# Secondary Hochschild homology and differentials

Jacob Laubacher  Department of Mathematics, St. Norbert College, De Pere, WI
54115<EMAIL_ADDRESS>
###### Abstract.

In this paper we study a generalization of Kähler differentials, which
correspond to the secondary Hochschild homology associated to a triple
$(A,B,\varepsilon)$. We establish computations in low dimension, while also
showing how this connects with the kernel of a multiplication map.

###### Key words and phrases:

Hochschild homology, cyclic homology, differentials.
_Corresponding author._ Jacob Laubacher ✉<EMAIL_ADDRESS>☎
920-403-2961.

###### 2020 Mathematics Subject Classification:

Primary 13D03; Secondary 16E40, 13N05

##  1\. Introduction

Gerhard Hochschild introduced the cohomology theory bearing his name in [10]
in order to study extensions of algebras over a field $\mathbbm{k}$. Later,
Gerstenhaber used this theory to study deformations in [9]. On the other hand,
Hochschild homology led to the study of cyclic homology, as well as a
stepping-stone towards de Rham cohomology.

Secondary Hochschild (co)homology, a generalization of Hochschild
(co)homology, has now been studied for some time. The cohomology originally
introduced in [15] by Staic, and the homology in [13], the theory involves a
triple $(A,B,\varepsilon)$ which consists of a commutative
$\mathbbm{k}$-algebra $B$ inducing a $B$-algebra structure on an associative
$\mathbbm{k}$-algebra $A$ by way of a morphism $\varepsilon:B\longrightarrow
A$. The corresponding secondary cyclic (co)homology was also studied in [13],
and properties, applications, and generalizations of the secondary Hochschild
(co)homology were investigated in [1], [2], [7], [8], [11], [12], and [16],
among others.

In this paper, our goal is to study how the secondary Hochschild homology
associated to the triple $(A,B,\varepsilon)$ interacts with a generalization
of Kähler differentials. In particular, we prove Theorem 4.3, which is a
generalized version of Theorem 2.9. In Section 2 we provide all necessary
background information, keeping this paper as self-contained as possible.
Section 3 introduces our notion of secondary Kähler differentials, leading to
an isomorphism between them and the secondary Hochschild homology under
certain conditions, as well as providing computations in low dimension. Next,
Section 4 discusses the kernel of the multiplication map from $A^{\otimes
2}\otimes B$ to $A$ and how this also corresponds to the aforementioned
differentials. We round out the paper in Section 5 by hypothesizing possible
avenues for future work.

##  2\. Preliminaries

We start by fixing $\mathbbm{k}$ to be a field containing $\mathbb{Q}$. We set
all tensor products over $\mathbbm{k}$, unless otherwise stated (so
$\otimes=\otimes_{\mathbbm{k}})$. Next, let $A$ be an associative
$\mathbbm{k}$-algebra with multiplicative unit. Most of our work is done over
a triple, the formal definition of which is below:

######  Definition 2.1.

([15]) We call $(A,B,\varepsilon)$ a triple if $A$ is a $\mathbbm{k}$-algebra,
$B$ is a commutative $\mathbbm{k}$-algebra, and $\varepsilon:B\longrightarrow
A$ is a morphism of $\mathbbm{k}$-algebras such that
$\varepsilon(B)\subseteq\mathcal{Z}(A)$. Call $(A,B,\varepsilon)$ a
commutative triple if $A$ is also commutative.

One can find examples of triples in [15] and [16], as well as more extensions
in [3] and [4].

###  2.1. The secondary Hochschild homology

Next we recall the secondary Hochschild homology associated to the triple
$(A,B,\varepsilon)$, which was introduced in [13].

Following notation in [13], we define
$\overline{C}_{n}(A,B,\varepsilon)=A^{\otimes n+1}\otimes
B^{\otimes\frac{n(n+1)}{2}}$ for all $n\geq 0$. As is customary in these
settings (see [5] for example), we can follow an organizational schematic to
represent the elements of $A^{\otimes n+1}\otimes B^{\otimes\frac{n(n+1)}{2}}$
as

$\otimes\begin{pmatrix}a_{0}&b_{0,1}&b_{0,2}&\cdots&b_{0,n-2}&b_{0,n-1}&b_{0,n}\\\
1&a_{1}&b_{1,2}&\cdots&b_{1,n-2}&b_{1,n-1}&b_{1,n}\\\
1&1&a_{2}&\cdots&b_{2,n-2}&b_{2,n-1}&b_{2,n}\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\
1&1&1&\cdots&a_{n-2}&b_{n-2,n-1}&b_{n-2,n}\\\
1&1&1&\cdots&1&a_{n-1}&b_{n-1,n}\\\ 1&1&1&\cdots&1&1&a_{n}\\\ \end{pmatrix},$

where $a_{i}\in A$, $b_{i,j}\in B$, and $1\in\mathbbm{k}$.

Next, we have that
$\partial_{n}:\overline{C}_{n}(A,B,\varepsilon)\longrightarrow\overline{C}_{n-1}(A,B,\varepsilon)$
is given by

$\displaystyle\partial_{n}$
$\displaystyle\Big{(}\otimes\begin{pmatrix}a_{0}&b_{0,1}&\cdots&b_{0,n-1}&b_{0,n}\\\
1&a_{1}&\cdots&b_{1,n-1}&b_{1,n}\\\ \vdots&\vdots&\ddots&\vdots&\vdots\\\
1&1&\cdots&a_{n-1}&b_{n-1,n}\\\ 1&1&\cdots&1&a_{n}\\\ \end{pmatrix}\Big{)}$
$\displaystyle=\sum_{i=0}^{n-1}(-1)^{i}\left(\otimes\begin{pmatrix}a_{0}&b_{0,1}&\cdots&b_{0,i}b_{0,i+1}&\cdots&b_{0,n-1}&b_{0,n}\\\
1&a_{1}&\cdots&b_{1,i}b_{1,i+1}&\cdots&b_{1,n-1}&b_{1,n}\\\
\vdots&\vdots&\ddots&\vdots&\ddots&\vdots&\vdots\\\
1&1&\cdots&a_{i}\varepsilon(b_{i,i+1})a_{i+1}&\cdots&b_{i,n-1}b_{i+1,n-1}&b_{i,n}b_{i+1,n}\\\
\vdots&\vdots&\ddots&\vdots&\ddots&\vdots&\vdots\\\
1&1&\cdots&1&\cdots&a_{n-1}&b_{n-1,n}\\\ 1&1&\cdots&1&\cdots&1&a_{n}\\\
\end{pmatrix}\right)$
$\displaystyle~{}+(-1)^{n}\left(\otimes\begin{pmatrix}a_{n}\varepsilon(b_{0,n})a_{0}&b_{1,n}b_{0,1}&\cdots&b_{n-2,n}b_{0,n-2}&b_{n-1,n}b_{0,n-1}\\\
1&a_{1}&\cdots&b_{1,n-2}&b_{1,n-1}\\\ \vdots&\vdots&\ddots&\vdots&\vdots\\\
1&1&\cdots&a_{n-2}&b_{n-2,n-1}\\\ 1&1&\cdots&1&a_{n-1}\\\
\end{pmatrix}\right).$

It was again shown in [13] that $\partial_{n-1}\circ\partial_{n}=0$, and that
the resulting complex is denoted by
$\overline{\mathbf{C}}_{\bullet}(A,B,\varepsilon)$.

######  Definition 2.2.

([13]) The homology of the chain complex
$\overline{\mathbf{C}}_{\bullet}(A,B,\varepsilon)$ is called the secondary
Hochschild homology associated to the triple $(A,B,\varepsilon)$, and this is
denoted by $HH_{*}(A,B,\varepsilon)$.

######  Remark 2.3.

([13]) By taking $B=\mathbbm{k}$, notice that the secondary Hochschild
homology associated to the triple $(A,B,\varepsilon)$ reduces to the usual
Hochschild homology associated to $A$. In notation, one has that
$HH_{n}(A,\mathbbm{k},\varepsilon)=HH_{n}(A)$ for all $n\geq 0$.

Since it will be used later in the paper, it may be helpful to view the
complex $\overline{\mathbf{C}}_{\bullet}(A,B,\varepsilon)$ in low dimension.
In particular, we have

$\ldots\xrightarrow{~{}\partial_{3}~{}}A^{\otimes 3}\otimes B^{\otimes
3}\xrightarrow{~{}\partial_{2}~{}}A^{\otimes 2}\otimes
B\xrightarrow{~{}\partial_{1}~{}}A\longrightarrow 0$

with

$\partial_{1}\Big{(}\otimes\begin{pmatrix}a&\alpha\\\ 1&b\\\
\end{pmatrix}\Big{)}=a\varepsilon(\alpha)b-b\varepsilon(\alpha)a,$

and

$\partial_{2}\Big{(}\otimes\begin{pmatrix}a&\alpha&\beta\\\ 1&b&\gamma\\\
1&1&c\\\
\end{pmatrix}\Big{)}=\otimes\begin{pmatrix}a\varepsilon(\alpha)b&\beta\gamma\\\
1&c\\\ \end{pmatrix}-\otimes\begin{pmatrix}a&\alpha\beta\\\
1&b\varepsilon(\gamma)c\\\
\end{pmatrix}+\otimes\begin{pmatrix}c\varepsilon(\beta)a&\alpha\gamma\\\
1&b\\\ \end{pmatrix}.$

######  Example 2.4.

([13]) For any triple $(A,B,\varepsilon)$, we have
$HH_{0}(A,B,\varepsilon)=HH_{0}(A)=\frac{A}{[A,A]}$.

###  2.2. The secondary cyclic homology

Connes introduced cyclic cohomology in [6], extensive details of which can be
found in [14]. Here we recall the analogous secondary homological version that
was established in [13]. We start by considering the permutation
$\lambda=(0,1,2,\ldots,n)$ and the cyclic group
$C_{n+1}=\langle\lambda\rangle$. Notice that $C_{n+1}$ has a natural action on
$\overline{C}_{n}(A,B,\varepsilon)$ given by

$\lambda\Big{(}\otimes\begin{pmatrix}a_{0}&b_{0,1}&\cdots&b_{0,n-1}&b_{0,n}\\\
1&a_{1}&\cdots&b_{1,n-1}&b_{1,n}\\\ \vdots&\vdots&\ddots&\vdots&\vdots\\\
1&1&\cdots&a_{n-1}&b_{n-1,n}\\\ 1&1&\cdots&1&a_{n}\\\
\end{pmatrix}\Big{)}=(-1)^{n}\left(\otimes\begin{pmatrix}a_{n}&b_{0,n}&\cdots&b_{n-2,n}&b_{n-1,n}\\\
1&a_{0}&\cdots&b_{0,n-2}&b_{0,n-1}\\\ \vdots&\vdots&\ddots&\vdots&\vdots\\\
1&1&\cdots&a_{n-2}&b_{n-2,n-1}\\\ 1&1&\cdots&1&a_{n-1}\\\
\end{pmatrix}\right).$

Consider the new complex built by setting
$\overline{C}_{n}^{\lambda}(A,B,\varepsilon)=\overline{C}_{n}(A,B,\varepsilon)/\operatorname{Im}(1-\lambda)$,
while continuing to use the maps $\partial_{n}$ from Section 2.1. It was
verified in [13] that $\partial_{n}$ maintains the necessary structure.

######  Definition 2.5.

([13]) The homology of the chain complex
$\overline{\mathbf{C}}_{\bullet}^{\lambda}(A,B,\varepsilon)$ is called the
secondary cyclic homology associated to the triple $(A,B,\varepsilon)$, and
this is denoted by $HC_{*}(A,B,\varepsilon)$.

######  Remark 2.6.

([13]) As before, taking $B=\mathbbm{k}$, we have that the secondary cyclic
homology associated to the triple $(A,B,\varepsilon)$ reduces to the usual
cyclic homology associated to $A$. That is,
$HC_{n}(A,\mathbbm{k},\varepsilon)=HC_{n}(A)$ for all $n\geq 0$.

