The wall is supposed to be attached to the steel frame that supports the floor. If there's a gap, you can easily fill it with foam but that's not a repair. The screws that hold the wall to the frame probably came loose which allowed the wall to pull free. When this occurs at the front of the RV, it allows the front of the coach to bounce up and down, independent of the lower frame which will only make the condition worse. The correct fix for this problem is to get the wall tightly attached to the steel framing so that it doesn't come loose again.
How's your repair holding up?
We are dealing with Forest River and our warranty dealer Grumbines in Harrisburg PA today with this issue of our 2007 GTX toy hauler. Seems that both left and right walls are detaching. They told us that this would be covered under warranty since it is structural.
We like your fix and thought about letting both these parties see your repair. We question the long term repair of just replacing the self taping screws that it came with from factory. It seems obvious that these screws are unable to hold up under the demands of weight and movement.
Though it has only been six months, we have driven about 3000 miles and it is still solid. I feel that I have fixed it for good. I wasn't going to have the same screws that the factory installed used again as the fix.
We too have this same problem AGAIN. 2009 Georgetown 36' toyhauler.
Left and right walls were repaired by dealer early in spring of this year.
Unfortunatly they repaired them the same way the factory built the RV even though we had asked for some upgrading. Well it's now fall and 3000 miles later the left wall is loose again. We are following Eric52's plan of repair ourselves. We are doing both sides. Thanks to the helpful Eric52. Hopefully get this repaired once, well, actually twice and done. I must add though that overall we are very happy with our RV. This is our second RV.
Mechanically it has been reliable and sound for 24,000 miles. I'll update on how we make out. Any suggestions for perfecting this repair?
The fix I made is solid. It has been nearly 18 mos. and it is still extremely solid.


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text:
\section{Introduction}

Weil representations have proven to be a powerful tool in the theory of group representations. They originate from a very general construction of A. Weil (\cite{weil}), which has as a consequence the existence of a projective representation of the group $\Sp(2n,K)$, $K$ a locally compact field.
Weil built this representation taking advantage of the representation theory of the related Heisenberg group, as described by the Stone-von Neumann theorem in the real case (\cite{lion}).
In particular, these representations have allowed to build in a universal and uniform way all irreducible complex linear representations of  the general linear group of rank $2$ over a finite field (\cite{soto}), and later over a local field, except in residual characteristic two (\cite{ku}).

Weil representations can be constructed in various ways. For instance, they can be constructed via Heisenberg groups, via constructions of equivariant vector bundles (\cite{gpsa}), via presentations or via dual pairs (\cite{aubert}), as we shall see later.
The method using presentations is accomplished by having a simple presentation of the group, and
then defining linear operators on a suitable vector space which preserve the relations among the generators of the presentation. This idea was originally suggested by Cartier, and used successfully by Soto-Andrade, for symplectic groups $\Sp(2n,\mathbb{F}_q)$ (\cite{soto}) and complex irreducible representations of $\SLc(2,\mathbb{F}_q)$.  In the case of $\Sp(2n,\mathbb{F}_q)$ he considered this group as a group "$\SLc (2)$" but with entries in the matrix ring $M_n(\mathbb{F}_q)$, and thus he obtained a suitable presentation for the group. In this way, he constructed the Weil representation for the symplectic groups $\Sp(2n,\mathbb{F}_q)$. In the case of $\Sp(4,\mathbb{F}_q)$, Soto-Andrade obtained all the irreducible representations of this group by decomposing the two Weil representations associated to the two isomorphy types of quadratic forms of rank $4$ over $\mathbb{F}_q$.

This point of view was generalized and gave rise to the groups $\SLc^{\varepsilon}_*(2,A)$ for $(A,*)$ an involution ring and $\varepsilon =\pm 1 \in A$.  These groups are a generalization of special linear groups $\SLc(2, K)$, where $K$ is a field. They were defined for $\varepsilon = -1$ by Pantoja and Soto-Andrade in \cite{jofal} and generalized to $\varepsilon = 1$ in \cite{comm}.

The groups $\SLc^{\varepsilon}_*(2,A)$ include, among others, symplectic and split orthogonal groups. For instance, if $K$ is a field, $A = M_n(K)$, $\diamond$ the transposition of matrices then $\SLc_{\diamond}^{\varepsilon = -1}(2,A)$ is the symplectic group $\Sp(2n,K)$ and $\SLc_{\diamond}^{\varepsilon = 1}(2,A)$ is the split orthogonal group $\ot_K(2n,2n)$. Thus, these groups allow us to look at higher rank classical groups as rank two groups, considering them with coefficients in a new ring.

The procedure used by Soto-Andrade in \cite{soto} was approached even in the case of a non semi-simple involutive ring $(A, *)$ with non-trivial nilpotent Jacobson radical. In fact, in \cite{lucho} Guti\'errez found a Bruhat presentation and a generalized Weil representation for $\SLc_*^{\varepsilon=-1}(2,A_m)$, where $A_m=\mathbb{F}_q[x]/ \left\langle x^m \right\rangle $.

In \cite{ongen} Guti\'errez, Pantoja and Soto-Andrade build Weil representations in a very general way via generators and relations, for the groups $\SLc^{\varepsilon}_*(2,A)$ for which a ``Bruhat" presentation analogue to the classical one holds. In this way, in order to use this method it is very important to have an adequate presentation of the group. In \cite{manusc}, Pantoja generalized the classical Bruhat presentation of $\SLc(2,K)$ to $\SLc_*^{\varepsilon=-1}(2, A)$, when A is simple artinian ring with involution.

In this work, we construct a generalized Weil representation of finite split orthogonal groups $\ot_{q}(2n,2n)$, using the method described in \cite{ongen}. As mentioned above, this group is naturally the group $\SLc_{\diamond}^{\varepsilon=1}(2,M_{2n}(\mathbb{F}_q))$. However one of the results of this paper is to realize this group as a $\SLc_{\sim}^{\varepsilon=-1}(2,M_{2n}(\mathbb{F}_q))$ group, where $\sim$ is a certain specific involution in $M_{2n}(\mathbb{F}_q)$, different from $\diamond$. This allows to use the Bruhat-like presentation that is exhibited in \cite{manusc} and facilitate significantly technical aspects of the construction. In fact, the result is more general, and provides an isomorphism between the groups $\SLc_*^+(2,M_2(A_0))$ and $\SLc_{\sim}^-(2,M_2(A_0))$, where $(A_0,*)$ is a unitary involutive ring and $\sim$ is another involution in $A_0$ obtained from $*$.

Also, we study the structure of the associated unitary group and using this group we get an initial decomposition of the representation.

In section 6 we show compatibility of the method of Guti\'errez, Pantoja and Soto-Andrade with theory of dual pairs. For this, we prove that the representation of $\ot_q(2n,2n)$ constructed using this method is equal to the restriction of the Weil representation to $\ot_q(2n,2n)$ for the dual pair $(\Sp(2,k),\ot_q(2n,2n))$. Also, we prove that the initial decomposition mentioned above is the same as decomposition with respect to the action of $\Sp(2,k)$ via the Weil representation.

