text:
\section{Introduction}
There is much work on quantum systems,
where the position and momentum take values
in the ring ${\mathbb Z}(d)$ (the integers modulo $d$).
Recent reviews have been presented in \cite{1,2,3,4,5}.

An important topic in this context is the mutually unbiased bases
\cite{m1,m2,m3,m4,m5,m7,m8,m9,m10,m11,m12,m13,m14}.
For a prime number $d$, the number of mutually unbiased bases is equal to $d+1$.
These results can be extended to
quantum systems where the position and momentum take values in a Galois fields $GF(p^n)$ \cite{Gal1,Gal2}
(where $p$ is a prime number).
Hamiltonians for the construction of such systems from $n$ component subsystems, each of which is $p$-dimensional,
have been discussed
in \cite{V}.
The number of mutually unbiased bases in these systems is equal to $p^n+1$.
There is currently much work on mutually unbiased bases in systems with dimension which is not a power of a
prime number
(e.g., for the case $d=6$\cite{stefan1,stefan2}).

In mutually unbiased bases the absolute value of the overlap of any two vectors in two different bases is
$d^{-1/2}$.
Let $d_i$ be the divisors of $d$, apart from $d$ and $1$.
In this paper we introduce the concept of weak mutually unbiased bases, where roughly speaking,
this absolute value is equal to $d^{-1/2}$ or alternatively to one of the $d_i^{-1/2},0$
(the precise definition is given below).
We show a correspondence (`duality') between the properties of the weak mutually unbiased bases and the
properties of the maximal lines in the
${\mathbb Z}(d)\times {\mathbb Z}(d)$ phase space of these systems.
This duality shows that our concept of weak mutually unbiased bases is intimately related and motivated
by the geometrical properties
of the phase space of these systems.
We note here that the term  weak mutually unbiased bases, has also been used in \cite{weak}
for a different concept.

The ${\mathbb Z}(d)\times {\mathbb Z}(d)$ phase space is a finite geometry \cite{f1,f2,f3}.
Most of the existing work on finite geometries is on near-linear geometries where
two points belong to at most one line.
In \cite{SV} we have pointed out that  ${\mathbb Z}(d)\times {\mathbb Z}(d)$ violates this axiom and
that it is not a near-linear geometry.
Only in the special case that $d$ is a prime number and ${\mathbb Z}(d)$ is a field, the
${\mathbb Z}(d)\times {\mathbb Z}(d)$ is a near-linear geometry (the $GF(p^n)\times GF(p^n)$ is also a near-linear geometry).

In section II we study lines in ${\mathbb Z}(d)\times {\mathbb Z}(d)$.
In particular we show how symplectic transformations map the various lines into other lines.
We also discuss a factorization of lines in ${\mathbb Z}(d)\times {\mathbb Z}(d)$
in terms of `component lines' in ${\mathbb Z}(d_i)\times {\mathbb Z}(d_i)$
where $d=\prod d_i$ and any $(d_i,d_j)$ are coprime.
This section extends considerably our previous results for lines in ${\mathbb Z}(d)\times {\mathbb Z}(d)$ in ref.\cite{SV}.

In section III we discuss briefly quantum systems with variables in ${\mathbb Z}(d)$ in order to establish the notation.
In section IV we define weak mutually unbiased bases.
For simplicity we consider the case with $d=d_1d_2$, where $d_1$ and $d_2$ are prime numbers
(where $d_1<d_2$).
In this case the divisors of $d$ (apart from $1,d$) are $d_1, d_2$.
This leads to bases which are tensor products of mutually unbiased bases in the Hilbert spaces
$H_{d_1}$ and $H_{d_2}$, describing systems with variables in
${\mathbb Z}(d_1)$ and ${\mathbb Z}(d_2)$, correspondingly.
We have considered recently such bases in the context of tomography \cite{SV},
but here we arrive at these bases starting from the above definition for
weak mutually unbiased bases.
We note that a tensor product of  mutually unbiased bases in two Hilbert spaces has also been used in \cite{K}.
In section V we discuss the duality between
the weak mutually unbiased bases and the lines in the
${\mathbb Z}(d)\times {\mathbb Z}(d)$ phase space of these systems.
We conclude in section VI with a discussion of our results.

\section{Lines in ${\mathbb Z}(d)\times {\mathbb Z}(d)$}

If $d_i$ is a divisor of $d$, then $(d/d_i){\mathbb Z}(d_i)$ is the subgroup of ${\mathbb Z}(d)$
which consists of the elements $0,d/d_i,2d/d_i,...,[(d_i-1)d/d_i]$.
We factorize $d$ as
\begin{eqnarray}
d=\prod _{i=1}^Np_i^{e_i},
\end{eqnarray}
where $p_i$ are prime numbers.
The Dedekind function is given by
\begin{eqnarray}
\psi(d)=d\prod _{i=1}^N\left (1+\frac{1}{p_i}\right ).
\end{eqnarray}
The Jordan totient function $J_2(d)$ is given by:
\begin{eqnarray}
J_2(d)=d^2\prod _{i=1}^N\left (1-\frac{1}{p_i^2}\right )=\psi(d)\varphi (d)
\end{eqnarray}
Here $\varphi (d)$ is the Euler totient function.
These functions have been used in the context of finite systems in \cite{VB,P}.

A line through the origin in ${\mathbb Z}(d)\times {\mathbb Z}(d)$ is the set
\begin{eqnarray}\label{l}
{\cal L}(\nu,\mu)=\{(\nu \alpha,\mu\alpha)\;|\; \alpha \in {\mathbb Z}(d)\}
\end{eqnarray}
Mathematically this is a cyclic module generated by $(\nu,\mu)$.
We use a more physical language and we refer to it as a line through the origin.

The matrices
\begin{eqnarray}\label{SY}
g(\kappa, \lambda |\mu , \nu)\equiv
\left(\begin{array}{cc}
\kappa& \lambda\\
\mu& \nu\\
\end{array}
\right)
;\;\;\;\;\;\;{\rm det} (g)=\kappa \nu-\lambda \mu=1\;({\rm mod}\; d);\;\;\;\;\;
\kappa, \lambda, \mu, \nu \in {\mathbb Z}(d).
\end{eqnarray}
form the group $Sp(2,{\mathbb Z}(d))$.
The cardinality of this group is $dJ_2(d)$.

Acting with the matrix $g(\kappa, \lambda |\mu , \nu)$
on a point $(\rho, \sigma)\in {\mathbb Z}(d)\times {\mathbb Z}(d)$
we get the point $(\kappa \rho +\lambda \sigma, \mu \rho+\nu \sigma)$.
Acting with the symplectic matrix $g(\kappa, \lambda |\mu , \nu)$
on all the points of a line  ${\cal L}(\rho, \sigma)$ we get all the points of the line
${\cal L}(\kappa \rho +\lambda \sigma, \mu \rho+\nu \sigma)$ for which we also use the notation
$g(\kappa, \lambda |\mu , \nu){\cal L}(\rho, \sigma)$.

In \cite{SV} we have proved  various properties for lines in
${\mathbb Z}(d)\times {\mathbb Z}(d)$.
For completeness below we give these properties together with several new properties, but we prove only the new ones.
\begin{definition}
Lines ${\cal L}(\nu,\mu)$ through the origin, with exactly $d$ points are called
maximal lines.
\end{definition}
\begin{proposition}\label{1234}
\begin{itemize}
\mbox{}

\item[(1)]
If $\lambda$ is an inverible element (unit) in ${\mathbb Z}(d)$
then ${\cal L}(\nu,\mu)={\cal L}(\nu \lambda,\mu \lambda)$, and if
$\lambda$ is a non-inverible element in ${\mathbb Z}(d)$ then
${\cal L}(\nu \lambda,\mu \lambda)\subset {\cal L}(\nu,\mu)$.

\item[(2)]
The number of points in ${\cal L}(\nu,\mu)$ is $d/{\mathfrak G}(\nu, \mu, d)$
(where ${\mathfrak G}(\nu, \mu, d)$ is the greatest common divisor of these numbers).
If $d$ is a prime number, all lines are maximal (apart from the ${\cal L}(0,0)$
which consists of the origin only).

\item[(3)]
Let $d_i$ be a divisor of $d$.
Lines through the origin with exactly $d_i$ points, consists of points $(\rho, \sigma)$
with coordinates $\rho,\sigma$ in the $(d/d_i){\mathbb Z}(d_i)$ subgroup of ${\mathbb Z}(d)$.

\item[(4)]
Let $d_i$ be a divisor of $d$.
There is a total number of $\psi(d_i)$ lines through the origin with $d_i$ points.
Consequently, there are $\psi(d)$ maximal lines through the origin.

\item[(5)]
The intersection of two lines ${\cal L}(\nu ,\mu )$ and ${\cal L}(\rho, \sigma)$
is a `subline' and the number of its points is a divisor of $d$.
When $d$ is a prime number, two lines can only have one point in common.
\item[(6)]
The lines ${\cal L}(\rho, \sigma)$ and $g(\kappa, \lambda |\mu , \nu){\cal L}(\rho, \sigma)$
have the same number of points.

\item[(7)]
In the special case that $d$ is a prime number,
all lines through the origin are given by
\begin{eqnarray}
{\cal L}(0,1);\;\;\;\;g(0,1|-1,-\lambda){\cal L}(0,1);\;\;\;\;\lambda=0,1,..., d-1
\end{eqnarray}
\end{itemize}
\end{proposition}

\begin{proof}
Parts (1),(2) and (5) have been proved in \cite{SV} (proposition 2.1).
\begin{itemize}
\item[(3)]
The number of points in a line ${\cal L}(\rho, \sigma)$
through the origin, is $d/{\mathfrak G}(\rho, \sigma, d)$.
Therefore for lines with $d_i$ points, ${\mathfrak G}(\rho, \sigma, d)=d/d_i$.
This means that $\rho=\rho 'd/d_i$ where $\rho '$ is an element of ${\mathbb Z}(d_i)$
(and similarly for $\sigma$).

\item[(4)]
We have proved above that the lines ${\cal L}(\nu,\mu)$ with $d_i$ points have points with
components $\nu,\mu$ in the $(d/d_i){\mathbb Z}(d_i)$ subgroup of ${\mathbb Z}(d)$.
In addition to that there are $\psi (d)$ lines
through the origin ${\cal L}(\rho, \sigma)$ where $\rho,\sigma \in {\mathbb Z}(d)$
with exactly $d$ points. Consequently, there are $\psi (d_i)$ lines through the origin with
$d_i$ points.

\item[(6)]
Since $det(g)\ne 0$ two different points on a line are mapped to two other points which are different
from each other. This proves the statement.

\item[(7)]
We first point out that the line ${\cal L}(0, \tau)$ (with $\tau \ne 0$) is the same as the line
${\cal L}(0, 1)$.
We next prove that any other line ${\cal L}(\rho, \sigma)$ (which will have $\rho \ne 0$)
can be written as $g(0,1|-1,-\lambda){\cal L}(0,1)$.
We need to prove that given a point $(\alpha \rho, \alpha \sigma)$  on the line
${\cal L}(\rho, \sigma)$ (where $\alpha \ne 0$ for points other than the origin),
there exist a point  $(0, \beta)$ on the line
${\cal L}(0, 1)$ such that $\beta=\alpha \rho$ and $-\lambda \beta =\alpha \sigma$.
But this is true for $\beta =\alpha \rho$ and $\lambda=-\rho ^{-1}\sigma$ (the $\rho ^{-1}$ exists because
$d$ is a prime number and ${\mathbb Z}(d)$ is a field).
This completes the proof.
\end{itemize}
\end{proof}

The $\psi(d)$ maximal lines through the origin, have $d$ points each.
Except the origin, these lines have $\psi(d) (d-1)$ points in total, some of which have been counted many times
simply because the geometry is not near-linear and two lines might have many points in common.
But except the origin, there are only $d^2-1$ points in ${\mathbb Z}(d)\times {\mathbb Z}(d)$.
As a measure of the deviation of our geometry from a near-linear geometry, we introduce the redundancy parameter
\begin{eqnarray}\label{red}
{\mathfrak r}=\frac{\psi(d)(d-1)}{d^2-1}-1=\frac{\psi(d)}{d+1}-1.
\end{eqnarray}
When $d$ is a prime number, two different lines have at most one point in common, the geometry is near-linear,
and ${\mathfrak r}=0$.