######  Example 2.7.

([13]) For any triple $(A,B,\varepsilon)$, we have that
$HC_{0}(A,B,\varepsilon)=HH_{0}(A,B,\varepsilon)$.

Connes famously established a long exact sequence bearing his name that
connected Hochschild homology with its cyclic counterpart. In [13] the
analogue of Connes’ long exact sequence for the secondary case was produced:

######  Theorem 2.8.

_([13])_ Let $\mathbbm{k}$ be a field of characteristic zero. For a triple
$(A,B,\varepsilon)$, we have the long exact sequence

$\ldots\xrightarrow{B_{*}}HH_{n}(A,B,\varepsilon)\xrightarrow{I_{*}}HC_{n}(A,B,\varepsilon)\xrightarrow{S_{*}}HC_{n-2}(A,B,\varepsilon)\xrightarrow{B_{*}}HH_{n-1}(A,B,\varepsilon)\xrightarrow{I_{*}}\ldots$

###  2.3. The kernel of a multiplication map

For a commutative algebra $A$, set $I$ to be the kernel of the multiplication
map from $A\otimes A$ to $A$. That is, consider $m:A\otimes A\longrightarrow
A$ given by $m(a\otimes b)=ab$, and then set $I=\operatorname{Ker}(m)$. One
can show that $I$ is generated by elements of the form $1\otimes a-a\otimes
1$, where $a\in A$, and that the $A$-bimodule $I/I^{2}$ is $A$-symmetric.
Next, recalling Kähler differentials $\Omega_{A|\mathbbm{k}}^{1}$, we have the
following classic result which unites our concepts of Hochschild homology,
Kähler differentials, and the set $I$ (see [17], for instance):

######  Theorem 2.9.

For a commutative algebra $A$, we have that

$HH_{1}(A)\cong\Omega_{A|\mathbbm{k}}^{1}\cong I/I^{2}.$

##  3\. Secondary Differentials

The main objective of this paper is to get a result like Theorem 2.9 that
corresponds to the secondary case. The goal of this section is to establish
the former isomorphism.

######  Definition 3.1.

For a commutative triple $\mathcal{T}=(A,B,\varepsilon)$, denote
$\Omega_{\mathcal{T}|\mathbbm{k}}^{1}$ to be the left $B\otimes A$-module of
secondary Kähler differentials generated by the $\mathbbm{k}$-linear symbols
$d(\alpha\otimes a)$ for $\alpha\in B$ and $a\in A$ with the module structure
of $(\alpha\otimes a)\cdot d(\beta\otimes
b)=a\varepsilon(\alpha)d(\beta\otimes b)$, along with the relations

  1. (i)

$d(\lambda(\alpha\otimes a)+\mu(\beta\otimes b))=\lambda d(\alpha\otimes
a)+\mu d(\beta\otimes b)$,

  2. (ii)

$d((\alpha\otimes a)(\beta\otimes b))=a\varepsilon(\alpha)d(\beta\otimes
b)+b\varepsilon(\beta)d(\alpha\otimes a)$, and

  3. (iii)

$d(\alpha\otimes 1)+d(\alpha\otimes 1)=d(1\otimes\varepsilon(\alpha))$.

for all $a,b\in A$, $\alpha,\beta\in B$, and $\lambda,\mu\in\mathbbm{k}$.

We note that of course $(\alpha\otimes a)(\beta\otimes b)=\alpha\beta\otimes
ab$. Furthermore, notice that as consequence of the above, one can write

$\displaystyle d(\alpha\beta\otimes ab)$
$\displaystyle=a\varepsilon(\alpha)d(\beta\otimes
b)+b\varepsilon(\beta)d(\alpha\otimes a),$ $\displaystyle d(1\otimes ab)$
$\displaystyle=ad(1\otimes b)+bd(1\otimes a),$ $\displaystyle
d(\alpha\beta\otimes 1)$ $\displaystyle=\varepsilon(\alpha)d(\beta\otimes
1)+\varepsilon(\beta)d(\alpha\otimes 1),\text{and}$ $\displaystyle
d(\alpha\otimes a)$ $\displaystyle=\varepsilon(\alpha)d(1\otimes
a)+ad(\alpha\otimes 1).$

######  Remark 3.2.

Observe that $d(1\otimes 1)=0$, and therefore $d(\lambda\otimes\mu)=0$ for all
$\lambda,\mu\in\mathbbm{k}$. We also note that secondary Kähler differentials
are in fact Kähler differentials of $B\otimes A$ as a commutative algebra (by
way of (i) and (ii)), with the addition of (iii).

######  Remark 3.3.

By taking $B=\mathbbm{k}$, notice that secondary Kähler differentials become
the usual Kähler differentials $\Omega_{A|\mathbbm{k}}^{1}$. More
specifically, (iii) gets suppressed in Definition 3.1 with the identification
of $B=\mathbbm{k}$.

######  Proposition 3.4.

For a commutative triple $\mathcal{T}=(A,B,\varepsilon)$, we have that

$HH_{1}(A,B,\varepsilon)\cong\Omega_{\mathcal{T}|\mathbbm{k}}^{1}.$

###### Proof.

Since $A$ is commutative, the map $\partial_{1}:A^{\otimes 2}\otimes
B\longrightarrow A$ is trivial. Therefore we have that
$HH_{1}(A,B,\varepsilon)$ is the quotient of $A^{\otimes 2}\otimes B$ by the
relation

(3.1) $\otimes\begin{pmatrix}ab\varepsilon(\alpha)&\beta\gamma\\\ 1&c\\\
\end{pmatrix}-\otimes\begin{pmatrix}a&\alpha\beta\\\
1&bc\varepsilon(\gamma)\\\
\end{pmatrix}+\otimes\begin{pmatrix}ca\varepsilon(\beta)&\alpha\gamma\\\
1&b\\\ \end{pmatrix}=0.$

The map
$HH_{1}(A,B,\varepsilon)\longrightarrow\Omega_{\mathcal{T}|\mathbbm{k}}^{1}$
sends the class of $\otimes\begin{pmatrix}a&\alpha\\\ 1&b\\\ \end{pmatrix}$ to
$ad(\alpha\otimes b)$. This map is well-defined because (3.1) maps to

$\displaystyle ab\varepsilon(\alpha)$ $\displaystyle d(\beta\gamma\otimes
c)-ad(\alpha\beta\otimes
bc\varepsilon(\gamma))+ca\varepsilon(\beta)d(\alpha\gamma\otimes b)$
$\displaystyle=ab\varepsilon(\alpha)\varepsilon(\beta\gamma)d(1\otimes
c)+ab\varepsilon(\alpha)cd(\beta\gamma\otimes
1)-a\varepsilon(\alpha\beta)d(1\otimes bc\varepsilon(\gamma))$
$\displaystyle~{}-abc\varepsilon(\gamma)d(\alpha\beta\otimes
1)+ca\varepsilon(\beta)\varepsilon(\alpha\gamma)d(1\otimes
b)+ca\varepsilon(\beta)bd(\alpha\gamma\otimes 1)$
$\displaystyle=ab\varepsilon(\alpha)\varepsilon(\beta\gamma)d(1\otimes
c)+ab\varepsilon(\alpha)c\varepsilon(\beta)d(\gamma\otimes
1)+ab\varepsilon(\alpha)c\varepsilon(\gamma)d(\beta\otimes 1)$
$\displaystyle~{}-a\varepsilon(\alpha\beta)bcd(1\otimes\varepsilon(\gamma))-a\varepsilon(\alpha\beta)b\varepsilon(\gamma)d(1\otimes
c)-a\varepsilon(\alpha\beta)c\varepsilon(\gamma)d(1\otimes b)$
$\displaystyle~{}-abc\varepsilon(\gamma)\varepsilon(\alpha)d(\beta\otimes
1)-abc\varepsilon(\gamma)\varepsilon(\beta)d(\alpha\otimes
1)+ca\varepsilon(\beta)\varepsilon(\alpha\gamma)d(1\otimes b)$
$\displaystyle~{}+ca\varepsilon(\beta)b\varepsilon(\alpha)d(\gamma\otimes
1)+ca\varepsilon(\beta)b\varepsilon(\gamma)d(\alpha\otimes 1)$
$\displaystyle=ab\varepsilon(\alpha)c\varepsilon(\beta)d(\gamma\otimes
1)-a\varepsilon(\alpha\beta)bcd(1\otimes\varepsilon(\gamma))+ca\varepsilon(\beta)b\varepsilon(\alpha)d(\gamma\otimes
1)$ $\displaystyle=0.$

In the other direction, the map
$\Omega_{\mathcal{T}|\mathbbm{k}}^{1}\longrightarrow HH_{1}(A,B,\varepsilon)$
sends $ad(\alpha\otimes b)$ to the class of $\otimes\begin{pmatrix}a&\alpha\\\
1&b\\\ \end{pmatrix}$, which is indeed a cycle because
$ab\varepsilon(\alpha)-ba\varepsilon(\alpha)=0$, relying on $A$ being
commutative. This is in fact a module homomorphism and we will now verify it
respects the relations from Definition 3.1.

Observe that (i) is immediate.

For (ii), notice that $d(\alpha\beta\otimes
bc)-b\varepsilon(\alpha)d(\beta\otimes c)-c\varepsilon(\beta)d(\alpha\otimes
b)$ maps to

$\otimes\begin{pmatrix}1&\alpha\beta\\\ 1&bc\\\
\end{pmatrix}-\otimes\begin{pmatrix}b\varepsilon(\alpha)&\beta\\\ 1&c\\\
\end{pmatrix}-\otimes\begin{pmatrix}c\varepsilon(\beta)&\alpha\\\ 1&b\\\
\end{pmatrix},$

which is $0$ by taking (3.1) with $\gamma=1_{B}$ and $a=1_{A}$.

Finally for (iii), we note that $d(\gamma\otimes 1)+d(\gamma\otimes
1)-d(1\otimes\varepsilon(\gamma))$ maps to

$\otimes\begin{pmatrix}1&\gamma\\\ 1&1\\\
\end{pmatrix}+\otimes\begin{pmatrix}1&\gamma\\\ 1&1\\\
\end{pmatrix}-\otimes\begin{pmatrix}1&1\\\ 1&\varepsilon(\gamma)\\\
\end{pmatrix},$

which is $0$ from (3.1) when we take $\alpha=\beta=1_{B}$ and $a=b=c=1_{A}$.