\section{The groups $\SLc_{*}^{\varepsilon}(2,A)$.}

Let $(A,*)$ be a unitary ring with an involution $*$. We can extend the involution $*$ in $A$ to the ring $M_2(A)$ putting
\[
T^*=\matriz{a}{b}{c}{d} ^*=\matriz{a^*}{c^*}{b^*}{d^*}.
\]

We consider $\varepsilon = \pm 1 \in A$ and $J_{\varepsilon}= \left(\begin{array}{cc}
0 & 1 \\
\varepsilon & 0
\end{array} \right)\in M_2(A).$ Let us denote by $H_{\varepsilon}$  the associated $\varepsilon$- hermitian form defined by the matrix $J_{\varepsilon}$.

\begin{defi}
\label{defSL}
The group $\SLc_*^{\varepsilon}(2,A)$ is the set of all automorphisms
$g$ of the $A$-module $M = A \times A$ such that $H_{\varepsilon} \circ (g \times g) = H_{\varepsilon}$.
 In matrix form:

$$\SLc_*^{\varepsilon}(2,A)=\lbrace T \in M_2(A) \mid T J_{\varepsilon}T^*=J_{\varepsilon}\rbrace$$

\end{defi}

\begin{rmk}
In \cite{comm} it is shown that a matrix $\matriz{a}{b}{c}{d} \in M_2(A)$ is in $\SLc^{\varepsilon}_*(2,A)$ if and only if the following equalities hold:
\begin{enumerate}
\item $ab^*=-\varepsilon ba^*$;
\item $cd^*=- \varepsilon dc^*$;
\item $a^*c=- \varepsilon c^*a$;
\item $b^*d= - \varepsilon d^*b$;
\item $ad^*+ \varepsilon bc^* = a^*d+ \varepsilon c^*b=1$.
\end{enumerate}

\end{rmk}

We note that if $(A,\diamond)$ is the matrix ring $M_m(\mathbb{F}_q)$ with the transpose involution $\diamond$, then
$\SLc_{\diamond}^{-1}(2,A)$ is the symplectic group $\Sp(2m,\mathbb{F}_q)$ defined over $\mathbb{F}_q$. On the other hand
$\SLc_{\diamond}^1(2,A)$ gives the split orthogonal group $\ot_q(m,m)$.

In what follows we put $\SLc_{*}^+(2,A)=\SLc_*^{1}(2,A)$, $\SLc_*^-(2,A)=\SLc_*^{-1}(2,A)$ and $\diamond$ always will be the transpose involution in a matrix ring.

Let $(A_0,*)$ be a unitary ring with involution $*$ and $A=M_2(A_0)$.
Let us consider the matrix
\mbox{ $J= \left( \begin{array}{cc}
0 & 1 \\
-1 & 0\\
\end{array}
\right) \in A^{\times}$}, which satisfies $J^{-1}=J^*=-J$. Using the matrix $J$ we can define a new involution in $A$, namely, $a^{\sim}=Ja^*J^{-1}$.

Let us consider $M_2(A)$ provided with the involutions $*$ and $\sim$ inherited from A.\\

\begin{thm}
\label{duality}
The groups $\SLc_{*}^{+}(2,A)$ and $\SLc_{\sim}^{-}(2,A)$ are isomorphic.
\end{thm}

\begin{Proof}  Let $U= \left( \begin{array}{cc}
J & 0 \\
0 & J\\
\end{array}
\right)  \in \GL_2(A) .$
A direct computation proves that $T^{\sim}U=UT^{*}$ for all $T \in M_2(A).$
It is clear that $TJ_-T^{\sim}=J_-$ if and only if $TJ_-UT^*=J_-U$. Then we must show that $J_-U$ and $J_+$ are equivalent. In fact, the (orthogonal) matrix $P=\left( \begin{array}{cc}
0 & J \\
1 & 0 \\
\end{array}
\right)\in M_2(A)$ satisfies $PJ_+P^{*}=J_-U $.
\end{Proof}

Although the split orthogonal group is naturally a ``$\SLc^+$''- group, in practical terms it is better to look at it as a ``$\SLc^-$''-group, because this fact will greatly facilitate technical aspects.\\

\begin{cor}
The split orthogonal group $\ot_q(2n,2n)$ is isomorphic to the group $\SLc_{\sim}^{-}(2,M_{2n}(\mathbb{F}_q))$, where the involution $\sim$ in $M_{2n}(\mathbb{F}_q)$is given by $a^{\sim}=J_{2n}a^{\diamond}J^{-1}_{2n}.$  ($J_{2n}= \left( \begin{array}{cc}
0 & I_n \\
-I_{n} & 0\\
\end{array}
\right) \in M_{2n}(\mathbb{F}_q)  ).$
\label{corolario}
\end{cor}

\begin{Proof} Taking $(A_0,*)$ as the involutive ring $(M_n(\mathbb{F}_q), \diamond)$ and using the theorem (\ref{duality}) we get that the groups $\SLc_{\diamond}^{+}(2,M_{2n}(\mathbb{F}_q))$ and $\SLc_{\sim}^{-}(2,M_{2n}(\mathbb{F}_q))$ are isomorphic.

\end{Proof}

\section{A Bruhat presentation for $\SLc^{\varepsilon}_*(2,A)$  }

Let $A$ be a unitary ring with involution $*$ .We will write $A^{s}_{\varepsilon,*}$ to denote the set of all $\varepsilon$- symmetric elements in A respect to the involution $*$. Namely;
 $$A^{s}_{\varepsilon,*} = \lbrace a \in A \mid a^*=-\varepsilon a  \rbrace.$$
In order to facilitate the notation, we put $A^{s}_{+,*}=A^{s}_{1,*}$ and $A^{s}_{-,*}=A^{s}_{-1,*}$.\\ Let us consider
\[
h_t=\left(\begin{array}{cc}
t & 0\\
0 & t^{* -1}\end{array} \right) (t \in A^{\times}),\quad  w= \left(\begin{array}{cc}
0 & 1\\
\varepsilon & 0\end{array} \right),\quad u_s=\left(\begin{array}{cc}
1 & s \\
0 & 1 \end{array} \right) (s \in A^{s}_{\varepsilon,*})
\]
\vspace{0.1cm}
\begin{defi}
\label{bruhatpres}
We will say that ${\SLc^{\varepsilon}_*(2,A)}$ has a Bruhat presentation if it is
generated by the above elements with defining relations:
\begin{enumerate}
\item $h_t h_{t'}=h_{tt'}$, $u_s u_{s'}=u_{s+s'}$;
\item $w^2= h_{\varepsilon}$;
\item$h_t u_s= u_{tst^*}h_t$;
\item $wh_t=h_{t^{*-1}}w$;
\item$wu_{t^{-1}}wu_{-\varepsilon t}wu_{t^{-1}}=h_{-\varepsilon t},\quad t \in A^{\times}\cap A^{s}_{\varepsilon,*}.$
\end{enumerate}
\end{defi}

\begin{rmk}
Observing the last relation, we note that in order to have a Bruhat presentation for ${\SLc^{\varepsilon}_*(2,A)}$ is necessary that $A^{\times}\cap A^{s}_{\varepsilon,*} \neq \emptyset.$\\
\end{rmk}

In \cite{manusc} it is proved that if $A$ is a simple artinian ring with involution $*$ that either is infinite or isomorphic to the full matrix ring over $\mathbb{F}_q$ with $q>3$, then the group $\SLc_*^-(2,A)$ has a Bruhat presentation.

Thus, the group $\SLc_{\sim}^-(2,M_{2n}(\mathbb{F}_q))$ mentioned in Corollary (\ref{corolario}) has a Bruhat presentation if $q>3$. We will use this fact to construct the desired representation.

\section{A generalized Weil representation for the split orthogonal group $\ot_q(2n,2n)$. \label{rep}}

In this section, our aim is to construct a Weil representation for the split orthogonal group seen as the group $\SLc_{\sim}^{-}(2,M_{2n}(\mathbb{F}_q))$. One way to this goal is to construct the representation using the Bruhat presentation.
For this purpose we will use the result that follows (\cite{ongen}).

