# The Theory of Fundamental Duality, Quantum Dualiton and Topological Dual
Invariance

B.T.T.Wong111CERN, The University of Hong Kong<EMAIL_ADDRESS>
###### Abstract

Fundamental duality is a concept which refers to two irreducible,
heterogeneous principles which are in opposite and complementary of each
other. The complementary principle in quantum mechanics is also praised by
Bohr. This important concept is known to appear in a lot of places in our
physical universe, however a rigorous mathematical definition and physics
theory has not yet ever developed in a formal way. In this paper, we establish
a formalism for fundamental duality and study its various properties and
theorems. One of the most profound results is that we establish a relation
between dual invariance and topological invariance, and we find that the
topological Chern-Simons form is a dual invariant action. Finally we apply the
concept of duality to study dual state oscillation, and predict a theoretical
new matter of state of dualiton, which is the particle excitation of the dual
field by second quantization. This new exotic quasi-particle state is expected
to have an impact in particle physics and condensed matter physics.

Keywords: Fundamental Duality, Klein-4 group, representation theory, quantum
field theory, Chern-Simons form, dual invariance, dualiton

##  1 Introduction

Fundamental duality 222The duality we refer here is different from the duality
in common physics terms, such as the duality in particle-wave duality and
T-duality, S-duality in string theory. To avoid confusion, we use the word
“fundamental”., conceptually is simply the study of elements with opposite
nature. For example, $+$ and $-$ is a dual pair in which $+$ and $-$ are
opposite; the two spin states of an electron, spin-up $\uparrow$ and spin-down
$\downarrow$ is also a dual pair. Macroscopic and microscopic world are dual
to each other that the former follows the law of general relativity and it is
absolutely deterministic while the latter follows the law of quantum mechanics
which is probabilistic. There are numerous examples that duality shows its
appearance.

A fundamental dual system contains two opposite elements, and the two opposite
elements form a fundamental dual pair. One element is dual to the other, and
vice versa. We can see that such duality is a fundamental property in nature
and it appears everywhere in our universe. Although the concept of fundamental
duality is simple, it can be extremely profound and far more difficult than
what it seems. It is a no easy task to define rigorously by means of
mathematical definition for each duality system.

It is well-agreed that two things which are dual to each other have opposite
properties, and they are never identical. We will never say an element is same
as the dual element. However, we will like to introduce a very important idea
for observable frame or observation perspective, that allows us to make
equivalent statement of dual elements in a system.

The next question for fundamental duality is very deep, it addresses the
problem of whether fundamental duality is always conserved. In most of the
cases, we may agree that duality is generally conserved where we have two
sides equally. However, in reality, especially from the perspective of
physics, this is not true. Here we will give profound examples from particle
physics. We know that a fermion (spin-$\frac{1}{2}$) must have its own
antiparticle. The electron particle $e^{-}$ has its own antiparticle positron
$e^{+}$ [1, 2, 3]. This particle-antiparticle pair share the same mass but
opposite charge. However, in our universe, matter dominates over antimatter
naturally by an order of parts per $\sim 10^{9}$, and this is the long-lasting
unsolved problem of matter antimatter asymmetry problem in physics [4]. Next,
we know that parity is conserved in electromagnetic force and strong force,
but is violated in weak force in a maximal manner due to the vertex minus
axial vector $V-A$ structure [5, 6]. In the Higgs mechanism, the process of
spontaneous symmetry breaking is to pick a positive vacuum $+v$ over a $-v$
one in which both happens to be equally probable, such that this allows
particles to gain mass, i.e. ‘to bring them into existence’ [7, 8, 9, 10, 11,
12]. If all particles have no mass, basically our universe literally cannot
exist realistically. Hence, in nature, fundamental duality sometimes
conserves, but sometimes not. And it is very difficult to answer when it is
conserved and when it is not. So often if it is not conserved it is maximally
violated.

The context of this subject becomes even more difficult when there are
different layers of fundamental duality superimposing all at once in one
collective framework, known as multi-duality. Then the extraction of
information in the framework is highly non-trivial.

In the first part of this paper, it is written in the aim of developing formal
mathematical formalism that can define fundamental duality properly, in
conjunction with observation perspective. We would like to transform the
abstract concept of fundamental duality into the language of rigorous
mathematics. The construction is based on the parity group $\mathbb{Z}_{2}$ ,
which is a dual symmetry. Much work is also established for
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$, which is the double duality symmetry
group, mathematically the Klein-4 group. It is then followed to study the
general multi-duality symmetry group
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}$.
Therefore, groups and representation theory are used throughout the text.
Secondly, the idea of dual symmetry is integrated with quantum mechanics. In
particular, in simple words we can formulate the 2 dual elements in a
fundamental duality system as two states $|0\rangle$ and $|1\rangle$. We will
give a thorough establishment for a novel theoretical development, and
construct a number of theorems on this subject. To study how we can interpret
information of dual systems, ideas from entropy in information theory is used,
and this give a useful way to study fundamental dual systems.

In the second part of the paper, we will apply the concept of duality and
multi-duality symmetry quantum field theories. First, for scalar field
theories, there are remarkable consequences of duality symmetries in high-
order interaction terms. This enriches the study of quantum field theory on
behalf to the traditional ones. Next, we discover a very important equivalent
relation between topological invariance and dual invariance in the Chern-
Simons field theory. Finally, we study the subject of duality wave, which is a
wave oscillation of dual states. Upon quantization, it promotes to the duality
field and gives a prediction of a new state of matter-the quanta of duality
wave known as _dualiton_. This quasi particle is expected to give rise a new
exotic matter state in particle physics and condensed matter physics.

##  2 The Theory of Duality and Fundamental Formalism

###  2.1 Single Duality Structure

We will begin by introducing a full set of definitions for duality.

######  Definition 2.1.

(I) Let $U$ be an element set and its dual $U^{*}$, where there exists a one-
to-one bijective map on $U$ and $U^{*}$. Define the duality map $*$, as a
function $*:U\rightarrow U^{*}$ which is a representation of the parity group
$\mathbb{Z}_{2}$. The inverse is just the map itself $*:U^{*}\rightarrow U$.
The double duality map $**$ is the identity map $I_{d}$, that $**:U\rightarrow
U$ and $**:U^{*}\rightarrow U^{*}$. $U$ and $U^{*}$ are said to be dual if
$U\cap U^{*}=\emptyset$ under the $*$ map. Define the zero set as $\\{0\\}$.
The complete duality set $W$ is defined by $W=U\cup U^{*}\cup\\{0\\}$. The
concept of zero is introduced such that $U=U^{*}$ is dual invariant if and
only if $U=U^{*}=\\{0\\}$. For $u_{i}\in U$ and $u_{i}^{*}\in U^{*}$, there
exist a map $\cdot$ such that $u_{i}\cdot u_{i}^{*}=0$, where $0$ here is the
zero element in the zero set.

(II) The duality set embedded in an extrinsic observer frame in $k$-dimension
is said to be a complete single duality structure. The observer’s frame of in
$k$ dimension forms a duality $S_{k}$ and $S_{k}^{\star}$. Let the duality
operator for observer’s frame be a map $\star:S_{k}\rightarrow S_{k}^{*}$,
which is a is a representation of the parity group $\mathbb{Z}_{2}$. The
inverse is the operator itself, $\star:S_{k}^{\star}\rightarrow S_{k}$, and
$\star\star$ is the identity map $I_{d}$. We define the zero set for observer
as $\\{0\\}_{S_{k}}$. For observer at $\\{0\\}_{S_{k}}$, it is defined as the
intrinsic observer of the duality system. We concern the extrinsic frame, and
define the complete observer’s as $B=S_{k}\cup S_{k}^{\star}$. There exists a
set in the duality set which is independent of the observer’s frame, which is
the zero set $\\{0\\}\in W$. The complete duality structure is defined as
$\\{W,B\\}$.

(III) Each set or dual set have to be observer’s frame specific. In the
$S_{k}$ observer’s frame, we specify the set and dual set under the observer
frame is denoted by $(U|S_{k})$ and $(U^{*}|S_{k})$ respectively. In the
$S_{k}^{*}$ observer’s frame, we have $(U|S_{k}^{\star})$ and
$(U^{*}|S_{k}^{\star})$ respectively.

(IV) The dual equivalence of two elements $a,b$, denoted by $a\equiv b$ (or
$a:=:b,a\,*=*\,b$ ) is defined by

$a\equiv b\,\text{if}\,\begin{cases}a=b\\\ \text{or}\\\ a\neq
b\,\,\text{but}\,a,b\,\text{are equivalent by some relation
establishment}\end{cases}$ (1)

(V) In complete duality, we have the following identity for dual equivalence,

$(U|S_{k})\equiv(U^{*}|S_{k}^{\star})\,\,\,\,{\rm
and}\,\,\,\,(U|S_{k}^{\star})\equiv(U^{*}|S_{k})\,.$ (2)

Element-wise, let $u\in U$ and $u^{*}\in U^{*}$, we have

$(u|S_{k})\equiv(u^{*}|S_{k}^{\star})\,\,\,\,{\rm
and}\,\,\,\,(u|S_{k}^{\star})\equiv(u^{*}|S_{k})\,.$ (3)

(VI) The duality operator of the element set $*$ acts as the following:

$*(u|S_{k})=(u^{*}|S_{k})\,\,\,\,,\,\,\,\,**(u|S_{k})=*(u^{*}|S_{k})=(u|S_{k})\,.$
(4)

(VII) The duality operator of observer set $*$ acts as the following:

$\star(u|S_{k})=(u|S_{k}^{\star})\,\,\,\,,\,\,\,\,\star\star(u|S_{k})=\star(u|S_{k}^{\star})=(u|S_{k})\,.$
(5)

(VIII) The dual map of the set and the dual operator can act together, which
is an identity map. The two different duality map commutes. In other words,
$\star\circ*=*\circ\star=I_{d}$. From example, we have,

$\star*(u|S_{k})\equiv\star(u^{*}|S_{k})\equiv(u^{*}|S_{k}^{\star})\,\,\,\,{\rm
and}\,\,\,\,*\star(u|S_{k})=*(u|S_{k}^{\star})=(u^{*}|S_{k}^{\star})\,.$ (6)

(IX) The $\\{I_{d},*\\}$ and $\\{I_{d}^{\prime},\star\\}$ are elements of two
parity groups $\mathbb{Z}_{2}$ under multiplication. The Klein-4 group, which
is the called the 4-duality group, is
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}=\\{I,*,\star,*\circ\star\\}$. The
$(u|S_{k}),(u^{*}|S_{k}),(u|S_{k}^{\star})$ and $(u^{*}|S_{k}^{\star})$ form a
4-representation of $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$, and can be
represented by a 4-tableau diagram,

Figure 1:  The 4-tableau representation of a complete duality structure. Boxes
with the same colour denote dual equivalence relation of each other.

The last definition (IX) is achieved by constructing an isomorphism from the
four cases to the basis of $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$, such we have
a one-to-one map identification as

$(u|S_{k})\rightarrow|00\rangle\,,\quad\,(u|S_{k}^{\star})\rightarrow|01\rangle\,,\quad\,(u^{*}|S_{k})\rightarrow|10\rangle\,,\quad\,(u^{*}|S_{k}^{\star})\rightarrow|11\rangle,.$
(7)

The $(u|S_{k})$ is called an identity element, in which no dual operation is
acted upon on it. Without loss of generality, we can also pick
$(u^{*}|S_{k}^{\star})$ as the identity. In terms of the number of dual
operation that act on the element and observer, we can write

$0+0\equiv 1+1\quad\text{and}\quad 0+1\equiv 1+0\,.$ (8)

The element-observer composite $(\,|\,)$ can also be viewed as a tensor
product, i.e. $(a|b)\equiv(a|\otimes|b)\equiv a\otimes b$. Consider the basis
of representation of $\mathbb{Z}_{2}$ be $u\oplus u^{*}$, and the basis of
representation of another $\mathbb{Z}_{2}$ as $S_{k}\oplus S_{k}^{\star}$.
Then we have

$\displaystyle(u\oplus u^{*})\otimes(S_{k}\oplus S_{k}^{\star})$
$\displaystyle=(u\otimes S_{k})\oplus(u\otimes
S_{k}^{\star})\oplus(u^{*}\otimes S_{k})\oplus(u^{*}\otimes S_{k}^{\star})$
(9)
$\displaystyle=(u|S_{k})\oplus(u|S_{k}^{\star})\oplus(u^{*}|S_{k})\oplus(u^{*}|S_{k}^{\star})\,,$

which is the basis of representation of $\mathbb{Z}_{2}\otimes\mathbb{Z}_{2}$
direct product group. But since
$\mathbb{Z}_{2}\otimes\mathbb{Z}_{2}\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$,
therefore the above serves as the basis of the 4-duality group. For
simplicity, using 7 we can write it as

$\psi=|00\rangle\oplus|01\rangle\oplus|10\rangle\oplus|11\rangle$ (10)

Since by recalling that 3, $(u|S_{k})\equiv(u^{*}|S_{k}^{\star})\,\,\,\,{\rm
and}\,\,\,\,(u|S_{k}^{\star})\equiv(u^{*}|S_{k})\,$, we have
$|00\rangle\equiv|11\rangle\,\,\,\,{\rm
and}\,\,\,\,|01\rangle\equiv|10\rangle$. This can be viewed as

Figure 2:  The two equivalence relationship can be viewed as
$++,--\rightarrow+$ and $+-,-+\rightarrow-$ In other words, considering the
backward argument, the conventional relationship of arithmetic can be viewed
as the equivalent relationships of tensor product of dual sets as demonstrated
above. .

Explicitly in the tensor product representation, we have the operators as

$\displaystyle(1\otimes 1)(u\otimes S_{k})=(u\otimes S_{k})$ (11)
$\displaystyle(*\otimes 1)(u\otimes S_{k})=(u^{*}\otimes S_{k})$
$\displaystyle(1\otimes\star)(u\otimes S_{k})=(u\otimes S_{k}^{\star})$
$\displaystyle(*\otimes 1)(1\otimes\star)(u\otimes
S_{k})=(*\otimes\star)(u\otimes S_{k})=(u^{*}\otimes S_{k}^{\star})$

We have the Klein-4 group as $\mathbb{Z}_{2}\otimes\mathbb{Z}_{2}=\\{1\otimes
1,*\otimes 1,1\otimes\star,*\otimes\star\\}$.

We group the terms as

$\psi=\big{[}(u|S_{k})\oplus(u^{*}|S_{k}^{\star})\big{]}_{D}\oplus\big{[}(u|S_{k}^{\star})\oplus(u^{*}|S_{k})\big{]}_{D^{*}}\,,$
(12)

or

$\psi=\big{[}|00\rangle\oplus|11\rangle\big{]}_{D}\oplus\big{[}|01\rangle\oplus|10\rangle\big{]}_{D^{*}}=|\psi_{D}\rangle\oplus|\psi_{D^{*}}\rangle\,,$
(13)

where the subscript $D$ and $D^{*}$ indicate the two dual partitions, and the
two states are orthogonal to each other
$\langle\psi_{D}|\psi_{D^{*}}\rangle=0.$ Now we will show that in fact the
basis of $\mathbb{Z}_{2}\otimes\mathbb{Z}_{2}$ can be separated into two dual
partitions, such that the blue boxes are dual to the green boxes in figure
2.1. Define the parity operator of the partition $D$ as follow,

$\hat{P}_{D}=(1\otimes\star)\oplus(1\otimes\star)\,.$ (14)

Then we have

$\displaystyle\hat{P}_{D}\big{[}|00\rangle\oplus|11\rangle\big{]}_{D}$
$\displaystyle=\big{[}(1\otimes\star)\oplus(1\otimes\star)\big{]}\big{[}|00\rangle\oplus|11\rangle\big{]}_{D}$
(15)
$\displaystyle=(1\otimes\star)(|0\rangle\otimes|0\rangle)\oplus(1\otimes\star)(|1\rangle\otimes|1\rangle)$
$\displaystyle=(|0\rangle\otimes|1\rangle)\oplus(|1\rangle\otimes|0\rangle)$
$\displaystyle=\big{[}|01\rangle\oplus|10\rangle\big{]}_{D^{*}}\,.$

It follows that

$\hat{P}_{D}\big{[}|01\rangle\oplus|10\rangle\big{]}_{D^{*}}=\hat{P}_{D}^{2}|00\rangle\oplus|11\rangle\big{]}_{D}=[|00\rangle\oplus[|11\rangle\big{]}_{D}\,.$
(16)

It can be also easily checked that

$\displaystyle\hat{P}_{D}^{2}$
$\displaystyle=[(1\otimes\star)\oplus(1\otimes\star)][(1\otimes\star)\oplus(1\otimes\star)]$
(17)
$\displaystyle=(1\otimes\star)(1\otimes\star)\oplus(1\otimes\star)(1\otimes\star)$
$\displaystyle=(1\otimes 1)\oplus(1\otimes 1)$ $\displaystyle=1\oplus 1$
$\displaystyle=I$

which is the identity matrix. Therefore, the two bases
$\big{[}|00\rangle\oplus|11\rangle\big{]}_{D}\,\,\,\,\text{and}\,\,\,\,\big{[}|01\rangle\oplus|10\rangle\big{]}_{D^{*}}$
are the basis of of the duality group $\mathbb{Z}_{2}$. Therefore, $\psi$ can
be decomposed into two EPR basis pair. Symbolically we can write

$\psi=2\oplus 2\,.$ (18)

Note that the choice of $\hat{P}_{D}$ is not unique, we can also define,

$\hat{Q}_{D}=(*\otimes 1)\oplus(*\otimes 1)\,.$ (19)

Then we have

$\displaystyle\hat{Q}_{D}\big{[}|00\rangle\oplus|11\rangle\big{]}_{D}$
$\displaystyle=\big{[}(*\otimes 1)\oplus(*\otimes
1)\big{]}\big{[}|00\rangle\oplus|11\rangle\big{]}_{D}$ (20)
$\displaystyle=(*\otimes 1)(|0\rangle\otimes|0\rangle)\oplus(*\otimes
1)(|1\rangle\otimes|1\rangle)$
$\displaystyle=(|1\rangle\rangle\otimes|0\rangle\rangle)\oplus(|0\rangle\otimes|1\rangle)$
$\displaystyle=\big{[}|01\rangle\oplus|10\rangle\big{]}_{D^{*}}\,.$

And similarly we have $\hat{Q}_{D}^{2}=I$ which is the identity map.