Much work in finite geometries is on near-linear geometries where
two points belong to at most one line \cite{f1,f2,f3}.
Our geometry is {\bf not} a near-linear geometry
because two lines can have $d_1$ points in common, where $d_1|d$ ($d_1$ is a divisor of $d$).
Only in the case of prime $d$, the geometry is a near-linear geometry.

Different aspects of the ${\mathbb Z}(d)\times {\mathbb Z}(d)$ as a finite geometry have been studied in
\cite{r1,r2,r3,r4,r5,r6}. In particular we note that the isotropic lines in \cite{r5} correspond to the
maximal lines in the present work. In the present paper we emphasize the fact that two lines may have
many points in common, and this leads to the concept of sublines and to the
`geometric redundancy' in Eq.(\ref{red}).
Later this geometric redundancy, will be related through duality to
redundancy in the weak mutually unbiased bases.

\subsection{Factorization}

Let $d=d_1...d_N$ where any pair of $d_i,d_j$ are coprime.
Based on the Chinese remainder theorem, we can introduce one-to-one maps between
${\mathbb Z}(d)$ and ${\mathbb Z}(d_1)\times...\times {\mathbb Z}(d_N)$.
Good used them in fast Fourier transforms\cite{Good}.

Below we consider the case $d=d_1d_2$ where $d_1,d_2$ are prime numbers.
Let
\begin{equation}\label{20}
r_1=d_2;\;\;\;\;r_2=d_1;\;\;\;\;
t_i r_i=1\;( mod\ d_i);\;\;\;\;\;
s_i=t_i r_i\in {\mathbb Z}(d).
\end{equation}
The first map is
\begin{eqnarray}\label{map1}
m\leftrightarrow (m_1,m_2);\;\;\;\;\;
m_i=m (mod\ d_i);\;\;\;\;m=m_1s_1+m_2s_2
\end{eqnarray}
The second map is
\begin{eqnarray}\label{map2}
m\leftrightarrow (\overline m_1,\overline m_2);\;\;\;\;\;\overline m_i=mt_i=m_it_i(mod\ d_i);\;\;\;\;\;
m=\overline m_1 r_1+ \overline m_2 r_2 (mod\ d)
\end{eqnarray}
We next introduce the following bijective map
\begin{eqnarray}
{\mathbb Z}(d)\times {\mathbb Z}(d)\;\;\leftrightarrow\;\;[{\mathbb Z}(d_1)\times{\mathbb Z}(d_2)]
\times [{\mathbb Z}(d_1)\times {\mathbb Z}(d_2)]
\end{eqnarray}
given by
\begin{eqnarray}\label{map100}
(m,n)\leftrightarrow (m_1,m_2,\overline n_1,\overline n_2).
\end{eqnarray}
We use here the first map of Eq.(\ref{map1}) for $m \leftrightarrow (m_1,m_2)$
and the second  map of Eq.(\ref{map2}) for $n \leftrightarrow (\overline n_1,\overline n_2)$.
In ${\mathbb Z}(d_i)\times{\mathbb Z}(d_i)$ we consider the lines
\begin{eqnarray}
{\cal L}^{(i)}(\nu _i,\mu _i)=\{(\nu _i\alpha _i,\mu _i\alpha _i)\;|\;
\alpha _i\in {\mathbb Z}(d_i)\}.
\end{eqnarray}
We can rewrite the
line  ${\cal L}(\nu,\mu)$ in ${\mathbb Z}(d)\times{\mathbb Z}(d)$ given in Eq.(\ref{l}), in terms
of the two lines ${\cal L}^{(i)}(\nu _i,\mu_i)$ in ${\mathbb Z}(d_i)\times{\mathbb Z}(d_i)$, as
\begin{eqnarray}\label{589}
{\cal L}(\nu,\mu)=
{\cal L}^{(1)}(\nu _1,\overline \mu _1)\times {\cal L}^{(2)}(\nu _2, \overline \mu _2)
\end{eqnarray}
To prove this we use the fact that if $\alpha \leftrightarrow (\alpha _1,\alpha _2)$ and
$\nu\leftrightarrow (\nu_1,\nu _2)$
then $\alpha \nu\leftrightarrow (\alpha _1\nu_1,\alpha _2\nu_2)$.
In the case $(\nu _1,\overline \mu _1)\ne(0,0)$ and $(\nu _2,\overline \mu _2)\ne(0,0)$
the lines ${\cal L}^{(1)}(\nu _1,\overline \mu _1)$ and ${\cal L}^{(2)}(\nu _2, \overline \mu _2)$
have $d_1$ and $d_2$ points correspondingly, and the line
${\cal L}(\nu,\mu)$ has $d_1d_2$ points.
We refer to ${\cal L}^{(1)}(\nu _1,\overline \mu _1)$ and ${\cal L}^{(2)}(\nu _2, \overline \mu _2)$
as the first component line and the second component line of ${\cal L}(\nu,\mu)$.

\begin{proposition}
\mbox{}
\begin{itemize}
\item[(1)]
The set of $\psi(d)$ maximal lines in ${\mathbb Z}(d)\times {\mathbb Z}(d)$ through the origin,
(where $d=d_1d_2$ and $d_1,d_2$ are prime numbers)
is given by
\begin{eqnarray}\label{AS}
&&{\cal L}_1={\cal L}^{(1)}(0,1)\times {\cal L}^{(2)}(0,1)
;\;\;\;\;\nonumber\\
&&{\cal L}_{2+\lambda _2}=
{\cal L}^{(1)}(0,1)\times [g(0,1|-1,-\lambda _2){\cal L}^{(2)}(0,1)]\nonumber\\
&&{\cal L}_{2+d_2+\lambda _1}=
[g(0,1|-1,-\lambda _1){\cal L}^{(1)}(0,1)]\times {\cal L}^{(2)}(0,1)\nonumber\\
&&{\cal L}_{2+d_1+d_2+\lambda _2+\lambda _1d_2}=
[g(0,1|-1,-\lambda _1){\cal L}^{(1)}(0,1)]\times [g(0,1|-1,-\lambda _2){\cal L}^{(2)}(0,1)]
\end{eqnarray}
where $0\le \lambda _1\le d_1-1$ and $0\le \lambda _2\le d_2-1$.
\item[(2)]
These maximal lines through the origin, can also be derived through symplectic transformations in ${\mathbb Z}(d)\times {\mathbb Z}(d)$
acting on the points of the line ${\cal L}_1$, as follows:
\begin{eqnarray}\label{AS1}
&&{\cal L}_{2+\lambda _2}=g(s_1,t_2s_2|-d_1,s_1-\lambda _2s_2){\cal L}_1\nonumber\\
&&{\cal L}_{2+d_2+\lambda _1}=g(s_2,t_1s_1|-d_2,s_2-\lambda _1s_1){\cal L}_1\nonumber\\
&&{\cal L}_{2+d_1+d_2+\lambda _2+\lambda _1d_2}=
g(0,\eta|-d_1-d_2,-\lambda _1s_1-\lambda_2s_2){\cal L}_1
\end{eqnarray}
Here $\eta=t_1^2d_2+t_2^2d_1$.

\item[(3)]
Acting with the matrices $g(\kappa, \lambda |\mu , \nu)$ on a maximal line ${\cal L}(\rho, \sigma)$
through the origin, we get all $\psi (d)$ maximal lines through the origin.

\end{itemize}
\end{proposition}
\begin{proof}
\mbox{}
\begin{itemize}
\item[(1)]
The lines ${\cal L} ^{(1)}(0,1)$, ${\cal L}^{(2)}(0,1)$ have $d_1, d_2$
points, correspondingly.
According to proposition \ref{1234} the lines
$g(0,1|-1,-\lambda _1){\cal L}^{(1)}(0,1)$, $g(0,1|-1,-\lambda _2){\cal L}^{(2)}(0,1)$ also have
$d_1, d_2$ points, correspondingly.
Consequently all lines in Eq.(\ref{AS}) have $d=d_1d_2$ points each, and they are maximal lines.
These lines are different from each other, and Eqs.(\ref{AS}) give
$\psi(d)=(d_1+1)(d_2+1)$ such lines.
Since the number of maximal lines in ${\mathbb Z}(d)\times {\mathbb Z}(d)$ through the origin
is $\psi(d)$, Eqs.(\ref{AS}) give all maximal lines.
\item[(2)]
We have proved in \cite{SV} that
\begin{eqnarray}\label{D175}
g(\kappa, \lambda |\mu, \nu)=g^{(1)}(\kappa _1, \lambda _1 r_1|\overline \mu _1, \nu _1)\otimes
g^{(2)}(\kappa _2, \lambda _2 r_2|\overline \mu _2, \nu _2)
;\;\;\;\;
\kappa, \lambda, \mu, \nu \in {\mathbb Z}(d)
\end{eqnarray}
Here the $\kappa _i, \lambda _i , \nu _i \in {\mathbb Z}(d_i)$ are the components of
$\kappa , \lambda , \nu $ correspondingly, according to the map of Eq.(\ref{map1}).
${\overline \mu} _i \in {\mathbb Z}(d_i)$ are the components of $\mu$
according to the map of Eq.(\ref{map2}).
Using this on the lines in Eq.(\ref{AS}) which are products of two lines in
${\mathbb Z}(d_1)\times {\mathbb Z}(d_1)$ and ${\mathbb Z}(d_2)\times {\mathbb Z}(d_2)$,
we get the lines in Eq.(\ref{AS1}) in ${\mathbb Z}(d)\times {\mathbb Z}(d)$.

\item[(3)]

For $d=d_1d_2$ where $d_1,d_2$ are prime numbers, given an arbitrary line factorized as in Eq.(\ref{589}),
we act with symplectic transformations and using Eq.(\ref{D175}) we get
\begin{eqnarray}\label{5890}
g(\kappa, \lambda |\mu, \nu){\cal L}(\nu,\mu)=g^{(1)}(\kappa _1, \lambda _1 r_1|\overline \mu _1, \nu _1)
{\cal L}^{(1)}(\nu _1,\overline \mu _1)\times
g^{(2)}(\kappa _2, \lambda _2 r_2|\overline \mu _2, \nu _2){\cal L}^{(2)}(\nu _2, \overline \mu _2)
\end{eqnarray}
For a prime number, Eq.(6) shows that we can act with symplectic transformations on
${\cal L}(0,1)$ and get all lines through the origin (these lines are maximal lines).
Therefore we can get any line by acting with symplectic transformations on any other line.
This in conjuction with Eq.(\ref{5890}) proves the statement.