Let $A$ be a ring with an involution $*$.
Let us suppose that the ring $A$ is finite and that the group $G= \SLc_*^{\varepsilon}(2, A)$ has a Bruhat presentation.
Let $M$ be a finite right $A$-module and let us consider the following data:

\begin{enumerate}
\label{genweilrep}
\item A bi-additive function $\chi: M \times M \longrightarrow \mathbb{C}^{\times}$ and a character $\alpha \in \widehat{A}^{\times}$ such that for all $x,y \in M, t \in A^{\times}$:
\begin{enumerate}
\item $\chi(xt,y)=\alpha(tt^*)\chi(x,yt^*)$
\item $\chi(y,x)=\chi(- \varepsilon x,y)$
\item $\chi(x,y)=1$ for all $x \in M \Rightarrow y=0$
\end{enumerate}

\item A function $\gamma: A^{s}_{\varepsilon,*} \times M \longrightarrow  \mathbb{C}^{\times}$ such that for all
$s,s' \in A^{s}_{\varepsilon,*}$, $x,z \in M$, $r\in A^{\times}$, $t \in A^{s}_{\varepsilon,*}\cap A^{\times} $ :
\begin{enumerate}
\item $\gamma(s+s',x)=\gamma(s,x)\gamma(s',x)$
\item $\gamma(s,xr)=\gamma(rsr^*,x)$
\item $\gamma(t,x+z)=\gamma(t,x)\gamma(t,z)\chi(x,zt)$
\end{enumerate}

\item $c\in \mathbb{C}^{\times}$ such that $c^2 \vert M \vert=\alpha(\varepsilon)$,
and for all $t \in A^{s}_{\varepsilon,*} \cap A^{\times}$ the following equality holds:\\
$$\sum_{y \in M}\gamma(t,y)=\dfrac{\alpha(\varepsilon t)}{c}$$
\end{enumerate}
\begin{thm}
{\em (Guti\'errez, Pantoja and Soto-Andrade, \cite{ongen})}
Let M be a finite right $A$-module. Denote $L^2(M)$ the vector space of all complex-valued functions on $M$, endowed with the inner product with respect to the counting measure on $M$. Set:
\begin{enumerate}
\item $\rho(h_t)(f)(x)=\overline{\alpha}(t)f(xt)$, $f\in L^2(M), t  \in A^{\times}, x \in M$;
\item $\rho(u_s)(f)(x)=\gamma(s,x)f(x)$, $f\in L^2(M), b \in A^{s}_{\varepsilon,*}, x \in M$;
\item $\rho(w)(f)(x)=c\sum_{y \in M}\chi(-\varepsilon x, y)f(y)$, $f\in L^2(M), x \in M$
\end{enumerate}

(where $\overline{\alpha}$ denotes the complex conjugate of the character $\alpha$).
These formulas define a unitary linear representation $(L^2(M), \rho)$ of $G$, called the generalized Weil representation of $G$ associated to the data $(M, \alpha, \gamma, \chi)$.

\label{gen}
\end{thm}
\vspace{0.3cm}
\begin{rmk}
Let us note that this definition contains the classic Weil representation of $\SLc_2(K)$, where $K$ is a field (see \cite{soto}, for instance).

\end{rmk}

\vspace{0.3cm}

In what follows, we will focus on finding the necessary data to construct a generalized Weil representation for $\ot_q(2n,2n)$.

From now on we put $k=\mathbb{F}_q$, $A=M_{2n}(k)$ and we consider $q$ odd greater than $3$.

We will apply Theorem (\ref{gen}) to the group $\SLc_{\sim}^{-}(2,A)\cong \ot_q(2n,2n)$.
To do this, we recall the following fact. Let $E$ be a finite dimensional vector space over a field $K$. In \cite{tignol}, the authors describe a correspondence between the linear anti-automorphisms of $End_K(E)$ and the equivalence classes of non degenerate bilinear forms on $E$ modulo multiplication by a factor in $K^{\times}$. Under this correspondence, $K$-linear involutions on $End_K(E )$ correspond to non degenerate bilinear forms which are either symmetric or skew-symmetric.
Let $B$ be a non degenerate bilinear form. The aforementioned correspondence associates $B$ with $\sigma_B$, where $\sigma_B$ is a linear anti-automorphism in $End_K(E)$ defined by the following equality;
\begin{equation}
B(f(x),y)=B(x,\sigma_B(f) y), \quad \quad  f \in End_K(E), x,y \in E.
\label{correspondencia}
\end{equation}
Now, let $V$ a vector space of $k$-dimension $2n$. We fix a basis for $V$ in order to put $M_{2n}(k)\simeq End_k(V).$ \newline
Let $<\quad,\quad>: V \times V \longrightarrow k$ be the non degenerate symmetric bilinear form given by the standard dot product. We consider the non degenerate skew-symmetric bilinear form
$[\quad,\quad]:V \times V \longrightarrow k$, given by
$$[x,y]=<x,yJ_{2n}>.$$
According to the correspondence between involutions and non degenerate bilinear forms described above, the symmetric bilinear form $<\quad,\quad>$ corresponds to the transpose involution $\diamond$. Similarly, the skew-symmetric bilinear form $[\quad,\quad]$ corresponds to the new involution $\sim.$ That is, for all $x,y \in V$ and $a \in A$;

\begin{equation}
 <xa,y>=<x,ya^{\diamond}>
\end{equation}
\begin{equation}
 [xa,y]=[x,ya^{\sim}]
 \label{invsimplec}
\end{equation}

Now, let $\psi$ be a non trivial character of $k^+$. Using the notation above let us consider:

 \begin{enumerate}
 \item
 $M$ the right $A-$module  $V^{2}$ with the following action:
 $$(x,y)a=(xa,ya) \quad \quad a\in A ,x,y \in V.$$
 \item
 $\chi:M \times M \longrightarrow {{\mathbb C}^\times},$
$\chi((x,y),(v,z))= \psi([x,z]-[y,v]).$
\item
$\alpha$ the trivial character of $A^{\times}.$
\item
$\gamma: A^{s}_{-,\sim}\times M \longrightarrow {{\mathbb{C}}%
^\times},$
$\gamma(u,(x,y))=\psi([xu,y]).$\\
\end{enumerate}

\begin{lem}
For all $u\in A^{\times } \cap A^{s}_{-,\sim}$, the map $Q_u:V^{2}\longrightarrow k$ given by \newline
 $Q_u((x,y))=[xu,y]$ is a non degenerate split quadratic form.
 Furthermore, for $u, u^{\prime} \in A^{\times } \cap A^{s}_{-,\sim} $ the quadratic forms $Q_u$ and $Q_{u^{\prime}}$ are equivalent.

 \label{lemafc}
 \end{lem}

 \begin{Proof}
Let $\lambda \in k, (x,y),(v,z)\in V^2$.
Clearly $Q_u(\lambda (x,y))=\lambda^{2}Q_u((x,y))$.
We will prove that
$$B((x,y),(v,z))=Q_u((x+v,y+z))-Q_u((x,y))-Q_u((v,z))$$
is a symmetric non degenerate bilinear form.
We have;
$$B((x,y),(v,z))= [xu+vu,y+z]-[xu,y]-[vu,z]=[xu,z]+[vu,y].$$
Now;
\begin{align*}
B((x,y)+(r,t),(v,z))=& [(x+r)u,z]+[vu,(y+t)]\\
=&[xu,z]+[ru,z]+[vu,y]+[vu,t]\\
=&B((x,y),(v,z))+B((r,t),(v,z))\\
\\
B(\lambda(x,y),(v,z))=&[\lambda xu,z]+[vu,\lambda y]\\
=&\lambda[xu,z]+\lambda [vu,y]\\
=&\lambda B((x,y),(v,z)).
\end{align*}

Then, $B$ is a symmetric bilinear form.

Let us suppose that $B((x,y),(v,z))=0$ for all $(v,z)\in V^2$. If we choose $v=0$, then $[xu,z]=0$ for all $z \in V$. Since $[\quad,\quad]$ is non degenerate and $u$ is invertible, we get $x=0$. Similarly $y=0$. Therefore, $B$ is non degenerate.