To complete the proof, we simply note that the two maps defined above are
inverses of each other. That is,

$\otimes\begin{pmatrix}a&\alpha\\\ 1&b\\\ \end{pmatrix}\longmapsto
ad(\alpha\otimes b)\longmapsto\otimes\begin{pmatrix}a&\alpha\\\ 1&b\\\
\end{pmatrix},$

and

$ad(\alpha\otimes b)\longmapsto\otimes\begin{pmatrix}a&\alpha\\\ 1&b\\\
\end{pmatrix}\longmapsto ad(\alpha\otimes b).$

The isomorphism follows. ∎

######  Corollary 3.5.

For a commutative triple $\mathcal{T}=(A,B,\varepsilon)$, we have that

$HC_{1}(A,B,\varepsilon)\cong\frac{\Omega_{\mathcal{T}|\mathbbm{k}}^{1}}{d(1\otimes
A)}.$

###### Proof.

First note that $HC_{0}(A,B,\varepsilon)=HH_{0}(A,B,\varepsilon)$ by Example
2.7, and then by Example 2.4 and since $A$ is commutative, we have that
$HH_{0}(A,B,\varepsilon)=\frac{A}{[A,A]}=A$. Hence
$HC_{0}(A,B,\varepsilon)=A$. Therefore, employing Theorem 2.8 in low
dimension, we get that

$\ldots\longrightarrow
A\xrightarrow{~{}B_{*}~{}}\Omega_{\mathcal{T}|\mathbbm{k}}^{1}\xrightarrow{~{}I_{*}~{}}HC_{1}(A,B,\varepsilon)\longrightarrow
0.$

Here $B:A\longrightarrow A^{\otimes 2}\otimes B$ is defined by

$B(a)=\otimes\begin{pmatrix}1&1\\\ 1&a\\\
\end{pmatrix}+\otimes\begin{pmatrix}a&1\\\ 1&1\\\ \end{pmatrix}.$

Then by exactness,

$HC_{1}(A,B,\varepsilon)=\operatorname{Im}(I_{*})\cong\frac{\Omega_{\mathcal{T}|\mathbbm{k}}^{1}}{\operatorname{Ker}(I_{*})}=\frac{\Omega_{\mathcal{T}|\mathbbm{k}}^{1}}{\operatorname{Im}(B_{*})}.$

Finally we observe that the image of the map $B$ has the form $d(1\otimes
a)+ad(1\otimes 1)=d(1\otimes a)$, and this is exactly $d(1\otimes A)$, which
was what we wanted. ∎

######  Remark 3.6.

Again, taking $B=\mathbbm{k}$, we see that both Proposition 3.4 and Corollary
3.5 reduce to their well-known results in the usual case. That is,
$HH_{1}(A)\cong\Omega_{A|\mathbbm{k}}^{1}$ and
$HC_{1}(A)\cong\Omega_{A|\mathbbm{k}}^{1}/d(A)$, respectively.

##  4\. The Kernel of a Multiplication Map

In this section, we establish the latter isomorphism corresponding to Theorem
2.9 for the secondary case.

For a commutative triple $(A,B,\varepsilon)$, set $J$ to be the kernel of the
multiplication map from $A^{\otimes 2}\otimes B$ to $A$. Explicitly, consider
$m:A^{\otimes 2}\otimes B\longrightarrow A$ given by
$m\Big{(}\otimes\begin{pmatrix}a&\alpha\\\ 1&b\\\
\end{pmatrix}\Big{)}=ab\varepsilon(\alpha)$, and set
$J=\operatorname{Ker}(m)$. It’s straightforward to see that $J$ is generated
by elements of the form

$\otimes\begin{pmatrix}1&\alpha\\\ 1&a\\\
\end{pmatrix}-\otimes\begin{pmatrix}a\varepsilon(\alpha)&1\\\ 1&1\\\
\end{pmatrix}.$

Notice that when $B=\mathbbm{k}$ we get that $J$ reduces to $I$, as described
in Section 2.3.

In order to capture the desired isomorphism, we must also introduce an
additional set. Define $\hat{J}$ to be the set consisting of all finite sums
of elements of the form

$\otimes\begin{pmatrix}1&\alpha\\\ 1&1\\\
\end{pmatrix}-\otimes\begin{pmatrix}\varepsilon(\alpha)&1\\\ 1&1\\\
\end{pmatrix}+\otimes\begin{pmatrix}1&\alpha\\\ 1&1\\\
\end{pmatrix}-\otimes\begin{pmatrix}1&1\\\ 1&1\varepsilon(\alpha)\\\
\end{pmatrix},$

where $\alpha\in B$. One can check that $\hat{J}$ is both an $A$-bimodule, as
well as a submodule of $J$.

######  Remark 4.1.

When $B=\mathbbm{k}$, observe that $\hat{J}=\\{0\\}\subseteq J^{2}=I^{2}$.

One can also see that the $A$-bimodule $J/(J^{2}+\hat{J})$ is $A$-symmetric.
That is, the two module structures agree. We are now ready for our sought-
after isomorphism:

######  Proposition 4.2.

For a commutative triple $\mathcal{T}=(A,B,\varepsilon)$, we have that

$\Omega_{\mathcal{T}|\mathbbm{k}}^{1}\cong J/(J^{2}+\hat{J}).$

###### Proof.

We first define the map $\Omega_{\mathcal{T}|\mathbbm{k}}^{1}\longrightarrow
J/(J^{2}+\hat{J})$ by sending $ad(\alpha\otimes b)$ to the class of
$\otimes\begin{pmatrix}a&\alpha\\\ 1&b\\\
\end{pmatrix}-\otimes\begin{pmatrix}a\varepsilon(\alpha)b&1\\\ 1&1\\\
\end{pmatrix}$. Notice that this element is indeed in $J$. Next we aim to show
that the map is well-defined and respects the relations found in Definition
3.1.

Of course, (i) is clear.

For (ii), notice that $d((\alpha\beta\otimes
ab))-a\varepsilon(\alpha)d(\beta\otimes b)-b\varepsilon(\beta)d(\alpha\otimes
a)$ maps to

$\displaystyle\otimes$ $\displaystyle\begin{pmatrix}1&\alpha\beta\\\ 1&ab\\\
\end{pmatrix}-\otimes\begin{pmatrix}ab\varepsilon(\alpha\beta)&1\\\ 1&1\\\
\end{pmatrix}-\otimes\begin{pmatrix}a\varepsilon(\alpha)&\beta\\\ 1&b\\\
\end{pmatrix}+\otimes\begin{pmatrix}ab\varepsilon(\alpha\beta)&1\\\ 1&1\\\
\end{pmatrix}$
$\displaystyle~{}-\otimes\begin{pmatrix}b\varepsilon(\beta)&\alpha\\\ 1&a\\\
\end{pmatrix}+\otimes\begin{pmatrix}ab\varepsilon(\alpha\beta)&1\\\ 1&1\\\
\end{pmatrix}$ $\displaystyle=\otimes\begin{pmatrix}1&\alpha\beta\\\ 1&ab\\\
\end{pmatrix}-\otimes\begin{pmatrix}a\varepsilon(\alpha)&\beta\\\ 1&b\\\
\end{pmatrix}-\otimes\begin{pmatrix}b\varepsilon(\beta)&\alpha\\\ 1&a\\\
\end{pmatrix}+\otimes\begin{pmatrix}ab\varepsilon(\alpha\beta)&1\\\ 1&1\\\
\end{pmatrix}\in J^{2}\subseteq J^{2}+\hat{J}.$

Finally, $d(\alpha\otimes 1)+d(\alpha\otimes
1)-d(1\otimes\varepsilon(\alpha))$ maps to

$\displaystyle\otimes$ $\displaystyle\begin{pmatrix}1&\alpha\\\ 1&1\\\
\end{pmatrix}-\otimes\begin{pmatrix}\varepsilon(\alpha)&1\\\ 1&1\\\
\end{pmatrix}+\otimes\begin{pmatrix}1&\alpha\\\ 1&1\\\
\end{pmatrix}-\otimes\begin{pmatrix}\varepsilon(\alpha)&1\\\ 1&1\\\
\end{pmatrix}-\otimes\begin{pmatrix}1&1\\\ 1&\varepsilon(\alpha)\\\
\end{pmatrix}+\otimes\begin{pmatrix}\varepsilon(\alpha)&1\\\ 1&1\\\
\end{pmatrix}$ $\displaystyle=\otimes\begin{pmatrix}1&\alpha\\\ 1&1\\\
\end{pmatrix}-\otimes\begin{pmatrix}\varepsilon(\alpha)&1\\\ 1&1\\\
\end{pmatrix}+\otimes\begin{pmatrix}1&\alpha\\\ 1&1\\\
\end{pmatrix}-\otimes\begin{pmatrix}1&1\\\ 1&\varepsilon(\alpha)\\\
\end{pmatrix}\in\hat{J}\subseteq J^{2}+\hat{J},$

and hence (iii) is satisfied.

Next we define the map
$J/(J^{2}+\hat{J})\longrightarrow\Omega_{\mathcal{T}|\mathbbm{k}}^{1}$ by
sending the class of $\otimes\begin{pmatrix}1&\alpha\\\ 1&a\\\
\end{pmatrix}-\otimes\begin{pmatrix}a\varepsilon(\alpha)&1\\\ 1&1\\\
\end{pmatrix}$ to $1d(\alpha\otimes a)-a\varepsilon(\alpha)d(1\otimes
1)=d(\alpha\otimes a)$. Notice this is well-defined because elements in
$J^{2}$, which are of the form

$\displaystyle\left(\otimes\begin{pmatrix}1&\alpha\\\ 1&a\\\
\end{pmatrix}-\otimes\begin{pmatrix}a\varepsilon(\alpha)&1\\\ 1&1\\\
\end{pmatrix}\right)\left(\otimes\begin{pmatrix}1&\beta\\\ 1&b\\\
\end{pmatrix}-\otimes\begin{pmatrix}b\varepsilon(\beta)&1\\\ 1&1\\\
\end{pmatrix}\right)$ $\displaystyle=\otimes\begin{pmatrix}1&\alpha\beta\\\
1&ab\\\ \end{pmatrix}-\otimes\begin{pmatrix}b\varepsilon(\beta)&\alpha\\\
1&a\\\ \end{pmatrix}-\otimes\begin{pmatrix}a\varepsilon(\alpha)&\beta\\\
1&b\\\
\end{pmatrix}+\otimes\begin{pmatrix}a\varepsilon(\alpha)b\varepsilon(\beta)&1\\\
1&1\\\ \end{pmatrix}$

map to

$d(\alpha\beta\otimes ab)-b\varepsilon(\beta)d(\alpha\otimes
a)-a\varepsilon(\alpha)d(\beta\otimes
b)+a\varepsilon(\alpha)b\varepsilon(\beta)d(1\otimes 1)=0$

by (ii) from Definition 3.1 and Remark 3.2. Likewise, elements in $\hat{J}$,
which are of the form

$\otimes\begin{pmatrix}1&\alpha\\\ 1&1\\\
\end{pmatrix}-\otimes\begin{pmatrix}\varepsilon(\alpha)&1\\\ 1&1\\\
\end{pmatrix}+\otimes\begin{pmatrix}1&\alpha\\\ 1&1\\\
\end{pmatrix}-\otimes\begin{pmatrix}1&1\\\ 1&1\varepsilon(\alpha)\\\
\end{pmatrix}$

map to

$d(\alpha\otimes 1)-\varepsilon(\alpha)d(1\otimes 1)+d(\alpha\otimes
1)-d(1\otimes\varepsilon(\alpha))=0$

by (iii) from Definition 3.1 and Remark 3.2.

To conclude, we notice that these maps are inverses of each other. That is,

$\otimes\begin{pmatrix}1&\alpha\\\ 1&a\\\
\end{pmatrix}-\otimes\begin{pmatrix}a\varepsilon(\alpha)&1\\\ 1&1\\\
\end{pmatrix}\longmapsto d(\alpha\otimes
a)\longmapsto\otimes\begin{pmatrix}1&\alpha\\\ 1&a\\\
\end{pmatrix}-\otimes\begin{pmatrix}a\varepsilon(\alpha)&1\\\ 1&1\\\
\end{pmatrix},$

and

$ad(\alpha\otimes b)\longmapsto\otimes\begin{pmatrix}a&\alpha\\\ 1&b\\\
\end{pmatrix}-\otimes\begin{pmatrix}a\varepsilon(\alpha)b&1\\\ 1&1\\\
\end{pmatrix}\longmapsto ad(\alpha\otimes b).$

The isomorphism follows. ∎

We can now state our main theorem in full:

######  Theorem 4.3.

For a commutative triple $\mathcal{T}=(A,B,\varepsilon)$, we have that

$HH_{1}(A,B,\varepsilon)\cong\Omega_{\mathcal{T}|\mathbbm{k}}^{1}\cong
J/(J^{2}+\hat{J}).$

###### Proof.

Unite Proposition 3.4 and Proposition 4.2. ∎

######  Remark 4.4.

Taking $B=\mathbbm{k}$, we get that Theorem 4.3 becomes Theorem 2.9.

