text:
Chloe Sevigny, who was once called the "coolest girl in the world", is an American actress, fashion designer and also a model in her earlier days. Her debut role was in the controversial film Kids (1995), which got her nominated for Independent Spirit Award. By 1999, Sevigny shot to recognition in the fact based drama, Boys Dont Cry, earning her nominations for Golden Globe and Academy Award for Best Supporting Actress. After her controversial film Kids, her role in The Brown Bunny, also stirred a controversy wherein she performed unsimulated fellatio. She is well known in the television series Big Love which got her the Golden Globe for Best Supporting Actress. She then became a popular face on the small screen. Recently she ventured into direction with her debut short film called Kitty.
Chloe Sevigny Religion
Chloe Sevigny was born as Chloe Stevens Sevigny, as a second child to Janine and David Sevigny. She has one older sibling. She was raised as a strict Catholic by her parents. She had rebelled against the family religion in her younger days but today she is a practicing Catholic. In one of her interviews, she stated that she also practicesYoga, as she suffers from scoliosis (a spinal deformity) and yoga give her the much needed relief.
Chloe Sevigny Political Views
Though Chloe keeps her political views to herself, she has openly stated that she is a keen follower of politics and current affairs. She follows outlets and figures opposite her political affiliation to understand their points of view. She is more into humanitarian activities and her top picks are Democratic Senator Elizabeth Warren, barrister Amal Clooney, actor and humanitarian Angelina Jolie and Palestinian-American-Muslim civil rights activist Linda Sarsour.
Who is she dating?
Chloe Sevigny has had many relationships with men. She has dated Jason Segel in a short relationship, Matt McAuley with whom she dated for nearly 8 years, Jarvis Cocker for an year and director Harmony Korine for another couple of years. She also had a brief fling with director James Ransone and rumoured to have been with Vincent Gallo for a couple of years.
Chloe Sevigny Hobbies
Chloe Sevigny expressed interest in acting right from her childhood days. She is very much like the person next door. She used to be model, she is into direction, she has been a fashion designer. She is ambitious but not ambitious enough to the extent that she wants to be a movie star or famous. She like to act in good movies and move some people. As of late, she accepts that she is a homebody, just like all of us. She had always aspired to be an actress and acting is her main hobby.
Originally hailing from Connecticut, Sevigny attributes weekend trips into nearby New York City in her teens as an important early saving grace. Impressed by her flair for street fashion, the editors of the progressive teen mag asked Sevigny to intern in their offices which led to a few modeling jobs.
"Chloë's not afraid to look different and in looking different, she looks very charismatic. No one in LA gets it. Her attitude is foreign to this city", said fashion historian Cameron Silver.Sources
https://en.wikipedia.org/wiki/Chlo%C3%AB_Sevigny#Awards_and_nominations
http://www.imdb.com/name/nm0001721/bio
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text:
Is This Going To Make Me A Different Person? Gender Transition Questions Answered
By Connor Donevan • Apr 18, 2017
Julian Harris (left) tells Tor Des Roches what he can expect as he begins medically transitioning from female to male.
Courtesy of Julian Harris; Andrew Jones/Courtesy of Tor Des Roches
When you're facing a major life change, it helps to talk to someone who's already been through it. All Things Considered is connecting people on either side of a shared experience, and they're letting us eavesdrop on their conversations in our series Been There.
Tor Des Roches is right in the middle of a long, life-changing process. It started years ago, when, as a teenager, he started to have the feeling that he didn't identify as female — his sex at birth.
At the time, he didn't want to say he was transgender. "I just always kept assuming that, oh, it's just a phase, it's just a phase," he says. "It was just one of those things that I was really hoping would go away."
The feeling didn't go away, and Tor struggled to accept it. He suffered from an eating disorder, partly, he says, because he didn't want to have a woman's figure.
Eventually, he did come to terms with his gender identity. He came out as trans to close friends and family, wore men's clothes and started going by Tor instead of the name he was given at birth.
But the person Tor saw in the mirror didn't match the person he felt like. And he decided that this year, at age 24, he would have sex reassignment surgery and start taking testosterone. It's a big step, and he was nervous, he says. "Is this gonna make me a different person? Or is my attitude gonna change?"
Tor talked about some of those concerns with Julian Harris. His transition from female to male began eight years ago. Julian says he remembers being frustrated that the changes didn't take effect immediately. "One day you're getting sir'd and he'd, and the next day you're getting she'd and her'd," he says.
Lessons from Julian Harris
On becoming less self-conscious after physically transitioning
I stopped trying to overcompensate. And what I mean by that is, before I started my physical transition, I went overboard with my mannerisms. I had to be this kind of ultra-masculine-presenting person: the way I walked, trying not to gesture too much. And once those changes took hold and I was being seen by the world as who I really was, I didn't have to do that anymore.
On keeping reasonable expectations
Be mindful of those triggers that can get you in that place of despair or suffering, because the things that you might have some insecurities about now may still be there. And so, you know, [be] able to take care of yourself, to make sure that you'll be OK regardless.
On discrimination he's faced since transitioning
People see me as a threat, you know, as a black man. I think I'm very non-threatening. I'm always smiling. I wear glasses. But sometimes I'll walk down the street and people will cross, or people will clutch their bags a little tighter, or pull their kids to them. It's funny, I can be on a packed bus sometimes, and no one will sit next to me.
You know, there's this male privilege, though, that's been bestowed upon me. So, if I go to the sports bar and I have my little sports analysis, people listen to me. You know, before they didn't. You know, safety — I feel if I'm walking home late at night, I'm not concerned about a sexual assault.
So you get a little, and then you lose a little.
Tor des Roches is right in the middle of a long, life-changing process. It began years ago when he was a teenager. He was starting to have the feeling that he didn't identify as female, his sex at birth.
TOR DES ROCHES: I just always kept assuming that, oh, it's just a phase; it's just a phase. I didn't want to say I was transgender. It was just one of those things I was really hoping that would go away. It was a real struggle to come to terms with it.
MCEVERS: Eventually he did come to terms with it. He came out as trans to close friends and family. He wore men's clothes. He started going by Tor instead of the name he was given at birth. But the person Tor saw in the mirror didn't match the person he felt like. And he decided that this year, at age 24, he would have sex reassignment surgery and start hormone therapy. It's a big step, and Tor wanted someone to talk to about all the physical and emotional changes he was about to go through.
ROCHES: Hi, Julian.
JULIAN HARRIS: Hey, Tor. How are you?
ROCHES: I'm doing great, yourself?
HARRIS: I'm doing pretty well.
MCEVERS: Julian Harris went through a similar process eight years ago when he started his physical transition from female to male. Julian sat down to talk about it with Tor for our series Been There, connecting people at either end of a shared experience, and they both had challenges.
Tor's family was supportive, but he struggled with an eating disorder. He says it was partly because he didn't want to have a woman's figure. Julian, who was raised as a Jehovah's Witness, was worried about coming out to his family.
HARRIS: Well, I told my cousin first. And she was really awesome. I mean she didn't quite get it. But you know, she was supportive of me. But when I told my mom, I told her through a letter. And you know, she had a hard time accepting it. She basically told me that if I did this, then I was no longer her child. So in essence...
ROCHES: Right.
HARRIS: ...Yeah, I lost my immediate family. But I gained a lot of chosen family. I'm really glad to hear, though, that your parents are supportive because it's...
ROCHES: Yeah, me too.
HARRIS: ...Yeah, because it's a difficult thing.
ROCHES: It definitely took me a good while to get to this point. And I am pretty nervous about the surgery not because I don't feel it's right, but it's definitely a big change for me. And I'm kind of nervous about going on hormones myself. I heard some friends tell me that, oh, you know, once you start taking testosterone, you'll feel a lot more energized, and you'll feel great. But at the same time, it makes me nervous, wondering, is this going to make me a different person, or is my attitude going to change? I'm just - I guess I'm wondering how it - that's been like for you?
HARRIS: For me, it gave me a little more energy. It increased my sex drive. It changed my body. My hips are small. My shoulders are more broad.
HARRIS: I got a goatee in three months. You know, your voice starts to crack like you're 13.
HARRIS: You know, it's like a second puberty, but you're an adult. But yeah, the biggest thing, you know, for me, my personality - I didn't change. The interesting thing is, though, I stopped trying to overcompensate. And what I mean by that is, before I started my physical transition, I went overboard with my mannerisms. I had to be this kind of ultra-masculine presenting person - the way I walked, trying not to gesture too much. And once those changes took hold and I was being seen by the world as who I really was, I didn't have to do that anymore.
ROCHES: I definitely relate to that. When I catch myself putting my hand on my hip, I am just like, no, don't do that; don't do that.
HARRIS: Yeah.
ROCHES: I get nervous even now when I speak or I go out in public because I'm fairly androgynous, but I know that if I speak, I feel like my voice gives it away. So I tend to mumble, or I avoid talking because I'm so worried, like, what if I get found out? It feels like I'm fake and I'm just faking it, so I definitely relate to that.
HARRIS: Yeah, and it's hard. And some people will treat you that way.
HARRIS: But you're not faking it. I mean no one does this because, oh, this sounds like something that might be interesting to do. (Laughter) Once your exterior matches their interior, it doesn't become that big of a worry. But I think that it's also important to kind of be mindful of those triggers that can get you in that place of despair or suffering because the things that you might have some insecurities about now may still be there. And so...
HARRIS: ...You know - being able to take care of yourself to make sure that you'll be OK regardless.
ROCHES: Right. I guess I was also wondering, do you face any discrimination? I mean I know from personal experience, I remember being at a convention with my friends, and I just had to go to the bathroom, and so I decided to go to the women's bathroom. And again, since I'm so androgynous, it's kind of - I'm in the middle where it's hard to tell whether I'm male or female from first glance.
And so I was walking to the bathroom, and I had some woman start yelling at me. What are you doing in the women's restroom? And at that point in time, I just got anxiety because it's, like, oh, my gosh, what am I supposed to do?
HARRIS: Right. You know, and when you're in that period in the - where you're, like you said, in the middle, one day you're getting sir'd and he'd, and the next day you're getting she'd and her'd. I remember when I first started, you know, hormones, I was just so impatient. I wanted them to take effect immediately. I wanted everybody to see me as male.
I was working at a coffee shop, and there was this man who would drink on the corner. And when I would walk by, he'd kind of eye me. And he'd always say, hey, baby. And I would get really, really upset, but I would just keep walking. I would ignore him. And you know, I started to grow facial hair, like I said, three months in. And then one day, he was standing there with another guy, and the man looked and said, hey, baby. And his friend looked at him and was like, why are you hitting on that guy?
HARRIS: And I felt really validated.
HARRIS: At that point I was like, yes. And then I kind of smirked and was like, yeah, why are you hitting on me?
HARRIS: So that was something. But it was - it's funny now, but it was really, really upsetting. Other things - you know, the biggest things I face now as far as discrimination is that people see me as a threat. You know, as a black man, I think I'm very non-threatening. I'm always smiling. I wear glasses. But sometimes I walk down the street, and people will cross, or people will clutch their bags a little tighter or pull their kids to them.
It's funny. I can be on a packed bus sometimes, and no one will sit next to me. You know, there's this male privilege, though, that's been bestowed upon me. So if I go to the sports bar and I have my little sports analysis, people listen to me (laughter).
ROCHES: Oh, wow.
HARRIS: You know, before they didn't. You know, safety I feel. You know, if I'm walking home late at, night I'm not concerned about a sexual assault. So you get a little, and then you lose a little.
ROCHES: I guess I wanted to ask you, Julian. So when you look in the mirror, are you amazed by the changes, and are you happy with them?
HARRIS: I am happy with the changes. You know, I didn't turn into Denzel Washington (laughter). But yeah, so I have to say, you know, I walk into a place. People don't look at me. I just - I guess I'm just like an average, short guy.
MCEVERS: That average, short guy was Julian Harris who underwent sex reassignment eight years ago. He was talking to Tor Des Roches for our series Been There. And if you're going through something big and life-changing and you want advice about it, let us know. That's how we heard from Tor. Email nprcrowdsource@npr.org with Been There in the subject line.
(SOUNDBITE OF CLAP YOUR HANDS SAY YEAH'S "BLUE TURNING GRAY") Transcript provided by NPR, Copyright NPR.

