text:
\section{Introduction}

Positive and completely positive maps play a central role  in the theory of operator algebras and in the related fields of dilation theory,
random matrix theory, free probability, semidefinite optimization theory and quantum information \cites{CMW,optimization,NC}.
In quantum theory, completely positive maps  are ubiquitous as they represent general  evolutions of quantum states,
while positive maps are a powerful tool for  characterizing quantum entanglement  \cites{peres1996separability,Horodeckis,horodecki1997separability,terhal2000schmidt,terhal2000bell,sanpera2001schmidt,clarisse2005characterization,ranade2007jamiolkowski,chruscinski2009spectral,skowronek2009cones} (see \cite{horodecki2009quantum} for a review).

Much of the  theory of positive maps  was originally developed in the case where  the underlying number field is the complex numbers.
In contrast, the case of  operator algebras over the reals  has not been looked at as extensively.
Notable exceptions are the works of Ruan \cites{Ruan,Ruan1}, and, very recently,  Blecher and  Tepsan  \cite{BlecherTepsan}.
The real variant quantum theory  was considered in an early work by St\"uckelberg  \cite{stueckelberg1960quantum}, and was later found to differ from standard complex quantum theory in a number of essential features.
Araki  \cite{araki1980characterization} observed that, while the state space of composite systems  in complex quantum theory  is equal to  the product of the dimension of  the components, in real quantum theory it is strictly larger.
Dually, this fact implies that product observables are not sufficient to identify general entangled states \cites{wootters1990local,hardy2012limited}.
Explicitly, Wootters \cite{wootters1990local} provided an example of two mixed states that have orthogonal support, and yet give rise to the same correlations for all possible local observables.
Building on this example, it was later shown that there exist  evolutions in real quantum theory  that are indistinguishable when acting on a single system, but can be  distinguished perfectly when applied locally on  an entangled state \cite{chiribella2010}.
These  phenomena  can be summarized by the statement: \textit{real quantum theory violates the Local Tomography Axiom} (see \cite{chiribella2021process} for a review), often assumed in axiomatic reconstructions of standard finite-dimensional quantum theory (see e.g. \cites{Hardy01,CDP11,DB11,MM11,Hardy11}).
Further differences between real and complex quantum theory were observed in \cites{wootters2012entanglement,aleksandrova2013real,chiribella2013quantum,wootters2014rebit}.
Very recently, an experimental test that distinguishes between complex quantum theory and real quantum theory was proposed \cite{Renou}.

In this paper  we develop  the theory of positive maps acting on a full matrix algebra  over the reals, and discuss their applications to the study of entanglement in real quantum mechanics.
We  characterize the subset of  real maps with  a positive complexification, providing explicit examples of positive real maps outside this subset.
See also our paper \cite{CDPR}.
Then, we prove that the existence of positive real maps with non-positive complexification is equivalent to the existence of quantum states that are entangled in real quantum theory, but are separable when regarded as elements of the larger state space of  complex quantum theory.
These states, of which we also provide explicit examples, cannot violate Bell inequalities  \cite{bell1964einstein}  (see \cite{brunner2014bell} for a review),
even if  pre-processing operations are allowed \cite{popescu1995bell} and if an arbitrarily large number of identically prepared systems is available \cites{navascues2011activation,palazuelos2012superactivation}.
Operationally,  the existence of real-entangled, but complex-separable states  can be interpreted in a resource-theoretic framework
\cite{wu2021operational}, as the ability to prepare a larger class of states by performing  a larger set of local operations that are not restricted to have real matrix elements in a given basis.

Finally, we characterize the real maps that break  quantum entanglement,  extending a classic characterization by Horodecki, Shor and Ruskai \cite{HSR} to the real domain, and exploring the relation between entanglement breaking maps and another class of maps, known as PPT.
In the complex case, Christandl  conjectured (see  \cite{RJKHW} ) that $\Phi \circ \Phi$ is entanglement breaking for every PPT map $\Phi$ and in \cite{CMW} (also see \cite{CYT}) it was shown that this conjecture (now known as the PPT-squared conjecture) holds in dimension $3$.
In stark contrast, here we show that the real version of the so called PPT-squared conjecture is false in even dimension 2.
Nevertheless, we show that the original  PPT-squared  conjecture (stated in the complex case) implies another conjecture for real maps satisfying a suitable strengthening of the  PPT property.
When this stronger property is assumed, we  prove that the real version of the PPT-squared conjecture holds asymptotically.
Moreover, when assumed trace preserving and unital, we prove that finite number of iterations of PPT maps become real entanglement breaking.

The rest of the paper is structured as follows.
In Section \ref{sec:defpos} we discuss the notion of positivity on real matrix algebras, pointing out the need to include a Hermiticity requirement in the definition of a positive map.
In Section \ref{sec:choi},  we  consider  $p$-positive maps and characterize their Choi matrices.
The results of this section are mostly adaptations of existing results in the complex case.
In Section \ref{sec:poscomplex} we characterize the set of $p$-positive real maps that admit a $p$-positive complexification,
and in Section \ref{sec:examples1} we provide concrete examples of $p$-positive real maps that violate this condition.
We then move to the study of entanglement on real Hilbert spaces, characterizing entanglement witnesses and showing that the existence of positive real maps with non-positive complexification is equivalent to the existence of mixed states that are entangled in real quantum theory but separable in the complex version (Section \ref{sec:witnesses}).
Examples of real-entangled but complex-separable quantum states are provided in Section \ref{sec:examples2}.
In Section \ref{sec:breaking} we consider the notion of entanglement $p$-breaking channels, which transform arbitrary entangled states into states with Schmidt rank at most $p$, and show an explicit example of a map that breaks complex-entanglement but preserves real-entanglement.
Finally, we conclude with a discussion of the relation between entanglement breaking channels and PPT channels, proposing a real version of the PPT-squared conjecture and showing that its asymptotic version holds true (Section \ref{sec:PPT}).

\section{Positivity and Hermiticity}\label{sec:defpos}

Let $\<\!\< K$ be either $\mathbb{C}$ or $\mathbb{R}$.
A matrix    $P \in M_n(\mathbb{K})$ is called positive (denoted $P\ge0$) if it is of the form $P  =  A^*  A$ for some $A\in  M_n (\mathbb{K})$, where $A^*$ is the adjoint of $A$
(for $\<\!\< K =  \mathbb{R}$, the adjoint  $A^*$ coincides with the transpose $A^t$).
 The set of all positive matrices in $M_n(\<\!\< K)$ will be denoted by $\operatorname{PSD}(\<\!\< K^n)$.

A matrix is Hermitian if $A=A^*$, and we write $\operatorname{Herm}(\<\!\< K^n)$ for the set of Hermitian matrices in $M_n(\<\!\< K)$.
For a Hermitian matrix $A \in \operatorname{Herm}(\<\!\< K^n)$, positivity is equivalent to the condition $\<  v  | A  v\>  \ge0$ for every $v\in  \mathbb{K}^n$.
Here, the product is defined as $  \<  v  |  w\>  =  v^*  w$, regarding the vectors $v,w \in  \mathbb{K}^n$ as column vectors.

We say that a linear map $\Phi :  M_n  (\mathbb{K})  \to  M_{m}  (\mathbb{K})$ is {\em positivity preserving} if $\Phi  (P)  \ge0$ for all $P\ge 0$,
and {\em Hermitian} if $\Phi  (A^*)   =  \Phi(A) ^*$ for all $A\in M_n (\mathbb{K})$.
If the map $\Phi$ is  positivity preserving and Hermitian, we call it  {\em positive}.

In the complex field, preservation of positivity  implies Hermiticity.  This implication, however, does not hold on the real field.
Counterexamples are abundant. For example, the map
\[
 \Phi   :     M_2 (\mathbb{R})  \to  M_1  (\mathbb{R}),   \,\,
 \begin{pmatrix} a  &  b  \\  c  &  d \end{pmatrix}   \to   (a+ d  + b-c )/2
\]
is positivity preserving but not Hermitian.

Let ${\operatorname{Sym}}_n(\mathbb{R}) = \{ A \in M_n(\mathbb{R}) : A^t = A \}$ be the space of all symmetric matrices, and let  ${\operatorname{Asym}}_n(\mathbb{R}) = \{ A \in M_n(\mathbb{R}) : A^t = -A \}$.
It is immediate to see that a map $\Phi  $ is Hermitian if and only if
\[
\Phi \left( {\operatorname{Sym}}_n(\mathbb{R})  \right)  \subseteq   {\operatorname{Sym}}_m(\mathbb{R})
\quad\text{and}\quad  \Phi   \left( {\operatorname{Asym}}_n(\mathbb{R})  \right)   \subseteq  {\operatorname{Asym}}_m(\mathbb{R})  .
\]

In special cases, Hermiticity follows from positivity preservation, plus additional properties, such as unitality and unit norm 1:

\begin{prop}
If $\Phi: M_n(\mathbb{R}) \to M_{m}(\mathbb{R})$ is positivity preserving, unital and norm 1, then $\Phi$  is Hermitian.
\end{prop}

\begin{proof}
Since $\Phi$ is positivity preserving,   $\Phi    \left( {\operatorname{Sym}}_n(\mathbb{R})  \right)  \subseteq   {\operatorname{Sym}}_m(\mathbb{R})$.
We now show that unitality and the norm-1 condition imply that   $\Phi   \left( {\operatorname{Asym}}_n(\mathbb{R})  \right)  \subseteq  {\operatorname{Asym}}_m(\mathbb{R}) $.

The proof is by contradiction. Let $C$ be an antisymmetric matrix, with $\|C\|  =1$ without loss of generality.
Suppose that $\Phi(C) = X$ is not antisymmetric. Then $B = \frac12(X+X^t) \ne 0$.
Since $B^t=B \ne0$, the spectrum of $B$ is real and not $\{0\}$.
Let $\lambda$ be an eigenvalue of $B$ with $|\lambda| = \|B\|$.

Since $C$ is antisymmetric and norm 1, we have  $-I_n \le C^2 \le 0$ and $-1$ is an eigenvalue of $C^2$.
For $t\in \mathbb{R}$,
\[
 \| I_n + tC \|^2 = \| I_n - t^2C^2\| = 1 + t^2.
\]

Now choose the sign of $t$ so that $|t\lambda|$ belongs to the spectrum of $tB$; then
\[
 \| \Phi(I_n + tC) \| = \| I_m + tX \| \ge   1 + |t\lambda| .
\]
For sufficiently small $t$, this yields a contradiction with the norm 1 property of $\Phi$.
\end{proof}

One may ask if Hermiticity follows from positivity preservation and from the property $\|  \Phi \|  =  \| \Phi  (I_n) \|$, without requiring unitality.
The following counterexample answers the question in the negative.

\begin{example}
We provide an example of a positivity preserving  map  $\Phi$   on $M_2 (\mathbb{R})$ that satisfies the condition  $\|  \Phi \|  =  \| \Phi  (I_2) \|$ but is not Hermitian.
Define
\[
 \Phi (A)  =  \Phi  \begin{pmatrix}  a& b  \\  c   & d\end{pmatrix} =    \begin{pmatrix}   \frac{a+d}2 &  0 \\  0  & \frac{b-c}2 \end{pmatrix}\,.
\]
 The map $\Phi$  is positivity preserving, since $\Phi  \begin{spmatrix} a &  b \\  b & d\end{spmatrix}  =  \frac  {a+d}2 \,  E_{11}$, for any positive matrix.  Moreover, it satisfies
\begin{align*}
 \left\|  \Phi  (A)  \right\| =  \max \left\{  \left| \frac{a+d}2\right|\, ,   \left|  \frac{b-c}2\right|\right\}
 \le  \left\|  A \right\|
\end{align*}
Hence, $\|  \Phi \|  = \|  \Phi  (I_2)\|=1$. However, $\Phi$ maps antisymmetric matrices into symmetric ones, and therefore is not Hermitian.
\end{example}

We conclude with an elementary lemma that will be useful later.

\begin{lemma}\label{lem:rankoneisenough}
A Hermitian map   $\Phi  :  M_n ({\mathbb{K}})  \to  M_m ({\mathbb{K}})$  is positive if and only if   $  \< w   |  \Phi  (  v v^*)   w\>\ge 0$ for every pair of vectors $  v\in  \mathbb{K}^n$ and $w\in\mathbb{K}^m$, where $vv^*$  is the rank-1 matrix with entries $  (vv^*)_{ij} :  =   v_i  \overline v_j$,  $\overline v_j$ denoting the complex conjugate of $v_j$.
\end{lemma}

\begin{proof}
Since every positive matrix $P$ is a sum of matrices of the form $vv^*$,  $\Phi$ is positive if and only if $\Phi  (vv^*)$ is positive for every $v\in  \mathbb{K}^n$.
Note that  $\Phi (vv^*)$ is Hermitian, because  $\Phi$ is a Hermitian.
Hence, $\Phi  (vv^*)$ is positive if and only if   $\<  w|  \Phi (vv^*)  w\>$ for all $w \in  \mathbb{K}^m$.
\end{proof}
\section{$p$-positive maps and their Choi representation}\label{sec:choi}

A linear map $\Phi  :  M_n ({\mathbb{K}})  \to  M_m ({\mathbb{K}})$ is called {\em $p$-positive} if  the map
$  \Phi  \otimes  \operatorname{\textsl{id}}_{p}:   M_n ({\mathbb{K}})   \otimes M_p ({\mathbb{K}})  \to  M_m ({\mathbb{K}}) \otimes M_p ({\mathbb{K}}) $ is positive, where $\operatorname{\textsl{id}}  _p$ is the identity map on $M_p ({\mathbb{K}})$.
Positive maps correspond to the special case $p=1$.   A map is {\em completely positive} if it is $p$-positive for every $p\in  \mathbb{N}$.

 A linear map $\Phi  :  M_n ({\mathbb{K}})  \to  M_m ({\mathbb{K}})$ is conveniently represented in terms of its Choi matrix~\cite{choi1975completely},  defined as
\begin{align*}
C_\Phi:=\sum_{i,j} E_{i,j}\otimes \Phi(E_{i,j}) \,,
\end{align*}
where $E_{i,j}$ are the standard matrix units of $M_n ({\mathbb{K}}) $.     A related representation is the Jamio\l kowski matrix~\cite{jamiolkowski1972linear}
$J_\Phi:=\sum_{i,j} E_{i,j}\otimes \Phi(E_{j,i})$, originally introduced in the study of positive maps.