##  5\. Future Work

There are rich properties involving $\Omega_{A|k}^{1}$ and $I/I^{2}$, such as
the second fundamental exact sequence for $\Omega$. It is reasonable to ask if
$\Omega_{\mathcal{T}|\mathbbm{k}}^{1}$ and $J/(J^{2}+\hat{J})$ also fit into
some exact sequence. Furthermore, one of the main applications of Hochschild
homology is that it leads to de Rham cohomology. It is appropriate that the
secondary Hochschild homology could lead to a secondary de Rham cohomology,
which could have interesting geometrical consequences and results. Ultimately,
one wonders if we could derive an analogue of the Hochschild-Kostant-Rosenberg
Theorem.

Finally, with the study of secondary Hochschild homology and secondary Kähler
differentials, a natural follow-up is to investigate the secondary Hochschild
cohomology and generalized derivations. We plan to delve into this more in the
future.

## References

  * [1] Mamta Balodi, Abhishek Banerjee, and Anita Naolekar. BV-operators and the secondary Hochschild complex. _C. R. Math. Acad. Sci. Paris_ , 358(11-12):1239–1258, 2020.
  * [2] Samuel Carolus, Samuel A. Hokamp, and Jacob Laubacher. Deformation theories controlled by Hochschild cohomologies. _São Paulo J. Math. Sci._ , 14(2):481–495, 2020.
  * [3] Samuel Carolus and Jacob Laubacher. Simplicial structures over the 3-sphere and generalized higher order Hochschild homology. _Categ. Gen. Algebr. Struct. Appl._ , 15(1):93–143, 2021.
  * [4] Samuel Carolus, Jacob Laubacher, and Mihai D. Staic. A simplicial construction for noncommutative settings. _Homology Homotopy Appl._ , 23(1):49–60, 2021.
  * [5] Samuel Carolus and Mihai D. Staic. $G$-algebra structure on the higher order Hochschild cohomology $H_{S^{2}}^{*}(A,A)$. _Algebra Colloq._ , 29(1):113–124, 2022.
  * [6] Alain Connes. Noncommutative differential geometry. _Inst. Hautes Études Sci. Publ. Math._ , (62):257–360, 1985.
  * [7] Bruce R. Corrigan-Salter and Mihai D. Staic. Higher-order and secondary Hochschild cohomology. _C. R. Math. Acad. Sci. Paris_ , 354(11):1049–1054, 2016.
  * [8] Apurba Das, Satyendra Kumar Mishra, and Anita Naolekar. Noncommutative differential calculus structure on secondary Hochschild (co)homology. _Comm. Algebra_ , 50(6):2349–2365, 2022.
  * [9] Murray Gerstenhaber. On the deformation of rings and algebras. _Ann. of Math. (2)_ , 79:59–103, 1964.
  * [10] Gerhard Hochschild. On the cohomology groups of an associative algebra. _Ann. of Math. (2)_ , 46:58–67, 1945.
  * [11] Jacob Laubacher. _Secondary Hochschild and cyclic (co)homologies_. OhioLINK Electronic Theses and Dissertations Center, Columbus, Ohio, 2017. Dissertation (Ph.D.)–Bowling Green State University.
  * [12] Jacob Laubacher. Properties of the secondary Hochschild homology. _Algebra Colloq._ , 25(2):225–242, 2018.
  * [13] Jacob Laubacher, Mihai D. Staic, and Alin Stancu. Bar simplicial modules and secondary cyclic (co)homology. _J. Noncommut. Geom._ , 12(3):865–887, 2018.
  * [14] Jean-Louis Loday. _Cyclic homology_ , volume 301 of _Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences]_. Springer-Verlag, Berlin, second edition, 1998.
  * [15] Mihai D. Staic. Secondary Hochschild cohomology. _Algebr. Represent. Theory_ , 19(1):47–56, 2016.
  * [16] Mihai D. Staic and Alin Stancu. Operations on the secondary Hochschild cohomology. _Homology Homotopy Appl._ , 17(1):129–146, 2015.
  * [17] Charles A. Weibel. _An introduction to homological algebra_ , volume 38 of _Cambridge Studies in Advanced Mathematics_. Cambridge University Press, Cambridge, 1994.


# An entanglement witness and an effective criterion for bipartite states of
different dimensions

Vahid Jannesary111email<EMAIL_ADDRESS>Department of Physics,
Sharif University of Technology, Tehran 14588, Iran Vahid Karimipour222email:
<EMAIL_ADDRESS>Department of Physics, Sharif University of Technology,
Tehran 14588, Iran

###### Abstract

We present a simple approach for generation of a diverse set of positive maps
between spaces of different dimensions. The proposed method enables the
construction of Entanglement Witnesses tailored for systems in $d_{1}\times
d_{2}$ dimensions. We then show how application of this entanglement witness
to a general density matrix leads to a simple calculable criterion in closed
form. As a support for the power of this criterion, we apply it to a variety
of (one-parameter) PPT entangled states and show that the ranges of the
parameters where these states are entangled are much larger than previously
reported.

## Introduction

Entanglement stands out as one of the most remarkable distinctions between the
quantum and classical realms, playing a pivotal role in numerous quantum
protocols, for review papers see [1, 2, 3]. As a result, the creation,
manipulation, application, and identification of entanglement have profound
significance. The identification of entangled states, in particular, presents
a formidable challenge, underscored by its classification as an NP-Hard
problem in mathematical terms [4]. Hence, the pursuit of viable techniques for
detecting entangled states takes on paramount importance.

Entangled states are referred to as states that are not separable, and
separable states refer to states that can be written as follows[5]:

$\rho=\sum p_{i}\rho_{i}^{(1)}\otimes\rho_{i}^{(2)},$ (1)

where $0\leq p_{i}\leq 1$ and $\sum p_{i}=1$. One strategy for identifying
entanglement involves investigating whether a state can be expressed in the
aforementioned form. This, however, poses a formidable challenge. As a result,
there is a pressing need for more effective methods for the detection of
entanglement. Over time, diverse techniques have emerged for this purpose. The
Peres criterion is considered one of the primary methods in this regard[6]:
It’s relatively straightforward to see that the partial transpose of all
separable states is positive. This is seen simply by noting that for any state
of the form (1), one has

$(I\otimes T)(\rho)\geq 0,$ (2)

where $T$ is the transpose, hence the name partial transpose. Therefore, if
the partial transpose of a quantum state is not positive, it must be
entangled. However, this criterion alone cannot identify all types of quantum
entangled states because there are entangled states known as PPT, whose
partial transpose are positive. In fact separable states are a subset of PPT
states and only when $d_{1}d_{2}\leq 6$, these two sets are equal. Here
$d_{1}$ and $d_{2}$ are the dimensions of the two quantum systems. The
effectiveness of the Peres criterion (2) hinges on a basic property of the
transpose map $T$ which is a positive and yet not completely positive map.
Therefore it was natural to search for more general types of positive maps
lacking the property of complete positivity and thus constructing more general
Entanglement Witnesses than the simple partial transpose. In fact, it has been
known that any entanglement witness $W$ can be written in the form

$W(M)=(I\otimes M)|\phi^{+}\rangle\langle\phi^{+}|,$

where $M$ is a positive map and
$|\phi^{+}\rangle=\frac{1}{\sqrt{d}}\sum_{\mu=0}^{d-1}|\mu\mu\rangle$, is a
maximally entangled state[7]. In other words, $W(M)$ is the Choi-matrix of the
map $M$, which due to its lack of complete positivity can have negative
eigenvalue. Such a witness has the property

$\operatorname{Tr}(W\sigma)\geq 0,\quad\forall\sigma\in S,$

where $S$ is the set of all separable states. Alternatively, one can think of
$W$ as a block-positive operator, i.e.

$\langle\psi,\phi|W|\psi,\phi\rangle,\ \forall\ \psi\in H_{1}\ ,\forall\
\phi\in H_{2},$

which has negative eigenvalue when acting on the whole Hilbert space
$H_{1}\otimes H_{2}$. Thus the problem of constructing entanglement witnesses
becomes equivalent to constructing positive but not completely positive maps.
Based on this equivalence, entanglement witnesses were introduced in many
works, see [7, 8, 9, 10] and references therein. For more recent works, see
[11, 12, 13]. In practical situations, where the two particles are far apart
in remote laboratories, application of any entanglement witnesses found so
far, requires a common reference frame between the two parties holding the
particles. It was recently shown in [14] that by a process called incomplete
teleportation, all these entangled states are still able to detect
entanglement of remote parties, even in the absence of a common frame of
reference.

In contrast with completely positive maps which due to the Kraus theorem can
be easily constructed, construction of positive maps is not a straightforward
task, and appropriate methods need to be proposed for this purpose. In
addition to the finding new classes of witnesses, the relation of EW’s and
positive maps, have led to many new insights on both [8]. For example, a basic
distinction refers to decomposable witnesses which are of the form

$W=P+Q^{\Gamma},$

where $P,Q$ are positive operators and $\Gamma$ is the operator of partial
transposition. Any witness that cannot be written in the above form is called
indecomposable. It can be proved that decomposable witnesses cannot identify
PPT entangled states. In this sense these are somewhat weaker than
indecomposable witnesses. One can also characterize witnesses by their
optimality and extremality [8], a subject which we will not deal with in this
work. The line that we follow is inspired by a number of recent works [15, 16,
17, 18, 19, 20] in which specific types of positive maps have been defined and
the corresponding witnesses have been constructed.

It is worth noting that in almost all the mentioned articles, the entanglement
witnesses constructed have been designed for two systems of equal dimensions.
In this paper, we propose a simple method for identification of entanglement
in systems with dimensions of $d_{1}\times d_{2}$. This dimension-specific
applicability enhances the versatility of the method, making it a valuable
tool in different practical scenarios. Often, due to practical constraints in
laboratory setups, certain measurements may be more accessible or accurate
than others. Our method elegantly accommodates this by allowing for the
construction of entanglement witnesses based solely on those restricted
measurements which are feasible in the given experimental context(See
examples).

The structure of this paper is as follows: In Section 1, we examine the main
idea of this article, which is to construct a new class of positive maps. In
Section 2, an alternative method is utilized to obtain the same results that
presented in Section 1. In section 3 we provide a simple criterion built with
our Entanglement Witness which easily detects entangled states. In Section 4,
we present several examples which shows the power of our criterion in
detecting PPT entangled states. We end up the paper with a conclusion.

##  1 Construction of positive maps for arbitrary different dimensions

Consider the Hilbert space $H_{d_{1}}$ of dimension $d_{1}$ and let a pure
state $X\in D(H_{d_{1}})$, (the space of states on $H_{d_{1}}$) be written as

$X=\frac{1}{d_{1}}I_{d_{1}}+\sum_{i=1}^{d_{1}^{2}-1}x_{i}\Gamma_{i},$

where $\\{\Gamma_{i},\ \ \ i=1\cdots d_{1}^{2}-1\\}$ are a set of orthonormal
traceless Hermitian operators $\in L(H_{d_{1}})$, i.e.

$\operatorname{Tr}(\Gamma_{i}\Gamma_{j})=\delta_{ij},\ \ \
\operatorname{Tr}\Gamma_{i}=0.$

Purity of the state $X$ (i.e. $\operatorname{Tr}X^{2}=1$) and the above
normalization conditions restricts the norm of the vector ${\bf
x}=(x_{1},\cdots x_{d_{1}^{2}-1})$ to

${\bf x}\cdot{\bf x}=\frac{d_{1}-1}{d_{1}},$ (3)

where ${\bf x}\cdot{\bf x}:=\sum_{i=1}^{d_{1}^{2}-1}x_{i}^{2}$.