\end{itemize}
\end{proof}

We note that $d_i$ are prime numbers and the ${\mathbb Z}(d_i)\times{\mathbb Z}(d_i)$
geometry is near-linear geometry, i.e., two lines have at most one point in common.
The following proposition explains how in ${\mathbb Z}(d)\times{\mathbb Z}(d)$
two lines may have one or $d_1$ or $d_2$ points in common.

\begin{proposition}
Let
${\cal L}(\nu,\mu)={\cal L} ^{(1)}(\nu _1,\overline \mu _1)\times {\cal L}^{(2)}(\nu _2, \overline \mu _2)$
and
${\cal L}(\nu',\mu')={\cal L}^{(1)}(\nu _1',\overline \mu _1')\times {\cal L}^{(2)}(\nu _2',\overline \mu _2')$
be a pair of maximal lines through the origin in ${\mathbb Z}(d)\times{\mathbb Z}(d)$.
There are $\psi(d)[\psi (d)-1]/2$ such pairs
(here the pairs are not ordered and each pair contains two lines which are different from each other).
Then one of the following holds:
\begin{itemize}
\item[(1)]
The two lines have $d_2$ points in common if and only if the second component line is the same
(i.e., $\nu_2=\alpha \nu_2'$ and $\overline \mu_2=\alpha \overline \mu_2'$ where
$\alpha\in {\mathbb Z}(d_2)$ and $\alpha \ne 0$).
There are $d_1\psi(d)/2$ such pairs of lines.

\item[(2)]
The two lines have $d_1$ points in common if and only if the first component line is the same
(i.e., $\nu_1=\alpha \nu_1'$ and $\overline \mu_1=\alpha \overline \mu_1'$ where
$\alpha\in {\mathbb Z}(d_1)$ and $\alpha \ne 0$).
There are $d_2\psi(d)/2$ such pairs of lines.

\item[(3)]
The two lines have only the origin in common, if and only if both component lines are different.
There are $d\psi(d)/2$ such pairs of lines.
\end{itemize}

\end{proposition}
\begin{proof}
We have $\psi(d)$ maximal lines and from this follows that there are
$\psi(d)[\psi (d)-1]$ ordered pairs of lines (where the lines in each pair are different from each other).
Therefore we have $\psi(d)[\psi (d)-1]/2$ pairs of lines which are not ordered.

We next consider three cases:
\begin{itemize}
\item[(1)]
The second component lines of the two lines are the same, and the first component lines
${\cal L}^{(1)}(\nu _1',\overline \mu _1')$ and ${\cal L}^{(1)}(\nu _1,\overline \mu _1)$
are maximal lines in ${\mathbb Z}(d_1)\times{\mathbb Z}(d_1)$
and they have only the origin $(0,0)$ in common (because $d_1$ is a prime number).
The $(0,0)$ combined with the $d_2$ points in ${\cal L}^{(2)}(\nu _2, \overline \mu _2)$
as described in Eq.(\ref{map100}), produce the $d_2$ common points between the lines
${\cal L} ^{(1)}(\nu _1,\overline \mu _1)\times {\cal L}^{(2)}(\nu _2, \overline \mu _2)$
and ${\cal L}^{(1)}(\nu _1',\overline \mu _1')\times {\cal L}^{(2)}(\nu _2,\overline \mu _2)$.

It is easily seen that there are
$[d_1(d_1+1)][d_2+1]/2$ pairs of lines, and this is equal to $d_1\psi (d)/2$.

\item[(2)]
The proof here is analogous to the previous case.

\item[(3)]
Both component lines in the first line, are different from their counterparts in the second line.
If the two lines had a point $(\nu,\mu)\ne (0,0)$ in common, then
the corresponding point $(\nu _1,\overline \mu _1)$ would be a common point of the first component lines
and the corresponding point $(\nu _2,\overline \mu _2)$ would be a common point of the second component lines.
But at least one of the $(\nu _1,\overline \mu _1)$
and $(\nu _2,\overline \mu _2)$ would be different from
$(0,0)$. This would mean that lines in ${\mathbb Z}(d_i)\times{\mathbb Z}(d_i)$ would have more
than one point in common, which is not possible because $d_i$ is a prime number.

There are $[d_1(d_1+1)][d_2(d_2+1)]/2$ pairs of lines in this case, and this is equal to $d\psi (d)/2$.

\end{itemize}

The inverse of these statements, follows immediately from the fact that above we have
exhausted all possible cases
where the two lines have $1,d_1,d_2$ points in common.

\end{proof}

\begin{remark}\label{Rema}
We consider the set ${\cal S}$ of all maximal lines through the origin in
${\mathbb Z}(d)\times {\mathbb Z}(d)$ (where $d=d_1d_2$ and $d_1,d_2$ are prime numbers with
$d_1<d_2$).
We will introduce a partition of this set such that any two lines in each subset intersect only at the origin.
In order to do this we introduce the notation
\begin{eqnarray}
{\mathfrak L}^{(1)}_{-1}&=&{\cal L}(0,1)\nonumber\\
{\mathfrak L}^{(1)}_{\lambda}&=&g(0,1|-1,-\lambda){\cal L}(0,1);\;\;\;\;\;\lambda=0,...,d_1-1
\end{eqnarray}
The lines ${\mathfrak L}^{(1)}_i$ are defined for $i=-1,...,d_1-1$.
It will be convinient to take $i\in {\mathbb Z}(d_1+1)$.
Using similar notation for the second subsystem, we consider the sets
\begin{eqnarray}
{\cal S}_n=\{{\mathfrak L}^{(1)}_i\times {\mathfrak L}^{(2)}_{i+n}\;|\;i\in {\mathbb Z}(d_1+1)\}
;\;\;\;\;\;n\in {\mathbb Z}(d_2+1)
\end{eqnarray}
The cardinality of ${\cal S}_n$ is $d_1+1$ and the cardinality of ${\cal S}$ is
$\psi(d)=(d_1+1)(d_2+1)$.
Any two lines in the same subset ${\cal S}_n$ intersect only at the origin.
Lines in different subsets might intersect at more points.
\end{remark}

\section{Quantum systems with variables in ${\mathbb Z}(d)$}

We consider a quantum system with positions and momenta in ${\mathbb Z}(d)$ (with odd $d$).
The corresponding Hilbert space $H_d$ is $d$-dimensional.
We consider two bases, the positions $|{X};m\rangle$ and
the momenta $|{P};m\rangle$ which are related through the Fourier transform:
\begin{equation}\label{PPPT}
|{P};n\rangle=F|{X};n\rangle;\;\;\;\;
F=d^{-1/2}\sum _{m,n}\omega (mn)\ket{X;m}\bra{X;n};\;\;\;\;
\omega(m)=\exp \left (i\frac {2\pi m}{d}\right ).
\end{equation}
We also consider the displacement operators
\begin{eqnarray}
&&{Z}^ \alpha =\sum _{n\in {\mathbb Z}_d}\omega (n \alpha)|{X};n\rangle \langle {X};n|;\;\;\;\;\;
{X}^\beta=\sum _{n\in {\mathbb Z}_d}\omega (-n \beta)|{P};n\rangle \langle {P};n|\nonumber\\
&&X^\beta \;Z^\alpha = Z^\alpha \;X^\beta \;\omega (-\alpha \beta);\;\;\;\;\;X^d=Z^d={\bf 1}
\end{eqnarray}
General displacement operators are
${D}(\alpha, \beta)={Z}^\alpha \;{X}^\beta \;\omega (-2^{-1}\alpha \beta)$.

We also define the symplectic transformations as
\begin{eqnarray}\label{DD175}
&&X'=S(\kappa, \lambda |\mu, \nu)\;X\;[S(\kappa, \lambda |\mu, \nu)]^{\dagger}
=D(\lambda ,\kappa)\nonumber\\
&&Z'=S(\kappa, \lambda |\mu, \nu)\;Z\;[S(\kappa, \lambda |\mu, \nu)]^{\dagger}=D(\nu,\mu)\nonumber\\
&&\kappa \nu -\lambda \mu=1;\;\;\;\;\;\kappa, \lambda, \mu, \nu \in {\mathbb Z}_d.
\end{eqnarray}
Explicit formulas for $S(\kappa, \lambda |\mu, \nu)$ have been given in \cite{1,Vie}.
Acting with them on the positions and momentum bases we get new bases:
\begin{eqnarray}\label{Sym7}
\ket{X(\kappa, \lambda|\mu,\nu);n} \equiv S(\kappa, \lambda|\mu,\nu)\ket{X;n};\;\;\;\;\;
\ket{P(\kappa, \lambda|\mu,\nu);n} \equiv S(\kappa, \lambda|\mu,\nu)\ket{P;n}
\end{eqnarray}

Mutually unbiased bases are a set
of orthonormal bases $\ket{{\mathfrak X}_i;n}$ where $n\in {\mathbb Z}(d)$
and the index $i$ takes values in some set $S$, such that
for all $i,j\in S$ with $i\ne j$
\begin{eqnarray}\label{8}
|\langle {\mathfrak X}_i;n\ket{{\mathfrak X}_j;m}|=d^{-1/2}
\end{eqnarray}
It is known that for a prime number $d$ the states
\begin{eqnarray}\label{S100}
\ket{X;m};\;\;\;\;\ket{X(0,1|-1,-\lambda );m};\;\;\;\;\;\lambda=0,1,...,d-1
\end{eqnarray}
are mutually unbiased.

Based on the maps  in Eqs(\ref{map1}), (\ref{map2})
we can factorize a system with variables in ${\mathbb Z}(d)$
in terms of two subsystems with variables in ${\mathbb Z}(d_1)$ and ${\mathbb Z}(d_2)$
(where $d=d_1d_2$ and $d_1,d_2$ are prime numbers)\cite{FA1,FA2}.
There is an isomorphism between the Hilbert space $H_d$
and the product of the Hilbert spaces of the subsystems
${H}_{d_1}\otimes {H}_{d_2}$, where
the position states in $H_d$ correspond to the products of position states in $H_{d_i}$, as follows:
\begin{equation}\label{nn}
|X; m\rangle \leftrightarrow |X^{(1)}; \overline m_1\rangle \otimes|X^{(2)}; \overline m_2\rangle.
\end{equation}
The momentum states in $H_d$ correspond to the products of
position states in $H_{d_i}$, as follows:
\begin{equation}\label{nn}
|P;m\rangle \leftrightarrow |P^{(1)}; m_1\rangle \otimes |P^{(2)}; m_2\rangle
\end{equation}

\section{Weak mutually unbiased bases}

Below $d=d_1d_2$ where $d_1,d_2$ are prime numbers (different than $2$).

\begin{definition}\label{def}
We consider a set $\ket{{\mathfrak B}_i;m}$ of $\ell$ orthonormal bases in the Hilbert space $H_d$,
where $m\in{\mathbb Z}(d)$ and $i=1,...,\ell$.
Let
\begin{eqnarray}
f_{ji}(n,m)=|\langle {\mathfrak B}_j;n \ket{{\mathfrak B}_i;m}|;\;\;\;\;\;f_{ij}(m,n)=f_{ji}(n,m)
\end{eqnarray}
We call them weak mutually unbiased bases if for any pair of different bases
($i\ne j$), one of the following holds:
\begin{itemize}
\item[(1)]
\begin{eqnarray}\label{B1}
f_{ji}(n,m)&=&d_1^{-1/2};\;\;{\rm for\;the}\;d_1d\;{\rm pairs}\; (n,m)\in {\mathbb Z}(d)\times {\mathbb Z}(d)\;\; {\rm such\;\; that}\;\;n=m\;({\rm mod}\;d_2)\;\;
\nonumber\\
f_{ji}(n,m)&=&0;\;\;\;{\rm for \;\;the \;\;rest}\;\;(n,m)\;\;
\end{eqnarray}

\item[(2)]
\begin{eqnarray}\label{B2}
f_{ji}(n,m)&=&d_2^{-1/2};\;\;{\rm for\;the}\;d_2d\;{\rm pairs}\;(n,m)\in {\mathbb Z}(d)\times {\mathbb Z}(d)\;\; {\rm such\;\; that}\;\;n=m\;({\rm mod}\;d_1)
\nonumber\\
f_{ji}(n,m)&=&0;\;\;\;{\rm for \;\;the \;\;rest}\;\;(n,m)
\end{eqnarray}

\item[(3)]
\begin{eqnarray}\label{B3}
f_{ji}(n,m)=(d_1d_2)^{-1/2};\;\;\;\;{\rm for\;\;all}\;\;(n,m)
\in {\mathbb Z}(d)\times {\mathbb Z}(d)
\end{eqnarray}

\end{itemize}
\end{definition}

\begin{theorem}
\begin{itemize}
\mbox{}
\item[(1)]
Any set of weak mutually unbiased bases in $H_d$ can be written as
$\ket{B_j^{(1)};{\overline m_1}}\otimes \ket{B_j^{(2)};{\overline m_2}}$
where $\ket{B_j^{(1)};{\overline m_1}}$ is a set of mutually unbiased bases in
$H_{d_1}$, and $\ket{B_j^{(2)};{\overline m_2}}$ is a set of mutually unbiased bases in
$H_{d_2}$.
Some of the  bases $\ket{B_i^{(1)};{\overline m_1}}$ with different index $i$ might be the same basis,
and similarly for $\ket{B_i^{(2)};{\overline m_2}}$.