We will  often use the following lemma,  the complex version of which is well known.

\begin{lemma}\label{lemma-pos}
A linear map $\Phi:M_n({\mathbb{K}})\rightarrow M_m({\mathbb{K}})$ is
\begin{enumerate}
\item Hermitian  if and only if its Choi matrix $C_\Phi$ is Hermitian,
\item $p$-positive if and only if $C_\Phi$ is Hermitian and $ \<  V  |  C_\Phi  V\>\ge 0$ for  every  vector  $V  \in  \mathbb{K}^n\otimes \mathbb{K}^m$ with Schmidt rank at most $p$.
\item $p$-positive if and only if its adjoint map $\Phi^*$ is $p$-positive.
\end{enumerate}
\end{lemma}

\begin{proof}
The complex version  of (1) was proven in \cites{de1967linear,ranade2007jamiolkowski},  while the complex version of (2) was proven in  \cite{jamiolkowski1972linear}
(for $p=1$, see also \cite{stormer-book}*{Proposition~4.1.11}) and \cites{ranade2007jamiolkowski,chruscinski2009spectral} (general $p$).
The proof of both items uses   the relation
\begin{align}\label{choiinverse}
 \operatorname{Tr} (C_\Phi (A^t\otimes B)) = \operatorname{Tr} (\Phi(A)B) ,
\end{align}
valid for arbitrary matrices  $A\in M_n({\mathbb{K}})$ and $B\in M_m({\mathbb{K}})$.

If $\Phi$ is Hermitian, then $C_\Phi^*  =  \sum_{i,j} E_{i,j}^*\otimes \Phi(E_{i,j})^* =  \sum_{i,j}  E_{j,i}   \otimes \Phi  (E_{i,j}^*)  =  C_\Phi$.
Conversely, if $C_\Phi$    is Hermitian, then one has
\begin{align*}
 (\Phi  (A)^*)_{i,j}   &= \operatorname{Tr}[  \Phi  (A)^*\, E_{j,i}   ]  =    \overline{\operatorname{Tr}[    \Phi  (A)\,  E_{j,i}^* ] }  = \overline{  \operatorname{Tr} (C_\Phi (A^t\otimes E_{j,i}^*))} \\
 &=  \operatorname{Tr} (C_\Phi^* (  (A^t)^*\otimes E_{j,i}))  =   \operatorname{Tr} (C_\Phi (  (A^*)^t\otimes E_{j,i}))  = \operatorname{Tr}[   \Phi  (A^*) \,   E_{j,i} ] \\
 &=\Phi  (A^*)_{i,j}    \,.
\end{align*}
This concludes the proof of (1).

 If $\Phi$ is $p-$positive, then $\Phi \otimes \operatorname{\textsl{id}}  _p$ is Hermitian.
Then,  Lemma \ref{lem:rankoneisenough} implies $\<  w  |     (\Phi \otimes \operatorname{\textsl{id}}  _p)   (vv^*)  w\>\ge 0$ for every $v  \in  \mathbb{K}^n\otimes \mathbb{K}^p$
and every $w\in  \mathbb{K}^m\otimes \mathbb{K}^p$.
Let us expand the vectors as $v  =  \sum_{i=1}^p    a_i \otimes e^{(p)}_i$ and $w  =  \sum_{j=1}^p    b_j \otimes e^{(p)}_j$, where $ \{e_i^{(p)}\}$
is the standard orthonormal basis in  $\mathbb{K}^p$, and $a_i  \in  \mathbb{K}^n$,  $b_j  \in  \mathbb{K}^m$ are arbitrary vectors.
With this expansion, we have
\begin{align}
 \nonumber  \<  w  |     (\Phi \otimes \operatorname{\textsl{id}}  _p)   (vv^*)  w\>      &  =   \sum_{i,j=1}^{p}   \<  b_i   |  \Phi   (a_i a_j^*  )   b_j\>
 =\sum_{i,j=1}^p   \operatorname{Tr}  (  \Phi  (  a_i  a_j^*)  b_j b_i^*)    \\
 \nonumber & = \sum_{i,j=1}^p   \operatorname{Tr} (  C_\Phi  \,  ( (  a_i  a_j^*)^t \otimes   b_j b_i^* ) ) \\
 \label{aaaa} &=   \sum_{i,j=1}^p       \<   \overline a_i \otimes b_i|    C_\Phi      (  \overline a_j  \otimes b_j)\>   =  \< V |C_\Phi  V\>\, ,
\end{align}
with $V:  = \sum_{i=1}^p     \overline a_i  \otimes b_i$.
Since the vectors $a_i$ and $b_i$ are arbitrary, $V$ is an arbitrary vector with Schmidt rank at most  $p$.
Summarizing,  the condition $\<  w  |  (\Phi \otimes \operatorname{\textsl{id}}  _p)   (vv^*)  w\>\ge 0$ implies  $\<   V |  C_\Phi  V\>  \ge 0 $ for every vector $V$ with Schmidt rank at most $p$.

Conversely, suppose that $C_\Phi$ is Hermitian and satisfies    $\<   V |  C_\Phi  V\>  \ge 0 $ for every vector $V$ with Schmidt rank at most $p$.
Since $C_\Phi$ and $\operatorname{\textsl{id}}_p$ are both Hermitian,  item 1 of this proof implies that  $ \Phi \otimes \operatorname{\textsl{id}}  _p$ is Hermitian.
Lemma \ref{lem:rankoneisenough} then  implies that $\Phi \otimes \operatorname{\textsl{id}}  _p$ is positive if   $ \<  w  |     (\Phi \otimes \operatorname{\textsl{id}}  _p)   (vv^*)  w\> \ge0$ for every $v$ and $w$.
But Eq. (\ref{aaaa})  guarantees that this condition follows from the condition  $\<   V |  C_\Phi  V\>  \ge 0 $ for every vector $V$ with Schmidt rank at most $p$.
This concludes the proof of (2).

To prove (3), note that, for every pair of matrices $A$ and $B$, one has
\begin{align*}
  \operatorname{Tr}  (    C_{\Phi^*}    (  B^t  \otimes A ))
  &=   \operatorname{Tr}  ( \Phi^*  (B)   A  )
  =  \operatorname{Tr}  (  B  \Phi  (A))   \\&
  =   \operatorname{Tr}(   C_\Phi   (   A^t \otimes B))
  =  \operatorname{Tr}  (     C_\Phi^t  \, S     (  B^t \otimes A) S^*  )\, ,
\end{align*}
where $S : \mathbb{K}^m \otimes \mathbb{K}^n \to \mathbb{K}^n \otimes \mathbb{K}^m $  is the linear operator defined by $S ( w\otimes v)  =  v\otimes w$ for all $v \in  \mathbb{K}^n$ and $w\in \mathbb{K}^m$.
Since $A$ and $B$ are arbitrary, we conclude
\begin{align}\label{choiadjoint}
 C_{\Phi^*} = S^* C_\Phi^t S .
\end{align}
Clearly, $C_{\Phi^*}$ is Hermitian  if and only if $C_\Phi$ is, and  $\<  W  |   C_{\Phi^*}  W\>  \ge 0 $   for every vector $W  \in  \mathbb{K}^m\otimes \mathbb{K}^n $
with Schmidt rank $\le p$ if and only if $\<  V  |   C_{\Phi}  V\>  \ge 0 $   for every vector $V  \in  \mathbb{K}^n\otimes \mathbb{K}^m $ with Schmidt rank $\le p$.
Hence, (2) implies that $\Phi^*$ is $p$-positive if and only if $\Phi$ is.
\end{proof}

\section{Real maps with $p$-positive complexification}\label{sec:poscomplex}

Given a linear map $\Phi: M_n(\mathbb{R}) \to M_m(\mathbb{R})$ its {\em complexification} is the  linear map
$\widetilde{\Phi}: M_n(\mathbb{C}) \to M_m(\mathbb{C}),$ defined by
\[
 \widetilde{\Phi}(X+iY) = \Phi(X) + i \Phi(Y) \quad\text{for all}\quad X, Y \in M_n(\mathbb{R}).
\]

We now characterize the set of  maps with a $p$-positive complexification.
The characterization uses  the correspondence between  matrices  $A =  (a_{ij}) \in  M_{n,m}  (\mathbb{K})$ and
vectors  in $\operatorname{Vec}  (A)  \in  \mathbb{K}^n\otimes \mathbb{K}^m$ given by
\begin{align*}
 \operatorname{Vec} (A)   : =   \sum_{i=1}^n \sum_{j=1}^m  \,  a_{ij}  \,  e^{(n)}_i\otimes e^{(m)}_j ,
\end{align*}
where, for $r \in  \{m,n\}$,   $\{e^{(r)}_i\}_{i=1}^r$  is the standard orthonormal basis for $\mathbb{K}^r$.

\begin{theo}\label{theo:ChoiCondition}
A map $\Phi  :  M_n (\mathbb{R})  \to  M_m  (\mathbb{R})$  has a $p$-positive complexification  if and only if its Choi matrix   $C_\Phi$ is Hermitian and satisfies the condition
$\<  \operatorname{Vec}  (X)  |  C_\Phi  \operatorname{Vec} (X)  \>    +  \<  \operatorname{Vec}  (Y)  |  C_\Phi  \operatorname{Vec} (Y)  \>  \ge 0$
for every pair of matrices $X  ,  Y  \in  M_{n,m} (\mathbb{R})$ such that $\operatorname{rank} (X+  i  Y)  \le p$.
\end{theo}

\begin{proof}
By Lemma \ref{lemma-pos},   $\widetilde \Phi$ is $p$-positive if and only if its Choi matrix $  C_{\widetilde \Phi}   =  C_\Phi$ is Hermitian
and satisfies the condition $\<   V  |  C_\Phi   V\> \ge0 $ for every vector $V  \in  \mathbb{C}^{n}\otimes \mathbb{C}^m$ with Schmidt rank at most $p$.
In the vector notation, this condition is equivalent to   $\<  \operatorname{Vec} (A)  |  C_\Phi  \operatorname{Vec}  (A)  \>  \ge 0 $ for every matrix
$A  \in  M_{n,m}  (\mathbb{C})$ with $\operatorname{rank}  (A) \le p$.
Writing $A   =  X  +  i  Y$, with $X,  Y  \in  M_{n,m }  (\mathbb{R})$, and using the fact that  $C_\Phi$ is a symmetric matrix, we obtain
\[
 \<  \operatorname{Vec} (A)  |  C_\Phi  \operatorname{Vec}  (A)  \>    =  \<  \operatorname{Vec} (X)  |  C_\Phi  \operatorname{Vec}  (X)  \>  +  \<  \operatorname{Vec} (Y)  |  C_\Phi  \operatorname{Vec}  (Y)  \> .  \qedhere
\]
\end{proof}

\begin{cor}\label{cor:2p}
If a  map  $\Phi: M_n(\mathbb{R}) \to M_{m}(\mathbb{R})$   is $2p$-positive, then its complexification is $p$-positive.
\end{cor}

\begin{proof}
Immediate from Theorem \ref{theo:ChoiCondition}, Lemma \ref{lemma-pos}, and the fact that, for $A  =  X+  i Y$,
$\operatorname{rank}  (X)  =  \operatorname{rank}  (A +  \overline A)   \le  2 \operatorname{rank} (A)$, and similarly for $\operatorname{rank}  (Y)$.
\end{proof}

Note that the corollary above establishes that if $\Phi$ is completely positive, then the complexification is also completely positive, a fact proved in Lemma 2.3 in \cite{BlecherTepsan}.
In particular, this corollary shows that  $2$-positivity on the real field is sufficient for positivity of the complexification.
In general, however, $2$-positivity is not necessary: for example, the maps   $\Lambda_q  :  M_n (\mathbb{R}) \to M_n(\mathbb{R})$ defined by
\begin{align*}
 \Lambda_q    (A)   =   \operatorname{Tr}(A)\,   I_n     -  q   A
\end{align*}
have $p$-positive complexification whenever $q  \le 1/p$ \cite{takasaki1983geometry}.
They are not $(p+1)$-positive (and therefore not $2p$ positive) for $q  > 1/(p+1)$, as one can verify from the relation
$\<  V_l |  C_{\Lambda_q}  V_l\>  =  1-  q l    $, with  $V_l  :=  \sum_{i=1}^{l}  \,  e_i^{(n)} \otimes e_i^{(n)}/\sqrt {l}$.
Similar counterexamples can be constructed using the spectral methods developed by Chru{\'s}ci{\'n}ski and Kossakowski in \cite{chruscinski2009spectral}.

\section{Examples of positive real maps with non-positive complexification}\label{sec:examples1}

In this section, we provide several examples of positive maps whose complexification is not.
The first deals with $p$-positivity for general $p$.
The others deal with $p=1$.

\begin{example}
We provide an example of a $(2p\!-\!1)$-positive real map such that its complefixication is not $p$-positive.
Define $\Gamma_q:M_{2p}(\mathbb{R}) \to M_{2p}(\mathbb{R})$ by
\begin{align*}
  \Gamma_q(A) =   \operatorname{Tr} ( A)   I_{2p}  -   q (     O_+  A  O_+  +  O_-  A  O_-  ) ,
\end{align*}
where $O_+   = \begin{spmatrix}   0  &  I_p  \\  I_p  &  0  \end{spmatrix}$ and $O_-   = \begin{spmatrix}    I_p  &  0  \\   0&   -I_p   \end{spmatrix} $.
The Choi matrix of this map is
\[
 C_{\Gamma_q}   =  I_{2p}    \otimes  I_{2p}   -   q  \big( \operatorname{Vec}  (O_+)   \operatorname{Vec}  (O_+)^t   +  \operatorname{Vec}  (O_-)   \operatorname{Vec}  (O_-)^t \big) .
\]
Note that the vectors  $\operatorname{Vec}  (O_+)$ and $\operatorname{Vec}  (O_-)$ are orthogonal and each has length $\sqrt{2p}$,
so that
\[
 \cl P = \frac1{2p} \big( \operatorname{Vec}  (O_+)   \operatorname{Vec}  (O_+)^t   +  \operatorname{Vec}  (O_-)   \operatorname{Vec}  (O_-)^t \big)
\]
is a projection.
Thus $C_{\Gamma_q}   =  I_{2p}    \otimes  I_{2p}   -   2pq \cl P$.
Moreover the unit vectors in $\operatorname{ran} \cl P = \operatorname{span}\{ O_+, O_- \}$ are $\{ \frac1{\sqrt{2p}} (\cos\theta\, O_+ + \sin\theta\, O_- ) : \theta\in [0, 2\pi] \}$.
Now $O_\theta := \cos\theta\, O_+ + \sin\theta\, O_- \in \cl O_{2n}$ is an orthogonal matrix for all $\theta$.