Let $\Phi:D(H_{d_{1}})\longrightarrow D(H_{d_{2}})$ be unital and trace-
preserving map which acts as follows:

$\Phi(\frac{1}{d_{1}}I_{d_{1}})=\frac{1}{d_{2}}I_{d_{2}}\ \ \ \ ,\ \ \
\Phi(\Gamma_{i})=\frac{1}{d_{2}}\sum_{k=1}^{d_{2}^{2}-1}\Lambda_{ik}\Omega_{k}.$
(4)

In fact, $\Lambda_{ik}$ are real numbers representing the characterization of
the map $\Phi$. Here, our definition remains consistent with the trace-
preserving property of the map, in contrast to the definition presented in
[18], which is grounded in the property of such a map that takes an identity
matrix in $D(H_{d_{1}})$ to a identity matrix in $D(H_{d_{2}})$. The factor
$\frac{1}{d_{2}}$ in the second equation is written for later convenience, and
$\\{\Omega_{k},\ \ \ k=1\cdots d_{2}^{2}-1\\}$ are a set of orthonormal
traceless Hermitian operators $\in L(H_{d_{2}})$, i.e.

$\operatorname{Tr}(\Omega_{k}\Omega_{l})=\delta_{kl},\ \ \
\operatorname{Tr}\Omega_{k}=0.$

Then $\Phi$ maps $X$ to

$\Phi(X)=\frac{1}{d_{2}}I_{d_{2}}+\frac{1}{d_{2}}\sum_{i=1}^{d_{1}^{2}-1}\sum_{k=1}^{d_{2}^{2}-1}x_{i}\Lambda_{ik}\Omega_{k}.$
(5)

###### Remark 1

Up to now we have assumed that $X$ is a pure state. This can be extended to
arbitrary operatores $X\in L(H_{d_{1}})$ as follows:

$\Phi(X)=\frac{\operatorname{Tr}(X)}{d_{2}}I_{d_{2}}+\frac{1}{d_{2}}\sum_{i=1}^{d_{1}^{2}-1}\sum_{k=1}^{d_{2}^{2}-1}x_{i}\Lambda_{ik}\Omega_{k}.$

To ensure positivity of the map $\Phi$, we now use Mehta’s Lemma [21] which we
state below:

Mehta’s Lamma (1989): Let $A$ be a Hermitian matrix of dimension $D$, if

$\operatorname{Tr}(A^{2})\leq\frac{(\operatorname{Tr}A)^{2}}{D-1},$ (6)

then $A$ is positive.

In view of linearity of the map $\Phi$, to prove positivity of the map $\Phi$,
it is sufficient to prove that $\Phi(X)$ is positive for any pure state $X$.
That is the map $\Phi$ is positive if the following condition holds

$\operatorname{Tr}(\Phi(X)^{2})\leq\frac{(\operatorname{Tr}\Phi(X))^{2}}{d_{2}-1}=\frac{1}{d_{2}-1}.$
(7)

To this end, we calculate $\operatorname{Tr}(\Phi(X)^{2})$, which in view of
orthonormality condition (4), turns out to be

$\operatorname{Tr}(\Phi(X)^{2})=\frac{1}{d_{2}}+\frac{1}{d_{2}^{2}}{\bf
x}^{T}{\Lambda}\Lambda^{T}{\bf x},$

where the matrix $\Lambda$ is composed of $\Lambda_{ij}$ elements. In view of
(6), we demand that

$\frac{1}{d_{2}}+\frac{1}{d_{2}^{2}}{\bf x}^{T}{\Lambda}\Lambda^{T}{\bf
x}\leq\frac{1}{d_{2}-1},\hskip 28.45274pt\forall{\bf x}.$

We can now replace this with an even stronger inequality, namely

$\frac{1}{d_{2}}+\frac{1}{d_{2}^{2}}\lambda_{max}(\Lambda{\Lambda^{T}}){\bf
x}\cdot{\bf x}\leq\frac{1}{d_{2}-1},$ (8)

where $\lambda_{max}:=\lambda_{max}(\Lambda\Lambda^{T})$, is the largest
eigenvalue of the matrix $\Lambda\Lambda^{T}$, or equivalently $\lambda_{max}$
is the square of the largest singular value of the matrix $\Lambda^{T}$.
Inserting (8) in (7) and using (3), yields the following constraint on
$\lambda_{max}$

$\lambda_{max}\leq\frac{d_{1}d_{2}}{(d_{1}-1)(d_{2}-1)}.$ (9)

The EW corresponding to the map (5) is given by [7]

$W=(I_{d_{1}}\otimes\Phi)|\beta\rangle\langle\beta|,$

where $|\beta\rangle=\frac{1}{\sqrt{d}}\sum_{\mu=0}^{d_{1}-1}|\mu,\mu\rangle$
is a maximally entangled state. To do this calculation, we extend the
orthonormal sets $\\{\Gamma_{i}\\}$ and $\\{\Omega_{k}\\}$ respectively to
orthonormal bases for the space of matrices $M_{d_{1}}$ and $M_{d_{2}}$,
namely to
$\\{\Gamma_{\mu}\\}=\\{\Gamma_{0}=\frac{I}{\sqrt{d_{1}}},\Gamma_{i}\\}$ and
$\\{\Omega_{\mu}\\}=\\{\Omega_{0}=\frac{I}{\sqrt{d_{2}}},\Omega_{i}\\}$ and
taking remark 1 into account, we write

$\displaystyle W$ $\displaystyle=$
$\displaystyle\frac{1}{d_{1}}\sum_{\mu,\nu}|\mu\rangle\langle\nu|\otimes\Phi(|\mu\rangle\langle\nu|)$
(10) $\displaystyle=$
$\displaystyle\frac{1}{d_{1}}\sum_{\mu,\nu}|\mu\rangle\langle\nu|\otimes\sum_{\alpha=0}^{d_{1}^{2}-1}\sum_{\beta=0}^{d_{2}^{2}-1}\Lambda_{\beta\alpha}\operatorname{Tr}(|\mu\rangle\langle\nu|\Gamma_{\beta})\Omega_{\alpha}$
$\displaystyle=$
$\displaystyle\frac{1}{d_{1}}\sum_{\alpha,\beta}\Lambda_{\beta\alpha}\Gamma^{T}_{\beta}\otimes\Omega_{\alpha}.$
(11)

where $\Lambda_{00}=\frac{1}{d_{2}},\Lambda_{0i}=0,\Lambda_{i0}=0$. After
multiplication by an overall factor $d_{1}d_{2}$, this leads to the following
witness

$W=I_{d_{1}}\otimes
I_{d_{2}}+\sum_{i=1}^{d_{1}^{2}-1}\sum_{k=1}^{d_{2}^{2}-1}{\Lambda}_{ik}\Gamma^{T}_{i}\otimes\Omega_{k}.$

Since the set $\Gamma^{T}_{i}$ is also an orthonormal set, we can rename it to
$\Gamma_{i}$. Thus, the following matrix can be a entanglement witness too:

$W=I_{d_{1}}\otimes
I_{d_{2}}+\sum_{i=1}^{d_{1}^{2}-1}\sum_{k=1}^{d_{2}^{2}-1}{\Lambda}_{ik}\Gamma_{i}\otimes\Omega_{k}.$
(12)

Therefore for any two dimensions, any operator of the above form can be an
entanglement witness, provided that the condition (9) holds. That condition in
fact proves Block positivity of this Witness. Whether or not a $W$ constructed
in this way, has also a negative eigenvalue depends on the matrix $\Lambda$
that we choose. In all previous papers that were mentioned in the
introduction, $\Lambda$ should be orthogonal. These comprise a much smaller
class compared to the $\Lambda$ matrices in this paper. Moreover, finding
orthogonal matrices in high dimensions is not a straightforward task. In this
paper, one can use any matrix with any properties to construct entanglement
witnesses, as long as the largest eigenvalue of $\Lambda\Lambda^{T}$ satisfies
eq. (9).

In the next section, we derive the same result of eq. (12) in an alternative
way by focusing on the EW itself, rather than the positive map which leads to
it via its Choi-matrix. In view of this alternative construction, one can
first choose the local operators (according to the experimental restrictions)
and then tuning the coefficients for defining the final entanglement witness.
This is a significant advantage compared to methods where a positive map may
lead to a witness which requires non-feasible measurements.

##  2 An alternative construction of Entanglement Witnesses

In this section we directly apply Mehta’s Lemma to a form of EW which we take
as follows:

$W=I_{d_{1}}\otimes
I_{d_{2}}+\sum_{i=1}^{d_{1}^{2}-1}\sum_{j=1}^{d_{2}^{2}-1}\Lambda_{ij}\Gamma_{i}\otimes\Omega_{j}.$

We demand that $W$ be block-positive. This means that for any vector
$|v\rangle\in H_{d_{2}}$, the following matrix should be positive

$B\equiv\langle
v|W|v\rangle=I_{d_{1}}+\sum_{i=1}^{d_{1}^{2}-1}\sum_{j=1}^{d_{2}^{2}-1}\Lambda_{ij}\Gamma_{i}\langle
v|\Omega_{j}|v\rangle.$

Now we use the following:

$\operatorname{Tr}(B)=d_{1},$

and

$\operatorname{Tr}(B^{2})=d_{1}+\sum_{i,j=1}^{d_{2}^{2}-1}(\Lambda^{T}\Lambda)_{ij}\langle
v|\Omega_{i}|v\rangle\langle v|\Omega_{j}|v\rangle,$

where we have used the relations $\operatorname{Tr}(\Gamma_{i})=0,\ \ \
\operatorname{Tr}(\Gamma_{i}\Gamma_{j})=\delta_{ij}$. We now resort to the
Mehta’s Lemma and demand that the following condition hold for all the vectors
$|v\rangle$

$\operatorname{Tr}(B^{2})\leq\frac{\operatorname{Tr}(B)^{2}}{d_{1}-1}.$

This leads to

$d_{1}+\sum_{i,j=1}^{d_{2}^{2}-1}(\Lambda^{T}\Lambda)_{ij}\langle
v|\Omega_{i}|v\rangle\langle
v|\Omega_{j}|v\rangle\leq\frac{d_{1}^{2}}{d_{1}-1}\hskip 28.45274pt\forall\
v.$

Defining the vector ${\bf q}$ with components $q_{i}:=\langle
v|\Omega_{i}|v\rangle$, this can be written as

$d_{1}+{\bf q}^{T}\Lambda\Lambda^{T}{\bf q}\leq\frac{d_{1}^{2}}{d_{1}-1}\hskip
28.45274pt\forall\ v,$

or in a stronger form

$d_{1}+\lambda_{max}(\Lambda^{T}\Lambda){\bf q}.{\bf
q}\leq\frac{d_{1}^{2}}{d_{1}-1}.$ (13)

It remains to calculate ${\bf q}.{\bf q}$, and this is easily found by noting
that

${\bf q}.{\bf q}=\sum_{i=1}^{d_{2}^{2}-1}\langle v|\Omega_{i}|v\rangle\langle
v|\Omega_{i}|v\rangle=\sum_{i=1}^{d_{2}^{2}-1}\langle
v,v|\Omega_{i}\otimes\Omega_{i}|v,v\rangle.$

Using the identity

$\frac{1}{d}I+\sum_{i}\Omega_{i}\otimes\Omega_{i}=P,$ (14)

where $P$ is the permutation operator, we find that

${\bf q}.{\bf q}=\langle v,v|P-\frac{1}{d_{2}}I|v,v\rangle=1-\frac{1}{d_{2}}.$

Inserting this in (13), leads to the final form for the upper bound of
$\lambda_{max}$, namely

$\lambda_{max}(\Lambda^{T}\Lambda)\leq\frac{d_{1}d_{2}}{(d_{1}-1)(d_{2}-1)},$
(15)

which is identical to the bound obtained by the positive map in section 1.
However we should also ensure that $W$ is not trivially positive, that is we
should remove the possibility that all its eigenvalues be positive. We again
use Mehta’s lemma. To this end we use

$\operatorname{Tr}(W)=d_{1}d_{2},$

and

$\operatorname{Tr}(W^{2})=d_{1}d_{2}+\operatorname{Tr}(\Lambda^{T}\Lambda),$

and ensure that the following inequality be satisfied

$\operatorname{Tr}(W^{2})>\frac{\operatorname{Tr}(W)^{2}}{d_{1}d_{2}-1},$

or

$\frac{d_{1}d_{2}}{d_{1}d_{2}-1}<\operatorname{Tr}(\Lambda^{T}\Lambda).$ (16)

Combining (15) and (16), we arrive at the final condition for the parameters
of the matrix $\Lambda$, in order that $W$ can be an entanglement witness:

$\frac{d_{1}d_{2}}{d_{1}d_{2}-1}<\operatorname{Tr}(\Lambda^{T}\Lambda),\ \ \ \
{\rm and}\ \ \ \lambda_{max}\leq\frac{d_{1}d_{2}}{(d_{1}-1)(d_{2}-1)}.$ (17)

Note that the second condition is sufficient for the block-positivity of $W$,
while the first condition is necessary for $W$, ensuring that it is not a
positive matrix.