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SmartStream Technologies, the financial Transaction Lifecycle Management specialist, today announced that it has won "Best Reconciliation and Exception Management Provider" in the 2009 Waters Rankings. The seventh annual Waters Rankings is an industry-respected award where the winners are chosen by the qualified readers of Waters magazine, a financial IT publication with more than 10,000 readers.
Smartstream won the newly-created category with 32.9 percent of votes, while the runner-up scored second place with 26 percent of votes. The votes for the 2009 Waters Rankings came from 1,000 readers of Waters who work inside major investment firms on both the buy side and sell side.
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text:
\section{Introduction}
\input{introduction}

\section{Preliminaries}
\label{preliminaries}
\input{preliminar}

\section{Product states}
\label{products}

\input{products}

\section{Separable states}
\label{separable}

\input{separable}

\section{Multiple copies}
\label{multiple}
\input{multiple}

\section{1$\times$1 modes}
\label{sec:1x1}
\input{1x1}

\subsection{Thermal states of fermionic chains}
\input{1x1particular}

\section{Detailed proofs}
\label{sec:proofs}
\input{tabsummary}
\input{proofs}

\subsection{$1\times1$-modes system}

\subsection{General states}

With these considerations, we may give the following definitions
of a product state. They are summarized in Table~\ref{tab:prod}.
\input{tabproduct}
\begin{itemize}
\item
We may call a state product if there exists some state
acting on the Fock space
of the form
$\tilde{\rho}=\tilde{\rho}_A\otimes\tilde{\rho}_B$,
and producing the same expectation values for all local observables.
Formally,~\cite{foot2}
\begin{eqnarray}
\nonumber
\product{0} &:=& \left\{ \rho :\,  \exists\tilde{\rho}_A,\,\tilde{\rho}_B, \, [\tilde{\rho}_{A(B)},\parity_{A(B)}]=0\quad \mathrm{s.t.} \right.\\
&&\left. \quad \rho(A_\pi \, B_\pi) =\tilde{\rho}_A(A_\pi)\tilde{\rho}_B(B_\pi) \right.\\
\nonumber
&&\left. \quad \quad\forall  A_\pi\in {\cal A}_{\pi},\, B_\pi\in {\cal B}_{\pi} \right\}.
\end{eqnarray}

\item
Alternatively, product states may be defined as those for which
the expectation value of products of local observables factorizes,
\begin{eqnarray}
\nonumber
\product{1} &:=& \left\{ \rho : \rho(A_\pi \, B_\pi) =\rho(A_\pi) \rho(B_\pi)
\right.\\
&&\left. \quad \quad  \forall A_\pi\in {\cal A}_{\pi},\, B_\pi\in {\cal B}_{\pi} \right\}.
\end{eqnarray}

\item
At the level of the Fock representation, a product state
can be defined as that writable as a tensor product,
\begin{equation}
\product{2} :=  \left\{ \rho : \rho=\rho^A \otimes \rho^B \right\}.
\end{equation}

\item
From the point of view of the subalgebras of observables for both
partitions, one may ignore the commutation with the parity operator
and require factorization of any product of observables for a product
state~\cite{moriya05}. This yields another set
\begin{equation}
\product{3} :=\left\{ \rho : \rho(A \, B) =\rho(A) \rho(B)
\quad  \forall A\in {\cal A},\, B\in {\cal B} \right\}.
\end{equation}

\end{itemize}

The two first definitions are equivalent,
$\product{0}\equiv\product{1}$.
They correspond to states
with a separable projection onto the diagonal blocks that preserve
parity in each of the subsystems.
This means that
$$\sum_{\alpha,\,\beta=e,\,o} \proj{\alpha}{A}\otimes \proj{\beta}{B}
\rho
 \proj{\alpha}{A}\otimes \proj{\beta}{B},
$$
is a product in the sense of \product{2}.

The three remaining sets are strictly different.
In particular
$\product{2} \subset \product{1}$
and
$\product{3} \subset \product{1}$,
but $\product{3} \neq \product{2}$.
The inclusion $\product{2},\product{3} \subseteq \product{1}$
is immediate from the definitions.
The non equality of the sets can be seen by explicit examples as those
shown in Table~\ref{tab:prod}.
The difference between $\product{3}$ and $\product{2}$, however,
is limited to non-physical states, i.e. those
not commuting with parity~\cite{moriya05}.

\subsection{Physical states}

Being parity a conserved quantity in the systems of interest,
the only physical states will be those commuting with $\parity$.
It makes then sense to restrict the study of entanglement to such
states.
By applying each of the above definitions to the physical states,
$\pset$, we obtain the following sets of physical product states.

\begin{itemize}
\item
$\productpar{1} := \product{1} \cap \pset= \product{0} \cap \pset$
\item
$\productpar{2} := \product{2} \cap \pset$
\item
$\productpar{3} := \product{3} \cap \pset$
\end{itemize}
We notice that
$\rho\in\productpar{2}$ is equivalent to $\rho=\rho_A\otimes\rho_B$
where both factors are also
parity conserving.

With the parity restriction, the three sets are related by
\begin{equation}
\productpar{3}=\productpar{2}\subset\productpar{1}.
\end{equation}

The proofs of all the relations above
are shown in section~\ref{subsec:prod}.

\subsection{Pure states}

For pure states, all $\productpar{i}$ reduce to the same set.
If the state vector is written in a basis of well-defined parity
in each subsystem,
it is possible to show that the condition of
$\productpar{1}$ requires that such expansion has a single
non-vanishing coefficient, and thus the state can be written as
a tensor product also with the definition of $\product{2}$.

\subsection{Product states}
\label{subsec:prod}

\newtheorem{incl}{}[subsection]

\begin{incl}
{$\product{0}\equiv\product{1}$}
\label{proof1}
\end{incl}
\begin{proof}
States in $\product{0}$ satisfy the restriction
that
$$
\rho(A_\pi \, B_\pi) =\tilde{\rho}_A(A_\pi)\tilde{\rho}_B(B_\pi)
$$
for some product state $\tilde{\rho}$ and all parity conserving operators
$A_\pi, \, B_\pi$.
Since the only elements or $\rho$ contributing to such expectation values
are in the diagonal blocks $\proj{\alpha}{A}\otimes \proj{\beta}{B}
\rho
 \proj{\alpha}{A}\otimes \proj{\beta}{B}$, ($\alpha,\,\beta=e,\,o$),
the condition is equivalent to saying that the sum of these blocks is equal to
the (parity commuting) product state
$\tilde{\rho}=\tilde{\rho}_A\otimes\tilde{\rho}_B$.

The condition for $\rho\in\product{1}$ turns out to be equivalent to this.
We may decompose the state as a sum
$$
\rho=\sum_{\alpha,\,\beta=e,o}\proj{\alpha}{A}\otimes \proj{\beta}{B}
\rho
 \proj{\alpha}{A}\otimes \proj{\beta}{B}+R := \rho'+R,
$$
where $\rho'$ is a density matrix commuting
with $\parity_A$ and $\parity_B$, and
$R$ contains only the terms that violate parity in some subspace.
It is easy to check that $R$ gives no contribution to expectation
values of the form $\rho(A_\pi \, B_\pi)$,
so that $\rho'(A_\pi \, B_\pi)=\rho'(A_\pi)\rho'(B_\pi)$.
On the other hand, an operator that is odd under parity has the form
$A_{\not\pi}=\proj{e}{A}A_{\not\pi}\proj{o}{A}+\proj{o}{A}A_{\not\pi}\proj{e}{A}$.
Therefore
$\rho'(A_{\not\pi} \, B_{\not\pi})=0=\rho'(A_{\not\pi})\rho'(B_{\not\pi})$.
Since $\rho'$ commutes with parity, and then odd observables give zero
expectation value,
we have checked that all expectation values $\rho'(A\,B)$ factorize
and then $\rho'$ is a product.
\end{proof}

\begin{incl}
{$\product{2}\subset\product{1}$}
\label{proof2}
\end{incl}
\begin{proof}
The inclusion $\product{2}\subseteq\product{1}$ is immediate
from the fact that the
products of even observables in the
$A_\pi\, B_\pi$
correspond, via a Jordan-Wigner transformation, to
products of local even operators $\tilde{A}_e\,\tilde{B}_e$
in the Fock representation, and thus they factorize for any state in
$\product{2}$.
The strict character of the inclusion is shown with an explicit example as
$\ensuremath{\rho_{\product{1}}}$, in Table~\ref{tab:prod}.
\end{proof}

\begin{incl}
{$\product{3}\subset\product{1}$}
\end{incl}
\begin{proof}
The  inclusion $\product{3}\subseteq\product{1}$ in immediate
from the definitions of both sets.
The example $\ensuremath{\rho_{\product{1}}}\notin\product{3}$
(Table~\ref{tab:prod}) shows it is strict.
\end{proof}

\begin{incl}
{$\product{2}\neq\product{3}$}
\end{incl}
\begin{proof}
The example
\begin{eqnarray}
\nonumber
\ensuremath{\rho_{\product{2}}} &=&\PiinPiii \\
\nonumber
&=&
\frac{1}{2}\left (
\begin{array}{r r}
1 & -1 \\
-1 & 1
\end{array}
\right ) \otimes
\frac{1}{2}\left (
\begin{array}{r r}
1 & 1 \\
1 & 1
\end{array}
\right ),
\end{eqnarray}
fulfills
$\ensuremath{\rho_{\product{2}}} \in \product{2}$, but $\ensuremath{\rho_{\product{2}}} \notin \product{3}$
because it has non vanishing expectation value for products of odd
operators, f.i.
$\langle c_2 c_3 \rangle_{\ensuremath{\rho_{\product{2}}}}=i\neq 0$.

On the other hand, it is also possible to construct a state as
$$
\ensuremath{\rho_{\product{3}}} =\PiiinPii,
$$
satisfying
$\ensuremath{\rho_{\product{3}}} \in \product{3}$ (it is easy to check the explicit
characterization for $1\times1$ modes of Table~\ref{tab:1x1}),
but $\ensuremath{\rho_{\product{3}}} \notin \product{2}$ because it is not possible
to write it as a tensor product.
\end{proof}

\begin{incl}
{$\productpar{2}\subset\productpar{1}$}
\end{incl}
\begin{proof}
The non strict inclusion is immediate from the result for general
states~(\ref{proof2}).
Actually, the same example $\ensuremath{\rho_{\product{1}}}$ is parity preserving and then
shows the non equivalence of both sets.
\end{proof}

\begin{incl}
{$\productpar{2}\equiv\productpar{3}$}
\end{incl}
\begin{proof}
For any physical state $[\rho,\,\parity]=0$, the expectation value of
any odd operator is null.
On the other hand, all $\product{3}$ states (in particular those in
$\productpar{3}$) fulfill
$\rho(A_{\not\pi} B_{\not\pi})=0$~\cite{moriya05}.
Since a state in $\productpar{2}$ can be written as a product of two factors
each of them commuting with the local parity operator,
then the only non--vanishing expectation values in these
sets of states correspond to
products of parity conserving local observables.
It is then enough to check that
$$
\rho(A_{\pi} B_{\pi})=\rho(A_{\pi})\rho(B_{\pi})
\iff
\rho=\rho_A \otimes \rho_B.
$$
Given the state $\rho$ we can look at the Fock
representation and write it as an expansion in the Pauli operator basis,
where coefficients correspond to expectation values of products
$\sigma_{a_1}^{(1)}\otimes \ldots \otimes \sigma_{a_m}^{(m)}$.