Given a unit vector $V = \operatorname{Vec}(A)$, so that $\operatorname{Tr} (A^tA) = 1$, one has
\begin{align*}
 \<  V   |  C_{\Gamma_q}  V\>  &= \|V\|^2 -  2pq \| \cl P V \|^2 \\&
 = 1  - 2p q  \max_{\substack{W = \cl P W\\ \|W\|=1}}   \<  V |  W\>^2 \\&
 = 1 - \sqrt{2p} q \max_{\theta \in [0,2\pi]} \operatorname{Tr}(A^tO_\theta) .
\end{align*}
If the unit vector $V$ has Schmid number at most $2p-1$, i.e., $\operatorname{rank}  (A)  \le 2p-1$, we obtain
\[
  \operatorname{Tr}(A^t O_\theta )  \le  \operatorname{Tr} |A|   \le  \sqrt{ 2p-1} .
\]
Hence, we obtain the inequality $\<  V   |  C_{\Gamma_q}  V \> \ge   1-  q  \sqrt{2p(2p-1)}$ for every vector $V$ with Schmidt rank $\le 2p-1$.
By Lemma \ref{lemma-pos}, we conclude that the map $\Phi$ is $(2p \!-\! 1)$-positive for every $\sqrt{ 2p(2p-1)}q \le 1$.

Take $q$ with $\frac1{2p} < q \le \frac 1{\sqrt{ 2p(2p-1)}}$.
Then $\Gamma_q$ is  $(2p \!-\! 1)$-positive, but its complexification $\widetilde {\Gamma_q}$ is not $p$-positive.
To see this, let
\[
 V = \frac1{2\sqrt{p}} \operatorname{Vec}(O_+ + i O_-) .
\]
Then $\operatorname{rank}(O_+ + i O_-) = p \operatorname{rank} \begin{spmatrix} i & 1 \\ 1 & -i \end{spmatrix} = p$.
Compute
\[
 \<  V |C_{\widetilde {\Gamma_q}}  |V\>   = \<  V |C_{\Gamma_q}  |V\> =  1 - 2pq < 0 .
\]
Therefore $\widetilde {\Gamma_q}$ is not $p$-positive by  Lemma \ref{lemma-pos} again.
\end{example}

In the rest of the section we focus on the $p=1$ case.

\begin{example}
Let us start with a simple example of positivity preserving real map with non-positive complexification.
Define $\Phi_s:M_n(\mathbb{R}) \to M_n(\mathbb{R})$ by
\begin{align*}
 \Phi_s(A) = A + s(A-A^t) .
\end{align*}
Then, $\Phi_s$ is a positivity preserving for all $s \in \mathbb{R}$.
However it is not Hermitian if $s \ne 0$.
Hence $\widetilde{\Phi}_s$ is not Hermitian, and thus is not positive.
Explicitly, if we set
\[
 P = E_{1,1}+E_{2,2} + i (E_{1,2} -  E_{2,1}) \ge 0 ,
\]
where $E_{i,j}$ are the standard matrix units, then
\[
 \widetilde{\Phi}_s(P) = E_{1,1} + E_{2,2} + (1+2s)i(E_{1,2}-E_{2,1}) ,
\]
which  has eigenvalue $-2s$, and therefore is not positive for $s>0$.
\end{example}

In the complex field, a result of   Russo and Dye  (Corollary 2.9 in \cite{paulsenbook})  states that a positive map
$\Phi$ on a unital C*-algebra satisfies the condition $\|\Phi\| = \| \Phi(I) \|$.
Hence, a necessary condition for positivity of the complexification  is that  $\Phi$  is positivity preserving, Hermitian,
and satisfies the Russo-Dye condition $\|  \Phi  \|   =  \|  \Phi  (I) \|$.
We now provide counterexamples, showing that the Russo-Dye condition is not sufficient for positive complexification.

\begin{example}
Here we provide an example of a Hermitian, positive, unital, norm one map on $M_2(\mathbb{R})$ whose complexification is not positive.
Define
\[
 \Phi(A) = \Phi \begin{pmatrix} a & b \\ c & d\end{pmatrix} =
 \begin{pmatrix} d & \frac{b-c}2 \\ \frac{c-b}2 & d\end{pmatrix} .
\]
This map is clearly unital and Hermitian.
It is positive since $\Phi \begin{spmatrix} a & b \\ b & d\end{spmatrix} = d I_2$.
Since the two columns of $\Phi(A)$ are orthogonal, we have
\[
 \| \Phi(A)\| = \left\|  \begin{pmatrix} d & \frac{b-c}2 \\ \frac{c-b}2 & d\end{pmatrix} \right\| = \sqrt{ d^2 + \frac{(b-c)^2}4 } .
\]
Now,
\begin{align*}
 \|A\| &\ge \frac12 \left\|  \begin{pmatrix} a & b \\c & d\end{pmatrix} + \begin{pmatrix} a & -c \\ -b & d\end{pmatrix}  \right\|
 = \left\|  \begin{pmatrix} a & \frac{b-c}2 \\ \frac{c-b}2 & d\end{pmatrix} \right\| \\&
 \ge \sqrt{ d^2 + \frac{(b-c)^2}4 } = \|\Phi(A)\|.
\end{align*}
Hence, $\|  \Phi  \|   =  1$.
Finally, the non-positivity of $\widetilde \Phi$ can be seen from the relation
$  \widetilde \Phi  \begin{spmatrix}         \lambda    &     i   \\   -i    &  1/\lambda   \end{spmatrix}
=   \begin{spmatrix}         1/\lambda    &     i   \\   -i    &  1/\lambda   \end{spmatrix}$ for   $\lambda  >  1$.
\end{example}

\begin{example}
Here we provide an example of a Hermitian, positive, trace-preserving  map on $M_2(\mathbb{R})$ that satisfies the Russo-Dye condition but has a non-positive complexification.
Define
\begin{align*}
 \Psi  (A)   &=  \Psi \begin{pmatrix}      a  &   b   \\   c    &  d      \end{pmatrix}
 :=    \begin{pmatrix}      0    &    \frac{b-c}2   \\   \frac{c-b} 2   &  a+d      \end{pmatrix} \\&
 = \frac12 \begin{pmatrix}      a  &   b   \\   c    &  d      \end{pmatrix}
 + \frac12 \begin{pmatrix}      d  &   -c   \\   -b    &  a      \end{pmatrix}
 + \begin{pmatrix}   -\frac{a+d}2  &   0   \\   0    &  \frac{a+d}2     \end{pmatrix} .
\end{align*}
This map is  clearly trace-preserving and Hermitian.
It is positive since $\Psi \begin{spmatrix} a & b \\ b & d\end{spmatrix} =    (a+d)\,  E_{22}  $.
Moreover, one has
\begin{align*}
  \|  \Psi (A)\|  &\le \frac{\|A\|}2 + \frac{\|A\|}2 + \frac{|\operatorname{Tr}(A)|}2  \le  2\|  A\| .
\end{align*}
Hence,  $\|  \Psi  \|    =  \|  \Psi  (I_2) \|  =  2$.
On the other hand,  $\widetilde \Psi$  is not  positive, as one can see from the relation
\[
  \widetilde \Psi   \begin{pmatrix}      1    &    i   \\   -i    &  1      \end{pmatrix}    =    \begin{pmatrix}      0    &    i   \\   -i    &  2      \end{pmatrix}    .
\]
\end{example}

\begin{example}
Finally, we provide an example of Hermitian, positive, trace-preserving and unital map on $M_3 (\mathbb{R})$ that
satisfies the Russo-Dye condition, but has a non-positive complexification.
Define the one-parameter family of maps for $t\in [0,1]$,
\begin{align*}
 \rho_t (A)   &
 = \rho_t   \begin{pmatrix}      a_{11}  &   a_{12}      & a_{13}    \\    a_{21}  &   a_{22}    & a_{23}   \\     a_{31}  &   a_{32}   &  a_{33}     \end{pmatrix}
 :=   \begin{pmatrix}
  \frac{  a_{11 }+  a_{22}  +  2  a_{33}  }4  &  \frac{  (a_{12}  -  a_{21})  t}{2 \sqrt 2}    &  0  \\
  \frac{  (a_{21}  -  a_{12})  t} {2 \sqrt 2}    &    \frac{  a_{11 }+  a_{22}  +  2  a_{33}  }4   &  0      \\
  0  &  0  &   \frac{  a_{11} +  a_{22} }2
 \end{pmatrix} .
\end{align*}
This map is  clearly unital, trace-preserving, positive, and Hermitian.

Set $x  = \frac{a_{11}  +  a_{22}}2$,   $y  = \frac{a_{12}  -  a_{21}}2$,  $z =  a_{33}$,  and  $r  =  \max  \{   \sqrt{x^2  +  y^2}, |z| \}$.
Compute
\begin{align*}
 \|   A  \|   &\ge  \left \| \begin{pmatrix} a_{11}  &  a_{12} & 0   \\  a_{21}  &  a_{22} & 0  \\  0  &  0  &  a_{33}  \end{pmatrix} \right\|  \\&
 \ge \max \Big\{ \left\| \frac12 \begin{pmatrix}  a_{11}  & a_{12} \\  a_{21}  &   a_{22}  \end{pmatrix}
 + \frac12 \begin{pmatrix}  a_{22}  & -a_{21} \\  -a_{12}  &   a_{11}  \end{pmatrix} \right\|, \ |z| \Big\} \\&
 = \max \Big\{ \left\| \begin{pmatrix}  x  & y \\  -y  &   x \end{pmatrix} \right\|, \ |z| \Big\}
=  \max  \left\{  \sqrt{ x^2   +  y^2 }  ,  |z| \right\}    = r   .
\end{align*}
The three columns of $\rho_t(A)$ are orthogonal. Hence
\begin{align*}
 \|  \rho_t  (A)\|  &= \left \| \begin{pmatrix} \frac{x+z}2  &  \frac{yt}{\sqrt2} & 0   \\  \frac{-yt}{\sqrt2}  &  \frac{x+z}2  & 0  \\  0  &  0  &  x  \end{pmatrix} \right\|
 \le  \max  \left\{  \sqrt{ \left(\frac{   x  +  z}2 \right)^2 +  \frac{ t^2\,  y^2}  2 }  ,  |x| \right\} \\&
 \le  \max  \left\{  \sqrt{ \left(\frac{   |x|  +  |z|}2 \right)^2 +  \frac{ y^2}  2 }  ,  r \right\} .
\end{align*}
Let $s = \frac{|x|}r$ and $c = \frac{|y|}r$ and note that $s^2+c^2 \le 1$.
Hence
\begin{align*}
 \Big(\frac{ |x| + |z|}2 \Big)^2 +  \frac{ y^2}  2  &\le \frac{r^2}4 \big( (s+1)^2 + 2c^2 \big)
 \le \frac{r^2}4 \big( s^2 + 2s + 1+ 2 - 2s^2 \big) \\&
 = \frac{r^2}4 \big( 4 - (1-s)^2 \big) \le r^2.
\end{align*}
Therefore
\[
 \|  \rho_t  (A)\| \le r \le \|A\|.
\]
Thus $\rho_t$ has norm 1 for every $t\in  [0,1]$.

On the other hand,  $\widetilde{ \rho_t}$  is not  positive for every $t  > 1/\sqrt 2$ because
\[
  \widetilde {\rho_t}   \begin{pmatrix}      1    &    i    &  0   \\   -i    &  1  &  0  \\   0& 0&  0      \end{pmatrix}
 = \frac 12 \begin{pmatrix}      1    &    i t \sqrt 2    &  0   \\   -i  t\sqrt 2    &  1  &  0  \\   0& 0&  0      \end{pmatrix}    .
\]
We   also observe that  $\widetilde{ \rho_t}$ is positive for $t \le \frac1{\sqrt2}$. Indeed, if $A = \big[ a_{ij} \big] \ge 0$,
\begin{align*}
  \widetilde{ \rho_t}(A) = \widetilde{ \rho_t} \begin{pmatrix}  a_{11}  & a_{12}  & 0 \\ \overline{a_{12}}  & a_{22} & 0 \\  0  &  0  &  a_{33}  \end{pmatrix}
 &=    \begin{pmatrix}  \frac{  a_{11 } +  a_{22}  +  2  a_{33}  }4  &  \frac{ i  {\operatorname{Im}}   (a_{12} )   t}{ \sqrt 2}    &  0  \\
 \frac{ -i  {\operatorname{Im}}   (a_{12} ) t}{ \sqrt 2}     &    \frac{  a_{11 }+  a_{22}  +  2  a_{33}  }4   &  0   \\  0  &  0  & \frac{  a_{11 }+  a_{22} }2  \end{pmatrix} .
\end{align*}
Since
\[
 \Big|  \frac{ i  {\operatorname{Im}}   (a_{12} )   \,  t}{ \sqrt 2}  \Big|^2 \le \frac{|a_{12}|^2}4 \le \frac{a_{11}a_{22} }4 \le \Big( \frac{a_{11} + a_{22}}4 \Big)^2 ,
\]
we obtain $\widetilde{ \rho_t}(A)  \ge 0$.

This example extends  to any dimension $n  >3$  by defining
\[
 \sigma_{n,t}  \begin{pmatrix}  A  &   B   \\   C   &    D   \end{pmatrix}   :  =     \begin{pmatrix}  \rho_t(A)   &   0  \\  0  &  D \end{pmatrix}
\]
for $A \in M_3 (\mathbb{R})$,  $B  \in  M_{2,n-3} (\mathbb{R})$,  $C\in  M_{n-3,2} (\mathbb{R})$ and $D  \in  M_{n-3,n-3}  (\mathbb{R})$.
The map $\sigma_{n,t}$ for $t \in(\frac1{\sqrt2},1]$ and $n>3$ is an example of a positive, unital, trace-preserving, norm-1 map on $M_n$ that has no positive complexification.
\end{example}

\section{Witnesses of real entanglement}\label{sec:witnesses}

We now show that the existence of positive real maps with non-positive complexification is equivalent to the existence of quantum states that are entangled in real Hilbert space quantum mechanics, and yet are separable when regarded as states on  complex Hilbert spaces.     The equivalence is established by proving a duality between the cone of real positive maps and the cone of separable states in real Hilbert space quantum mechanics.