##  3 A simple criterion

Consider now the form of the witness in equation (12) subject to the condition

$\lambda_{max}\leq\frac{d_{1}d_{2}}{(d_{1}-1)(d_{2}-1)},$ (18)

which is sufficient for its block-positivity. Let an arbitrary bi-partite
state be expressed as follows in an orthogonal trace-less basis:

$\rho=\frac{1}{d_{1}d_{2}}I_{d_{1}}\otimes
I_{d_{2}}+\sum_{i=1}^{d_{1}^{2}-1}{a}_{i}\Gamma_{i}\otimes
I_{d_{2}}+\sum_{j=1}^{d_{2}^{2}-1}{b}_{j}I_{d_{1}}\otimes\Omega_{j}+\sum_{i=1}^{d_{1}^{2}-1}\sum_{j=1}^{d_{2}^{2}-1}{r}_{ij}\Gamma_{i}\otimes\Omega_{j}.$

It is readily seen that

$\operatorname{Tr}(W\rho)=1+\sum_{ij}\Lambda_{ij}r_{ij}=1+\operatorname{Tr}(R\Lambda^{T}),$
(19)

where the matrix $R$ is composed of $r_{ij}$ elements. In order to make
$\operatorname{Tr}(W\rho)<0$, which at the same time proves that $W$ is indeed
an entanglement witness (i.e. has a negative eigenvalue) and detects the
entanglement of $\rho$, we try to choose $\Lambda$ in such a way to minimize
$tr(R\Lambda^{T})$, while maintaining the condition (18). To achieve this, we
first express the matrix $R$ using singular value decomposition:

$R=USV.$

Here we have used the fact that $R$ is a real matrix, which implies the
matrices $U\in M_{d_{1}\times d_{1}}$ and $V\in M_{d_{2}\times d_{2}}$ are
real orthogonal matrices. The matrix $S\in M_{d_{1}\times d_{2}}$ is a
rectangular diagonal matrix with non-negative entries on the main diagonal.
Now we choose the matrix $\Lambda^{T}$ as follows:

$\Lambda^{T}=V^{T}LU^{T},$

where $L\in M_{d_{2}\times d_{1}}$ is chosen to be a rectangular diagonal
matrix whose entries on its main diagonal are equal to
$l_{i}=-\sqrt{\frac{d_{1}d_{2}}{(d_{1}-1)(d_{2}-1)}}$. This leads to

$\Lambda^{T}\Lambda=V^{T}(LL^{T})V,$

where $LL^{T}\in M_{d_{2}\times d_{2}}$ is a matrix whose diagonal elements
are either zero or equal to $\frac{d_{1}d_{2}}{(d_{1}-1)(d_{2}-1)}$. This
means that the condition (9) on the maximum eigenvalue of $\Lambda^{T}\Lambda$
is satisfied. We then find that

$\operatorname{Tr}(\Lambda^{T}R)=\operatorname{Tr}(SL)=-\sqrt{\frac{d_{1}d_{2}}{(d_{1}-1)(d_{2}-1)}}\operatorname{Tr}(S).$

In this way we have obtained the lowest value of
$\operatorname{Tr}(\Lambda^{T}R)$, while maintaining the condition (18) of
block-positivity of $W$. Therefore, if this value is less than $-1$, which in
view of (19) is equivalent to $\operatorname{Tr}(W\rho)<0$, we conclude that
the state $\rho$ is entangled. In addition, one can show that
$Tr(S)=Tr(\sqrt{RR^{T}})$. Therefore, we have the following simple criterion:

${\rm If\ \
}\gamma:=\operatorname{Tr}(S)>\sqrt{\frac{(d_{1}-1)(d_{2}-1)}{d_{1}d_{2}}},\ \
\ {\rm the\ state\ is\ entangled.}$ (20)

The condition that we derived does not require finding the elements of the
matrix $\Lambda$, and only requires recasting the density matrix in terms of a
basis chosen by orthogonal trace-less matrices, and calculating the singular
values of the matrix $R$.

It should be mentioned that the singular values of $R$ and hence the value of
$\gamma$ do not depend on the bases which we choose for the two parts for
expanding the density matrix $\rho$, as long as these bases are orthonormal
and traceless.

Remark: Interestingly, this criterion is a special case of Theorem 1 in [22],
where the authors show that this is a powerful criterion and unifies several
criteria known before.

We will now show by several examples that this criterion, originated from the
EW defined in (12), is indeed powerful enough to detect entanglement of a
class of PPT entangled states which has not been previously detected.

##  4 Examples

In this section, we consider a few examples and show the power of our
criterion. In all the following examples, except example 4.2, the states are
PPT entangled states reported in previous works. We will show that the
criterion given in (20), can detect entanglement in a larger interval of the
parameters previously reported, and it also can detect PPT entangled states
with different dimensions, i.e. $d_{1}\neq d_{2}$. For checking the following
examples, the reader should note that we have used the following order for
Gellmann matrices:

$\displaystyle\textrm{for }\ \ \ \ \ \ 1\leq i<\frac{1}{2}d(d-1):\hskip
28.45274pt\hskip 28.45274pt\ \Gamma_{i}\in\Gamma_{x},$
$\displaystyle\textrm{for }\ \ \ \ \ \ \frac{1}{2}d(d-1)\leq i<d(d-1):\ \ \ \
\ \ \Gamma_{i}\in\Gamma_{y},$ (21) $\displaystyle\textrm{for }\ \ \ \ \ \
d(d-1)\leq i<d^{2}:\hskip 28.45274pt\hskip 28.45274pt\ \
\Gamma_{i}\in\Gamma_{z},$ (22)

where for $1\leq m<n\leq d$,

$\Gamma_{x}=\big{\\{}\ \ \frac{1}{\sqrt{2}}(|m\rangle\langle
n|+|n\rangle\langle m|)\ \big{\\}},$ $\Gamma_{y}=\big{\\{}\ \
\frac{i}{\sqrt{2}}(|m\rangle\langle n|-|n\rangle\langle m|)\ \ \big{\\}},$

and for $1\leq n\leq d-1$

$\Gamma_{z}=\big{\\{}\frac{1}{\ \
\sqrt{n(n+1)}}(\sum_{m=0}^{n-1}|m\rangle\langle m|-n|n\rangle\langle n|)\ \
\big{\\}}.$

Moreover in all those cases where the state is a PPT state and its partial
transpose equals to itself, i.e. $\rho=\rho^{\Gamma}$, we can use the anti-
symmetricity of the Gell-Mann matrices in $\Gamma_{y}$ to simplify the
calculations in the following way:

Let at least one of the matrices $\Gamma_{i}$ or $\Omega_{j}$ (say
$\Gamma_{i}$) be anti-symmetric, then we have

$r_{ij}=\operatorname{Tr}(\rho(\Gamma_{i}\otimes\Omega_{j}))=\operatorname{Tr}(\rho^{\Gamma}(\Gamma_{i}\otimes\Omega_{j}))=\operatorname{Tr}(\rho.(\Gamma_{i}^{T}\otimes\Omega_{j}))=-\operatorname{Tr}(\rho(\Gamma_{i}\otimes\Omega_{j}))=0.$

The same is true when $\Omega_{j}$ is anti-symmetric.

###  4.1 Example 1:

In this example, we demonstrate that the method explained in this paper can
identify PPT entangled states, and thus, it can generate indecomposable EWs.
It has been shown in [23] that the following state is PPT entangled state when
$x\neq 1$ (for $x=1$ this is a separable state):

$\rho=\frac{1}{N}\left(\begin{array}[]{ccc|ccc|ccc}1&.&.&.&1&.&.&.&1\\\
.&\frac{1}{x}&.&.&.&.&.&.&.\\\ .&.&x&.&.&.&.&.&.\\\
\hline\cr.&.&.&x&.&.&.&.&.\\\ 1&.&.&.&1&.&.&.&1\\\
.&.&.&.&.&\frac{1}{x}&.&.&.\\\ \hline\cr.&.&.&.&.&.&\frac{1}{x}&.&.\\\
.&.&.&.&.&.&.&x&.\\\ 1&.&.&.&1&.&.&.&1\\\ \end{array}\right).$ (23)

Here dots represent zeros, $N=3+x+x^{-1}$ is the normalization factor and
$x>0$. The above state can be written in the basis of normalized Gell-Mann
matrices as follows:

$\displaystyle\rho_{x}=\frac{1}{9}I\otimes I+\frac{1}{N}$
$\displaystyle\Big{[}$
$\displaystyle\Gamma_{1}\otimes\Gamma_{1}+\Gamma_{2}\otimes\Gamma_{2}+\Gamma_{3}\otimes\Gamma_{3}-\Gamma_{4}\otimes\Gamma_{4}-\Gamma_{5}\otimes\Gamma_{5}-\Gamma_{6}\otimes\Gamma_{6}$
(24) $\displaystyle+$
$\displaystyle\left(1-\frac{\xi}{2}\right)(\Gamma_{7}\otimes\Gamma_{7}+\Gamma_{8}\otimes\Gamma_{8})-\frac{\sqrt{3}}{2}\eta(\Gamma_{7}\otimes\Gamma_{8}-\Gamma_{8}\otimes\Gamma_{7})\Big{]},$

where $\xi=x+x^{-1}$ and $\eta=x-x^{-1}$. The eigenvalues of the matrix $R$
are easily found to be

$\lambda(R)=\frac{1}{N}\\{1(3),-1(3),(1-\frac{\xi}{2})\pm
i\frac{\sqrt{3}}{2}\eta\\},$

where the numbers in parentheses indicate the multiplicity of each singular
value. Thus, taking into account the normalization factor $N$, we find

$Tr(S)=\frac{6+2\sqrt{(1-\frac{\xi}{2})^{2}+\frac{3}{4}\eta^{2}}}{3(1+\xi)}.$

Condition (20) for entanglement of the state now amounts to

$\frac{1}{4}(\xi^{2}-\eta^{2})<\xi-1,$

or after simplification (inserting $\xi=x+x^{-1}$ and $\eta=x-x^{{}_{1}}$)
leads to the following condition for the state (23) to be entangled:

$2<x+x^{-1}.$

This shows that the state $\rho_{x}$ is entangled for all values of $x$,
except for $x=1$, which coincides with the result of [23].

