# Error Basis and Quantum Channel

B. V. Rajarama Bhat  Indian Statistical Institute,8th Mile, Mysore Road, RVCE
Post, Bangalore 560 059, India<EMAIL_ADDRESS>,  Purbayan
Chakraborty  Laboratoire de mathématiques de Besançon, Université de Franche-
Comté, 16, route de Gray, F-25 030 Besançon cedex, France
<EMAIL_ADDRESS>and  Uwe Franz  Laboratoire de
mathématiques de Besançon, Université de Franche-Comté, 16, route de Gray,
F-25 030 Besançon cedex, France<EMAIL_ADDRESS>http://lmb.univ-
fcomte.fr/uwe-franz

###### Abstract.

The Weyl operators give a convenient basis of $M_{n}(\mathbb{C})$ which is
also orthonormal with respect to the Hilbert-Schmidt inner product. The
properties of such a basis can be generalised to the notion of a nice error
basis(NEB), as introduced by E. Knill [3]. We can use an NEB of
$M_{n}(\mathbb{C})$ to construct an NEB for $\mathrm{Lin}(M_{n}(\mathbb{C}))$,
the space of linear maps on $M_{n}(\mathbb{C})$. Any linear map will then
correspond to a $n^{2}\times n^{2}$ coefficient matrix in the basis
decomposition with respect to such an NEB of
$\mathrm{Lin}(M_{n}(\mathbb{C}))$. Positivity, complete (co)positivity or
other properties of a linear map can be characterised in terms of such a
coefficient matrix.

##  1\. Introduction

The Pauli matrices had the key role for modelling quantum error correction for
one qubit system. E. Knill generalised the properties of Pauli matrices in
higher dimension to introduce the notion of nice error basis in 1996 [3].
Since then it has played a fundamental important role in quantum error
correction. A nice error basis(NEB) gives a convenient orthonormal basis of
$M_{n}(\mathbb{C})$ constructed from a projective representation of a group
known as an index group. Knill in his paper showed that the existence of such
a basis of $M_{n}(\mathbb{C})$ is equivalent to the existence of an
irreducible character of a suitable group which vanishes outside its center.
The discrete Weyl relation [5, Equation (46)] $UV=qVU$, where $q$ is an $n$-th
root of unity, and its n-dimensional realisation

$Ue_{k}=e_{k+1\,{\rm mod}\,n},\quad Ve_{k}=q^{k}e_{k},\quad k=0,1,\ldots,n-1,$

with $\\{e_{0},\ldots,e_{n-1}\\}$ an orthonormal basis of $\mathbb{C}^{n}$,
were discussed by Hermann Weyl when he studied the equivalence between
Heisenberg’s matrix mechanics and Schrödinger’s wave mechanics in 1927. Knill
used these discrete Weyl type operators to construct an NEB from the group
$\mathbb{Z}^{2k}_{p}$ where $p$ is prime. Later Klappenecker and Rötteler [1]
and then Parthasarathy [9] constructed a similar NEB from the index group
$\mathbb{Z}_{n}\times\mathbb{Z}_{n}$ and Parthasarathy used them for quantum
error correcting codes.

Recently, X. Huang, T. Zhang et al [12] have used the Weyl operators as a
basis of $M_{n}(\mathbb{C})$ to describe a quantum state and give a necessary
condition for separability of a bipartite state. In this paper we take
advantage of an NEB of $M_{n}(\mathbb{C})$ in particular, Weyl operators on
$M_{n}(\mathbb{C})$, to construct a convenient basis of
$\mathrm{Lin}(M_{n}(\mathbb{C}))$, the set of all linear maps on
$M_{n}(\mathbb{C})$. Such a basis of $\mathrm{Lin}(M_{n}(\mathbb{C}))$ will be
used to decompose a linear map on $M_{n}(\mathbb{C})$, then positive and
completely positive maps will be characterised with respect to the
$n^{2}\times n^{2}$ coefficient matrix found in such decompositions. Moreover,
we will see that the product of linear maps in
$\mathrm{Lin}(M_{n}(\mathbb{C}))$ translates into the convolution type product
of the corresponding coefficient matrices in $M_{n^{2}}(\mathbb{C})$, which
will be used to give a characterisation of completely co-positive maps on
$M_{n}(\mathbb{C})$. Such an exploitation of isomorphims between
$\mathrm{Lin}(M_{n}(\mathbb{C}))$ and $M_{n^{2}}$ or $M_{n}(\mathbb{C})\otimes
M_{n}(\mathbb{C})^{*}$ may lead to an alternate approach to study different
cones of positive maps (e.g. $k$-positive map, $k$-superpositive maps)
compared to the existing theory using the Choi-Jamiolkowski isomorphism (cf.
[11]).

In section $2$, we present the results of E. Knill[3], A. Klappenecker and
Martin Rötteler[1][2] regarding the nice error basis and its construction from
a group of central type. As a very useful example of NEB for computation, we
will discuss the Weyl operators. In section $3$, we will take advantage of the
NEB to construct a convenient basis of $\mathrm{Lin}(M_{n}(\mathbb{C}))$ with
respect to which we will decompose any linear map
$\alpha\in\mathrm{Lin}(M_{n}(\mathbb{C}))$ to obtain the corresponding
$n^{2}\times n^{2}$ coefficient matrix $D_{\alpha}$. We will compute the
coefficient matrix for few maps which are important in quantum information
e.g. depolarising channel, transposition and conditional expectation onto
diagonals. Section $4$ will be on the relation between the coefficient matrix
$D_{\alpha}$ and the Choi matrix $C_{\alpha}$ corresponding to a linear map
$\alpha$ on $M_{n}(\mathbb{C})$. Finally, section $5$ will be devoted to the
characterisation of positivity properties of linear maps in terms of their
coefficient matrices. More precisely, we will characterise the positive,
completely positive (in general quantum channel), completely co-positive maps
on $M_{n}(\mathbb{C})$ and a special case, $1$-super positive or entanglement
breaking map in $M_{2}(\mathbb{C})$.

##  2\. Nice error basis and ONB of $M_{n}(\mathbb{C})$

In this section we briefly present the results of E. Knill [3], Klappenecker
and Rotteler [1] on nice error basis. We recall the definition of a nice error
basis as introduced by E. Knill [3].

######  Definition 2.1.

Let $G$ be a group of order $n^{2}$ for some natural number $n$. A nice error
basis(NEB) on $\mathbb{C}^{n}$ [1] is a set of unitary operators
$E=\\{\rho_{g}\in U(g):g\in G\\}$ such that

  1. (1)

$\rho(1)$ is the identity matrix, where $1$ denotes the identity element of
the group $G$.

  2. (2)

${\rm Tr}(\rho_{g})=n\delta_{g,1}$.

  3. (3)

$\rho_{g}\rho_{h}=\omega(g,h)\rho_{gh}$, where $\omega(g,h)\in\mathbb{C}$ is a
scalar.

For such a set of operators the labelling group $G$ is called the index group
of the corresponding NEB.

Conditions (1) and (3) simply tell us that the representation $\rho$ is a
projective representation. If we equip $M_{n}(\mathbb{C})$ with the inner
product $\langle A,B\rangle:={\rm Tr}(A^{*}B)$ then conditions (1)-(3) ensure
that $E$ is an orthonormal set since

$\displaystyle\langle\rho_{g},\rho_{h}\rangle={\rm
Tr}(\rho_{g}^{*}\rho_{h})=\omega(g^{-1},g)^{-1}{\rm
Tr}(\rho_{g^{-1}}\rho_{h})$
$\displaystyle=\omega(g^{-1},g)^{-1}\omega(g^{-1},h){\rm Tr}(\rho_{g^{-1}h})$
$\displaystyle=\omega(g^{-1},g)^{-1}\omega(g^{-1},h)n\delta_{g^{-1}h}.$

Comparing the dimension it follows that $E$ is an orthogonal basis of
$M_{n}(\mathbb{C})$. It is also easy to see that the associativity of the
group $G$ implies that function $\omega:G\times G\rightarrow\mathbb{C}$ is a
2-cocycle. If we normalise each $\pi_{g}$ so that $\rm{det}(\pi_{g})=1$ then
$\omega(g,h)$ becomes an $n$-th root of unity for any $g,h\in G$.