Furthermore, we can have rectangular duality. We consider $\psi$ with the
following partitions,

$\psi=\big{[}|00\rangle\oplus|01\rangle\big{]}_{P}\oplus\big{[}|11\rangle\oplus|10\rangle\big{]}_{P^{*}}=|\psi_{P}\rangle\oplus|\psi_{P^{*}}\rangle\,.$
(21)

Clearly partitions $P$ and $P^{*}$ are dual to each other, and the two dual
basis are orthogonal to each other $\langle\psi_{P}|\psi_{p^{*}}\rangle=0$.
This is referred as the the vertical rectangular duality. Similarly, we can
have

$\psi=\big{[}|00\rangle\oplus|10\rangle\big{]}_{Q}\oplus\big{[}|11\rangle\oplus|01\rangle\big{]}_{Q^{*}}=|\psi_{Q}\rangle\oplus|\psi_{Q^{*}}\rangle\,,$
(22)

where $Q$ and $Q^{*}$ are dual to each other, and the two dual basis are
orthogonal to each other $\langle\psi_{Q}|\psi_{Q^{*}}\rangle=0$. This is
referred as the horizontal rectangular duality. The idea is illustrated as
follow.

Figure 3:

Explicitly, we can draw out the whole idea of duality for illustration. The
dual elements and dual observers form a generic 4-dual diagram as follow.

Figure 4: In the diagram, we can see that $U$ observed by $S_{k}$ is
equivalent to $U^{*}$ observed by $S_{k}^{\star}$ such that
$(U|S_{k})\equiv(U^{*}|S_{k}^{\star})$; and $U^{*}$ observed by $S_{k}$ is
equivalent to $U$ observed by $S_{k}^{\star}$ such that
$(U^{*}|S_{k})\equiv(U|S_{k}^{\star})$.

The above abstract definition can be easily understood by some examples. The
simplest case for a dual system would be positive and negative numbers. Let
start from the most fundamental case. Let $U=\\{-1\\}$ and $U^{*}=\\{+1\\}$,
and the zero set $\\{0\\}$. Consider a pair of dual observers living on a 2D
manifold, the one in front of the two numbers is $S_{2}$, and the one behind
the two numbers is $S_{2}^{*}$. Let’s use the normal convention of a number
line, the left is -1 and the right is +1, then we have,

$(-1|S_{2})\equiv(+1|S_{2}^{*})\,\,\,\,{\rm
and}\,\,\,\,(-1|S_{2}^{*})\equiv(+1|S_{2})\,.$ (23)

And we have the map $\cdot=+$ such that $(-1)+(+1)=0$. If we let
$U=\mathbb{R}^{-}$ and $U^{*}=\mathbb{R}^{+}$ and the zero set, the you have
the duality for the real number system. Another example would be spin. Let
$U=\\{\downarrow\\}=\begin{pmatrix}0\\\ 1\end{pmatrix}$ and
$U^{*}=\\{\uparrow\\}=\begin{pmatrix}1\\\ 0\end{pmatrix}$, and
$*=\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}$ with $**=\mathbb{I}$, where we
observe in a 3 dimensional space. Then we have

$(\downarrow|S_{3})\equiv(\uparrow|S_{3}^{*})\,\,\,\,{\rm
and}\,\,\,\,(\downarrow|S_{3}^{*})\equiv(\uparrow|S_{3})\,.$ (24)

This is demonstrated in figure 5. And we have the $\cdot$ map as the inner
product $\langle\uparrow|\downarrow\rangle=0$.

Figure 5:

Although not obviously noticed, spontaneously duality symmetry breaking of
choice is always implicitly inferred. For example we define left-hand side as
negative in our observation perspective, but this is equivalently to a
positive right-hand side in the dual perspective. However we often make a
particular choice of representation so that at the end only one representation
out of the two equivalence is used. Without the loss of generality we can pick
the dual one, but for realistic observable we must pick a particular one. In
quantum mechanics terms, this is a state collapse of a dual state. Explicitly,

$|\Psi\rangle=\frac{1}{\sqrt{2}}\Big{(}|(U,U^{*}|S_{k})\rangle+|(U^{*},U|S_{k}^{\star})\rangle\Big{)}\,,$
(25)

where we can map $|(U,U^{*}|S_{k})\rangle\rightarrow|01\rangle$ and
$|(U^{*},U|S_{k}^{\star}\rangle\rightarrow|10\rangle)$ respectively, with
equal probability of $1/2$. This is an EPR pair and an entangled state. In
general we can write

$|\Psi\rangle=\cos\theta|(U,U^{*}|S_{k})\rangle+\sin\theta|(U^{*},U|S_{k}^{\star}\rangle\,,$
(26)

When the phase is at $\pi/4$, we have both the probabilities as $1/2$. At
$\theta=0$ or $2\pi$, we have a deterministic state for
$|(U,U^{*}|S_{k})\rangle$ and at $\theta=\pi$, we have a deterministic state
for $|(U^{*},U|S_{k}^{\star})\rangle$.

Next we would like to promote the idea into a more abstract way. We can call
$U$ as a left dual ${}^{*}U$. In figure 4, if we slice along the
$S_{k},S_{k}^{\star}$ frame, we can see the pair of element $U^{*}{}^{*}U$ and
its dual ${}^{*}UU^{*}$. We identify as follow:

$U^{*}{}^{*}U\text{as}\,\,RL\quad\text{and}\quad^{*}UU^{*}\text{as}\,\,LR\,.$
(27)

In the case of of RL, the two $*$s are in the inner side and we term this as
“bonding” denoted as $\rightarrow\leftarrow$, while in the case of LR, the two
$*$s are at the outer side and we term this as “anti-bonding” denoted as
$\leftarrow\rightarrow$. Thus the ”bonding” and ”anti-bonding” representation
is a dual representation. And the two objects $U^{*}{}^{*}U$ and
${}^{*}UU^{*}$ form a basis of irreducible representation of $\mathbb{Z}_{2}$.
We can go in the other way that if there exist such a dual pair, then the
notion of observation frame is implied.

The role of element and observer is interchangeable in a 4-duality system. Now
we can treat the element as observer and observer as element, this is known as
element-observer duality.

There are several more important examples for duality. We would like to show
that the set of odd numbers and even numbers are duality. Let the odd number
set be $O=\\{-(2k-1),\cdots,-5,-3,-1,1,3,5,\cdots,(2k-1)\\}$ and the even
number set be $\\{-2k,-4,-2,0,2,4,\cdots,2k\,\\}$. There is a one-to-one
bijective map from the even set to the odd set. We also have $O\cap
E=\emptyset$ Consider the $*$ function as adding $+1$ to each of the number in
the set. We have

$*O=O+1=E\,.$ (28)

It follows that

$**O=(O+1)+1=E+1=\\{-(2k+1),\cdots,-3,-1,1,3,5,7,\cdots,(2k+1)\\}=O\,.$ (29)

Therefore $**=1$ is the identity map. Hence the odd number set and the even
number is a duality.

Next we would like to show that momentum and position are duality. Let
$U=\\{x\\}$ and $U^{*}=\\{1/x\\}$ excluding $x=1$. We see $U\cap
U^{*}=\emptyset$. There is a one-to-one correspondence between the two sets.
Now consider

$*(x)=x^{-1}\,,$ (30)

then

$**(x)=*(x^{-1})=(x^{-1})^{-1}=x\,.$ (31)

Therefore $**=1$ is the identity map. Hence $x$ and $1/x$ are dual to each
other. Also we have the $\cdot$ map as multiplication $\times$ such that
$\frac{1}{x}\times x=1$. One important consequence is that for $x=0$, we have
$1/0=\infty$ thus $0$ and $\infty$ are dual to each other. Mathematically we
write

$*0=\infty\quad\text{and}\quad*\infty=0\,.$ (32)

Meaning-wise, we say nothing $(0)$ is dual to everything $(\infty)$, or
extremely small is dual to extremely large. One important property is that we
see when $x=1$, we get the same values that $*1=1$. This is the dual invariant
number. This serves as the zero number $\\{0\\}$. Therefore we have $W=U\cup
U^{*}\cup\\{1\\}$ as the complete set of duality. Now returning to physics,
consider $x$ as the wavelength $\lambda$, the momentum is $p=h/\lambda$. where
$h$ is the planck’s constant and can be regarded as 1 in the natural unit.
Hence it follows that $x$ and $p$ are dual to each other, therefore position
and momentum are dual to each other.

We can construct a general dual invariant function. The following function

$f(x)=\bigg{(}x+\frac{1}{x}\bigg{)}^{n}$ (33)

for $n$ is any positive integer greater than 1 is dual invariant that
$f(x)=f(\frac{1}{x})$, so this function remains the same for the exchange of
$x\leftrightarrow\frac{1}{x}$. A special attention goes to the case for $n=2$,
for which

$f(x)=x^{2}+\frac{1}{x^{2}}+2=x^{2}+\frac{1}{x^{2}}+\text{constant}\,.$ (34)

In string theory, the mass spectrum for a closed bosonic string with $26$
dimensions has a mass spectrum as [13, 14]

$M^{2}=\frac{n^{2}}{R^{2}}+\frac{m^{2}R^{2}}{\alpha^{\prime
2}}+\frac{2}{\alpha^{\prime}}(N_{L}+N_{R}-2)\,,$ (35)

where $N_{L}$ is the number of left-moving modes, $N_{R}$ is the number of
right moving modes, $n$ is the quantized number of Kaluza-Kelin momentum mode
and $m$ is the winding number. The $\alpha^{\prime}$ is the string’s length
scale and is related to the tension of the string. Also, $N_{R}-N_{L}=nm.$ The
mass spectrum in 35 is invariant under the interchange $n\leftrightarrow m$
and $R\leftrightarrow\tilde{R}=\frac{\alpha^{\prime}}{R}$. This is known as
the T-duality [13, 14]. In particular, when $n=m$ and in generic natural
length unit $\alpha^{\prime}=1$, we have

$M^{2}=\frac{n^{2}}{R^{2}}+n^{2}R^{2}+\text{constant}\,,$ (36)

which is a $\mathbb{Z}_{2}$ invariant.

#### The dual equation

Finally we would like to write out the duality theory in a compact form. Let
$a=(u|S_{k})$, and the full negation operator be $!=\star*$, then we nicely
obtain the follow equation,

$a\equiv\,!a\,.$ (37)

The negation of this equation is

$!a\equiv\,a\,.$ (38)

This is because

$!!=\star*\star*=(\star\otimes*)(\star\otimes*)=(\star\star\otimes**)=(I\otimes
I)=I.$ (39)

Equation 37 is the same as 38. Thus both equations 37 and 38 are dual
invariant. On the other hand, let $\bar{a}=(u^{*}|S_{k})$ then we have

$\bar{a}\equiv\,!\bar{a}\,.$ (40)

The negation of this equation is

$!\bar{a}\equiv\,a$ (41)

which is same as 40. Thus both equations 40 and 41 are dual invariant.
Therefore, the dual equation consists of 4 equations 37, 38, 40 , 41,

$\begin{cases}&a\equiv\,!a\,\,\text{and}\,\,!a\equiv\,a\,\,\text{for
partition}\,\,D\\\
&\bar{a}\equiv\,!\bar{a}\,\,\text{and}\,\,!\bar{a}\equiv\,\bar{a}\,\,\text{for
partition}\,\,D^{*}\end{cases}\,.$ (42)

Diagramatically,

Figure 6:

###  2.2 Representation of dual operators

In the section we will construct matrix representations of the dual operators.
Let $|0\rangle$ and $|1\rangle$ be a pair of dual states, satisfying
$*|0\rangle=|1\rangle$ and $*|1\rangle=|0\rangle$ where $*=\hat{P}$ is the
dual operator, satisfying the orthogonal relation $\langle 0|1\rangle=\langle
1|0\rangle=0$.

The identity element can be constructed by

$\boldsymbol{1}=|0\rangle\langle 0|+|1\rangle\langle 1|\,,$ (43)

and the parity element can be constructed by

$\hat{P}=|0\rangle\langle 1|+|1\rangle\langle 0|\,.$ (44)

This can be proven easily by

$\displaystyle\hat{P}^{2}$ $\displaystyle=(|0\rangle\langle
1|+|1\rangle\langle 0|)(|0\rangle\langle 1|+|1\rangle\langle 0|)$ (45)
$\displaystyle=|0\rangle\langle 1|0\rangle\langle 1|+|0\rangle\langle
1|1\rangle\langle 0|+|1\rangle\langle 0|0\rangle\langle 1|+|1\rangle\langle
0|1\rangle\langle 0|$ $\displaystyle=|0\rangle\langle 0|+|1\rangle\langle 1|$
$\displaystyle=\boldsymbol{1}\,.$

It can also be checked that

$\displaystyle\boldsymbol{1}^{2}$ $\displaystyle=(|0\rangle\langle
0|+|1\rangle\langle 1|)(|0\rangle\langle 0|+|1\rangle\langle 1|)$ (46)
$\displaystyle=|0\rangle\langle 0|0\rangle\langle 0|+|0\rangle\langle
0|1\rangle\langle 1|+|1\rangle\langle 1|0\rangle\langle 0|+|1\rangle\langle
1|1\rangle\langle 1|$ $\displaystyle=|0\rangle\langle 0|+|1\rangle\langle 1|$
$\displaystyle=\boldsymbol{1}\,.$

Next, to construct matrix representation for $\boldsymbol{1}$ and $\hat{P}$,

$\boldsymbol{1}_{ab}=\langle a|\boldsymbol{1}|b\rangle\,.$ (47)

Explicitly,

$\boldsymbol{1}=\begin{pmatrix}\langle 0|\boldsymbol{1}|0\rangle&\langle
0|\boldsymbol{1}|1\rangle\\\ \langle 1|\boldsymbol{1}|0\rangle&\langle
1|\boldsymbol{1}|1\rangle\end{pmatrix}\,.$ (48)

To evaluate the matrix element, we compute

$\displaystyle\boldsymbol{1}_{ab}$ $\displaystyle=\langle a|0\rangle\langle
0|b\rangle+\langle a|1\rangle\langle 1|b\rangle$ (49)
$\displaystyle=\delta_{a0}\delta_{0b}=\delta_{a1}\delta_{1b}\,.$

Therefore as expected,

$\boldsymbol{1}=\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}\,.$ (50)

For $\hat{P}$,

$\hat{P}_{ab}=\langle a|\hat{P}|b\rangle\,.$ (51)

Explicitly,

$\hat{P}=\begin{pmatrix}\langle 0|\hat{P}|0\rangle&\langle
0|\hat{P}|1\rangle\\\ \langle 1|\hat{P}|0\rangle&\langle
1|\hat{P}|1\rangle\end{pmatrix}\,.$ (52)

To evaluate the matrix element,

$\displaystyle\hat{P}_{ab}$ $\displaystyle=\langle a|0\rangle\langle
1|b\rangle+\langle a|1\rangle\langle 0|b\rangle$ (53)
$\displaystyle=\delta_{a0}\delta_{1b}+\delta_{a1}\delta_{0b}\,.$

Therefore we have

$\hat{P}=\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}\,.$ (54)

which is called the $\boldsymbol{M}$ matrix. The matrix representation in 54
can be diagonalized, since the eigenvalue is $1$ or $-1$, we have

$\hat{P}^{\prime}=\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}\,.$ (55)

which is another representation of the parity operator. And there exists a
similarity transformation such that $U\hat{P}U^{-1}=\hat{P}^{\prime}$.