Making use of the Jordan-Wigner transformation~(\ref{eq:JW}),
any product
of even observables in the Fock space is mapped to a product of even
operators in the subalgebras $\cal{A}$, $\cal{B}$.
So it is easy to see that the property of factorization is equivalent
in both languages and thus
$$
\rho\in\productpar{2}\iff\rho\in\productpar{3}.
$$

This equivalence implies also that of the convex hulls,
$\separpar{2}\equiv\separpar{3}$.
\end{proof}

\subsubsection{Pure states}

\begin{incl}
{For pure states $\productpar{1}\iff\productpar{2}$}
\end{incl}
\begin{proof}

A pure state $|\Psi\rangle\langle\Psi| \in \pset$ is
such that $\parity \Psi= \pm\Psi$.
We consider the even case (the same reasoning applies for the odd one).
Since such a state vector is a direct sum of two components,
one of them even with respect to both $\parity_A,\,\parity_B$
and the other one odd
with respect to both local operations,
and applying the Schmidt decomposition to each of those components,
it is always possible to write the state as
$$
|\Psi \rangle =\sum_i \alpha_i |e_i\rangle |\varepsilon_i\rangle +
\sum_i \beta_i|o_i\rangle |\theta_i\rangle,
$$
where $\{|e_i\rangle\}$ ($\{|\varepsilon_i\rangle\}$) are mutually
orthogonal states with $\parity_A |e_i\rangle=+|e_i\rangle$
($\parity_B |\varepsilon_i\rangle=+|\varepsilon_i\rangle$)
and $\{|o_i\rangle\}$ ($\{|\theta_i\rangle\}$)
are mutually
orthogonal states with $\parity_A |o_i\rangle=-|o_i\rangle$
($\parity_B |\theta_i\rangle=-|\theta_i\rangle$).

The condition of $\productpar{1}$ imposes that
$\langle \Psi |A_\pi\,B_\pi|\Psi\rangle=
\langle \Psi |A_\pi|\Psi\rangle \langle \Psi |B_\pi|\Psi\rangle$ for all
parity preserving observables.
In particular, we may consider those of the form
\begin{eqnarray}
A_\pi&=&\sum_k A_k^e|e_k\rangle\langle e_k|+A_k^o |o_k\rangle\langle o_k|,
\nonumber
\\
B_\pi&=&\sum_k B_k^e|\varepsilon_k\rangle\langle \varepsilon_k|+B_k^o |\theta_k\rangle\langle \theta_k|.
\nonumber
\end{eqnarray}
On these observables the restriction reads
\begin{eqnarray}
\left(
\sum_i |\alpha_i|^2 A_i^e
\right.\!\!
&+& \!\!\left.
\sum_i |\beta_i|^2 A_i^o
\right)
\!\left(
\sum_i |\alpha_i|^2 B_i^e\! +\! \sum_i |\beta_i|^2 B_i^o
\right)
\nonumber
\\
&&=\sum_i |\alpha_i|^2 A_i^e B_i^e + \sum_i |\beta_i|^2 A_i^o B_i^o
\nonumber
\end{eqnarray}
Let us assume that the state $\Psi$ has more than one term in the
even-even sector, i.e. $\alpha_1\neq 0$ and $\alpha_2\neq 0$ (we may
reorder the sum, if necessary).
Then we apply the condition to $A=A_1^e|e_1\rangle\langle e_1|$,
$B=B_2^e|\varepsilon_2\rangle\langle \varepsilon_2|$, and applying the equality we deduce
$|\alpha_1|^2 A_1^e |\alpha_2|^2 B_2^e=0$, and thus $|\alpha_1| |\alpha_2|=0$,
so that there can only be a single term in the
$|e_i\rangle |\varepsilon_i\rangle$ sum.
An analogous argument shows that also the sum of
$|o_i\rangle |\theta_i\rangle$ must have at most one single
contribution, for the state to be in $\productpar{1}$.

By applying the equality to operators
$A=A_1^o|o_1\rangle\langle o_1|$  and
$B=B_1^e|\varepsilon_1\rangle\langle \varepsilon_1|$
we also rule out the possibility that $\Psi$ has a contribution from
each sector.
Then, if $\Psi\in\productpar{1}$, it has one single term in the Schmidt
decomposition, and therefore it is a product in the sense of $\productpar{2}$.
\end{proof}

\subsection{Separable states}
\label{subsec:separ}

\begin{incl}
{$\separ{2}\subset\separ{1}$ and $\separpar{2'}\subset\separpar{1}$}
\label{proof4}
\end{incl}
\begin{proof}
The first (non strict) inclusion is immediate from the relation between
product states~\ref{proof2}.
To see that both sets are not equal, we use again an explicit example.
It is possible to construct a state
in $\productpar{1}\subset\separpar{1}$ which has non-positive partial
transpose and is thus not in $\separ{2}$.
However, this has to be found in bigger systems than the previous
counterexamples, as in a 2-mode system the conditions for $\separpar{1}$ and
$\separpar{2}$ are identical, as shown in Table~\ref{tab:1x1}.

By constructing random matrices $\rho_A \otimes \rho_B$ in the
parity preserving sector, and adding off-diagonal terms $R$ which are
also randomly chosen, we
find a counterexample $\ensuremath{\rho_{\separpar{1}}}$ in a $2\times 2$-system
such that $\ensuremath{\rho_{\separpar{1}}} \in \productpar{1}$ by construction, but
its partial transposition with respect to the subsystem B,
$\ensuremath{\rho_{\separpar{1}}}^{T_B}$, has a negative eigenvalue.

When taking intersection with the set of physical states, the inclusion
still holds, and it is again strict, since the counterexample
$\ensuremath{\rho_{\separpar{1}}}$ is in particular in $\productpar{1}$.
\end{proof}

\begin{incl}
{$\separpar{1}\equiv\separ{1} \cap \pset$}
\end{incl}
\begin{proof}
Obviously,
$\separpar{1}\subseteq\separ{1} \cap \pset$.
To see the converse direction of the inclusion, we consider a state
$\rho\in\separ{1}\cap\pset$. Then there is a decomposition
$\rho=\sum_i\lambda_i\rho_i$ with $\rho_i\in\product{1}$, but not
necessarily in $\pset$.
We may split the sum into the even and odd terms under the parity
operator,
$$
\rho=\rho_{\pi}+\rho_{\not\pi}:=\sum_i \lambda_i \frac{1}{2}(\rho_i+\parity \rho_i \parity)
+\sum_i \lambda_i \frac{1}{2}(\rho_i-\parity \rho_i \parity).
$$
The second term, $\rho_{\not\pi}$, gives no contribution to operators that commute
with $\parity$. Since $\rho$ is physical, this term also gives zero
contribution to odd observables, so that
$$
\rho=\sum_i \lambda_i \frac{1}{2}(\rho_i+\parity \rho_i \parity).
$$
It only remains to be shown that each
$\rho_{i_\pi}:= \frac{1}{2}(\rho_i+\parity \rho_i \parity)$ is still a product state
in $\productpar{1}$.
But since for parity commuting observables all the contributions come
from the symmetric part of the density matrix,
$\rho_i(A_\pi B_\pi)=\rho_{i_\pi}(A_\pi B_\pi)$,
and the condition for $\productpar{1}$ holds for $\rho_{i_\pi}$.
Therefore we have found a convex decomposition of $\rho$ in terms of
product states all of them conforming to the symmetry.

The analogous relation for $\separpar{2}$ was shown in~\cite{moriya05}.
\end{proof}

\begin{incl}
{$\separpar{2}\subset\separpar{2'}$}
\end{incl}
\begin{proof}
Since $\productpar{2}=\product{2}\cap\pset$, taking convex hulls and
intersecting again with $\pset$ implies that
$\separpar{2}\subseteq\separpar{2'}$.
However, not all separable states can be decomposed as a convex sum of
product states all of them conforming to the parity symmetry.
In particular, the state
$$
\ensuremath{\rho_{\separpar{2'}}}=\SiipnSii,
$$
which has PPT and is thus in $\separpar{2'}$, is not in $\separpar{2}$
(recall that for the $1\times1$-system, only density matrices  which are
diagonal in the number basis are in $\separpar{2}$).
\end{proof}

\begin{incl}
{$\equivsep{1}\equiv\equivsep{2'}\equiv\equivsep{2}$}
\end{incl}
\begin{proof}
From the relations
$\separpar{2}\subset\separpar{2'}\subset\separpar{1}$ and the definition
of the equivalence classes it is evident that
$\equivsep{2}\subseteq\equivsep{2'}\subseteq\equivsep{1}$.
To show the equivalence of all sets it is enough to prove that any
 state $\rho \in \equivsep{1}$ is also in $\equivsep{2}$, i.e. that
there exists a state in $\separpar{2}$ equivalent to $\rho$.

For $\rho\in\equivsep{1}$, there is a $\tilde{\rho} \in \separpar{1}$,
i.e. $\tilde{\rho}=\sum\lambda_k\tilde{\rho}_k$ with each
$\tilde{\rho}_k \in \productpar{1}$,
producing identical expectation values for products of even operators
$A_\pi B_\pi$.
If we define
$$
\rho_k':= \sum_{\alpha,\,\beta=e,\,o}\proj{\alpha}{A}\otimes\proj{\beta}{B}
\tilde{\rho}_k \proj{\alpha}{A}\otimes\proj{\beta}{B},
$$
it is evident that $\rho':=\sum_k\lambda_k \rho_k'$ produces the
same expectation values as $\tilde{\rho}$ for the
relevant operators (see proof~\ref{proof1}).
Therefore, $\rho \sim \rho'$.
Moreover, since $\rho_k'(A_{\not\pi}B_{\not\pi})=0$ for all odd-odd
products, every $\rho_k'\in\productpar{2}$, and so
$\rho'\in\separpar{2}$.
\end{proof}

