next

{\bf Lemma 2:} For $r=1,\ldots, l,$
we have $$\Pi(\varpi_r) \supseteq \{\sum_{s=1}^r\pm\epsilon^{i_s}|1\leq i_1 < \ldots < i_r  \leq l\}.$$

{\it Proof.} According to the Corollary  5.1.11. pp. 237 and Theorem 5.1.8. (3) pp. 236 in Goodman, Wallach \cite{GW}, the
$\mathfrak{g}$-module $F(\varpi_r)$ is isomorphic to the linear span
of isotropic $r$-vectors in $\mathbb{V}^{\mathbb{C}}$, i.e., multi-vectors $v=f_1\wedge\ldots\wedge f_r,$ where $\omega(f_i,f_j)=0$ for $i,j=1,\ldots, l,$ $r=1,\ldots, 2l.$ Second, it is easy to realize that one can choose a Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{g}$ in a way that the following is true. For $i=1,\ldots, l,$ the basis vector $e_i \in \mathbb{V}^{\mathbb{C}}$ is a weight vector of the weight $\epsilon_i$ and the  vector $e_{i+l}$ is a  weight vector of the weight $-\epsilon_i,$ both for the defining representation $\mathfrak{g}$ on $\mathbb{V}^{\mathbb{C}}.$ Using this, the result follows easily.
$\Box$

Now, we  introduce the module $\mathbb{W}$ we shall be mostly concern with in this paper. As a vector space
$$\mathbb{W}=\bigwedge^{\bullet}(\mathbb{V}^{*})^{\mathbb{C}}\otimes \mathbb{S}.$$
The representation $ \rho: \mathfrak{g} \to \mbox{End}(\mathbb{W})$ of $\mathfrak{g}$ on ${\mathbb W}$ we are interested in, is defined by the prescription $$\rho(X)(\alpha \otimes s):= \mbox{ad}(\mbox{X})^{* \wedge r}(\alpha) \otimes L(X)s$$ for
 $X\in \mathfrak{g},$ $\alpha\in \bigwedge^r (\mathbb{V}^*)^{\mathbb{C}},$ $r=0,\ldots, 2l$ and $s\in \mathbb{S}.$ The module $\mathbb{W}$ will be   called the
module of (exterior) forms with values in symplectic spinors.

Now, we can state the decomposition theorem. The proof of this theorem is based on a direct application of Theorem 1 and Lemma 2.

{\bf Theorem 3:} For $i=0,\ldots,2l,$ the following decomposition into irreducible $\mathfrak{g}$-modules
$$\bigwedge^i (\mathbb{V}^{*})^{\mathbb{C}}\otimes \mathbb{S}_{\pm} \simeq \bigoplus_{j, (i,j)\in\Xi} \mathbb{E}_{ij}^{\pm} \quad \mbox{ holds.}$$

{\it Proof.}
Using Theorem 5.1.8. pp. 236 and Corollary 5.1.9. pp. 237 in Goodman, Wallach \cite{GW}, we have for
$i=2k,$ $k\in \mathbb{N}_0$
\begin{eqnarray}
\bigwedge^i (\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}_{\pm}=(F(\varpi_0)\oplus F(\varpi_2)\oplus \ldots \oplus F(\varpi_i))\otimes \mathbb{S}_{\pm}. \label{1}
\end{eqnarray}
Using the cited theorems in Goodman, Wallach \cite{GW}  again, we have have for $i= 2k+1,$ $k \in \mathbb{N}_0$
\begin{eqnarray}
\bigwedge^i (\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}_{\pm}=(F(\varpi_1)\oplus F(\varpi_3)\oplus \ldots \oplus F(\varpi_i))\otimes \mathbb{S}_{\pm}. \label{2}
\end{eqnarray}

We shall consider the mentioned tensor products  for $i=0,\ldots,l$ only, because the result for  $i=l+1,\ldots,2l$ follows from that one for $i=0,\ldots,l$ immediately
due to the  $\mathfrak{g}$-isomorphism $\bigwedge^i (\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}_{\pm} \simeq \bigwedge^{2l-i} (\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}_{\pm}$ and the definition of $\mathbb{E}_{ij}^{\pm}$ for
$i=l+1,\ldots,2l$ and $j=0,\ldots, m_i.$

Let us consider the tensor products by $\mathbb{S}_+$ and $\mathbb{S}_-$ separately.

\begin{itemize}
\item[1)] First, let us consider the tensor product $\bigwedge^i (\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}_+$
for $i=0,\ldots,l.$

Using Lemma 2 and Theorem 1, we can easily compute that for $j=1,\ldots, l,$
$T_{\varpi_j}^0=\{\epsilon_1+\ldots+\epsilon_j, \epsilon_1 + \ldots +\epsilon_{j-1}- \epsilon_l\} \subseteq \Pi(\varpi_j)$ and thus,
$F(\varpi_j)\otimes \mathbb{S}_+=L(\underbrace{\frac{1}{2},\ldots,\frac{1}{2}}_j,\underbrace{-\frac{1}{2},\ldots,-\frac{1}{2}}_{l-j})\oplus
L(\underbrace{\frac{1}{2},\ldots,\frac{1}{2}}_{j-1},\underbrace{-\frac{1}{2},\ldots,-\frac{1}{2}}_{l-j},-\frac{3}{2})$ using the relation $\omega_j=\sum_{i=1}^j\epsilon_i.$
Summing up these terms according to (\ref{1}) and (\ref{2}), we obtain the statement of the theorem for both of the cases $i$  is odd and $i$ is even.

\item[2)]  Now, let us consider the tensor product $\bigwedge^i(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}_-$ for $i=0,\ldots,l.$

 Using Lemma 2, we can easily compute that for $j=1,\ldots, l-1,$ we have
$T_{\varpi_j}^1=\{\epsilon_1+\ldots+\epsilon_j,\epsilon_1+\ldots+\epsilon_{j-1}+\epsilon_l\}\subseteq \Pi(\omega_j)$ and $T^1_{\varpi_l}=\{\epsilon_1+\ldots + \epsilon_l, \epsilon_1+\ldots + \epsilon_{l-1}-\epsilon_l\}\subseteq \Pi(\varpi_l).$
Therefore using Theorem 1, we have
$F(\omega_j)\otimes \mathbb{S}_-=L(\underbrace{\frac{1}{2},\ldots,\frac{1}{2}}_{j-1},\underbrace{-\frac{1}{2},\ldots,-\frac{1}{2}}_{l-j+1})\oplus L(\underbrace{\frac{1}{2},\ldots,\frac{1}{2}}_{j},\underbrace{-\frac{1}{2},\ldots,-\frac{1}{2}}_{l-j-1},-\frac{3}{2})$ for $j=1,\ldots, l-1.$ For $j=l,$ we obtain
$F(\varpi_l)\otimes \mathbb{S}_+=L(\frac{1}{2} \ldots \frac{1}{2} -\frac{3}{2})\oplus L(\frac{1}{2} \ldots \frac{1}{2}-\frac{5}{2})$ using Theorem 1 again.
Summing up these terms according to (\ref{1}) and (\ref{2}), we obtain the statement of the theorem for both cases
$i$ is odd and $i$ is even.
\end{itemize}
$\Box$

{\bf Remark:} Due to Theorem 3 and definitions of $\mathbb{E}_{ij}$, we know that for $i=0,\ldots, 2l$ the $\mathfrak{g}$-module $\bigwedge^{i}(\mathbb{V}^{*})^{\mathbb{C}}\otimes \mathbb{S}$ is
multiplicity-free.

\section{The commutant $\hbox{End}_{\mathfrak{g}}(\mathbb{W})$ and its supersymmetric realization}

In this section, we shall prove that the associative algebra $\hbox{End}_{\mathfrak{g}}(\bigwedge^{\bullet}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S})$ is generated  by the below introduced four explicitly given elements (a "raising", a "lowering" operator and two projections).

 For $r=0,\ldots, 2l$ and $\alpha \otimes s = \alpha \otimes (s_+, s_-) \in \bigwedge^r
(\mathbb{V}^*)^{\mathbb{C}}\otimes (\mathbb{S}_+ \otimes \mathbb{S}_-),$  we set
$$F^+:\bigwedge ^{r}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S} \to
\bigwedge^{r+1}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S},\, F^+(\alpha \otimes
s)=\frac{\imath}{2}\sum_{i=1}^{2l}\epsilon^i\wedge \alpha \otimes e_i.s,$$
$$F^-:\bigwedge ^{r}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S} \to \bigwedge ^{r-1}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}, \,
F^-(\alpha \otimes s)=\frac{1}{2}\sum_{i=1}^{2l}\omega^{ij}\iota_{e_i}\alpha \otimes
e_j.s,$$
$$p_{\pm}:\bigwedge ^{r}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S} \to \bigwedge ^{r}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}, \, p_{\pm}(\alpha\otimes s)=\alpha \otimes s_{\pm}$$
and extended linearly to the whole $\mathbb{W}.$ Next, consider the operator $H$ defined by the formula
$$H=2 (F^+ F^- + F^-F^+).$$

 The values of the operator
$H$ are determined in the next

{\bf Lemma 4:} Let $(\mathbb{V},\omega)$ be a
symplectic vector space of dimension $2l.$   Then for   $r=0,\ldots, 2l,$ we have
$$H_{|\bigwedge^r(\mathbb{V}^*)^{\mathbb{C}} \otimes \mathbb{S}}=\frac{1}{2}(r-l)\hbox{Id}_{|\bigwedge^r(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}}.$$

{\it Proof.} The proof is straightforward, see Kr\'ysl \cite{SK}. $\Box$

{\bf Lemma 5:} The maps $F^{\pm}, p_{\pm}$ and $H$ are $\mathfrak{g}$-equivariant
with respect to the representation $\rho$ of $\mathfrak{g}$ on $\mathbb{W}.$

{\it Proof.} The operators $p_{\pm}$ are clearly $\mathfrak{g}$-equivariant. The $\mathfrak{g}$-equivariance of $F^{\pm}$ and $H$
can be verified straightforwardly. See Kr\'ysl \cite{SK} for a proof.
$\Box$

{\bf Definition 1:} Let us denote the associative algebra generated by $F^{\pm}$ and $p_{\pm}$   by $\mathfrak{C}$.

Due to the previous lemma, we already know that $\mathfrak{C} \subseteq \mbox{End}_{\mathfrak{g}}(\mathbb{W}).$  Now, we shall prove that $\mathfrak{C}$ exhausts the whole
commutant $\mbox{End}_{\mathfrak{g}}(\mathbb{W}).$ For convenience, let us define the following two subsets $\Xi_+$ and $\Xi_-$ of the set $\Xi.$ We define $\Xi_-=\Xi \setminus \{(j,2l-j)| j=l,\ldots, 2l\}$ and $\Xi_+=\Xi \setminus \{(j,j)| j=0,\ldots, l\}.$ The sets $\Xi_+$ and $\Xi_-$ can be visualized as the left hand side and the right hand part of the 'triangular shape' of $\Xi,$ respectively.