\item[(2)]
The maximum number of weak mutually unbiased bases is $\psi(d)$.
In this case, there are $\psi(d)[\psi(d)-1]/2$ sets of values $f_{ji}(n,m)$
with $i,j=1,..., \psi (d)$ and $i\ne j$.
From them, $d_1\psi(d)/2$ belong to the first category of Eq.(\ref{B1}),
$d_2\psi(d)/2$ belong to the second category of Eq.(\ref{B2}),
and $d\psi(d)/2$ belong to the third category of Eq.(\ref{B3}).

\end{itemize}
\end{theorem}

\begin{proof}
\mbox{}
\begin{itemize}

\item[(1)]
We are given a set $\ket{{\mathfrak B}_i;m}$ of $\ell$ orthonormal bases in the $d$-dimensional
Hilbert space $H_d$,
where $m\in{\mathbb Z}(d)$ and $i=1,...,\ell$.
As explained above, using Eq.(\ref{map2}) this can be factorized as
\begin{eqnarray}
\ket{{\mathfrak B}_i;m}=\ket{B_i^{(1)};{\overline m_1}}\otimes \ket{B_i^{(2)};{\overline m_2}}
;\;\;\;\;\;i=1,...,\ell
\end{eqnarray}
where $\ket{B_i^{(1)};{\overline m_1}}$ is an orthonormal basis in $H_{d_1}$ and
$\ket{B_i^{(2)};{\overline m_2}}$ is an orthonormal basis in $H_{d_2}$.
We need to prove that the set $\{\ket{B_i^{(1)};{\overline m_1}}|i=1,...,\ell\}$,
is a set of mutually unbiased bases in $H_{d_1}$.
The same applies to the second subsystem.

We consider the following three cases:
\begin{itemize}
\item[(a)]
For $i\ne j$ we consider the case of Eq.(\ref{B1}).
According to the map of Eq.(\ref{map2}), $n\leftrightarrow (\overline n_1,\overline n_2)$
and $m\leftrightarrow (\overline m_1,\overline m_2)$ and for $n=m\;({\rm mod}\;d_2)$ we get
\begin{eqnarray}\label{qa1}
|\langle B_i^{(1)};{\overline n_1}\ket{B_j^{(1)};{\overline m_1}}|=\frac{1}{d_1^{1/2}
|\langle B_i^{(2)};{\overline n_2}\ket{B_j^{(2)};{\overline m_2}}|}
\end{eqnarray}

The condition $n=m\;({\rm mod}\;d_2)$ gives $\overline n_2=\overline m_2$.
As $(n,m)$ take all values in ${\mathbb Z}(d)\times {\mathbb Z}(d)$ such that $n=m\;({\rm mod}\;d_2)$,
the $(\overline n_1,\overline m_1)$ take all values in ${\mathbb Z}(d_1)\times {\mathbb Z}(d_1)$.

From Eq.(\ref{qa1}) follows that
$d_1^{-1/2}\le |\langle B_i^{(1)};{\overline n_1}\ket{B_j^{(1)};{\overline m_1}}|\le 1$.
In addition to that
\begin{eqnarray}
\sum _{{\overline m_1}\in {\mathbb Z}(d_1)}|\langle B_i^{(1)};{\overline n_1}\ket{B_j^{(1)};{\overline m_1}}|^2=1
\end{eqnarray}
From this we conclude that $|\langle B_i^{(1)};{\overline n_1}\ket{B_j^{(1)};{\overline m_1}}|=d_1^{-1/2}$, for
all $(\overline n_1,\overline m_1)$ in ${\mathbb Z}(d_1)\times {\mathbb Z}(d_1)$.
Therefore the $\ket{B_i^{(1)};{\overline n_1}}$ are mutually unbiased bases in $H_{d_1}$.
Since there are $d_1+1$ mutually unbiased bases in $H_{d_1}$,
some of the bases $\ket{B_i^{(1)};{\overline n_1}}$ with different index $i$, might be the same basis.

In this case $|\langle B_i^{(2)};{\overline n_2}\ket{B_j^{(2)};{\overline m_2}}|=1$
and therefore
$\ket {B_i^{(2)};{\overline m_2}}$ and $\ket{B_j^{(2)};{\overline n_2}}$
are the same basis in $H_{d_2}$.
Here ${\overline n_2}={\overline m_2}$ because $n=m\;({\rm mod}\;d_2)$.

\item[(b)]
The second case corresponding to Eq.(\ref{B2}) is similar to the above case.
\item[(c)]
For $i\ne j$ we consider the case of Eq.(\ref{B3}).
According to the map of Eq.(\ref{map2}), $n\leftrightarrow (\overline n_1,\overline n_2)$
and $m\leftrightarrow (\overline m_1,\overline m_2)$, and we get
\begin{eqnarray}\label{qa2}
|\langle B_i^{(1)};{\overline n_1}\ket{B_j^{(1)};{\overline m_1}}|
|\langle B_i^{(2)};{\overline n_2}\ket{B_j^{(2)};{\overline m_2}}|=(d_1d_2)^{-1/2}
\end{eqnarray}
We will prove that $|\langle B_i^{(2)};{\overline n_2}\ket{B_j^{(2)};{\overline m_2}}|$
is constant for all $\overline m_2 \in {\mathbb Z}(d_2)$.
We consider the overlap of the vector  $\ket{{\mathfrak B}_j;n}$ in the $i$-basis as in Eq.(\ref{qa2}),
with another vector $\ket{{\mathfrak B}_j;m'}$ in the $j$-basis such that $m=m'\;({\rm mod}\; d_1)$.
Then $\overline m_1=\overline m_1'$ and as
$m'$ takes all values in ${\mathbb Z}(d)$ such that $m=m'\;({\rm mod}\; d_1)$,
the $\overline m_2'$ takes all values in ${\mathbb Z}(d_2)$.
We get
\begin{eqnarray}\label{qa3}
|\langle B_i^{(1)};{\overline n_1}\ket{B_j^{(1)};{\overline m_1}}|
|\langle B_i^{(2)};{\overline n_2}\ket{B_j^{(2)};{\overline m_2}'}|=(d_1d_2)^{-1/2}
\end{eqnarray}
From Eqs.(\ref{qa2}),(\ref{qa3}) we see that $|\langle B_i^{(2)};{\overline n_2}\ket{B_j^{(2)};{\overline m_2}}|$
is constant for all $\overline m_2 \in {\mathbb Z}(d_2)$.
This in conjuction with the relation
\begin{eqnarray}\label{qa4}
\sum _{{\overline m_2}\in {\mathbb Z}(d_2)}|\langle B_i^{(2)};{\overline n_2}\ket{B_j^{(2)};{\overline m_2}}|^2=1
\end{eqnarray}
leads to the result that $|\langle B_i^{(2)};{\overline n_2}\ket{B_j^{(2)};{\overline m_2}}|=d_2^{-1/2}$.
In a similar way we prove that
$|\langle B_i^{(1)};{\overline n_1}\ket{B_j^{(1)};{\overline m_1}}|=d_1^{-1/2}$.
Therefore the $\ket{B_i^{(1)};{\overline m_1}}$ are mutually unbiased bases in $H_{d_1}$ and
$\ket{B_i^{(2)};{\overline m_2}}$ are mutually unbiased bases in $H_{d_2}$.

\end{itemize}
\item[(2)]
Since the number of mutually unbiased bases in $H_{d_1}$ is $d_1+1$ and in $H_{d_2}$ is $d_2+1$, we conclude that
the maximum number of weak mutually unbiased bases is $\psi (d)=(d_1+1)(d_2+1)$.
It is now clear that there are $\psi(d)[\psi(d)-1]/2$ sets of values $f_{ji}(n,m)$
with $i,j=1,..., \psi (d)$ and $i\ne j$ and we now
prove that $d_1\psi(d)/2$ of them belong to the first category of Eq.(\ref{B1}).

We proved earlier that in the case of Eq(\ref{B1}),
\begin{eqnarray}
\ket{{\mathfrak B}_i;m}=\ket{B_i^{(1)};{\overline m_1}}\otimes \ket{B_i^{(2)};{\overline m_2}}
;\;\;\;\;\;\ket{{\mathfrak B}_j;m}=\ket{B_j^{(1)};{\overline n_1}}\otimes \ket{B_j^{(2)};{\overline n_2}}
\end{eqnarray}
where $\ket{B_i^{(1)};{\overline m_1}}$ and $\ket{B_j^{(1)};{\overline n_1}}$
are mutually unbiased bases in $H_{d_1}$ and
$\ket{B_i^{(2)};{\overline m_2}}$ and $\ket{B_j^{(2)};{\overline n_2}}$ are the same basis in $H_{d_2}$.
We have $(d_1+1)^2-(d_1+1)=d_1(d_1+1)$ pairs of
$\ket{B_i^{(1)};{\overline m_1}}$,$\ket{B_j^{(1)};{\overline n_1}}$
bases, with $i\ne j$. We multiply this with the number $d_2+1$ of mutually unbiased bases in $H_{d_2}$
and we prove that there are $d_1(d_1+1)(d_2+1)/2=d_1\psi(d)/2$ sets of values $f_{ji}(n,m)$
which belong to the first category of Eq.(\ref{B1}).
The denominator $2$ is due to the symmetry $f_{ij}(m,n)=f_{ji}(n,m)$.

In a similar way we prove that
$d_2\psi(d)/2$ sets of values $f_{ji}(n,m)$ belong to the second category of Eq.(\ref{B2}).
The total number of sets of values $f_{ji}(n,m)$ is $\psi(d)[\psi(d)-1]/2$ and therefore
$d\psi(d)/2$ of them belong to the third category of Eq.(\ref{B3}).