\input{proofs_mult}
\subsection{Multiple copies}
\label{subsec:mult}

\begin{incl}
{$\rho^{\otimes 2}\in\separpar{1} \Rightarrow \rho\in\separpar{1}$}
\label{mult1}
\end{incl}
\begin{proof}
An arbitrary state can be decomposed in two terms,
$\rho=\rho_E+\rho_O$, where
$$
\rho_E:=
\sum_{\alpha,\,\beta=e,\,o} \proj{\alpha}{A}\otimes \proj{\beta}{B}
\rho
 \proj{\alpha}{A}\otimes \proj{\beta}{B},
$$
and
$$
\rho_O:=
\sum_{
\substack{
{\alpha,\,\beta,\,\gamma,\,\delta=e,\,o}\\
{(\alpha,\,\beta)\neq (\gamma,\,\delta)}}
} \proj{\alpha}{A}\otimes \proj{\beta}{B}
\rho
 \proj{\gamma}{A}\otimes \proj{\delta}{B}.
$$
For any state in $\separpar{1}$, there exists a decomposition
$\rho_E=\sum_i\lambda_i\rho_E^i$,
$\rho_O=\sum_i\lambda_i\rho_O^i$,
such that $\rho_E^i+\rho_O^i\in\productpar{1}$.
Let us consider two copies of a state such that
$\tilde{\rho}:=\rho^{\otimes 2}\in\separpar{1}$.
Then, using the above decomposition of $\tilde{\rho}$,
and taking the partial trace with respect to the second
system,
we obtain a decomposition of the single copy,
$\rho=\rho_E+\rho_O=\sum_i \lambda_i \mathrm{tr}_2(\tilde{\rho}_E^i)
+\sum_i \lambda_i \mathrm{tr}_2(\tilde{\rho}_O^i)$.
Since $\tilde{\rho}_E^i$ was a tensor product,
$\tilde{\rho}_E^i=\tilde{\rho}_{\tilde{A}}\otimes\tilde{\rho}_{\tilde{B}}$,
with $\tilde{A}\equiv A_1A_2$, $\tilde{B}\equiv B_1B_2$,
so is  $\mathrm{tr}_2(\tilde{\rho}_E^i)$, and therefore
$\rho\in\separpar{1}$.
\end{proof}

\begin{incl}
{$\rho^{\otimes 2}\in\equivsep{1} \Rightarrow \rho\in\equivsep{1}$}
\label{mult2}
\end{incl}
\begin{proof}
Using the same decomposition as above,
$\rho=\rho_E+\rho_O$,
a state $\rho\in\equivsep{1}$ satisfies $\rho_E\in\separpar{2'}$.
If we consider
$\tilde{\rho}:=\rho^{\otimes 2}=\tilde{\rho}_E+\tilde{\rho}_O$,
the condition $\equivsep{1}$ on the state of the two copies reads
$$
\tilde{\rho}_E=\rho_E\otimes\rho_E+\rho_O\otimes\rho_O\in\separpar{2'},
$$
in terms of the components of the single copy state.
Taking the trace with respect to one of the copies, then,
and using the fact that $\rho_O$ is traceless,
$\rho_E\in\separpar{2'}$,
so that $\rho\in\equivsep{1}$.
\end{proof}

\begin{incl}
{$\rho\  \mathrm{NPPT} \Rightarrow \rho^{\otimes 2}\notin\equivsep{1}$}
\label{mult4}
\end{incl}
\begin{proof}
We may restrict the proof to states such that  $\rho \in \equivsep{1}$.
In other case the implication follows immediately
from  the previous result (\ref{mult2}).
Written in a basis of well-defined local parities,
any density matrix that commutes with the parity operator has a block
structure (analogous to that of~(\ref{gen1x1even}) for the $1\times 1$
case).
\begin{equation}
\rho=\left(
\begin{array}{c c c c}
\rho_{ee} & 0 & 0 & C\\
0 & \rho_{eo} & D & 0 \\
0 & D^{\dagger} & \rho_{oe} & 0 \\
C^{\dagger} & 0 & 0 & \rho_{oo}
\end{array}
\right).
\label{geneven}
\end{equation}
The diagonal blocks correspond to the projections onto simultaneous
eigenspaces of both parity operators,
$\rho_{\alpha\beta}=\proj{\alpha}{A}\otimes \proj{\beta}{B} \rho \proj{\alpha}{A}\otimes \proj{\beta}{B}$,
whereas
$C=\proj{e}{A}\otimes \proj{e}{B} \rho \proj{o}{A}\otimes \proj{o}{B}$
and $D=\proj{e}{A}\otimes \proj{o}{B} \rho \proj{o}{A}\otimes \proj{e}{B}$.

From the characterization~(\ref{charactZ1}) of separability,
the state is in \equivsep{1} iff
all the diagonal blocks
$\rho_{\alpha \beta}$ are in \separpar{2'}.
It is then enough to prove that the partial transpose of $\rho$
is positive iff
$\proj{e}{A}\otimes \proj{e}{B}
\rho\otimes\rho
 \proj{e}{A}\otimes \proj{e}{B}$
has PPT.
Non positivity of the partial transpose of $\rho$ implies then
the non separability (\separpar{2'})
of the one of the diagonal blocks of $\rho\otimes\rho$.

The partial transposition of the above matrix yields
\begin{equation}
\rho^{T_B}=\left(
\begin{array}{c c c c}
\rho_{ee}' & 0 & 0 & D'\\
0 & \rho_{eo}' & C' & 0 \\
0 & (C'^{\dagger}) & \rho_{oe}' & 0 \\
(D')^{\dagger} & 0 & 0 & \rho_{oo}'
\end{array}
\right),
\label{genevenPT}
\end{equation}
where $X':= X^{T_B}$, and the $T_B$ operation acts on each block
transposing the last $m_B-1$ indices.

If we take two copies of the state, we find for the corresponding
uppermost diagonal block
$\tilde{\rho}_{ee}:= \proj{e}{A}\otimes \proj{e}{B}
\rho\otimes\rho
 \proj{e}{A}\otimes \proj{e}{B}$,
\begin{equation}
\tilde{\rho}_{ee}=\left(
\begin{array}{c c c c}
\rho_{ee}\otimes\rho_{ee} & 0 & 0 & C\otimes C\\
0 & \rho_{eo}\otimes\rho_{eo} & D\otimes D & 0 \\
0 & D^{\dagger}\otimes  D^{\dagger} & \rho_{oe}\otimes\rho_{oe} & 0 \\
C^{\dagger}\otimes C^{\dagger} & 0 & 0 & \rho_{oo}\otimes\rho_{oo}
\end{array}
\right),
\label{rho2ee}
\end{equation}
and for the partial transposition
\begin{equation}
(\tilde{\rho}_{ee})^{T_B}=\left(
\begin{array}{c c c c}
\rho_{ee}'\otimes\rho_{ee}' & 0 & 0 & D'\otimes D'\\
0 & \rho_{eo}'\otimes\rho_{eo}' & C'\otimes C' & 0 \\
0 & C'^{\dagger}\otimes  C'^{\dagger} & \rho_{oe}'\otimes\rho_{oe}' & 0 \\
D'^{\dagger}\otimes D'^{\dagger} & 0 & 0 & \rho_{oo}'\otimes\rho_{oo}'
\end{array}
\right).
\label{rho2eePT}
\end{equation}
The matrices (\ref{genevenPT}) and (\ref{rho2eePT}) are the direct sum of two
blocks. Thus they are positive definite iff each such block is positive
definite.
Let us consider one of the blocks of (\ref{rho2eePT}), namely
\begin{equation}
\left(
\begin{array}{c c}
\rho_{ee}'\otimes\rho_{ee}'& D'\otimes D'\\
D'^{\dagger}\otimes D'^{\dagger} & \rho_{oo}'\otimes\rho_{oo}'
\end{array}
\right).
\label{B1rho2ee}
\end{equation}
Let us first assume that $\rho_{oo}'$ is non-singular.
Applying a standard theorem in matrix analysis
and making use of the fact that our
$\rho \in \equivsep{1}$, so that each diagonal block is PPT,
we obtain that~(\ref{B1rho2ee}) is positive iff
$$
\rho_{ee}'\otimes\rho_{ee}'\geq (D'\otimes D') (\rho_{oo}'^{-1}\otimes
\rho_{oo}'^{-1}) (D'^{\dagger}\otimes D'^{\dagger}),
$$
which holds iff
$$
\rho_{ee}' \geq D' (\rho_{oo}')^{-1}D'^{\dagger}.
$$
Reasoning in the same way for the second block of (\ref{rho2eePT}),
one gets that
\begin{equation}
(\tilde{\rho}_{ee})^{T_B} \geq 0 \Leftrightarrow \rho^{T_B}\geq 0.
\end{equation}
The result holds also if the assumption of non-singularity
of $\rho_{oo}$ ($\rho_{oe}$ for the second block) is not valid.
In that case, we may take $\rho_{oo}$ diagonal and then,
by positivity of $(\tilde{\rho}_{ee})^{T_B}$ (or $\rho^{T_B}$
for the reverse implication), find that $D'$ must have some
null columns. This allows us to reduce both matrices to a similar
block structure, where the reduced  $\rho_{oo}$ ($\rho_{oe}$) is
non-singular.
\end{proof}

\begin{incl}
{For $1\times1$ systems, $\rho^{\otimes 2}\in\equivsep{1} \iff \rho\in\separpar{2'}$}
\label{mult3}
\end{incl}
\begin{proof}
One of the directions is immediate, and valid for an arbitrarily large system,
since $\rho\in\separpar{2'}$ implies
$\rho^{\otimes 2}\in\separpar{2'}\subset\separpar{1}\subset\equivsep{1}$.
On the other hand, if we take
$\tilde{\rho}:=\rho^{\otimes 2}\in\equivsep{1}$,
then the diagonal blocks of this state are separable,
 in particular
$\proj{e}{\tilde{A}}\otimes\proj{e}{\tilde{B}}\tilde{\rho}
\proj{e}{\tilde{A}}\otimes\proj{e}{\tilde{B}}\in\separpar{2'}$,
which was calculated in (\ref{rho2ee}).
For the case of $1\times1$ modes, with
$\rho$ given by~(\ref{gen1x1even}), this block reads
$$
\left(
\begin{array}{c c c c}
(1-x-y+z)^2 & 0 & 0 & r^2\\
0 & (x-z)^2 & s^2 & 0 \\
0 & (s^*)^2 & (y-z)^2 & 0 \\
(r^*)^2 & 0 & 0 & z^2
\end{array}
\right).
$$
This is in $\separpar{2'}$ iff it has PPT, and
this happens if and only if $\rho$ has PPT, i.e. $\rho\in\separpar{2'}$.
\end{proof}

\subsection{General states}

Taking the convex hull of the general product states,
we define the sets
\begin{itemize}
\item
$\separ{1}:=\mathrm{co}\left(\product{1}\right)$,
\item
$\separ{2}:=\mathrm{co}\left(\product{2}\right)$,
\item
$\separ{3}:=\mathrm{co}\left(\product{3}\right)$.
\end{itemize}

These contain both physical states, commuting with $\parity$,
and non-physical ones.
It can be shown that
$\separ{3} \subset \separ{2} \subset \separ{1}$.

The non-strict inclusion $\separ{2} \subseteq \separ{1}$
is immediate from the inclusion between product sets.
The strict character can be seen with an example, in particular
in the subset of physical states, $\ensuremath{\rho_{\separpar{1}}}$.
$\separ{3}\subset \separ{2}$ was proved in~\cite{moriya05}.

\subsection{Physical states}

From the physical sets of product states we define the
following sets of separable states,
\begin{itemize}
\item
$\separpar{1}:=\mathrm{co}\left(\productpar{1}\right)$,
\item
$\separpar{2}:=\mathrm{co}\left(\productpar{2}\right)$,
\end{itemize}
Obviously, the corresponding $\separpar{3}\equiv\separpar{2}$.
The inclusion relations among product states imply
$\separpar{2}\subseteq\separpar{1}$.
It is easy to see with an example that this inclusion is also strict.
Table~\ref{tab:separ} summarizes the definitions and mutual relations
of the various separability sets.
\input{tabsepar}

As shown in Fig.~\ref{fig:scheme}, we may take the physical states
that satisfy the definitions for separability introduced in
the previous subsection, and hence use
$\separ{i}\cap\pset$ as the definition of separable states.
This yields the sets
\begin{itemize}
\item
$\separ{1}\cap\pset\equiv\separpar{1}$,
\item
$\separpar{2'}:=\separ{2}\cap\pset$,
\item
$\separ{2}\cap\pset\equiv\separpar{2}$.
\end{itemize}

Only $\separpar{2'}$ is different from the
separable sets defined above.
Actually, given an $\separ{1}$ state that commutes with $\parity$,
it is possible to construct a decomposition according to $\separpar{1}$
by taking the parity preserving part of each term in the original
convex combination.
Therefore $\separ{1}\cap\pset\subseteq\separpar{1}$, while the
converse inclusion is evident.
For $\separ{3}\cap\pset$, on the other hand, it was shown in~\cite{moriya05}
that any parity preserving state in $\separ{3}$ has a decomposition in
terms of only parity preserving terms, and is thus in $\separpar{3}$.