######  Theorem 2.2.

Let $\mathcal{E}=\\{\pi(g);g\in G\\}$ be a set of unitary matrices indexed by
a finite group $G$. Then $\mathcal{E}$ is a NEB iff $\pi$ is a unitary
faithful irreducible projective representation of order $|G|^{1/2}$.

###### Proof.

See theorem 1, [1]. ∎

###  2.1. NEB from Groups of Central Type:

######  Definition 2.3.

A group H is called of central type if there exists an irreducible character
$\chi$ such that $\chi(1)^{2}=|H:Z(H)|$, where $Z(H)$ is the center of the
group.

It is also equivalent to the existence of an irreducible character which
vanishes outside $Z(H)$(cf. p. 28, [6]). Let $H$ be a group of central type
with a unitary irreducible character $\chi$ verifying the above condition.
Suppose, $\mathfrak{X}_{\chi}$ is the unitary irreducible representation
corresponding to the character $\chi$. Recall that a representation
$\mathfrak{X}$ is called faithful iff $\mathrm{Ker}(\mathfrak{X})=\\{Id\\}$.
If we define the kernel of a character $\chi$ as $\rm{Ker}\chi:=\\{h\in
H;\chi(h)=\chi(1)\\}$ then it is well known that
$\rm{Ker}(\mathfrak{X}_{\chi})=\rm{Ker}(\chi)$. We take the quotient group
$H/\text{Ker }\chi$ so that the natural representation
$\tilde{\mathfrak{X}}_{\chi}(\bar{h}):=\mathfrak{X}_{\chi}(h)$ of $H/{\rm
Ker}\chi$ becomes faithful.

######  Lemma 2.4.

Let $H$ be a group and $\chi$ is an irreducible character of $G$. Then
$Z(H/\rm{Ker}\chi)=Z(\chi)/\rm{Ker}\chi$, where $Z(\chi):=\\{h\in
H;|\chi(h)|=\chi(1)\\}$ is called the center of the representation. Moreover,
$Z(\chi)/\rm{Ker}\chi$ is a cyclic group.

###### Proof.

Lemma 2.27, [6] ∎

######  Lemma 2.5.

If $\chi$ is an irreducible character of the group $H$ then
$\chi(1)^{2}\leq|H:Z(\chi)|$. Equality occurs if and only if $\chi$ vanishes
on $G-Z(\chi)$.

###### Proof.

Corollary 2.30, [6]. ∎

######  Theorem 2.6.

Let $H$ be a group of central type. Then the group
$(H/\rm{Ker}\chi)/(Z(H)/\rm{Ker}\chi)\cong H/Z(H)$ is an index group.

###### Proof.

Let $H$ be a group of central type with an irreducible character $\chi$ which
satisfies $\chi(1)^{2}=|H:Z(H)|$ then $Z(\chi)=Z(H)$ [cf. p. 28, [6]].
Therefore we have that $Z(H/{\rm Ker}\chi)=Z(H)/{\rm Ker}\chi$ by Lemma 2.4.
For each $h\in(H/{\rm Ker}\chi)/(Z(H)/{\rm Ker}\chi)$ we choose a coset
representative $\phi(h)$ in $H/{\rm Ker}\chi$. Let us denote
$\pi_{h}=\tilde{\mathfrak{X}}_{\chi}(\phi(h))$. Therefore we have

$\pi_{h}\pi_{k}=\tilde{\mathfrak{X}}(\phi(h))\tilde{\mathfrak{X}}(\phi(k))=\tilde{\mathfrak{X}}(\phi(h)\phi(k))=\tilde{\mathfrak{X}}(\phi(hk)z_{h,k}),$

where $z_{h,k}\in Z(H)/\rm{Ker}\chi$. $\tilde{\mathfrak{X}}_{\chi}(Z(H)/{\rm
Ker}\chi)$ consists of scalar multiples of identity only. So we obtain
$\pi_{h}.\pi_{k}=\tilde{\mathfrak{X}}(\phi(hk))\tilde{\mathfrak{X}}(z_{h,k})=\omega(h,k)\pi_{hk}$,
where $\omega(h,k)\in\mathbb{C}$. Since the representation is irreducible, all
the $\pi_{h}$ spans $M_{n}(\mathbb{C})$ for some $n\in\mathbb{N}$. Using
isomorphism theorem we get $(H/\text{Ker }\chi)/(Z(H)/\text{Ker }\chi)\cong
H/Z(H)$. As we know the character $\chi$ vanishes outside $Z(H)$, it follows
that ${\rm Tr}(\pi_{h})=0$ except at the identity. So we find that $(H/\rm
Ker\chi)/(Z(H)/\rm Ker\chi)\cong H/Z(H)$ is an index group if H is a group of
central type. ∎

Example: We present here two examples to briefly clarify the previous result-
one with an Abelian index group and another with a non-Abelian index group.

  * i.

The group of unit quaternions $Q=\\{\pm 1,\pm i,\pm j,\pm k\\}$ (with
multiplication as the group operation) has eight elements and five irreducible
representations (up to equivalence), which we can choose as

$\displaystyle\varepsilon$ $\displaystyle:$
$\displaystyle\varepsilon(i)=\varepsilon(j)=1,$ $\displaystyle\sigma_{i}$
$\displaystyle:$ $\displaystyle\sigma_{i}(i)=1,\quad\sigma_{i}(j)=-1,$
$\displaystyle\sigma_{j}$ $\displaystyle:$
$\displaystyle\sigma_{j}(i)=-1,\quad\sigma_{j}(j)=1,$
$\displaystyle\sigma_{k}$ $\displaystyle:$
$\displaystyle\sigma_{k}(i)=-1=\sigma_{k}(j),$ $\displaystyle\pi$
$\displaystyle:$ $\displaystyle\pi(i)=\left(\begin{array}[]{cc}0&i\\\
i&0\end{array}\right),\quad\pi(j)=\left(\begin{array}[]{cc}0&-1\\\
1&0\end{array}\right).$

For the character table we get

$\begin{array}[]{c||c|c|c|c|c|c|c|c||c}&1&-1&i&-i&j&-j&k&-k&\text{dim}\\\
\hline\cr\chi_{\varepsilon}=\varepsilon&1&1&1&1&1&1&1&1&1\\\
\chi_{\sigma_{i}}=\sigma_{i}&1&1&1&1&-1&-1&-1&-1&1\\\
\chi_{\sigma_{j}}=\sigma_{j}&1&1&-1&-1&1&1&-1&-1&1\\\
\chi_{\sigma_{k}}=\sigma_{k}&1&1&-1&-1&-1&-1&1&1&1\\\
\chi_{\pi}&2&-2&0&0&0&0&0&0&2\end{array}$

We see that

$\displaystyle{\rm ker}(\chi_{\pi})$ $\displaystyle=$ $\displaystyle\\{1\\},$
$\displaystyle Z(\chi_{\pi})$ $\displaystyle=$ $\displaystyle\\{1,-1\\}=Z(Q).$

$Q$ is a group of central type: its center is $Z(Q)=\\{-1,1\\}$ and it has a
$|Q/Z(Q)|^{1/2}$-dimensional irreducible representation. We have
$Q/Z(Q)\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$.