One can observe that because

$\boldsymbol{1}\cdot\boldsymbol{1}=\boldsymbol{1}\quad,\quad\hat{P}\cdot\hat{P}=\boldsymbol{1}\quad,\quad\boldsymbol{1}\cdot\hat{P}=\hat{P}\quad,\quad\hat{P}\cdot\boldsymbol{1}=\hat{P}\,,$
(56)

this is isomorphic to the law of arithmetic multiplication that
$++=+,--=+,+-=-,-+=-$. Thus we can identify
$+\rightarrow\boldsymbol{1},\,\,-\rightarrow\hat{P}$. Also, this structure is
isomorphic to $\mathrm{even}+\mathrm{even}=\mathrm{even}$,
$\mathrm{odd}+\mathrm{odd}=\mathrm{even}$,
$\mathrm{even}+\mathrm{odd}=\mathrm{odd}$ ,
$\mathrm{odd}+\mathrm{even}=\mathrm{odd}$. In addition, notice that
$|00\rangle$ and $|11\rangle$ states are dual invariant (D.I) states, as no
matter you look from the left or right they remains the same,

$(|00\rangle|L)\equiv(|00\rangle|R)\quad\text{and}\quad(|11\rangle|L)\equiv(|11\rangle|R)\,,$
(57)

we call them full states. While $|01\rangle$ and $|10\rangle$ states are non-
dual invariant (non-D.I) states, we have

$(|01\rangle|L)\equiv|10\rangle|R)\quad\text{and}\quad(|10\rangle|L)\equiv(|01\rangle|R)\,.$
(58)

We regard these two states as the half states. Therefore the full states are
dual even invariant, while the half states are non-dual invariant. Now we see
that

$\displaystyle|00\rangle\,(\mathrm{D.I})+|11\rangle\,(\mathrm{D.I})$
$\displaystyle=\,(\mathrm{D.I})$ (59)
$\displaystyle|01\rangle\,(\mathrm{non-D.I})+|10\rangle\,(\mathrm{non-D.I})$
$\displaystyle=\,(\mathrm{D.I})$
$\displaystyle|00\rangle\,(\mathrm{D.I})+|10\rangle\,(\mathrm{non-D.I})$
$\displaystyle=\,(\mathrm{non-D.I})$
$\displaystyle|01\rangle\,(\mathrm{non-D.I})+|11\rangle\,(\mathrm{D.I})$
$\displaystyle=\,(\mathrm{non-D.I})$

In words, we can write,

$\displaystyle\mathrm{full}+\mathrm{full}$ $\displaystyle=\mathrm{full}$ (60)
$\displaystyle\mathrm{half}+\mathrm{half}$ $\displaystyle=\mathrm{full}$
$\displaystyle\mathrm{half}+\mathrm{full}$ $\displaystyle=\mathrm{half}$
$\displaystyle\mathrm{full}+\mathrm{half}$ $\displaystyle=\mathrm{half}$

With such isomorphism, we can refer $\mathrm{full}\cong\mathrm{even}$ and
$\mathrm{half}\cong\mathrm{odd}$.

###  2.3 Mirror duality and identities

It is interesting to ask a question: does there exist a structure which is
dual to the above, such that $++=-,+-=+,-+=+,--=-$? The answer is yes when we
consider matrix operation towards the left. Normally we multiply matrix
towards the right, here we introduce a new idea-left matrix multiplication.
Consider

$\begin{pmatrix}0&1\\\
1&0\end{pmatrix}\overset{\mathrm{left\,\,multiplication}}{=}\begin{pmatrix}0&1\\\
1&0\end{pmatrix}\begin{pmatrix}0&1\\\
1&0\end{pmatrix}\overset{\mathrm{right\,\,multiplication}}{=}\begin{pmatrix}1&0\\\
0&1\end{pmatrix}\,,$ (61)

similarly, we have

$\begin{pmatrix}1&0\\\
0&1\end{pmatrix}\overset{\mathrm{left\,\,multiplication}}{=}\begin{pmatrix}0&1\\\
1&0\end{pmatrix}\begin{pmatrix}1&0\\\
0&1\end{pmatrix}\overset{\mathrm{right\,\,multiplication}}{=}\begin{pmatrix}0&1\\\
1&0\end{pmatrix}\,$ (62)

and similar for the others. We have the following,

$(\boldsymbol{1}\cdot\boldsymbol{1})_{L}=\hat{P}\quad,\quad(\hat{P}\cdot\hat{P})_{L}=\hat{P}\quad,\quad(\boldsymbol{1}\cdot\hat{P})_{L}=\boldsymbol{1}\quad,\quad(\hat{P}\cdot\boldsymbol{1})_{L}=\boldsymbol{1}\,,$
(63)

where the subscript label refers that we are multiplying matrix to the left.
The dualities involving left matrix multiplication is called the mirror
duality.

####  2.3.1 Association dualities

The general dualities here are set on the following grounds on dual studies:

$\hat{P}_{+}=\begin{pmatrix}1&0\\\
0&0\end{pmatrix}\,,\,\hat{P}_{-}=\begin{pmatrix}0&0\\\
0&1\end{pmatrix}\,,\,\hat{P}_{+}^{*}=\begin{pmatrix}0&1\\\
0&0\end{pmatrix}\,,\,\hat{P}_{-}^{*}=\begin{pmatrix}0&0\\\
1&0\end{pmatrix}\,,\,$ (64)

They can be constructed by the following,

$\hat{P}_{+}=|0\rangle\langle 0|\,,\quad\hat{P}_{-}=|1\rangle\langle
1|\,,\quad\hat{P}^{*}_{+}=|0\rangle\langle
1|\,,\quad\hat{P}^{*}_{-}=|1\rangle\langle 0|\,.$ (65)

The generic form is

$\hat{P}_{ab}=|a\rangle\langle b|\quad\text{where}\,\,a,b=0,1\,,$ (66)

the matrix element of the association dualities are

$\langle c|\hat{P}_{ab}|d\rangle=\langle c|a\rangle\langle
b|d\rangle=\delta_{ca}\delta_{bd}\quad\text{where}\,\,a,b,c,d=0,1\,.$ (67)

We have the dual operations of

$\hat{P}_{+}^{*}=\hat{P}_{+}*=\hat{P}_{+}\boldsymbol{M}\,,\quad,\quad\hat{P}_{-}=\hat{P}_{+}^{(R)}\,\,,\,\,\hat{P}_{-}^{*}=*\hat{P}_{+}=\boldsymbol{M}\hat{P}_{+}\,.$
(68)

Now we compute the product of these operators. $L$ means we multiply the
matrix towards the left direction, while $R$ means we multiply the matrix
towards the normal right direction.

$(\hat{P}_{+}\hat{P}_{-}|L)=P_{-}^{*}\quad\text{and}\quad(\hat{P}_{-}\hat{P}_{+}|L)=\hat{P}^{*}_{+}\,,$
(69) $(\hat{P}_{+}\hat{P}_{-}|R)=(\hat{P}_{-}\hat{P}_{+}|R)=\boldsymbol{0}\,,$
(70)

where for the last equation we have dual invariance for the operator element
over the right action perspective. Since we have
$(\hat{P}_{+}\hat{P}_{-})^{\star}=(\hat{P}_{-}\hat{P}_{+}),$ which is left-
land action invariant. Next we have

$(\hat{P}^{*}_{+}\hat{P}_{-}^{*}|R)=\hat{P}_{+}\quad\text{and}\quad(\hat{P}^{*}_{-}\hat{P}_{+}^{*}|R)=\hat{P}_{-}\,,$
(71)
$(\hat{P}_{+}^{*}P_{-}^{*}|L)=(\hat{P}_{-}^{*}P_{+}^{*}|L)=\boldsymbol{0}\,.$
(72)

To simply the expression for clear demonstration process, let
$A=\hat{P}_{+}\hat{P}_{-}$, and its dual $A^{\star}=\hat{P}_{-}\hat{P}_{+}$,
we have

$(A|R)\equiv(A^{\star}|R)\,.$ (73)

Thus $A=A^{\star}=\boldsymbol{0}$ , thus the zero matrix is a dual invariant
under left or right matrix operation. We further have the following
properties, consider the anticommutators of the operators:

$(\\{\hat{P_{+}},\hat{P}_{-}\\}|L)=\boldsymbol{M}\quad,\quad(\\{\hat{P_{+}},\hat{P}_{-}\\}|R)=\boldsymbol{0}\quad;\quad$
(74)

while

$(\\{\hat{P}_{+}^{*},\hat{P}^{*}_{-}\\}\
|L)=\boldsymbol{0}\quad,\quad(\\{\hat{P}_{+}^{*},\hat{P}^{*}_{-}\\}|R)=\boldsymbol{1}\,.$
(75)

Now consider the following comparison,

$\displaystyle(\hat{P}_{+}\hat{P}_{+}+\hat{P}_{-}\hat{P}_{-}|R)=\boldsymbol{1}\quad\text{and}\quad(\hat{P}_{+}\hat{P}_{-}+\hat{P}_{-}\hat{P}_{+}|R)=\boldsymbol{0}\,.$
(76)
$\displaystyle(\hat{P}_{+}\hat{P}_{+}+\hat{P}_{-}\hat{P}_{-}|L)=\boldsymbol{0}\quad\text{and}\quad(\hat{P}_{+}\hat{P}_{-}+\hat{P}_{-}\hat{P}_{+}|L)=\boldsymbol{M}\,.$

And their dual,

$\displaystyle(\hat{P}_{+}^{*}\hat{P}_{+}^{*}+\hat{P}_{-}^{*}\hat{P}_{-}^{*}|R)=\boldsymbol{0}\quad\text{and}\quad(\hat{P}_{+}^{*}\hat{P}_{-}^{*}+\hat{P}_{-}^{*}\hat{P}_{+}^{*}|R)=\boldsymbol{1}\,.$
(77)
$\displaystyle(\hat{P}_{+}^{*}\hat{P}_{+}^{*}+\hat{P}_{-}^{*}\hat{P}_{-}^{*}|L)=\boldsymbol{M}\quad\text{and}\quad(\hat{P}_{+}^{*}\hat{P}_{-}^{*}+\hat{P}_{-}^{*}\hat{P}_{+}^{*}|L)=\boldsymbol{0}\,.$

Here we define the sum dualities, as we can see

$\displaystyle*(\hat{P}_{+}\hat{P}_{-}+\hat{P}_{-}\hat{P}_{+}|R)=(\hat{P}_{+}^{*}\hat{P}_{-}^{*}+\hat{P}_{-}^{*}\hat{P}_{+}^{*}|R)\,.$
(78)
$\displaystyle*(\hat{P}_{+}\hat{P}_{+}+\hat{P}_{-}\hat{P}_{-}|R)=(\hat{P}_{+}^{*}\hat{P}_{+}^{*}+\hat{P}_{-}^{*}\hat{P}_{-}^{*}|R)\,.$

Therefore we have for $R$,

$(*\boldsymbol{0}=\boldsymbol{1}|R)\quad\text{and}\quad(*\boldsymbol{1}=\boldsymbol{0}|R)\,.$
(79)

Hence in fact the zero matrix and the identity matrix is dual to each other
under the right matrix operation. Then we also have the perspective duality

$\displaystyle\star(\hat{P}_{+}\hat{P}_{+}+\hat{P}_{-}\hat{P}_{-}|R)=(\hat{P}_{+}\hat{P}_{+}+\hat{P}_{-}\hat{P}_{-}|L)\,.$
(80)
$\displaystyle\star(\hat{P}_{+}^{*}\hat{P}_{-}^{*}+\hat{P}_{-}^{*}\hat{P}_{+}^{*}|R)=(\hat{P}_{+}^{*}\hat{P}_{-}^{*}+\hat{P}_{-}^{*}\hat{P}_{+}^{*}|L)\,.$

Therefore we have

$\star\boldsymbol{1}=\boldsymbol{0}\quad\text{and}\quad\star\boldsymbol{0}=\boldsymbol{1}\,.$
(81)

Now for $L$,

$\displaystyle*(\hat{P}_{+}\hat{P}_{-}+\hat{P}_{-}\hat{P}_{+}|L)=(\hat{P}_{+}^{*}\hat{P}_{-}^{*}+\hat{P}_{-}^{*}\hat{P}_{+}^{*}|L)\,.$
(82)
$\displaystyle*(\hat{P}_{+}\hat{P}_{+}+\hat{P}_{-}\hat{P}_{-}|L)=(\hat{P}_{+}^{*}\hat{P}_{+}^{*}+\hat{P}_{-}^{*}\hat{P}_{-}^{*}|L)\,.$

Therefore we have for $L$,

$(*\boldsymbol{M}=\boldsymbol{0}|L)\quad\text{and}\quad(*\boldsymbol{0}=\boldsymbol{M}|L)\,.$
(83)

Hence in fact the zero matrix and the identity matrix is dual to each other
under the left matrix operation. Then we also have the perspective duality

$\displaystyle\star(\hat{P}_{+}\hat{P}_{-}+\hat{P}_{-}\hat{P}_{+}|R)=(\hat{P}_{+}\hat{P}_{-}+\hat{P}_{-}\hat{P}_{+}|L)\,.$
(84)
$\displaystyle\star(\hat{P}_{+}^{*}\hat{P}_{+}^{*}+\hat{P}_{-}^{*}\hat{P}_{-}^{*}|R)=(\hat{P}_{+}^{*}\hat{P}_{+}^{*}+\hat{P}_{-}^{*}\hat{P}_{-}^{*}|L)\,,$

Therefore for the $*$ case we have

$\star\boldsymbol{M}=\boldsymbol{0}\quad\text{and}\quad\star\boldsymbol{0}=\boldsymbol{M}\,.$
(85)

If we consider the collection representation matrices of 76 and 77

$\begin{pmatrix}\boldsymbol{1}&\boldsymbol{0}\\\
\boldsymbol{0}&\boldsymbol{M}\end{pmatrix}\quad\text{and}\quad\begin{pmatrix}\boldsymbol{0}&\boldsymbol{1}\\\
\boldsymbol{M}&\boldsymbol{0}\end{pmatrix}\,,$ (86)

which is left-right dual to each other.

Next we define the following 4 association matrices

$\hat{-}=\begin{pmatrix}0&0\\\
1&1\end{pmatrix}\quad,\quad\hat{+}=\begin{pmatrix}1&1\\\
0&0\end{pmatrix}\quad,\quad\hat{\\#}=\begin{pmatrix}1&0\\\
1&0\end{pmatrix}\quad,\quad\hat{\flat}=\begin{pmatrix}0&1\\\
0&1\end{pmatrix}\quad,\quad$ (87)

Next we can check that the following identities hold:

$\begin{aligned}
&\hat{-}=(\hat{P}_{+}\hat{P}_{+}^{\star}|R)+(\hat{P}_{+}\hat{P}_{+}^{\star}|L)\\\
&\hat{+}=(\hat{P}_{-}\hat{P}_{-}^{\star}|R)+(\hat{P}_{-}\hat{P}_{-}^{\star}|L)\\\
&\hat{\\#}=(\hat{P}_{-}^{*}\hat{P}_{+}|R)+(\hat{P}^{*}_{-}\hat{P}_{+}|L)\\\
&\hat{\flat}=(\hat{P}_{+}^{*}\hat{P}_{-}|R)+(\hat{P}^{*}_{+}\hat{P}_{-}|L)\end{aligned}\,.$
(88)

Note that $\\{\hat{+},\hat{-}\\}$ and $\\{\hat{\\#},\hat{\flat}\\}$ are dual
pair such that

$*\hat{-}=\hat{+}\quad,\quad\hat{*}+=\hat{-}\quad;\quad\hat{\\#}=\flat*\quad,\quad\flat=\hat{\\#}*\,.$
(89)

Now we will investigate representations of
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ based on these operators under left or
right matrix operations. The matrix operators are regarded as the
perspectives. For $\\{\hat{+},\hat{-}\\}$ It is easy to prove the following,

$\begin{aligned} &(\hat{-}\hat{+}|R)\equiv(\hat{+}\hat{-}|L)=\hat{+}\\\
&(\hat{-}\hat{+}|L)\equiv(\hat{+}\hat{-}|R)=\hat{-}\end{aligned}\,\,\,\,.$
(90)

And for $\\{\hat{\\#},\hat{\flat}\\}$,

$\begin{aligned}
&(\hat{\\#}\hat{\flat}|R)\equiv(\hat{\flat}\hat{\\#}|L)=\hat{\flat}\\\
&(\hat{\\#}\hat{\flat}|L)\equiv(\hat{\flat}\hat{\\#}|R)=\hat{\flat}\end{aligned}\,\,\,\,.$
(91)

Therefore for $\\{\hat{+},\hat{-}\\}$, and $\\{\hat{\\#},\hat{\flat}\\}$ two
dual sets respectively, we have 90 and 91

It takes the generic form of

$\displaystyle(AB|R)\equiv(BA|L)$ (92)

If $A$ and $B$ are dual to each other, i.e. $B=*A$ and $A=*B$. then the
Klein-4 group $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ is preserved. However if
they are not, we will see that this leads to breaking down of the Klein-4
group to its subgroup $\mathbb{Z}_{2}$. Let’s us check,

$(\hat{-}\hat{\\#}|R)\equiv(\hat{\\#}\hat{-}|L)=2\hat{P}_{+}\,.$ (93)

However,

$(\hat{-}\hat{\\#}|L)=\begin{pmatrix}0&0\\\
0&0\end{pmatrix}\quad\text{and}\quad(\hat{\\#}\hat{-}|R)=\begin{pmatrix}1&1\\\
1&1\end{pmatrix}$ (94)

in which they are not equivalent,

$(\hat{-}\hat{\\#}|L)\neq(\hat{\\#}\hat{-}|R)\,.$ (95)

Similarly we have

$(\hat{-}\hat{\flat}|R)\equiv(\hat{\flat}\hat{-}|L)=2\hat{P}_{+}^{*}\,,$ (96)

but

$(\hat{-}\hat{\flat}|L)=\begin{pmatrix}1&1\\\
1&1\end{pmatrix}\quad\text{and}\quad(\hat{\flat}\hat{-}|R)=\begin{pmatrix}0&0\\\
0&0\end{pmatrix}\,.$ (97)

Therefore

$(\hat{-}\hat{\flat}|R)\neq(\hat{\flat}\hat{-}|L).$ (98)

We also have

$(\hat{+}\hat{\\#}|R)\equiv(\hat{\\#}\hat{+}|L)=2\hat{P}_{-}^{*}\,.$ (99)

but

$(\hat{+}\hat{\\#}|L)=\begin{pmatrix}1&1\\\
1&1\end{pmatrix}\quad\text{and}\quad(\hat{\\#}\hat{+}|R)=\begin{pmatrix}0&0\\\
0&0\end{pmatrix}\,.$ (100)

Therefore,

$(\hat{+}\hat{\\#}|L)\neq(\hat{\\#}\hat{+}|R)\,.$ (101)

Finally we also have

$(\hat{+}\hat{\flat}|R)\equiv(\hat{\flat}\hat{+}|L)=2\hat{P}_{-}$ (102)

but

$(\hat{+}\hat{\flat}|L)=\begin{pmatrix}0&0\\\
0&0\end{pmatrix}\quad\text{and}\quad(\hat{\flat}\hat{+}|R)=\begin{pmatrix}1&1\\\
1&1\end{pmatrix}\,.$ (103)

Therefore we have,

$(\hat{+}\hat{\flat}|L)\neq(\hat{\flat}\hat{+}|R)\,.$ (104)

The whole idea is fairly abstract, we can construct diagramatic approach to
illustrate the full idea. Let’s divide them into two big cases, for $A,B$ that
are dual to each other, and $A,B$ which are non-dual to each other. There are
two sub-cases for the case which $A,B$ are dual to each other, while there are
4 sub-cases for the case which $A,B$ are not dual to each other. In other
words, there are six sub-cases. We first consider the case with
$\\{\hat{+},\hat{-}\\}$, we map $R\rightarrow 0$ and $L\rightarrow 0$, ;
$\hat{-}\hat{+}\rightarrow 0$ and $\hat{+}\hat{-}\rightarrow 1$ (see figure
7). Similarly for the $\\{\hat{\\#},\hat{\flat}\\}$, we have in figure 8.