###  4.2 Example 2:

This example demonstrates that our method can identify the entanglement of all
NPT states in the following form in any dimension $d$:

$\rho=p|\psi\rangle\langle\psi|+(1-p)\frac{I}{d},\quad|\psi\rangle=\frac{1}{\sqrt{d}}\sum_{i=0}^{d-1}|i,i\rangle.$

To calculate $\gamma$ in equation (20), we need to obtain the coefficients
$r_{ij}$:

$r_{ij}=p\operatorname{Tr}[\frac{1}{d}(\sum_{k,l}|kk\rangle\langle
ll|).(\Gamma_{i}\otimes\Gamma_{j})]=\frac{p}{d}\operatorname{Tr}(\Gamma_{i}\Gamma_{j}^{T})=\pm\frac{p}{d}\delta_{ij}.$

Here we have noticed that the expression
$\operatorname{Tr}(\Gamma_{i}\Gamma_{j}^{T})$ equals $\delta_{ij}$ when
$\Gamma_{j}^{T}=\Gamma_{j}^{\dagger}$ and equals $-\delta_{ij}$ when
$\Gamma_{j}^{\dagger}=-\Gamma_{j}^{T}$. As it is evident, this state is
diagonal in the Gell-Mann matrix basis, and the absolute value of all its
singular values is equal to $\frac{p}{d}$, and the number of its singular
values is also equal to $d^{2}-1$. Therefore, we find for this state

$\gamma=\operatorname{Tr}(S)=\frac{p}{d}(d^{2}-1),$

which, according to eq (20), should be greater than $\frac{d-1}{d}$ so the
state is entangled. Thus the range of $p$ for the state to be entangled is
$p>\frac{1}{d+1}$. This is the same value obtained from the Peres criterion
which in this case follows from

$\rho^{\Gamma}=(1-p)\frac{I}{d^{2}}+p\frac{1}{d}F,$

where $F$ is the swap operator. The eigenstates are the symmetric and
antisymmetric states and the smallest eigenvalue, namely
$\frac{1-p}{d^{2}}-\frac{p}{d}$, is negative when $p>\frac{1}{d+1}$ and hence
the state is entangled. For this example the two methods give equivalent
results. This result has already been reported in [24].

###  4.3 Example 3

In this example, we provide a PPT entangled state which can be detected by the
criterion in equation (20). To construct a PPT entangled state we define the
following Unextendible Product Basis in $C_{3}\otimes C_{4}:$

$\displaystyle|\phi_{1}\rangle=\frac{1}{\sqrt{2}}|0\rangle|1-2\rangle,\quad|\phi_{2}\rangle=\frac{1}{\sqrt{6}}|0\rangle(2|0\rangle-|1+2\rangle),\quad|\phi_{3}\rangle=\frac{1}{\sqrt{2}}|1\rangle|1-2\rangle,\quad$
$\displaystyle|\phi_{4}\rangle=\frac{1}{\sqrt{2}}|2\rangle|1-2\rangle,\quad|\phi_{5}\rangle=\frac{1}{\sqrt{6}}|2\rangle(|1+2\rangle-2|3\rangle),\quad|\phi_{6}\rangle=\frac{1}{\sqrt{2}}|1-2\rangle|0\rangle,\quad$
(25)
$\displaystyle|\phi_{7}\rangle=\frac{1}{\sqrt{2}}|0-1\rangle|3\rangle,\quad|\phi_{8}\rangle=\frac{1}{\sqrt{12}}|0+1+2\rangle|0+1+2+3\rangle,$
(26)

and construct the PPT entangled state
$\rho=I-\sum_{i}|\phi_{i}\rangle\langle\phi_{i}|$ [8], whose explicit form is
given by

$\rho=\frac{1}{48}\left(\begin{array}[]{c|c|c|c}3\
\epsilon&-\epsilon&-\epsilon&-\epsilon\\\
\hline\cr-\epsilon&D&D^{\prime}&-\epsilon\\\ \hline\cr-\epsilon&D^{\prime
T}&D&-\epsilon\\\ \hline\cr-\epsilon&-\epsilon&-\epsilon&3\ \epsilon\\\
\end{array}\right),$

where $\epsilon$ is the 3 dimensional square matrix all of its elements are
equal to 1, and

$D=\left(\begin{array}[]{ccc}5&-1&-1\\\ -1&5&-1\\\
-1&-1&5\end{array}\right),\quad
D^{\prime}=\left(\begin{array}[]{ccc}-1&5&-1\\\ -1&-1&5\\\
5&-1&-1\end{array}\right).$

The matrix $R$ is simply obtained by noting that its elements are given by
$r_{mn}=\operatorname{Tr}(\rho(\Gamma_{m}\otimes\Omega_{n}))=-\sum_{i}\langle\phi_{i}|\Gamma_{m}\otimes\Omega_{n}|\phi_{i}\rangle$.
The result is

$R=\frac{1}{24}\left(\begin{array}[]{c|c|c}-\epsilon_{3\times 6}&0_{3\times
6}&A_{3\times 3}\\\ \hline\cr 0_{3\times 6}&0_{3\times 6}&0_{3\times 3}\\\
\hline\cr B_{2\times 6}&0_{2\times 6}&C_{2\times 3}\end{array}\right),$

where $\epsilon_{3\times 6}$ is a $3\times 6$ matrix all of whose elements are
1, and similarly ${0}_{3\times 6}$ is a matrix all of whose elements are zero.
Also

$\displaystyle A$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccc}0&0&-3\sqrt{\frac{3}{2}}\\\ 0&0&0\\\
3&\sqrt{3}&\sqrt{\frac{3}{2}}\end{array}\right),$ (30) $\displaystyle B$
$\displaystyle=$
$\displaystyle\left(\begin{array}[]{cccccc}{2}&{2}&-1&0&0&0\\\
\frac{2}{\sqrt{3}}&\frac{2}{\sqrt{3}}&\frac{1}{\sqrt{3}}&0&\frac{-4}{\sqrt{3}}&\frac{-4}{\sqrt{3}}\end{array}\right),$
(33) $\displaystyle C$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccc}0&0&-\frac{1}{2}\sqrt{\frac{3}{2}}\\\
-\frac{1}{\sqrt{3}}&-\frac{1}{3}&-\frac{5}{6\sqrt{2}}\end{array}\right).$ (36)

The singular values of this matrix are obtained numerically and one finds that

$\gamma=\operatorname{Tr}(S)=1.429>\sqrt{\frac{3\times 4}{2\times
3}}=\sqrt{2},$

which shows that while $\rho$ is a PPT entangled state, it can be detected by
the criterion (20).

###  4.4 Example 4

In [23], among other things, the following family of PPT entangled states in
$C_{4}\times C_{4}$ are introduced

$\rho=\frac{1}{N}\left(\begin{array}[]{cccc|cccc|cccc|cccc}1&.&.&.&.&1&.&.&.&.&1&.&.&.&.&1\\\
.&\frac{1}{a}&.&.&.&.&.&.&.&.&.&.&.&.&1&.\\\
.&.&1&.&.&.&.&.&1&.&.&.&.&.&.&.\\\ .&.&.&a&.&.&1&.&.&.&.&.&.&.&.&.\\\
\hline\cr.&.&.&.&a&.&.&.&.&.&.&1&.&.&.&.\\\ 1&.&.&.&.&1&.&.&.&.&1&.&.&.&.&1\\\
.&.&.&1&.&.&\frac{1}{a}&.&.&.&.&.&.&.&.&.\\\
.&.&.&.&.&.&.&1&.&.&.&.&.&1&.&.\\\ \hline\cr.&.&1&.&.&.&.&.&1&.&.&.&.&.&.&.\\\
.&.&.&.&.&.&.&.&.&a&.&.&1&.&.&.\\\ 1&.&.&.&.&1&.&.&.&.&1&.&.&.&.&1\\\
.&.&.&.&1&.&.&.&.&.&.&\frac{1}{a}&.&.&.&.\\\
\hline\cr.&.&.&.&.&.&.&.&.&1&.&.&\frac{1}{a}&.&.&.\\\
.&.&.&.&.&.&.&1&.&.&.&.&.&1&.&.\\\ .&1&.&.&.&.&.&.&.&.&.&.&.&.&a&.\\\
1&.&.&.&.&1&.&.&.&.&1&.&.&.&.&1\\\ \end{array}\right),$ (37)

where $N=4a+\frac{4}{a}+8$, and it is shown that their entanglement can be
detected by an entanglement witness, called $W_{ext}$, which shows that these
states are entangled in the interval $0<a<1$. We now show that the interval of
entanglement of these states is in fact much larger and covers the whole
positive line $0<a<\infty$, except the point $a=1$. To show this, the $R$
matrix corresponding to (37) is calculated, which turns out to be

$R=\frac{1}{(a+1)^{2}}\big{(}A\oplus A^{\prime}\oplus B\big{)},$

where

$A=\left(\begin{array}[]{cccccc}\frac{a}{4}&.&.&.&.&\frac{a}{4}\\\
.&\frac{a}{2}&.&.&.&.\\\ .&.&\frac{a}{4}&\frac{a}{4}&.&.\\\
.&.&\frac{a}{4}&\frac{a}{4}&.&.\\\ .&.&.&.&\frac{a}{2}&.\\\
\frac{a}{4}&.&.&.&.&\frac{a}{4}\end{array}\right),\ \ \ \ \
A^{\prime}=\left(\begin{array}[]{cccccc}-\frac{a}{4}&.&.&.&.&\frac{a}{4}\\\
.&.&.&.&.&.\\\ .&.&-\frac{a}{4}&-\frac{a}{4}&.&.\\\
.&.&-\frac{a}{4}&-\frac{a}{4}&.&.\\\ .&.&.&.&.&.\\\
\frac{a}{4}&.&.&.&.&-\frac{a}{4}\end{array}\right),$

and

$B=\left(\begin{array}[]{ccc}-\frac{(a-1)^{2}}{8}&-\frac{a^{2}+2a-3}{8\sqrt{3}}&-\frac{(a-1)a}{2\sqrt{6}}\\\
\frac{3a^{2}-2a-1}{8\sqrt{3}}&-\frac{(a-1)^{2}}{24}&-\frac{a^{2}+a-2}{6\sqrt{2}}\\\
\frac{a-1}{2\sqrt{6}}&\frac{2a^{2}-a-1}{6\sqrt{2}}&-\frac{(a-1)^{2}}{12}\end{array}\right),$

with singular values given as

$\Big{\\{}\frac{\left|a-1\right|}{4\left|a+1\right|}\ (2),\
\frac{\left|a\right|}{2(a+1)^{2}}\ (6),\ \frac{(a-1)^{2}}{4(a+1)^{2}}\
(1)\Big{\\}}.$

This leads to

$\gamma=\frac{1}{4}\frac{(a-1)^{2}}{(a+1)^{2}}+3\frac{|a|}{(a+1)^{2}}+\frac{1}{2}\left|\frac{a-1}{a+1}\right|.$

Figure 1 shows that the value of $\gamma$ is always larger than
$\sqrt{\frac{(d_{1}-1)(d_{2}-1)}{d_{1}d_{2}}}=\frac{3}{4}$ except for $a=1$.

Figure 1:  Using the criterion intruduced in this paper, i.e. eq (17), we can
see that state (37) is entangled for $0<a<\infty;a\neq 1$, which is much
larger the range intruduced in [23].

Note that by $a<\infty$, we are emphasizing that $a$ should be a finite value.
In fact, in the limit $a\longrightarrow\infty$, the state approaches to

$\rho_{\infty}=\frac{1}{4}\Big{(}|03\rangle\langle 03|+|10\rangle\langle
10|+|11\rangle\langle 11|+|22\rangle\langle 22|\Big{)},$

which is obviously separable. Incidentally, this is the limiting case where
$\gamma$ approaches the value $\frac{3}{4}$.