For $\mathbb{K}  \in  \{\mathbb{R}, \mathbb{C}\}$,  the cone of separable matrices is

\[
 \operatorname{SEP}(\mathbb{K}^n\otimes \mathbb{K}^m)=
 \Big\{   \sum_{i} A_i\otimes B_i  :  A_i\in \operatorname{PSD}(\mathbb{K}^n), \,  B_i\in \operatorname{PSD}(\mathbb{K}^m) \Big\} .
\]
The elements of $\operatorname{SEP}(\mathbb{K}^n\otimes \mathbb{K}^m)$ will be called {\em $\mathbb{K}$-separable}, and the elements of
$\operatorname{PSD} (\mathbb{K}^n\otimes \mathbb{K}^m) \setminus   \operatorname{SEP}(\mathbb{K}^n\otimes \mathbb{K}^m)$ will be called {\em $\mathbb{K}$-entangled}.

More generally, we consider the cone of positive matrices  with Schmidt rank at most $p$ \cite{terhal2000schmidt}.
We call these matrices  {\em $\mathbb{K}$-$p$-separable}  and denote them by
\[
 \operatorname{SEP}_p(\mathbb{K}^n\otimes \mathbb{K}^m) =
 \Big\{ \sum_{j}  \operatorname{Vec}(A_j)  \operatorname{Vec}(A_j)^*  :  A_j  \in  M_{n,m}  (\mathbb{K}) \, , \operatorname{rank} ( A_j) \le p \Big\} .
\]
Note that $ \operatorname{SEP}_1(\mathbb{K}^n\otimes \mathbb{K}^m)  =   \operatorname{SEP}(\mathbb{K}^n\otimes \mathbb{K}^m)$.
The elements of   $\operatorname{PSD} (\mathbb{K}^n\otimes \mathbb{K}^m) \setminus   \operatorname{SEP}_p(\mathbb{K}^n\otimes \mathbb{K}^m)$ will be called {\em $\mathbb{K}$-$p$-entangled}.

We also define the cone consisting  of all real positive  matrices that are   $\mathbb{C}$-$p$-separable, denoted as
\begin{align*}
 \operatorname{CSEP}_p(\mathbb{R}^n \otimes \mathbb{R}^m)   &:= \operatorname{SEP}_p(\mathbb{C}^n \otimes \mathbb{C}^m) \cap \operatorname{PSD}(\mathbb{R}^n \otimes \mathbb{R}^m)   \\
   &  = \Big\{   \sum_{j}    \operatorname{Vec}(A_j)  \operatorname{Vec}(A_j)^* +  \operatorname{Vec}(\overline A_j)  \operatorname{Vec}(\overline A_j)^*   : \\[-2.0ex]
     &  \qquad   \qquad  A_j  \in  M_{n,m}  (\mathbb{C})  \text{ and } \operatorname{rank} ( A_j) \le p\,  \:\text{ for all }\:  j    \Big\} .
\end{align*}
For $p=1$, we use the notation  $ \operatorname{CSEP}(\mathbb{R}^n \otimes \mathbb{R}^m) :  =  \operatorname{CSEP}_1(\mathbb{R}^n \otimes \mathbb{R}^m) $.

We now provide an equivalent characterization of $ \operatorname{CSEP}_p(\mathbb{R}^n \otimes \mathbb{R}^m)$, which shows that every $\mathbb{C}$-$p$-separable real matrix is also $\mathbb{R}$-$2p$-separable:

\begin{prop}\label{prop:csepp}   One has
\begin{align*}
 \operatorname{CSEP}_p(\mathbb{R}^n \otimes \mathbb{R}^m)     &  = \Big\{   \sum_{j}    \operatorname{Vec}(X_j)  \operatorname{Vec}(X_j)^t +  \operatorname{Vec}(Y_j)  \operatorname{Vec}(Y_j)^t   : \\[-2ex]
     &  \qquad \qquad     X_j, Y_j    \in  M_{n,m}  (\mathbb{R})    , \,  \operatorname{rank} ( X_j+  i  Y_j  ) \le p\, , \forall j   \Big \}
\end{align*}
and $\operatorname{CSEP}_p(\mathbb{R}^n \otimes \mathbb{R}^m) \subset \operatorname{SEP}_{2p}(\mathbb{R}^n \otimes \mathbb{R}^m)$.
\end{prop}

\begin{proof}
The characterization follows from decomposing  $A_j$ as $A_j  =  X_j   +  i  Y_j$,  with $X_j,  Y_j  \in  M_{m,n}  (\mathbb{R})$,  and from the relation
\[
 \operatorname{Vec}(A_j)  \operatorname{Vec}(A_j)^* +  \operatorname{Vec}(\overline A_j)  \operatorname{Vec}(\overline A_j)^* =   \operatorname{Vec}(X_j)  \operatorname{Vec}(X_j)^t +  \operatorname{Vec}(Y_j)  \operatorname{Vec}(Y_j)^t   .
\]
The inclusion equality follows from the first and from the fact that
$\operatorname{rank}  (A_j) \le p$ implies  $\operatorname{rank}  (X)\le 2p$ and $\operatorname{rank}(Y) \le 2p$.
\end{proof}

In the next section we will provide an example showing that the inclusion $\operatorname{CSEP}_p(\mathbb{R}^n \otimes \mathbb{R}^m) \subset \operatorname{SEP}_{2p}(\mathbb{R}^n \otimes \mathbb{R}^m)$ is strict.

\begin{rem}
Note that the cone  $\operatorname{SEP}_p(\mathbb{K}^n \otimes \mathbb{K}^m)$ is closed.
To see this it is enough to show that the convex set consisting of the elements in these cones of trace 1 is a closed convex set.
But the elements in $\operatorname{SEP}_p(\mathbb{K}^n \otimes \mathbb{K}^m)$ of trace one is the convex hull of the set of  the matrices of the form $A =  V V^*$ where  $V$ is a unit vector with Schmidt rank $\le p$.
The set of such matrices is compact  and, in finite dimensions,
the convex hull of a compact set is again a compact set by Caratheodory's Theorem.
The cone $\operatorname{CSEP}_p(\mathbb{R}^n \otimes \mathbb{R}^m)$ is also closed since it is the intersection of two closed sets.
\end{rem}

The complex version of part (1) of the next theorem was proven in \cite{skowronek2009cones}.

\begin{thm}\label{dualcones}
Let $\Phi : M_n(\mathbb{R}) \to M_m(\mathbb{R})$ be a Hermitian  map and let $C_{\Phi}$ be its Choi matrix. Then,
\begin{enumerate}
\item $\Phi$ is $p$-positive if and only if
\[
 \operatorname{Tr} (C_{\Phi} \,  P) \ge 0 \quad\text{for all}\quad P \in \operatorname{SEP}_p(\mathbb{K}^n \otimes \mathbb{K}^m),
\]

\item The complexification $\widetilde{\Phi}:M_n(\mathbb{C}) \to M_m(\mathbb{C})$ is  $p$-positive if and only if
\[
 \operatorname{Tr} (C_{\Phi} \,P) \ge 0 \quad\text{for all}\quad P \in \operatorname{CSEP}_p(\mathbb{R}^n \otimes \mathbb{R}^m).
\]
\end{enumerate}
\end{thm}

\begin{proof}
The set $\operatorname{SEP}_p(\mathbb{K}^n \otimes \mathbb{K}^m)$ is the convex hull of its rank-one  elements, of the form $ \operatorname{Vec} (A) \operatorname{Vec}  (A)^*$ with $\operatorname{rank}  (A) \le p$.
The condition   $\operatorname{Tr} (C_{\Phi}P) \ge 0 \quad\text{for all}\quad P\in \operatorname{SEP}_p(\mathbb{K}^n \otimes \mathbb{K}^m)$ is equivalent to
\[
 0  \le    \operatorname{Tr} ( C_\Phi  \,   \operatorname{Vec} (A) \operatorname{Vec}  (A)^*  )  = \<  \operatorname{Vec} (A)|  C_\Phi   \operatorname{Vec}  (A)\>
\]
for all matrices with rank at most $p$.
In turn, this condition is equivalent to $\<  V  |  C_\Phi  V\>  \ge0$ for every vector $V$ with Schmidt number at most $p$.
By Lemma \ref{lemma-pos}, this condition is equivalent to $p$-positivity of $\Phi$.  This proves (1).

We now  prove (2).
By Proposition \ref{prop:csepp},    the condition   $\operatorname{Tr} (C_{\Phi}  P) \ge 0$ for all  $P \in \operatorname{CSEP}_p(\mathbb{R}^n \otimes \mathbb{R}^m)$ is equivalent to
\begin{align*}
 0    &\le    \operatorname{Tr}   ( C_\Phi  \,     (  \operatorname{Vec} (X) \operatorname{Vec}  (X)^t  +  \operatorname{Vec} (Y) \operatorname{Vec}  (Y)^t   )  )    \\
   &   = \<  \operatorname{Vec} (X)|  C_\Phi   \operatorname{Vec}  (X)\>  +   \<  \operatorname{Vec} (Y)|  C_\Phi   \operatorname{Vec}  (Y)\> \, ,
\end{align*}
for all real matrices  $X,Y$  such that $\operatorname{rank} (X+  i Y)\le p$.
By Theorem \ref{theo:ChoiCondition}, this condition is equivalent to $p$-positivity of $\widetilde \Phi$.
\end{proof}

\begin{cor}\label{cor:duality}
Let $P\in \operatorname{PSD}(\mathbb{K}^n \otimes \mathbb{K}^m)$. Then:
\begin{enumerate}
\item $P$ is $\mathbb{K}$-$p$-separable if and only if $\operatorname{Tr} (C_\Phi P)\geq 0$, for all $p$-positive $\Phi:M_n(\mathbb{K})\rightarrow M_m(\mathbb{K}),$
\item for $\mathbb{K}  = \mathbb{R}$,  $P$ is $\mathbb{C}$-$p$-separable if and only if $\operatorname{Tr} (C_{\Phi}P) \geq 0$ for all $\Phi: M_n(\mathbb{R}) \to M_m(\mathbb{R})$ such that $\widetilde{\Phi}$ is $p$-positive.
\end{enumerate}
\end{cor}

\begin{proof}
Use the fact that these cones are closed and apply the standard Hahn-Banach results for separating a point from a closed cone.
\end{proof}

This corollary implies that, if a real positive matrix $P$  is $\mathbb{R}$-$p$-entangled, then there must exist
a $p$-positive map $\Phi$ such that $\operatorname{Tr} (C_\Phi P)<0$.
We say that $\Phi$  is a \emph{witness of  $\mathbb{R}$-$p$-entanglement} for $P$.
Similarly, if a real positive matrix $P$ is $\mathbb{C}$-$p$-entangled, then there must exist a positive map $\Psi$ with positive complexification  such that $\operatorname{Tr} (C_\Psi P)<0$.
We say that $\Psi$ is  a {\em witness of $\mathbb{C}$-$p$-entanglement} for  $P$.
In the literature, the term entanglement witness was typically used for $p=1$, while  witnesses for $p>1$ were sometimes called Schmidt number witnesses  \cites{terhal2000bell,sanpera2001schmidt}.

Before proceeding to the next result we need a lemma for which we refer to \cite{CMW}. The assertion holds for any field.
\begin{lem}\label{schmidt-dec}
Every vector $V\in  \mathbb{K}^n\otimes \mathbb{K}^m$  with Schmidt rank $\le p$ can be written as $$V =   (I_n  \otimes S) W=(T\otimes I_m)Z ,$$ where $W$ is a vector in $  \mathbb{K}^n\otimes \mathbb{K}^p$, $Z$ is a vector in $\mathbb{K}^p\otimes \mathbb{K}^m$ with  $S:  \mathbb{K}^p  \to  \mathbb{K}^m$  and $T:  \mathbb{K}^p  \to  \mathbb{K}^n$ are partial isometries.
\end{lem}

The fact that $p$-positive maps are witnesses of $p$-entanglement is equivalently expressed by the following corollary,  the complex version of which was established  in \cite{Horodeckis} for  $p=1$ and in \cite{terhal2000schmidt} for general $p$.

\begin{cor}\label{horodeckis}
For a matrix $P\in \operatorname{PSD}(\mathbb{K}^n \otimes \mathbb{K}^m)$, the following are equivalent
\begin{enumerate}
\item $P$ is $\mathbb{K}$-$p$-separable,
\item $\Phi \otimes \operatorname{\textsl{id}} _r(P) \geq 0$ for every   $r\in  \mathbb{N}$ and for every $p$-positive map $\Phi: M_n(\mathbb{K}) \to M_r(\mathbb{K})$,
\item $\Phi \otimes \operatorname{\textsl{id}} _m(P) \geq 0$ for every $p$-positive map $\Phi: M_n(\mathbb{K}) \to M_m(\mathbb{K})$.
\item $\operatorname{\textsl{id}}_n \otimes \Phi(P)\geq 0$ for every $p$-positive map $\Phi:M_m(\mathbb{K})\rightarrow M_l(\mathbb{K})$
\end{enumerate}
\end{cor}

\begin{proof}

$(1)  \Rightarrow (2)$.  If $P$ is $\mathbb{K}$-$p$-separable, it can be written as $P  =  \sum_i   V_i  V_i^*$, where each $V_i$ is a vector in $\mathbb{K}^n\otimes \mathbb{K}^m$ with Schmidt rank at most $p$.    Using lemma \ref{schmidt-dec} we know that every vector $V_i \in  \mathbb{K}^n\otimes \mathbb{K}^m$  with Schmidt rank $\le p$ can be written as $V_i  =   (I_n  \otimes S_i ) W_i$, where $W_i$ is a vector in $  \mathbb{K}^n\otimes \mathbb{K}^p$ and  $S_i:  \mathbb{K}^p  \to  \mathbb{K}^m$ is a partial isometry.
Then, for every map $\Phi: M_n(\mathbb{K}) \to M_r(\mathbb{K})$, we have
\[
 \Phi \otimes \operatorname{\textsl{id}} _m(P) =  \sum_i   \,   (I_r\otimes   S_i)        ~(\Phi \otimes \operatorname{\textsl{id}} _p(   W_i W_i^*  )) ~  (I_r\otimes   S_i^*)\ge 0 .
\]
If $\Phi$ is $p$-positive, then each term in this sum is positive; so $\Phi \otimes \operatorname{\textsl{id}} _m(P) \ge 0$.

$(2)  \Rightarrow (3)$.  Take $ r  = m$.