Now, if $u,u^{\prime}\in A^{\times} \cap A^{s}_{-,\sim}$ then
 $uJ_{2n}^{\diamond}$ and $u^{\prime}J_{2n}^{\diamond}$ are invertible skew symmetric matrices. In fact if $u \in A^{s}_{-,\sim} $ then $u^{\sim}=J_{2n}u^{\diamond}J_{2n}^{-1}=u$. Also $J_{2n}^{\diamond}=-J_{2n}$, so we get that $(uJ_{2n}^{\diamond})^{\diamond}=J_{2n}u^{\diamond}=uJ_{2n} =-uJ_{2n}^{\diamond}$.
 Thus $uJ_{2n}^{\diamond}$ and $u^{\prime}J_{2n}^{\diamond}$ represent a non degenerate skew symmetric bilinear form, therefore they are equivalent. So, there exists $j \in A^{\times}$ such that $uJ_{2n}^{\diamond}=ju^{\prime}J_{2n}^{\diamond}j^{\diamond}$. Thus,
\begin{align*}
Q_{u^{\prime}}((xj,yj))&=[xju^{\prime},yj]\\
&=Q_u((x,y))
\end{align*}

If we choose $u=I_{2n}$, the quadratic form $Q_u$ is represented by the matrix $\left(\begin{array}{cc}
0 & -J_{2n}\\
J_{2n} & 0\end{array} \right)$. Thus, $Q_u$ is split.
\end{Proof}

 \begin{thm}

The data $(M,\alpha,\gamma,\chi)$ describe a Generalized Weil Representation for $G=\ot_q(2n,2n)$. Furthermore, this representation is independent of the choice of the character $\psi$.
\label{weilreport}
\end{thm}
\begin{Proof}
We will check that $\chi$ satisfies the corresponding conditions.\\ Let $(x,y), (v,z) \in M, a\in A.$
\begin{align*}
\textit{(a)} \quad \chi ((x,y)a,(v,z))=& \psi
([xa,z]-[ya,v]) \\
=&\psi([x,za^{\sim}]-[y,va^{\sim}]) \qquad ,\textit{by (\ref{invsimplec}) }\\
=& \chi ((x,y),(v,z)a^{\sim}).\\
\\
\textit{(b)}\quad \chi ((v,z),(x,y))=& \psi
([v,y]-[z,x]) \\
=&\psi([x,z]-[y,v])\\
=&\chi((x,y),(v,z)).
\end{align*}%

\textit{(c)}  Let us suppose that $\chi ((x,y),(v,z))=1$ for all $(v,z) \in M$. If $v=0$, then $\psi([x,z])=1$ for all $z\in V$. If $x \neq 0$, then $[x
,\; \cdot \; ]:V \longrightarrow k$ is a non trivial linear functional. Therefore it is surjective.
Let $\lambda \in k$ such that $\psi (\lambda)\neq 1$, and $t=t(\lambda)\in V$ such that
$\lambda=[x,t]$, then we get the following contradiction:
\begin{equation*}
1=\psi ([x,t])=\psi (\lambda).
\end{equation*}

Therefore $x=0$, and similarly $y=0$.
\newpage
Now, we will prove that $\gamma$ satisfies the corresponding properties.
Let $u,u^{\prime }\in A^{s}_{-,\sim}$, $a\in A^{\times}$, $(x,y),(v,z) \in M;$

\begin{align*}
\textit{(a)}\quad  \gamma(u+u^{\prime},(x,y))
=&\psi([xu+xu^{\prime},y]) \\
=&\psi([xu,y])\psi([xu^{\prime},y])\\
=&\gamma(u,(x,y)) \gamma(u^{\prime },(x,y)).\\
\\
\textit{(b)} \quad \gamma(u,(x,y)a)=& \psi([xau,ya]) \\
=&\psi([xaua^{\sim},y]) \\
=&\gamma(aua^{\sim},(x,y)).\\
\\
\textit{(c)} \quad \gamma(u,(x,y)+(v,z))=&\psi([(x+v)u,y+z]) \\
=&\psi([xu,y])\psi([xu,z])\psi([vu,y])\psi([vu,z]) \\
=&\gamma(u,(x,y))\gamma(u,(v,z))\psi([x,zu]-[y,vu]) \\
=& \gamma(u,(x,y))\gamma(u,(v,z))\chi((x,y),(v,z)u).
\end{align*}

Now, we must choose $c \in {{\mathbb C}^\times}$ satisfying $c^{2}|M|=1$ and show that for $u\in A^{\times }\cap A^{s}_{-,\sim}$ the following equality holds:

\begin{equation*}
\sum_{(x,y)\in M}\gamma (u,(x,y))=\sum_{x,y\in V}\psi ([xu,y])=\dfrac{1}{c}.
\end{equation*}

 According to the lemma \ref{lemafc}, we know that $\sum_{(x,y) \in M}\gamma(u,(x,y))$ is a Gauss sum associated to a split quadratic form in a vector space of even dimension $4n$. This sum is calculated, for instance, in \cite{soto}.
 In fact, $$\sum_{(x,y) \in M}\gamma(u,(x,y)) = q^{2n},$$

 We choose $c=\dfrac{1}{q^{2n}}$. Thus,
$$ \sum_{(x,y)\in M}\gamma(u,(x,y))=\dfrac{1}{c}.$$

 Now, let $\psi_1$ and $\psi_2$ be two non trivial characters of $k^+$. Let us prove that the corresponding representations are isomorphic.

 Let $\lambda \in k^{\times}$ such that $\psi_2(r)=\psi_1(\lambda r)$ for all $r \in k$.
 Let $(L^2(M),\rho_1)$ and $(L^2(M), \rho_2)$ the Weil representations obtained from $\psi_1$ and $\psi_2$ respectively. Then, the linear automorphism \newline
 $\Psi:L^2(M) \longrightarrow L^2(M)$ given by $(\Psi f)(x,y)=f(x,\lambda y)$ is a isomorphism between the representations $(L^2(M),\rho_1)$ and $(L^2(M), \rho_2)$.

 \end{Proof}

\section{An initial decomposition.}
\label{decomp}

\begin{defi}
\label{unit}
The group $\uni(\gamma, \chi)$ is the group of all $A$-linear automorphisms $\beta$ of $M$ such that:
\begin{enumerate}
\item $\gamma(u,\beta(x,y))=\gamma(u,(x,y))$ for all $u \in A^{s}_{\varepsilon,*},(x,y)\in M.$
\item $\chi(\beta(x,y),\beta(v,z))=\chi((x,y),(v,z))$ for all $(x,y),(v,z) \in M.$
\end{enumerate}
In what follows we will denote $\uni(\gamma, \chi)$ simply by $\uni$.
\end{defi}

 \newpage
Following the idea of \cite{ongen}, if we know the structure of the group $\uni$ and the set of its irreducible representations, we can find an \textit{initial} decomposition of the Weil Representation in the sense that we do not know if the components obtained are irreducible. In what follows, we make this decomposition explicit.

For $\beta \in \uni$ and $x \in M$ we put $ \beta.x=\beta(x)$.
 The group $\uni$ acts naturally on $L^2(M)$. That is to say  the action is given by:

\begin{align*}
&\sigma: \uni \longrightarrow Aut_{\mathbb{C}}(L^2(M)),\\
&\sigma_{\beta}(f)(x)=f(\beta^{-1}.x)
\end{align*}

In \cite{ongen} it is shown that the natural action of $\uni$ on $L^2(M)$ commutes with the action of the Weil Representation.

\vspace{0.2cm}

Let $\widehat{\uni}$ be the set of the irreducible representations of $\uni$.
We consider the isotypic decomposition of $L^2(M)$ with respect to $\uni$:
\[
L^2(M) \cong \bigoplus_{(V_{\pi},\pi) \in \widehat{\uni}} n_{\pi}V_{\pi}.
\]
Since $n_{\pi}=dim_{\mathbb{C}}(Hom_{\uni}(V_{\pi},L^2(M))) =dim_{\mathbb{C}}(Hom_{\uni}(L^2(M),V_{\pi}))$, we can write this decomposition in the following  way:

\[
L^2(M) \cong \bigoplus_{(V_{\pi},\pi) \in \widehat{\uni}} (Hom_{\uni}(L^2(M),V_{\pi})\otimes_{\mathbb{C}} V_{\pi}.)
\]
If we put $m_{\pi}=dim_{\mathbb{C}}(V_{\pi})$, we get;
\[
L^2(M) \cong \bigoplus_{(V_{\pi},\pi) \in \widehat{\uni}} m_{\pi}Hom_{\uni}(L^2(M),V_{\pi}.)
\]
If $(V_{\pi},\pi)\in \widehat{\uni}$ and $\beta \in \uni$, we denote by $\pi_{\beta }$ the map $\pi(\beta):V_{\pi} \longrightarrow V_{\pi}$.
The space $Hom_{\uni}(L^2(M),V_{\pi})$ is formed by linear functions $\Theta:L^2(M) \longrightarrow V_{\pi}$ such that for any $\beta \in \uni$
\begin{equation}
\Theta \circ \sigma_{\beta} =\pi_{\beta} \circ \Theta .
\label{entre}
\end{equation}

Let us consider the Delta functions $\lbrace e_x \mid x \in M \rbrace$ and the map  $\theta:M \longrightarrow V_{\pi}$  such that $\theta(x)=\Theta(e_x)$ for all $x \in M$. Since $\sigma_{\beta}(e_x)=e_{\beta.x}$, condition (\ref{entre}) becomes:

\begin{equation}
\theta(\beta.x)=\pi_{\beta} \circ \theta(x).
\label{casi}
\end{equation}
Conversely, let $\theta:M \longrightarrow V_{\pi}$ satisfying (\ref{casi}). We extend linerarly and we get a map $\Theta:L^2(M) \longrightarrow V_{\pi}$ such that (\ref{entre}) holds.