{\bf Lemma 6:} For each $(i,j)\in \Xi,$ we have
\begin{eqnarray*}
F^{+}_{|\mathbb{E}_{ij}^{\pm}}&:& \mathbb{E}_{ij}^{\pm}\stackrel{\sim}{\rightarrow}
\mathbb{E}_{i + 1,j}^{\mp} \qquad \mbox{ if } (i,j) \in \Xi_{-} \mbox{ and}\\
F^{-}_{|\mathbb{E}_{ij}^{\pm}}&:& \mathbb{E}_{ij}^{\pm}\stackrel{\sim}{\rightarrow}
\mathbb{E}_{i - 1,j}^{\mp} \qquad \mbox{ if } (i,j) \in \Xi_{+}.
\end{eqnarray*}

{\it Proof.} First, let us prove by induction the following formulas. For $(i,j) \in \Xi$ we have
\begin{eqnarray} \label{formule}
 F^-F^+_{|\mathbb{E}_{ij}}= \left\{\begin{array}{l}
                                                       \frac{1}{4}\left(\frac{1+i-j}{2}\right)\mbox{Id}_{|\mathbb{E}_{ij}} \quad  \hbox{ if } \ i+j \hbox{ is odd}   \hbox{,} \\
                                                      \frac{1}{4}\left(\frac{i+j}{2} -l\right)\mbox{Id}_{|\mathbb{E}_{ij}}  \quad {\hbox{if} \  i+j \hbox{ is even.}
                                                       }
                                                       \end{array}
                                                       \right.
                                                       \end{eqnarray}

Let us fix an integer $j \in \{0,\ldots, l\}$ and proceed by the induction on the form degree $i.$
\begin{itemize}
\item[I.] For $i =j $  and $\phi \in \mathbb{E}_{ii},$ let us compute $F^-F^+\phi = (\frac{1}{2} H -  F^+F^-)\phi= \frac{1}{4}(i-l)\phi- F^+F^-\phi$ due to the definition of $H$ and Lemma 4. We have $F^-_{|\mathbb{E}_{ii}}=0$ because $F^-$ is $\mathfrak{g}$-equivariant (Lemma 5), lowers the form degree by one and there is no summand isomorphic
to $\mathbb{E}_{ii}$ in $\bigwedge^{i-1}(\mathbb{V}^{*})^{\mathbb{C}}\otimes \mathbb{S}$
 (see the Theorem 3). Summing up, we have $F^-F^+\phi=\frac{1}{4}(i-l)$ according to (\ref{formule}).
\item[II.] Now, let us suppose the statement is true for $(i,j) \in \Xi,$ $i+j$ is odd. For $(i+1,j)\in \Xi$ and $\phi \in \mathbb{E}_{i+1,j},$ let us compute $F^-F^+\phi = \frac{1}{2} H \phi - F^+F^-\phi = \frac{1}{4}(i+1-l)\phi - F^+F^-\phi$  due to the definition of $H$ and Lemma 4.
Using the induction hypothesis, we  have that $F^-F^+_{|\mathbb{E}_{ij}}=\frac{1}{4}(\frac{1+i-j}{2})\mbox{Id}_{|\mathbb{E}_{ij}}.$ Thus $F^+_{|\mathbb{E}_{ij}}$ is injective. Because $F^+$ is $\mathfrak{g}$-equivariant,
raises the form degree by one and there is no other summand in $\bigwedge^{i+1}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}$ isomorphic to $\mathbb{E}_{ij}$ than $\mathbb{E}_{i+1,j},$ we see that $F^+_{|\mathbb{E}_{ij}}: \mathbb{E}_{ij} \to \mathbb{E}_{i+1,j}.$ Because of the proved injectivity and the Schur lemma for complex highest weight modules (see Diximier \cite{Dixmier}), we know that $F^{+}_{|\mathbb{E}_{ij}}$ is actually an isomorphism. Thus there exists $\tilde{\phi}\in \mathbb{E}_{ij}$ such that $\phi=F^+\tilde{\phi}.$
We may compute $F^+F^-\phi=F^+F^-(F^+\tilde{\phi})=F^+(F^-F^+\tilde{\phi})=\frac{1}{4}(\frac{1+i-j}{2})F^+\tilde{\phi}=\frac{1}{4}(\frac{1+i-j}{2})\phi$ by the induction hypothesis. Substituting this relation into the already derived $F^-F^+\phi=\frac{1}{4}(i+1-l)\phi - F^+F^-\phi,$ we get $F^-F^+\phi = \frac{1}{4}(i+1-l)\phi - \frac{1}{4}(\frac{1+i-j}{2})\phi= \frac{1}{4}(\frac{i +1 + j}{2}-l)\phi$ according to the formula (\ref{formule}).

Now, let us suppose the statement is true for $(i,j) \in \Xi$ and $i+j$ is even.  For $(i+1,j)\in \Xi$ and $\phi \in \mathbb{E}_{i+1,j},$ we can compute $F^-F^+\phi=\frac{1}{2}H\phi-F^+F^-\phi = \frac{1}{4}(i+1-l)\phi-F^+F^-\phi$ due to the definition of $H$ and  Lemma 4.
By parallel lines of reasoning, we get the existence of $\tilde{\phi} \in \mathbb{E}_{ij}$ such that $\phi = F^+\tilde{\phi}.$  Using the induction hypothesis, we may write $F^+F^-\phi=F^+F^-(F^+\tilde{\phi})=F^+(F^-F^+\tilde{\phi})=\frac{1}{4}(\frac{i-j +1}{2}-l)F^+\tilde{\phi}=\frac{1}{4}(\frac{i+j}{2}-l) \phi$ by the induction hypothesis. Substituting this expression into the computation above, we get $F^-F^+\phi=\frac{1}{4}(i+1-l)\phi - \frac{1}{4}(\frac{i+j}{2}-l)\phi =\frac{1}{4}(\frac{1 + (i +1) -j}{2})\phi.$ Thus, the formula follows.
\end{itemize}

Using the formula (\ref{formule}), we see that
$F^-F^+_{|\mathbb{E}_{ij}}$ is injective iff $i+j \neq 2l$ and $j \neq i+1$, i.e., $(i,j)\in \Xi_-,$ the second condition being empty. Especially, $F^+$ is injective for $(i,j)\in \Xi_-.$ Using the Schur lemma, we get that $F^+$ is an isomorphism. From this, we also know that $F^-$ is injective on the image of $F^+,$ i.e., it is an isomorphism for $(i,j)\in \Xi_+.$
$\Box$

{\bf Remark:} It is  easy to see that $F^-$ is zero when restricted to
$\mathbb{E}_{ii}^{\pm},$ $i=0,\ldots,l.$  Namely, we know that $F^-$ is lowering the form degree by one, it is a $\mathfrak{g}$-equivariant map and  there is no submodule of the module of symplectic spinor valued exterior forms of form degree  $i-1$ isomorphic to $\mathbb{E}_{ii}^{\pm}$ (see Theorem 3).
A similar discussion can be made for $F^+$ restricted to $\mathbb{E}_{im_i},$ $i=l,\ldots, 2l.$

For $(i,j)\in \Xi,$ let us denote the projection operators from the space $\mathbb{W}$ to the submodule
$\mathbb{E}_{ij}^{\pm}$ by $S_{ij}^{\pm},$ i.e.,
$$S_{ij}^{\pm}:\bigwedge^{\bullet}(\mathbb{V}^*)^{\mathbb{C}} \otimes \mathbb{S}_{\pm}\to \mathbb{E}_{ij}^{\pm} \subseteq \bigwedge^i (\mathbb{V}^*)^{\mathbb{C}} \otimes \mathbb{S}_{\pm}.$$

{\bf Lemma 7:} For each $(i,j)\in \Xi,$ the projection operators
$S_{ij}^{\pm}$ are elements of the associative algebra $\mathfrak{C} = \langle F^+, F^-, p_+, p_- \rangle.$

{\it Proof.} For each $i = 0, \ldots, 2l,$ let us  define the following operators $S_i^{\pm}:
\bigwedge^{\bullet}(\mathbb{V}^*)^{\mathbb{C}} \otimes \mathbb{S}_{\pm} \to
\bigwedge^{i}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}_{\pm}$ by the formula
$$S_i^{\pm}:=\left( \prod_{k=0, k\neq i}^{2l}\frac{2H-k+l}{i-k}\right) p_{\pm}.$$
Using Lemma 4, we see that the image of each $S_i^{\pm}$ lies in the
prescribed space and the normalization is correct. Recall that due to its definition, $H$ can be expressed using  the operators $F^{+}$ and
$F^-$ only and thus for $i=0,\ldots, 2l,$ the projection $S_i^{\pm}$ lies in
the algebra $\mathfrak{C}$. Now let us fix an integer $i\in \{0,\ldots, 2l\}.$ We prove that for
each $j$ such that $(i,j)\in \Xi,$ the projection
$S_{ij}^{\pm}:\bigwedge^{i}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}_{\pm} \to \mathbb{E}_{ij}^{\pm} \in \mathfrak{C}.$
  We proceed by
induction (on $j$).
\begin{itemize}
\item[I.] For $j=0,$ we can define $S_{i0}'':=(F^+)^{i}(F^-)^{i}.$ Using
the fact that applying the $F^-$ (or $F^{+}$) lowers (or raises) the form
degree by $1,$ we see that $S_{i0}'': \bigwedge^{i}(\mathbb{V}^*)^{\mathbb{C}}\otimes
\mathbb{S}_{\pm}\to \mathbb{E}_{i0}^{\pm}$. Using the Schur lemma
for complex irreducible highest weight modules (see Dixmier \cite{Dixmier}), we see that there exists a complex number $\lambda_{i0}\in
\mathbb{C}$ such that ${S_{i0}''}_{|\mathbb{E}^{\pm}_{i0}}=\lambda_{i0}\mbox{Id}_{|\mathbb{E}_{i0}^{\pm}}.$
Due to  Lemma 6, we know that
$\lambda_{i0}\neq 0.$ Thus defining
$S_{i0}^{\pm}:=\frac{1}{\lambda_{i0}}S_{i0}''\circ S_i^{\pm},$ we get a projection from $\bigwedge^{\bullet}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}_{\pm}$ onto
$\mathbb{E}_{i0}^{\pm}$ expressed as linear combinations of compositions of
$F^{\pm}$ and $p_{\pm}.$