All the considerations above leave us with three strictly different
sets of separable physical states,
\begin{equation}
\separpar{2} \subset \separpar{2'} \subset \separpar{1}
\end{equation}

From the definitions, it
is immediate that $\separpar{2} \subseteq \separpar{2'}$.
The inclusion is strict because
not every state $\rho \in\separpar{2'}$
has a decomposition in terms of products of even states
(see example $\ensuremath{\rho_{\separpar{2'}}}$ in Table~\ref{tab:separ}).
The condition for  $\separpar{2}$ is then more restrictive.

From the relation between product sets,
$\separ{2} \subseteq \separ{1}$,
and $\separpar{2'} \subseteq \separpar{1}$.
The strict inclusion can be shown by constructing an explicit example
of a $\productpar{1}$ state without positive
partial transpose (PPT)~\cite{peres96}
in the 2$\times$2-modes system.

The detailed proofs of
the equivalences and inclusions above
are shown in section~\ref{subsec:separ}.

\subsection{Equivalence classes}

If one is only interested in the measurable correlations of the
state, rather than in its properties after further evolution or
processing, it makes sense to define an equivalence relation
between states by
$$
\rho_1 \sim \rho_2 \quad \mathrm{if} \quad
\rho_1(A_\pi  B_\pi)=\rho_2(A_\pi B_\pi) \ \ \forall A_\pi\!\in\! \cal{A}_\pi,\,
B_\pi\!\in\!\cal{B}_\pi,
$$
i.e. two states are equivalent if they produce the same
expectation values for all physical local operators. Therefore,
two states that are equivalent cannot be distinguished by means of
local measurements.

With the restriction of parity conservation, the states that can be
locally prepared are of the form $\separpar{2}$, i.e.
$\rho=\sum_k \rho_k^A \otimes \rho_k^B$, where
$[\rho_k^{A(B)},\parity_{A(B)}]=0$.
Since the only
locally accessible observables are local, parity preserving
operators, i.e. quantities of the form $\rho(A_\pi B_\pi)$,
it makes sense to say that a given
state is separable if it is equivalent to a  state that can
be prepared locally.
With this definition, the set of separable states is equal to
the equivalence class of $\separpar{2}$ with respect to the
equivalence relation above.

Generalizing this concept, we may construct the equivalence classes
for each of the relevant separability sets,
$$
\equivsep{i}:= \{\rho\,:\, \exists \tilde{\rho}\in\separpar{i},\,
\rho\sim\tilde{\rho}\},\quad i=1,2',2.
$$
From the inclusion relation among the separability sets,
$\equivsep{2}\subseteq \equivsep{2'}\subseteq\equivsep{1}$.
And, obviously, $\separpar{i}\subseteq\equivsep{i}$.

On the other hand, any state $\rho\in\equivsep{1}$ has also an equivalent state
in $\separpar{2}$ (see section~\ref{subsec:separ}), so that
$$
\equivsep{2}=\equivsep{2'}=\equivsep{1}.
$$

This equivalence class includes then all the separability sets
described in the previous subsection.
However, it is strictly larger, as can be seen by the explicit example
$\ensuremath{\rho_{\equivsep{1}}}$ in Table~\ref{tab:separ}.

\subsection{Characterization}

It is possible to give a characterization of the previously defined
separability sets in terms of the usual mathematical concept of
separability, i.e. with respect to the tensor product.
This allows us to use standard separability criteria
(see~\cite{horodecki07review} for a recent review) in order to decide
whether a given state is in each of these sets.

The definition $\separpar{2'}$ corresponds
to the separability
in the sense of the tensor product, i.e. the standard notion~\cite{Werner1},
applied to parity preserving states.

As convex hull of $\product{2}\cap\pset$, the
set $\separpar{2}$ consists of states with a decomposition
in terms of tensor products, with the additional restriction that
every factor commutes with the local version of the parity operator.
Using the block diagonal structure
$\proj{e}{} \rho \proj{e}{} +\proj{o}{} \rho \proj{o}{}$
of any parity preserving state,
each block
must have independent decompositions in the sense of the tensor product.
Then a state will be in $\separpar{2}$ iff both
$\proj{e}{} \rho \proj{e}{}$ and $\proj{o}{} \rho \proj{o}{}$
are in $\separpar{2'}$.

A state $\rho$ is in $\productpar{0}$ if its diagonal blocks
are a tensor product,
\begin{equation}
\sum_{\alpha,\,\beta=e,\,o} \proj{\alpha}{A}\otimes \proj{\beta}{B}
\rho
 \proj{\alpha}{A}\otimes \proj{\beta}{B}=
\tilde{\rho}^A \otimes \tilde{\rho}^B \in \productpar{2}.
\label{charactP0}
\end{equation}
The set $\separpar{1}$ is characterized as
the convex hull of $\productpar{1}\equiv\productpar{0}$, i.e.
it is formed by convex combinations of states that
can be written as the sum of a parity preserving tensor product
plus some off--diagonal terms.

Finally, the equivalence class $\equivsep{1}\equiv\equivsep{2}$
is completely defined in terms of the expectation values of observable
products $A_\pi B_\pi$.
These have no contribution from off-diagonal blocks in $\rho$,
so the class can be characterized in terms of
the diagonal blocks alone.
Therefore a state is in $\equivsep{1}$ iff
\begin{equation}
\sum_{\alpha,\,\beta=e,\,o} \proj{\alpha}{A}\otimes \proj{\beta}{B}
\rho
 \proj{\alpha}{A}\otimes \proj{\beta}{B}
\in \separpar{2'}.
\label{charactZ1}
\end{equation}
Since the condition involves only the block diagonal part of the state,
it is equivalent to the individual separability
(with respect to the tensor product) of each of the blocks.

redpajama_set_name: RedPajamaArXiv


text:
Loza Abera Geinore (Amharic:ሎዛ አበራ) is an Ethiopian professional footballer who currently plays for Nigd Bank of the Ethiopian Women's Premier League and the Ethiopia women's national team as a forward.

Early life
She was the third child of her family and the only female player in her neighborhood, she used to play football with mainly male composed teams both in school football matches and  neighborhood football matches in her town Durame, Loza started playing football at the age of six. In 2011, she participated in the All Ethiopian games representing the SNNP regional state, a tournament in which she scored seven goals and that become her initial momentum to get noticed by Ethiopian Woman National Team recruiters.

Club career

Ethiopia
Abera joined her first professional club, Hawassa City S.C. Women, in 2012. Spending two years at the club, she helped the club to two 3rd-place finishes in the league while being the team's top goal scorer for both seasons.

After an invitation from then coach Asrat Abate, Abera joined Dedebit F.C. Women (now defunct). She spent four years at Dedebit F.C., winning the League and top goal scorer honors in all four years. She finished the 2015–16 EWPL season with 47 goals during season and 10 goals in the playoff competition.

Despite lucrative offers from various teams, she made a promise to herself to finish high school. She did, and went on to Addis-Abeba University school of commerce while signing contracts from the area.

Abera signed with Adama City Women midway through the 2018–19 EWPL season after her brief stint in Sweden with Kungsbaka DFF. Continuing her success in her home country, she helped Adama City win its first ever EWPL title.

After six seasons playing in the EWPL, Abera has scored over 200 goals and holds the record for most goals in the league's history. In November 2020, she signed for Nigd Bank of the Ethiopian Women's Premier League.

Abroad
Abera has had trials with Antalyaspor (Turkey) and Kungsbaka DFF (Sweden). During her stay in Kungsbaka, she helped the team win a regional title and gain promotion to the top tier. However, her stay with the club ended after the season due to financial reasons.

In September 2019, she signed for Birkirkara F.C. (women) Women of the Malta Women's Premier League. She finished her first season as the Maltese league's top goal scorer with 30 goals in 12 appearances.

Awards
Abera was included on the list of the BBC's list of 100 Women announced on 23 November 2020.

References

ሎዛ አበራ

Living people
Women's association football forwards
Ethiopian women's footballers
Adama City F.C. players
1997 births
BBC 100 Women

redpajama_set_name: RedPajamaWikipedia


text:
Tomorrica er en flod i det sydøstlige Albanien . Floden Tomorrica er en venstre biflod til Devoll, der løber ud i Adriaterhavet i det sydøstlige Albanien. Floden, der er 42 kilometer lang, har sit udspring i 1800 meters højde i   Tomorr-bjergenes sydlige skråninger på  Mietabjerget (Maja e Mietës, 2.024 m).  Dens afvandingsområde er 376 km², og samler vandet fra  Tomorr-bjergene i vest og Ostrovicë-bjergene i øst. I  den  nedre del af flodenkommer  bifloderne fra venstre: Velica, Leshnica, Kushova og Bregas.

Kilder og henvisninger

Floder i Albanien


Error Code 110: The Endless Wave drivers have not successfully initalized. Please try reinstalling the GigaStudio software.
Uninstall Yoke. Reboot. MIDI Yoke is entirely gone now.. Giga works fine. Can someone please explain what I\'m doing wrong?
Make sure you start the tool after you have loaded up GS. Also, make sure the inputs are all correct.
Can someone please explain what I\'m doing wrong?
Sam: Are you working w/ a sequencer on the same machine as the GS?
If you go directly from your keyboard you won\'t need MIDI YOKE.
However, in order to work MIDI Yoke is prone to MIDI feedback. This happens the moment you choose the same port for In/Out in a sequencer. This can easily happen for instance in CubaseSX where the InPort default is AllPorts. The moment you change the outport to a MIDIYoke Port you\'ll have feedback.
Example Setup for Sequencer - LegatoTool - GS configuration.
Typically you will launch the LegatoTool last.
Yes, everything\'s on the same machine.. one computer only. So when I record, the MIDI goes straight from the keyboard to Sonar. And of course Sonar triggers the sounds in Giga. And for playback, the MIDI just goes from Sonar to Giga.
So I don\'t need MIDI Yoke? Or I do? If I do, how do I set it up? Every time I install MIDI Yoke (then reboot of course), Giga locks up when loading and never loads. What do I need to do to get this stuff working? Same deal with VOTA Utility. It says it needs a MIDI loopback driver.. so I have not been able to utilize VOTA to its fullest extent either.
Thanks, I will try that out! Too bad it\'s $50 though. VOTA should\'ve included it. I\'m unable to use the VOTA Utility, and unable to use MIDI Yoke. So I\'m kinda stuck here.
Midi yoke never worked here, try the Sonic Foundry virtual Midi ports instead.
I\'m not familiar with the Sonic Foundry MIDI ports. Any info on this?
if its 98/98SE try Hubi\'s Loopback.