Thus, we can see the space $Hom_{\uni}(L^2(M),V_{\pi})$ as the function space formed by maps
$\theta:M \longrightarrow V_{\pi}$ such that $\theta(\beta.x)=\pi_{\beta}\circ \theta(x)$ for all $\beta \in \uni, x \in M$. The group $G=\SLc^{\varepsilon}_*(2,A)$ acts on this space via the Weil representation, using the same formulas as defined in Theorem (\ref{gen}). Similarly, it is possible to define the natural action of the group $\uni$ in this space, because- like $L^2(M)$- it is formed for functions with domain $M$.

\newpage
Let $\rho$ denote the Weil action of $G$ on $L^2(M)$ and $\widehat{\rho}$ the Weil action of $G$ on $\bigoplus_{(V_{\pi},\pi) \in \widehat{\uni}} m_{\pi}Hom_{\uni}(L^2(M),V_{\pi})$.
Because of how we define the Weil representation, there exist scalars $K_g(x,y)\in \mathbb{C}$ depending only on $g \in G$ and $x,y \in M$ such that for all $f \in L^2(M)$, $\Lambda \in \bigoplus_{(V_{\pi},\pi) \in \widehat{\uni}} m_{\pi}Hom_{\uni}(L^2(M),V_{\pi})$ the following statements holds:

\begin{align}
\rho_g(f)&= \sum_{y \in M}K_g(\cdot,y)f(y);\\
\widehat{\rho}_g(\Lambda)&=\sum_{y \in M}K_g(\cdot,y)\Lambda(y). \label{a}
\end{align}
In this way, we get:\\

\begin{lem}
 $(L^2(M),\rho)$ and $(\bigoplus_{(V_{\pi},\pi) \in \widehat{\uni}} m_{\pi}Hom_{\uni}(L^2(M),V_{\pi}),\widehat{\rho})$ are isomorphic representations of $G$.
\end{lem}

\begin{Proof}
The linear isomorphism between $L^2(M)$ and $\bigoplus_{(V_{\pi},\pi) \in \widehat{\uni}} m_{\pi}Hom_{\uni}(L^2(M),V_{\pi})
is an isomorphism between representations.
\end{Proof}

Finally, we have:\\

\begin{prop}
{\em The space $Hom_{\uni}(L^2(M),V_{\pi})$ is invariant under the Weil action of $G$.}
\end{prop}

\begin{Proof}
 Let $g \in G$, $\theta \in Hom_{\uni}(L^2(M),V_{\pi})$, $\beta \in \uni$, $x\in M$:
\begin{align*}
(\widehat{\rho}_g \theta) (\beta.x)&= \sigma_{\beta^{-1}}(\widehat{\rho}_g \theta)(x),\quad \textit{(definition of } \sigma_{\beta})\\
&=\widehat{\rho}_g(\sigma_{\beta^{-1}}\theta)(x)\\
&=\widehat{\rho}_g(\pi_{\beta} \circ \theta)(x), \quad (\sigma_{\beta^{-1}}(\theta)=\pi_{\beta} \circ \theta)\\
&= \pi_{\beta}(\widehat{\rho}_g \theta(x).)
\end{align*}
 The last equality holds because (\ref{a}).
\end{Proof}
\vspace{0.5cm}

Now, having made the decomposition above explicit, our purpose is to obtain an initial decomposition for our particular case $G=\ot_q(2n,2n) \cong \SLc_{\sim}^{-}(2,M_{2n}(k))$. For this it is enough to know the structure of the group $\uni$ and the set of irreducible representations.\\

\begin{rmk}
We note that since in our case $A=M_{2n}(k)$ and $A^{s}_{-,\sim}\cap A^{\times} \neq \emptyset $, the first condition in definition (\ref{unit}) implies the second one (see \cite{ongen}).\\
\end{rmk}

\begin{thm}
Let $\gamma$ and $\chi$ be the functions defined above. Then, $$\uni(\gamma,\chi)\cong \SLc_2(k)$$.
\end{thm}

\noindent{\em Proof.} Let $\beta \in \uni(\gamma, \chi)$. In particular $\beta$ is  $k-$ linear, therefore we can suppose that $\beta \in M_{4n}(k)$. We can write the action of $A$ on $M$ in matrix language as follows:

 \begin{equation*}
  (x \quad y) \matriz{a}{0}{0}{a}=(xa \quad ya)\qquad x,y \in V, a \in A.
 \end{equation*}

Since $\beta$ is $A-$linear we have that $\beta(x,y)a=\beta(xa,ya)$. In matrix language;
\begin{equation}
\beta \matriz{a}{0}{0}{a}=\matriz{a}{0}{0}{a} \beta.
\label{eq}
\end{equation}

Let $\beta_1, \beta_2, \beta_3, \beta_4 \in A$ such that $\beta = \matriz{\beta_1}{\beta_2}{\beta_3}{\beta_4}$. Then, using (\ref{eq}) we get that each of these blocks must be scalar. Thus, there are $b_1,b_2,b_3,b_4 \in k$ such that  $\beta=\matriz{b_1 I_{2n}}{b_2 I_{2n}}{b_3 I_{2n}}{b_4 I_{2n}}$ and hence
\begin{equation*}
\gamma (u,\beta(x,y))=\psi([(b_1x+b_3y)u,b_2x+b_4y]).
\end{equation*}

Let us note that the bilinear form  $(x,y)\mapsto[xu, y]$ is skew symmetric for all $u  \in A^{s}_{-,\sim}$, hence $[xu,x]=0$ for all $x \in V, u \in A^{sym}$. Thus for all $x,y \in V, u \in A^{s}_{-,\sim}$
\begin{align*}
\gamma (u,\beta(x,y))=&\psi((b_1b_4-b_2b_3)[xu,y])\\=&\psi([xu,y])\\=&\gamma (u,(x,y)).
\end{align*}

Consequently
$\psi((b_1b_4-b_2b_3-1)[xu,y])=1$ for all $x, y \in V, u \in A^{s}_{-,\sim}.$

From this last equality it follows that $b_1b_4-b_2b_3=1$. In fact, let $u \in A^{s}_{-,\sim} \cap A^{\times}$, $x \neq 0$ and let us suppose $b_1b_4-b_2b_3-1 \neq 0$.

The map $F_{x,u}:V \longrightarrow k$ given by $F_{x,u}(z)=[(b_1b_4-b_2b_3-1)xu,z]$ is a non trivial linear functional and therefore is surjective.
Let $\lambda \in k$ such that $\psi (\lambda) \neq 1$ and $z=z(\lambda) \in V$ such that \newline $\lambda=[(b_1b_4-b_2b_3-1)xu,z]$. Then
$\psi(\lambda)= \psi([(b_1b_4-b_2b_3-1)xu,z])=1. $ This contradicts our assumption and therefore our result follows.
$\qed$
\vspace{0.6cm}

Thus, for our case, we get an initial decomposition of the Weil Representation $(L^2(M),\rho)$. We expect to address the question about irreducibility elsewhere.