$(3)  \Rightarrow (1)$.  Given a map $\Phi$, the condition   $\Phi \otimes \operatorname{\textsl{id}} _m(P) \geq 0$ implies
\begin{align*}
 0  &\le     \sum_{i,j}   \operatorname{Tr}  ( ( E_{i,j}  \otimes E_{i,j})    \Phi \otimes \operatorname{\textsl{id}}_m  (P) )
 =     \sum_{i,j}   \operatorname{Tr}  (( \Phi^*( E_{i,j})  \otimes E_{i,j}) \, P ) \\
 & =     \operatorname{Tr}  ( S C_{\Phi^*}  S^* P )   =  \operatorname{Tr}  ( C_{\Phi^*}  S^* P  S) ,
\end{align*}
where $S : \mathbb{K}^m \otimes \mathbb{K}^n \to \mathbb{K}^m \otimes \mathbb{K}^n$ is the linear operator defined by $S( w\otimes v) =  v\otimes w$ for all  $v\in  \mathbb{K}^n$ for all $w\in  \mathbb{K}^m$.
If $\Phi:  M_n  (\mathbb{K}) \to  M_m(\mathbb{K})$ is an arbitrary $p$-positive map, then $\Phi^*:  M_m  (\mathbb{K}) \to  M_n(\mathbb{K})$ is an arbitrary $p$-positive map by Lemma \ref{lemma-pos}.
For $\mathbb{K}  = \mathbb{R}$,  Corollary \ref{cor:duality} implies that $S^*PS$, and therefore $P$, is $\mathbb{R}$-$p$-separable.

The equivalence of $(1)$ and $(4)$ follows exactly like the equivalence of $(1)$ and $(3)$ using the second equality of Lemma \ref{schmidt-dec}.
\end{proof}

\begin{cor}
For every $p\in  \mathbb{N}$, the following are equivalent
\begin{enumerate}
\item  there exists a $p$-positive map $\Phi:  M_n  (\mathbb{R}) \to M_m (\mathbb{R})$ with non-$p$-positive complexification,
\item there exists an $\mathbb{C}$-$p$-separable matrix  $P   \in  \operatorname{PSD} (  \mathbb{R}^n\otimes \mathbb{R}^m)$ that is not $\mathbb{R}$-$p$-separable.
\end{enumerate}
\end{cor}

\begin{proof}
Immediate from Corollary \ref{cor:duality}.
\end{proof}

\section{Examples of $\mathbb{R}$-entangled states that are $\mathbb{C}$-separable}\label{sec:examples2}

In this section, we use the distinctions between the various cones of positive maps to provide explicit examples.
The first example is an $\mathbb{R}$-$(2p-1)$-entangled but $\mathbb{C}$-$p$-separable matrix.
The remaining examples focus on the $p=1$ case, providing examples of quantum states that are entangled in
real Hilbert space quantum mechanics, but separable in the complex version.
The key observation is that real separable states must be invariant under partial transposition (IPT).

\begin{example}\label{ex:pentangled}
Here we provide an example of a matrix $P  \in   \operatorname{PSD} ( \mathbb{R}^{2p} \otimes \mathbb{R}^{2p})$  that is $\mathbb{R}$-$(2p-1)$-entangled, but $\mathbb{C}$-$p$-separable.  The matrix is
\[
 P  =    \frac{V V^*   +   \overline V  V^t}2 ,
\]
where  $V  \in   \mathbb{C}^{2p}  \otimes  \mathbb{C}^{2p} $ is the complex vector    $V   =   \sum_{j=1}^p   \alpha_j \otimes \alpha_j/\sqrt p$,
with $\alpha_j   =  (  e_j^{(2p)}  +i  \, e_{j+p}^{(2p)})/\sqrt 2$.
It is evident from the definition  that $P$ is $\mathbb{C}$-$p$-separable.    On the other hand,  $P$ can be equivalently rewritten as
\[
 P  =  \frac{ \operatorname{Vec}(  O_+)   \operatorname{Vec}(  O_+)^t  +   \operatorname{Vec}(O_-)   \operatorname{Vec}(  O_-)^t}{4p}  ,
\]
with $O_+   = \begin{spmatrix}   0  &  I_p  \\  I_p  &  0  \end{spmatrix}$ and $O_-   = \begin{spmatrix}    I_p  &  0  \\   0&   -I_p   \end{spmatrix} $.
If $P$ were $\mathbb{R}$-$(2p-1)$-separable, there should exist at least one real vector with Schmidt rank $\le 2p-1$  in the linear span of $\operatorname{Vec}(O_+)$ and $\operatorname{Vec} (O_-)$.
But this is not possible, because the real span of $O_+$ and $O_-$ contains only full-rank matrices.
\end{example}

\begin{lemma}
Every  $P  \in   \operatorname{SEP}  (\mathbb{R}^n\otimes \mathbb{R}^m)$ must satisfy the condition $P   =     \tau_n \otimes \operatorname{\textsl{id}}_m  (P)  =  \operatorname{\textsl{id}}_n \otimes \tau_m  (P)$, where  $\tau_r:   M_r  (\mathbb{R})  \to  M_r  (\mathbb{R})  ,  \,    A \mapsto  A^t $ is the transpose map on $M_r  (\mathbb{R})$.
\end{lemma}

\begin{proof}
By definition every  $P  \in   \operatorname{SEP}  (\mathbb{R}^n\otimes \mathbb{R}^m)$  can be decomposed as $P = \sum_i A_i\otimes B_i $ for some $A_i \in \operatorname{PSD}(\mathbb{R}^n)$ and  $B_i\in \operatorname{PSD}(\mathbb{R}^m)$.
Applying the transpose map to the first  factor, one gets
\[
  \tau_n \otimes \operatorname{\textsl{id}}_m(P) = \sum_i A_i^t\otimes B_i = \sum A_i\otimes B_i = X  ,
\]
and similarly when the transpose is applied to the second factor.
\end{proof}

Invariance under partial transpose is a necessary condition for separability of real quantum states, similarly to the positive partial transpose (PPT)  criterion for separability in  the complex case \cites{peres1996separability,Horodeckis,horodecki1997separability}.  Based on this analogy, we call the invariance under partial transpose the IPT criterion for separability of real quantum states.

\begin{prop} \label{P:strict inclusion}
For $n,m \ge 2$, the inclusions
\[
 \operatorname{SEP}(\mathbb{R}^n \otimes \mathbb{R}^m) \subsetneq \operatorname{CSEP}(\mathbb{R}^n \otimes \mathbb{R}^m) \subsetneq \operatorname{PSD}(\mathbb{R}^n \otimes \mathbb{R}^m),
\]
are both strict.
\end{prop}

\begin{proof}
Strictness of the second inclusion is trivial: for example, the matrix
$\sum_{i,j=1}^{min(n,m)} E_{i,j} \otimes E_{i,j}$
is in  $ \operatorname{PSD}(\mathbb{R}^n \otimes \mathbb{R}^m)$, but it is not $\mathbb{C}$-separable, and therefore is not in $ \operatorname{CSEP}(\mathbb{R}^n \otimes \mathbb{R}^m)$.

To prove strictness of the first inclusion, we show that there exist elements of   $\operatorname{CSEP}(\mathbb{R}^n \otimes \mathbb{R}^m)$  that are not invariant under  partial transpose.    For example, take the matrix
$A=\begin{pmatrix}
0 & -i\\
i & 0
\end{pmatrix}$
and consider the Hermitian operator
\[
 P = I_2\otimes I_2+ A\otimes A
 = \left(   \begin{array}{cc|cc}  1 & 0 & 0 & -1\\  0 & 1 & 1 & 0\\ \hline 0 & 1 & 1 & 0\\ -1 & 0 & 0 &1  \end{array} \right)
  \in \operatorname{PSD}(\mathbb{R}^2\otimes \mathbb{R}^2)
\]
Since $I_2 \pm A \ge 0$ and
\[
 2P = (I_2 +A)\otimes (I_2+A) + (I_2-A) \otimes (I_2-A) ,
\]
$P$ belongs to $\operatorname{CSEP}(\mathbb{R}^n \otimes \mathbb{R}^m)$.
However, $\operatorname{\textsl{id}} \otimes T(P)\neq P$ and hence $P\not \in  \operatorname{SEP}(\mathbb{R}^n \otimes \mathbb{R}^m)$.

This takes care of the case $n=m=2$. For the general case, one can  form matrices $A_n= A \oplus 0_{n-2} \in M_n(\mathbb{C})$ and
$A_m= A \oplus 0_{m-2} \in M_m(\mathbb{C})$ and observe that the same argument shows that
$I_n \otimes I_m + A_n \otimes A_m$ is $\mathbb{R}\mathbb{C}$-separable but not $\mathbb{R}$-separable.
\end{proof}

Other examples of $\mathbb{C}$-separable but  are not equal to their partial transpose (and therefore are $\mathbb{R}$-entangled) are provided in the following.

\begin{example}
Consider the real positive matrix
\begin{align*}
 P_s &=  \frac{s}{n(n+1)}(I_n\otimes I_n+W) + \frac{1-s}{n(n-1)}(I_n\otimes 1_n-W) \\
 &= \frac{n\!+\!1 - 2s}{n^3-n} \Big( I_n\otimes I_n + \big(2ns - (n\!+\!1)\big) W \Big) \quad\text{for}\quad  s\in  [0,1] \, ,
\end{align*}
where $W=\sum_{i,j}E_{i,j}\otimes E_{j,i}\in M_n(\mathbb{C})\otimes M_n(\mathbb{C})$.
It is known \cite{werner-89} (see also chapter 6 of \cite{watrous}) that  $P_s$ is $\mathbb{C}$-separable for $s  \in  [\frac{1}{2},  1]$.
On the other hand,  $P_s$ is not equal to its partial transpose unless $s=  \frac{n+1}{2n}$,
and therefore $P_s$  is  $\mathbb{R}$-entangled for $s$ in $\big[ \frac 12, 1 \big] \setminus \big\{ \frac{n+1}{2n} \big\}$. \\
\end{example}

The above example can be generalized as follows:

\begin{prop}
For every  symmetric matrix $A\in   M_n(\mathbb{R})\otimes M_m(R)$ with $\operatorname{\textsl{id}} \otimes T(A)\neq A$, there exists an  $s_* >  0$ such that  the matrix
\begin{align*
 P_s =(1-s) I_n \otimes I_m + s\,  A
\end{align*}
is $\mathbb{R}$-entangled and $\mathbb{C}$-separable for every $s\in  (0,s_*]$.
\end{prop}

\begin{proof}
A  result of Gurvits and Barnum \cite{gur-bar}  states that, for every Hermitian matrix $H\in \operatorname{Herm}(\mathbb{C}^n\otimes \mathbb{C}^m)$ with  $\|H\|_2\leq 1$,
the matrix   $I_n\otimes I_m +H$ is $\mathbb{C}$-separable.
If $ A$ is a real symmetric matrix, then this result implies that the matrix $I_n\otimes I_m + \frac{s}{1-s} A$ is $\mathbb{C}$-separable whenever $s/(1-s)  \le \|  A\|_2$.
Therefore the matrix $P_s$ is $\mathbb{C}$-separable whenever $s \le  s_*:  =\|A \|_2/(\|  A\|_2  + 1)$.
On the other hand,  $\operatorname{\textsl{id}} \otimes T(P_s)\neq P_s$ for every $s>0$, and therefore $P_s$ is $\mathbb{R}$-entangled for every $s>0$.
\end{proof}

\section{Entanglement $p$-breaking maps}\label{sec:breaking}

In \cite{HSR},  Horodecki, Shor, and Ruskai  introduced the notion of {\em entanglement breaking} maps,
that is, maps that transform every entangled state into a separable state when acting on one of the components.
In general, we say that a  map $\Phi: M_n(\mathbb{K})\rightarrow M_m(\mathbb{K})$ is  \emph{$\mathbb{K}$-entanglement $p$-breaking} if
\begin{align*}
  \Phi\otimes \operatorname{\textsl{id}} _r(P) \in \operatorname{SEP}_p(\mathbb{K}^m\otimes \mathbb{K}^r) \quad\text{for all}\quad P \in \operatorname{PSD}(\mathbb{K}^n\otimes \mathbb{K}^r) \text{ and }  r \ge1.
\end{align*}
For $p=1$, we just call the map $\Phi$ $\mathbb{K}$-entanglement breaking.
In the complex case, entanglement $p$-breaking maps had been considered in \cite{skowronek2009cones}, where they were called $p$-superpositive maps.

The cone of $\mathbb{K}$-entanglement breaking ($\mathbb{K}$-entanglement $p$-breaking) maps $\Phi: M_n(\mathbb{K})\rightarrow M_m(\mathbb{K})$  will be denoted  as  $\mathbb{K}\text{-}\operatorname{EB} (n,m)$  ($\mathbb{K}\text{-}\operatorname{EB}_p  (m,n)$).  We now provide a characterization of the $\mathbb{K}$-entanglement $p$-breaking channels.
Note that since the cone $\operatorname{SEP}_p(\mathbb{K}^m\otimes \mathbb{K}^r)$ is larger for larger $p$, meaning that $\mathbb{K}$-entanglement $p$-breaking is a weaker condition than $\mathbb{K}$-entanglement breaking when $p\ge2$.

In the following result, the complex versions of (1)-(3) and  (4) were provided for $p=1$ in \cite{HSR} and \cite{JKPP}, respectively.
The complex version of (5) and (6) was provided in \cite{skowronek2009cones} for general $p$.

\begin{theo}\label{Rentanglementbreaking characterization}
Let $\Phi: M_n(\mathbb{K}) \to M_m(\mathbb{K})$ be a linear map. Then the following are equivalent.
\begin{enumerate}
 \item $\Phi$ is $\mathbb{K}$-entanglement $p$-breaking.
 \item $C_\Phi \in \operatorname{SEP}_p(\mathbb{K}^n \otimes \mathbb{K}^m)$,
 \item there exist matrices $(C_i)_{i=1}^k  \subset  M_{m,n}(\mathbb{K})$ such that  $\operatorname{rank}  (C_i) \le p$ for every $i$ and
 \[
  \Phi(X) = \sum_{i=1}^k C_i X  C_i^*,
 \]
 \item $\Phi = \Delta \circ \Gamma$ where $\Gamma: M_n(\mathbb{K}) \to l^\infty_k(\mathbb{K})  \otimes M_p  (\mathbb{K})$ and $\Delta : l^\infty_k(\mathbb{K}) \otimes M_p  (\mathbb{K})\to M_m(\mathbb{K})$ are completely positive maps for some $k\ge1$.
 \item for every $r$ and every $p$-positive map $\Psi: M_m(\mathbb{K}) \to M_r(\mathbb{K})$ the map $\Psi \circ \Phi$ is completely positive,
 \item for every $r$ and every $p$-positive map $\Psi: M_r(\mathbb{R}) \to M_n(\mathbb{R})$ the map $\Phi \circ \Psi$ is completely positive.
\end{enumerate}
\end{theo}

\begin{proof}
$(1) \Rightarrow (2)$: Applying $\operatorname{\textsl{id}}_n \otimes \Phi$ to $X=  \sum_{i,j}  E_{ij}  \otimes E_{ij}$, we obtain  that
$C_\Phi = (\operatorname{\textsl{id}}_n \otimes \Phi) \big( \sum_{ij} E_{ij}\otimes E_{ij} \big)$ belongs to  $ \operatorname{SEP}_p(\mathbb{K}^n \otimes \mathbb{K}^m)$.