The 2-cocycle $\omega:\mathbb{Z}_{2}\times\mathbb{Z}_{2}\to\mathbb{T}$ is
given by the relations

$\pi(g_{1})\pi(g_{2})=\omega(g_{1},g_{2})\pi(g_{1}g_{2}),$

for $g_{1},g_{2}\in\mathbb{Z}_{2}$. If we write
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}=\\{(\pm 1,\pm 1)\\}$ multiplicatively and
choose

$\displaystyle\pi(+1,+1)$ $\displaystyle=$ $\displaystyle\pi(1)=I_{2},$
$\displaystyle\pi(+1,-1)$ $\displaystyle=$
$\displaystyle\pi(i)=\left(\begin{array}[]{cc}0&i\\\ i&0\end{array}\right),$
$\displaystyle\pi(-1,+1)$ $\displaystyle=$
$\displaystyle\pi(j)=\left(\begin{array}[]{cc}0&-1\\\ 1&0\end{array}\right)$
$\displaystyle\pi(-1,-1)$ $\displaystyle=$
$\displaystyle\pi(k)=\left(\begin{array}[]{cc}i&0\\\ 0&-i\end{array}\right).$

We get

$\begin{array}[]{c|cccc}\omega&(+1,+1)&(+1,-1)&(-1,+1)&(-1,-1)\\\
\hline\cr(+1,+1)&1&1&1&1\\\ (+1,-1)&1&-1&1&-1\\\ (-1,+1)&1&-1&-1&1\\\
(-1,-1)&1&1&-1&-1\\\ \end{array}$

  * ii.

Klappenecker and Rötteler construceted [1] an example of NEB corresponding to
a non-commutative index group which we briefly mention here. Consider the
group $H_{n}$ for some $n\in\mathbb{N}$, generated by the composition of the
maps

$\tau:x\mapsto x+1(\mathrm{mod}\quad 2^{n})\quad\text{and}\quad\alpha:x\mapsto
5x(\mathrm{mod}\quad 2^{n}).$

If $A:=\langle\tau\rangle$ and $B:=\langle\alpha\rangle$ then $H_{n}=A\rtimes
B$.

######  Theorem 2.7.

The group $H_{n}$ is a group of central type of order $2^{2n-2}$ with cyclic
center $Z(H_{n})=\langle\tau^{2^{n-2}}\rangle$. The index group
$H_{n}/Z(H_{n})$ is non-Abelian for $n\geq 5$.

###### Proof.

See theorem 5, [1]. ∎

Let $\phi:\mathbb{Z}/2^{n}\mathbb{Z}\rightarrow\mathbb{C}$ be a map defined by

$\phi(x)=\exp{\left(\frac{2\pi i5^{x}}{2^{n}}\right)}.$

Then the diagonal matrix

${\pi}(\tau)=\mathrm{diag}(\phi(0),\phi(1),\ldots,\phi(2^{n-1}-1))$

and the shift

${\pi}(\alpha)=\begin{bmatrix}0&1&0&\ldots&0\\\ 0&0&1&\ldots&0\\\
\vdots&&&\ddots&\vdots\\\ 0&0&0&\ldots&1\\\ 1&0&0&\ldots&0\end{bmatrix}$

define an ordinary faithful irreducible representation of $H_{n}$. The NEB
corresponding to the index group $H_{n}/Z(H_{n})$ is given by

$\mathcal{E}=\\{\pi(\tau)^{k}\pi(\alpha)^{l}:0\leq k,l<2^{n-2}-1\\}.$

For more examples of NEB and index groups, see the website
https://people.engr.tamu.edu/andreas-klappenecker/ueb/ueb.html

###  2.2. NEB from Discrete Weyl Relation

We define a bicharacter
$\varkappa:\mathbb{Z}_{n}\times\mathbb{Z}_{n}\longrightarrow\mathbb{C}$ such
that

  * (i)

$|\varkappa(x,y)|=1$, for all $x,y\in G$,

  * (ii)

Symmetry: $\varkappa(x,y)=\varkappa(y,x)$, for all $x,y\in G$,

  * (iii)

Non-degeneracy: $\varkappa(x,y)=1$ for all $y\in G$ iff $x=0$,

  * (iv)

Character: $\varkappa(x,y+z)=\varkappa(x,y)\cdot\varkappa(x,z)$.

For example we can take

$\varkappa(k,\ell)=\exp\left(\frac{2\pi ik\ell}{n}\right),\qquad
k,\ell\in\mathbb{Z}_{n}.$

We fix an orthonormal basis, $\\{|x\rangle;x\in\mathbb{Z}_{n}\\}$ of
$\mathbb{C}^{n}$. We define two unitary representation $U,V$ of
$\mathbb{Z}_{n}$ on $\mathbb{C}^{n}$, by the relation

$\displaystyle U_{a}|x\rangle$ $\displaystyle=$ $\displaystyle|x+a\rangle,$
$\displaystyle V_{a}|x\rangle$ $\displaystyle=$
$\displaystyle\varkappa(a,x)|x\rangle,$

The operators $U_{a},V_{b}$, $a,b\in Z_{n}$, satisfy the Weyl commutation
relations

$U_{a}U_{b}=U_{a+b},\qquad V_{a}V_{b}=V_{a+b},\qquad
V_{b}U_{a}=\varkappa(a,b)U_{a}V_{b}$

for $a,b\in\mathbb{Z}_{n}$.

We define the Weyl operators

$W_{a,b}=U_{a}V_{b}$

for $a,b\in\mathbb{Z}_{n}$ (see p. 212, [8]). The matrix coefficients of a
Weyl operator $W_{a,b}$ w.r.t. to the basis
$\\{|x\rangle;x\in\mathbb{Z}_{n}\\}$ are given by

$\langle y|W_{a,b}|x\rangle=\varkappa(b,x)\delta_{y,x+a},\qquad
x,y\in\mathbb{Z}_{n},$

or, equivalently,

(2.5) $W_{a,b}=\sum_{x\in G}\varkappa(b,x)\,|x+a\rangle\langle x|.$

These Weyl operators $\\{W_{a,b};(a,b)\in Z_{n}\times\mathbb{Z}_{n}\\}$ form a
nice error basis and $\mathbb{Z}_{n}\times\mathbb{Z}_{n}$ becomes its index
group. To check that we can verify some straightforward relations

  * (i)

(Unitary)$W_{a,b}^{*}=W_{a,b}^{-1}=\varkappa(a,b)W_{-a,-b}$.

  * (ii)

(Projective representations) $W_{a,b}W_{x,y}=\varkappa(b,x)W_{a+x.b+y}$.

If we compute the trace of the Weyl operators with respect to the Hilbert-
Schmidt inner product on $M_{n}$

${\rm Tr}(W_{a,b})=\sum_{x\in G}\langle
x|\underbrace{U_{a}V_{b}|x\rangle}_{=\varkappa(b,x)|x+a\rangle}=\left\\{\begin{array}[]{ll}n&\mbox{
if }(a,b)=(0,0),\\\ 0&\mbox{ else.}\end{array}\right.$

Here we used the orthogonality of the characters,

$\sum_{x\in
G}\underbrace{\overline{\varkappa(a,x)}\varkappa(b,x)}_{=\varkappa(b-a,x)}=\\#G\,\delta_{a,b}.$

##  3\. Convenient Basis of $\mathrm{Lin}(M_{n}(\mathbb{C}))$

For a pair of matrices $A,B\in M_{n}$ we define a linear map $T_{A,B}:M_{n}\to
M_{n}$ by

$T_{A,B}(X)=AXB^{*},\qquad\text{ for }X\in M_{n}.$

######  Proposition 3.1.

The map $M_{n}(\mathbb{C})\times M_{n}(\mathbb{C})\ni(A,B)\mapsto
T_{A,B}\in\mathrm{Lin}(M_{n}(\mathbb{C}))$ extends to an unique isomorphism of
*-algebras $T:M_{n}\otimes M_{n}^{*}\ni A\otimes B^{*}\to
T_{A,B}\in\mathrm{Lin}(M_{n},M_{n})$, which is also an isomorphism of Hilbert
spaces.