Figure 7:  Figure 8:

However, for the remaining case for which $A$ and $B$ are not dual to each
other, things are much more complicated, the structural of
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ duality breaks down, we only have one
statement for duality equivalence out of two.

The full idea can be represented by a diagram as follow:

Figure 9: In the diagram, the arrow denotes the direction of the matrix
multiplication, $R$ denotes right and $L$ denotes left. A special note is
remarked for the navy-blue oval with dotted boundary, this corresponds to the
dual equivalence of $(\hat{+}\hat{\\#}|R)\equiv(\hat{\\#}\hat{+}|L)$. Here for
this case we read from the back of the page, so we have the red $R$ and red
$L$. For the others we read the direction from above the page.

From equation 94, 97, 100 and 103, we see that,

$\displaystyle(\hat{\\#}\hat{-}|R)=*\star(\hat{-}\hat{\\#}|L)=!(\hat{-}\hat{\\#}|L)\,,$
(105)
$\displaystyle(\hat{\flat}\hat{-}|R)=*\star(\hat{-}\hat{\flat}|L)=!(\hat{-}\hat{\flat}|L)\,,$
$\displaystyle(\hat{\\#}\hat{+}|R)=*\star(\hat{+}\hat{\\#}|L)=!(\hat{+}\hat{\\#}|L)\,,$
$\displaystyle(\hat{\flat}\hat{+}|R)=*\star(\hat{+}\hat{\flat}|L)=!(\hat{+}\hat{\flat}|L)\,,$

where $!=*\star$. Therefore from these results we have

$\begin{pmatrix}1&1\\\ 1&1\end{pmatrix}=!\begin{pmatrix}0&0\\\
0&0\end{pmatrix}\quad\text{and}\quad\begin{pmatrix}0&0\\\
0&0\end{pmatrix}=!\begin{pmatrix}1&1\\\ 1&1\end{pmatrix}$ (106)

with $!!=\boldsymbol{1}$. Therefore, $\begin{pmatrix}0&0\\\ 0&0\end{pmatrix}$
and $\begin{pmatrix}1&1\\\ 1&1\end{pmatrix}$ is dual to one another. Define
$\mathbb{I}=\begin{pmatrix}1&1\\\ 1&1\end{pmatrix}$ as the dual invariant
identity matrix, we write 106 as

$\mathbb{I}=!\,\boldsymbol{0}\quad\text{and}\quad\boldsymbol{0}=!\,\mathbb{I}\,.$
(107)

In addition, we also have

$\displaystyle(\hat{-}\hat{\\#}|L)=(*_{1}\hat{+}*_{2}\hat{\flat}|L)=(\hat{+}\hat{\flat}|L)=\boldsymbol{0}$
(108)
$\displaystyle(\hat{\flat}\hat{-}|R)=(*_{1}\hat{\flat}*_{2}\hat{-}|R)=(\hat{\\#}\hat{+}|R)=\boldsymbol{0}\,,$
$\displaystyle(\hat{-}\hat{\flat}|L)=(*_{1}\hat{-}*_{2}\hat{\flat}|L)=(\hat{+}\hat{\\#}|L)=\mathbb{I}\,,$
$\displaystyle(\hat{\\#}\hat{-}|R)=(*_{1}\hat{\\#}*_{2}\hat{-}|R)=(\hat{\flat}\hat{+}|R)=\mathbb{I}\,.$

This can be further grouped into $L$ and $R$ category as follows :

Figure 10:

###  2.4 Addition and pointwise multiplication duality

Since we know that $\boldsymbol{0}$ and $\mathbb{I}$ is dual to each other,
now we would like to see how to construct these dual matrices from their
progenitor dual matrices. First consider

$\boldsymbol{0}+\mathbb{I}=\mathbb{I}+\boldsymbol{0}=\mathbb{I}\quad\text{and}\quad\boldsymbol{0}\bullet\mathbb{I}=\mathbb{I}\bullet\boldsymbol{0}=\boldsymbol{0}\,,$
(109)

where $\bullet$ is the pointwise matrix product (Hadamard product). Therefore
in terms of operator, we say the matrix addition and the matrix pointwise
operator is dual to each other. Let $\hat{O}$ be $+$ and $\hat{O}^{*}$ be
$\bullet$, we have the dual operator set as $\\{+,\bullet\\}$. Element-wise,
we define

$\begin{pmatrix}1&1\\\ 1&1\end{pmatrix}=!\begin{pmatrix}0&0\\\
0&0\end{pmatrix}=\begin{pmatrix}!0&!0\\\
!0&!0\end{pmatrix}\quad\text{and}\quad$ (110) $\begin{pmatrix}0&0\\\
0&0\end{pmatrix}=!\begin{pmatrix}1&1\\\
1&1\end{pmatrix}=\begin{pmatrix}!1&!1\\\ !1&!1\end{pmatrix}\quad\,,$ (111)

therefore

$1=!\,0\quad\text{and}\quad 0=!\,1\,.$ (112)

Hence 0 and 1 are dual to each other, we can form the dual set $\\{0,1\\}$.
This 0,1 duality has many applications, for example,

$\begin{pmatrix}1\\\ 0\end{pmatrix}=!\begin{pmatrix}0\\\
1\end{pmatrix}=\begin{pmatrix}!0\\\ !1\end{pmatrix}\,,$ (113)
$\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}=!\begin{pmatrix}0&1\\\
1&0\end{pmatrix}=\begin{pmatrix}!0&!1\\\ !1&!0\end{pmatrix}\,,$ (114)
$\begin{pmatrix}1&1\\\ 0&0\end{pmatrix}=!\begin{pmatrix}0&0\\\
1&1\end{pmatrix}=\begin{pmatrix}!0&!0\\\ !1&!1\end{pmatrix}\,.$ (115)

So we have duality sets, for examples$\\{\begin{pmatrix}1&0\\\
0&1\end{pmatrix},\begin{pmatrix}0&1\\\
1&0\end{pmatrix}\\}\quad,\quad\\{\begin{pmatrix}1&1\\\
0&0\end{pmatrix},\begin{pmatrix}0&0\\\ 1&1\end{pmatrix}\\}\,$ , etc.

###  2.5 Column and row duality for $\hat{+}$ and $\hat{-}$ operators

Finally, we would like to investigate the column and row duality for the
$\hat{+}$ and $\hat{-}$ operators. Consider their products

$\begin{aligned}
&(\hat{+}\hat{+}|R)=\hat{+}\quad\text{and}\quad(\hat{+}\hat{+}|L)=\hat{+}\\\
&(\hat{-}\hat{-}|R)=\hat{-}\quad\text{and}\quad(\hat{-}\hat{-}|L)=\hat{-}\\\
&(\hat{+}\hat{-}|R)=\hat{+}\quad\text{and}\quad(\hat{+}\hat{-}|L)=\hat{-}\\\
&(\hat{-}\hat{+}|R)=\hat{-}\quad\text{and}\quad(\hat{-}\hat{+}|L)=\hat{+}\end{aligned}\,\,\,\,.$
(116)

Diagramatically,

Figure 11: This is from the R-Perspective, which is a row duality. Figure 12:
This is from the L-Perspective, which is a column duality.

####  2.5.1 Duality equivalence in left and right transpose

Next we will introduce the concept of transpose and dual transpose. Transpose
of a matrix is given by the usual definition, which swap elements by the
diagonal. Given a matrix $A$, $(A^{T})_{ij}=A_{ji}$. We call this left
transpose. Now we also define its dual operation of right transpose, which
swap elements along the off-diagonal, for example,

$\begin{pmatrix}a&b\\\ c&d\end{pmatrix}^{R}=\begin{pmatrix}d&b\\\
c&a\end{pmatrix}\,.$ (117)

We have

$\hat{+}^{T}=\hat{\flat}\quad,\quad\hat{-}^{T}=\hat{\sharp}\quad,\quad\hat{+}^{R}=\hat{\sharp}\quad,\quad\hat{-}^{R}=\hat{\flat}\,.$
(118)

Therefore we have

$\hat{+}^{T}=\hat{-}^{R}\quad\text{and}\quad\hat{+}^{R}=\hat{-}^{T}\,.$ (119)

In perspective representation, we can write it as

$(\hat{+}|T)\equiv(\hat{-}|R)\quad\text{and}\quad(\hat{+}|R)\equiv(\hat{-}|T)$
(120)

For simplicity, if we write $\hat{+}\rightarrow 0,R\rightarrow 0$ and
$\hat{-}\rightarrow 1,L\rightarrow 1$ , we have

$(0|0)\equiv(1|1)\quad\text{and}\quad(0|1)\equiv(1|0),$ (121)

which are representations of $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ as before.
We also have the following

$\hat{+}^{(TR)}=\hat{+}^{(RT)}=\hat{-}\quad\text{and}\quad\hat{-}^{(TR)}=\hat{-}^{(RT)}=\hat{+}\,.$
(122)

Therefore the $RT$ (or $TR$) operations act as the dual matrix
$\boldsymbol{M}$.

####  2.5.2 Complementary matrices

Next we define complementary matrices, these matrices have three 1 entries and
one 0 entry, denoted by $A_{ij}$, where $ij$ is the position of the 0,

$A_{11}=\begin{pmatrix}0&1\\\ 1&1\end{pmatrix}\quad,\quad
A_{22}=\begin{pmatrix}1&1\\\ 1&0\end{pmatrix}\quad,\quad
A_{12}=\begin{pmatrix}1&0\\\ 1&1\end{pmatrix}\quad,\quad
A_{21}=\begin{pmatrix}1&1\\\ 0&1\end{pmatrix}\quad\,.$ (123)

and we have their pointwise-element dual,

$!A_{11}=\hat{P}_{+}=\begin{pmatrix}1&0\\\
0&0\end{pmatrix}\quad,\quad!A_{22}=\hat{P}_{-}=\begin{pmatrix}0&0\\\
0&1\end{pmatrix}$ (124) $!A_{12}=\hat{P}_{+}^{*}=\begin{pmatrix}0&1\\\
0&0\end{pmatrix}\quad,\quad!A_{21}=\hat{P}_{-}^{*}=\begin{pmatrix}0&0\\\
1&0\end{pmatrix}\,.$ (125)

Since

$!A_{ij}+A_{ij}=A_{ij}+!A_{ij}=\mathbb{I}\quad\text{and}\quad!A_{ij}\bullet
A_{ij}=A_{ij}\bullet!A_{ij}=\boldsymbol{0}\,,$ (126)

therefore it follows that $!A_{ij},A_{ij}$ are dual to each other. Now we have
the following identities,

$\displaystyle(A_{11}\hat{P}_{+}|R)\equiv(\hat{P}_{-}A_{22}|R)=\hat{P}^{*}_{-}\,,$
(127)
$\displaystyle(\hat{P}_{+}A_{11}|R)\equiv(A_{22}\hat{P}_{-}|R)=\hat{P}^{*}_{+}\,,$

or

$\displaystyle(A_{11}!A_{11}|R)\equiv(!A_{22}A_{22}|R)=!A_{21}\,,$ (128)
$\displaystyle(!A_{11}A_{11}|R)\equiv(A_{22}!A_{22}|R)=!A_{12}\,.$

Since $\hat{P}_{-}^{*}$ and $\hat{P}_{+}^{*}$ are dual to each other (by right
transpose),

$*^{\prime}(\hat{P}_{-}^{*})=(\hat{P}_{-}^{*})^{R}=\hat{P}_{+}^{*}\,.$ (129)

We then have

$\displaystyle*^{\prime}(A_{11}\hat{P}_{+}|R)=(\hat{P}_{+}A_{11}|R)\,,$ (130)
$\displaystyle*^{\prime}(\hat{P}_{-}A_{22}|R)=(A_{22}\hat{P}_{-}|R)\,.$

We can also see that

$A_{11}=\hat{*}_{R}A_{22}=A^{R}_{22}\quad\text{and}\quad\hat{*}_{R}A_{11}=A^{R}_{11}=A_{22}\,.$
(131)

Next we have the following identities

$\displaystyle(A_{21}\hat{P}_{-}^{*}|L)\equiv(P_{+}^{*}A_{12}|L)=\hat{P}_{-}\,,$
(132)
$\displaystyle(\hat{P}_{-}^{*}A_{21}|L)\equiv(A_{12}P_{+}^{*}|L)=\hat{P}_{+}\,,$

or

$\displaystyle(A_{21}!A_{21}|L)\equiv(!A_{12}A_{12}|L)=!A_{22}\,,$ (133)
$\displaystyle(!A_{21}A_{21}|L)\equiv(A_{12}!A_{12}|L)=!A_{11}\,,$

Since $\hat{P}_{-}$ and $\hat{P}_{+}$ are dual to each other (by left
transpose),

$*^{\prime\prime}\hat{P}_{-}=\hat{P}_{-}^{L}=\hat{P}_{+}\,.$ (134)

We then have

$\displaystyle*^{\prime\prime}(A_{21}\hat{P}_{-}^{*}|L)=(\hat{P}_{-}^{*}A_{12}|L)\,,$
(135)
$\displaystyle*^{\prime\prime}(\hat{P}_{+}^{*}A_{21}|L)=(A_{12}P_{+}^{*}|L)\,.$

We can also see that

$A_{21}=\hat{*}_{L}A_{12}=A^{T}_{12}\quad\text{and}\quad\hat{*}_{L}A_{21}=A^{T}_{21}=A_{12}\,.$
(136)

Therefore from 131 and 136 we obtain the following generalization. For
$i,j=1,2$, $*i=j$ and $*j=i$ with $**=\boldsymbol{1}$. We have

$*A_{ij}=A_{*i*j}=A_{ji}\quad\text{and}\quad*A_{ii}=A_{*i*i}=A_{jj}\,.$ (137)

Next, we continue to have the following identities,

$\displaystyle(A_{22}\hat{P}_{-}|L)\equiv(\hat{P}_{-}^{*}A_{21}|R)=\hat{+}\,,$
(138)
$\displaystyle(A_{11}\hat{P}_{+}|L)\equiv(\hat{P}^{*}_{+}A_{12}|R)=\hat{-}\,,$

or

$\displaystyle(A_{22}!A_{22}|L)\equiv(!A_{21}A_{21}|R)=\hat{+}\,,$ (139)
$\displaystyle(A_{11}!A_{11}|L)\equiv(!A_{12}A_{12}|R)=\hat{-}\,,$

Therefore we have

$\displaystyle\hat{*}_{R}*^{\prime\prime}(A_{22}\hat{P}_{-}|L)=(A_{11}\hat{P}_{+}|L)\,,$
(140)
$\displaystyle*^{\prime}\hat{*}_{L}(\hat{P}_{-}^{*}A_{21}|R)=(\hat{P}^{*}_{+}A_{12}|R)\,.$

Hence, $\hat{*}_{R}*^{\prime\prime}$ and $*^{\prime}\hat{*}_{L}$ are dual to
each other. Finally we have the identities of,

$\displaystyle(\hat{P}_{-}A_{22}|L)\equiv(A_{12}\hat{P}_{+}^{*}|R)=\hat{\flat}\,,$
(141)
$\displaystyle(\hat{P}_{+}A_{11}|L)\equiv(A_{21}\hat{P}_{-}^{*}|R)=\hat{\\#}\,,$

or

$\displaystyle(!A_{22}A_{22}|L)\equiv(A_{12}!A_{12}|R)=\hat{\flat}\,,$ (142)
$\displaystyle(!A_{11}A_{11}|L)\equiv(A_{21}!A_{21}|R)=\hat{\\#}\,.$

Therefore we have

$\displaystyle*^{\prime\prime}\hat{*}_{R}(\hat{P}_{-}A_{22}|L)=(\hat{P}_{+}A_{11}|L)\,,$
(143)
$\displaystyle\hat{*}_{L}*^{\prime}(A_{12}\hat{P}_{+}^{*}|R)=(A_{21}\hat{P}_{-}^{*}|R)\,.$

Hence, $*^{\prime\prime}\hat{*}_{R}$ and $\hat{*}_{L}*^{\prime}$ are dual to
each other.