text:
Resignation Of Cuyahoga Election Board Demanded
Tuesday, 20 March 2007, 11:49 am
Opinion: Bradblog.Com
Ohio SOS Demands Immediate Resignation Of All Cuyahoga County Election Board Members
- Robert Bennett, Ohio's GOP Chair and President of the Board of Elections in Ohio's Most Democratic County, Vows to Stay On
- Growing Battle Underscores a Host of Other Myths, Disinfo Campaigns Concerning Buckeye State Election Administration...
BLOGGED BY Brad Friedman ON 3/19/2007 12:21PM
Just out from AP...
CLEVELAND (AP) - All four election board members for Cuyahoga County, troubled by recount rigging charges and voting machine problems, have been told to resign or face being fired, a state official said Monday.
Secretary of State Jennifer Brunner said she called the four board members of Ohio's most populous county late Sunday, asking them to leave by close of business Wednesday.
Brunner, a Democrat, said new leadership is needed in the county before the 2008 presidential election. She said that if the board members don't resign by her deadline, they would face public hearings and removal.
The board's chairman, Robert Bennett, is also chairman of the state GOP in Ohio. His term on the board ends in 2010. The other board members are two Democrats and one Republican.
Bennett said Monday he plans to finish his term, which expires in 2010.
"We are making progress on our major goal, which is to restore voter confidence in Cuyahoga County elections," he said at a news conference.
The issue comes in the wake of two Cuyahoga County Election Officials recently found guilty of rigging the 2004 Presidential Election recount and subsequently sentenced to the maximum 18 months in prison after they refused to cooperate with the state's special prosecutor, Kevin Baxter.
Both Baxter and the district court judge, Peter J. Corrigan, indicated they believed the conspiracy to game the recount was part of a larger, higher-reaching conspiracy. "It seems unlikely that your superiors wouldn't know what you were doing," Corrigan declared during the sentencing.
Brunner told the Plain-Dealer this morning that the convictions were "so serious that I don't think I had any alternative" but to request the resignation of the entire board. "It has to do with public confidence and perception," she said.
In a news conference this morning, Bennett vowed to fight Brunner in court rather than leave on his own accord.
Cuyahoga County's then Election Director, Michael Vu, who was not charged in the incident, but who has since resigned, had previously admitted that sealed ballot boxes were opened and secretly counted prior to the state mandated recount in order to assure the recount would not reveal any inconsistencies with the officially reported results. He has since recanted his original story.
After the recent sentencing, Bennett, who was ultimately responsible for overseeing the convicted elections officials --- supposedly one Republican and one Democrat --- who rigged a presidential election recount in his county said, incredibly enough, "the convictions are a travesty of justice."
The issue also underscores an important point we've been trying to make for years: Despite the claims by disgraced former Buckeye State SoS, J. Kenneth Blackwell and his operatives (including Bennett) for the travesty that occurred in 2004, which suggest that Ohio's supposedly "bi-partisan" structure of county elections boards --- theoretically all comprised of two Republicans and two Democrats each --- prevents any partisan malfeasance, those boards and the officials who work under them, have always served ultimately at the pleasure of the state SoS. That power and pressure was wielded frequently by Blackwell during his tenure, resulting in a host of not-so-bi-partisan decisions and oversight from Ohio elections officials in 2004 and beyond.
Find more from Bradblog.Com on InfoPages.


Join the Boston Sunday Night Film Club this Sunday, Nov 20th at 7p for The Edge of Seventeen at the Regal Fenway Stadium 13. Look for Sean wearing a nametag in the theatre lobby about 15 minutes before the film. As always, after the film we will descend on a local establishment for dinner/drinks/discussion.
Note: This is a reserved-seating showing, the host will be sitting in Row F.


text:
Robert Markland the younger
In 1811, Robert Markland the younger 'late a resident of the island of Jamaica' and Thomas Markland, sons of Robert Markland, formed a new partnership at Manchester under Markland Brothers, in succession to the one that had existed between their father Robert Markland and Thomas Markland at Manchester. Robert Markland, presumably the younger, had been the counterpart in the conveyance of estates by Robert Kerr (q.v.) in 1805.
The Markland family provide an example of the inter- connectedness of manufacturing and the slave economy in Manchester. Robert Markland I (c.1747–1828) was a check and fustian manufacturer and heir to an estate at Pemberton, near Wigan. His brother Edward was part of the partnership Markland, Cookson and Fawcett, the owners of the largest cotton mill in Leeds. He [Robert] married Elizabeth Hibbert (1753–1838) in 1776. The couple resided at Mabfield just outside of Manchester. Two of their sons became involved in the West India trade. Robert Markland II (c.1782–1825) formed a slave-trading partnership with Simon Taylor, the richest man in Jamaica, and the Hibberts. Markland was responsible for the Hibberts' plantations after they left the island until his own departure from Jamaica in 1808. Robert's younger brother James Heywood Markland (1788–1864) (q.v.) became an important proslavery lobbyist. Having trained as a solicitor in Manchester, he moved to London and joined the legal partnership Markland and Wright. The firm acted as solicitors for the Society of West India Planters and Merchants.
London Gazette 16440 01/01/1811 p. 21; The National archives, PWLB - Records of the Loan Commissioners and the Public Works Loan Board PWLB 11 - West India Relief Commissioners: Mortgage Indentures PWLB 11/79 - Jamaica: WELCOME, HADDINGTON, LEYDEN, ROXBURGH, BURRELL'S PENN, MOUNT PARNASSUS, WELCOME BARGUADIER PENN, STAPLETON MOUNT and LEMON HALL estates.
http://www.ucl.ac.uk/lbs//media-new/pdfs/lbsmanchester2.pdf [accessed 27/02/2017]; Obituary of Robert Markland junior from The Gentleman's Magazine Vol. 95 (Jan-June 1825) Part 1, p.185.
1823 [EA] - → Not known
Norwich Castle and Campbell's Mountain [ Jamaica | St Andrew ]
Name partner
Markland Brothers
South American merchant
In 1811, Robert Markland the younger 'late a resident of the island of Jamaica' and Thomas Markland, sons of Robert Markland, formed a new partnership at Manchester under Markland Brothers, in...
Thomas Hibbert of Chalfont
Robert Markland the older had married Thomas Hibbert of Chalfont's sister...
Robert Kerr
Tentative association...
James Heywood Markland
Dorothy Diana Yates (née Smellie)
???? - 29th Aug 1873
'Robert Markland the younger', Legacies of British Slave-ownership database, http://wwwdepts-live.ucl.ac.uk/lbs/person/view/2146650641 [accessed 17th January 2020].


text:
TelevisionComic Book Shows
Riverdale Showrunner Shares More on Luke Perry's 'Important' Tribute Episode
Filed to:riverdale
Luke Perry in an earlier episode of Riverdale.
Photo: Jack Rowand (The CW)
Luke Perry's death was shocking and tragic, and it left a hole—not only with fans who appreciated him, but in the CW series Riverdale. The show chose to wait to address Perry's death until after season three was over. Now, they've unveiled plans for the season four premiere, an episode that showrunner Roberto Aguirre-Sacasa calls one of the most important Riverdale may ever do.
Aguirre-Sacasa announced that the season four premiere would be called "In Memoriam"—honoring both Perry and his character, Archie's dad Fred Andrews. He hasn't given any more details yet on how the show is going to portray Perry's death in the series, but Aguirre-Sacasa did call the episode "a tribute to a fallen friend," who passed away on March 4 after a massive stroke.
Perry's death happened while Riverdale was filming season three, and fans were a bit surprised that it wasn't addressed in the story before the end of the season. Instead, Fred was sent away on business, and Archie's mom (played by Molly Ringwald) stepped in as Archie's parental figure. There's a reason for that, according to the showrunner.
Batwoman's Beginnings and Arrow's End Start This October
Looking forward to basically every show in the DC-CWniverse except Legends of Tomorrow (although,…
In a previous interview with TV Line, Aguirre-Sacasa said his team had already figured out how they wanted to address his death, but wanted to make sure the tribute was given the time and space it deserved. "We just didn't want to rush in and do something before we had fully figured it out. We'd already written basically the end of the season, so rather than try to squeeze in something that's so monumental and so important, we thought, 'Let's take the time to think about what we want to do, and do it right," he said.
Riverdale returns with "In Memoriam" on October 9.
For more, make sure you're following us on our new Instagram @io9dotcom.
More from io9
The Riverdale Finale Spoiler Q&A: What The Hell Just Happened?
Lucy Hale Will Lead the CW's Riverdale Spinoff Katy Keene
Luke Perry Tributes, From His Riverdale and 90210 Cast Members, and Beyond
Video Editor and Staff Writer at io9. My doppelganger is that rebelling greeting card from Futurama.


text:
Two Shot and Killed in Barry County, One Arrested
Police are investigating a double murder that occurred Friday afternoon in Barry County.
Wood TV 8 reports that the shootings happened around 3p.m. between Martin and Delton in the area of Lewis and Lindsey roads in Orangeville Township.
According to the Barry County Sheriff's Office two men were shot and killed, one in a residence and one by the road.
Passers-by spotted a man bleeding in a ditch. When they tried to help, a man holding a pink handgun approached them.
David Harrison, who lives about two miles from the incident, told Wood TV 8, "He pointed a gun in our face and told us to get going."
Another man, Jason Wyatt, says the gunman approached him as well, "He walked up to my window and put his gun in my face and told me that I need to leave. I tried seeing if I could help again, and then he got more agitated and put a gun closer toward me and told me that we had to get out of there."
The men called 911 and tried to keep other people from entering the area.
They say it was a scary situation and they are grateful to be alive.
Police say the shooter is in custody and there is no threat to the public.
The names of the victims or the shooter have not yet been released.
Source: Two Shot and Killed in Barry County, One Arrested
Filed Under: Barry County, Police, shooting
Categories: West Michigan News