The above proposition shows that if $\\{B_{i};1\leq i\leq n^{2}\\}$ is a basis
of $M_{n}$ and we define $T_{ij}(X)=B_{i}XB_{j}^{*}$ for any $X\in M_{n}$ then
$\\{T_{ij}\in\rm{Lin}(M_{n});1\leq i,j\leq n^{2}\\}$ is a basis of
$\rm{Lin}(M_{n})$. Taking a NEB as a basis of $M_{n}$ has the added advantage
that in that case $T_{ij}$ also becomes NEB in $\rm{Lin}(M_{n})$.

######  Lemma 3.2.

Let $\\{\frac{1}{\sqrt{n}}\pi_{g};g\in G\\}$ be a NEB of $M_{n}$ corresponding
to an index group $G$. Define the linear map
$T_{x,y}:M_{n}(\mathbb{C})\rightarrow M_{n}(\mathbb{C})$

$T_{x,y}(X):=\pi_{x}X\pi_{y}^{*}$

Then the $\\{\frac{1}{n}T_{x,y};x,y\in G\\}$ is a NEB of $\textrm{Lin}(M_{n})$
with index group $G\times G$ and 2-cocycle $\omega^{L}:(G\times
G)\times(G\times G)\rightarrow\mathbb{T}$ given by

$\omega^{L}((x^{\prime},y^{\prime}),(x,y))=\frac{\omega(x^{\prime},x)}{\omega(y^{\prime},y)}.$

###### Proof.

It is trivial to check that $T_{1,1}=\textrm{Id}$ where $1$ is the identity
element of $G$. To check the trace condition we compute

$\displaystyle\mathrm{Tr}_{\mathrm{Lin}(M_{n})}(T_{x,y})=\sum\Big{\langle}|i\rangle\langle
j|,T_{xy}|i\rangle\langle
j|\Big{\rangle}=\sum\mathrm{Tr}_{M_{n}}\Big{(}|j\rangle\langle
i|\pi_{x}|i\rangle\langle j|\pi_{y}^{*}\Big{)}$ $\displaystyle=\sum\langle
i|\pi_{x}|i\rangle\langle
j|\pi_{y}^{*}|j\rangle=\mathrm{Tr}(\pi_{x})\mathrm{Tr}(\pi_{y}^{*})=n^{2}\delta_{1,x}\delta_{1,y},$

which shows that trace of each operator $T_{x,y}$ is zero except for the
identity. For any $X\in M_{n}$ and $x,y,x^{\prime},y^{\prime}\in G$ we see
that

$T_{x^{\prime},y^{\prime}}\circ
T_{x,y}(X)=\pi_{x^{\prime}}\pi_{x}X\pi_{y}^{*}\pi_{y^{\prime}}^{*}=\frac{\omega(x^{\prime},x)}{\omega(y^{\prime},y)}\pi_{x^{\prime}x}X\pi_{y^{\prime}y}^{*}=\frac{\omega(x^{\prime},x)}{\omega(y^{\prime},y)}T_{x^{\prime}x,y^{\prime}y}(X),$

which proves the claim about the two cocycle $\omega^{L}$. ∎

The next proposition follows immediately since any NEB forms an ONB of the
associated space of linear maps.

######  Proposition 3.3.

The set $\\{\frac{1}{n}T_{x,y};x,y\in G\\}$ forms an orthonormal basis of
$\mathrm{Lin}(M_{n}(\mathbb{C}))$ with respect to Hilbert-Schmidt inner
product.

Let $\alpha\in\mathrm{Lin}(M_{n}(\mathbb{C}))$ and $\\{B_{i};1\leq i\leq
n^{2}\\}$ be a basis of $M_{n}(\mathbb{C})$. Since $T_{ij}$(defined above )
forms a basis of $\rm{Lin}(M_{n})(\mathbb{C})$ we can decompose $\alpha$ as

(3.1) $\alpha(X)=\sum_{1\leq i,j\leq n^{2}}D_{\alpha}(i,j)B_{i}XB_{j}^{*}.$

In particular if we take a NEB $\\{\frac{1}{\sqrt{n}}\pi_{x};x\in G\\}$ as
basis of $M_{n}(\mathbb{C})$ we can explicitly compute the coefficient
$D_{\alpha}$.

(3.2)
$\alpha(X)=\sum_{x,y}D_{\alpha}(x,y)T_{x,y}(X)=\sum_{x,y}D_{\alpha}(x,y)\pi_{x}X\pi_{y}^{*}.$

for all $X\in M_{N}(\mathbb{C})$. Using the orthonormality of the basis
$T_{x,y}$ and NEB $\\{\frac{1}{\sqrt{n}}\pi_{g};g\in G\\}$ of
$M_{n}(\mathbb{C})$, we have $D_{\alpha}(x,y)=\frac{1}{n}\langle
T_{x,y},\alpha\rangle_{\rm{Lin}(M_{N})}$ i.e.

(3.3) $\displaystyle D_{\alpha}(x,y)$ $\displaystyle=$
$\displaystyle\frac{1}{n^{2}}{\rm Tr}(T_{x,y}^{{\dagger}}\alpha)$
$\displaystyle=$ $\displaystyle\frac{1}{n}\sum_{g\in G}\langle
T_{x,y}(\pi_{g}),\alpha(\pi_{g})\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{n^{2}}\sum_{g\in G}{\rm
Tr}(\pi_{y}\pi_{g}^{*}\pi_{x}^{*}\alpha(\pi_{g})).$

Here $T^{{\dagger}}$ denotes the involution applied on $T$ w.r.t the Hilbert-
Schmidt inner product on $\mathrm{Lin}(M_{n}(\mathbb{C}))$. If we use the Weyl
operators as the nice error basis for defining $T_{x,y}$ i.e.
$T_{x,y}(X)=W_{x}XW_{y}^{*}$ for $x,y\in\mathbb{Z}_{n}\times\mathbb{Z}_{n}$
then we can compute the coefficient $D_{\alpha}$ in (3.3), using
$\\{|a\rangle\langle b|;a,b\in\mathbb{Z}_{n}\\}$ as a o.n.b of
$M_{n}(\mathbb{C})$

(3.4) $\displaystyle D_{\alpha}(x,y)$ $\displaystyle=$
$\displaystyle\frac{1}{n}\sum_{a,b\in G}{\rm Tr}\Big{(}W_{y}|b\rangle\langle
a|W_{x}^{*}\alpha\big{(}|a\rangle\langle b|\big{)}\Big{)}$ $\displaystyle=$
$\displaystyle\frac{1}{n}\sum_{a,b\in
G}\frac{\varkappa(y_{2},b)}{\varkappa(x_{2},a)}\Big{\langle}a+x_{1}\Big{|}\alpha\big{(}|a\rangle\langle
b|\big{)}\Big{|}b+y_{1}\Big{\rangle},$

for $x=(x_{1},x_{2}),y=(y_{1},y_{2})$.