####  2.5.3 Non-dual invariant identity and dual invariant identity

In matrix operation, we are already familiar with the usual identity matrix
$\boldsymbol{1}$. However as we know, it is only the identity under right
matrix operation, we say it is non-dual invariant identity. Let’s see for the
following illustration

Figure 13:

It is only the identity matrix in the $S_{k}$ perspective, but if we view from
the other side, i.e. $S_{k}^{\star}$ perspective, it is the dual matrix
$\boldsymbol{M}$. Mathematically we have

$(\boldsymbol{1}|S_{k})\equiv(\boldsymbol{M}|S_{k}^{\star})\,.$ (144)

And similarly we also have

$(\boldsymbol{M}|S_{k})\equiv(\boldsymbol{1}|S_{k}^{\star})\,.$ (145)

To construct a dual invariant identity matrix $\mathbb{I}$, we can simply add
up the identity matrix and its dual, so that

$\boldsymbol{1}+\boldsymbol{M}=*(\boldsymbol{1}+\boldsymbol{M})=\begin{pmatrix}1&1\\\
1&1\end{pmatrix}=\mathbb{I}\,,$ (146)

in which this matrix remains the same no matter which perspective do we look
at. Also we have

$\hat{+}+\hat{-}=*(\hat{+}+\hat{-})=\mathbb{I}\,,$ (147)

where $\hat{-}=*\hat{+}=!\hat{+}$. Also,

$\hat{\\#}+\hat{\flat}=!(\hat{\\#}+\hat{\flat})=\mathbb{I}\,,$ (148)

where $\hat{\\#}=!\hat{\flat}$. Also

$A_{ij}+!A_{ij}=!(A_{ij}+!A_{ij})=\mathbb{I}\,.$ (149)

And while $\boldsymbol{1}^{n}=\boldsymbol{1}$, but for $\mathbb{I}$ we have

$\mathbb{I}^{n}=2^{n-1}\mathbb{I}\,.$ (150)

In addition, since

$\hat{+}\bullet\hat{-}=!(\hat{+}\bullet\hat{-})=\hat{-}\bullet\hat{+=}\boldsymbol{0}\,,$
(151)
$\hat{\\#}\bullet\hat{\flat}=!(\hat{\\#}\bullet\hat{\flat})=\hat{\flat}\bullet\hat{\\#}=\boldsymbol{0}\,,$
(152) $A_{ij}\bullet!A_{ij}=!(A_{ij}\bullet!A_{ij})=\boldsymbol{0}\,.$ (153)

and $\mathbb{I}$, $\boldsymbol{0}$ are dual to each other, thus this further
generalize $+,\bullet$ are dual to each other for all cases.

###  2.6 General Analysis and Topological duality

In this section, we wil carry out a combined analysis of duality of the
previous sections discussed above. Notice that from 65 we assume working on
right operation. The result is different if we work on the left operation. The
following diagram illustrates the different outcomes:

Figure 14: The association dualities in $R$ and $L$ matrix operation
perspective. The $\sigma_{v}$ is the reflection operator along the vertical
plane and the $\sigma_{h}$ is the reflection operator along the horizontal
plane. The $\hat{P}_{+},\hat{P}_{-},\hat{P}_{+}^{*},\hat{P}_{-}^{*}$ form the
matrix basis of the $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ group in the matrix
space.

We see that

$(|0\rangle\langle 0|+|1\rangle\langle 1||R)\equiv(|1\rangle\langle
0|+|0\rangle\langle 1||L)=\boldsymbol{1}$ (154)

and

$(|0\rangle\langle 1|+|1\rangle\langle 0||R)\equiv(|0\rangle\langle
0|+|1\rangle\langle 1||L)=\boldsymbol{M}$ (155)

Therefore we see that in fact $u=|0\rangle\langle 0|+|1\rangle\langle 1|$ and
$u^{*}=|0\rangle\langle 1|+|1\rangle\langle 0|$ are dual to each other. Also
as $L$ and $R$ is dual to each other, this shows the follow structure of

$(u|S_{k})=(u^{*}|S_{k}^{\star})=\boldsymbol{1}\quad\text{and}(u^{*}|S_{k})=(u|S_{k}^{*})=\boldsymbol{M}$
(156)

for $S_{k}=R$ and $S_{k}^{\star}=L$. Notes that $u$ is a dual invariant, while
$u^{*}$ is another dual invariant, this is simply both $u$ and $u^{*}$ are
complete,

$u=\hat{P_{+}}+\hat{P_{-}}\,,\quad u^{R}=\hat{P_{-}}+\hat{P_{+}}=u\,,$ (157)

and

$u^{*}=\hat{P^{*}_{+}}+\hat{P^{*}_{-}}\,,\quad
u^{T}=\hat{P^{*}_{-}}+\hat{P^{*}_{+}}=u^{*}\,.$ (158)

So we can see that two different dual invariants can be dual to each other.

Next, we consider the row and column dualities in both $R$ and $L$
perspectives. First for convenience, we address the decimal number
representation on the upper-right corner in each box of 14 for both $R$ and
$L$ perspective. For example we denote $(0|R)=(|0\rangle\langle
0||R),\,\,(1|R)=(|0\rangle\langle 1||R)$, etc. Then we have the following
equivalence,

$\displaystyle(0|R)=(2|L)=\hat{P}_{+}$ (159)
$\displaystyle(1|R)=(0|L)=\hat{P}_{+}^{*}$
$\displaystyle(2|R)=(3|L)=\hat{P}_{-}^{*}$
$\displaystyle(3|R)=(1|L)=\hat{P}_{-}\,.$

Note that $\\{\hat{P}_{+},\hat{P}_{-}\\}$ is a dual pair and
$\\{\hat{P}_{+}^{*},\hat{P}_{-}^{*}\\}$, this is arise from the fact that
$00,11$ are dual pair and $01,10$ is another dual pair. It follows that 0 is
dual to 3 and 1 is dual to 2. Now we add up up two association dualities to
form $\hat{-},\hat{+},\hat{\sharp},\hat{\flat}$. We have the following, for
the right perspective,

$\displaystyle\hat{+}$ $\displaystyle=(|0\rangle\langle
0||R)+(|0\rangle\langle 1||R)=(0|R)+(1|R)=(1|R)$ (160) $\displaystyle\hat{-}$
$\displaystyle=(|1\rangle\langle 1||R)+(|1\rangle\langle
0||R)=(3|R)+(2|R)=(5|R)$

We know that $\\{\hat{+},\hat{-}\\}$ is a dual pair under the $\sigma_{h}$
reflection and this can be confirmed by seeing $0,1$ duality, i.e.

$*[(|0\rangle\langle 0||R)+(|0\rangle\langle 1||R)]=(|1\rangle\langle
1||R)+(|1\rangle\langle 0||R)\,.$ (161)

Therefore, 1 and 5 are dual to each other. Since 1 and 5 are both odd, we call
$\\{\hat{+},\hat{-}\\}$ the odd association dual pair under the right
perspective. Next we also have

$\begin{aligned} \hat{\sharp}&=(|0\rangle\langle 0||R)+(|1\rangle\langle
0||R)=(0|R)+(2|R)=(2|R)\\\ \hat{\flat}&=(|1\rangle\langle
1||R)+(|1\rangle\langle 0||R)=(3|R)+(1|R)=(4|R)\end{aligned}\,.$ (162)

We know that $\\{\hat{\sharp},\hat{\flat}\\}$ is a dual pair under the
$\sigma_{v}$ reflection and this can be confirmed by seeing $0,1$ duality,
i.e.

$*[(|0\rangle\langle 0||R)+(|1\rangle\langle 0||R)]=(|1\rangle\langle
1||R)+(|0\rangle\langle 1||R)\,.$ (163)

Therefore, 2 and 4 are dual to each other. Since 2 and 4 are both even, we
call $\\{\hat{\sharp},\hat{\flat}\\}$ the even association dual pair.

For the left perspective,

$\displaystyle\hat{+}$ $\displaystyle=(|0\rangle\langle
0||L)+(|1\rangle\langle 0||L)=(0|L)+(2|L)=(2|L)$ (164) $\displaystyle\hat{-}$
$\displaystyle=(|1\rangle\langle 1||L)+(|0\rangle\langle
1||L)=(3|L)+(1|L)=(4|L)\,.$

Again, we see that these two matrices are dual to each other by
$0\leftrightarrow 1$ duality. As $2$ and $4$ are even, we say the dual pair
$\\{\hat{+},\hat{-}\\}$ is even association dual pair under the left
perspective. Next we also have

$\displaystyle\hat{\sharp}$ $\displaystyle=(|1\rangle\langle
1||L)+(|1\rangle\langle 0||L)=(3|L)+(2|L)=(5|L)$ (165)
$\displaystyle\hat{\flat}$ $\displaystyle=(|0\rangle\langle
0||L)+(|0\rangle\langle 1||L)=(0|L)+(1|L)=(1|L)\,.$

Again, we see that these two matrices are dual to each other by
$0\leftrightarrow 1$ duality. As $1$ and $5$ are odd, we say the dual pair
$\\{\hat{\sharp},\hat{\flat}\\}$ is odd association dual pair under the left
perspective. Therefore, for a dual pair, whether it is odd or even parity
subjects to $L/R$ operation perspective. The above demonstrates nicely the
following fact that

$\displaystyle(\mathrm{even}|R)$ $\displaystyle\equiv(\mathrm{odd}|R)$ (166)
$\displaystyle(\mathrm{odd}|R)$ $\displaystyle\equiv(\mathrm{even}|R)\,.$

Diagramatically, we can express $\\{\hat{+},\hat{-}\\}$ in rows and
$\\{\hat{\sharp},\hat{\flat}\\}$ in columns in the right perspective, vice
versa. This will give the concept of row and column dualities,

Figure 15:

In general, the following rule is satisfied,

$\displaystyle(\mathrm{row}|R)$ $\displaystyle\equiv(\mathrm{column}|L)$ (167)
$\displaystyle(\mathrm{column}|R)$ $\displaystyle\equiv(\mathrm{row}|L)$

Therefore, in general a row is dual to a column and a column is dual to a row.
However, there is still a degree of freedom when we carry out transformation
between row and columns. A row (column) can be converted to a a column (row)
by either left or right transpose. In here, for the $\\{\hat{+},\hat{-}\\}$
dual pair in right respective, it is brought to the left perspective by left
transpose operation $T$ (anti-clockwise rotation). Also, from the fact that
$(1|R)\equiv(2|L)$, the change of perspective involves an increment of decimal
representation by 1; while $(5|R)\equiv(4|L)$, the change of perspective
involves an decrement of decimal representation by 1. For a general form of
$(a_{R}|R)\equiv(a_{L}|L)$, in the $\hat{+}$ case, we have
$\Delta_{+}=a_{L}-a_{R}=2-1=+1$. For the $\hat{-}$ case,we have
$\Delta_{-}=a_{L}-a_{R}=4-5=-1$. Therefore we have $\Delta_{-}=-\Delta_{+}$.
So in fact $\Delta_{-}$ is dual to $\Delta_{+}$. It follows that

$(\Delta_{+}|\hat{+})\equiv(\Delta_{-}|\hat{-})$ (168)

which is just

$(+1|\hat{+})\equiv(-1|\hat{-})\,.$ (169)

The same idea goes for the $\\{\hat{\sharp},\hat{\flat}\\}$ dual pair, where
we have

$(\Delta_{\sharp}|\hat{\sharp})\equiv(\Delta_{\flat}|\hat{\flat})$ (170)

which is just

$(+3|\hat{\sharp})\equiv(-3|\hat{\flat})\,.$ (171)

In here, for the $\\{\hat{\sharp},\hat{\flat}\\}$ dual pair in right
respective, it is brought to the left perspective by right transpose operation
$R$ (clockwise rotation), which is dual to $T$ (anti-clockwise direction) in
$\\{\hat{+},\hat{-}\\}$. This affirms $\\{\hat{+},\hat{-}\\}$ and
$\\{\hat{\sharp},\hat{\flat}\\}$ are indeed dual to each other.

Generally speaking, the difference between $\Delta$ values of the dual
representations characterize the specific dual representation. For the above
case, $\\{\hat{+},\hat{-}\\}$ has $\\{\Delta_{+},\Delta_{-}\\}=\\{+1,-1\\}$,
and representation $R$ is brought to representation $L$ by left transpose $T$.
$\\{\hat{\sharp},\hat{\flat}\\}$ has
$\\{\Delta_{\sharp},\Delta_{\flat}\\}=\\{+3,-3\\}$, and representation $R$ is
brought to representation $L$ by right transpose $R$. And we must have
$|\Delta|=|\Delta^{*}|$ so the $\Delta$ values of the two dual representation
must be conserved.

Now we consider a special case,

$\boldsymbol{1}=(|0\rangle\langle 0|+|1\rangle\langle
1||R)=(0|R)+(3|R)=(3|R)\,,$ (172)

the same goes for

$\boldsymbol{1}=(|1\rangle\langle 0|+|0\rangle\langle
1||L)=(2|L)+(1|L)=(3|L)\,.$ (173)

Therefore we have

$(3|R)=(3|L)\,.$ (174)

Therefore $3=3^{*}$, 3 is a dual invariant number. Therefore we have
$\Delta_{\boldsymbol{1}}=0$. Similarly we also have the case of its dual
$\mathbb{M}$,

$\boldsymbol{M}=(|0\rangle\langle 1|+|1\rangle\langle
0||R)=(1|R)+(2|R)=(3|R)\,,$ (175)

the same goes for

$\boldsymbol{M}=(|0\rangle\langle 0|+|1\rangle\langle
1||L)=(0|L)+(3|L)=(3|L)\,.$ (176)

Again $\Delta_{\boldsymbol{M}}=0$. This implies that both $\boldsymbol{1}$ and
$\boldsymbol{M}$ are dual invariant. However, what does it mean? We know that
$\\{\boldsymbol{1},\boldsymbol{M}\\}$ forms a dual pair, how can they be dual
invariant? It seems that we arrive at some contradiction. Yet if we consider
the diagonal perspective, dual invariance is conserved.

Figure 16:

In general, we expect that for some representation $X$ and its dual $X^{*}$ if
$\Delta_{X}=\Delta_{X^{*}}=0$, then they are dual invariant under the diagonal
perspective.

Finally, we can construct the full dual invariant $\mathbb{I}$. Notice that
for the $R$-perspective

$\mathbb{I}=(1|R)+(5|R)=(2|R)+(4|R)=(3|R)+(3|R)=(6|R)\,.$ (177)

And for the $L$ perspective,

$\mathbb{I}=(2|L)+(4|L)=(1|L)+(5|L)=(3|L)+(3|L)=(6|L)\,.$ (178)

Therefore we have

$(6|R)=(6|L)\,.$ (179)

Thus $6$ is also a dual invariant number, and it represents the $\mathbb{I}$
matrix. Hence the above expression $6=1+5=2+4=3+3$ can be expressed in matrix
form.

From the above, we have shown that 1 is dual to 5, 2 is dual to 4, and 3 is
dual invariants. This means in the number line 1,2,3,4,5 we have $3$ that
partitions this set into two halves. The number 3 acts like a mirror such that
the number in two opposite sides are dual to each other. The following
illustrates this important result

Figure 17:

In addition, we know that 6 is a dual invariant, and of course 0 is a dual
invariant number. To complete the whole construction, we assign $0$ as the
zero matrix $\boldsymbol{0}$. This makes sense because as in section 2.1.4, we
have $!\boldsymbol{0}=\mathbb{I}$, so since $3$ is the duality mirror, and
each opposite side of $3$ have matrices exchanging $0\leftrightarrow 1$. The
following illustration shows the whole idea:

Figure 18: Representation of numbers 0 to 6 by duality matrices. 0, 3,6 are
dual invariant matrices in $R$ and $L$ operation perspective. Others are non-
dual invariant under $R$ and $L$ operation perspective.

Finally, we would like to express the $R$ and $L$ perspective graphically.
According to 159, if we arrange $0-4$ to be the corners of a square, we would
like to work out the path order of $L$-perspective and $R$\- perspective. For
the $R$-perspective, let’s start from the order of 0123, then it is a square
with one hole with zero node. For $L$-perspective, it is 2031, which is a
twisted square with two holes and one node. Hence, a square representation in
$R$-perspective is a twisted square representation in $L$-perspective. On the
other way, starting from the $L$-perspective, if we choose the order of 0123,
then it is a square; while it would be 1302 for the $R$-perspective, which is
a twisted square. Thus we have the following duality equations,

$\displaystyle\bigg{(}\begin{gathered}\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{topo1.png}\end{gathered}\bigg{|}R\bigg{)}$
$\displaystyle\equiv\bigg{(}\begin{gathered}\includegraphics[trim=0.0pt 0.0pt
0.0pt 0.0pt,clip,scale={0.5}]{topo2.png}\end{gathered}\bigg{|}L\bigg{)}$ (180)
$\displaystyle\bigg{(}\begin{gathered}\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{topo2.png}\end{gathered}\bigg{|}R\bigg{)}$
$\displaystyle\equiv\bigg{(}\begin{gathered}\includegraphics[trim=0.0pt 0.0pt
0.0pt 0.0pt,clip,scale={0.5}]{topo1.png}\end{gathered}\bigg{|}L\bigg{)}\,.$

It is easier to visualize in 2D. If we consider 2D, the square corresponds a
topological torus, and the twisted square corresponds to a topological surface
of genius equal to 2.