labbaART has started a donation pool!
Hey! I'm labbaART, I live in Italy and I'm a 24 year old aspiring comic book artist. I love drawing giantesses (bigger is better~), chatting with people and making new friends!
Favorite books Lockhart's Lament. Favorite writers Steven Strogatz. Favorite games http://mathoverflow.net/questions/93276/a-game-on-noetherian-rings Tools of the Trade Pencil, pens and paper. Other Interests Mathematics.
Interested in some cool manga featuring big girls?
Before getting to the title, you've probably noticed the Journal Skin I used is slightly different this time. Indeed, I've just passed the one thousand watchers milestone... and I wanted to thank you all for this!
When I opened this account more than one year ago, I had already been drawing giantesses for at least one year; but I didn't want to upload them online, because I didn't think anyone would care. If DeferredGalaxy3 hadn't persuaded me to open this account, maybe you wouldn't even know Ester, Miriam and all the other big girls you can now find in my gallery! Apart from him who got me to open this account in the first place, I don't want to mention anyone else: because there would be so many people I'd like to thank for their friendship and their support that I'm definitely going to forget at least one of them, and that would be unforgivable.
So... thank you all—friends or simple acquaintances, giantess aficionados or people who simply find my big girls curious, commenters or lurkers. If I'm still here drawing after so much time, I have to thank you for this!
So... what was I saying? Ah right, the 1000 watchers thing. To celebrate this milestone, I'd like to share with you four manga featuring giantesses/shrinking that I've read in the past (or that I'm still reading, since two of them are still ongoing!) and that you might not know about. I hope you have fun reading them!
Sayonara Peter Pan (ONGOING) — It's a truly amazing manga about a group of teens who study at a prestigious school in a secretive island... only to find out that they're actually just one or two inch tall (everything is crafted to be their size), and that they're all part of an experiment. A friend from Japan showed me a few scans of some chapters that aren't available here yet (but that will soon be translated into English), and wow, there's everything one could have ever hoped to find in a manga involving size difference stuff!
Gigant (ONGOING) — written and drawn by the same author of Gantz, if I'm not mistaken. It's about a girl who can grow in size to enormous heights, and she has to fight some aliens who want to destroy Japan for some reason. Do I really need to say more?
Giant Warrior Nagi (completed) — from the actual description that can be found on mangadex: "In a world where traditional warfare has been replaced by proxy battles fought by extremely powerful, and extremely LARGE, individual representatives chosen from each country, one rookie warrior faces her introduction to the international stage". Lovely, right?
And what about you? Do you know some cool manga/anime/other piece of media involving macro/micro themes that you think most people don't know about? If yes, please share them here: let's put our links together, and let's have some fun discovering cool GTS stuff we were totally unaware of!
On an unrelated note, are you a professional/semi-professional comic artist/mangaka? If yes, please send me a note: I need advice on a certain something, and I'd be incredibly grateful if you wanted to share a bit of your knowledge and experience with me!
For those of you who don't know what I'm talking about, shipgirls are characters from the Kantai Collection (Kancolle) anime; they're personifications of ships, which makes them perfect giantesses since the right way of drawing them is technically by making them as big as the ship they're supposed to represent! This one here is Hatsuyuki: a friend of mine is really into her, so I made this picture as a gift for him! Anyway, I hope you all like it!
Time for a class picture!~ This was taken more or less two months ago, just before the beginning of summer vacation. This was her last year in middle school, so she'll be a high schooler starting from September... but thanks to this photo, she'll never forget about her former classmates!
It's about time I made another pic of Miriam alone, without any other (big) characters appearing alongside her (considering that she'll also be with someone the next time I draw her *cough* right, DeferredGalaxy3 *cough*?). My first picture of this OC dates back to a year and an half ago (link)... I may not draw her very often, but she's still my favourite character of mine together with Ester!
Height: 73.76 m (242 ft.).
Personality: sweet and innocent, cannot stand bullies... but can also be quite harsh and careless when playing innocently with tinies.
Likes: theme parks, outdoor activities, playing with tiny cars and dolls.
Grazie mille per il llama!!


As Turkuvaz, we process your personal data and make every effort to protect your data by taking technical and administrative measures prescribed by law. You can access the explanatory text for your processed personal data by visiting the data policy page.
The area controlled by the Islamic State of Iraq and al-Sham (ISIS) expanded suddenly after ISIS made a move into central Iraq from the east of Syria. While the world spoke about this expansion, weapons that ISIS seized, many of which were U.S. weapons given to the Iraqi army, became the main point of discussion.
Under normal circumstances, planes, helicopters, tanks and heavy guns that the Iraqi army could not take with them while pulling back should have been destroyed. The U.S. calculated that weapons might fall into the hands of al-Qaeda and the Taliban if requests for combat aircraft were granted, but the U.S. still agreed to sell these to Iraq. Nouri al-Maliki's government not only asked the U.S.to assist in creating an effective air force but also the Czech Republic, which sold Baghdad 12 L-159T1 fighter aircraft.
After ISIS' expansion in the region, Maliki realized that the air force was not effective and immediately bought second-hand Suoki 25s from Russia.
As this unfolded on the Iraqi front, ISIS maintained control over parts of Iraq and Syria. We will have a chance to see if ISIS will continue the Taliban's passiveness with regard to using air force and sophisticated weapons, although the Taliban has experience in governing in Afghanistan.
ISIS added new Russian T55 tanks to previous ones that they acquired from the Syrian opposition before entering Mosul and Tikrit in Iraq. It is known that ISIS has 30 of these tanks, and they have used some in Syria, Deir ez-Zor, rural Humus and some in Mosul and Tikrit. It has been reported that some other models of Russian tanks used by the Iraqi Army, the T72, are in the hands of ISIS, and they likely possess five to 10.
According to some reports, although it is not confirmed, ISIS seized one or two U.S.- made M1 Abrams tanks. The Iraqi Army has 140 M1 Abrams tanks. In October 2012, Iraq bought six more of these M1 Abrams tanks. The M1 Abrams tanks that the Iraqi Army has are technologically inferior to the ones the U.S. uses and closer to the M1 Middle East model, manufactured in Egypt. It is known that M1 tanks are mostly located along Kuwait and Saudi Arabia borders in the south and some in Bagdad and its surrounding areas. Because of this, the chances of ISIS taking control of these tanks are slim.
Actually, the number of tanks that ISIS has seized is more significant than them taking over American-made Hummers. Regardless, the world is more focused on the military jeeps and vans. U.S.-made Hummer GMC and M1088 trucks that ISIS confiscated have garnered more media coverage. While ISIS upgraded to U.S.-made jeeps and trucks from Toyota and Nissan, the number of vehicles that ISIS has taken reached more than 200, and some of these are sent to Syria. Moreover, ISIS has scorpion or cobra four by four tactical armored vehicles by Otokar used in Mosul by the Iraqi Army and English Land Rover jeeps.
Although ISIS reportedly took control of some U.S.-made Black Hawk UH60 and Bell-IA-407 helicopters when they occupied Mosul Airport, authorities from the U.S. and Iraq have denied that these helicopters were in ISIS' inventory. In Syria, some MI8 and MI17 helicopters became unfit to fly and were confiscated by the Free Syrian Army and then ISIS. The Syrian opposition has not been equipped to use any air vehicles in their fight against the government for more than three years. It seems like this situation will continue.
While they were effective in Syria, ISIS was accustomed to fighting with Kalashnikovs, although they use very few American made M16 automatic rifles. It is known that the number of M16 and M16A4 rifles confiscated from the Iraqi Army is slight and that ISIS has distributed these weapons to many groups with night vision equipment and rifle bombs. The number of RPG7 rocket-propelled grenades ISIS possesses increased after their offensive in Iraq. Although the effectiveness of new armored vehicles is in question, the Syrian opposition destroyed or disabled many regime tanks using RPG7. Although ISIS has problems supplying ammunition for sophisticated weapons, it seems that machine guns and vehicles for transportation are not problems.
The American-made low attitude air defense system FIM92 Stringer MANPAD and the Soviet model, as well as Polish- and Bulgarian- made ZU23-2 and ZU23-4 anti-aircraft guns are now in ISIS' inventory. Despite that, there is no certain information about the number of guns and whether ISIS has an air system to defend them at the moment. BM21 Grad multiple barreled bombardier rockets, 155mm M198 howitzers and Type 59 130 mm cannons have been added to ISIS' inventory in Iraq. The large number of weapons in ISIS' inventory prompts the question of whether the organization has enough trained members or not. In Libya and Syria, the opposition learned to use tanks, cannons and multiple barreled bombardier rockets, RPGs and other weapons in a very short time. ISIS shoots with tanks and cannons, which it confiscated in Mosul and Tikrit, in some areas.
ISIS is known to have experience with using tanks in Syria. ISIS portrays itself as a group that has adopted a policy of not carrying heavy weapons as a principle. Bringing new weapons into their inventory comes with new disadvantages. Transportation of heavy weapons is difficult and means that the organization's convoys are now targets. ISIS has shown with SCUD missiles in Raqqa -- thought to have been taken from the Syrian regime -- that the organization is aiming to acquire a diverse inventory of weapons.


A folk icon since it was first introduced in 1965, The Fender Villager 12-string Acoustic Electric is an outstanding instrument at an amazing price point that continues to inspire players of various styles over 50 years later. It features a solid spruce top, laminated mahogany back and sides for full acoustic projection and rich, full-bodied tone. The maple "C"-shaped neck with walnut fingerboard is beautifully playable and responsive and is topped by a classic "hockey-stick" headstock. The cutaway dreadnought body gives you full access to this great neck — a feature that guitarists here at Sweetwater really appreciate.
The Fender Villager 12-string comes loaded with Fishman's Presys pickup system. The powerful Presys system features treble, mid, bass, volume, and phase controls that let you dial in that perfect sound. Another bonus is the built-in tuner that lets you tune without complicated and confusing handheld tuners. With the Fishman Presys pickup system, you get stellar amplified acoustic tone that translates all the sparkle, jangle, and shine of a 12-string!
The Fender Villager 12-string Acoustic Electric has many great appointments including scalloped X-bracing that adds further tonal support to the solid spruce top. There's a classic gold pickguard, bone nut, and a vintage-inspired rosewood Fender "Viking" bridge with a compensated bone saddle that not only looks great but adds even further tone-enhancement to this wonderful instrument.


Senior Dalton Shew examines the up close details of a cow heart.
On Mar. 1 the Contagion A course had the chance to dissect a cow heart, lungs, and tongue. Northwest Meat Market had supplied Contagion teacher Mrs. Rhonda Drumm with all of the fresh cow parts at a discounted rate to allow her student the learning opportunity.
"I think the opportunity is pretty cool since we get to learn hands on, and I really enjoyed doing the dissection," said senior Jenna Maske.
Drumm dissected a small portion first. She then allowed the students to do other dissections of their own on each of the parts to advance the learning opportunity.
With doing the dissection in Contagion, it gives students considering a career in the medical field a chance to make sure that this is something they are interested in.
Contagion teacher Mrs. Rhonda Drumm dissects the lungs of a cow for the class.


MinaKato . sexyjennifer1. MishelleGold. AdorableOskar.
Natusika12IsabelePervQueenEmmaMadissonJULIANANOCHETY9 .NightPrincess4UDanaZetaJasminee77hotfibi .Jasminee77IsabelePervQueenariadnaandloganAchillesMuscle .AaronLynnwild9catAngelOfSeaXEmmaMadisson .Av6ry123EternityLoveKitaSongXjasonclay .RitulalyaDirtyMsAnneJULIANANOCHETY9shylaOconer .liveboy81DanaZetaRoseJennyAchillesMuscle .hotpinayanneTHEMISTRESSLAVESophieFreshAchillesMuscle .LovingKattiaCuteAndSexyOneCARAMELSTICSWEETnNaughty19x .AmberNicoleXTwerkySexyAsscumlover24FelixStrong .EternityLoveAndrewWatsonJuanita6WhitneyRose .
TrixyPixiPrincessBedLustXJuanita6LawrenceX .THEMISTRESSLAVEBrun3tteGirlsPascalSharpWhitneyRose .NadyaNadya2CuteAndSexyOneAndrewWatsonMillerManEaterX .IsabelePervQueenKarlaDirty4UXIsabelePervQueenLeoniHope .EvasoloAchillesMuscleKarlaDirty4UXAmeliaJohnson .AelisAchillesMuscleliveboy81LovingKattia .DanaZetaRitulalyaAchillesMusclelililatinxx .SashaBarkerAmberNicoleXKitaSongXVictoriaRoose .JULIANANOCHETY9AelisJessicaEdmondsDanaZeta .MissStunningMARRYm3BABYAlexSHUFFETEternityLove .LittleCuteJanenatashapleasuremelanie25Xhotpinayanne .


text:
@class CMHSearchHistoryHeaderView;

@protocol CMHSearchHistoryHeaderViewDelegate <NSObject>
/// 删除按钮
- (void)searchHistoryHeaderViewDidClickedDeleteItem:(CMHSearchHistoryHeaderView *)searchHistoryHeaderView;
@end

@interface CMHSearchHistoryHeaderView : UICollectionReusableView

/// delegate
@property (nonatomic , readwrite , weak) id <CMHSearchHistoryHeaderViewDelegate>delegate;