######  Lemma 3.4.

Let $\alpha,\beta\in\mathrm{Lin}(M_{n}(\mathbb{C}))$ with coefficients
$(D_{\alpha}(x,y))_{x,y\in G}$, as defined in the equation (3.2). Then the
coefficient of their composition $\alpha\circ\beta$ are given by

$D_{\alpha\circ\beta}(x,y)=\sum_{p,q\in
G}\omega(p,xp^{-1})\overline{\omega(q,yq^{-1})}D_{\alpha}(p,q)D_{\beta}(p^{-1}x,q^{-1}y).$

###### Proof.

We have

$\displaystyle\alpha\circ\beta(X)$ $\displaystyle=$ $\displaystyle\sum_{p,q\in
G}D_{\alpha}(p,q)\pi_{p}\left(\sum_{p^{\prime},q^{\prime}\in
G}D_{\beta}(p^{\prime},q^{\prime})\pi_{p^{\prime}}X\pi_{q^{\prime}}^{*}\right)\pi_{q}^{*}$
$\displaystyle=$ $\displaystyle\sum_{p,p^{\prime},q,q^{\prime}\in
G}\omega(p,p^{\prime})\overline{\omega(q,q^{\prime})}D_{\alpha}(p,q)D_{\beta}(p^{\prime},q^{\prime})\pi_{pp^{\prime}}X\pi_{qq^{\prime}}^{*}$
$\displaystyle=$ $\displaystyle\sum_{x,y\in G}\underbrace{\left(\sum_{p,q\in
G}\omega(p,p^{-1}x)\overline{\omega(q,q^{-1}y)}D_{\alpha}(p,q)D_{\beta}(p^{-1}x,q^{-1}y)\right)}_{D_{\alpha\circ\beta}}\pi_{x}X\pi_{y}^{*},$

which completes the proof. ∎

We can define two different involutions on $\mathrm{Lin}(M_{n}(\mathbb{C}))$.
The first comes from the Hilbert-Schmidt inner product and is characterised by
the condition

$\langle X,\alpha(Y)\rangle=\langle\alpha^{\dagger}(X),Y\rangle$

for all $X,Y\in M_{n}(\mathbb{C})$.

The second is inherited from the involution in $M_{n}(\mathbb{C})$ and defined
by $\alpha^{\\#}(X)=\alpha(X^{*})^{*}$.

Both involutions are conjugate linear, but only the first is anti-
multiplicative, whereas the second is multiplicative, i.e., we have

$(\alpha\circ\beta)^{\dagger}=\beta^{\dagger}\circ\alpha^{\dagger},\qquad(\alpha\circ\beta)^{\\#}=\alpha^{\\#}\circ\beta^{\\#}$

for $\alpha,\beta\in\rm{Lin}(M_{n}(\mathbb{C}))$.

######  Proposition 3.5.

We have

$D_{\alpha^{\dagger}}(x,y)=\frac{\omega(x,x^{-1})}{\omega(y,y^{-1})}\overline{D_{\alpha}(x^{-1},y^{-1})}$

and

$D_{\alpha^{\\#}}(x,y)=\overline{D_{\alpha}(y,x)}$

for $x,y\in G$.

###### Proof.

It is easy to see that for any $x,y\in G$ we have
$T_{x,y}^{{\dagger}}=\frac{\omega(x,x^{-1})}{\omega(y,y^{-1})}T_{x^{-1},y^{-1}}$.
Applying the involution ${\dagger}$ on the decomposition of $\alpha$ in (3.2)

$\alpha^{{\dagger}}=\sum_{x,y\in
G}\overline{D_{\alpha}(x,y)}T_{x,y}^{{\dagger}}=\sum_{x,y\in
G}\frac{\omega(x,x^{-1})}{\omega(y,y^{-1})}\overline{D_{\alpha}(x,y)}T_{x^{-1},y^{-1}}$

the first claim follows. Similarly, we can trivially check that
$T_{x,y}^{\\#}=T_{y,x}$ for any $x,y\in G$. Then the second claim follows by
applying $\\#$ on the decomposition (3.2)

$\alpha^{\\#}=\sum_{x,y\in
G}\overline{D_{\alpha}(x,y)}T_{x,y}^{\\#}=\sum_{x,y\in
G}\overline{D_{\alpha}(x,y)}T_{y,x}$

∎

###  3.1. Examples

Here we compute the kernel or the $n^{2}\times n^{2}$ matrix $D_{\alpha}$
corresponding to different positive maps
$\alpha\in\mathrm{Lin}(M_{n}(\mathbb{C}))$ which are important in quantum
information.
Identity map: The identity map corresponds to the kernel
$D_{\text{Id}}(x,y)=\delta_{1,x}\delta_{1,y}$ for $x,y\in G$ as we can write

$X=D_{1,1}\pi_{1}X\pi_{1}^{*}=\sum_{x,y\in
G}\delta_{1,x}\delta_{1,y}\pi_{x}X\pi_{y}^{*}.$

Depolarising Channel: Let $P\in\mathrm{Lin}(M_{n}(\mathbb{C}))$ be the
diagonal sum $P=\sum_{g\in G}T_{g,g}$. For any $h\in G$ and $X\in
M_{n}(\mathbb{C})$ we have

$\pi_{h}P(X)=\sum_{g}\omega(h,g)\pi_{hg}X\pi_{g}^{*}\quad\text{and}\quad
P(X)\pi_{h}=\sum_{g}\frac{\omega(h,h^{-1})}{\omega(h^{-1},g)}\pi_{g}X\pi_{h^{-1}g}^{*}.$

After a change of variable we can write
$P(X)\pi_{h}=\sum_{g}\frac{\omega(h,h^{-1})}{\omega(h^{-1},hg)}\pi_{hg}X\pi_{g}^{*}$.
Using the defintion of 2-cocycle

$\frac{\omega(h,h^{-1})}{\omega(h^{-1},hg)}=\frac{\omega(h,h^{-1})}{\omega(h,h^{-1})\omega(1,g)\overline{\omega(h,g)}}=\omega(h,g).$

Thus we see that $P(X)$ commutes with every basis element $\pi_{h}$ on
$M_{n}$. So we conclude that $P(X)=cI_{n}$ for some $c\in\mathbb{C}$.
Computing the trace of both sides we find that $c=\rm{Tr}(X)$. Therefore we
see that the map $P$, defined as the diagonal sum of the operators $T_{g,g}$s,
is actually the depolarising channel which corresponds to the identity matrix
$D_{P}(x,y)=\delta_{x,y}$ where $x,y\in G$.