Figure 19:

Therefore, a surface of genus 1 is dual to a surface of genus 2. A surface of
one hole is dual to the surface to two holes. This can be further entails
that, if $u=u^{*}$, it should be a topological invariant. This has important
implication in the application of topological quantum field theory in later
sections.

###  2.7 Arithmetic Duality

In this section, we will study dualities in arithmetic operators. First let
$x$ and $x^{\prime}$ be elements, then we define a function $f(x,x^{\prime})$.
If $f(x,x^{\prime})=f(x^{\prime},x)$ then it is symmetric and it is dual
invariant. In particle physics term, it is bosonic. If
$f(x,x^{\prime})=*f(x,x^{\prime})$ where $*$ is some dual operator with
$**=1$, then it is anti-symmetric, or non-dual invariant. In particle physics
term, it is fermonic.

First, we study the addition operator $+$. We have the following,

$f(x,x^{\prime})=x+x^{\prime}=x^{\prime}+x=f(x^{\prime},x)\,.$ (181)

Thus the addition function is dual invariant (which is same when looking from
the left or looking from the right ) , it is symmetric and hence addition
corresponds to bosonic operator. Explicitly,

$(x+x^{\prime}|R)\equiv(x^{\prime}+x|L)=(x+x^{\prime}|L)\equiv(x^{\prime}+x|R)\,.$
(182)

An implication for this result is that the cosine function is symmetric, and
hence bosonic,

$\cos x=\frac{e^{ix}+e^{-ix}}{2}=\frac{e^{-ix}+e^{ix}}{2}\,.$ (183)

Next, we study the multiplication operator $\times$, We have the following,

$f(x,x^{\prime})=x\times x^{\prime}=x^{\prime}\times x=f(x^{\prime},x)\,.$
(184)

Thus the multiplication function is dual invariant,

$(x\times x^{\prime}|R)\equiv(x^{\prime}\times x|L)=(x\times
x^{\prime}|L)\equiv(x^{\prime}\times x|R)\,.$ (185)

It is also symmetric and bosonic.

Now, consider the subtraction operator $-$. We have the following,

$f(x,x^{\prime})=x-x^{\prime}=-(x^{\prime}-x)=-f(x^{\prime},x)\,.$ (186)

Thus the subtraction function is anti-symmetric , it is not dual invariant in
the left/right observer perspective, thus subtraction is fermonic. We have
$*=-$ and $--=+1$. Notice that

$(x-x^{\prime}|R)=x-x^{\prime}\quad,\quad(x-x^{\prime}|L)=x^{\prime}-x\quad,\quad(x^{\prime}-x|R)=x^{\prime}-x\quad,\quad(x^{\prime}-x|L)=x-x^{\prime}\,.$
(187)

Therefore,

$(x-x^{\prime}|R)\equiv(x^{\prime}-x|L)\quad\text{and}\quad(x-x^{\prime}|L)\equiv(x^{\prime}-x|R)\,.$
(188)

We can express this in terms of the standard
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ 4-tableau,

Figure 20:

in which we map $x-x^{\prime}\rightarrow 0,\,\,x^{\prime}-x\rightarrow 1$ and
$R\rightarrow 0,\,\,L\rightarrow 1$. An implication for this result is that
the sine function is symmetric, and hence fermonic,

$\sin x=\frac{e^{ix}-e^{-ix}}{2i}=-\frac{e^{-ix}-e^{ix}}{2i}\,.$ (189)

Finally we consider the division operator $\div$. We have the following,

$f(x,x^{\prime})=x\div x^{\prime}=(x^{\prime}\div
x)^{-1}=f^{-1}(x^{\prime},x)$ (190)

Thus the subtraction function is anti-symmetric, it is not dual invariant in
the left/right observer perspective, We have $*f=f^{-1}$ and
$**f=(f^{-1})^{-1}=f$ thus $**=1$. Notice that

$(x\div x^{\prime}|R)=x\div x^{\prime}\quad,\quad(x\div
x^{\prime}|L)=x^{\prime}\div x\quad,\quad(x^{\prime}\div x|R)=x^{\prime}\div
x\quad,\quad(x^{\prime}\div x|L)=x\div x^{\prime}\,.$ (191)

Therefore,

$(x\div x^{\prime}|R)\equiv(x^{\prime}\div x|L)\quad\text{and}\quad(x\div
x^{\prime}|L)\equiv(x^{\prime}\div x|R)\,.$ (192)

We can express this in terms of the standard
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ 4-tableau,

Figure 21:

in which we map $x\div x^{\prime}\rightarrow 0,\,\,x^{\prime}\div x\rightarrow
1$ and $R\rightarrow 0,\,\,L\rightarrow 1$.

Therefore, all in all, we have both addition and multiplication $+,\times$ as
one category-the dual category; and subtraction and division as the other
category, the non-dual category. We can define $W=\\{+,\times\\}$ and
$W^{*}=\\{-,\div\\}$.

Next, we will show that $+$ and $\times$ operators are dual to each other,
despite that they are both symmetric and bosonic. This can be shown by the
$0,1$ duality we showed in previous section. We consider two function $F(0,1)$
and $F(1,0)$. We then define the perspective as a map such that for some
$a,b$,

$L:F(a,b)\rightarrow a\quad\text{and}\quad R:F(a,b)\rightarrow b\,,$ (193)

where $L:F(a,b)=:(F(a,b)|L)$ and $R:F(a,b)=:(F(a,b)|R)$. We then have the
following,

$(F(0,1)|L)=(F(1,0)|R)=0\quad\text{and}\quad(F(0,1)|R)=(F(1,0)|L)=1\,.$ (194)

First we have

$(F(0,1)|L)=0\times 1=0=1\times 0=(F(1,0)|R)\,,$ (195)

then we have

$(F(0,1)|R)=0+1=1=1+0=(F(1,0)|L)\,.$ (196)

But by the result in 112, $!0=1$ and $!1=0$ where 0 and 1 are dual to each
other, so $(F(0,1)|L)$ and $(F(1,0)|R)$ corresponds to the multiplication
operation $\times$; while $(F(0,1)|R)$ and $(F(1,0)|L)$ corresponds to the
addition operation $+$. In terms of the $\mathbb{Z}\times\mathbb{Z}$ tableau,

Figure 22:

Therefore $+$ and $\times$ are dual to each other as $0$ and $1$ are dual to
each other.

###  2.8 Arithmetic Expression Duality and the Cyclic Rule

Now, we establish the duality on arithmetic expression under left, right
observation perspective. We will show that positive numbers are left-right
dual invariant, while negative numbers are not. Recalling that expression of
$f(x,x^{\prime})=x+x^{\prime}$, which is bosonic, which implies
$f(x,x^{\prime})=f(x^{\prime},x)$ . This also means that when we look from the
$R$ perspective, this is $x+x^{\prime}$. If we look from the $L$ prespective,
this is $x^{\prime}+x$. Thus bosonic also means that the expression is right-
left perspective dual invariant,

$(x+x^{\prime}|R)\equiv(x+x^{\prime}|L)\,.$ (197)

Now consider another bosonic expression $f(x,x^{\prime})=-x-x^{\prime}$. This
expression is simply $-x-x^{\prime}$ when looking from the right perspective,
but how about looking from the left perspective? We are running into a problem
as we have $x^{\prime}-x-$ which is an operator but not an element. However,
as we can obviously see that $f(x,x^{\prime})=f(x^{\prime},x)$ for this case,
so we expect it is left right dual invariant.

Let’s see for the fermionic case. For $f(x,x^{\prime})=x-x^{\prime}$, when we
look from the right perspective, it is $x-x^{\prime}$. If we look from the
left perspective, it becomes $x^{\prime}-x$. This is totally fine. However, if
we have $f(x,x^{\prime})=-x+x^{\prime}$, we know that again it satisfies
$f(x,x^{\prime})=-f(x^{\prime},x)$, which is not dual invariant. But when we
look from the right perspective, we get $x^{\prime}+x-$ which is problematic.
To counter this problem we have to rearrange the terms as
$-x+x^{\prime}=x^{\prime}-x$, such that when we look from the left
perspective, it becomes $x-x^{\prime}$ which is
$-(-x+x^{\prime})=-f(x,x^{\prime})$. Although the $-+$ case can be solved by
rearranging the terms, we see that the $--$ case, $-x-x^{\prime}$ cannot be
solved by rearranging the terms. This urges us that we demand a more generic
rule to define arithmetic expression duality.

The generic rule is as follow, which we call the cyclic rule. We wind up any
arithmetic expressions in a close ring.

Figure 23: We define $f_{++}(x,x^{\prime})=+x+x^{\prime}$,
$f_{--}(x,x^{\prime})=-x-x^{\prime}$, $f_{+-}(x,x^{\prime})=+x-x^{\prime}$ and
$f_{-+}(x,x^{\prime})=-x+x^{\prime}$.

To read these ring diagrams we have the following rules:

  1. 1.

The order goes with operator(sign)-element-operator(sign)-element.

  2. 2.

The value obtained along a particular cyclic direction must be the same, no
matter the starting point.

  3. 3.

The anti-clockwise direction denotes the one of the perspective, while the
clockwise direction denotes the dual perspective.

Using the $f_{--}$ as the example, using the first point , we begin with the
$-$ sign on the top, going towards the anti-clockwise direction we have
$-x-x^{\prime}$ for one$(R)$ perspective. For the dual$(L)$ perspective, start
from the $-$ sign on the top, going with the clockwise direction, we have
$-x^{\prime}-x$ for one$(L)$ perspective. The similar idea holds all for
$f_{++},f_{+-}$ and $f_{-+}$ cases. In terms of the 4-tableau, we have

Figure 24: In this diagram, D.I. stands for left-right perspective dual
invariant.

From the $f_{++}$ case, if we pick $x=0$, then we have
$(0+x^{\prime}|R)\equiv(+x^{\prime}+0|L)\implies(+x^{\prime}|R)\equiv(+x^{\prime}|L)$.
Therefore positive numbers are dual invariant. However, for the $f_{+-}$ case
and the $f_{-+}$ case, as $(f_{+-}|R)\equiv(f_{-+}|L)$, i.e.
$(x-x^{\prime}|R)\equiv(x^{\prime}-x|L)$, if we pick $x=0$, this gives
$(-x^{\prime}|R)\equiv(x^{\prime}|L)$, so negative numbers are not dual
invariant.

###  2.9 Cyclic rule on tensor product states

In this section, we will investigate the power of cyclic rule on tensor
product state. First notice the following identity:

$\displaystyle|0\rangle\otimes\langle 0|$ $\displaystyle=\langle
0|\otimes|0\rangle=\hat{P}_{+}$ (198) $\displaystyle|1\rangle\otimes\langle
1|$ $\displaystyle=\langle 1|\otimes|1\rangle=\hat{P}_{-}\,.$

On the other hand,

$\displaystyle|0\rangle\otimes\langle 1|$ $\displaystyle=\langle
1|\otimes|0\rangle=\hat{P}^{*}_{+}$ (199)
$\displaystyle|1\rangle\otimes\langle 0|$ $\displaystyle=\langle
0|\otimes|1\rangle=\hat{P}^{*}_{-}\,.$

We can illustrate such results by the cyclic rule.

Figure 25: Tensor product states with the anticlockwise perspective

For the first diagram $|0\rangle\langle 0|$, following the lower half anti-
clockwise direction, we have $|0\rangle\otimes\langle 0|$. Following the upper
anti-clockwise direction, we have $\langle 0|\otimes|0\rangle$. Since along
the same direction, the expression must be the same, thus by the cyclic rule
we must have $|0\rangle\otimes\langle 0|=\langle 0|\otimes|0\rangle$. The same
idea goes with the second digram for the $|1\rangle\langle 1|$ case. For the
third diagram, following the lower half anti-clockwise direction, we have
$|0\rangle\otimes\langle 1|$. Following the upper anti-clockwise direction, we
have $\langle 1|\otimes|0\rangle$. Since along the same direction, the
expression must be the same, thus by the cyclic rule we must have
$|0\rangle\otimes\langle 1|=\langle 1|\otimes|0\rangle$. The same idea goes
with the fourth digram for the $|1\rangle\langle 0|$ case.

Now consider the dual perspective, which is the clockwise operation.

Figure 26: Tensor product states with the dual clockwise perspective

For the first two diagrams, they remain the same. Therefore, as expected,
$|0\rangle\langle 0|$ and $|1\rangle\langle 1|$ states are dual invariant. We
have

$\displaystyle(|0\rangle\otimes\langle 0||\mathrm{anticlockwise})$
$\displaystyle\equiv(|0\rangle\otimes\langle 0||\mathrm{clockwise})$ (200)
$\displaystyle(|1\rangle\otimes\langle 1||\mathrm{anticlockwise})$
$\displaystyle\equiv(|1\rangle\otimes\langle 1||\mathrm{clockwise})\,,$

while the $|0\rangle\langle 1|$ case and $|1\rangle\langle 0|$ case are non-
dual invariant.

###  2.10 Element-Operator duality

In the final part, we study element and operator duality. Let $a$ be an
element (a number), $+$ be an operator. We have $+a=a$ is an element, but $a+$
is an operator. Define also the $L$ and $R$ perspective, then we have

$(+a|R)=\text{element}\,\,,\,\,(+a|L)=\text{operator}\,,\,\,,(a+|R)=\text{operator}\,\,\,,\,\,(a+|L)=\text{element}\,.$
(201)

Therefore we have

$(+a|R)\equiv(a+|L)\quad\text{and}\quad(a+|R)\equiv(+a|L)\,.$ (202)

In terms of $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ 4-tableau, we have

Figure 27:

Next, we would study the fundamental element-operator duality in
multiplication. Consider the multiplication of two elements, $A_{1}\cdot
A_{2}$. Now consider inserting an identity element in the middle,
$A_{1}(1)A_{2}=A_{1}A_{2}$. Explicitly, there are two ways to for insertion
and two forms to insert. Let $1$ be the element and the multiplication $\cdot$
be the operator. Define the element operator (EO) form as $1\cdot$, and the
operator element form (OE) form as $\cdot 1$. We immediately have

$(1\cdot|R)\equiv(\cdot 1|L)\quad\text{and}\quad(1\cdot|L)\equiv(\cdot
1|R)\,.$ (203)

Therefore $1\cdot$ and $\cdot 1$ are dual to each other,

$1\cdot=*(\cdot 1)\quad\text{and}\quad\cdot 1=*(1\cdot)\,.$ (204)

Or equivalently EO form and OE form are dual to each other,
$\mathrm{EO}=*\mathrm{OE}$ and $\mathrm{OE}=*\mathrm{EO}$. The two ways for
insertion are

$A_{1}\,{}_{\wedge}\,\,\,\,{}_{\wedge}\cdot A_{2}\quad\text{and}\quad
A_{1}\cdot\,_{\wedge}\,\,\,\,{}_{\wedge}A_{2}\,\,\,,$ (205)

where the first one is the left insertion denoted by $l$ perspective and the
second one is the right insertion by $r$ perspective. To restore the product
of $A_{1}A_{2}$ , we need to have

$A_{1}\,{}_{\wedge}(\cdot 1)\,_{\wedge}\cdot A_{2}\quad\text{and}\quad
A_{1}\cdot\,_{\wedge}(1\cdot)\,_{\wedge}A_{2}\,\,\,,$ (206)

so both of them looks like $A_{1}\cdot 1\cdot A_{2}$, we notice that the OEO
form $\cdot 1\cdot$ is dual invariant with respect to the right observer and
left observer, i.e. $(\cdot 1\cdot|L)\equiv(\cdot 1\cdot|R)$. The OEO form
here is referred as the connector. Therefore we have

$(\cdot 1|l)\equiv(1\cdot|r)\,,$ (207)

or

$(\mathrm{OE}|l)\equiv(\mathrm{EO}|r)\,.$ (208)

Next, we have another form of insertion,

$A_{1}\,{}_{\wedge}(1\cdot)\,_{\wedge}\cdot A_{2}\quad\text{and}\quad
A_{1}\cdot\,_{\wedge}(\cdot 1)\,_{\wedge}A_{2}\,\,\,,$ (209)

This gives,

$A_{1}\,\,\,1\cdot\cdot A_{2}\quad\text{and}\quad A_{1}\cdot\cdot
1\,\,\,A_{2}\,.$ (210)

The EOO form $1\cdot\cdot$ and OOE form $\cdot\cdot 1$ is not dual invariant,
in particular we see that

$(1\cdot\cdot|R)\equiv(\cdot\cdot
1|L)\quad\text{and}\quad(1\cdot\cdot|L)\equiv(\cdot\cdot 1|R)\,.$ (211)

Such EOO form and OOE form are referred as the disconnector, they will break
the product into two separate elements $A_{1},A_{2}$ and does not return to a
value. In terms of $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ tableau, we have the
following diagram representation,

Figure 28:

###  2.11 Multi-duality Structure

The theory of a single duality system can be generalized into multi-duality
system, in which more than one complete duality structure is concerned.