\section{Dual Pairs.}

In this section we will prove that the representation $(L^2(M),\rho)$ of $\ot_q(2n,2n)$ constructed in section \ref{rep} is equal to the restriction of the Weil representation to $\ot_q(2n,2n)$ for the dual pair $(\Sp(2,k),\ot_q(2n,2n))$. Also, we will prove that the initial decomposition described above is the same as decomposition with respect to the action of $\Sp(2,k)$ via the Weil representation.

Let $J_{2n}= \left(\begin{array}{cc}
0 & I_n \\
-I_n & 0
\end{array} \right)\in M_{2n}(k)$ and $F= \left(\begin{array}{cc}
0 & J_{2n} \\
-J_{2n} & 0
\end{array} \right)\in M_{4n}(k).$
The matrix $F$ defines the following non-degenerate split symmetric bilinear form in $V_1=k^{4n}$
$$(u,v)=v^tFu \qquad u,v \in V_1$$
The group $G$ of isometries of this form is isomorphic to the split orthogonal group $\ot_q(2n,2n)$.
As before, set
$$a^{\sim}=J_{2n}a^tJ_{2n}\qquad a \in M_{2n}(k).$$

A direct calculation shows that the following matrices belong to the group $G$:
\[
h_a=\left(\begin{array}{cc}
a & 0\\
0 & (a^{\sim})^{ -1}\end{array} \right) ,\quad  w= \left(\begin{array}{cc}
0 & I_{2n}\\
-I_{2n} & 0\end{array} \right),\quad u_s=\left(\begin{array}{cc}
I_{2n} & s \\
0 & I_{2n} \end{array} \right)
\]
$(a \in M_{2n}(k)^{\times}, s =s^{\sim} \in M_{2n}(k))$.

Therefore $G=\SLc_{\sim}^{-}(2,M_{2n}(k)$.

\newpage
Let $V_2= k^2$ and $W=Hom(V_1,V_2)$. The following formula defines a non-degenerate symplectic form on $W$.
$$\ll w_1,w_2 \gg =tr(w_1Fw_2^tJ_2)\quad \quad (w_1,w_2 \in W )$$
The group $G$ acts on $W$ by
$$g(w_1)=w_1g^{-1} \quad \quad (g \in G, w_1 \in W).$$
This action preserves the symplectic form $\ll\cdot,\cdot \gg$. In fact, since $g \in G$,
$$\ll w_1g^{-1}, w_2g^{-1} \gg =tr(w_1g^{-1}F(g^{-1})^tw_2^tJ_2)=tr(w_1Fw_2^tJ_2)= \ll w_1, w_2 \gg.$$

Let
$$X= \lbrace(x,0) \mid x \in M_{2,2n}(k) \rbrace ;\quad Y= \lbrace (0,y)\mid y \in M_{2,2n}(k) \rbrace.$$
Then $W=X\oplus Y$ is a complete polarization. We will consider the Schr\"odinger model of the Weil representation of $\Sp(W)$ attached to the above complete polarization realized on $L^2(X)$ as in \cite{aubert}. Let $(L^2(X),\omega)$ such representation.

We identify $X$ with $M_{2,2n}(k)$ in the canonical way
 $$X \ni (x,0) \leftrightsquigarrow x \in M_{2,2n}(k).$$
 \begin{rmk}
 Let us note that the module $M$ in section \ref{rep} is canonically isomorphic to $X$. Consequently the spaces $L^2(M)$ and $L^2(X)$ are also isomorphic.
 \end{rmk}

Let $\psi$ a non-trivial character of the additive group $k^+$.
For all $x \in X$, $y \in Y$ it is clear that $h_a(x)=xh_a^{-1} \in X$ and $h_a(y)=yh_a^{-1} \in Y$, then the matrix $h_a$ preserves $X$ and $Y$. Also, $det(h_a \vert_X)\in k^{\times 2}$. Thus, proposition 34 in \cite{aubert} shows that:
$$\omega (h_a)f(x)=f(xa)  \quad \quad (f \in L^2(X))$$
Thus, $\omega(h_a)=\rho(h_a).$

Now, let us see the action of $\omega$ on $u_s$. The matrix $u_s$ acts trivially on $Y$ and on $W/Y$. Therefore, proposition 35 in \cite{aubert} shows that:
$$\omega(u_s)f(x)=\psi(\ll xc(-u_s),x \gg)f(x),$$
where $c(-u_s)= \left(\begin{array}{cc}
0 & -s/2\\
0 & 0\end{array} \right) \in M_{4n}(k)$ is the Cayley transform for $-u_s$.

Let $x= \left(\begin{array}{cc}
x_1 & 0\\
x_2 & 0\end{array} \right)\in X $, $x_1,x_2 \in k^{2n}$. Then,
 $$\ll xc(-u_s),x \gg=x_2sJ_{2n}x_1^t.$$
 In order to prove that $\omega(u_s)=\rho(u_s)$ we have to check that
 $$[x_1s,x_2]=\ll xc(-u_s),x \gg,$$
 where $[\quad,\quad]$ is the symplectic form defined in section $4$. In fact,
 $$[x_1s,x_2]=[x_1,x_2s]=-[x_2s,x_1]=x_2sJ_{2n}x_1^t$$
\newpage
 It is clear that the matrix $w$ maps $X$ bijectively onto $Y$ and $Y$ onto $X$, and $w^2=-1$. Then, using proposition 36 of \cite{aubert} we get:
 $$\omega(w)f(x)=\dfrac{1}{\sqrt{\vert X \vert}}\sum_{x'\in X}\psi (\ll w(x),x' \gg)f(x')$$
 Thus, in order to prove that $\omega(w)=\rho(w)$ we have to check that
 $$\chi(x,x')=\psi (\ll xw^{-1},x' \gg).$$
 Let $x= \left(\begin{array}{cc}
x_1 & 0\\
x_2 & 0\end{array} \right) $,$x'= \left(\begin{array}{cc}
x'_1 & 0\\
x'_2 & 0\end{array} \right) $, $x_1,x_2, x'_1, x'_2 \in k^{2n}$. So,
$$\psi (\ll xw^{-1},x' \gg)=\psi(x_2J_{2n}(x_1')^t-x_1J_{2n}(x_2')^t).$$
On the other hand,
\begin{align*}
\chi(x,x')&=\psi([x_1,x'_2]+[x'_1,x_2])\\
&=\psi (\left( x_1 \quad x_2 \right) \left(\begin{array}{cc}
0 & -J_{2n}\\
J_{2n} & 0\end{array} \right) \left(\begin{array}{c}
(x'_1)^t \\
(x'_2)^t \end{array} \right))\\
&=\psi(x_2J_{2n}(x_1')^t-x_1J_{2n}(x_2')^t)).
\end{align*}
Thus, we have showed that the representation constructed in section \ref{rep} is equal to the restriction of the Weil representation to $O_q(2n, 2n)$ for the dual pair $(\Sp(2,k),\ot_q(2n,2n))$.

Furthermore, since an element $g \in \Sp(2,k)=\SLc_2(k)$ preserves $X$ and $Y$ and $det(g\vert_X)\in k^{\times 2}$, using proposition 34 in \cite{aubert} we get that the group $\SLc_2(k)$ acts on $L^2(X)$ as follows:
$$\omega(g)f(x)=f(g^{-1}x) \quad g \in \SLc_2(k), f \in L^2(X), x \in X.$$

Therefore, the initial decomposition in section \ref{decomp} is the same as the decomposition with respect to the action of $\SLc_2(k)$ via the Weil representation.