Transposition: Let $T\in\mathrm{Lin}(M_{n}(\mathbb{C}))$ be the transposition
map given by $X\mapsto X^{t}$. We compute the $n^{2}\times n^{2}$ matrix
$D_{T}$ corresponding to the transposition map. For any$x,y\in G$ we have

$\displaystyle D_{T}(x,y)=\langle T_{x,y}|T\rangle$ $\displaystyle=$
$\displaystyle\sum_{1\leq i,j\leq n}\mathrm{Tr}\Big{(}(T_{x,y}|i\rangle\langle
j|)^{*}T|i\rangle\langle j|\Big{)}$ $\displaystyle=$ $\displaystyle\sum_{1\leq
i,j\leq n}\mathrm{Tr}\Big{(}\pi_{y}|j\rangle\langle
i|\pi_{x}^{*}|j\rangle\langle i|\Big{)}$ $\displaystyle=$
$\displaystyle\sum_{1\leq i,j\leq n}\langle
i|\pi_{x}^{*}|j\rangle\mathrm{Tr}\Big{(}\pi_{y}|j\rangle\langle i|\Big{)}$
$\displaystyle=$ $\displaystyle\sum_{1\leq i,j\leq n}\langle
i|\pi_{x}^{*}|j\rangle\langle
i|\pi_{y}|j\rangle=\mathrm{Tr}(\overline{\pi}_{x}\pi_{y})$

In particular, if we take the Weyl operators
$\\{W_{a,b};a,b\in\mathbb{Z}_{n}\\}$ as the chosen NEB and
$\\{|i\rangle;i\in\mathbb{Z}_{n}\\}$ as the standard basis of $\mathbb{C}^{n}$
then

$\displaystyle D_{T}((a,b),(c,d))=\sum_{i,j\in\mathbb{Z}_{n}}\langle
i|W_{a,b}^{*}|j\rangle\langle
i|W_{c,d}|j\rangle=\sum_{i,j\in\mathbb{Z}_{n}}{\chi(a,b)}\chi(-b,j)\chi(d,j)\delta_{i,j-a}\delta_{i,c+j}.$

So we have

$D_{T}{((a,b),(c,d))}=\left\\{\begin{array}[]{cc}\sum_{i\in\mathbb{Z}_{n}}{\chi(a,b)}\chi(d-b,i+a)&\text{
if }c=-a,\\\ 0&\text{otherwise.}\end{array}\right.$

Conditional Expectation onto Diagonal: Consider the linear map
$C:M_{n}(\mathbb{C})\rightarrow M_{n}(\mathbb{C})$,
$C(X)=(\delta_{ij}x_{ij})_{1\leq i,j\leq n}$ for $X=(x_{ij})_{1\leq i,j\leq
n}\in M_{n}(\mathbb{C})$. This map is a conditional expectation onto the
*-subalgebra of diagonal matrices with respect to the standard basis. We
compute the coefficient matrix $D_{C}$ of the map $C$. For any $x,y\in G$

$\displaystyle D_{C}(x,y)=\langle T_{x,y}|C\rangle$ $\displaystyle=$
$\displaystyle\sum_{1\leq i,j\leq n}\mathrm{Tr}\Big{(}(T_{x,y}|i\rangle\langle
j|)^{*}C|i\rangle\langle j|\Big{)}$ $\displaystyle=$ $\displaystyle\sum_{1\leq
i,j\leq n}\delta_{ij}\mathrm{Tr}(\pi_{y}|j\rangle\langle
i|\pi_{x}^{*}|i\rangle\langle j|)$ $\displaystyle=$ $\displaystyle\sum_{1\leq
j\leq n}\delta_{ij}\langle i|\pi_{x}^{*}|i\rangle\langle
j|\pi_{y}|j\rangle=\mathrm{Tr}(C(\pi_{x}^{*})C(\pi_{y})).$

In particular, taking the Weyl operators as NEB gives

$D_{C}((a,b),(c,d)=\left\\{\begin{array}[]{cc}\sum_{j\in\mathbb{Z}_{n}}\chi(d-b,j)&\text{
if }c=a=0,\\\ 0&\text{otherwise.}\end{array}\right.$

##  4\. Correspondence between Choi matrix $C_{\alpha}$ and $D_{\alpha}$

Recall that the Choi-Jamiołkowski matrix of a map
$\alpha\in\mathrm{Lin}(M_{n}(\mathbb{C}))$ is the $n^{2}\times n^{2}$-matrix
defined by

$C_{\alpha}=\sum_{j,k=1}^{n}E_{jk}\otimes\alpha(E_{jk})\in
M_{n}(\mathbb{C})\otimes M_{n}(\mathbb{C})\cong M_{n^{2}}(\mathbb{C}).$

It is known that $\alpha\in\mathrm{Lin}(M_{n}(C))$ is completely positive (CP)
iff $C_{\alpha}$ is positive. Furthermore,
$\alpha\in\mathrm{Lin}(M_{n}(\mathbb{C}))$ is $k$-positive if and only if

$\langle v,C_{\alpha}v\rangle\geq 0$

for all $v\in\mathbb{C}^{n}\otimes\mathbb{C}^{n}$ with Schmidt rank not more
than $k$. For any completely positive map $\alpha$ we have the Kraus
decompositon

$\alpha=\sum_{j=1}^{r}\mathrm{Ad}_{L_{j}}$

for some matrices $L_{j}\in M_{n}(\mathbb{C})$, where for any $X\in
M_{n}(\mathbb{C})$ the conjugate map $\mathrm{Ad}_{L_{j}}$ is given by
$\mathrm{Ad}_{L_{j}}(X)=L_{j}XL_{j}^{*}$. We call $\alpha$ $1$-super positive
or entanglement breaking iff $\text{rank}(L_{j})=1$ for any $j$. It is called
completely co-positive iff $T\circ\alpha$ is completely positive, where $T$ is
the transposition map, see [10, 11].

We can switch from $C_{\alpha}$ to $D_{\alpha}$ by a change of basis.

######  Proposition 4.1.

If $T_{x,y}$ is defined with respect to the Weyl operators then the Choi-
Jamiołkowski matrix of $T_{x,y}$ is given by

$C_{T_{x,y}}(v,w)=\frac{\varkappa(x_{2},v_{1})}{\varkappa(y_{2},w_{1})}\delta_{v_{1}+x_{1},v_{2}}\delta_{w_{1}+y_{1},w_{2}},$

for
$x=(x_{1},x_{2}),y=(y_{1},y_{2}),v=(v_{1},v_{2}),w=(w_{1},w_{2})\in\mathbb{Z}_{n}\times\mathbb{Z}_{n}$.
More generally, if $\alpha$ is of the form
$\alpha=\sum_{x,y\in\mathbb{Z}_{n}\times Z_{n}}D_{\alpha}(x,y)T_{x,y}$, then
its Choi-Jamiołkowski matrix is given by

$C_{\alpha}(v,w)=\sum_{x_{2},y_{2}\in
G}\frac{\varkappa(x_{2},v_{1})}{\varkappa(y_{2},w_{1})}D_{\alpha}\big{(}(v_{2}-v_{1},x_{2}),(w_{2}-w_{1},y_{2})\big{)},$

for $v,w\in\mathbb{Z}_{n}\times\mathbb{Z}_{n}$. Conversely, the coefficients
from the equation (3.4) can be computed from the Choi-Jamiołkowski matrix via

$D_{\alpha}(x,y)=\frac{1}{n}\sum_{a,b\in
G}\frac{\varkappa(y_{2},b)}{\varkappa(x_{2},a)}C_{\alpha}\big{(}(a,a+x_{1}),(b,b+y_{1})\big{)}$

for $x,y\in\mathbb{Z}_{n}\times\mathbb{Z}_{n}$.

###### Proof.

Using Formula (2.5), we get

$W_{y}^{*}=\sum_{d\in\mathbb{Z}_{n}}\frac{1}{\varkappa(y_{2},d)}|d\rangle\langle
d+y_{1}|$

and

$T_{x,y}\big{(}|a\rangle\langle
b|\big{)}=\frac{\varkappa(x_{2},a)}{\varkappa(y_{2},b)}|a+x_{1}\rangle\langle
b+y_{1}|,$

for $a,b\in\mathbb{Z}_{n}$, $x,y\in\mathbb{Z}_{n}\times\mathbb{Z}_{n}$. So, if
we choose $\\{|a\rangle;a\in\mathbb{Z}_{n}\\}$ as a basis of $\mathbb{C}^{n}$,
we can write the corresponding matrix units as $|a\rangle\langle b|$,
$a,b\in\mathbb{Z}_{n}$, and we get

$\displaystyle C_{T_{x,y}}$ $\displaystyle=$
$\displaystyle\sum_{a,b\in\mathbb{Z}_{n}}|a\rangle\langle b|\otimes
T_{x,y}\big{(}|a\rangle\langle b|\big{)}$ $\displaystyle=$
$\displaystyle\sum_{a,b\in\mathbb{Z}_{n}}\frac{\varkappa(x_{2},a)}{\varkappa(y_{2},b)}|(a,a+x_{1})\rangle\langle(b,b+y_{1})|,$

which proves the first claim of the proposition.