\item[II.] Let us suppose that for $k=0,\ldots, j,$ the operators $S_{ik}^{\pm}$
can be written as linear combinations of compositions of the
operators $F^{\pm}$ and $p_{\pm}.$ Now, we shall use the
operators $S_{i0}^{\pm},\ldots, S_{ij}^{\pm}$ in order to define $S_{i,j+1}^{\pm}.$
Let us take an element $\xi \in \bigwedge^{i}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}_{\pm}$
and form an element $\zeta:= S_{i,j+1}'\xi:=\xi -
\sum_{k=0}^jS_{ik}^{\pm}\xi \in \bigoplus_{k=j+1}^{m_i} \mathbb{E}_{ik}^{\pm}.$ Now, form an element
$\zeta':=S_{i,j+1}''\zeta:=(F^+)^{i-j}(F^-)^{i-j}\zeta.$ In the
same way as in the item I., we conclude that $\zeta' \in \mathbb{E}_{i,j+1}^{\pm}.$
Form the homomorphism $S_{i,j+1}^{'''}:=S_{i,j+1}^{''}\circ
S_{i,j+1}'.$
Using the Schur lemma in the case of $S^{'''}_{i,j+1 |\mathbb{E}_{i,j+1}^{\pm}}: \mathbb{E}_{i,j+1}^{\pm} \to \mathbb{E}_{i,j+1}^{\pm},$ we
conclude that there is a complex number $\lambda_{i,j+1} \in
\mathbb{C}$ such that $S^{'''}_{i,j+1|\mathbb{E}_{i,j+1}^{\pm}}=\lambda_{i,j+1}
\mbox{Id}_{|\mathbb{E}_{i,j+1}^{\pm}}.$  Due to Lemma 6, we know that
$\lambda_{i,j+1}\neq 0.$ Now
$S_{i,j+1}^{\pm}:=\frac{1}{\lambda_{i,j+1}}S_{i,j+1}'''\circ S_i^{\pm}$ is the desired projection. Going through the construction back, we see that  we have used the operators $F^{\pm}$ and $p_{\pm}$ only. $\Box$
\end{itemize}

Now, we prove that the algebra $\mathfrak{C}$ exhausts the whole commutant $\mbox{End}_{\mathfrak{g}}(\mathbb{W}).$

{\bf Theorem 8:} We have $$\mbox{End}_{\mathfrak{g}}(\mathbb{W}) = \mathfrak{C}.$$

{\it Proof.}
Due to  Lemma 5, we know that $\mathfrak{C} \subseteq
\hbox{End}_{\mathfrak{g}}(\bigwedge^{\bullet}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}).$ We prove the opposite inclusion.
Obviously, we have $\mbox{End}_{\mathfrak{g}}(\mathbb{W})=
\mbox{End}_{\mathfrak{g}}(\bigwedge^{\bullet}{(\mathbb{V}^*)}^{\mathbb{C}}\otimes \mathbb{S}_+)\oplus
\mbox{End}_{\mathfrak{g}}(\bigwedge^{\bullet}{(\mathbb{V}^*)}^{\mathbb{C}}\otimes \mathbb{S}_-)\oplus
\mbox{Hom}_{\mathfrak{g}}(\bigwedge^{\bullet}{(\mathbb{V}^*)}^{\mathbb{C}}\otimes \mathbb{S}_+,\bigwedge^{\bullet}{(\mathbb{V}^*)}^{\mathbb{C}}\otimes \mathbb{S}_-)
\oplus \mbox{Hom}_{\mathfrak{g}}(\bigwedge^{\bullet}{(\mathbb{V}^*)}^{\mathbb{C}}\otimes \mathbb{S}_-,\bigwedge^{\bullet}{(\mathbb{V}^*)}^{\mathbb{C}}\otimes \mathbb{S}_+).$ Let us
prove $\mbox{Hom}_{\mathfrak{g}}(\bigwedge^{\bullet}{(\mathbb{V}^*)}^{\mathbb{C}}\otimes \mathbb{S}_+,\bigwedge^{\bullet}{(\mathbb{V}^*)}^{\mathbb{C}}\otimes \mathbb{S}_-)\subseteq \mathfrak{C}$ only.
Due to the structure of the decomposition of $\bigwedge^{\bullet}(\mathbb{V}^{*})^{\mathbb{C}}\otimes \mathbb{S}$  (Theorem 3), we see that
$\hbox{Hom}_{\mathfrak{g}}(\bigwedge^{\bullet}{(\mathbb{V}^*)}^{\mathbb{C}}\otimes \mathbb{S}_+,
\bigwedge^{\bullet}{(\mathbb{V}^*)}^{\mathbb{C}}\otimes \mathbb{S}_-)\simeq \bigoplus_{(i,j), (k,m)\in \Xi}
\hbox{Hom}_{\mathfrak{g}}(\mathbb{E}_{ij}^+, \mathbb{E}_{km}^-).$
Regarding the highest weights of the irreducible $\mathfrak{g}$-modules $\mathbb{E}_{ij}^{+}$ and $\mathbb{E}_{km}^{-}$ and using the
 the   Schur lemma, we see that the following is true. If there exists a non trivial $\mathfrak{g}$-equivariant map between $\mathbb{E}_{ij}^+$ and $\mathbb{E}_{km}^{-},$ then $j=m$ and $k= 2r+i+1$ for some $r \in \mathbb{Z}.$
Suppose $r\geq 0.$
Due to Lemma 7,  Lemma 6 and the remark below this lemma,
we know that $(F^{+})^{1+2r}S_{ij}^{+}: \mathbb{E}_{ij}^+
\to \mathbb{E}_{i+2r+1,j}^-$  for each $r\in \mathbb{N}_0$ and $(i,j)\in \Xi.$
Due to  Lemma 7 again, we know that the mapping is nontrivial whenever $(i,j)\in \Xi_+$ and $(i+2r+1,j) \in \Xi.$
Thus $\mathbb{C}(F^{+})^{1+2r}S_{ij}^+=\mbox{Hom}_{\mathfrak{g}}(\mathbb{E}_{ij}^+,\mathbb{E}_{i+1+2r,j}^-)$ due to the   Schur lemma.
Similarly one can proceed for $r<0$ and $F^-.$
Using the fact that $S_{ij}^{+}\in \mathfrak{C}$ (Lemma 7), the inclusion in the considered special case follows. Tracing the remaining cases,
we get $\mathfrak{C}= \mbox{End}_{\mathfrak{g}}(\mathbb{W})$.

$\Box$

{\bf Remark:} Let us remark that more conceptually, we could have defined the projections $S_{ij}^{\pm}$ onto the submodules $\mathbb{E}_{ij}^{\pm}$ using the Casimir operator of the super Lie algebra $\mathfrak{osp}(1|2).$

Now, we  introduce a representation of the super ortho-symplectic Lie algebra $\mathfrak{osp}(1|2)=:\mathfrak{g}'=\mathfrak{g}_0'\oplus \mathfrak{g}_1'$ on $\mathbb{W}$
 in order to decompose the space $\mathbb{W}$   as a $(\mathfrak{g} \times \mathfrak{g}')$-module.
The super Lie bracket of two $\mathbb{Z}_2$-homogeneous elements $u, v \in \mathfrak{g}'$ will be denoted by $[u,v]$ if and only if at least one of them is from the even part $\mathfrak{g}'_0.$ In the other cases, we will denote it by $\{u,v\}.$ Further there exists a basis $\{ h, e^+, e^-, f^+, f^-\}$ of $\mathfrak{g}',$ such that the set $\{e^+,h,e^-\}$ spans the even part $\mathfrak{g}'_0,$ the set $\{f^+,f^-\}$ spans the odd part $\mathfrak{g}'_1$ and
the only nonzero relations among the basis elements are
\begin{equation}
[h,e^{\pm}]= \pm e^{\pm} \qquad [e^+,e^-]=2h \label{he}
\end{equation}
\begin{equation}
[h,f^{\pm}]=\pm \frac{1}{2}f^{\pm} \qquad \{f^{+},f^-\}=\frac{1}{2}h \label{hf}
\end{equation}
\begin{equation}
[e^{\pm},f^{\mp}]=-f^{\pm} \qquad \{f^{\pm},f^{\pm}\}=\pm \frac{1}{2}e^{\pm}. \label{ef}
\end{equation}

For $r=0,\ldots, 2l,$ let us introduce the operators
$$E^{\pm}:\bigwedge ^{r}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S} \to \bigwedge ^{r\pm 2}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}, \,
E^{\pm}=\pm 2 \{F^{\pm},F^{\pm}\}$$ where $\{,\}$ denotes the anticommutator in the associative algebra $\mbox{End}(\mathbb{W}).$
Consider the following
mapping $\sigma: \mathfrak{osp}(1|2) \to  \hbox{End}(\bigwedge^{\bullet}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S})$ defined by
$$\sigma(e^{\pm}):=E^{\pm}, \, \mbox{ } \sigma(f^{\pm}):=F^{\pm} \, \mbox{ and } \sigma(h):=H$$
and extend it linearly to the whole $\mathfrak{g}'=\mathfrak{osp}(1|2).$

We prove that $\sigma$ is a super Lie algebra representation.
To do it, let us introduce a $\mathbb{Z}_2$-grading on the vector space $\mathbb{W}.$
We set $\mathbb{W}_0 :=(\bigoplus_{i = 0}^{l}\bigwedge^{2i}(\mathbb{V}^*)^{\mathbb{C}}) \otimes \mathbb{S}$ and $\mathbb{W}_1 :=(\bigoplus_{i=0}^{l-1}\bigwedge^{2i+1}(\mathbb{V}^*)^{\mathbb{C}}) \otimes \mathbb{S}.$ The vector space $\mbox{End}(\bigwedge^{\bullet}(\mathbb{V}^{*})^{\mathbb{C}} \otimes \mathbb{S})$ will be considered with the super Lie algebra structure inherited from the super vector space structure $\mathbb{W}=\mathbb{W}_0\oplus \mathbb{W}_1.$

{\bf Theorem 9:} The mapping $$\sigma: \mathfrak{osp}(1|2)\to \hbox{End}(\bigwedge^{\bullet}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S})$$
is a super Lie algebra representation.