\end{itemize}
\end{proof}

\begin{remark}
A pair of bases which satisfy Eq.(\ref{B3}) are mutually unbiased.
But bases which belong to different such pairs
are not necessarily mutually unbiased.
\end{remark}

\subsection{An explicit construction of weak mutually unbiased bases}

There exists at least one set of weak mutually unbiased bases.
In \cite{SV}, we have introduced a set of $\psi (d)$ orthonormal bases in $H_d$ which are tensor products of
$d_1+1$ mutually unbiased bases in $H_{d_1}$  with $d_2+1$ mutually unbiased bases in $H_{d_2}$.
We have shown that this set of bases obeys the requirements in definition 4.1.
Here we summarize briefly this construction in order to present
in the next section, the duality between
lines in ${\mathbb Z}(d)\times {\mathbb Z}(d)$ and weak mutually unbiased bases.
\begin{eqnarray}\label{S}
&&\ket{{\mathfrak B}_1;m}=\ket{X^{(1)};{\overline m}_1}\otimes \ket{X^{(2)};{\overline m}_2}=\ket{X(1,0|0,1);{m}}\nonumber\\
&&\ket{{\mathfrak B}_{2+\lambda _2};m}=\ket{X^{(1)};{\overline m}_1}\otimes \ket{X^{(2)}(0,1|-1,-\lambda _2);{\overline m}_2}
=\ket{X(s_1, t_2s_2|-d_1, s_1-\lambda _2s_2);{m}}\nonumber\\
&&\ket{{\mathfrak B}_{2+d_2+\lambda _1};m}=
\ket{X^{(1)}(0,1|-1,-\lambda _1);{\overline m}_1}\otimes \ket{X^{(2)};{\overline m}_2}=
\ket{X(s_2,t_1s_1|-d_2,-\lambda _1s_1+s_2);{m}}\nonumber\\
&&\ket{{\mathfrak B}_{2+d_1+d_2+\lambda _2+\lambda _1d_2};m}=
\ket{X^{(1)}(0,1|-1,-\lambda _1);{\overline m}_1}\otimes \ket{X^{(2)}(0,1|-1,-\lambda _2);{\overline m}_2}
\nonumber\\&&=
\ket{X(0,\eta, |-d_1-d_2, -\lambda_1s_1-\lambda _2 s_2);{m}}
\end{eqnarray}
Here $0\le \lambda _1\le d_1-1$, $0\le \lambda _2\le d_2-1$, and the
variables $s_1,t_1,s_2,t_2$ have been defined in Eq.(\ref{20}).

\begin{remark}
In \cite{Kantor} it has been shown that there many sets of unitarily inequivalent mutually unbiased bases.
This will lead to many unitarily inequivalent weak mutually unbiased bases.
Below we show the duality between lines in ${\mathbb Z}(d)\times {\mathbb Z}(d)$ and
the weak mutually unbiased bases constructed explicitly through symplectic transformations, above.
\end{remark}

\begin{remark}\label{Rema1}
This remark is analogous to remark \ref{Rema} for lines.
We use the notation
\begin{eqnarray}\label{AS1}
\ket{{\mathfrak X}_{-1}^{(1)};{\overline m_1}}&=&\ket{X^{(1)};\overline m_1}\nonumber\\
\ket{{\mathfrak X}_{\lambda _1}^{(1)};{\overline m_1}}&=&\ket{X^{(1)}(0,1|-1,-\lambda _1);
{\overline m}_1};\;\;\;\;\lambda _1=0,...,p_1-1,
\end{eqnarray}
and define a partition of the set of weak mutually unbiased bases , as follows:
\begin{eqnarray}
{\cal T}_\ell&=&\{\ket{{\mathfrak X}_{i_1}^{(1)};{\overline m}_1}\otimes \ket{{\mathfrak X}^{(2)}_{i_1+\ell};
{\overline m}_2}\;|\;
i_1\in {\mathbb Z}_{p_1+1}\} ;\;\;\;\ell\in {\mathbb Z}_{p_2+1}
\end{eqnarray}
We have explained in \cite{SV} that
the bases in the same set ${\cal T}_\ell$ are mutually unbiased.
\end{remark}

\section{Duality between lines in ${\mathbb Z}(d)\times {\mathbb Z}(d)$ and weak mutually unbiased bases}

There exists a correspondence (duality) between the lines in ${\mathbb Z}(d)\times {\mathbb Z}(d)$ and the
weak mutually unbiased bases.
The `dictionary' for this duality is as follows:
\begin{itemize}
\item
The line ${\cal L}_i$
corresponds to the basis $\ket{{\mathfrak B}_i;m}$.
We note that the same parameters are used in the symplectic transformations in Eq.(\ref{AS1})
for ${\cal L}_i$, and in the
symplectic transformations in Eq.(\ref{S}) for $\ket{{\mathfrak B}_i;m}$.
\item
The $\psi (d)$ maximal lines through the origin, correspond to the $\psi(d)$ weak mutually unbiased bases.
\item
A pair of maximal lines through the origin which have only the origin as common point,
correspond to a pair of weak mutually unbiased bases with absolute value of the overlap equal to $d^{-1/2}$ (Eq.(\ref{B3})). In this case the pair of bases is mutually unbiased.
We have seen that there are $d\psi(d)/2$ pairs of maximal lines through the origin
with this property, and also
$d\psi(d)/2$ pairs of weak mutually unbiased bases with the corresponding property.
\item
A pair of maximal lines through the origin which have $d_1$ points in common,
correspond to a pair of weak mutually unbiased bases with absolute value of the overlap equal to $d_2^{-1/2}$ (Eq.(\ref{B2}))
We have seen that there are $d_2\psi(d)/2$ pairs of maximal lines through the origin
with this property, and also
$d_2\psi(d)/2$ pairs of weak mutually unbiased bases with the corresponding property.
\item
A pair of maximal lines through the origin which have $d_2$ points in common,
correspond to a pair of weak mutually unbiased bases with absolute value of the overlap equal to $d_1^{-1/2}$
(Eq.(\ref{B1}))
We have seen that there are $d_1\psi(d)/2$ pairs of maximal lines through the origin
with this property, and also
$d_1\psi(d)/2$ pairs of weak mutually unbiased bases with the corresponding property.
\item
In Eq.(\ref{red}) we introduced a redundancy parameter that measures the deviation of our finite geometry from the near-linear geometries.
In Eq.(72) of ref.\cite{SV} we introduced another redundancy parameter for the bases that
we now call weak mutually unbiased bases, and we
have explained that it measures the `tomographical overcompleteness' of these bases.
These two parameters are equal, which indicates that the concept of weak mutually unbiased bases is
taylored for the geometry of ${\mathbb Z}(d)\times {\mathbb Z}(d)$.
In the case of fields (prime $d$), the geometric redundancy is ${\mathfrak r}=0$ and the weak
mutually unbiased bases, become mutually unbiased bases.

The concept of mutually unbiased bases, is intimately linked to the number of degrees of freedom in
the density matrix, without any redundancy.
In tomography experiments, probabilities are measured along lines in ${\mathbb Z}(d)\times {\mathbb Z}(d)$.
The fact that lines in this geometry may intersect at many points, leads to the geometric
redundancy of Eq.(\ref{red}). This makes necessary a redundancy in the bases linked with such measurements,
and this leads to the concept of weak mutually unbiased bases.
\end{itemize}

\subsection{Example for the case $d=15$}

We consider the case $d=15$.
In table I we show the $\psi (15)$ maximal lines ${\cal L}(\rho,\sigma)$ through the origin in ${\mathbb Z}(15)\times {\mathbb Z}(15)$ factorized in terms of
the lines ${{\cal L}^{(1)}}(\rho_1,\overline \sigma _1)$ and ${{\cal L}^{(2)}}(\rho_2,\overline \sigma _2)$, in
${\mathbb Z}(3)\times {\mathbb Z}(3)$ and ${\mathbb Z}(5)\times {\mathbb Z}(5)$, correspondingly.
In table II we show the weak mutually unbiased bases $\ket{B_j;m}$ in $H_{15}$
factorized in terms of the mutually unbiased bases
$\ket{B_j^{(1)};\overline m_1}$ and $\ket{B_j^{(2)};\overline m_2}$ in $H_3$ and $H_5$, correspondingly.

As an example we compare the line ${\cal L}_4={\cal L}(3,7)=g(10,12|13,13){\cal L}_1$ with the basis
$\ket{B_4;m}=\ket{X(10,12|12,13);m}$.
It is seen that symplectic transformations with the same parameters appear in both cases.

According to the factorization in Eq.(\ref{589}), the line ${\cal L}_4={\cal L}(3,7)$ is factorized as
${\cal L}_4={{\cal L}^{(1)}}(0,2)\times {{\cal L}^{(2)}}(3,4)$.
We have explained earlier that ${\cal L}(\nu,\mu)={\cal L}(\nu \lambda,\mu \lambda)$
for any invertible element in ${\mathbb Z}(d)$.
Therefore ${{\cal L}^{(1)}}(\rho_1,\overline \sigma _1)=
{{\cal L}^{(1)}}(\lambda _1\rho_1,\lambda _1\overline \sigma _1)$ for any $\lambda _1\in {\mathbb Z}(3)$
with $\lambda _1\ne 0$, and similarly  ${{\cal L}^{(2)}}(\rho_2,\overline \sigma _2)=
{{\cal L}^{(2)}}(\lambda _2\rho_2,\lambda _2\overline \sigma _2)$ for any $\lambda _2\in {\mathbb Z}(5)$
with $\lambda _2\ne 0$.
From this follows that ${\cal L}_4={{\cal L}^{(1)}}(0,1)\times {{\cal L}^{(2)}}(1,2)$, which appears in the table.
The basis $\ket{B_4;m}$ is factorized as $\ket{B_4;m}=\ket{X^{(1)};\overline m_1}\times \ket{X^{(2)}(0,1|-1,-2);
\overline m_2}$. Details of this factorization have been given in \cite{SV}.

In table III we present a partition of the set of the maximal lines through the origin in
${\mathbb Z}(15) \times {\mathbb Z}(15)$ (as discussed in remark \ref{Rema}).
All the lines in the set ${\cal S}_j$ intersect only in the origin.
Lines in different sets may have many points in common.
In table IV we present a partition of the set of the weak mutually unbiased bases (for the case $d=15$).
Bases in the same column are mutually unbiased bases. (as discussed in remark \ref{Rema1}).

\section{Discussion}

The concept of mutually unbiased bases, is related to the number of degrees of freedom in
the density matrix. It appears to be an ideal concept for optimal tomography, because it has no redundancy.
On the other hand we have explained in this paper that in ${\mathbb Z}(d)\times {\mathbb Z}(d)$
phase space, there is a geometric redundancy in the sense that lines may have many points in common.
The geometry is {\bf not} a near-linear geometry and this leads to the concept of sublines.
Probabilities measured along different lines in tomography experiments, are not independent
(they should obey the constraints in Eq.(34) in ref.\cite{SV}).

The concept of weak mutually unbiased bases incorporates this redundancy.
There is a duality in the sense that the properties of maximal lines have counterparts in the properties of bases.
For simplicity the work has been presented for the special case that $d$ is the product of two prime numbers.
The generalization to a product of many prime numbers is straightforward. For example
if $d=d_1d_2d_3$ where $d_1, d_2, d_3$ are prime numbers (different from each other), the definition \ref{def}
will contain all divisors $d_1, d_2, d_3, d_1d_2, d_2d_3, d_1d_3$ (to the power $-1/2$).
In this case the  weak mutually unbiased bases are tensor products of
mutually unbiased bases in three subsystems with dimensions $d_1,d_2,d_3$.
In the case that $d$ contains powers of prime numbers, we will use Galois fields for the
labelling of the states in the component subsystems.

Quantum tomography is an important technique for state reconstruction in finite quantum systems
\cite{W1,W2,W3,W4,W5,W6,W7,W8,W9,W10,W11}.
The weak mutually unbiased bases can be used in this context.
Other techniques in this general context are symmetric informationally complete
positive operator valued measures \cite{POVM1,POVM2} and designs\cite{d1,d2,d3,d4}.