For
$\alpha=\sum_{x,y\in\mathbb{Z}_{n}\times\mathbb{Z}_{n}}D_{\alpha}(x,y)T_{x,y}$,
this yields

$C_{\alpha}=\sum_{x,y\in\mathbb{Z}_{n}\times\mathbb{Z}_{n}}\sum_{a,b\in\mathbb{Z}_{n}}D_{\alpha}(x,y)\frac{\varkappa(x_{2},a)}{\varkappa(y_{2},b)}|(a,a+x_{1})\rangle\langle(b,b+y_{1})|$

or

$C_{\alpha}(v,w)=\sum_{x_{2},y_{2}\in\mathbb{Z}_{n}}\frac{\varkappa(x_{2},v_{1})}{\varkappa(y_{2},w_{1})}D_{\alpha}\big{(}(v_{2}-v_{1},x_{2}),(w_{2}-w_{1},y_{2})\big{)}.$

For the converse we use the equation (3.4),

$\displaystyle D_{\alpha}(x,y)$ $\displaystyle=$
$\displaystyle\frac{1}{N}\sum_{a,b\in
G}\frac{\varkappa(y_{2},b)}{\varkappa(x_{2},a)}\Big{\langle}a+x_{1}\Big{|}\alpha\big{(}|a\rangle\langle
b|\big{)}\Big{|}b+y_{1}\Big{\rangle}$ $\displaystyle=$
$\displaystyle\frac{1}{N}\sum_{a,b\in\mathbb{Z}_{n}}\frac{\varkappa(y_{2},b)}{\varkappa(x_{2},a)}\Big{\langle}a+x_{1}\Big{|}\left(\sum_{g,h\in\mathbb{Z}_{n}}C_{\alpha}\big{(}(a,g),(b,h)\big{)}|g\rangle\langle
h|\right)\Big{|}b+y_{1}\Big{\rangle}$ $\displaystyle=$
$\displaystyle\sum_{a,b\in\mathbb{Z}_{n}}\frac{\varkappa(y_{2},b)}{\varkappa(x_{2},a)}C_{\alpha}\big{(}(a,a+x_{1}),(b,b+y_{1})\big{)},$

where we used the identity

$\alpha\big{(}|a\rangle\langle b|\big{)}=\sum_{g,h\in
G}C_{\alpha}\big{(}(a,g),(b,h)\big{)}|g\rangle\langle h|.$

∎

##  5\. Characterisation of positive and completely positive maps

######  Theorem 5.1.

Let $\\{B_{x}\\}_{x=1,2...n^{2}}$ be a basis of $M_{n}(\mathbb{C})$. Consider
a linear map $\alpha\in L(M_{n}(\mathbb{C}))$ of form
$\alpha(X)=\sum_{x,y=1}^{n^{2}}D_{\alpha}(x,y)B_{x}XB_{y}^{*}$. Then $\alpha$
is

  1. i.

Hermitianity preserving if and only if $D_{\alpha}$ is Hermitian.

  2. ii.

positive if and only if for any $v,w\in\mathbb{C}^{n}$,

$\langle v\otimes w,\tilde{\alpha}(v\otimes w)\rangle\geq 0$

where
$\tilde{\alpha}=\tau\circ\sum_{x,y=1}^{n^{2}}D_{\alpha}(x,y)(B_{x}\otimes
B_{y}^{*})$ and $\tau(u\otimes v)=v\otimes u$ is the flip operator.

###### Proof.

$\alpha\in\rm{Lin}(M_{n}(\mathbb{C}))$ is Hermitianity preserving iff
$\alpha(X^{*})^{*}=\alpha(X)$ i.e. $\alpha^{\\#}=\alpha$. Comparing the
coefficient matrix of both sides the first claim follows directly from the
proposition 3.5.

On the other hand, $\alpha$ positive if and only if it maps rank one
projection to positive operators. i.e. for all $v,w\in\mathbb{C}$

$\displaystyle 0\leq\left\langle v,\alpha(|u\rangle\langle u|)v\right\rangle$
$\displaystyle=\left\langle v,\sum D_{\alpha}(x,y)B_{x}|u\rangle\langle
u|B_{y}^{*}v\right\rangle$ $\displaystyle=\sum D_{\alpha}(x,y)\langle
v,B_{x}u\rangle\langle u,B^{*}_{y}v\rangle$ $\displaystyle=\left\langle
u\otimes v,\tau\circ\sum D_{\alpha}(x,y)B_{x}\otimes B_{y}^{*}(u\otimes
v)\right\rangle$ $\displaystyle=\langle u\otimes v,\tilde{\alpha}(u\otimes
v)\rangle.$

∎

######  Theorem 5.2.

A linear map $\alpha\in\mathrm{Lin}(M_{n})$ is a completely positive map with
Kraus rank $r$ then if and only if the corresponding coefficient matrix
$D_{\alpha}\in M_{n^{2}}$ as defined in (3.1), is positive semi-definite of
rank $r$.

###### Proof.

If $\alpha\in L(M_{n})$ is CP with Kraus rank $r$ then there exist
$\\{L_{j}\in M_{n};1\leq j\leq r\\}$ such that $\alpha$ can be written as
Kraus decomposition

$\alpha=\sum_{j=1}^{r}Ad_{L_{j}},$

where $Ad_{L_{j}}$ is the conjugate map given by
$Ad_{L_{j}}(X)=L_{j}XL_{j}^{*}$ for any matrix $X\in M_{n}$. Since the map
$L(M_{n})\ni\alpha\mapsto D_{\alpha}\in M_{n^{2}}$ is a linear isomorshism, we
have $D_{\alpha}=\sum_{j=1}^{r}D_{Ad_{L_{j}}}$. We can expand each $L_{j}$
with respect to the NEB $\\{B_{x};x\in G\\}$ and write
$L_{j}=\sum_{z}l_{j}(z)B_{z}$. Therefore we get

$\displaystyle Ad_{L_{j}}(X)=L_{j}XL_{j}^{*}=\left(\sum_{z\in
G}l_{j}(z)B_{z}\right)X\left(\sum_{z^{\prime}\in
G}l_{j}(z^{\prime})B_{z^{\prime}}\right)^{*}$
$\displaystyle=\sum_{z,z^{\prime}\in
G}l_{j}(z)\overline{l_{j}(z^{\prime})}B_{z}XB_{z^{\prime}}^{*}$

We find that $D_{Ad_{L_{j}}}$ is a rank one operator given by

$D_{Ad_{L_{j}}}=|l_{j}\rangle\langle l_{j}|$

where $l_{j}=(l_{j}(1),l_{j}(2),\ldots,l_{j}(n^{2}))^{t}$ is a vector in
$\mathbb{C}^{n^{2}}$. Thus $D_{\alpha}=\sum_{j=1}^{r}|l_{j}\rangle\langle
l_{j}|$ is a positive semi-definite operator of rank $r$.