$(2) \Rightarrow (3)$: If  $C_\Phi  \in  \operatorname{SEP}_p(\mathbb{K}^n \otimes \mathbb{K}^m)$, then it can be written as
$C_\Phi   =  \sum_{i=1}^k     \operatorname{Vec}(A_i)  \operatorname{Vec}( A_i)^*$, with $\operatorname{rank} (A_i)  \le p$ for every $i$.
Using Eq. (\ref{choiinverse}), we obtain
\begin{align*}
  \operatorname{Tr}  [\Phi  (    X )  \, Y]    &= \operatorname{Tr}  (   C_\Phi   (X^t \otimes  Y))     \\&
  =    \sum_{i=1}^k    \<  \operatorname{Vec}(A_i) ,   (X^t\otimes Y)  \operatorname{Vec}( A_i)\>  \\&
  =     \sum_{i=1}^k   \operatorname{Tr}  (A_i^t    X \overline A_i   Y)\, ,
\end{align*}
for arbitrary matrices $X$ and $Y$.  Since  $Y$ is arbitrary, we conclude $\Phi  (X)   =  \sum_{i=1}^k  C_i  X  C_i^*$ with $C_i  : =  A_i^t$.

$(3) \Rightarrow (1)$:     If   $\Phi  (X)   =  \sum_{i=1}^k  C_i  X  C_i^*$ with $\operatorname{rank}  (C_i) \le p$, then for every $A  \in   M_{n,m}(\mathbb{K})$, one has
\[
 (\Phi \otimes \operatorname{\textsl{id}}_n)   (\operatorname{Vec} (A)  \operatorname{Vec}(A)^*) =  \sum_i  \operatorname{Vec}  (C_i  A)   \operatorname{Vec} ( C_i  A)^* .
\]
Since $\operatorname{rank}  (C_i  A)  \le \operatorname{rank}  (C_i) \le p$, we deduce that  $(\Phi \otimes \operatorname{\textsl{id}}_n)   (\operatorname{Vec} (A)  \operatorname{Vec}(A)^*)$   is $\mathbb{K}$-$p$-separable.
Since $A$ is arbitrary,  we conclude that  $(\Phi \otimes \operatorname{\textsl{id}}_n)   V V^* $   is $\mathbb{K}$-$p$-separable for every vector $V$,
and therefore $ (\Phi \otimes \operatorname{\textsl{id}}_n) (X)$ is   $\mathbb{K}$-$p$-separable for every positive $X$.
This proves the equivalence of (1), (2), and (3).

$(3) \Rightarrow (4)$:   Since $\operatorname{rank}  (C_i)\le p$, there exists an isometry  $S_i:   \operatorname{ran}  (C_i) \to   \mathbb{K}^p$.
 Let $\delta_1,\dots,\delta_k$ be the standard basis for $l^\infty_k(\mathbb{K})$. Define
\[
 \Gamma(X) = \sum_{i=1}^k  \delta_i  \otimes  S_i C_i X  C_i^*  S_i^*  \quad\text{and}\quad \Delta\big( \sum_{i=1}^k   \delta_i \otimes Y_i\big) = \sum_{i=1}^k   S_i^* Y_i  S_i .
\]
The form of $\Gamma$ and $\Delta$ guarantees  that they are completely positive. By construction, $\Phi = \Delta \circ \Gamma$.

$(4) \Rightarrow (5)$:  The action of  $\Delta$  on a generic element of $l^\infty_k  (\mathbb{K})\otimes M_p$ can be expressed as
$\Delta  ( \sum_i  \delta_i \otimes  Y_i )  =  \sum_i       \Delta_i   (Y_i)$, with $\Delta_i:  M_p   (\mathbb{K}) \to M_m(\mathbb{K})$ completely positive.
Since the domain of  $\Delta_i$   is $M_p  (\mathbb{K})$,  $\Delta_i$ is $\mathbb{K}$-entanglement $p$-breaking.
Hence $\Delta$ is also $\mathbb{K}$-entanglement $p$-breaking.
Using the implication  $(1) \Rightarrow (2)$, we conclude that each $C_{\Delta_i}$ is $\mathbb{K}$-$p$-separable.
For an arbitrary $p$-positive map  $\Psi :  M_m(\mathbb{K}) \to M_r  (\mathbb{K})$,  $(\Psi\circ \Delta_i)$ is completely positive.
Indeed,    $C_{  \Psi\circ\Delta_i}    =  (\operatorname{\textsl{id}}_p  \otimes \Psi)   C_{\Delta_i}$, and Corollary  \ref{horodeckis} implies that   $(\operatorname{\textsl{id}}_p  \otimes \Psi)   C_{\Delta_i}$ is positive whenever $C_{\Delta_i}$ is $\mathbb{K}$-$p$-separable.
We conclude that  $\Psi\circ\Phi$ is completely positive, as it factors as $\Psi\circ\Phi = (\Psi\circ \Delta)\circ\Gamma$, which is the composition of two completely positive maps.

$(5) \Rightarrow (6)$:   If  $\Psi \circ \Phi$ is completely positive, then  $C_{\Psi\circ \Phi}    =   ( \operatorname{\textsl{id}}_m\otimes \Psi)  (C_\Phi)\ge 0$.
If $\Psi$ is an arbitrary $p$-positive map, this implies that $C_\Phi$ is $\mathbb{K}$-$p$-separable, by Corollary \ref{horodeckis}.
Hence, also $C_{\Phi^*}$ is $\mathbb{K}$-$p$-separable.
Using the implication  $(2)\Rightarrow (1)$, we obtain that $\Phi^*$ is $\mathbb{K}$-entanglement $p$-breaking.
Using the implication  $(1)\Rightarrow (5)$,  we conclude that  $   \Psi  \circ \Phi^*$ is completely positive for every $p$-positive $\Psi$.
Now,  a generic map $\Gamma$  is completely positive if and only if its adjoint  $\Gamma^*$ is completely positive  (by Lemma \ref{lemma-pos}).
Hence, the map $\Phi  \circ \Psi^*$ is completely positive for every $p$-positive $\Psi$.
But for an arbitrary $p$-positive map $\Psi$, $\Psi^*$ is an arbitrary $p$-positive map (by Lemma \ref{lemma-pos})).
Hence,  $\Phi  \circ \Psi$ is completely positive for every $p$-positive $\Psi$.

$(6) \Rightarrow (2)$:   For every map $\Psi$, the map $\Phi \circ \Psi$  is completely positive if and only if its adjoint $\Psi^*  \circ \Phi^*$ is completely positive.
If $\Psi$ is an arbitrary $p$-positive map, then $\Psi^*$ is an arbitrary $p$-positive map (by Lemma \ref{lemma-pos}).
Therefore  $C_{\Phi^*}$ is $\mathbb{K}$-$p$-separable, by Corollary \ref{horodeckis}.
Recalling Eq. (\ref{choiadjoint}),  we conclude that $C_\Phi$ is $\mathbb{K}$-$p$-separable.
\end{proof}

\begin{rem}
If $\Phi : M_n(\mathbb{R}) \to M_m(\mathbb{R})$, then the complexification $\widetilde\Phi$ is $\mathbb{C}$-entanglement $p$-breaking
if and only if $C_{\Phi} \in \operatorname{CSEP}_p(\mathbb{R}^n \otimes \mathbb{R}^m)$.
Since  we have the strict inclusion $\operatorname{SEP}_p (\mathbb{R}^n\otimes \mathbb{R}^m) \subsetneq  \operatorname{CSEP}_p(\mathbb{R}^n \otimes \mathbb{R}^m)$,
the complexification can be   $\mathbb{C}$-entanglement  $p$-breaking even if the original map $\Phi$ is not $\mathbb{R}$-entanglement $p$-breaking.
In other words,  there exist maps that break $\mathbb{C}$-entanglement without breaking $\mathbb{R}$-entanglement.
\end{rem}

\begin{example}
Here we provide an example of a map  $\Phi : M_{2p}(\mathbb{R}) \to M_{2p}(\mathbb{R})$ that has a $\mathbb{C}$-entanglement $p$-breaking complexification,
but is not $\mathbb{R}$-entanglement $(2p-1)$-breaking (and therefore is also not entanglement $p$-breaking).    The map is
\[
 \Phi  (A) =        O_+   A  O_+  +  O_-  A  O_-  \, ,
\]
where  $O_\pm$ are the orthogonal matrices
$O_+   = \begin{spmatrix}   0  &  I_p  \\  I_p  &  0  \end{spmatrix}$ and $O_-   = \begin{spmatrix}    I_p  &  0  \\   0&   -I_p   \end{spmatrix} $.
Its Choi matrix $C_{\Phi}$ was studied in example \ref{ex:pentangled}, where we showed that $C_\Phi$ is $\mathbb{C}$-$p$-separable, but $\mathbb{R}$-$(2p-1)$-entangled.
By Theorem \ref{Rentanglementbreaking characterization}, this implies that $\widetilde\Phi$ is $\mathbb{C}$-entanglement $p$-breaking, while $\Phi$ is not $\mathbb{R}$-entanglement $(2p-1)$-breaking.
\end{example}

For $p=1$, the above example shows that a completely positive map can break $\mathbb{C}$-entanglement without breaking $\mathbb{R}$-entanglement.

\section{Real version of the PPT-squared  conjecture}\label{sec:PPT}

A positive matrix $P  \in  \operatorname{PSD}  (\mathbb{K}^n\otimes \mathbb{K}^m)$ is called
\begin{enumerate}
\item  {\em PPT  $($positive partial transpose$)$}  if   $ (   \tau_n\otimes \operatorname{\textsl{id}}_m)    (P)  \ge  0$,
\item  {\em IPT  $($invariant under partial transpose$)$}  if   $ (   \tau_n\otimes \operatorname{\textsl{id}}_m)    (P)  =  P$.
\end{enumerate}
A  completely positive map   $\Phi: M_n(\mathbb{K}) \to M_m(\mathbb{K})$ is called PPT (IPT)   if its  Choi matrix is PPT  (IPT).

Clearly, every entanglement breaking map is PPT:  if the Choi matrix is separable,   then it is also PPT by the Peres-Horodecki criterion \cites{peres1996separability,Horodeckis,horodecki1997separability}.
In the real case, entanglement breaking maps and their complexifications are IPT.

\begin{prop} \label{C:RPPT equals transpose}
If $\Phi: M_n(\mathbb{R}) \to M_m(\mathbb{R})$ is $\mathbb{R}$-entanglement breaking, then its complexification $\widetilde \Phi$ is  IPT,
and   $\tau_m \circ \widetilde \Phi =  \widetilde \Phi  \circ \tau_n  = \widetilde \Phi$.
\end{prop}

\begin{proof}
Since $C_{\Phi} \in \operatorname{SEP}(\mathbb{R}^n \otimes \mathbb{R}^m)$, we can write $C_\Phi = \sum_k A_k \otimes B_k$ for $A_k \in \operatorname{PSD}(\mathbb{R}^n)$ and $B_k\in \operatorname{PSD}(\mathbb{R}^m)$.
Recalling that $C_{\widetilde \Phi}  = C_\Phi$, we obtain
\[
 \tau_n \otimes \operatorname{\textsl{id}}_m(C_{\widetilde \Phi}) = \sum_k A_k^t \otimes B_k = \sum_k A_k \otimes B_k = C_{\widetilde \Phi } \, ,
 \]
that is  $\widetilde \Phi$ is IPT.    Moreover, one has
\[
 C_{\tau_m \circ \widetilde \Phi} =  \operatorname{\textsl{id}}_m\otimes  \tau_m (C_{\widetilde \Phi})  =   \sum_k A_k \otimes B_k^t = \sum_k A_k \otimes B_k = C_{\widetilde \Phi} \,,
\]
which implies $\tau_m \circ \widetilde \Phi  =  \Phi$, and
and
\[
 C_{ \widetilde \Phi\circ \tau_n} =  C_{ \tau_m\circ \widetilde \Phi\circ \tau_n}
 =     \sum_{i,j}    E_{i,j}  \otimes   (\widetilde \Phi   ( E_{i,j}^t))^t =   \sum_{i,j}    E_{i,j}  \otimes   (\Phi   ( E_{i,j}^t))^t =  C_{\Phi} \, ,
\]
where the last equality follows from the Hermiticity of $\Phi$.
\end{proof}

In a presentation at the workshop  ``Operator structures in quantum information theory''  (Banff Research Station, February 26-March 2 2012) \cite{RJKHW}, Christandl raised what has later  become known as the PPT-squared conjecture.

\begin{conj}[\textbf{PPT-squared}] \label{PPT2C}
For every PPT map $\Phi:M_n(\mathbb{C})\rightarrow M_n(\mathbb{C})$, $\Phi\circ\Phi$ is $\mathbb{C}$-entanglement breaking.
\end{conj}

In \cite{CMW}, Christandl, Muller-Hermes, and Wolf show that if this conjecture is true, then the composition $\Phi \circ \Psi$
is entanglement breaking for any two PPT maps with compatible domain and range.
They also show that the conjecture is true for $n=3$ \cite{CMW}*{Corollary 3.1}.

The example below shows that a direct real analogue of Conjecture~\ref{PPT2C} is false. Namely, there exist real PPT maps whose square is not $\mathbb{R}$-entanglement breaking, even for $n=2$.

In \cite{KMP} it is shown that if $\Phi: M_n(\mathbb{C}) \to M_n(\mathbb{C})$ is idempotent, i.e., $\Phi \circ \Phi = \Phi$,
unital, i.e, $\Phi(I_n) = I_n$ and PPT, then $\Phi \circ \Phi$ is $\mathbb{C}$-entanglement breaking.
In \cite{KMP} it is also shown that if $\Phi$ is only unital and PPT, then
every limit point of the sequence of iterates $\Phi^k = \Phi\circ \dots \circ \Phi$ is a $\mathbb{C}$-entanglement breaking map.
In \cite{RJP} it is shown that if $\Phi: M_n(\mathbb{C}) \to M_n(\mathbb{C})$ is unital, trace-preserving and PPT,
then $\Phi^k$ is $\mathbb{C}$-entanglement breaking for some power of $\Phi$.