\vspace{1cm}
\begin{flushleft}
\textbf{Acknowledgments:} The author would like to thank Jos\'e Pantoja, Jorge Soto, Luis Guti\'errez and Anne-Marie Aubert for their never ending support and willingness.
\end{flushleft}

\newpage

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20th Century Fox and Chernin Entertainment have tapped author Dennis Lehane to put together a script for Silk Road.
To the surprise of no one who has followed one of the most incredible stories of the year, the Deep Web black market known as Silk Road is going to Hollywood.
It's only been two weeks since the FBI shut down the the hugely successful online drug marketplace, but speculation about a silver screen version has been going on since the handcuffs closed around alleged mastermind Ross William Ulbricht's wrists.
20th Century Fox and Chernin Entertainment have tapped author Dennis Lehane to put together a script for Silk Road based on an unreleased piece of longform journalism by Joshua Davis for Wired, reports Defamer.
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text:
def is_leap_year(year):
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text:
Q: Location of policy file on Debian I'm working on an application/library, under DebianOS, that requires the presence of certain standard policy files (related to unlimited strength ciphers, 192, 256 bit AES keys).
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A: Try with 'find':
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text:
« SW's Top 10 | Main | Narcissism in the Real World II »
Narcissism in the Real World
During the course of the last week we have seen the full range of Narcissism on display. The Oscars have always been a celebration of Narcissism (who can ever forget Sally Field's famous squeals of delight, "they like me, they really like me") but recently Hollywood has outdone themselves. Ben Stein points out the striking omission of any mention, even a dismissive perfunctory mention, of our men and women in uniform overseas helping to protect the very freedoms the glitterati take for granted; instead they applaud their faux bravery, which costs them nothing:
The idea that it is brave to stand up for gays in Hollywood, to stand up against Joe McCarthy in Hollywood (fifty years after his death), to say that rich white people are bad, that oil companies are evil -- this is nonsense. All of these are mainstream ideas in Hollywood, always have been, always will be. For the people who made movies denouncing Big Oil, worshiping gays, mocking the rich to think of themselves as brave -- this is pathetic, childish narcissism.
Hollywood is above all about self: self-congratulation, self-promotion, and above all, self-protection. This is human and basic, but let's not kid ourselves. There is no greatness there in the Kodak theater. The greatness is on patrol in Kirkuk. The greatness lies unable to sleep worrying about her man in Mosul. The greatness sleeps at Arlington National Cemetery and lies waiting for death in VA Hospitals. God help us that we have sunk so low as to confuse foolish and petty boasting with the real courage that keeps this nation and the many fools in it alive and flourishing on national TV.
The MSM take their Narcissistic grandiosity to another level. The New York Times, among others, have proclaimed, while defending their publication of top secret information, that they believe they are better suited to determine what is harmful to our national security than the professionals.
MacRanger offers another example and is more caustic about it than I am; his take on the MSM is worth reading:
Aside from the fact that the vaunted and out of control - and stragegically dangerous - MSM is out of about to get their ass handed to them, it's fun to see them whine about their percieved self-importance.
Get over yourself, you lost that importance when you stopped reporting the news and began commenting, bloviating and outright manufacturing the the news. You lost credibility a long time ago. Yet like the stuck pig you are you just have to squeal about it.
He also offers an example of exactly what he is describing. While you are at his site, you might want to check in on what happens when a United Sates Senator puts his own Narcissistic needs ahead of his duties and responsibilities to his country. It is not a pretty sight.
Both Dr. Sanity and Neo-neocon point out that the left and too many Democrats rationalize their unpatriotic behavior which is in the service of their narcissistic needs. Dr. Sanity says:
You also see the pursuit of cheap glory in the left's sneering contempt and devaluation of their fellow citizen siblings who are willing to defend freedom in the real world, instead of the fantasyland that exists only in the windmills of the leftist mind.
Meanwhile, in that real world, there exists real evil, as well as real threats to our freedoms. People are dying to defend Freedom of Speech. People are being murdered for desiring freedom and democracy. Standing up against those real world villians, though, is far too frightening for the narcissistically preoccupied (why, they might get hurt! And they want to make sure they stand on the correct side of history, after all).
Neo-neocon points out the infantile roots of their behavior:
I'm not saying leftists--or those liberals who join the "blame America first, often, loudly, and last" chorus--are children. They are not. But in their relationship to their own government they seem to be acting out a similar dynamic.
Infants who have tantrums can be ignored. Unfortunately, we cannot ignore our MSM and the left-wing of our polity; they control organs of information dissemination, control our schools, and are damaging our war efforts. This could all be dismissed as just noxious and annoying if not for the real world consequences.
The most dangerous form of Narcissism is on display almost everyday. While it is always problematic to diagnose from a distance, Mahmoud Ahmadinejad offers all the signs of being a Malignant Narcissist. He is grandiose, heedless of other people's lives and desires, and believes he is divinely inspired and chosen. People who try to paint President Bush's as a theocrat and a fascist in their inane confusion try to obscure the differences between our President and a true fascist theocrat; Mahmoud Ahmadinejad is the "real thing." Dinocrat has more:
We are pretty much obsessed with Mahmoud Ahmadinejad because we take him at his word. He has famously said that Israel should be wiped off the map, and that the EU should relocate the Jews back to Europe if they feel so guilty about the Holocaust. Here's his latest, from Tass (HT, Captain Ed):
Islam will soon be the domineering force in the world, placing first in the number of its followers among all other religions. Iranian President Mahmoud Ahmadinejad expressed this confidence here at the end of his state visit to Malaysia. Following a meeting with Sultan Jamalullail I, the supreme head of the federation of nine states where Islam was proclaimed the state religion, he pontificated: "The world will be in the hands of Islam over the next few years."
Dinocrat also wonders how Ahmadinejad can bring his dream of murder in the name of his own grandiose vision to fruition; he paints a frightening and plausible scenario.
All of which brings me to the critical question:
At such a dangerous time, when we face implacable enemies who wish to destroy us and our freedoms, why do so many bright people continue to do things that place our nation and our friends at great risk for what seems to be the most minimal, narrow and partisan of advantages?
As luck would have it, a partial answer has been offered by those on the farthest reaches of the American left and tomorrow I will use their words to explore what motivates them and their witting and unwitting allies.

redpajama_set_name: RedPajamaCommonCrawl


When my patients are fully informed about their cancer and treatment options, they become empowered partners in shared decision-making.
Dr. Jeffrey Herman is a board-certified radiation oncologist who grew up in the Twin Cities and returned to his roots. He joined MRO as Director of Minneapolis Radiation Oncology at MRO Unity in 1992 to lead its expansion as the first radiation center in the northern metro area.
Dr. Herman uses sophisticated techniques in radiation technology, including 3-Dimensional Radiation Therapy, Intensity Modulated Radiation Therapy (IMRT), Stereotactic Ablative Body Radiotherapy (SABR), and Stereotactic Radiosurgery (SRS), among others.
Dr. Herman has served on several boards for advancing cutting-edge cancer treatment and prevention, including the Allina Breast Cancer and Lung Cancer boards.
When he is not working, he enjoys photography, computer technology, music, bicycling, and playing with his dog.