######  Definition 2.2.

Let the $i^{\rm{th}}$ unit of complete duality be $D_{i}=\\{W_{i},B_{i}\\}$.
The full set of total duality is defined as $D_{tot}$ with $|D_{tot}|=n$,
which is the union of all $U_{i}$. Each $D_{i}$ is a partition set of
$D_{tot}$. Mathematically,

$D_{tot}=\bigcup_{i=1}^{n}D_{i}\,\,\,\,{\mathrm{and}}\,\,\,\,W_{tot}=\bigcup_{i=1}^{n}W_{i}=\bigcup_{i=1}^{n}U_{i}\cup
U_{i}^{*}\cup\\{0\\}_{i}\,\,,\,\,B_{tot}=\bigcup_{i=1}^{n}B_{i}=\bigcup_{i=1}^{n}S_{k,i}\cup
S_{k,i}^{\star}\,.$ (212)

$\bigcap_{i=1}^{n}D_{i}=\emptyset\,\,\,\,{\mathrm{and}}\,\,\,\,\bigcap_{i=1}^{n}U_{i}\cup
U_{i}^{*}=\emptyset\,\,,\,\,\bigcap_{i=1}^{n}S_{k,i}\cup
S_{k,i}^{\star}=\emptyset\,.$ (213)

Next we would like to study the algebra of the multi-duality. We promote the
definition of dual set to dual space. (Remark the dual space here is not
referring to that in differential Geometry). The dual spaces are related by
tensor product.

######  Definition 2.3.

(I) Let $\mathcal{V}_{i}$ be a $i^{\mathrm{th}}$ of two dimensional space
which can be partitioned into two one dimensional sub-space $V_{i}$ and its
dual $V^{*}_{i}$ such that $\mathcal{V}_{i}=V_{i}\oplus V_{i}^{*}$. The multi-
dual spaces satisfy the following algebra. Define each partition as

$\bigotimes^{p,q}_{i,j}v_{i,j}:=\bigotimes_{i}^{p}V_{i}\otimes\bigotimes_{j}^{q}V_{j}^{*}\,.$
(214)

(II) Let $\mathcal{W}$ be the full multi-duality space. The multi-duality
system if multiplicity $n$ possess the natural decomposition algebra,

$\mathcal{W}=\bigotimes_{i=1}^{n}\mathcal{V}_{i}=\bigoplus_{p+q=n}\bigotimes^{p,q}_{i,j}v_{i,j}\,.$
(215)

(III) Let $\mathcal{S}_{k,i}$ be the $i^{\rm th}$ observer’s space
corresponding to $\mathcal{V}_{i}$ which can be partitioned into
$\mathcal{S}_{k,\,\,i}$ and its dual $\mathcal{S}_{k,\,\,i}^{\star}$. The
multi-dual observer’s space satisfies

$\bigotimes^{p,q}_{i,j}s_{k,\,\,i,j}:=\bigotimes_{i}^{p}S_{k\,\,i}\otimes\bigotimes_{j}^{q}S_{k\,\,j}^{\star}\,.$
(216)

and

$\mathcal{B}=\bigotimes_{i=1}^{n}\mathcal{S}_{k,\,\,i}=\bigoplus_{p+q=n}\bigotimes^{p,q}_{i,j}s_{k,\,\,i,j}\,.$
(217)

(IV) The algebra of the element space and the algebra of the observer space is
an isomorphism
$\bigotimes_{i=1}^{n}\mathcal{V}_{i}\cong\bigotimes_{i=1}^{n}\mathcal{S}_{k,\,\,i}$
for each $i$. There exists a bijective map for the two algebras.

(V) The complete dual space of multi-duality $\mathcal{C}$ is a natural
duality if $\mathcal{D}=\mathcal{W}\oplus\mathcal{B}$, where $\mathcal{D}$
induces a new observer’s space $\mathcal{O}_{l}$ and its dual
$\mathcal{O}_{l}^{\star}$ in dimension $l$ such that

$(\mathcal{W}|\mathcal{O}_{l})\equiv(\mathcal{B}|\mathcal{O}_{l}^{\star})\,\,\,\,{\mathrm{and}}\,\,\,\,(\mathcal{W}|\mathcal{O}_{l}^{\star})\equiv(\mathcal{B}|\mathcal{O}_{l})$
(218)

where

$\mathcal{B}=\mathcal{W}^{*}\,\,\,\,{\mathrm{and}}\,\,\,\,\mathcal{W}=\mathcal{B}^{*}\,.$
(219)

The role of element space and observer’s space is interchangeable under such
duality system.

(VI) Let the dual map for element space $*$ such that
$*_{V_{i}}:V_{i}\rightarrow V_{i}^{*}$ and $*_{V_{i}^{*}}:V_{i}^{*}\rightarrow
V_{i}$ where $**$ is the identity map $I_{d}(V_{i})$. Similarly define dual
map for observer space $\star$ such that
${\star}_{S_{k,\,\,i}}:S_{k\,\,i}\rightarrow S_{k,\,\,i}^{\star}$ and
${\star}_{S_{k,\,\,i}^{\star}}:S_{k\,\,i}^{\star}\rightarrow S_{k,\,\,i}$
where $\star\star$ is the identity map $I_{d}(S_{k,i})$. The maps
$*_{V_{i}}\circ\star_{S_{k,\,\,i}}=I_{d}(V_{i},S_{k,\,\,i})$ and
$\star_{S_{k,\,\,i}}\circ*_{V_{i}}=I_{d}(S_{k,\,\,i},V_{i})$ are identity
maps. For each partition, we define the collective duality map as the
following:

$*_{p,q}=\prod_{i}^{p}*_{V_{i}}\prod_{j}^{q}*_{V_{j}^{*}}\,\,\,\,{\mathrm{and}}\,\,\,\,\star_{p,q}=\prod_{i}^{p}\star_{S_{k,\,\,i}}\prod_{j}^{q}\star_{S_{k,\,\,j}^{\star}}\,,$
(220)

where the product sign here for notation simplicity denotes the operation of
composite maps.

(VII) With all these maps defined we have the following theorems. The
arbitrary number of $r\leq p$ $*_{V_{i}}$ maps and the arbitrary number of
$s\leq q$ $*_{V_{j}^{*}}$ maps acting on any partition must return to any
other partition.

$*_{r,s}\bigotimes_{i,j}^{p,q}v_{i,j}=\bigotimes_{i,j}^{p-r+s,q+r-s}v_{i,j}$
(221)

for $p-r+s=p^{\prime}$ and $q+r-s=q^{\prime}$. The same theorem holds for
observer spaces and its dual operators.

(VIII) The identity map
$*_{r,s}\circ\star_{r,s}=\star_{r,s}\circ*_{r,s}=I_{d}$ acts on the partition
as

$*_{r,s}\star_{r,s}\left(\bigotimes_{i,j}^{p,q}v_{i,j}\bigg{|}\bigotimes_{i,j}^{p,q}s_{k,\,\,i,j}\right)=\left(\bigotimes_{i,j}^{p-r+s,q+r-s}v_{i,j}\bigg{|}\bigotimes_{i,j}^{p-r+s,q+r-s}s_{k,\,\,i,j}\right)=\left(\bigotimes_{i,j}^{p,q}v_{i,j}\bigg{|}\bigotimes_{i,j}^{p,q}s_{k,\,\,i,j}\right)\,.$
(222)

The duality operators can be viewed as discrete parity symmetry and they form
a parity group. Define the parity group for elements as
$\rho_{i}(V_{i})=\\{I_{d}(V_{i}),\,*_{V_{i}}\\}$ and for observer as
$\rho_{i}(S_{k,\,\,i})=\\{I_{d}(S_{k,\,\,i}),\star_{S_{k,\,\,i}}\\}$. These
parity groups are isomorphic to the group $\mathbb{Z}_{2}$. The multi-duality
of the first class is the study of tensor product representations of the
parity groups of elements and observers. The $\mathcal{V}_{i}$ and
$\mathcal{S}_{k,\,\,i}$ spaces are representation vector spaces. Equations 215
and 217 show the reducible representation tensor product vector spaces as the
direct sum of each irreducible representations.

The irreducible tensor product vector spaces can recombine to form duality
systems. One grouping criteria is according to binomial coefficients. It can
be seen that the number of ways of the $p$ and $q$ indices combine follow the
binomial distribution. The total number of irreducible representation vector
spaces is,

$\sum_{i=0}^{n}{\mathrm{C}}_{i}^{n}=\sum_{i=0}^{n}\frac{n!}{i!(n-i)!}=2^{n}\,.$
(223)

If viewing the binomial coefficients as the pascal triangle, one sees that
${\mathrm{C}}_{i}^{n}$ and ${\mathrm{C}}_{n-i}^{n}$ is symmetric. Thus the
irreducible representation spaces can be naturally group as $2^{n-1}$ sub-
duality systems, and they can be further re-group into one large duality
system. The one large duality system is as follow,

$\mathcal{W}=\mathcal{V}_{1}\otimes\cdots\otimes\mathcal{V}_{n}=\mathcal{A}\oplus\mathcal{A^{*}}=\bigoplus_{l=1}^{2^{n-1}}(\mathcal{K}_{l}\oplus\mathcal{K}_{l}^{*})$
(224)

where

$\mathcal{A}=\bigoplus_{p+q=n,\,\,p\geq
q}\bigotimes_{i,j}^{p,q}v_{i,j}\,\,\,\,{\mathrm{and}}\,\,\,\,\mathcal{A}^{*}=*\mathcal{A}=\bigoplus_{p+q=n,\,\,p\leq
q}\bigotimes_{i,j}^{p,q}v_{i,j}\,,$ (225)

and this is the one large duality system. For the sub-$2^{n-1}$ duality
systems,

$\mathcal{K}_{l}=\bigoplus_{i,j}^{p,q}(v_{i,j})_{l}\,\,\,\,{\mathrm{and}}\,\,\,\,\mathcal{K}_{l}^{*}=*_{p,q}\mathcal{K}_{l}=\bigoplus_{i,j}^{q,p}(v_{i,j})_{l}\,.$
(226)

Any partition with its full dual forms the $K$ space. Alternatively, the
$K_{l}$ dual spaces can be explicitly defined through the pair-wise dual
operations $*_{V_{j}}*_{V_{j}^{*}}$. The $K_{l^{\prime}}\oplus
K_{l^{\prime}}^{*}$ dual space serves as the role of generating other $K$-dual
spaces under the pair-wise dual operators.

(IX) Each $K_{l}\oplus K_{l}^{*}$ dual space is a generator space, which can
generate other $K_{l^{\prime}}\oplus K_{l^{\prime}}^{*}$ dual spaces under the
pair-wise dual operators $*_{V_{j}}*_{V_{j}^{*}}$. Mathematically,

$*_{V_{j}}*_{V_{j}^{*}}(K_{l}\oplus K_{l}^{*})=(K_{l^{\prime}}\oplus
K_{l^{\prime}}^{*})\,.$ (227)

If we apply fully the $*_{V_{j}}*_{V_{j}^{*}}$ operators for all $j$ on a
$(K_{l}\oplus K_{l}^{*})$, then we obtain all other $(K_{l^{\prime}}\oplus
K_{l^{\prime}}^{*})$ except itself.

$\sum_{j=1}^{n}*_{V_{j}}*_{V_{j}^{*}}(K_{l}\oplus
K_{l}^{*})=\bigoplus_{l^{\prime}\neq l}(K_{l^{\prime}}\oplus
K_{l^{\prime}}^{*})\,.$ (228)

(X) The $\mathcal{W}$ space is the representation space of the multi-duality
group $\mathbb{Z}_{2}\otimes\cdots\otimes\mathbb{Z}_{2}$. The map $\rho$

$\rho:\mathbb{Z}_{2}\otimes\mathbb{Z}_{2}\otimes\cdots\otimes\mathbb{Z}_{2}\rightarrow\mathcal{V}\otimes\mathcal{V}\otimes\cdots\otimes\mathcal{V}\,.$
(229)

Next we would like to study some conserved dual operations under some
circumstances of invariance in duality system.

###  2.12 Duality Transformation and Duality Symmetry

Since a complete duality system bases on both the element space of observer’s
space, when we consider dual action we have the following circumstances. (1)
The element space is transformed while keeping the observer space constant.
(2) The observer space is transformed while keeping the element space
constant. (3) Both the element space and observer space transform. First we
consider the (1) case.

####  2.12.1 Local duality transformation

######  Definition 2.4.

If $\mathcal{W}=\mathcal{V}_{1}\otimes\cdots\otimes\mathcal{V}_{n}$ is an
invariant, then any dual operations on a particular partition or re-partitions
in $\mathcal{W}$ must induce a simultaneous same dual operation on that
original partition, such that $\mathcal{W}$ remains unchanged.

The $\mathcal{W}$ has $2^{n}$ distinct partitions. Suppose
$P_{1}=\bigotimes_{i,j}^{p_{1},q_{1}}v_{i,j}$ where $p_{1}+q_{1}=n$ is one of
the partitions. Now we act dual operators $*_{r_{1},s_{1}}$ where $r_{1}\leq
p_{1}$ and $s_{1}\leq q_{1}$ on $P_{1}$. Then the original $P_{1}$ will become
another partition $P_{2}^{\prime}$ in $\mathcal{W}$,

$*_{r_{1},s_{1}}P_{1}=P_{2}^{\prime}\,.$ (230)

The prime on $P_{2}$ denotes that $P_{2}$ is transformed from $P_{1}$.
However, if $\mathcal{W}$ is an invariant under any dual operations, then
$P_{1}\rightarrow P_{2}^{\prime}=*_{r_{1},s_{1}}P_{1}$ breaks the invariant as
$P_{1}\notin\mathcal{W}$, and now we have two $P_{2}$s, one from the original
one in $\mathcal{W}$ and the new one $P_{2}^{\prime}$ from $P_{1}$. To
preserve $\mathcal{W}$ we have to transform the original $P_{2}$ to
$P_{1}^{\prime}$, $*_{r_{2},s_{2}}^{-1}P_{2}=P_{1}^{\prime}$. But since the
inverse is just the same as itself in the parity group, thus in fact
$*_{r_{2},s_{2}}^{-1}=*_{r_{1},s_{1}}$. Therefore we just demand
$*_{r_{1},s_{1}}P_{2}=P_{1}^{\prime}$ using the same dual operator. Hence,
before dual transformation, we have $P_{1}$ and $P_{2}$ $\in\mathcal{W}$,
after local dual transformation, we have $P_{2}^{\prime}$ and $P_{1}^{\prime}$
$\in\mathcal{W}$ such that $\mathcal{W}$ remains unchanged. We call such dual
transformation local as it just operates on a particular partition. One
important note is the instantaneous induction on $P_{2}$ transforming back to
$P_{1}^{\prime}$. The two operations must have to be synchronized, as the
invariance of $\mathcal{W}$ must be conserved at any time. This will have
essential physics interpretation later.

The definition for the $K_{l}$ space in 227 is also naturally a local duality
transformation. The above concept applies similarly.

####  2.12.2 Global duality transformation

######  Definition 2.5.

Global duality transformation is a dual transformation of all partitions and
all elements in the partition in $\mathcal{W}$. The full transformation is
simply denoted as $*$.

####  2.12.3 The Dual Symmetry

We define a system to have dual symmetry, or called $D$-symmetry if the system
is invariant under the transformation of the dual operator.

######  Definition 2.6.

Let $U$ be some element space which $U\subseteq W$. If $U$ can be partitioned
into one space and is dual space, $U=X\oplus X^{*}$, then $U$ possesses dual
symmetry such that $*U=U$.

Since $\mathcal{W}=\mathcal{A}\oplus\mathcal{A}^{*}$, it is trivial to see
that $*\mathcal{W}=\mathcal{W}$, the full element space is global dual
symmetry invariant. The partition spaces
$\mathcal{K}_{l}\oplus\mathcal{K}_{l}^{*}$ for each $l$ is also a dual
symmetry invariant.