{\it Proof.} First, it is easy to see that $\sigma(\mathfrak{g}'_i) \subseteq \mbox{End}_i(\mathbb{W}),$ $i=0,1.$
Second, we shall verify that the operators $E^{\pm}, H, F^{\pm}$ satisfy the same relations as the ones for $e^{\pm}, h, f^{\pm}$ written in the rows (\ref{he}), (\ref{hf}), (\ref{ef}) above.
For $r=0,\ldots, 2l$ and $\alpha \otimes s \in \bigwedge^r (\mathbb{V}^*)^{\mathbb{C}} \otimes \mathbb{S},$ we have
\begin{eqnarray*}
[H,F^{+}](\alpha \otimes s)&=& H F^+ (\alpha \otimes s) - F^+ H(\alpha \otimes s)\\
&=& H (\frac{\imath}{2}\sum_{i=1}^{2l}\epsilon^i\wedge \alpha \otimes e_i .s) - F^+ \frac{1}{2}(r-l)(\alpha \otimes s)\\
&=& \sum_{i=1}^{2l}\left[\frac{1}{2}\frac{\imath}{2} (r+1-l)\epsilon^i \wedge \alpha \otimes e_i.s  - \frac{\imath}{2}\frac{1}{2}(r-l)\epsilon^i \wedge \alpha \otimes e_i.s\right]\\
&=&\frac{\imath}{4} \sum_{i=1}^{2l}\epsilon^i \wedge \alpha \otimes e_i.s = \frac{1}{2}F^+ (\alpha \otimes s).
\end{eqnarray*}
Thus we got the relation in the form of the $+$ version of the first equation written in the row $(\ref{hf})$ as required. Similarly, one can prove the $-$ version of the first equation written in that row. The second relation written in the row $(\ref{ef})$ and  the second relation in the row $(\ref{hf})$ follow easily from the definitions of $E^{\pm}$ and $H,$ respectively. The remaining relations, i.e., that ones in the row (\ref{he}) and the first relation in the row $(\ref{ef})$ can be proved using just the derived ones and by expanding the commutator and anticommutator of compositions of endomorphisms. We shall show explicitly how to prove the first relation in the row (\ref{ef}) only. Using   the definitions of the considered mappings, we may write
$[E^+,F^-]=[2\{F^+,F^+\}, F^-]=4[F^+F^+, F^-]=4 (F^+F^+F^- - F^-F^+F^+) = 4 [F^+(-F^-F^+ + \frac{1}{2}H)-F^-F^+F^+]=
4 (F^-F^+F^- - \frac{1}{2}HF^+ + \frac{1}{2}F^+H- F^-F^+F^+)=2 [F^+,H] = - F^-.$
$\Box$

Summing up, we have the following

{\bf Corollary 10:} The super Lie algebra representation $$\sigma: \mathfrak{osp}(1|2) \to \mbox{End}(\bigwedge^{\bullet}{(\mathbb{V}}^*)^{\mathbb{C}}\otimes \mathbb{S})$$ maps the super Lie algebra
$\mathfrak{osp}(1|2)$ into the commutant algebra  $\mbox{End}_{\mathfrak{g}}(\mathbb{W})$ of $\mathfrak{g}$-invariants.

{\it Proof.} Follows from Lemma 5 and Theorem 9. $\Box$

\section{Howe duality for $Mp(\mathbb{V},\omega)$ acting on $\mathbb{W}$}

 We start by defining a certain family $\{\sigma_j\}_{j=0}^l$ of finite dimensional  irreducible representations of the super Lie ortho-symplectic algebra $\mathfrak{osp}(1|2).$
   For $j=0,\ldots, l,$ let $\mathbb{G}^j$ be a complex vector space of complex dimension $2l-2j+1$ and consider a basis $\{f_{i}\}_{i=j}^{2l-j}$ of $\mathbb{G}^j.$
The super vector space structure on $\mathbb{G}^j$ is defined as follows. For $j=0,\ldots, l,$ we set  $(\mathbb{G}^j)_0:=\hbox{Span}_{\mathbb{C}}(\{f_i| i \in \{j,\ldots, 2l-j\}\cap 2\mathbb{N}_0  \})$ and complementarly, we define $(\mathbb{G}^j)_1:=\hbox{Span}_{\mathbb{C}}(\{f_i| i \in \{j,\ldots, 2l-j\} \cap (2\mathbb{N}_0 + 1)\}).$
For convenience, we set $f_{k}:=0$ for $k \in \mathbb{Z} \setminus \{j,\ldots, 2l-j\}.$
We will not  denote the dependence of the basis elements on the number $j$ explicitly.   As a short hand for each $(i,j)\in \Xi,$ we introduce the
$\mathbb{Q}$-numbers
$$A(l,i,j):=\frac{(-1)^{i-j}+1}{16}(i-j) + \frac{(-1)^{i-j+1}+1}{16}(i + j - 2l - 1).$$

For $j=0,\ldots, l,$ we define
the mentioned representations
$$\sigma_j: \mathfrak{osp}(1|2)\to \hbox{End}(\mathbb{G}^j)$$
\begin{eqnarray*}
\sigma_j(f^+)(f_i)&:=&A(l,i+1,j)f_{i+1},\, i = j,\ldots, 2l-j,\\
\sigma_j(f^-)(f_i)&:=&f_{i-1}, \, i=j,\ldots, 2l-j,\\
\sigma_j(h)&:=&2\{\sigma_j(f^+), \sigma_j(f^{-})\}, \\
\sigma_j(e^{\pm})&:=&\pm 2\{\sigma_j(f^{\pm}),\sigma_j(f^{\pm})\}.
\end{eqnarray*}

We have the following

 {\bf Lemma 11:} For $j=0,\ldots, l,$ the mapping $\sigma_j: \mathfrak{osp}(1|2)\to \hbox{End}(\mathbb{G}^j)$ is an irreducible representation of the super Lie algebra $\mathfrak{osp}(1|2).$

{\it Proof.}
First, we prove that for $j=0,\ldots, l,$ the mapping  $\sigma_j$ is a   representation of the super Lie algebra $\mathfrak{osp}(1|2).$
It is easy to see that  whereas the even part of $\mathfrak{g}'$ acts by transforming the even part of $\mathbb{G}^j$ into itself and the odd part into itself  as well,  the odd part of $\mathfrak{g}'$ acts by interchanging the mentioned two parts of $\mathbb{G}^j.$

Now, we should check weather the relations (\ref{he}), (\ref{hf}) and (\ref{ef}) are preserved by the mapping $\sigma_j$ for $j=0,\ldots, l.$
The  last relation in the row (\ref{ef}) and the last relation in (\ref{hf}) are satisfied due the definitions of $\sigma_j(e^{\pm})$ and $\sigma_j(h),$ respectively.
Let us prove the $+$ version of the first relation written in the row (\ref{hf}).
For $i=j,\ldots, 2l-j,$ the left hand side of the mentioned relation reads.
\begin{multline*}
 [\sigma_j(h)\sigma_j(f^+)-\sigma_j(f^+)\sigma_j(h)]f_i\\
 \begin{aligned}
= & 2[(\sigma_j(f^+)\sigma_j(f^-)+ \sigma_j(f^-)\sigma_j(f^+))\sigma_j(f^+)\\
  & - \sigma_j(f^+)(\sigma_j(f^+)\sigma_j(f^-)+\sigma_j(f^-)\sigma_j(f^+)])f_i\\
= & 2[\sigma_j(f^-) \sigma_j(f^+)\sigma_j(f^+) -\sigma_j(f^+) \sigma_j(f^+) \sigma_j(f^-)]f_i\\
= & 2[A(l,i+1,j)\sigma_j(f^-)\sigma_j(f^+)f_{i+1}-\sigma_j(f^+)\sigma_j(f^+) f_{i-1}]\\
= & 2[A(l,i+2,j)A(l,i+1,j) - A(l,i,j)A(l,i+1,j)]f_{i+1}\\
= & 2A(l,i+1,j)(A(l,i,j) - A(l,i+1,j))f_{i+1}\\
= & \frac{1}{2}A(l,i+1,j)f_{i+1}=\frac{1}{2}\sigma_j(f^+)f_{i}.
\end{aligned}
\end{multline*}
The $-$ version of this relation can be proved in a similar way. To verify the relations written in the row (\ref{he}) and the first relation in the row (\ref{ef}), it is sufficient to use the already proved relations and expand the definition of commutator and anticommutator of compositions of endomorphisms.

  Now, we prove that for $j=0,\ldots, l,$ the representation $\sigma_j$ is irreducible.
Let us suppose, there exists a non-trivial proper invariant subspace $X \subseteq \mathbb{G}^j$ of dimension $0<k<2l-2j+1.$
Let us chose a basis $\{v_j,\ldots, v_{j+k-1}\}$ of $X.$  Let $Y:=\hbox{Span}_{\mathbb{C}}(\{f_j,\ldots, f_{j+k-1}\})$ and define a vector space isomorphism $T: X \to Y$ by the prescription
$T(v_i):=f_i,$ $i=j,\ldots, j+k-1$ and extend it first linearly and then to an automorphism $\tilde{T}$ of $\mathbb{G}^j.$
Now, for each $j=0,\ldots,l$, consider a representation $\tilde{\sigma}_j: \mathfrak{osp}(1|2) \to \hbox{End}(\mathbb{G}^j)$ given by $\tilde{\sigma}_j(X):=\tilde{T}^{-1}\sigma_j(X)\tilde{T}$ for each $X\in \mathfrak{osp}(1|2).$ This representation is clearly equivalent to $\sigma_j.$
 Using the definition of $\sigma_j,$  we get $\tilde{\sigma_j}(f^+)v_{j+k-1}=\tilde{T}^{-1}\sigma_j(f^+)\tilde{T} v_{j+k -1}
=\tilde{T}^{-1}\sigma_j(f^+)f_{j+k-1} = \tilde{T}^{-1}f_{j-k} \notin X,$  i.e.,  $X$ is not invariant.
Thus we see, that $\tilde{\sigma}_j$ is irreducible for $j = 0 ,\ldots, l.$ Because of the equivalence
of $\tilde{\sigma}_j$ and $\sigma_j,$ the lemma follows.
$\Box$

Now, we prove a technical but quite important

{\bf Lemma 12:} For each $k \in \mathbb{N}_0$ and $i=0,\ldots, 2l,$ we have
\begin{eqnarray*}
&&(F^-)^k F^+=\\
&&(-1)^k F^+(F^-)^k + \left[\frac{(-1)^k+1}{16}k+\frac{(-1)^{k+1}+1}{16}(2i-2l-k+1)\right](F^{-})^{k-1},
\end{eqnarray*}
when acting on $\bigwedge^i(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}.$