@end

redpajama_set_name: RedPajamaGithub


They're coming to kill you. What you gonna do?
DM/Tag #bikeandbabes your photos to be featured!
für kleines Geld könnt ihr bei ihn ein shooting buchen, falls ihr nicht zu weit weg von Kerpen/Bergheim kommt.
We spend our whole lives worrying about the future, planning for the future, trying to predict the future, as if figuring it out will cushion the blow. But the future is always changing. The future is the home of our deepest fears and wildest hopes.
But one thing is certain when it finally reveals itself.
Happy Easter Weekend lively and lovely people❤.
My happy place is watching @bprolifik practice! .
Chegou pessoal mais prescisao,na hora da reposição de rolamentos.
Nachdem wir gestern sehr spät von Ansbach, unserer halben Bayernweltreise, bei meiner Oma ankamen, weil wir einfach nicht mehr weiterfahren konnten, war ich auch nicht mehr in der Lage, den Beitrag zu veröffentlichen, was ich ja jetzt nachholen werde 😉. Warum waren wir zur Hölle genau dort?
Damn nice day to joy ride.
Great days ride with @sportstersicknessuk at the annual "big one". This year held at donnington park race circuit. We enjoyed a leisurely ride there stopping off at various point on the way to grow our numbers as we progressed.
With a final congregation point of donnington services, the shear scale of this ride became apparent. Everyone had time for a brief chat before being marshalled onto the final ride to donnington park.
Outriders blocked all routes so we enjoyed an un-interrupted procession through junctions stop lights and roundabouts. A carefully selected and scenic ride led us through villages where local residents lined the streets to see huge numbers of Harley Davidson Sportsters. With an encore of loud exhaust we proceeded through scenic twist and turns before arriving at the track.
Covering over 400milea today there and back including the final route to donnington was an epic journey and something special to be part of. Cruising the motorway on the final stretch was an amazing sight to see the whole foreseeable lane filled with sportster and friends in formation. It was something that would only ever expect to see in the USA 🇺🇸. Finally if you were on the ride and have any footage please tag @houseofthunderuk and send us a few clips as-well. Especially if they feature me 🤘.
These shirts are now in stock and look awesome!!!
They are on our website so go and check them out now!
This track suit makes me look like an oompa loompa 😑😑😑 if anyone has a size 38 or xsmall track suit for sale, please let me know! .
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Sometimes you have to play the role of a fool to fool the fool who think they're fooling you !
My students found out that I ride a motorbike and asked, "Are you going to join a gang?" to which I queried, "What makes you think I'm not already in one?" They were delighted. College students are weird.


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Dave East in 'Survival' mode for debut album
Actor and recording artist Dave East is making his official debut with "Survival," even though he's well-known to the hip-hop community. He also starred in "Wu-Tang: An American Saga" and hopes to make his mark in Hollywood. (Nov. 21)


1 bedroom apartments in Rogers Park, Chicago - how much will I be paying per month?
Rogers Park apartments are fairly reasonable considering how nice the area is, so I'd guess you'll pay around $800 for a one bedroom place.
A 1 bedroom apartment in Rogers Park will cost you between $750-$950 per month.


Das Motiv ist aus der neuen "Lovely Emoji Collection" von Magnolia. Wieder eine ganz tolle neue Kollektion. Diese süße Tilda heißt "Cause I Love You Tilda". Und der passende Text ist aus dem "You Are A Star Kit". Verwendete Stanzen sind DooHickey Bride Bow und Tilda Tag. Das Designpapier ist vom Maja Design aus der "I Wish Collection"
The motif is from the new "Lovely Emoji Collection" by Magnolia. Again a really great new collection. This sweet tilda is called "Cause I Love You Tilda". And the text is from the "You Are A Star Kit". Used punches are DooHickey Bride Bow and Tilda Tag. The design paper is from the Maja Design from the "I Wish Collection"
Vielen Dank das ihr vorbei geschaut habt! Ich wünsche euch eine tolle Woche!
Thank you for stopping by! Wish you a nice week!
Kommentare on "♥ All About Magnolia Challenge # 20 - All About Love ♥"
O my.....this is soooo beautiful.....This makes my heart beat faster....so gorgeous and sweet. I loveeeeeeeeee this.
Thanx for joining us at LHOSM and Simply Magnolia.
Merci pour ta participation sur Magnolia forever!
Oohhhhhh Tilda is so beautiful.
DT Magnolia Forever Challenge Blog.
WOW.......so incredibly beautiful sweetie, your colouring is perfect & I love the colour scheme. I also love the shabby chic feel to it, stunning hun.
Thanks for joining us over at LHOSM, hope you will join us again. Good luck hun.


Company Central Carolina Neurology & Sleep is located in the city of Salisbury. The legal address of the company is 911 W Henderson St L30. Other interesting information can be found on working days by phone (704) 637-1779. The scheme of travel to the company Central Carolina Neurology & Sleep can be viewed on the interactive map.


Our most prestigious collection highlighting only the most mesmerizing larger sized center stones, this makes the fantasy of owning a piece fit for royalty a reality.
There is no better way to express everlasting love than by offering your partner a gift of jubilation. Whether it be an engagement, an anniversary or a simple "I Love You" this will serve for any festivity.
An alexandrite is eternal. This timeless collection highlights designs that are everlasting, a piece of jewelry that can be passed down from generation to generation.
Mark Henry's bread & butter, this collection boasts some of our most recognizable & best selling items. Over the years, these jewels have become some of the brands most classic & sought after ornaments.
A collection crafted with him in mind, these pieces cater to the everyday man.
Petite alexandrite set in clusters to capture a grandiose appearance for those wanting to embellish themselves in practical & timeless jewels.
Who says a pinky is not deserving of proper marveling? Tantalize recognizable attention to your most underappreciated finger with a ring from our Pinky Allure Collection.
Honor a marvelous life spent together with a bauble that symbolizes both affection and devotion.


theft-prevention program was set up as well as the number of bicycle sheds increased.
Amsterdam is undertaking an ambitious capital-improvement program that includes building a 10,000-bike parking garage at the main train station.
Highlights: Designated street lanes and traffic signals. Bicycle safety concerns. Biking is core to its transportation infrastructure. Rent public bicycles. Underground sheds and outside racks, which hold thousands of bikes under guard. Large bicycle culture.
Wish we could emulate some of that here in the US.


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Sign of the Zodiak - 1961
The Barbara Stanwyck Show - Sign of the Zodiak - Broadcast April 3, 1961 on NBC. Directed by Jacques Tourneur
Joan Blondell (as Helene Terry) drags her sister-in-law Madge (Stanwyck) to see a fortune teller (Pierre, played by Dan Duryea). Cynical and bitter after the death of her husband, Madge arrives at the occult parlor under protest and interrupts the proceedings when the fortune-teller begins to claim to see Madge's dead husband trying to communicate with them. Angry words are exchanged and Madge leaves in a rage. Later she realizes she left a prized sentimental pocket watch at the parlor, an item that had belonged to the dead husband. She returns the next day and the smooth-talking Pierre claims he kept the watch in a safe place, but when he goes to retrieve it for her, it's gone. Later, Madge discovers the watch is back at her home already...
Barbara Stanwyck and Joan Blondell work together well (they teamed up in the 1930s films Illicit and Night Nurse) and they're joined by Dan Duryea (who worked with Stanwyck on Ball of Fire) in a nicely done three-way vehicle about madness and revenge, though the compacted runtime (30 minutes) doesn't allow director Jacques Tourneur to explore more than the relationship between Blondell and Stanwyck's characters, which is radically different from what it first appears as the story starts.
The camera-work tries to fit faces into the screen as largely as possible, perhaps to compensate for the small-sized TV screens of that era, but it is sometimes distracting since Stanwyck's face at times has her eyebrows as the center of the frame (instead of her eyes). But Tourneur otherwise gives us good work, considering the small-TV-show sets, and Stanwyck, Duryea and Blondell are as good as ever, with better than average TV level writing and performances which makes for the obvious question: why didn't The Barbara Stanwyck Show last longer than one season, with 37 episodes, between Sept 1960 to July 1961?
The Barbara Stanwyck Show, Vol. 1 - Amazon
More Barbara Stanwyck
Original Page July 2013


The 2013 Symetra Tour schedule continues to look bright with the announcement today that Volvik, known as the leader in the colored performance golf ball market, will sponsor the inaugural Volvik Championship. Set to take place September 16-22 at Reunion Resort, a Salamander Golf & Spa Resort, in Orlando, Fla., the Volvik Championship will feature some of the best up-and-coming stars on the Road to the LPGA competing for a $100,000 purse.
The Volvik Championship further expands Volvik's partnership with the Symetra Tour as they are also the "Official Ball" of the tour and title partner of the "Volvik Race for the Card'.
The 2013 Volvik Championship will take place on Reunion's Palmer Course and serve as the penultimate event on the Symetra Tour schedule and could potentially play a pivotal role in determining the top-10 on the final Volvik Race for the Card standings. The year-long race will end the week following at the Symetra Tour Championship in Daytona Beach, Fla. where the top-10 at the conclusion of the tournament will earn fully-exempt status on the 2014 LPGA Tour.
"With Volvik being the #1 color golf ball on tour, it makes perfect sense to not only introduce this new event to the Symetra Tour schedule but to play it at Reunion Resort in Orlando, just a short drive from Volvik's headquarters," said Volvik USA President, Don Shin.
This year marks the second year in the Tour's 33-year history that the top-10 Symetra Tour players on the season-ending Volvik Race for the Card standings will earn fully-exempt LPGA Tour status for the following year.
Reunion Resort, which is part of Salamander's new Grand Golf Resorts of Florida collection, has a rich history of hosting professional golf tournaments, including an LPGA Tour event from 2006 to 2008 and the Legends Tour's 2012 ISPS Handa Cup. Site of the ANNIKA Academy, it is the only location in the world which features Signature course designs by Arnold Palmer, Jack Nicklaus and Tom Watson.
The Symetra Tour – Road to the LPGA, is the official developmental tour of the LPGA and enters its 33rd competitive season in 2013. With the support of its umbrella partner, Symetra, the Tour's mission is to prepare the world's best young women professional golfers for a successful career on the LPGA Tour. 2013 marks the second season that the top-10 money winners will receive full, equal status for the following season on the LPGA Tour. With more than 500 alumnae moving on to the LPGA, former Symetra Tour players have won a total of 371 LPGA titles, including 40 major championships. Symetra Tour events have contributed more than $5 million to charitable organizations throughout the nation since 1989. Follow the Symetra Tour on the web at www.SymetraTour.com, Facebook.com/RoadtotheLPGA, Twitter.com/Road2LPGA, and YouTube.com/Road2LPGA.
Managed by Salamander Hotels & Resorts, the AAA Four Diamond-rated Reunion Resort located in Orlando offers premier multi-bedroom villas, highlighted by the 96-suite main hotel with a stunning lobby; and luxury private homes, each with lavish appointments and exquisite decor. Part of the Grand Golf Resorts of Florida collection, Reunion Resort is the only location in the United States with golf courses designed by the golf greats Jack Nicklaus, Arnold Palmer and Tom Watson and also features the exclusive ANNIKA Academy by Annika Sörenstam. Reunion Resort features 330 resort accommodations including luxury condominiums, private villas and vacation homes surrounded by 2,300 acres of lush landscape and nature preserve, elegantly appointed meeting space, a full-service boutique spa, multiple pools, slides, water playgrounds and a 1,000-foot lazy river. For additional information call 407-662-1000 or visit www.reunionresort.com.