###  4.5 Example 5

This example is a continuation of the previous example and demonstrates that
for the second PPT entangled state introduced in [23], the interval of
entanglement (indicated as $0<b<1$), is indeed much larger and as in the
previous example covers all positive and finite values of $b$ different from
$1$. The state is given by

$\rho=\frac{1}{N}\left(\begin{array}[]{cccc|cccc|cccc|cccc}1&.&.&.&.&1&.&.&.&.&1&.&.&.&.&1\\\
.&\frac{1}{b}&.&.&1&.&.&.&.&.&.&.&.&.&.&.\\\
.&.&1&.&.&.&.&.&1&.&.&.&.&.&.&.\\\ .&.&.&b&.&.&.&.&.&.&.&.&1&.&.&.\\\
\hline\cr.&1&.&.&b&.&.&.&.&.&.&.&.&.&.&.\\\ 1&.&.&.&.&1&.&.&.&.&1&.&.&.&.&1\\\
.&.&.&.&.&.&\frac{1}{b}&.&.&1&.&.&.&.&.&.\\\
.&.&.&.&.&.&.&1&.&.&.&.&.&1&.&.\\\ \hline\cr.&.&1&.&.&.&.&.&1&.&.&.&.&.&.&.\\\
.&.&.&.&.&.&1&.&.&b&.&.&.&.&.&.\\\ 1&.&.&.&.&1&.&.&.&.&1&.&.&.&.&1\\\
.&.&.&.&.&.&.&.&.&.&.&\frac{1}{b}&.&.&1&.\\\
\hline\cr.&.&.&1&.&.&.&.&.&.&.&.&\frac{1}{b}&.&.&.\\\
.&.&.&.&.&.&.&1&.&.&.&.&.&1&.&.\\\ .&.&.&.&.&.&.&.&.&.&.&1&.&.&b&.\\\
1&.&.&.&.&1&.&.&.&.&1&.&.&.&.&1\\\ \end{array}\right),$

where $N=4b+\frac{4}{b}+8$, and its corresponding matrix is

$R=\frac{1}{(1+b)^{2}}\left(\frac{b}{2}I_{6}\oplus 0_{6}\oplus A\right),$

where $I_{6}$ and $0_{6}$ are respectively the identity matrix and the zero
matrix in $C_{6}$ and

$A=\left(\begin{array}[]{ccc}-\frac{1}{8}(b-1)^{2}&-\frac{b^{2}+2b-3}{8\sqrt{3}}&-\frac{(b-1)b}{2\sqrt{6}}\\\
\frac{3b^{2}-2b-1}{8\sqrt{3}}&-\frac{1}{24}(b-1)^{2}&-\frac{b^{2}+b-2}{6\sqrt{2}}\\\
\frac{b-1}{2\sqrt{6}}&\frac{2b^{2}-b-1}{6\sqrt{2}}&-\frac{1}{12}(b-1)^{2}\\\
\end{array}\right),$

with singular values given as

$\Big{\\{}\frac{\left|b-1\right|}{4\left|b+1\right|}\ (2),\
\frac{\left|b\right|}{2(b+1)^{2}}\ (6),\ \frac{(b-1)^{2}}{4(b+1)^{2}}\
(1)\Big{\\}},$

which is the same as the previous example and gives the same result.

###  4.6 Example 6

In [25], Terhal introduced a family of UPB for $3\times n$ systems. Here we
take $n=4$ and show that our criterion can detect the PPT entangled state
built by them in these dimensions.

The UPB states for the $3\times 4$ system are:

$\displaystyle|\psi_{0}\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}|0\rangle\otimes(|1\rangle-|3\rangle),\ \ \ $
$\displaystyle|\psi_{1}\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}|1\rangle\otimes(|2\rangle-|3\rangle),\ \ \ $
(38) $\displaystyle|\psi_{2}\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}|2\rangle\otimes(|0\rangle-|3\rangle),\ \ \ $
(39) $\displaystyle|\psi_{3}\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}(|0\rangle-|1\rangle)\otimes|0\rangle,$ (40)
$\displaystyle|\psi_{4}\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}(|1\rangle-|2\rangle)\otimes|1\rangle,$ (41)
$\displaystyle|\psi_{5}\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}(|2\rangle-|0\rangle)\otimes|2\rangle,$ (42)
$\displaystyle|\psi_{6}\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{12}}\sum_{i=0}^{2}\sum_{j=0}^{3}|i\rangle\otimes|j\rangle.$
(43)

The PPT entangled state $\rho=I-\sum_{0}^{6}|\psi_{i}\rangle\langle\psi_{i}|$
turns out to be

$\displaystyle\rho=\frac{1}{60}\left(\begin{array}[]{ccc|ccc|ccc|ccc}5&-1&-1&-1&5&-1&-1&-1&-1&-1&-1&-1\\\
-1&5&-1&5&-1&-1&-1&-1&-1&-1&-1&-1\\\ -1&-1&5&-1&-1&-1&-1&-1&-1&-1&5&-1\\\
\hline\cr-1&5&-1&5&-1&-1&-1&-1&-1&-1&-1&-1\\\
5&-1&-1&-1&5&-1&-1&-1&-1&-1&-1&-1\\\ -1&-1&-1&-1&-1&5&-1&-1&-1&5&-1&-1\\\
\hline\cr-1&-1&-1&-1&-1&-1&5&5&-1&-1&-1&-1\\\
-1&-1&-1&-1&-1&-1&5&5&-1&-1&-1&-1\\\ -1&-1&-1&-1&-1&-1&-1&-1&5&-1&-1&5\\\
\hline\cr-1&-1&-1&-1&-1&5&-1&-1&-1&5&-1&-1\\\
-1&-1&5&-1&-1&-1&-1&-1&-1&-1&5&-1\\\ -1&-1&-1&-1&-1&-1&-1&-1&5&-1&-1&5\\\
\end{array}\right).$ (56)

The corresponding $R$ matrix is given by

$R=\frac{1}{30}\left(\begin{array}[]{c|c|c}-\epsilon_{3\times 6}&0_{3\times
6}&A\\\ \hline\cr 0_{3\times 6}&0_{3\times 6}&0_{3\times 3}\\\ \hline\cr
B&0_{2\times 6}&0_{2\times 3}\end{array}\right),$

where

$A=\left(\begin{array}[]{ccc}3&\sqrt{3}&\sqrt{\frac{3}{2}}\\\
0&-2\sqrt{3}&\sqrt{\frac{3}{2}}\\\
-3&-\sqrt{3}&\sqrt{\frac{3}{2}}\end{array}\right),$

and

$B=\left(\begin{array}[]{cccccc}0&0&0&0&3&-3\\\
0&0&0&-2\sqrt{3}&\sqrt{3}&\sqrt{3}\\\ \end{array}\right).$

The singular values of $R$ are found to be

$\left\\{\frac{1}{2\sqrt{10}},\frac{1}{5\sqrt{2}},\frac{1}{5\sqrt{2}},\frac{1}{5\sqrt{2}},\frac{1}{5\sqrt{2}}\right\\},$

and we find

$\gamma=\frac{1}{2\sqrt{10}}+\frac{4}{5\sqrt{2}}=\big{(}\frac{8+\sqrt{5}}{10}\big{)}\frac{1}{\sqrt{2}},$

which is greater than
$\sqrt{\frac{(d_{1}-1)(d_{2}-1)}{d_{1}d_{2}}}=\frac{1}{\sqrt{2}}$. This shows
that the PPT state (56) is entangled.

##  5 Conclusion

In this paper, a class of Entanglement Witnesses for two particle systems with
different dimensions was presented. Specifically we showed in section (3), how
the use of this class of EW’s can be reduced to a simple criterion, (equation
(20)), which can be checked on the form of the density matrix. We then showed
by checking several one-parameter family of PPT entangled states that this
criterion can detect their entanglement in a region of parameters much larger
than previously thought.

## References

  * [1] Yue Yu.  Advancements in applications of quantum entanglement.  Journal of Physics: Conference Series, 2012(1):012113, sep 2021\.
  * [2] Nanxi Zou.  Quantum entanglement and its application in quantum communication.  Journal of Physics: Conference Series, 1827(1):012120, mar 2021\.
  * [3] Aditi Sen.  Quantum entanglement and its applications.  Current Science, 112:1361, 2017.
  * [4] Leonid Gurvits.  Classical complexity and quantum entanglement.  Journal of Computer and System Sciences, 69(3):448–484, 2004.  Special Issue on STOC 2003.
  * [5] Reinhard F. Werner.  Quantum states with einstein-podolsky-rosen correlations admitting a hidden-variable model.  Phys. Rev. A, 40:4277–4281, Oct 1989.
  * [6] Asher Peres.  Separability criterion for density matrices.  Physical Review Letters, 77(8):1413–1415, aug 1996.
  * [7] Michał Horodecki, Paweł Horodecki, and Ryszard Horodecki.  Separability of mixed states: necessary and sufficient conditions.  Physics Letters A, 223(1-2):1–8, nov 1996.
  * [8] Dariusz Chruściński and Gniewomir Sarbicki.  Entanglement witnesses: construction, analysis and classification.  Journal of Physics A: Mathematical and Theoretical, 47(48):483001, nov 2014.
  * [9] Barbara M. Terhal.  Detecting quantum entanglement.  Theoretical Computer Science, 287(1):313–335, 2002.  Natural Computing.
  * [10] Otfried Gühne and Géza Tóth.  Entanglement detection.  Physics Reports, 474(1):1–75, 2009.
  * [11] Joonwoo Bae, Dariusz Chruściński, and Beatrix C. Hiesmayr.  Mirrored entanglement witnesses.  npj Quantum Information, 6(1):15, 2020.
  * [12] Bihalan Bhattacharya, Suchetana Goswami, Rounak Mundra, Nirman Ganguly, Indranil Chakrabarty, Samyadeb Bhattacharya, and A S Majumdar.  Generating and detecting bound entanglement in two-qutrits using a family of indecomposable positive maps.  Journal of Physics Communications, 5(6):065008, jun 2021.
  * [13] M. Weilenmann, B. Dive, D. Trillo, E. A. Aguilar, and M. Navascués.  Entanglement detection beyond measuring fidelities.  Phys. Rev. Lett., 124:200502, May 2020.
  * [14] Vahid Jannessary, Fatemeh Rezazadeh, Sadegh Raeisi, and Vahid Karimipour.  Witnessing entanglement of remote particles with incomplete teleportation.  Phys. Rev. A, 108:042421, Oct 2023.
  * [15] Katarzyna Siudzińska and Dariusz Chruściński.  Entanglement witnesses from mutually unbiased measurements.  Scientific Reports, 11(1):22988, 2021.
  * [16] Tao Li, Le-Min Lai, Deng-Feng Liang, Shao-Ming Fei, and Zhi-Xi Wang.  Entanglement witnesses based on symmetric informationally complete measurements.  International Journal of Theoretical Physics, 59(11):3549–3557, oct 2020.
  * [17] Kun Wang and Zhu-Jun Zheng.  Constructing entanglement witnesses from two mutually unbiased bases.  International Journal of Theoretical Physics, 60(1):274–283, 2021\.
  * [18] Dariusz Chruściński and Andrzej Kossakowski.  Spectral conditions for positive maps and entanglement witnesses.  Journal of Physics: Conference Series, 284(1):012017, mar 2011.
  * [19] Dariusz Chruściński, Gniewomir Sarbicki, and Filip Wudarski.  Entanglement witnesses from mutually unbiased bases.  Physical Review A, 97(3), mar 2018.
  * [20] Anindita Bera, Gniewomir Sarbicki, and Dariusz Chruściński.  A class of optimal positive maps in m.  Linear Algebra and its Applications, 668:131–148, jul 2023.
  * [21] M.L. Mehta.  Matrix Theory: Selected Topics and Useful Results.  Hindustan Publishing Corporation, 1989.
  * [22] Gniewomir Sarbicki, Giovanni Scala, and Dariusz Chruściński.  Family of multipartite separability criteria based on a correlation tensor.  Physical Review A, 101(1), January 2020.
  * [23] Joonwoo Bae, Anindita Bera, Dariusz Chruś ciński, Beatrix C Hiesmayr, and Daniel McNulty.  How many mutually unbiased bases are needed to detect bound entangled states?  Journal of Physics A: Mathematical and Theoretical, 55(50):505303, dec 2022.
  * [24] Julio I. de Vicente.  Separability criteria based on the bloch representation of density matrices, 2007.
  * [25] Barbara M. Terhal.  A family of indecomposable positive linear maps based on entangled quantum states, 2000.