In summary we have introduced the concept of weak mutually unbiased bases, which is motivated by the
geometrical properties of the ${\mathbb Z}(d)\times {\mathbb Z}(d)$ phase space.

redpajama_set_name: RedPajamaArXiv


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The choirs of Quincy University will host two Christmas concerts on Sunday, December 6 entitled "The Light of Christmas". Concerts begin at 3:00 and 7:30
QU Garden Provides Unique Service Learning Opportunity
Seeking opportunities to give back to local communities, service learning students at Quincy University continue a "growing" tradition started in 2012. QU Garden Service Learning
QU Announces 25 Million Campaign
QUINCY — Quincy University announced the launch of a $25 million comprehensive campaign titled "Forever Forward" at a private gala reception on October 24
New Scholarship Program, Raise.me, Now at QU
Quincy University is now offering a new scholarship program (Raise.me) to support current high school students interested in pursuing post-secondary education. The QU Trust scholarship
Quincy University Projects Largest Incoming Class in Ten Years
Quincy University announced its 2015-16 enrollment figures and projects its largest incoming class in a decade. A total of 281 freshmen students are enrolled for
Quincy University Announces Record QU Fund Goal Results
Quincy University exceeded its $1 million QU Fund goal for the fifth straight year. The fund raised more than $1,093,000 for the university's endowment and
QU Hosts 152nd Commencement Ceremony
Quincy University hosted its 152nd Commencement ceremony on Sunday, May 17 in Quincy University's Pepsi Arena. The 2015 graduating class is comprised of 295 undergraduates
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redpajama_set_name: RedPajamaCommonCrawl


Set in Los Ajaches Nature Reserve near the southern tip of Lanzarote is El Papagayo beach. Actually there are four beaches collectively termed the Coasta de Papagayo but it is Playa de la Cera that is the best known. Frequently described as the most beautiful beach on Lanzarote Papagayo comprises a small crescent of white sand protected by cliffs to either side. Unlike many of the islands beaches Playa de la Cera has a cove like feeling. Whilst this is generally a good thing it can make it feel a little claustrophobic in the peak season when visitors from the nearby Playa Blanca resort pile down to the beach. Still on the right day it's a great spot to lounge around on the beach looking across to the neighbouring island of Fuerteventura and cooling off in the clear blue sea.


Turn up the heat and get sweaty in the hotel sauna, and on a perfect day you can have a perfect picnic at the picnic area. There's nothing more relaxing than soaking in the Marriott's Harbour Club jacuzzi, and the concierge is always at your service.
These are some other hotels near Marriott's Harbour Club, Hilton Head Island.
These outdoors activities are available near Marriott's Harbour Club.


The Continental Buffet & The Marine Breakfast Special this April at The Marine, Hermanus. Make April special by breakfasting at The Marine, Hermanus. Start your day with a lavish cold Continental Buffet or the hot Breakfast Special.
The Continental Buffet is set up each morning and guests can sit inside or take their pick from two al fresco options – either the courtyard by the pool, or the outside pavilion overlooking Walker Bay.
An impeccable selection of local cheeses and sliced cold meats awaits. Tea bread and health loaf make up the breakfast bread options. There will also be freshly baked pastries such as muffins, scones and croissants, which will be accompanied by homemade preserves like jam and marmalade.
There is a daily selection of fresh fruit juices and a choice of sliced fruit, fruit salad and poached fruit. Yoghurt flavours include low-fat fruit or summer fruit as well as unsweetened double-thick Greek-style yoghurt. For cereal lovers, cornflakes, muesli and bran flakes are provided.
The last option to choose from is the Franschhoek smoked salmon trout served with cream cheese, capers and lemon wedges. The cost for the buffet is R145 per person and includes tea and coffee.
If you prefer a cooked English-style breakfast over a Continental one, then opt for The Marine Breakfast Special instead. This includes a choice of eggs (poached, fried, boiled, scrambled), pork sausage, streaky bacon, mushrooms, baked beans, tomato and fried bread. End this scrumptious breakfast off with a lovely cup of tea and/or coffee. The cost for the buffet is R145 per person.
The Continental Buffet & The Marine Breakfast special are both available daily from 07h00 to 10h30 from 01-30 April 2019. Booking is essential.


Experience "Roo's Place", a peaceful retreat nestled in the forest. Make yourself at home in this private bright and cozy ski house located close to the best skiing, hiking, and craft breweries in Stowe, Vermont!
If your vision of the perfect getaway includes a warm, crackling fire, your search is over. After a day of fun in the fresh air build a roaring fire and settle in with a a good book, gather the group for a lively game of Codenames, or watch the latest movie on the flat screen TV.


Facebook is now become a primary source to engage with people or audience around the globe. Every single brands or events now prefer Facebook Campaign to start with. Marketing on Facebook is now consider the effective mean as compare to any other service either its Social Campaign, Commercial Campaign, Hollywood Campaign, Business Campaign or either its Political Campaign. Every campaign engages global audience with itself. So if you don't know how Political Election Campaign engage global audience and how it gets start and ends with positive or negative approach than the following Infographic about "How Political Election Campaigns Influence Voters on Facebook" will helps you to understand it.
Primary factor to engage global audience for Political Election is "Targeting", because 100 million potential voters are using social networks worldwide. So if any candidate wants to win Political Election in any country they must target their voters potential areas like mentioned in following Inforaphic.


After eight years of twice-a-week week blogging, Strobist was in 2014 restructured into a information archive. Travel Republic Limited is an appointed representative of ITC Compliance Limited which is authorised and regulated by the Financial Conduct Authority (their registration quantity is 313486) and which is permitted to advise on and arrange basic insurance coverage contracts. If you are already purchasing a travel blanket, and you like your accessories coordinated, then acquiring a travel blanket and pillow set makes a lot of senses.
Registering www.organization.travel as an alternative of (or an Internet country code prime-level domain such as orde) shows the planet that you are a travel and tourism related enterprise and recognized as portion of that neighborhood. Just a reminder that (a) not every little thing you study on the net is true, (b) google is not a substitute for an knowledge of a travel advisor, and (c) a great guide can add so a lot to your travel knowledge.
Time travel might very easily make them disappear or at least be radically different. It is usually a good practice to acquire travel insurance even if the person is not pregnant. If you're positive you need a large suitcase for extended overseas travel (or just to pack for numerous members of the family members), this bag is a great option. They readily adapted to living in the RV even though it took a tiny time for them to get used to traveling in a moving car.
I get lots of questions about living in Florence and emails asking for ideas for traveling in Italy. Though it is not a travel necessity, our ENO hammock has brought us a lot of excellent instances. We bought a travel trailer to travel and see new areas, but have been not enjoying the trials of trailering our delicate rig in its restricted range at its limited speeds. Travelling has constantly been a passion for me.I have visited about 20 nations and I am nevertheless not tired travelling whenever I differences in culture ,meals ,individuals makes me excited to travel far more. My young children are older now and it is considerably less complicated to travel with them I didn't do numerous trips alone with them when they had been tiny. A travel wardrobe need to be cautiously capsuled and as you know the perfect color scheme comes from 3 colors rather than two.
Becoming a travel nurse, then, can provide a safer place, much less anxiety, a more flexible schedule, and a richer personal life free of charge of the be concerned of higher-crime regions and high death prices among one's individuals. Mango and Sure Travel make no representation or warranty, whether or not express or implied, as to the operation, integrity, availability or functionality of the competitors information or linked sites or as to the accuracy, completeness or reliability of any information within this competitors details. In these cities, which must be on the quick list of any traveling photog, we cover almost everything you need to have to know before heading to one of these destinations with your camera.
Of course, it really is not always attainable to group up with an individual else, but there are approaches to make the journey less complicated. Just search for sites like Ibiza.travel, ThinkDigital.travel or Colombia.travel to see what we mean. I'm with Ardie – my husband is also a perform a holic….whilst I think it is fine I do like to travel very a bit. Travel when pregnant can be unpredictable, so you constantly want to be within attain. The approach of getting a short-term license varies amongst the states as effectively and a travel company recruiter will generally assist the traveling nurse to navigate this program successfully and act as a coach and advocate. I use to travel ten hours by automobile with two modest ones, seventeen months apart, to check out relatives. Provides ample coverage and only weighs eleven ounces, it is possibly one of the most compact travel blankets you will uncover, it fits into a cylinder seven inches extended by 4 inches in diameter. When traveling by auto, you will most likely want to keep your cat safely in her crate even though moving.
Bear in mind, every single individual in the traveling celebration must fill out and sign their own form, like those traveling as a household. Some people like to travel more than other people and I think it is in their gene pool- the predisposition to travel and explore, the adventure the conquest. Travel is such a wonderful adventure based on exactly where you are traveling to information improves along the way and to see various places is a wonderful concept. It's on the ground in front of the left-hand corner of the Lake Macalania Travel Agency, when facing the front door.
If you need to drag us cats on the airplane, we insist that you carry us in a sturdy airline-authorized cat carrier like my Sherpa Deluxe Bag. As an interested RN, LPN or Tech, you can search the Net internet sites of agencies that employ and represent travel nurses and determine if each business makes use of only RNs or includes LPNs, Techs and other experts in traveling health care service. My point is, if you travel by way of this vast nation you will encounter different levels of culture shock and you will love some areas and hate other individuals and that is ok! Thank you for supplying such a wealth of healthcare travel tips for your readers with chronic back pain or sciatica. Hi, my cat will be traveling internationally for the initial time (10 hour flight) and are trying to decide which size of Sherpa is greater- M -18x11x11 or L – 20×11.5×11.5. She is ten.five pounds.
Time travel may possibly effortlessly make them disappear or at least be radically distinct. It is often a good practice to get travel insurance even if the particular person is not pregnant. If you happen to be sure you require a huge suitcase for extended overseas travel (or just to pack for a number of members of the family), this bag is a fantastic alternative. They readily adapted to living in the RV even though it took a little time for them to get used to traveling in a moving automobile.
After eight years of twice-a-week week blogging, Strobist was in 2014 restructured into a knowledge archive. This video was taken around eight:20AM and was uploaded on The Travel Junkie Facebook page Our estimated time of arrival in Sapporo Station was around 5:30PM but there was a bit of a delay (SURPRISINGLY!) and we arrived at our final location ( which was our Airbnb ) around 7:30PM. A very good good quality travel blanket could last for a very long time, so do not forget to hold that in mind even though you are shopping. Just be respectful when it comes to peoples cultures and religion when traveling.
There are also travel agencies that serve as common sales agents for foreign travel companies, allowing them to have offices in countries other than exactly where their headquarters are positioned. If you are a U. S. citizen and want to travel to a single of its territories, you will not need to have a U. S. Passport. To that final item, several photo enthusiasts who travel do so overburdened both with camera gear and photo expectations when they hit the road.
But the reality is that only a extremely modest percentage of individuals who attempt will in fact make it as travel photographers. Good if you have a separate place for your tiny one or are traveling with a partner.) or just grab what you need and modify the baby in the bathroom. By participating in this competition, all entrants agree to be bound by these terms and situations and indemnify Mango and Certain Travel and their successors, employees, officers, suppliers, contractors, agents, consultants, directors and shareholders. Positive Travel common terms and conditions and Mango terms and circumstances apply.
Offering imported tea, espresso drinks, specialty coffee and handmade sandwiches, in a comforting atmosphere. I noticed that a lot of of the pet owners who have bought carriers on this web page have been purchasing these collapsible travel bowls for their pets. I am positive the day will come when I will have grandchildren and will really possibly be traveling with them the guidance will still apply. Search engines assign a value to thetravel extension due to the fact it has meaning through its direct connection to travel and tourism, unlikecom.
For European travel (my geographic region of expertise), the very best size depends on what kind of trip you've got planned. For any extended travel, nothing at all beats the Okkatots Travel Depot in my opinion. I have worked with and assistaed 1000's of traveling nurses with short-term housing. Neighborhood nursing schools and nursing licensing boards also sustain lists of neighborhood travel companies that hire nurses and these schools and boards are listed in regional telephone directories and in directories available on the Web. Most major airlines only permit you a single totally free checked bag (when traveling internationally), so decide on a size large sufficient to hold almost everything you require to take, even though keeping the weight beneath 50 lbs.
A single Friday evening during my years waiting tables, I picked up a shift for a co-worker. We just enjoy to pay a visit to our (grand)kids, we like to go to locations close to us, and we take pleasure in to be at residence, where we have got every thing we want: Garden, bed, fridge, kitchen, you name it. 1-day traveling is fine for us. And if we want to have a look at locations far away, we just do a left-click (or is it a right-click?) on Due to the fact there we'll find numerous extremely interesting – even thrilling – reports written by people, who love to travel.
Any participant in the travel and tourism business is eligible to register atravel name. From the airport, you have the exclusive yellow taxis that cater to airport travelers taking you into the distinct parts of Nairobi. I am a nurse cardiac technician from the Philippines and I am interested in operating as a travel nurse. We give Virtuoso and other Preferred Partner amenities as an affiliate of Brownell Travel. When we are on vacation or if we have to travel on enterprise, there are so a lot of things that we want to do. Nonetheless, keep in mind that pregnant ladies have a tendency to tire simply. If one strategy every thing all in advance, the travel experience will be much less stressful to the pregnant women. Our quite gray Shadow passed away a couple of years ago, but Gemini who is featured at the prime of the page, nonetheless travels with us. It's surprising to see the number of individuals who travel with cats these days.