The example below also shows that the direct real analogues of these results are false, even for $n=2$.

\begin{example} \label{E:idem PPT not EB}
Define $\Phi : M_2(\mathbb{R}) \to M_2(\mathbb{R})$ with
\[
  C_\Phi = \frac12 \begin{bmatrix} I_2 & \gamma \\ -\gamma & I_2 \end{bmatrix}
  \quad\text{where}\quad
  \gamma =\begin{bmatrix} \phantom{-} 0 & 1 \\ -1 & 0 \end{bmatrix}.
\]
Then $\Phi$ is a unital, completely positive, trace preserving map which is idempotent, $\Phi^2=\Phi$.
Moreover, $\operatorname{ran} \Phi = \operatorname{span}\{ I_2,\gamma \}$ is a real abelian C*-algebra with the usual product.
Since $C_\Phi$ and its partial transpose are positive, $\Phi$ is PPT.
Since $C_\Phi$ is not equal to its partial transpose, Corollary~\ref{C:RPPT equals transpose} shows that $\Phi$ is not $\mathbb{R}$-entanglement breaking.

However, $C_\Phi$ is in $\operatorname{CSEP}(\mathbb{R}^2\otimes \mathbb{R}^2)$: let $A_\pm = \frac12 (I_2 \pm i\gamma ) \ge 0$. Then
\[
 A_+ \otimes A_- + A_- \otimes A_+ = \frac12(I_2 \otimes I_2 + \gamma \otimes \gamma) = C_\Phi
\]
and
\[
 A_+ \otimes A_+ + A_- \otimes A_- = \frac12(I_2 \otimes I_2 - \gamma \otimes \gamma) = C_{\Phi \circ T}
\]
Therefore $\widetilde\Phi$ is $\mathbb{C}$-entanglement breaking.
\end{example}

With Corollary~\ref{C:RPPT equals transpose} and Example~\ref{E:idem PPT not EB} in mind,  we formulate a real version of the PPT-squared conjecture, which we call the IPT-squared conjecture.
This conjecture also makes sense for $\mathbb{C}$.
We formulate both versions.

\begin{conj}[\textbf{IPT-squared for $\mathbb{K}$}] \label{IPT2}
For $\mathbb{K} \in \{ \mathbb{R}, \mathbb{C}\}$ and every IPT map $\Phi:M_n(\mathbb{K})\rightarrow M_n(\mathbb{K})$, $\Phi\circ\Phi$ is $\mathbb{K}$-entanglement breaking.
\end{conj}

The following gives a further indication that the IPT-squared conjecture is the correct real counterpart of the original PPT-conjecture.

\begin{prop}
If the $\ensuremath{\operatorname{PPT}}^2$ conjecture is true, then $\ensuremath{\operatorname{IPT}}^2$ Conjecture is true for $\mathbb{C}$.
If the $\ensuremath{\operatorname{IPT}}^2$ Conjecture is true for $\mathbb{C}$, then it is true for $\mathbb{R}$.

In particular, in dimension three, the above conjectures are true.
\end{prop}

\begin{proof}
The first statement is obvious, since Conjecture~\ref{IPT2} for $\mathbb{C}$ is a special case of Conjecture~\ref{PPT2C}.

Assume that Conjecture~\ref{IPT2} is true for $\mathbb{C}$. Let $\Phi: M_n(\mathbb{R}) \to M_n(\mathbb{R})$ be IPT.
Then $\widetilde\Phi: M_n(\mathbb{C}) \to M_n(\mathbb{C})$ is IPT and hence $\widetilde\Phi \circ \widetilde\Phi = \widetilde{\Phi \circ \Phi}$ is $\mathbb{C}$-entanglement breaking.
This means that we can write
\[
 C_{\Phi \circ \Phi} = C_{\widetilde\Phi \circ \widetilde\Phi} = \sum_k P_k \otimes Q_k,
\]
with $P_k, Q_k \ge 0$ complex matrices.
Write $P_k = A_k + iB_k, \, Q_k = C_k + i D_k,$ where $A_k,B_k,C_k, D_k \in M_n(\mathbb{R})$ with $A_k, C_k$ symmetric and $B_k, D_k$ skew-symmetric.
Also note that $A_k, C_k \ge 0$.

Since $\Phi \circ \Phi$ is IPT,
\[
 C_{\Phi \circ \Phi} = C_{T \circ \Phi \circ \Phi} = \sum_k P_k \otimes Q_k^t = \sum_k P_k \otimes (C_k - i D_k).
\]
Averaging these two expressions yields
\[
 C_{\Phi \circ \Phi} = \sum_k P_k \otimes C_k = \sum_k A_k \otimes C_k + i \sum_k B_k \otimes C_k.
\]
Since $C_{\Phi \circ \Phi}$ is a real matrix, we have that
\[
 C_{\Phi \circ \Phi} =  \sum_k A_k \otimes C_k.
\]
Hence,  $\Phi \circ \Phi$ is $\mathbb{R}$-entanglement breaking.

Note that, it is known that in demension three (complex case), the PPT-sqaured conjecture is true (see \cite{CMW} and also \cite{CYT}). Hence using the earlier proof method, one can show that the $\ensuremath{\operatorname{IPT}}^2$ Conjecture is true for $\mathbb{R}$ in demension 3.
\end{proof}

We do not know if these conjectures are equivalent.

We can however establish the three results from \cite{KMP} and \cite{RJP} cited above for IPT maps.

\begin{prop}
Let $\Phi: M_n(\mathbb{R}) \to M_n(\mathbb{R})$ be unital, trace-preserving and IPT. Then there is a power $k$ such that $\Phi^k$ is $\mathbb{R}$-entanglement breaking.
\end{prop}

\begin{proof}
Since $\widetilde\Phi$ is unital, trace-preserving and PPT, there is some $k$ so that $(\widetilde\Phi)^k = \widetilde{\Phi^k}$ is $\mathbb{C}$-entanglement breaking.
Using the fact that $\tau_n \circ \Phi = \Phi$ implies $\tau_n \circ \Phi^k = \Phi^k$ and arguing as in the previous proof, we obtain that $C_{\Phi^k}$ is $\mathbb{R}$-separable.
\end{proof}

Before proving the analogues of the results in \cite{KMP}, an analogue of \cite{KMP}*{Lemma~3.1} is required.
However it is much more difficult in the real case.

When $\Phi:M_n(\mathbb{C}) \to M_n(\mathbb{C})$ is idempotent and unital, i.e., $\Phi\circ\Phi=\Phi$ and $\Phi(I_n) = I_n$,
then the range of $\Phi$ is an operator system which becomes a C*-algebra when endowed with the Choi-Effros product $A \star B = \Phi(AB)$.
In \cite{KMP}, it is shown that this C*-algebra is abelian.
They then use the fact that every finite dimensional complex abelian C*-algebra is of the form $\ell^{\infty}_k(\mathbb{C})$ and the complex analogue of \ref{Rentanglementbreaking characterization}(5), to deduce that $\Phi$ is $\mathbb{C}$-entanglement breaking.

However, not every finite dimensional real abelian C*-algebra is of the form $\ell^{\infty}_k(\mathbb{R})$.
In particular, in Example~\ref{E:idem PPT not EB}, the range of $\Phi$ is $*$-isomorphic to the real abelian C*-algebra
\[
  \ff A = \{  f \in C(\{a,b\}) : f(b) = \overline{f(a)} \} .
\]
This is a 2-dimensional abelian real C*-algebra which is not isomorphic to $l^\infty_2(\mathbb{R})$ or $l^\infty_1(\mathbb{C})$.
Note that $\Phi$ is PPT but not IPT.

\begin{thm}
Let $\Psi : M_n(\mathbb{R}) \to M_n(\mathbb{R})$ be a unital IPT idempotent map.
Then the operator system $\cl R := \Psi(M_n(\mathbb{R}))$ with the Choi-Effros product
is order isomorphic to $l^\infty_k(\mathbb{R})$ for some $k\ge1$, and $\Psi$  is $\mathbb{R}$-entanglement breaking.
\end{thm}

\begin{proof}
The Choi-Effros product on $\cl R$ is given by $A \star B := \Psi(AB)$,
and this makes $\cl R$ into a real C*-algebra $\ff A$.
The complexification $\widetilde\Psi$ is PPT, unital and idempotent.
By \cite{KMP}*{Lemma~3.1}, the Choi-Effros product on $\widetilde\Psi(M_n(\mathbb{C}))$ is abelian;
and hence $\ff A$ is abelian.
Let $\Gamma : \cl R \to \ff A$ be this identification, which is a complete isometry; so $\Gamma^{-1}$ is also completely isometric.

An old result of Arens and Kaplansky shows that if $\ff A$ is an abelian real C*-algebra,
then there is a locally compact Hausdorff space $Z$ and a homeomorphism $\tau:Z \to Z$ with $\tau^2=\operatorname{\textsl{id}}_Z$
such that
\[
 \ff A \simeq \{ f \in C_0(Z) : f(\tau(z)) = \overline{f(z)} \:\text{ for all }\: x \in Z \} .
\]
See \cite{rosenberg}*{Theorem 1.9} for an easy proof.
In our case, since $\ff A$ is finite dimensional, $X$ is a finite set with the discrete topology.
Hence
\[
  Z = \{ x_j, y_k, z_k: 1 \le j \le J, 1 \le k \le K \}
\]
such that $\tau(x_j)=x_j$, $\tau(y_k) = z_k$ and $\tau(z_k) = y_k$. Thus
\[
  \ff A = \Big\{ \sum_j a_j \delta_{x_j} + \sum_k b_k \delta_{y_k} + \overline{b_k} \delta_{z_k}: a_j \in \<\!\< R, \, b_k \in \<\!\< C \Big\}.
\]
A basis for $\ff A$ as a real vector space is given by $\{ \delta_{x_j}, \delta_{y_k} + \delta_{z_k}, i(\delta_{y_k} - \delta_{z_k}) \}$.
Write these vectors as $\{ \alpha_j, \beta_k, \gamma_k \}$.
Then $\alpha_j^*= \alpha_j$, $\beta_k^* = \beta_k$ and $\gamma_k^* = - \gamma_k$.

The map $\Phi = \Gamma \Psi : M_n(\<\!\< R) \to \ff A$ can be written as
\[
  \Phi(X) = \sum_j \operatorname{Tr}(XA_j) \alpha_j + \sum_k \operatorname{Tr}(XB_k) \beta_k + \operatorname{Tr}(XG_k) \gamma_k
\]
for matrices $A_j, B_k, G_k \in M_n(\<\!\< R)$.
Since $\Phi$ is CP,
\[
 X \ge 0 \implies \operatorname{Tr}(XA_j) \ge 0,\ \operatorname{Tr}(X B_k) \ge 0 \text{ and } \operatorname{Tr}(XG_k) =0 .
\]
Therefore $A_j, B_k \in \operatorname{PSD}(\mathbb{R}^n)$ and $G_k^t = -G_k$, since $G_k$ is orthogonal to all positive matrices, and hence to all symmetric matrices.

Let $P_j = \Gamma^{-1}(\alpha_j)$, $Q_k = \Gamma^{-1}(\beta_k)$ and $R_k = \Gamma^{-1}(\gamma_k)$.
Then $P_j, Q_k$ are positive, and $R_k^t = - R_k$ because $\Gamma$ and $\Gamma^{-1}$ are positive.
This is a real basis for $\cl R$.
Now we have
\[
 \Psi(X) = \Gamma^{-1} \Phi(X) = \sum_j \operatorname{Tr}(XA_j) P_j + \sum_k \operatorname{Tr}(XB_k) Q_k + \operatorname{Tr}(XG_k) R_k .
\]
The Choi matrix is
\[
 C_\Psi = \sum_j A_j \otimes P_j + \sum_k B_k \otimes Q_k + \sum_k G_k \otimes R_k.
\]
The partial transpose of this map is
\[
 C_{T \circ\Psi} = \sum_j A_j \otimes P_j + \sum_k B_k \otimes Q_k - \sum_k G_k \otimes R_k.
\]
However, $C_{T \circ\Psi} = C_\Psi$ since $\Psi$ is IPT. So $G_k = 0$.
Therefore $C_\Psi$ is separable, and hence $\Psi$ is $\mathbb{R}$-entanglement breaking by Theorem~\ref{Rentanglementbreaking characterization}.

Now $\Gamma$ is an isomorphism, and in particular it is surjective.
Therefore $K=0$ and $Z = \{ x_j : 1 \le j \le J \}$.
Hence $\ff A \simeq C_\mathbb{R}(Z) \simeq l^\infty_J(\mathbb{R})$.
\end{proof}

Now the asymptotic result follows as in \cite{KMP} by applying Ellis's Theorem that every compact semigroup contains idempotents (and its proof).

\begin{thm} \label{T:asymptotic REB}
Let $\Phi:M_n(\mathbb{R})\rightarrow M_n(\mathbb{R})$ be a unital IPT  map.
Then
\[
 \lim_{k\to\infty} \operatorname{dist}(\Phi^k, \mathbb{R}\text{-}\operatorname{EB}  (n,n)) = 0 .
\]
\end{thm}
\bigskip

\section*{Acknowledgments}

GC  was supported by the Hong Kong Research Grant Council through grant 17300918  and through
the Senior Research Fellowship Scheme via SRFS2021-7S02, and by the Croucher foundation.
KRD and VIP were partially supported by the Natural Sciences and Engineering Research Council of Canada (NSERC).
MR is supported by the European Research Council (ERC Grant Agreement No. 851716).