{\it Proof.} We will not write explicitly that we are considering the action of the considered operators on the space $\bigwedge^{i}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}$ and proceed by induction.
\begin{itemize}
\item[I.] For $k=0$ the lemma holds obviously.
\item[II.]
\begin{itemize}
\item[a.]We suppose that the lemma holds for an even integer $k\in \mathbb{N}_0.$
We have
\begin{eqnarray*}
(F^-)^{k+1}F^+&=& F^-(F^-)^kF^+\\
              &=& F^-[F^+(F^-)^k+\frac{k}{16}((-1)^k+1)(F^-)^{k-1}]\\
              &=& -(F^+)(F^-)^{k+1}+\frac{1}{2}H(F^-)^k+2\frac{k}{16}(F^{-})^k\\
              &=& -F^+(F^-)^{k+1}+\frac{1}{4}(i-k-l)(F^{-})^{k}+\frac{k}{8}(F^{-})^k\\
              & & -F^+(F^-)^{k+1}+\frac{2}{16}(2i-(k+1)+1-2l)(F^{-})^k,
\end{eqnarray*}
where we have  used the induction hypothesis, commutation relation $2\{F^{+}, F^{-}\}=H$
and  Lemma 4 on the value of $H.$ The last written expression coincides with that one of statement of the lemma for $k+1$ odd.
\item[b.] Now, suppose $k$ is odd.
We have
\begin{eqnarray*}
&&(F^-)^{k+1}F^{+}=F^-(F^-)^{k}F^{+}\\
&=&F^-[-F^+(F^-)^{k}+\frac{(-1)^{k+1}+1}{16}(2i-2l-k+1)(F^{-})^k]\\
&=&+F^+(F^-)^{k+1}-\frac{1}{2}H(F^{-})^k+\frac{1}{8}(2i-2l-k+1)(F^{-})^k\\
 &=&F^+(F^-)^{k+1}-\frac{1}{8}(2i-2k-2l)(F^-)^k+\\
&&+\frac{1}{8}(2i-2l-k+1)(F^{-})^k =F^+(F^-)^{k+1}-\frac{2}{16}(k+1)(F^{-})^k,
\end{eqnarray*}
where  we have again used the same tools as in the previous item.
\end{itemize}
\end{itemize}
$\Box$

Now let us define a family $\{\rho_j^{\pm} | j=0,\ldots, l\}$ of representations of the symplectic Lie algeba $\mathfrak{g}$ on $\mathbb{E}_{jj}^{\pm}$  by  the prescription
$$\rho_j^{\pm}: \mathfrak{g} \to \mbox{End}(\mathbb{E}_{jj}^{\pm}) \mbox{  } \, ,
\rho_j(X)v:=\rho(X)v$$ for $X\in \mathfrak{g}$ and $v \in \mathbb{E}_{jj}^{\pm}.$
  Let us introduce the following mapping $\mbox{Sgn}: \{\pm\} \times \mathbb{N}_0 \to \{\pm\}$ given by the prescription $\mbox{Sgn}(\pm, 2k):=\pm$ and $\mbox{Sgn}(\pm,2k+1)=\mp$ for each $k\in \mathbb{Z}.$
Now for $(i,j) \in \Xi,$ let us define a mapping
$\psi_{ij}^{\pm}: \mathbb{E}_{ij}^{\pm} \to \mathbb{E}_{jj}^{\mbox{Sgn}(\pm,i-j)} \otimes \mathbb{G}^j$
by the formula
$$\psi_{ij}^{\pm} v: = (F^{-})^{i-j}v\otimes f_i$$ for
an element $v \in \mathbb{E}_{ij}^{\pm}.$ Further we set $\psi:=\bigoplus_{(i,j)\in \Xi}(\psi_{ij}^+ \oplus \psi_{ij}^-).$
Let us  consider $\bigwedge^{\bullet}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S}$ with the action $L = \rho \otimes \sigma$ of $\mathfrak{g}\times \mathfrak{g}'$ and  the space $\bigoplus_{j=0}^l(\mathbb{E}_{jj}^{+}\oplus \mathbb{E}_{jj}^-) \otimes \mathbb{G}^j$ with the action $R =  \bigoplus_{j=0}^l(\rho_j^+ \oplus \rho_j^-) \otimes \sigma_j$ of $\mathfrak{g} \times \mathfrak{g}'.$

In the next theorem, the aforementioned Howe duality is stated.

{\bf Theorem 13:} We have the following $(\mathfrak{sp}(\mathbb{V}^{\mathbb{C}},\omega) \times \mathfrak{osp}(1|2))$-module isomorphism
$$\bigwedge^{\bullet}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S} \simeq \bigoplus_{j=0}^{l}
(\mathbb{E}_{jj}^+ \oplus \mathbb{E}_{jj}^-)\otimes \mathbb{G}^j.$$
{\it Proof.}
Due to Theorem 3 we know that
$\bigwedge^{\bullet}(\mathbb{V}^*)^{\mathbb{C}}\otimes \mathbb{S} = \bigoplus_{(i,j)\in \Xi}(\mathbb{E}_{ij}^+\oplus \mathbb{E}_{ij}^-).$
It is evident, that $\psi$ is a vector space isomorphism.
We prove that for each $(i,j)\in \Xi,$ the mapping $\psi_{ij}^{\pm}: \mathbb{E}_{ij}^{\pm} \to \mathbb{E}_{jj}^{\mbox{Sgn}(\pm,i-j)}\otimes \mathbb{G}^j$ is $(\mathfrak{g}\times \mathfrak{g}')$-equivariant. The $\mathfrak{g}$-equivariance follows easily because $F^-$ in the definition of $\psi_{ij}^{\pm}$ commutes with the representation $\rho$ of $\mathfrak{g}$, see Lemma 5. We shall check the $\mathfrak{g}'$-equivariance.
For each $(i,j)\in \Xi$ and $v \in \mathbb{E}_{ij}^{\pm},$ consider
$\psi_{ij}^{\pm}\sigma(f^-)v=\psi_{ij}^{\pm}F^- v=(F^-)^{i-1-j}F^-v \otimes f_{i-1}=(F^-)^{i-j} v \otimes f_{i-1}.$
On the other hand, we have
$\sigma_j(f^-)(\psi_{ij}^{\pm}v)=F^-((F^{-})^{i-j}v\otimes f_i)=(F^{-})^{i-j}v\otimes f_{i-1}.$
Now, we check the $\mathfrak{g}'$-equivariance in the case of $F^+.$
We shall use Lemma 12 to compute $\psi_{ij}^{\pm}\sigma_j(f^+)v=\psi_{ij}^{\pm}F^+v=(F^-)^{i+1-j}F^+v\otimes f_{i+1}=[(-1)^{i+1-j}F^+(F^-)^{i+1-j}v+A(l,i+1,j)(F^-)^{i-j}v]\otimes f_{i+1}=
A(l,i+1,j)(F^{-})^{i-j}v \otimes f_{i+1},$ where we have used the fact that $(F^-)^{i+1-j}v=0$ because $v\in \mathbb{E}_{ij}^{\pm}.$
Let us compute $\sigma(f^+)\psi_{ij}^{\pm}v=F^+\psi_{ij}^{\pm}v=(F^-)^{i-j}v\otimes A(l,i+1,j)f_{i+1}.$ Thus the equivariance with respect to $F^+$ is proved.
Because the operators $H,$ $E^+$ and $E^-$ are linear combinations of compositions of the operators $F^+$ and $F^-,$ the $\mathfrak{g}'$-equivariance of $\psi_{ij}^{\pm}$ follows. $\Box$

{\bf Remark:} One may call this Howe correspondence 2-folded or $2:1$ because the pairing of $\mathbb{E}_{jj}^+$ and $\mathbb{E}_{jj}^{-}$ with $\mathbb{G}^j$ is not one to one. Namely, to $\mathbb{G}^j$ the module $\mathbb{E}_{jj}^+$ as well as the module $\mathbb{E}_{jj}^-$ are paired in the corresponding tensor product.
That is basically  due to the fact that we are decomposing with respect to
$\mathfrak{g} \times \mathfrak{g}'$ only not concerning the whole commutant of $\mathfrak{g}$ acting on $\mathbb{W}$.

{\bf Remark:} Due to the fact that the category of admissible finite length Harish-Chandra modules is a full subcategory of the category of $\mathcal{U}(\mathfrak{g})$-modules and some
basic  properties of the minimal globalization functors, the  results of the paper have their appropriate minimal globalization counterparts.

\vskip 1cm

redpajama_set_name: RedPajamaArXiv


Spousal Social Security benefits are not denied automatically in the event you and your spouse opt to divorce. There are circumstances under which you remain eligible to receive your spousal benefits from your ex's Social Security. Thankfully, the criteria are straightforward, and as long as you qualify, there should not be an issue in collecting what is rightfully yours. Even though you've dissolved your marriage, you may still be eligible for Social Security benefits after your divorce.
Being formerly married will not automatically entitle you to your ex-spouse's Social Security benefits; you must meet certain requirements before the Social Security Administration will honor your request for Spousal Social Security. Specifically, you must have divorced after at least 10 years or marriage or longer; be at least 62 years old; have a lesser amount of Social Security benefits accumulated during your employment than your spouse's benefit amount; have been married to a spouse who is qualified to receive Social Security and/or disability benefits him or herself.
It does not matter if your spouse has remarried. If you meet the requirements above, you can file to receive Spousal Social Security, and your benefit amount will be 50 percent of your ex-spouse's amount once you reach your full retirement age, which is calculated based upon your birth year.
There are some situations where the above won't be quite as cut and dry. If you remarry, you will not be able to collect Spousal Social Security from your ex-spouse until such time as your current marriage ends and you are single again. In addition, if your ex-spouse does not yet qualify, or has not applied, for retirement benefits but you have - say your ex-spouse is younger and not of retirement age yet - you may still apply for your spousal benefits provided your divorce has been finalized for two years or longer.
Naturally, in this scenario the Social Security Administration will administer your retirement benefits to you first; however, if your spousal benefits are greater than yours are, you will receive a portion of those benefits, as well, even though your ex has not yet retired, so that the combined total equals the higher amount. You may also choose to receive only one or the other at this time and delay the additional benefit if you have reached full retirement age. Of course, all traditional rules apply if you continue working and desire to collect Social Security alongside your income.
Your bank account information (this is so the SSA can deposit your benefits monthly.
Once you've completed your application, the SSA will review it and consider your request for spousal benefits. How long this takes is dependent upon numerous factors, including your age and date of eligibility to receive retirement benefits. Expect to be contacted if the SSA has any questions or concerns about your application, and if, for some reason, your application is rejected, you should receive an explanation behind the administration's decision.
Hopefully this won't happen, particularly if you meet all the requirements for eligibility. If it does, however, you have the right to attest the decision and demand a hearing. Don't go this process alone. Rather, ensure that you have proper legal counsel backing you up as prepare to fight for your entitled benefits. After all, you will be fighting against an agency of the federal government and you should be represented by qualified people focused on getting you the benefits to which you are entitled.


Ticketmaster Platinum offers fans in demand tickets for an event direct from the artist. Platinum Tickets enable market-based pricing (adjusting prices according to supply and demand) for live event tickets. The goal is to give fans fair and safe access to in demand areas, while enabling artists and other people involved in staging live events to price tickets closer to their true value.
Please note that Platinum Tickets are not part of a VIP package and therefore do not include any extras, such as merchandise items or special entertainment services during the event.