Michigan Warbirds is located at the South Haven Area Regional Airport (LWA), South Haven, Michigan. South Haven is midway between Chicago Illinois and Grand Rapids MI off Interstate 196.
Please fill in the form below to submit your question or email me directly at michiganwarbirds@hotmail.com. I will get back to you as soon as possible. Thanks!


To Dust's Shawn Snyder on how the serendipitous discovery of the 1937 film The Dybbuk had a transformative effect on his new movie.
"There is a kind of wisdom one only gains through pain or loss."
Steven Sheil looks at why people watch and make horror films, and the profound emotions and questions that are inherently rooted in the genre.


text:
Ливърпул () е лек крайцер на Британския Кралски флот, от времето на Втората световна война от втората серия крайцери тип "Таун". Заложен е на 17 февруари 1936 г., спуснат на вода на 24 март 1937 г. и влиза в експлоатация на 2 ноември 1938 г. Корабът е 8-ми поред, който носи това име. Кръстница на кораба става мисис Монтагю Норман – съпругата на управителя на Банк Оф Ингланд.

Крайцерът участва във Втората световна война. В патрулите в Далечния Изток, сраженията в Средиземно море, Арктическите конвои. В хода на боевете на два пъти е сериозно повреден от италианските торпедоносци. Изваден от състава на флота през април 1952 г. и продаден за скрап.

Девизът на кораба е: "Deus nobis haec otia fecit" – Господ ни е дал свободата.

Втора световна война
Началото на Втората световна крайцерът посреща на Източноиндийската станция в състава на 5-та ескадра крайцери. На 6 септември той излиза да посрещне плаващата база "Лучия" в Оманския залив. От 10 септември е базиран в Аден, изпълнявайки функциите на прихващач на немски рейдери и блокадопробивачи. Тази задача изпълнява през целите септември и октомври. На 14 ноември крайцерът в преведен в състава на Китайската станция и прави преход до Хонконг с отбивки в Сингапур и Сайгон, където пристига на 10 декември. От 14 декември действа в китайски води, осъществявайки защита на търговията и прихващане на вражески съдове.

Инцидентът с "Асама-Мару"
Продължавайки през януари да патрулира във водите на Китай и Япония, на 21 число и на 35 мили (56 километра) от остров Нииджиима прихваща и обискира японския лайнер "Асама-Мару". По данни на Британското ръководство на борда му се намират немските моряци от търговския съд "Кълъмбъс", потопен до Източното крайбрежие на САЩ, които плават към Германия. Съответстно, командващия Китайската станция изпраща "Ливърпул" за прехващане на японския лайнер. Крайцерът пресича курса на лайнера на близко разстояние, карайки го да спре, след което изпраща обискираща партия в състав от 10 души, която, след като оглежда съда, сваля от него 21 пътника, предположително от състава на екипажа на "Кълъмбъс".

Четири дни по-късно ръководството на Nippon Yusen изпраща в отставка капитана на "Асама-Мару" – Ватанабе, инкриминирайки му "недостойно поведение". Въпреки това, японската страна отправя дипломатически протест, обявявайки действията на Великобритания за злоупотреба на воюваща страна по отношение на неутрална такава, което служи за повод за нарастване на напрежението между двете страни. Независимо от негативно отношение на японската общественост, двете правителства се стремят да урегулират конфликта чрез преговори. На 5 февруари страните стигат до договореност за освобождаването на 9 немци и задължения от страна на Япония да не допуска на свои съдове немски граждани на военнонаборна възраст.

На 28 януари крайцерът влиза за ремонт в Хонконг, с края на който е насочен към обичайната си служба.

Действия в Червено море
През април заедно с австралийския крайцер "Хобарт" е преведен в състава на морските сили в Червено море, където поддържа военните операции в Сомалиленд.

На 13 май заедно с линкора "Рамилиес", крайцерите "Кент" и "Хобарт", френския тежък крайцер "Сюфрен", разрушителите "Декой", "Дефендер" и шлюпа "Шоръм" съпровожда войсковия конвой US2, превозващ войски на ANZAC към Близкия изток, при неговия преход през Червено море. На 17 май крайцерът пристига в Порт Судан.

В Средиземно море
През юни крайцера влиза в състава на 7-ма крайцерска ескадра на Средиземноморския флот. В състава на флота от линкорите "Уорспайт" и "Малая", самолетоносача "Игъл", крайцерите на ескадрата и разрушители излиза за атаки над италианските северноафрикански конвои на позиция южно от Крит.

На 12 юни съвместно с крайцера "Глостер" и четири разрушителя прави обстрел на Тобрук. Крайцерите обстрелват вражеските тралчици и стария броненосен крайцер "Сан Джорджо", докато огъня на бреговите батареи не ги кара да отстъпят. Въпреки това, тралчикът "Джовани Берта" е потопен.

Бой с конвоя Есперо

На 27 юни 1940 г. крайцера под флага на вицеадмирал Джон Тови заедно с крайцерите "Кент", "Глостър", "Нептун", "Орион" и "Сидни" отплава за прикритие на крайцерите "Каледон" и "Кейптаун", съпровождащи конвой от Егейско море за Египет. По това време хидроплан "Съндърланд" открива западно от Занте италиански конвой в състав от 3 разрушителя, доставящ подкрепления за Тобрук. 7-ма крайцерска ескадра се отделя за прихващане на това съединение.

Италианският конвой е засечен от "Ливърпул" на 60 мили югозападно от нос Матапан в 18:30. В последващия бой, минималните разстояния в който надвишават 13 километра, е потопен разрушителя "Есперо". Двата други успяват да се откъснат и пристигат в Бенгази заедно със своя товар. "Ливърпул" в този бой получава попадение от 120 mm снаряд в кърмата, което не причинява значителни повреди. Британските крайцери при това изразходват над 5000 снаряда. На "Ливърпул" остават около 40 снаряда на ствол. Толкова големия разход на снаряди командващия флота адмирал Кънингам обяснява пред Адмиралтейството неопитността на екипажа и воденето на боя при настъпващ здрач. Разходът на боеприпаси води до отмяна на операцията "MA3" – провеждането от Малта на 2 конвоя: MF1 и MS1.

Боя при Калабрия

Операция "MA3" е повторена едва на 7 юли вече с друго име – "MA5", когато крайцерът в състава на съединение "A", заедно с линкорите "Уорспайт", "Малая", "Роял Соверен", самолетоносача "Игъл", крайцерите "Нептун", "Глостър", "Орион" и "Сидни" и разрушителите "Дейнти", "Дифендър", "Хести", "Хостайл", "Хиперион", "Айлекс", "Импириал", "Джуно", "Вемпайър" и "Вояджър", излизат като далечно прикритие на конвоите MF1 и MS1. На 8 юли съединението е подложено на бомбардировка от хоризонтални бомбардировачи, в хода на която попадение получава крайцерът "Глостър".

На 9 юли съединението участва в боя при Калабрия. "Ливърпул" и "Нептун" откриват огън в 15:22, 8 минути по-късно огън откриват италианските крайцери от разстояние 23 600 ярда (13,4 мили). В хода на продължилата престрелка линкорът "Уорспайт" постига попадение в италианския линкор "Джулио Чезаре" в 16:00, след което италианския флот напуска полето на боя. На 11 юли последват силни въздушни атаки. На 13 юли крайцерът получава повреди от поредните атаки.

На 21 юли крайцерът заедно с крайцера "Кейптаун", разрушителите "Стюарт", "Дейнти", "Дифендър" и "Даймънд" съпровожда конвоя в пристанища на Егейско море и се връща с конвоя в Александрия.

На 29 юли крайцерът получава попадение от бомба в предния мостик, която не се взривява, но пробива по пътя си през палубата кула "B". Това попадение за крайцера е могло да стане фатално.

На 30 юли във връзка с реорганизацията на леките сили на флота крайцера заедно с крайцерите "Глостър" и "Кент" е преведен в 3-та крайцерска ескадра на контраадмирал Edward de Faye Renouf и влиза за ремонт отнел целия август.

Последващи действия
В края на август, на 30-о число крайцерът излиза в морето заедно с линкорите "Уорспайт", "Малая", самолетоносача "Игъл", крайцерите "Кент", "Глостър", "Орион" и "Сидни", разрушителите "Декой", "Айлекс", "Империъл", "Янус", "Нубиън", "Вемпайър", "Вендета" и полския ORP "Харланд", като част от далечния ескорт на конвоя MS 2 за Малта (Операция "MB3"). На 2 септември флота се среща при Пантелерия със съединение в състав линкора "Валиънт", самолетоносача "Иластриъс", крайцерите "Калкута" и "Ковънтри", являващи се подкрепление за флота (Операция "HATS"). На 4 септември флота предприема въздушни атаки против Родос, а се връща в Александрия на 5 число.

На 17 септември "Иластриъс" предприема въздушна атака над Бенгази. Той е съпровождан, освен от "Ливърпул", от линкора "Валиънт", тежкия крайцер "Кент", крайцерите "Глостър", "Орион" и разрушителите "Нубиън", "Мохавк", "Хироу", "Хести", "Хиперион", "Хиуорд", "Декой", "Джарвис" и австралийския "Уотърхен". Самолетите на самолетоносача атакуват съдовете в залива, а също поставят мини. В резултат – с торпеда са потопени разрушителят "Борея" и параходите "Глория Стела" (5490 брт) и "Мария Евгения" (4702 брт), а разрушителят "Аквилон" се натъква на мина и също потъва.

След това крайцерът "Кент" и разрушителите "Нубиън" и "Мохавк" се отделят, за да атакуват в ранното утро италианските позиции в Бардия. Малко преди полунощ "Кент" е торпилиран в кърмата от италиански торпедоносец и получава тежки повреди. След това "Кент" е отбуксиран в Александрия от разрушителя "Нубиън" под прикритието на крайцера "Орион", крайцера за ПВО "Калкута" и разрушителите "Мохавк", "Джарвис", "Джуно", "Янус" и австралийския "Вендета".