\bibliographystyle{amsplain}

redpajama_set_name: RedPajamaArXiv


The Massachusetts men's basketball team will be without one of its starting big men next season, as Malik Hines will utilize the NCAA graduate transfer exemption rule and not return for his senior year with the Minutemen.
The junior was a key player for UMass this past season, starting 32 of 33 games and averaging nine points and six rebounds. His sophomore year, Hines started five games and appeared in 16 games as a freshman.
The announcement was made via tweet from the UMass men's basketball Twitter account.
Malik Hines to utilize graduate transfer exception rule; will not return to UMass for 2018-19 season.
Hines plans to earn his undergraduate degree in sociology this summer, allowing him to exert the transfer exemption.
As of now, the Minutemen will return Rashaan Holloway, Chris Baldwin and Khalea Turner-Morris to the big man rotation next season. Rutgers transfer Jonathan Laurent will help pick up the minutes lost from Hines.


text:
Michael Beale out to right Celtic injustice as he eyes Rangers success
It was the one that got away. Of all the failures and all the disappointments, it was the one that hurt the most and that still ranks to this day.
Whats done is done, of course. Michael Beale cannot rewrite history but he can shape the future as he prepares for another Hampden head-to-head with Celtic.
The build-up to the Viaplay Cup final next month will naturally involve talk of the last time the Old Firm met when the silverware was on the line. Beale needs no reminding of it more than three years on.
On a day that Rangers dominated, Celtic celebrated. There was no technology to determine that Christopher Jullien's strike was offside and should not have stood and an Alfredo Morelos penalty miss summed up an occasion to forget on what could, and should, have been one to savour.
"Yeah, they ended up with the trophy," Beale said as he reflected on the 2-1 win over Aberdeen and cast his mind back to his first Hampden final. "Well, if we had VAR… Are we using VAR at Hampden for this final?
"That was a sore one at the time. After that game, it was more frustration because we had played very well. That's a trophy that they've got that's a disappointment for us.
"Look, I'm just delighted that we have come back to another final. A lot has been said about this group but all they can do is keep winning football matches – and they keep finding a way. That's the sign of a good team ."
The win over Aberdeen on Sunday was the most significant of Beale's tenure as Rangers manager. If it can be topped next month, the Englishman will have his maiden medal as a boss in his own right and will have made the first inroads into righting the record that plagued his time alongside Steven Gerrard.
READ MORE: Rangers keeper Allan McGregor is the best in Scotland
The failure to win a knock-out trophy was the black mark on record of the management team that delivered title 55 and restored Rangers as a name and a force in Europe. Just weeks after his return to Ibrox, the silverware is within reach for Beale.
Success is ultimately what he will be judged on. As he embraced his players and applauded the supporters at the end of an extra-time victory, Beale was more concerned about what the achievement and the occasion meant for others.
"It means everything to me, but it's only a small thing when I look at all of those fans and the players as well," Beale said. "I think it means a lot to the group. I'm delighted for them.
"For me, as a coach, at the end of the season I'll look back and see if I've done a good for Rangers and then I'll bite down the gum shield and go again for 12 months. I really feel the responsibility, of wanting to do well for other people rather than myself."
Beale has spoken about being performance happy and results happy in the weeks since he succeeded Giovanni van Bronckhorst. He is still searching for the complete showing, but this win – secured thanks to goals from Ryan Jack and Kemar Roofe – was another triumph to go alongside the handful that have been earned in the Premiership.
Given their advantage in terms of personnel and the scoreline, the closing stages were perhaps more fraught than they needed to be as Aberdeen attempted to salvage their cup dreams while captain Anthony Stewart sat in the stands and contemplated his red card moment of madness.
This was a win that will do Rangers the world of good in terms of confidence and the buy-in to the Beale blueprint. It will also further allay the anxiety that the Englishman has addressed on a couple of occasions.
"We have to keep talking about it," Beale said. "It's natural when you try to hold on to something, you drop deep.
"On Sunday, I thought we were in control of the game. We started well, we should score with Fashion. Aberdeen were always in the game, I'm not saying we were dominant, but we were the stronger team in the game.
"We had the ball, we had chances, we had good areas to cross and then, from once isolated moment, we've conceded. And you saw it at a couple of set plays, when we are looking a little bit anxious.
READ MORE: Borna Barisic details Rangers mentality ahead of Celtic shot at glory
"And even then, at half time, we spoke about getting ourselves in front, we're again really dominant, then Allan has to make a big save out of nowhere. There are some things for me to work on.
"It's fair to say that in the time I've been back in, we've had more games than training sessions, in terms of we've been at it every two or three days. We will be until another six or seven days time, but at this moment in time, their reaction to the advice I'm giving, the people coming off the bench, is fantastic.
"The game's not always plain sailing because two teams play it. In this moment in time, I think that's a really good Aberdeen team. It's young, it's got legs, it's trying to play the right way.
"I know Jim will be disappointed tonight but he can be pleased with the direction they are moving in. I want to give them some credit as well."
Aberdeen emerged from a compelling cup tie with a pat on the back but it was Rangers who deserved the victory and who can now look forward to a return to the National Stadium.
It was done the hard way after Beale side's recovered from the loss of the opening goal to Bojan Miovski just before the break. Not for the first time, the manager had his say and earned his win with his messages at the interval and changes during the match.
"It's a group of which 75 per cent I've worked with for a long time, so there's a huge amount of trust in some of the advice I give them," Beale said. "I think that's a good sign, that you go in and are able to talk things through and stay on it.
READ MORE: Ben Davies eyes Celtic triumph and makes Kemar Roofe Rangers call
"But, I think also, you're doing me more service than maybe I deserve. I think sometimes the game has just got to settle down. Maybe teams get tired later on and maybe our style just gets them.
"I think it was a much more mature performance than when we lost to Aberdeen maybe a few years ago, when in the second half we rushed the game and played more long balls.
"On Sunday we kept playing. And at times our fans want us to play a little bit quicker, but when it nearly comes off – we might have a fantastic move for fashion to have one, and Tav to hit the bar.
"If one of them goes in, we're talking about a wonderful goal with lots of passes. So, I'm seeing really good signs.
"I was pleased with a lot, but for a long period of the game you are losing and there's that horrible feeling in your stomach, but fair play to the players, they keep finding a way."
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redpajama_set_name: RedPajamaCommonCrawl


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Dress as your favorite character from the new Emoji movie in this Gene Child Costume.
Includes a tunic, shirt, and pants.


CNHL has become insanely popular, and sitting at the top of their totem pole for FPV racing are the CNHL (China Hobby Line) 1300mah 100C 4s packs. Tested and loved, you can be assured that these batteries have the punch you crave and the speed you demand while racing. The 100c variant has impressive power, and a noticeable increase in juice/punch over the 1300mah 70c CNHL variant. With motors and props which continually demand more and more from batteries which each release, these batteries stand up to the task. CNHL 1300Mah Batteries now in stock!


We prepared Amendment C241 after the Heritage Council of Victoria recommended that the former Palace Theatre site could be of local heritage significance.
What does the amendment apply to?
What is the history of the former Palace Theatre site to date?
What other planning matters relate to this precinct?
The planning scheme amendment applies to the former Palace Theatre site, located at 20-30 Bourke Street, Melbourne.
The amendment seeks to introduce a heritage Statement of Significance for the site.
If the amendment is adopted, the Statement of Significance will be recognised under the Melbourne Planning Scheme. This will ensure the heritage attributes of the building are used to guide decisions regarding development of the former Palace Theatre site into the future.
For more information, call the City of Melbourne's Strategic Planning team on 9658 8404.
3 July 2014 – TheHeritage Council of Victoria resolved that the Palace Theatre was not of State significance but agreed "that the Palace Theatre is of cultural heritage significance for its social and historical values at a local level" and suggested "that inclusion of the Palace Theatre in the overlay as an individual listing may be warranted".
27 October 2014* –The City of Melbourne arranged for an independent heritage review which supported the view that the site was of individual significance at a local level and recommended that external and internal heritage controls be applied. Consequently a Statement of Significance* was prepared stating that 'The Palace Theatre is historically and socially significant and of architectural interest to the City of Melbourne'. (*A Statement of Significance details the main heritage characteristics of a precinct or site and is important when making decisions about development proposals affecting a heritage place).
1 December 2014 – After receiving reports that parts of the internal fabric of the Palace Theatre had been removed, the independent heritage consultant was granted permission to undertake another inspection of the building. Their recommendation previously put forward in the heritage study was revised to reflect the altered condition of the building. This is outlined in a supplementary report which recommends that internal heritage controls are no longer required.
12 February 2015 – Planning Scheme Amendment C241 the Palace Theatre Heritage Review is placed on public exhibition until 20 March 2015.
There is currently a planning application before VCAT which seeks to partly demolish and develop the former Palace Theatre site. The application will be heard on 30 March 2015.
The Palace Theatre building sits within Heritage Overlay precinct HO500 (Bourke Street Hill Precinct). In October 2014, the Minister for Planning placed Planning Scheme Amendment C240 Bourke Hill on exhibition until 6 February 2014. Amendment C240 proposes, among other things, to replace the interim mandatory height controls with a combination of mandatory and discretionary height controls.
Both matters (the planning permit application and Amendment C240) are being considered separately to Planning Scheme Amendment C241 Palace Theatre Heritage Review.


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Two years ago, the Department of Education (DOE) announced plans to reform special education in New York City. The stated goals of the reform were to improve access to more challenging academic curricula and to achieve better outcomes for students with disabilities by enrolling more special education students in community schools. Piloted in 260 public schools for the past two years, the reform was extended this month to most of the city's public schools, which are now expected to serve and meet the needs of considerably more special education students than in the past.
There is no doubt that New York City needs to change the way we educate and support our students with disabilities. For too long, students with disabilities have been sent to schools far away from home, where they have been segregated in classes with little access to challenging academic curricula and abysmal graduation rates. Black boys, in particular, have been disproportionately represented in the most segregated special education classes. This administration deserves credit for facing some of the most intractable problems with special education and making an attempt at reform.
The reform is now focused primarily on assuring that students with disabilities have access to their community schools. That is an important start, as we have seen some schools turn away students with disabilities for years, without making any attempt to serve them. However, the DOE must ultimately make sure that all students, including students with disabilities, receive effective instruction and all the supports and services they need to make that instruction successful, regardless of where they are seated. Based on our conversations with teachers and principals, many schools do not feel prepared to provide a quality education to students across the broad spectrum of disability. The DOE will have to make a significant investment in the classroom experiences of students with disabilities before we will begin to see real and meaningful changes to their daily experiences and long-term outcomes.
As the reform has rolled out, we have seen families caught between the DOE's insistence that students with disabilities attend their community schools, and the failure to prepare some of those schools adequately to welcome and support children with a wide variety of needs. Over the summer, AFC received calls from more than 40 parents of incoming kindergarten students with disabilities whose zoned schools made clear they would not be able to provide the type of class recommended for the individual children by the DOE for kindergarten. When we brought those cases to the attention of the DOE, they eventually resolved a majority of them. However, only a small percentage of families find their way to organizations like ours; we worry that the DOE has not put out the information and the contacts that families require to assert their rights and stand up for what their children need.
So where does this all leave us?
In order for the reform to boost achievement, the DOE needs to shift its focus from enrolling students in their community schools to providing them support once they get there. In particular, we call upon the DOE to:
Provide on-going, intensive, at-the-elbow support for school-level staff to meet the additional needs of increasing numbers of students with disabilities in their schools – including professional development on the delivery of effective, research-based instruction to students with learning and behavioral challenges;
Develop and implement immediately a better plan for engaging parents in the reform at the local level and giving parents information about what to do if their schools lack the expertise or capacity to meet their children's needs;
Complete and share publicly a detailed review of the first phase of the reform. Parents, educators, and the public need to know what happened in the 260 schools where this reform was piloted. What were best practices identified in those schools? Who were the students most affected? Which research-based interventions were used to provide support in reading, math, and behavior, and how successful were those interventions? What was the true impact of the pilot on the individual students with and without disabilities in those schools?
Special education reform is very much a work in progress. To succeed, the DOE will need to embrace parents as stakeholders in this process and double down on its commitment to build capacity and expertise at the school level.
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6 Stories That Will Renew Your Faith and Hope
The MSM likely missed some of these powerful stories about kindness and goodwill.
By Billy Hallowell Published on January 10, 2022 •
Billy Hallowell
A lot is going on across America — and the world — right now, but one of the biggest blessings of the past few weeks has been finding stories about kindness and goodwill.
These are uplifting stories that remind us of the good in the world (acts of kindness often overlooked by the mainstream media). I'll lead with those before diving into a few of the important faith, culture, and news headlines you'll want to be aware of (in case you missed them).
Also, before we get into the stories: listen to my latest podcast episode:
190K+ Young People Accepted Jesus
First, Christian ministry Groundwire saw more than 190,000 young people turn to Christ last year, and the positive pattern is showing no signs of slowing down.
They're using a unique way of reaching Millennials and Gen Zers with the gospel — check it out.
Pastor Raises $327K+ for Orphaned Kids
Hands down, this is my favorite story of the week. An Iowa pastor has raised more than $327,000 in just a few days to support five orphans whose parents tragically died.
Pastor Eugene Kiruhura of Shalom Covenant Church in Urbandale, Iowa, launched a GoFundMe last week for the family of Bazirake Kayira, who died in a tragic car accident. His wife also died a few months back during childbirth.
Read why the pastor is taking action to help.
Brave Firefighters Rescue a Family
Next comes a story about brave firefighters in Atlanta who raced to rescue three adults trapped in a burning home. One of their relatives later went on-air with a local outlet to thank God for the positive outcome.
"I'm a Christian, blessed," he said. "I thank God."
Loaves of Bread Amid Chaos
The now-infamous crisis on Virginia's I-95 last week left scores of travelers stranded for up to a day or more and captured a plethora of headlines. But now that the cars have been cleared, it's the stories of kindness that are starting to grab attention.
What one bakery did to help those stranded left a woman claiming the bakery's response "was one of the kindest moments" she has ever witnessed. Check it out.
A WWII Vet Dies at Age 112 — But What a Story He Had
Lawrence Brooks was the oldest-known World War II veteran in the U.S. when he died Wednesday at the amazing age of 112. And as his life is being celebrated and commemorated, friends and loved ones are revealing that Brooks had a secret to his longevity: actively loving God and loving others.
"When asked about the secret of living a long life, he would say: 'Serve God and be nice to people,'" the Rev. Morris Thompson Jr., bishop of the Episcopal Diocese of Louisiana, wrote to church members, according to Nola.com.
'Jesus Loves You': Trapped Woman's Rescue
This story is probably my second favorite of the week. A woman hit a patch of ice, lost control, and her car landed vertically in the woods; she was trapped inside until a good Samaritan happened to see the vehicle — and saved her.
"God sent you to rescue me," Yata Yeawolo told Dylan Dauphinee after she reunited with him following the rescue.
And the best part: the hood of her car had a giant sticker that read, "Jesus Loves You." It was an inspiring message seen as rescue workers helped her escape the vehicle. Check the story out here.
Now, for Other News…
As we come to a close, some other headlines you won't want to miss:
An NFL star is giving God all the glory.
Israel expert unveils what most people get wrong about Bible prophecy…
Politician and bishop in Finland face criminal charges for public stance on same-sex marriage.
Kathie Lee Gifford says Westerners are "too terrified" of gifts of the Spirit…
Nancy Pelosi ignores atheists' prayer demands…
Billy Hallowell has worked as a journalist and commentator for more than a decade. He is the director of communications and content for PureFlix.com, the former senior editor at Faithwire.com and the former faith and culture editor at TheBlaze. Article originally published at billyhallowell.substack.com.
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