- Control sensitivity might have changed a bit, new recommended settings?
Autothrottle not working well - undershoots powersettings and hunts back and forth.
- For me the altitude callouts are very low. Cant hear them very well.
- BAT switch needs to be ON to access the X-Plane menu.
I dont think this is related to XP11. This can happend if you dont have takeoff runway selected. Can you confirm that ?
Another test flight. I went ahead and setup take off page before INS aligned. No error whatsoever. Seems like the issue is intermittent, but I will page attention next time when it happens.
Also I had little fun disengage the autothrottle. I assigned button on my stick. But it seems like I need to move my hardware to idle in order to truly disengage AT.
P.S. Despite that is very great airplane to fly!
How is everyone flying IEXG in XP 11? When I load into it, switches are not moving, there's no side menu and more.
Update: Reinstall fixed the major issues.
Bug 1: Known, will be fixed with upcoming patch. Resolution for now: Don´t access page 2 before selecting departure runway.
if I select " start cold and dark " from XP11 menu, the aircraft starts always with the engines running.
Is a XP11 problem or IXEG bug ?
Hi, have you unchecked "Starts with engine running" in XP11 (where you select your aircraft) ?
And you selected "Cold and dark" from the IXEG menu (preflight)?
I dont have any problems here to start the aircraft in "Cold and Dark".
Probably it's a my mistake .... I've checked only " cold and dark " from IXEG menu and I haven't unchecked "start with engine running" in XP11 menu !
I'll try this evening at home.
When starting cold and dark, the aircraft tends to roll backwards until I can get the APU started and the hydraulic pumps turned on. Am I missing a setting for chocks somewhere to prevent a rolling aircraft? This only happens in XP11.
Yes, I notes the same think.
There may be a dataref changed - usually the brake cylinder holds enough pressure to power the brakes. However - on a real aircraft - this can also get depleted during longer ground times, and then there is nothing to stop the aircraft.
We don´t have "chocks" implemented yet, but are looking to do that in the future. This problem with the pressure being 0 after initial load, however, will be adressed in the XP11 conversion (or it might be a bug in XP11pb3).
Would this affect the frame rates or cause a complete crash? The reason I ask is that randomly, sometimes an hour into the flight, or in tonight's case, still taxiing for take off, frame rates drop from almost 30FPS to 1.5-2 FPS. The only way to resolve it is to shut down XP11 and restart the flight. This has happened on each of my last four attempts to complete a flight from KCHS to KBNA. I have no crashes, just sudden and dramatic drop in frame rates.
Disregard my previous post. I think I messed something up while updating Navigraph, because now, XP10 does not work at all! I'm reinstalling them from scratch and will repost if I end up with the same issue.
As for number 2 I don't have any aniti-virus programs. I'm running WIN 10 64 bit. WIN 10 however has stock app "windows defender". Could it be it?
It definitely was my issue. I added the exceptions for the entire XP10 and XP11 folders. Now the FMC is quick and crisp, no delays at all!


Virgin Media has expressed concerns about the Government's plans for a new £400m Digital Infrastructure Investment Fund, announced in today's Autumn Statement.
New research by Survation for Virgin Media found that just 4% of Britons believe that major public investment is the best way to stimulate investment in broadband. Three-quarters (75%) believe that there are better ways to stimulate investment than using taxpayers' money.
The broadband poll was prepared by Survation on behalf of Virgin Media. A representative sample of 1,021 UK adults were surveyed between 5-6 November 2016. Full data available on request.


TEMO is pleased to partner with numerous sunroom dealers, and we're looking for more home improvement companies to join our team. If you want to grow your business – and partner with one of the most reputable outdoor living manufacturers in the country – then contact TEMO today. We want to work with you!
In addition, our sunroom dealers are provided with all of the trainings and certifications they need to ensure their technicians can properly install our products. And, we also work with you on leads, which means you don't need to worry so much about hustling after new business. We'll do all the hard work for you; all you have to do is make the sale and install the sunroom!
If you'd like to join a team that is dedicated to manufacturing the highest-quality sunrooms, then become one of our sunroom dealers. Contact TEMO today and one of our friendly representatives can tell you more about our partnership opportunities.


Our ears perked up a few months ago at a posting on LTHForum about Bells Brewery reintroducing themselves into the market under a different label. Hopes were raised that maybe this winter we'd get a Two-Hearted Ale, a Cherry Stout, that maybe we'd have a replacement for Bell's Oberon this summer.
On the day of the rollout of the new Kalamazoo brew, we got in touch with Larry Bell about the pullout of Bell's Beer from the Chicago market, where we can find the new Kalamazoo offering, and more about wholesaler law than we ever thought we knew.
Where exactly could an intrepid beer explorer find this new beer? Will we get an unfiltered wheat ale, (coughOBERONcough) next year? Cross your fingers and read more after the jump.
Chicagoist: Before I ask you about the new beer, give me the Cliffs Notes on your pullout from the Chicago market. It's basically a dispute with your distributor, correct?
Larry Bell: Pretty much. National Wines and Spirits last year was trying to sell the Bells brand, and we did not come to an agreement on the brand being sold. And at that point we really couldn't afford to stay in the Chicago market.
C: Are there any plans to bring back the Bell's label itself back to Chicago?
LB: I don't forsee that right now. I can't see that right now. Right now National Wine and Spirits owns the rights to Bells forever. We are unable to reach any agreement with them, and so nothing happens with it.
C: You know, it doesn't seem right that you don't have control over your own product. I mean…forever?
LB: Well, yeah...you know, there are reasons for the law (ed. note: the law Bell is talking about is the Beer Industry Fair Dealing Act of 1982, the details of which, in complicated legalese, can be read here). Those laws for the wholesalers were written so that wholesalers were protected from big breweries crushing the wholesaler. And that's certainly understandable. Hopefully over time, we can look at some ways that maybe some smaller breweries can have a little bit more protection. But like I say, we've dealt with these laws in other states, and it's been fine. It's unfortunate that our relationship with National Wines and Spirits has degenerated so bad that no deal can be done.
Y'know, the thing is, the law isn't so much different than a lot of states that we do business in. But unfortunately, National Wines and Spirits wanted to use some real hardball tactics, and wasn't interested in working with their so-called brewery partners when they were doing all this change last year. Because our brand has been bought and sold any number of times.
C: Is it still just business at this point, or is it personal?
LB: I don't think there's any love lost between the two companies. [laughs] We are really grateful for these first two wholesalers to pick us up, and we're generally pretty supportive of what's called the three tier system of beer distribution in the US. I don't want people to get the impression that I'm out there bashing wholesalers, because I'm not.
Unfortunately, I got into a relationship with a wholesaler that didn't have our best interest at heart.
C: So is this a sort of re-entry of the market for Kalamazoo, and by proxy Bells?
LB: We're not re-entering the market as Kalamazoo Brewing. Kalamazoo Brewing is the former name of the corporation. The name of the corporation is Bells Brewery Incorporated now. We are re-entering the market with a new brand called Kalamazoo.
C: This seems like something of a legal loophole.
LB: Well, the law says that you can assign a new brand to whatever wholesaler you want. There was in this last year a case concerning brand extension in Illinois, where Heineken had come into Illinois with Heineken Premium Light, had assigned it to a different wholesaler, and the Heineken wholesaler sued saying they had a right to it. Well, they lost. Illinois then changed their law to say that a brand extension has to go to the wholesaler of that brand. Well, that's fine. What we're coming back with is not a brand extension, but actual new brands.
C: So you of course have a different distributor for this product.
LB: That's correct. We have assigned two distributors, Central Beverage and Schamberger.
C: I'm guessing you plan on getting sued over this.
LB: We don't know. In October in a conversation with National Wines and Spirits, at that point they told me their intention was to sue us, and make it as lengthy and costly a court battle as they could. Whether or not that's still their opinion, I don't know.
C: Was this based on the dispute over the Bells brand, or did it have to do with this new Kalamazoo label?
LB: No, that was [for] Kalamazoo, they're certainly aware of our intentions.
C: So, about this new beer – what exactly is the deal?
LB: Well, it actually shipped from the brewery [December 3rd]…and goes on tap [Tuesday, December 4th].
C: I've seen info on a few different beer labels for the Kalamazoo Brewery - what are you rolling out?
LB: We are rolling out one beer in about ten accounts, on draft only. That is the Kalamazoo Royal Amber Ale.
C: Is there any equivalent to an existing Bell's product?
LB: These are new products. They're different recipes, they have their own unique flavor profiles.
C: Do you have plans to bring anything else to the market?
LB: Basically, we've got to wait and see what's going to happen. If National decides that they do indeed want to sue, then I suspect they'll have to do that fairly quickly, and we'll all wind up in court to see what the judge has to say.
When I spoke to them in October, they had told me that they intended to get a temporary restraining order against us. So, we'll see. And then go from there. We've got some legal opinions that say that the law is on our side. we believe that we would prevail in a court of law for the right to sell this brand, but we are moving ahead very slowly...and deliberately.
Again, if National wants to do something they'll have to do it relatively soon, they'll have to let us know that they're going for the restraining order and we would all go to court together. And I would suspect that if that were to happen then I think it would happen next week. It would have to happen relatively soon.
C: So, in a perfect world, is there any chance we'll get something like an Oberon next summer?
LB: We have registered a trade name for a Kalamazoo unfiltered wheat ale, so there's a possibility that an unfiltered wheat beer could see Chicago.
If we do prevail, there's one other name that we filed for that we don't have label approval for, but I was able to make a special purchase of some European hops this last year, a relatively small quantity...but we've been holding on to those, and if everything goes our way then we will make a special double IPA for the Chicago market using those.
Want to get your hands on the new Kalamazoo? The taps that'll have the brew in Chicago are: Clark Street Ale House, Lemmings, Silver Cloud, Northside Bar & Grill, The Handlebar and Twisted Spoke.
Outside the city, Kalamazoo will be at Brixie's, Kendall Pub, Chef Paul's Bavarian Lodge, Jimmy's Grill, Old Town Pub - Geneva and Wasco, and will also be available at Durty Nellies next week. Larry Bell image via Southwest Michigan First.


Extensions and DID's can be add/edit/remove extensions and DID's from your PBX. The list must be created inside a CSV file, if you do not have one created you can download the Template CSV file or if you would like to make changes to your current Extensions/DID's by selecting Export Extensions or Export DID's.
To begin make sure you are in the Tools menu, then click on the Bulk Extensions tab. Here you will see a guideline at the bottom of the page of how to setup the CSV file.
After you have created the file click choose file then load file.
After the file has been uploaded you can go back to the Setup and select Extensions from there you will see all extensions that have been created on the right hand side along with their extension number.
If you have multiple extensions that need to be edited, you can make a change to the file and upload it to the Bulk Extensions.
Bulk DID's may also be created and edited by using the same process. Through the use of Bulk Extensions, and Bulk DID's you can streamline the process of adding/editing/removing extensions and DID's.