text:
Justice Committee on Feb. 23rd, 2017
Evidence of meeting #48 for Justice and Human Rights in the 42nd Parliament, 1st Session. (The original version is on Parliament's site, as are the minutes.) The winning word was crimes.
A recording is available from Parliament.
MPs speaking
Anthony Housefather
Ron McKinnon
Alistair MacGregor
Chris Bittle
Marco Mendicino
Ted Falk
Randy Boissonnault
Rob Nicholson
Also speaking
Ihsaan Gardee Executive Director, National Council of Canadian Muslims
Glenn Gilmour Counsel, Criminal Law Policy Section, Department of Justice
Lyne Casavant Committee Researcher
The Chair Liberal Anthony Housefather
Okay. I appreciate that.
We'll go to short questions, for anybody who has them.
Mr. Cooper.
Michael Cooper Conservative St. Albert—Edmonton, AB
Just to follow up the line of questioning by Mr. Housefather, based upon what I'm hearing you say, you would support encompassing all types of property. You make no distinction at all between whether it is a school, a synagogue, or a mosque. If the act is an act of mischief that is motivated by hate and that expresses hate, your position is that, regardless of what property specifically is targeted, this section of the act should encompass all of those acts of mischief. Is that, in essence, your position?
Executive Director, National Council of Canadian Muslims
Ihsaan Gardee
That's correct. You have to remember, as I mentioned, that these are message crimes, and that the message is still heard, even if it's targeted at a place that is not traditionally frequented by a particular faith group. It's still heard by the people from that group, as well as by others within society. It reinforces the "other-ization" of particular groups and creates a sense of alienation and marginalization in those groups, even if there is no formal direct connection to that particular place.
Thank you. I think that's well understood.
Are there any other questions?
Do we have unanimous consent for Mr. Mendicino to ask a question?
Go ahead, Mr. Mendicino.
February 23rd, 2017 / 4:40 p.m.
Marco Mendicino Liberal Eglinton—Lawrence, ON
Thank you, Mr. Chair.
Just building on the line of questioning by both the chair and Mr. Cooper, the two categories that seem to engage the section that is the subject of this act are hate speech and buildings and structures.
Based on your evidence, and as I understand your answers to the chair and Mr. Cooper, if we do away with any kind of defining of the categories of structures, does that render that second category essentially nugatory? What's the point of having any category?
I ask this because, if you go back to the original intent of Parliament when it created this section, it originally dealt specifically with building structures that are obviously and publicly identifiable as being used primarily for religious purposes, structures in those explicit categories that are already there. If we do away with that, what comments can you offer about what that does to the original intent of the section?
I think it would seem to broaden it and it would certainly seem to be an attempt to capture more of what is actually happening out there. If you maintain it as it is, you could have incidents occurring that are increasing the sense of alienation and marginalization of specific groups while appropriate attention and resources are not being provided by government through policies, as I mentioned in my submission, that specifically combat Islamophobia, anti-Semitism, racism, homophobia, or any type of xenophobia.
Also, building on the existing policies that we have to promote things like multiculturalism, inclusion, and so forth would seem to me to be important to consider as well.
Just to get back to the premise of my question, does it make sense to your mind to still have buildings and structures as a second criterion that engages this section for the purposes of special condemnation of hate speech expressed in certain places that relate to the identifiable protected group of those who are religious?
I think it makes sense to maintain continuity, to certainly have that there. But at the same time, it should recognize that these spaces are not the only ones that are targeted and that there is a growing trend of xenophobia targeting various groups. Currently Muslims are being targeted, as I mentioned. Tomorrow it will be another group. If you're not actually capturing that, then you don't have a real sense of what's actually going on.
For those other protected groups, I don't know if you're familiar with the Criminal Code or not, but under section 718.2, there are other sentencing principles that a trial judge can look at, including whether or not the offence is motivated by hatred based on all of the other groups.
Are you familiar with that section?
As I mentioned, I don't come from a legal background but I am surrounded by lawyers. I'm sure I can get back to you with a more qualified statement regarding your question.
Thanks, Mr. Chair.
Any there other short comments?
If not, I thank you very much for being here with us today. It's very much appreciated.
We'll take a short break while the next witness comes up.
Ladies and gentlemen, we're going to reconvene.
We are very lucky to be joined by Mr. Glenn Gilmour, counsel from the criminal law policy section of the Department of Justice, who is here to answer questions from the committee relating to this bill.
Mr. Gilmour, welcome.
Thank you very much for inviting me.
Did you have anything you wanted to say before we ask questions?
Counsel, Criminal Law Policy Section, Department of Justice
Glenn Gilmour
I have a couple of comments I would like to make first just for the record.
A part of it might be a bit of a history lesson for people around the table, but I thought I'd start with the discussion about the provision dealing with sentencing, subparagraph 718.2(a)(i) of the Criminal Code.
That was originally created in 1995 by the then-Liberal government under Jean Chrétien. I believe it was a campaign promise that had been made in the 1993 election. I thought I'd quote something that Allan Rock had said at that time. In his appearance before the Standing Committee on Justice and Legal Affairs in 1994, he discussed the hate crime sentencing provision that was proposed in the bill at that time, which subsequently enacted 718.2(a)(i). He said:
Why is it there? I think all of us are aware of the appalling increase in recent years in the incidence of hate crimes in our society. Every party that ran in the last election expressed its concern about that phenomenon. I think we join together regardless of party stripe in agreeing that we cannot tolerate hate crimes in Canadian society. It's there because of certain commitments made by the government, of which I'm a member, during the election and since it. It's there because B'nai Brith, for example, has told the Department of Justice that there are now over forty organized hate groups in Canada actively at work every day of the week.
He went on to say:
When someone goes onto my property and spray-paints graffiti on the side of my house, that is a crime that should be dealt with accordingly. I am the victim. But if they walk into the grounds of a synagogue and spray-paint a swastika on the side of the wall, the attack is not only against that property and that owner; it's against the Jewish faith as a whole. Every member of the Jewish faith is intended to feel intimidated and more vulnerable because of it. That is what distinguishes crimes motivated by hate
—presumably from other regular crimes that are not so motivated.
I put that on the record to indicate that when the sentencing provision was enacted in 1995, it was because of a strong feeling that Canada needed to enact special legislation to deal with hate crimes and to more appropriately denounce the seriousness of those kinds of crime.
Therefore, you have 718.2(a)(i), which reads:
A court that imposes a sentence shall also take into consideration the following principles: (a) a sentence should be increased or reduced to account for any relevant aggravating or mitigating circumstances relating to the offence or the offender, and, without limiting the generality of the foregoing, (i) evidence that the offence was motivated by bias, prejudice or hate based on race, national or ethnic origin, language, colour, religion, sex, age, mental or physical disability, sexual orientation, or any other similar factor
You have a very strong sentencing provision in the Criminal Code that was originally created way back in 1995. The concerns for putting it in there, I think, very much mirror the sorts of concerns that have been expressed around the table by some of the witnesses who have appeared before this committee.
The next change that happened was the creation of section 430(4.1) of the Criminal Code, the current hate crime of mischief directed against property "primarily used for religious worship". It's motivated by hatred based on various criteria.
The reason for the limiting of that offence to the concept of protecting property primarily used for religious worship was that it was thought that that particular kind of mischief would create a chilling effect on those who wanted to practise their religion. Therefore, it was designed specifically to protect that kind of property and not any other kind of property, even though when the bill that the offence was part of—the Anti-terrorism Act of 2001—was being debated in Parliament. There were some organizations that came before the House of Commons and the Senate and argued that it should be expanded to include other kinds of property.
I'd like to correct what appears to be a misconception that occurred in the testimony given the other day by Mr. Arya. There was a question asked about a house of worship being vandalized and the maximum punishment of 10 years in jail for that. If a Jewish community said it had been vandalized, the maximum punishment would be two years in jail. In case there is any misconception, that is incorrect. The way that the general mischief offence works in the Criminal Code is that it can either be prosecuted by way of indictment with a maximum penalty of 10 years in jail, or simply by way of summary conviction, which is a maximum of two years in jail.
The way the choice is made whether to proceed by indictment or by summary conviction depends not on the value of the damage to the property, but on the value of the property itself. Under the general mischief offence, if I were to vandalize a home, on the assumption that most homes these days cost more than $5,000 or less, it would be the general mischief offence that would apply, which has a maximum of 10 years in jail. That is the same penalty that is proposed in the private member's bill C-305. It's also the same penalty that currently exists for the hate crime mischief offence.
I also want to briefly mention that there had been mention of some statistics published in recent years of hate crimes that have been committed. According to my analysis of the testimony, the most recent statistics quoted for the committee were those from the year 2013. In fact, last year there was a table published by Statistics Canada that gave hate crime statistics for the year 2014. It was just a table; it was not in the form of a regular report with analysis. According to those tables, in the year 2014, the total police-reported hate crimes—and these are reports that are made by the police to Statistics Canada—was 1,295. Of those, in terms of the violations of the criminal law that occurred, the total of all violations to the criminal law that the police categorized as hate crimes was 1,170. Of those, 523 were mischief, and mischief in relation to religious property motivated by hate was a total of 89.
I just wanted to bring those particular issues to the attention of the committee.
Thank you very much for bringing those.
I just want to clarify one point in your correction. The difference, as I understand it, would be that in a summary conviction, if it's mischief related to religious property, it's an 18-month offence, and it's only six months at maximum if it's a property valued at more than $5,000.
I think the hate crime mischief offence, as it is currently worded, would allow for a maximum punishment of 10 years in jail.
But the minimum is six months in one, and 18 months in another, is it not?
There is that.
That's the only distinction, essentially that—
That's the only distinction.
—the minimum sentence is six months in one and 18 months in another.


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