Conversely, assume that $D_{\alpha}$ is positive semi-definite with rank $r$.
So there exists vectors $v_{1},v_{2},\ldots,v_{r}\in\mathbb{C}^{n^{2}}$ such
that $D_{\alpha}=\sum_{j=1}^{r}|v_{j}\rangle\langle v_{j}|$. If we denote
$\\{|x\rangle;x\in G\\}$ the standard basis on $\mathbb{C}^{n^{2}}$ then

$D_{\alpha}(x,y)=\sum_{j=1}^{r}\left\langle x|v_{j}\rangle\langle
v_{j}|y\right\rangle=\sum_{j=1}^{r}\langle x|v_{j}\rangle\overline{\langle
y|v_{j}\rangle}.$

Therefore we can write the equation (3.1) as

$\displaystyle\alpha(X)$ $\displaystyle=$ $\displaystyle\sum_{x,y\in
G}\sum_{j=1}^{r}\langle x|v_{j}\rangle\overline{\langle
y|v_{j}\rangle}B_{x}XB_{y}^{*}$ $\displaystyle=$ $\displaystyle\sum_{x,y\in
G}\sum_{j=1}^{r}\big{(}\langle x|v_{j}\rangle B_{x}\big{)}X\big{(}\langle
y|v_{j}\rangle B_{y}\big{)}^{*}$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{r}\left(\sum_{x}\langle x|v_{j}\rangle
B_{x}\right)X\left(\sum_{y}\langle y|v_{j}\rangle B_{y}\right)^{*}.$

If we denote $L_{j}=\sum_{x\in G}\langle x|v_{j}\rangle B_{x}$ then we get
$\alpha(X)=\sum_{j=1}^{r}L_{j}XL_{j}^{*}$, which shows that $\alpha$ is
completely positive with Kraus rank $r$. ∎

We remark here that the similar result was obtained by Poluikis and Hill from
a different approach [7]. But our approach has the advantange that coefficient
matrix corresponding to the composition of two linear maps becomes the
convolution type product of their individual coefficient matrices. This
composition law can be used to characterise the completely co-positive maps
too.

######  Corollary 5.3.

A linear map $\alpha\in\mathrm{Lin}(M_{n}(\mathbb{C}))$ is completely co-
positive iff the convolution product

$\sum_{p,q\in\mathbb{Z}_{n}\times\mathbb{Z}_{n}}\frac{\chi(p,x-p)}{\chi(q,y-q)}\textrm{Tr}(\overline{W}_{p}W_{q})D_{\alpha}(x-p,y-q)$

is positive semidefinite.

###### Proof.

Using the Theorem 5.2 a linear map $\alpha\in\mathrm{Lin}(M_{n}(\mathbb{C}))$
is co-CP iff in the decomposition $T\circ\alpha$ w.r.t the Weyl operators

$T\circ\alpha(X)=\sum_{x,y\in\mathbb{Z}_{n}\times
Z_{n}}D_{T\circ\alpha}W_{x}XW_{y}^{*},$

the coefficient matrix $D_{T\circ\alpha}$ is positive semi-definite. We use
the Lemma 3.4 and the coefficients we have found for transposition in example
3.1 to compute the coefficient $D_{t\circ\alpha}$ which complete the claim. ∎

######  Corollary 5.4.

A linear map $\alpha\in\mathrm{Lin}(M_{2})$ is $1$-super positive iff
$D_{\alpha}=\sum_{j=1}^{r}|l_{j}\rangle\langle l_{j}|$ where
$l_{j}=(l_{j}(1),\ldots,l_{j}(4))^{t}$ is a vector in $\mathbb{C}^{4}$
satisfying $l_{j}(1)^{2}=\sum_{k=2}^{4}l_{j}(k)^{2}$.

###### Proof.

Since $\alpha$ is $1$-super positive in $M_{2}$ there exists matrices
$L_{1},L_{2},\ldots,L_{r}\in M_{2}$ of rank $1$ such that
$\alpha=\sum_{j}^{r}Ad_{L_{j}}$. We can decompose each $L_{j}$ w.r.t the Pauli
basis

$\begin{array}[]{cccc}\sigma_{1}=\frac{1}{2}\begin{bmatrix}1&0\\\
0&1\end{bmatrix},&\sigma_{2}=\frac{1}{2}\begin{bmatrix}0&1\\\
1&0\end{bmatrix},&\sigma_{3}=\frac{1}{2}\begin{bmatrix}0&-i\\\
i&0\end{bmatrix},&\sigma_{4}=\frac{1}{2}\begin{bmatrix}1&0\\\
0&-1\end{bmatrix}.\end{array}$

to obtain $L_{j}=\sum_{i=1}^{4}l_{j}(k)\sigma_{k}$. After this decomposition
$L_{j}$ has the form

$\frac{1}{2}\begin{bmatrix}l_{j}(1)+l_{j}(4)&l_{j}(2)-il_{j}(3)\\\
l_{j}(2)+il_{j}(3)&l_{j}(1)-l_{j}(4)\end{bmatrix}.$

$L_{j}$ has rank $1$ iff $\textrm{det}(L_{j})=0$ i.e.
$l_{j}(1)^{2}=l_{j}(2)^{2}+l_{j}(3)^{2}+l_{j}(4)^{2}$. We have already seen
that in the proof of 5.2 that $D_{Ad_{L_{j}}}=|l_{j}\rangle\langle l_{j}|$,
which completes the the claim. ∎

######  Proposition 5.5.

A linear map $\alpha:M_{n}(\mathbb{C})\longrightarrow M_{n}(\mathbb{C})$

is trace preserving if and only if

$\sum_{x\in G}\omega(x,g)D_{\alpha}(x,xg)=\delta_{1,g}$

for all $g\in G$.

is unit preserving if and only if

$\sum_{x\in
G}\frac{\omega(x,x^{-1}z)}{\omega(z^{-1}x,x^{-1}z)}D_{\alpha}(x,z^{-1}x)=\delta_{1,z}$

for all $z\in G$.

###### Proof.

Since $\\{\pi_{g};g\in G\\}$ forms a basis of $M_{n}$ the map $\alpha$ is
trace preserving iff ${\rm Tr}(\alpha(\pi_{g}))={\rm Tr}(\pi_{g}$) for all
$g\in G$. Now

${\rm
Tr}\alpha(\pi_{g})=\sum_{x,y}D_{\alpha}(x,y)\frac{\omega(x,g)\omega(xg,y^{-1})}{\omega(y^{-1},y)}{\rm
Tr}(\pi_{xgy^{-1}})$

Substituting $xg=y$ we get

$\sum_{x\in G}\omega(x,g)D_{\alpha}(x,xg)={\rm Tr}(\pi_{g})=\delta_{1,g}$

for all $g\in G$.

By definition $\alpha$ is unit preserving iff $\alpha(I_{n})=I_{n}$.

$\displaystyle\alpha(I_{n})=\sum_{x,y\in G}D_{\alpha}(x,y)\pi_{x}\pi_{y}^{*}$
$\displaystyle=\sum_{x,y\in
G}\frac{\omega(x,y^{-1})}{\omega(y,y^{-1})}D_{\alpha}(x,y)\pi_{xy^{-1}}$
$\displaystyle=\sum_{x,z\in
G}\frac{\omega(x,x^{-1}z)}{\omega(z^{-1}x,x^{-1}z)}D_{\alpha}(x,z^{-1}x)\pi_{z}.$

Comparing the coefficients we get

$\sum_{x\in
G}\frac{\omega(x,x^{-1}z)}{\omega(z^{-1}x,x^{-1}z)}D_{\alpha}(x,z^{-1}x)=\delta_{1,z}$

for all $z\in G$. ∎

## Acknowledgements

P.C. and U.F. were supported by the ANR Project No. ANR-19-CE40-0002. B.V.R.
Bhat is suppported by the J C Bose Fellowship JBR/2021/000024 of SERB(India)

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