My Favorite Rooms Series continues today with view into some beautiful bedrooms. Each example is different from the next — inspiration for just about any sense of style.
1. Studies show that the more you can reduce light and noise in a bedroom, the better you'll sleep. This monochromatic design offers a cocoon of stylish comfort with chocolate upholstered walls that minimize both sound and light.
2. For those who prefer a brighter, lighter, and more sophisticated bedroom, this design is just the ticket. Form the stunning green wallpaper to the chic black-and-white bed, this is a bedroom that is the perfect escape from the world.
3. Shape, texture, and color are perfectly blended in this master bedroom, offering shades of blue and coral in dusty tones. I love the circle shapes and the mix of classic design and boho chic.
4. Layers of neutrals always make for a dreamy escape, and I love the tone-on-tone walls and linens with that standout of an upholstered headboard.
5. With just the right amount of whimsy and color, this kids' room is clean and clever, perfect for siblings to enjoy and pleasing to a parent's aesthetic as well! Who says kids' rooms can't have a touch of sophistication and high style?
Want to see what other ideas I've pulled for beautifying your bedroom? Check out my Sweet Dreams pinboard.


text:
Posted on May 15, 2017 by Gallipolis Daily Tribune
Rio Grande blanks Tigers in NAIA tourney opener
By Randy Payton - For Ohio Valley Publishing
GRAND RAPIDS, Mich. — Kelsey Conkey tossed a two-hit shutout and hit a home run to lead the University of Rio Grande in a 6-0 win over Olivet Nazarene (IL) University in the Grand Rapids Bracket of the NAIA Softball National Tournament Opening Round, Monday afternoon, at Farmer's Insurance Athletic Complex.
Rio Grande, ranked No. 17 nationally and seeded second in the four-team bracket, improved to 46-8 with the victory and will face either top-seeded and eighth-ranked Davenport University or Hastings (Neb.) College in Tuesday's winner's bracket final in the double-elimination tourney.
Olivet Nazarene, the No. 3 seed in the tourney, slipped to 31-13 with the loss and will play the Davenport-Hastings loser in a 3 p.m. elimination contest on Tuesday.
"When you come into a four-team, double-elimination tournament, I think that's very important that you win the first game," said Rio Grande head coach Chris Hammond. "It's all about getting off to a good start. We didn't do that last year, but we were able to today. It's a good spot to be in."
Conkey, a sophomore from Minford, Ohio, took a no-hitter into the sixth inning before ONU's Emily Carstens tripled along the right field line with one out. Conkey retired each of the next two batters on routine groundouts to end the threat.
The Tigers' other hit was a one-out double by Miranda Southall in the seventh inning but, again, Conkey retired each of the next two batters to secure her 10th shutout of the year.
Conkey also plated what proved to be the only run she would need with a leadoff home run in the home half of the second inning – tying sophomore Carly Skeese (Newark, OH) for the team lead with her 13th long ball of the season.
"It was a changeup and, actually, was a pretty good pitch," said Hammond. "She was able to hold back and then went down and got good bat on it. There wasn't much doubt that it was gone."
Rio Grande added three more runs in the inning.
Sophomore MacKenzie Nichols (Columbus, OH) followed Conkey's home run with a walk and was replaced by senior pinch-runner Jenna Jones (Lancaster, OH). One out later, senior Kari Jenkins (Jackson, OH) walked, a passed ball sent Jones to third and junior Gabby Gregg (Ashville, OH) had a sacrifice fly to make it 2-0.
Freshman Michaela Criner (Lancaster, OH) kept the inning going with a single, setting the stage for a two-run triple to right by senior Alex Stevens (Oak Hill, OH) which extended the cushion to 4-0.
That's how things stayed until the RedStorm added on a pair of insurance markers in the sixth inning thanks to a two-out, bases loaded single by Nichols and a subsequent walk to senior Cheyenne Hamaker (Hillard, OH).
Rio had plenty of opportunities before adding on the late runs, stranding single runners in each of the first three innings, leaving the bases loaded in the fourth and stranding a pair in the fifth.
Criner finished with three of the RedStorm's 12 hits, while Skeese, Conkey and Nichols all had two hits each.
Carstens suffered her sixth loss in 25 decisions, allowing all 12 hits and six runs over 5-2/3 innings.
Randy Payton is the Sports Information Director at the University of Rio Grande.
By Randy Payton
For Ohio Valley Publishing
Hi! A visitor to our site felt the following article might be of interest to you: Rio Grande blanks Tigers in NAIA tourney opener. Here is a link to that story: https://www.mydailytribune.com/sports/15462/rio-grande-blanks-tigers-in-naia-tourney-opener

redpajama_set_name: RedPajamaCommonCrawl


paper addresses more specifically is the validation of the algorithm using fixed point arithmetic for the whole optimization process. The results show how high quality transforms are evolved from scratch with limited precision arithmetic. Also, preliminary results of the implementation in an FPGA device are included.


BeautyBridge.com is one of the world's leading online retailers of beauty, skincare, and bath and body products. Thousands of devoted followers have already crossed over to BeautyBridge.com, saving time and money shopping for hard-to-find skincare products and cosmetics at competitive prices.
With an inventory of over 10,000 products from over 150 top-shelf brands like Youngblood, Le Couvent des Minimes, Exuviance, smashbox and BABOR, BeautyBridge.com customers can choose from industry-leading anti-aging products, luxurious lotions and scrubs, and soothing body and bath products, or refresh their tired makeup routine with tantalizing shadows, luscious lip colors, and pretty polishes. If a customer has a specific skincare concern, BeautyBridge.com has it covered with groundbreaking products for everything from acne and rosacea to eczema and cellulite. Best of all, BeautyBridge.com provides free consultations with a medical director and professional video tutorials for the products it carries, as well as industry news and an informative blog to help customers reveal their natural beauty.


Win £1,000 for a Berkshire Charity!
Three £1,000 prizes are up for grabs as Berkshire Community Foundation (BCF) launches its 2019 Vital Signs survey. BCF is asking people in Berkshire to spend five minutes giving their opinion of life in the county. As a thank you, they can nominate a charity to win one of the £1,000 prizes. Please consider nominating Connecting Communities in Berkshire!
Last year three local groups, Autism Berkshire, Family Friends Windsor and Maidenhead and Interakt, each received one of the £1,000 prizes.
Now in its sixth year, Vital Signs uses the views of the local community, along with other research, to enable BCF to produce a community 'health-check' report. This report reveals priorities, needs and opportunities for action by measuring the quality of life in local areas.
The results drive BCFs fundraising. Last year's Vital Signs reported that a safe home, good health and secure employment are all linked and underpin wellbeing so BCF launched Vital for Berkshire – homes, Health and Work. This campaign has already funded over 20 local groups that provide vital support including counselling, assessable transport, crisis support, education or training, health coaching, supporting people back into back into employment and tackling homelessness.
Recently our national network Action with Communities in Rural England (ACRE) worked with the BBC and gained powerful coverage of issues for young people in rural areas.
Transport and Digital connectivity issues were discussed on a BBC 5 Live programme which Flick Humphrey, ACRE Public Affairs and Communications Manager co-hosted with presenter Nihal Arthanayake.
Also on the show was Rural Services Network CEO Graham Biggs and Wheels to Work user from Shropshire RCC Jamie Yale. The show was a result of a film Flick put together which you can watch below.
Robert Jenerick and Lord Gardiner of Kimble Launch a £3m grant for village halls.
The grant is now open for applications and will be managed by CCB's national network Action for Communities in Rural England (ACRE) on behalf of Defra.
We will be running our Thyme Together sessions for older people starting Thursday 4th April and are appealing for volunteers to assist our session leader Debs to support participants. The sessions run weekly in the afternoons from 1-3pm on Thursdays through to October 31st 2019 and are an opportunity for older people to come to the garden and take part in gardening and art and craft activities using natural and recycled materials. There may be some extra sessions to support to fulfil our funder's requirements. If you are interested or have taken part before and would like to do so again please email Wendy at wendy@fiveaday.org.uk expressing an interest.
We are looking for volunteers for the Green Gardening/Green Art sessions for Adult Learners on Wednesdays 11am – 2pm. You will support our Tutor Debs to deliver a course for 10 weekly sessions. You will help participants with learning disabilities to learn how to grow plants, participate in practical wildlife activities including pond dipping, sweep netting and log turning and create individual and group artwork and sculptures in the Green Art sessions.
Sessions are adapted to individual learners.
You will need to be DBS checked as you will be working with vulnerable adults. These are fun sessions to deliver and very rewarding.
If you are interested please email Wendy at wendy@fiveaday.org.uk expressing an interest and provide details of any previous experience.
Company Number 3212736. Registered office address: Unit 11 Diddenham Business Park, Diddenham Court, Lambwood Hill, Grazeley, Reading, Berkshire RG7 1JQ.


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Michelle du Fresne dite de Saubonne (1485-1549), dame de Soubise, est la femina cordatissima de Guillaume Budé. Fille de Denis de Saubonne, seigneur de Fresnes-Coudray et humaniste protestante, elle a fait connaître Jean Marot, père de Clément Marot, à la reine Anne de Bretagne auprès de qui elle servit de secrétaire, puis de chambellan. Elle sera l'une des premières protectrices de Bernard Palissy.

Biographie

L'amie d'Anne de Bretagne

En 1503, sa présence est attestée à la cour de Louis XII par un acte faisant état de parties livrées à Mademoiselle Michelle de Saubonne pour envoyer à la reine de Hongrie.

En 1505, elle fut choisie pour servir de fille d'honneur à la reine. Chargée des bagues, du linge et des bijoux d'Anne de Bretagne, elle en devint bientôt le secrétaire, rôle qu'elle partagea avec Hélène de Laval, puis organisa pour la souveraine des joutes oratoires galantes.

Vers 1506, elle fit entrer au service de la reine Jean Marot, dont elle avait apprécié les poésies. Son fils gardera toute sa vie le souvenir qu'elle fut une des premières à accueillir son père vivant, quand à la Cour du Roy fut arrivant.

Michelle de Saubonne partageait les goûts de la reine et, comme le dit Marot,

Aimant littérature, Savoir exquis, vertus qui le ciel percent, Arts libéraux & ceux qui s'y exercent,

Elle avait introduit auprès de la reine Anne un des écrivains de ce temps, l'historien et poète Jean Le Maire de Belges.

La gouvernante de Renée de France

En 1507, elle se maria avec Jean IV de Parthenay, seigneur de Soubise, dont elle était la seconde épouse, et qui mourut cinq ans plus tard, la laissant enceinte avec trois filles. C'est par son fils, Jean V de Parthenay, seigneur de Soubise, né posthume en 1512, qu'elle sera la grand-mère de Catherine de Parthenay.