На 22 септември крайцерът съпровожда самолетоносача "Иластриъс" по време на поредното излизане на флота в Източното Средиземноморие. "Ливърпул" се отделя заедно с крайцерите "Глостър", "Кент", и разрушителите "Мохавк" и "Нубиън" за съпровождение на конвоя от Александрия за Нафплион.

На 28 септември "Ливърпул" съвместно с крайцера "Глостър" прехвърля в Малта 1200 души от състава на RAF и подкрепления за острова, при преходе корабите са прикривани от крайцера "Сидни".

Първо торпедно попадение
На 8 октомври крайцерът отплава в състава на флота, състоящ се от самолетоносачите "Игъл" и "Иластриес", линкорите "Малая", "Ремилис" и "Уорспайт", крайцерите "Аякс", "Глостър", "Орион", "Йорк" и "Сидни", разрушителите "Дейнти", "Декой", "Янус", "Джарвис", "Джуно", "Стюарт", "Вемпайър" и полския "Харланд" за превеждане на конвой MW 2 и настрещния конвой ME 4 от празни съдове в Египет. На 11 октомври съединението прикрива въздушните атаки на самолетите на "Игъл" и "Иластриес" над Лерос. На 12 октомври съединението е атакувано от италиански миноносци, един от които, "Артилиере", е потопен от огъня на крайцера "Йорк".

На 14 октомври, рано вечерта, след поредните атаки на Лерос, флота се насочва вкъщи, в Александрия. В това време крайцера е атакуван от два италиански торпедоносеца, бавно летящи на ниска височина. В 18:50 той получава попадение на торпедо в носовата част по десния борд. Взрива довежда до разлив на бензин от резервоар с авиационно гориво. На кораба има около 5000 галона високооктаново гориво за корабните хидроплани и гориво за корабните катери. Разлялото се гориво се смесва със 70 тона задбордна вода и протича на жилищната палуба. Независимо от издадената заповед за забрана за тютюнопущене и палене на огън, горивото се възпламенява от късо съединение. В 19:20 последва мощен взрив, който причинява големи разрушения в носовата част на кораба и изважда от строй кула "A", като откъсва нейния покрив. Повредите са толкова сериозни, че екипажът започва да подготвя спасателните средства.

В това време крайцерът "Орион" под прикритие на крайцерите за ПВО "Калкута" и "Ковънтри", а също и разрушителите "Дейнти", "Декой", "Даймънд" и "Вемпайър" вземат повредения крайцер на буксир и се насочва за Александрия на скорост 8 възела.

На 15 октомври при буксировката на крайцера се откъсва и потъва носовата част. През същия ден в морето погребват 12 загинали члена на екипажа, един от които е неразпознат. Още трима умират през нощта преди крайцерите да достигнат Александрия на 16 октомври. След това "Ливърпул" влиза за дълъг предварителен ремонт до март 1941 г. Командира на кораба Артър Дънкан Рид в края на октомври е преведен на линкора "Ремилис", а него заменят с Албърт Лорънс Поланд.

През ноември адмиралтейството обявява, че в резултат на атаката са загинали 3 офицера и 27 матроса. Още 35 души са ранени. Радарите наистина са засекли приближаващите се самолети и до флота даже е изпратено предупреждение, но неопитните вахтени оператори не успяват точно да идентифицират целите и не предупреждават капитана на мостика за заплахата.

В хода на предварителния ремонт е произведена и поставена врееменна носова част на кораба, след което той на 30 април 1941 г. през Индийския и Тихия океани отплава за ремонт в САЩ отбивайки се в Аден (5 – 6 май), Коломбо (12 – 13 май), Сингапур (17 – 20 май), Манила (24 – 25 май), Хонолулу (7 – 9 юни) и Сан Франциско (16 юни), пристигайки в корабостроителницата Мер Айлънд в Калифорния. Ремонтът продължава до октомври, но до септември присътствието на кораба се пази в тайна, докато Министерството на военноморския флот не публикува списък от 12 кораба, намиращи се в пристанища на САЩ.

Крайцерът напуска корабостроителницата на 3 ноември, пристигайки в Норфолк на Източното крайбрежие на САЩ на 17 ноември. След това крайцерът се отправя за вкъщи със спиране на Бермудите, пристигайки в Грийнок едва на 5 декември 1941 г., за да завърши поставянето на новите радиолокационни системи и провеждане на довършителните работи на отминалия ремонт.

През януари на крайцера са монтирани радарите: Type 273 за надводно следене, Type 281 за въздушно, Type 284 – артилерийски радар на главния калибър и Type 285 – артилерийски радар за 4-дюймовите зенитни оръдия.

Северните конвои
На 6 февруари крайцерът пристига в Скапа Флоу, влизайки в състава на флота на метрополията.

През март крайцера влиза в състава на 18-а ескадра крайцери, изпълняваща защитата на конвоите и носене на служба в северозападните подходи. Излиза заедно с крайцерите "Тринидад", "Лондон" и "Кент" в патрул източно от острова Ян Майен за прикритие на прехода на конвоя PQ-12, насочващ се към Колския залив.

На 12 април влиза в състава на ескорта на обратния конвой QP-10 заедно с разрушителите "Пенджаби", "Ориби", "Марн", "Фюри" и "Еклипс", прикриван от основните сили на Хоум Флийта. При прехода конвоя в течение на три дни е подложен на атаки от подводници и авиация, губейки от атаките три съда потоплени и един повреден. На 18 април крайцерът се отделя от конвоя, който пристига на 21 число в Рейкявик.

На 13 май крайцерът в състава на част от флота отплава за прикритие на завръщащия се от СССР повреден крайцер "Тринидад". Групата от линкора "Дук оф Йорк", самолетоносача "Викториъс", крайцерите "Кент", "Лондон", "Нигерия" и "Норфолк" се разполага западно от острова Медвежий в Баренцево море, но така и не успяват да посрещнат завръщащия се крайцер. На 14 май той е поразен от бомба, и екипажа му го изоставя. На обратния път съединението също е подложеное на атаки.

На 23 май крайцерът отново е в морето, този път в състава на съединение с крайцерите "Кент", "Нигерия" и "Норфолк" и разрушителите "Икарус", "Марн", "Онслоу" и "Ориби". Те прикриват поредния конвой – PQ-16 в състав от 35 съда. Започвайки от 25 число последват въздушни атаки на луфтвафе, повредили парахода "Карлтон". Съвместните атаки на подводници и самолети, направили 242 полета водят до загубата за 26 – 27 число на 7 съда. Но "Ливърпул" се отделя от конвоя на 26 число, за съпровождение на настрешния конвой QP-12, а се връща в базата си на 28 май.

Отново в Средиземно море
На 5 юни крайцера излиза в морето като част от ескорта на специалния конвой ws19s с крайцера "Кения" и разрушителите "Бедуин", "Ескапейд", "Икарус", "Марн", "Мачлес", "Онслоу", и ескортните миноносци "Бедсуорт", "Бленкни", "Мидълтън" и полския "Куявяк". Конвоя е предназначен за превоз на припаси за Малта – операция "Харпун". На 11 юни, в Гибралтар, крайцера влиза в състава на съединение W, прикриващ конвоя в прехода му през Средиземно море. В състава на съединението, освен "Ливърпул", влизат линкора "Малая", самолетоносачите "Игъл" и "Аргус" и крайцерите "Кения" и "Харибдис". На 13 юни корабите на съпровождението провеждат зареждане с гориво. в 17:25 се показва и първият самолет на противника – само разуснавач.

Второ торпедно попадение
Вече на 14 юни конвоя югозападно от Сардиния е подложен на атака от 38 самолета на Оста. В хода на атаката е потопен транспортът "Танимбар", а "Ливърпул" е атакуван от торпедоносци и получава попадение от авиоторпедо. Първите две торпеда пресичат курса на кораба и минават по левия борд, още едно торпедо минава зад кърмата, но 4-тото торпедо от италиански торпедоносец уцелва десния борд в района на машинното отделение и отново осакатява крайцера. Това се случва в 11:25 в точката с координати . Нахлуването на вода предизвиква крен от седем градуса и блокира рулевото управление. Скоростта на кораба пада до 4 възела. Също са блокирани подемниците за боеприпаси на кули "X" и "Y". На крайцера има 15 убити и 22 ранени.

В 12:40 крайцерът е взет на буксир от разрушителя "Антилопа" и под ескорт на разрушителя "Уесткот" и корветите "Джонквил" и "Спири" и се насочва за Гибралтар, но още в 16:26 буксира отдава конците, тъй като в 16:31 последва атака от 5 изтребител-бомбардировача Fiat CR.42. Но попадение не последва. В 17:28 крайцерът отново е взет на буксир, но буквално за няколко минути. В 18:00 последва поредната атака, този път от хоризонтални бомбардировачи: 1 Piaggio P.108 и 8 CANT Z.1007 Alcione. Отново без попадения, отново е взет на буксир едва в 19:45. В 20:15 последва втора атака от хоризонтални бомбардировачи – 5 Cant Z.1007. Без попадения. в 22:25 крайцера е подложен на атака 6 торпедоносеца. Без последствия.

На 15 юни в 14:22 крайцера отбива атаката на поредните три торпедоносеца. В 15:35 към отряда се присъединява буксира "Салвония", на който "Антилопа" и предава крайцера вечерта.

На 16 юни отряда е подложен на поредното нападение на вражеската авиация, но допълнителните повреди крайцера получава погрешка от интензивния зенитен огън на разрушителя "Уесткот". На крайцера има много ранени, в т.ч. и трима убити.

На 17 число в 17:15 крайцерът пристига в Гибралтар и още в 19:20 към обследване на подводната част пристъпват водолазите. На крайцера е направен подготовителен ремонт за прехода му през океана във Великобритания. За там крайцера потегля едва на 5 август. С пристигането му във Великобритания крайцера влиза за повече от двегодишен ремонт в корабостроителницата "Dockyard" в Розайт. В бойни действия крайцера няма възможност повече да участва.

Коментари

Литература

 Патянин С. В., Дашьян А. В. и др. Крейсера Второй мировой. Охотники и защитники. – М.: Коллекция, Яуза, ЭКСМО, 2007. – 362 с. – (Арсенал коллекция). – ISBN 5-699-19130-5.

Външни препратки
 GLOUCESTER light cruisers (1938 – 1939)

Л

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text:
Q: Send object to Java EE-server by Json I'm trying to send a simple object from an IOS-client to a JAX-RS-resource by Json. The json at the client-side look like this:
{"longitude": -122.136034834497, "latitude": 37.3666157838699}

And the JAX-RS-method looks like this:
@Path("/save")
@POST
@Produces("application/json")
@Consumes("application/json")
public MyObject saveQuizWalk(MyObject obj)
        throws JSONException {

    return myBean.saveObject(obj);
}

The java class looks like this:
public class MyObject {

private double longitude, latitude;

-----get/set------

If i convert an object at the server side to json, the json-string is formatted like this:
{"longitude":11.11,"latitude":12.11}

The request from the client looks like this:
        // Convert object to json
    let json = JSONSerializer.toJson(myObject.createJsonFriendlyObject()                   )
    print(json)
    // create the request & response
    let request = NSMutableURLRequest(URL: NSURL(string: "someurl")!)
    var response: NSURLResponse?

    // create some JSON data and configure the request

    request.HTTPBody = json.dataUsingEncoding(NSUTF8StringEncoding, allowLossyConversion: true)
    request.HTTPMethod = "POST"
    request.setValue("application/json", forHTTPHeaderField: "Content-Type")

    do {
        // send the request
        try NSURLConnection.sendSynchronousRequest(request, returningResponse: &response)
    }
    catch {

    }

The response from the server is 400. Why is that?
Hank

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