Vers 1510, elle devint la gouvernante de la fille cadette de Louis XII et Anne de Bretagne, Renée de France, dont elle fit l'éducation. Anne de Bretagne voulait qu'elle fut pour la princesse comme une seconde mère.

Dans les querelles qui divisaient la cour, Saubonne prit le parti d'Anne de la Bretagne contre celui de Louise de Savoie, qui, à la mort de la reine, voulut dépouiller l'enfant de son héritage.

En 1515, elle fut éloignée de Paris par François I car elle s'opposait au rattachement de la Bretagne à la couronne de France et qu'en outre, elle manifestait des sympathies pour l'église réformée.

Exil avec ses enfants

Elle se retira dès lors avec ses trois filles et son fils dans sa maison du Parc-Soubise, où elle les éleva, leur faisant apprendre les langues grecque et latine. Budé affirme qu'elle avait connaissance de la vraie religion et y instruisit tous ses enfants.

À la cour de Ferrare

En 1528, Renée de France ayant épousé Hercule II d'Este, cette princesse fit admettre le retour de sa gouvernante à la cour, puis l'emmena  en Italie, à Ferrare avec deux de ses filles et son fils. Peu après Clément Marot vint les y retrouver.

À l'arrivée de Calvin à la cour de Ferrare, sa fille ainsi que son gendre prirent la résolution d'abandonner l'Eglise catholique.

En 1536, le duc d'Este, qui se rapprochait de Charles Quint depuis la mort de son père, Alphonse Ier d'Este, voulut chasser Madame de Saubonne de Ferrare. il lui en voulait particulièrement des reproches qu'elle faisait quant à la façon dont il traitait son épouse.

Renée de France protesta de son renvoi, auprès de François I et du pape Paul III. Le cardinal du Bellay s'en mêla. Pendant une absence du duc, à Rome auprès du pape, Madame de Saubonne organisa pour le compte de François I une réunion du parti français. Hercule d'Este prit peur, retourna immédiatement à Ferrare, et la renvoya.

Clément Marot nous a laissé quelques vers  où il annonce le départ de Madame de Soubise, partant de Ferrare pour revenir en France :

Le clair soleil sur les champs puisse luyre
Dame prudente et te vueille conduire
Jusques au pied de ta noble maison
Il est certain que plus tost oraison
 Pour ta demeure à Dieu je vouldrois faire
 Mais puis que luy et le temps et l affaire
 Veulent tous trois que ta bonté desplace
Montz et torrens te puissent faire place
Dieu tout au long de ton allee entiere

et aussi ce distique, lors de son retour en France :
Viens le temps doux. Retire-toi la bise ! Ne fache point Madame de Soubise !

Retour en Saintonge

Après leur retour en Saintonge, on la trouve, avec sa fille, son gendre, le sire de Pons, et son fils, à l'origine de la diffusion du protestantisme dans les terres de Saintonge et du Bas-Poitou.  Quelques jours après le décès de sa fille Anne, Madame de Saubonne mourut en recommandant à sa famille le jeune Bernard Palissy; elle s'éteignit en professant la nouvelle religion ; elle avait toujours eu le prêche sur ses terres, et les édits du roi protégeaient encore les ministres. C'est alors que se réalisa la prophétie de Marot :

Et nous en bief, saurons, de ton absence,
 de quoi servait par deça ta présence.
Sa vie et sa généalogie nous sont connues au travers des écrits du mathématicien François Viète, qui fut le secrétaire de son fils à ses débuts, au parc Mouchamps.

Notes

Bibliographie
 Ouvrages sur les Parthenay et sur Michelle de Saubonne disponibles en ligne :
 Marcel Giraud-Mangin, "Michelle de Saubonne, dame d'atour d'Anne de Bretagne", dans Bulletin de la Société d'histoire et d'archéologie de Bretagne

Personnalité des guerres de religion
Personnalité française du protestantisme
Dame de compagnie française du XVIe siècle
Cour de François Ier
Cour de Louis XII
Décès en Saintonge
Gouvernante des enfants royaux

redpajama_set_name: RedPajamaWikipedia


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text:
EU Neighbours south Stories EU in action Stories Inventiveness…
Inventiveness at the service of children's education
Camille Dupire
If one word could describe young Mariam Konan Rasheed, it would be resourceful. Stuck at home during the COVID-19 pandemic, the 33-year-old Jordanian came up with the most entertaining way to educate her younger peers: online educational videos.
Prior to the pandemic, the graduate from the EU-funded EDU-SYRIA II scholarship had been volunteering with the Lutheran World Federation (LWF) in the field of nursery. "I cared for children between 2 and 5 years old from Zarqa Governorate by organising a lot of different interactive activities that also served to educate them," Mariam explains, noting that it helped her develop an understanding of the children's needs according to their age. "Because of the pandemic though, I was not able to go to work anymore so I thought about a way to continue working despite the lockdown."
She looked around her house and gathered numerous everyday life objects and material, including rope, pins, needles and rubber balls, among others. "Of course, it was very challenging, because I could not go out to buy the material I needed." But this did not stop her ingenious mind. She used her smartphone to record the videos, and shared them on YouTube and WhatsApp groups with beneficiaries of the LWF.
©EDUSYRIA
No longer than one-minute-long, the videos are tailored to the young children's attention span, with a playful background music to retain their interest. "I wanted to continue raising children's awareness on topics I care about, while helping them learn new things, such as geometric shapes, visual memory, etc." Designed to different age categories, the activities also reach out to children with behavioural issues through cognitive improvement games.
The videos came as a welcome distraction at a time when no movement was allowed around the Kingdom, and still continue to yield success outreach on social media. Mariam benefited from the EDU Syria II scholarship funded by the European Union under the EU Regional Trust Fund in Response to the Syrian Crisis (EUTF Syria), the EU Madad Fund.
EDU-SYRIA
EDU-SYRIA is a host of projects that chronologically extended over the last five years and still going, 2015 and further. The projects are EU funded via MADAD in response to the Syrian crises. Hence, the primary beneficiaries are Syrian refugee and underprivileged Jordanian youths. The ultimate objective of the projects is to improve the livelihood of those beneficiaries by providing them with higher education opportunities through a cluster of scholarships that are granted every cohort of high school graduate since 2015.
EDU-SYRIA I was launched in late 2015 where a fund of 4 million Euros supported a total of 390 higher education accredited degree, master's, bachelor's, and vocational programs. Another EUTF grant was bequeathed, 11 million Euros, inaugurating EDU-SYRIA II, the second wave of the project, October 2016. The project was the largest in Jordan in terms of the number of the higher education degree scholarships that was awarded, specifically, 1000. The EU Regional Trust Fund in Response to the Syrian Crisis program presented a smaller endowment of 2.6 million Euros in January 2019 allowing for additional 200 higher education degree scholarships. EDU-SYRIA III was launched in January 2020 with a fund of 15 million Euros. The components/actions of the last wave were more diversified in nature and design where some aiming to pick up the young adults who dropped out or leaked out of school. The direct total of beneficiaries is 2245 Syrian refugees and underprivileged Jordanians.
©EU Regional Trust Fund in Response to the Syrian Crisis
EU Regional Trust Fund in Response to the Syrian Crisis
Since its establishment in December 2014, a significant share of the EU's non-humanitarian aid for Syria's neighbouring countries is provided through the EU Regional Trust Fund in Response to the Syrian Crisis, the EU 'Madad' Fund. The Trust Fund brings a more coherent and integrated EU aid response to the crisis and primarily addresses economic, educational, protection, social, and health needs of refugees from Syria in neighbouring countries such as Jordan, Lebanon, Turkey and Iraq, and supports overstretched local communities and their administrations.
EU Delegation to Jordan website
EDU-SYRIA website
EU Regional Trust Fund in Response to the Syrian Crisis website
Civil society Cross-cutting areas Economic and social cooperation Education & vocational training Employment and entrepreneurship Gender issues Inclusive economy & sustainable growth Jordan South Syria * Youth


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Starting 2010 With Doctor Who and Sadie Hell – S05E30
January 1, 2010 Lenny Wu	Leave a comment
It's now 2010 and I started off the new year by watching Doctor Who. It was part 2 of The End of Time and also the end of David Tennant's stint as the Time Lord and the show's head writer, Russell T. Davies. Ever since Doctor Who returned to the airwaves in 2005, they have been coming out with the Christmas specials. The Christmas specials have become a tradition for myself because there's not much happening on Christmas day. This one is very important because in 2009, there was no full series of Doctor Who because Tennant was off acting on stage. There were the four specials and the final two were David Tennant's final adventure and swan song. Here's an image I put up and it's from the ending where he regenerates. It's a sad scene watching the Doctor go into regeneration. Doctor Who fans like myself have been fond of Tennant, it's just hard to see him leave the role. I'm going to miss the brown pinstripe suit (there's also the blue pinstripe suit), the Converse Chuck Taylor sneakers (I will continue to wear them no matter what), the glasses he occasionally wore and the sonic screwdriver.
Overall, the 2nd part of The End of Time was excellent. As non-spoiler as I can be, the acting among the cast was superb. Seeing the emotions of Tennant as The Doctor clings to his death was just strong and emotional. It was touching to watch the companions of previous seasons mark their farewell as they won't be coming back for future episodes. The words, I don't want to go will probably linger on for awhile because it's so heartbreaking and us fans will have to move on that Tennant is no longer the Doctor.
I can't wait for the Spring when Matt Smith (Eleventh Doctor) takes over for Doctor Who.
How could I not forget to mention that it's wonderful to see another "iconic" British character (well, actor) appear in the show. I know Timothy Dalton did two James Bond movies. It was nice to see two famous British character sort of meet face to face. I know it's silly of me to imagine James Bond meeting Doctor Who.
Since I wanted to embed the YouTube clip, here's the link to the preview of Doctor Who Series 5.
It looked like the finale scored 10.4 million viewers in the UK. Thus ends the David Tennant years on Doctor Who.
The First Radio Show of 2010 – Another Themed Show to Wrapup the Holidays
Here is the first radio show of 2010 (on January 6). Started with Sadie Hell, played a new Basia Bulat song (album to be released on January 26, 2010), and almost ending with Doctor Who (again). I played the best of the 2000s. It was basically my picks throughout 2000 and I had to change my song choices due to the mainstream music that I'm not supposed to play regularly. In irony, I played several British hits from the 2000s and it seems like a few people are starting to remember S Club 7. I will resume the interviews for next Wednesday.
Playlist is up on CKCU's website at this link.
Press play to listen
2000s2010Basia BulatBBCBritish musicCanadian independent musicClub SawconcertsConverseDavid TennantDoctor WhoEleventh DoctorI don't want to goIndieindie musiclocal bandsMatt SmithMusicNew Year's DayOttawaradio showS Club 7Sadie HellTenth DoctorThe End of TimeThe Love Machinetv shows
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