Next, we would like to show that the full $\mathcal{W}$ space is invariant
under the sub-dual symmetry of full dual operations
$*_{V_{j}}\oplus*_{V_{j}^{*}}$, i.e.,

$(*_{V_{j}}\oplus*_{V_{j}^{*}})W=W\,.$ (231)

This is equivalent to say, the $*_{V_{i}}\oplus*_{V_{i}^{*}}$ acting on all
partitions remain the same, which is an identity map $I_{d}$. We also need

$*_{V_{i}}V_{i}^{*}=0\,\,\,\,{\text{and}}\,\,\,\,*_{V_{i}^{*}}V_{i}=0\,.$
(232)

The proof is straight forward by going the opposite way,

$\displaystyle(*_{V_{j}}\oplus*_{V_{j}^{*}})\mathcal{W}$
$\displaystyle=(*_{V_{j}}\oplus*_{V_{j}^{*}})\bigoplus_{p+q=n}\bigotimes_{s,t}^{p,q}v_{s,t}$
(233)
$\displaystyle=(*_{V_{j}}\oplus*_{V_{j}^{*}})\bigotimes_{i=1}^{n}(V_{i}\oplus
V_{i}^{*})$ $\displaystyle=(*_{V_{j}}\oplus*_{V_{j}^{*}})\big{(}(V_{1}\oplus
V_{1}^{*})\otimes\cdots\otimes(V_{j}\oplus
V_{j}^{*})\otimes\cdots\otimes(V_{n}\oplus V_{n}^{*})\big{)}$
$\displaystyle=(*_{V_{j}}\oplus*_{V_{j}^{*}})\big{(}(V_{1}\oplus
V_{1}^{*})\otimes\cdots\otimes V_{j}\otimes\cdots\otimes(V_{n}\oplus
V_{n}^{*})\big{)}$
$\displaystyle\,\,\,\,\,\,\oplus(*_{V_{j}}\oplus*_{V_{j}^{*}})\big{(}(V_{1}\oplus
V_{1}^{*})\otimes\cdots\otimes V_{j}^{*}\otimes\cdots\otimes(V_{n}\oplus
V_{n}^{*})\big{)}$ $\displaystyle=\big{(}(V_{1}\oplus
V_{1}^{*})\otimes\cdots\otimes V_{j}^{*}\otimes\cdots\otimes(V_{n}\oplus
V_{n}^{*})\big{)}$ $\displaystyle\,\,\,\,\,\,\oplus\big{(}(V_{1}\oplus
V_{1}^{*})\otimes\cdots\otimes V_{j}\otimes\cdots\otimes(V_{n}\oplus
V_{n}^{*})\big{)}$ $\displaystyle=(V_{1}\oplus
V_{1}^{*})\otimes\cdots\otimes(V_{j}^{*}\oplus
V_{j})\otimes\cdots\otimes(V_{n}\oplus V_{n}^{*})$
$\displaystyle=\mathcal{W}\,.$

All of the above theorems apply to the observer space, since the element space
and the observer space themselves are a duality system. By definition, the two
spaces are isomorphic. Thus all theorems for one space apply to the other.
Therefore the (2) case would be the same as the (1) case, but just a change of
notations.

####  2.12.4 Local duality transformation

######  Definition 2.7.

If $(\mathcal{W}|\mathcal{B})$ is an invariant, then any dual operations on a
particular partition or re-partitions in $(\mathcal{W}|\mathcal{B})$ must be
invariant. If a partition in the element space is transformed by $*_{r,s}$,
the corresponding observer space is transformed by $\star_{r,s}$, such that
the overall change $*_{r,s}\circ\star_{r,s}=I_{d}$ is an identity, vice versa.

This is just the consequence of 2.3. One may think of whether we need to do
the same transformation for the original partition back to a new one just like
the case (1). The answer is no, because the change of observer’s space at the
same has compensated the issue. Let $\mathcal{P_{1}}$ be the corresponding
partition for the element space $P_{1}$, together as
$(P_{1}|\mathcal{P}_{1})$. For case (1) we are doing
$*_{r_{1},s_{1}}(P_{1}|\mathcal{P}_{1})=(P_{2}^{\prime}|\mathcal{P}_{1})$,
holding $\mathcal{P}_{1}$ constant. But in this case
$*_{r_{1},s_{1}}\star_{r_{1},s_{1}}(P_{1}|\mathcal{P}_{1})=(P_{2}^{\prime}|\mathcal{P}_{2}^{\prime})$
but this new $(P_{2}^{\prime}|\mathcal{P}_{2}^{\prime})$ is the same as the
original $(P_{1}|\mathcal{P}_{1})$. The $*_{r_{1},s_{1}}\star_{r_{1},s_{1}}$
is just the identity map.

####  2.12.5 Global duality transformation

######  Definition 2.8.

Define the complete global duality transformation as dual transformation of
all partitions in $\mathcal{W}$, denoted as $*$, and dual transformation of
all partitions in $\mathcal{B}$, denoted as $\star$. The composite map is an
identity map.

$*\star(\mathcal{W}|\mathcal{B})=(\mathcal{W}^{*}|\mathcal{B}^{\star})\equiv(\mathcal{W}|\mathcal{B})\,.$
(234)

####  2.12.6 The Duality Symmetry

The idea of duality symmetry for case (3) would be similar to case (1). The
$(\mathcal{W}|\mathcal{B})$ is a full duality invariant under the action of
pair-wise operators for both element space,

$(*_{V_{j}^{*}}\oplus*_{V_{j}^{*}})(\star_{S_{k,\,\,j}}\oplus\star_{S_{k,\,\,j}^{\star}})(\mathcal{W}|\mathcal{B})=(\mathcal{W}|\mathcal{B})\,.$
(235)

The proof will be the same as case(1) but just include the observer’s space.
Note that since the two different maps commute, we can write,

$(*_{V_{j}^{*}}\oplus*_{V_{j}^{*}})(\star_{S_{k,\,\,j}}\oplus\star_{S_{k,\,\,j}^{\star}})(\mathcal{W}|\mathcal{B})=(\star_{S_{k,\,\,j}}\oplus\star_{S_{k,\,\,j}^{\star}})(*_{V_{j}^{*}}\oplus*_{V_{j}^{*}})(\mathcal{W}|\mathcal{B})\,.$
(236)

We will demonstrate all the above abstract definitions of the duality space
above using a duality system with multiplicity $n=3$ as an example. Let there
be three duality units, so we have three parity groups and three
representation vector spaces together with three corresponding observer
spaces. Then the tensor product space is,

$\displaystyle\big{(}\mathcal{V}_{1}\otimes\mathcal{V}_{2}\otimes\mathcal{V}_{3}|\mathcal{S}_{k,\,\,1}\otimes\mathcal{S}_{k,\,\,2}\otimes\mathcal{S}_{k,\,\,3}\big{)}$
(237) $\displaystyle=$ $\displaystyle\big{(}(V_{1}\oplus
V_{1}^{*})\otimes(V_{2}\oplus V_{2}^{*})\otimes(V_{3}\oplus
V_{3}^{*})|(S_{k,\,\,1}\oplus S_{k,\,\,1}^{\star})\otimes(S_{k,\,\,2}\oplus
S_{k,\,\,2}^{\star})\otimes(S_{k,\,\,3}\oplus S_{k,\,\,3}^{\star})\big{)}$
$\displaystyle=$ $\displaystyle\big{(}(V_{1}\otimes V_{2}\otimes
V_{3})\oplus(V_{1}^{*}\otimes V_{2}\otimes V_{3})\oplus(V_{1}\otimes
V_{2}^{*}\otimes V_{3})\oplus(V_{1}\otimes V_{2}\otimes
V_{3}^{*})\,\,|\,\,S_{k}\cdots\big{)}\oplus$
$\displaystyle\big{(}(V_{1}^{*}\otimes V_{2}^{*}\otimes
V_{3}^{*})\oplus(V_{1}\otimes V_{2}^{*}\otimes
V_{3}^{*})\oplus(V_{1}^{*}\otimes V_{2}\otimes
V_{3}^{*})\oplus(V_{1}^{*}\otimes V_{2}^{*}\otimes V_{3})|S_{k}\cdots)\,,$

where the $S_{k}\cdots$ is the expansion counterparts for observer spaces (as
we are running out of space). We can see by definition 2.3 any dual maps on a
particular partition will give you another partition. For example,

$*_{V_{1}}*_{V_{2}^{*}}(V_{1}\otimes V_{2}^{*}\otimes V_{3})=V_{1}^{*}\otimes
V_{2}\otimes V_{3}$ (238)

is another partition. By definition 2.3 we see that for example,

$\displaystyle*_{V_{1}}\star_{S_{k,\,\,1}}*_{V_{2}^{*}}\star_{S_{k,\,\,1}^{\star}}\big{(}V_{1}\otimes
V_{2}^{*}\otimes V_{3}|S_{k,\,\,1}\otimes S_{k,\,\,2}^{\star}\otimes
S_{k,\,\,3}\big{)}$ (239) $\displaystyle=$
$\displaystyle\big{(}V_{1}^{*}\otimes V_{2}\otimes
V_{3}|S_{k,\,\,1}^{\star}\otimes S_{k,\,\,2}\otimes S_{k,\,\,3}\big{)}$
$\displaystyle\equiv$ $\displaystyle\big{(}V_{1}\otimes V_{2}^{*}\otimes
V_{3}|S_{k,\,\,1}\otimes S_{k,\,\,2}^{\star}\otimes S_{k,\,\,3}\big{)}\,.$

We identify the the second last line of 237 as $(A|A_{S_{k}})$ and the last
line as $(A^{\star}|A^{\star}_{S_{k}})$. Finally as $n=3$ then we have
$2^{3-1}=4$ sub-duality system, which is identified as follow:

$\displaystyle(\mathcal{K}_{1}|\mathcal{K}_{S_{k}})=\big{(}V_{1}\otimes
V_{2}\otimes V_{3}|S_{k,\,\,1}\otimes S_{k,\,\,2}\otimes S_{k,\,\,3}\big{)}$
(240)
$\displaystyle(\mathcal{K}_{1}^{*}|\mathcal{K}^{\star}_{S_{k}^{\star}}\big{)}=\big{(}V_{1}^{*}\otimes
V_{2}^{*}\otimes V_{3}^{*}|S_{k,\,\,1}^{\star}\otimes
S_{k,\,\,2}^{\star}\otimes S_{k,\,\,3}^{\star}\big{)}$ $\displaystyle\vdots$

and similarly for the 2,3 and 4 cases. We can see that,

$\displaystyle(*_{V_{1}}\oplus*_{V_{1}^{*}})(K_{1}\oplus K_{1}^{*})$
$\displaystyle=(K_{2}\oplus K_{2}^{*})$ (241)
$\displaystyle(*_{V_{2}}\oplus*_{V_{2}^{*}})(K_{1}\oplus K_{1}^{*})$
$\displaystyle=(K_{3}\oplus K_{3}^{*})$
$\displaystyle(*_{V_{3}}\oplus*_{V_{3}^{*}})(K_{1}\oplus K_{1}^{*})$
$\displaystyle=(K_{4}\oplus K_{4}^{*})$

thus this is an example demonstration of 227 and 228.

##  3 Construction of the diagramatic basis representation of 4-duality group

In this chapter we study the construction of basis of irreducible
representation of the 4-duality group $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. We
would extensively use the diagramatic representation of the 4-box tableaux,
which is called the 4-fundamental tableaux representation,

Figure 29: 4-box tableaux representation of the 4-duality group

Recalling the definition of the dual space, which consists of two element
vector space of $V$ and $V^{*}$ which are isomorphic to each other 333In the
most general general definition of dual space the isomorphism is not
necessarily imposed.. Here each coloured box, (red (R), blue (B), magenta (M)
and white (W) ) represent the following,

${\rm R}:=V,\quad{\rm{B}}:=V^{*},\quad{\rm{M}}:=V\oplus V^{*},\quad W=0.$
(242)

If one consider dual set then

${\rm R}:=U,\quad{\rm B}:=U^{*},\quad\rm{M}:=U\cup U^{*},\quad\rm{W}=U\cap
U^{*}=\emptyset\,.$ (243)

And in particular, we have $W=M^{*}=*(U\cup U^{*})=U^{*}\cap U=\emptyset$ and
$*\emptyset=*(U^{*}\cap U)=U\cup U^{*}=W$, where we have used de-Morgan’s
theorem. Thus $W$ and $M$ is dual to each other. The full union is sometimes
written as ‘All’ , while the null intersection is sometimes written as ‘Null’
or ‘None’. This can be understood diagramatically by the four colour in the
4-fundamental tableaux representation. The magenta is the mixing of red and
blue, while the white has no overlap between them. The origin, which is
defined as the central zero, can be omitted at the moment. We also define each
coloured box to have unity unit of area, thus the 4-fundamental tableau is a
4-unit object. Next we define the 4 quadrants for the 4-fundamental tableaux
representation. The four quadrants correspond to the 4 boxes, for which each
quadrant is a vector space $Q$. The quadrant number is defined by indexing the
coloured boxes by the following binary number $q_{Q}$:

$\text{W}:(00),\,0;\quad\text{R}:(01),\,1;\quad\text{B}:(10),\,2;\quad\text{M}:(11),3\,.$
(244)

In terms of vectors,

${\rm R}:=v,\quad{\rm{B}}:=v^{*},\quad{\rm{M}}:=v+v^{*},\quad
W=\boldsymbol{0}\,,$ (245)

Suppose $|0\rangle\in V$ and $|1\rangle\in V^{*}$ where $V$ and $V^{*}$ are
one dimensional vector spaces. Now consider the qubit
$|\psi(\theta)\rangle=\cos\theta|0\rangle+\sin\theta|1\rangle$. For
convenience, we consider the labelling of the zero vector,

$0|u\rangle=\boldsymbol{0}_{u}=\boldsymbol{0}\,,\quad\,0|u^{*}\rangle=\boldsymbol{0}_{u^{*}}=\boldsymbol{0}\,.$
(246)

We also take

$*\boldsymbol{0}_{u}=\boldsymbol{0}_{u^{*}}\,,\quad\,*\boldsymbol{0}_{u^{*}}=\boldsymbol{0}_{u}$
(247)

But note that the zero vector itself has no parity, we just introduce it for
convenience, so notice that

$\boldsymbol{0}_{u}=\boldsymbol{0}_{u^{*}}=\boldsymbol{0}\,.$ (248)

Then we have

$|\psi(0)\rangle=|0\rangle+0|1\rangle=|0\rangle+\boldsymbol{0}_{1}\,,$ (249)
$|\psi(\pi/2)\rangle=0|0\rangle+|1\rangle=\boldsymbol{0}_{0}+|1\rangle\,,$
(250)

The fundamental four-box tableaux now reads

Figure 30:

We can see that

$|\psi(\pi/2)\rangle=*|\psi(0)\rangle$ (251)

The Null state and the All state are dual invariant,

$*\boldsymbol{0}=\boldsymbol{0}_{1}+\boldsymbol{0}_{0}=\boldsymbol{0}\quad\text{and}\quad*(|0\rangle+|1\rangle)=|1\rangle+|0\rangle=|0\rangle+|1\rangle\,,$
(252)

while the half states are not. We can further check that states in the dual
partition are orthogonal. For R and B, we have $\langle v|v^{*}\rangle=\langle
0|1\rangle=0$ which satisfies. So clearly they are orthogonal. For M and W, it
is clearly that $\langle\boldsymbol{0}|(|0\rangle+|1\rangle)=0$ so the full
vector and the null vector are orthogonal and dual to each other. Notice that
the role of zero vector is to ensure
$|v\rangle+\boldsymbol{0}=\boldsymbol{0}+|v\rangle=|v\rangle$, this is in
analogy to the role of empty set that $U\cup\emptyset=\emptyset\cup U=U$.

Then in dimensional representation we have

Figure 31:

The dimensional representation is isomorphic to the basis of
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. Thus we have constructed the basis
representation of $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ by the dimensional
representation of the 4-box tableaux.

Now we would like to construct another new representation in terms of direct-
sum vector space.

${\rm
R}:=V\oplus\boldsymbol{0}_{V^{*}},\quad{\rm{B}}:=\boldsymbol{0}_{V}\oplus
V^{*},\quad{\rm{M}}:=V\oplus V^{*},\quad
W=\boldsymbol{0}_{V}\oplus\boldsymbol{0}_{V^{*}}\,,$ (253)

where

$\boldsymbol{0}_{V}=\boldsymbol{0}_{V^{*}}=\boldsymbol{0}=|\boldsymbol{0}\rangle\,.$
(254)

In terms of element,

Figure 32:

The zero vector is the only dual invariant element in the representation, such
that

$*\boldsymbol{0}=\boldsymbol{0}\,.$ (255)

Next we can check that the vectors in the same dual partition are orthogonal.
We need to use the following identity,

$\bigg{\langle}\bigoplus_{i=1}^{n}v_{i}\bigg{|}\bigoplus_{i=1}^{n}u_{i}\bigg{\rangle}=\sum_{i=1}^{n}\langle
v_{i}|u_{i}\rangle\,.$ (256)

Let $|w_{\mathrm{full}}\rangle=|u\rangle\oplus|u^{*}\rangle$,
$|w_{\mathrm{+}}\rangle=|u\rangle\oplus|\boldsymbol{0}_{u^{*}}\rangle$,
$|w_{\mathrm{-}}\rangle=|\boldsymbol{0}_{u}\rangle\oplus|u^{*}\rangle$ and
$|w_{\mathrm{null}}\rangle=\boldsymbol{0}$, we have

$\displaystyle\langle
w_{\mathrm{null}}|w_{\mathrm{full}}\rangle=\langle\boldsymbol{0}|u\oplus
u^{*}\rangle=\langle\boldsymbol{0}_{u}|u\rangle+\langle\boldsymbol{0}_{u^{*}}|u^{*}\rangle=0$
(257) $\displaystyle\langle w_{\mathrm{+}}|w_{\mathrm{-}}\rangle=\langle
u\oplus\boldsymbol{0}_{u^{*}}|\boldsymbol{0}_{u}\oplus u^{*}\rangle=\langle
u|\boldsymbol{0}_{u}\rangle+\langle\boldsymbol{0}_{u^{*}}|u^{*}\rangle=0\,.$

Furthermore, we can check that we can generate dual vector simply by acting
the $*$ on the vector that satisfies orthogonality. The dual construction of
32 is illustrated as follows in 33:

Figure 33:

We can check that,

$\displaystyle\langle w^{*}_{\mathrm{full}}|w_{\mathrm{full}}\rangle=\langle
u^{*}\oplus u|u\oplus u^{*}\rangle=\langle u^{*}|u\rangle+\langle
u|u^{*}\rangle=0\,.$ (258) $\displaystyle\langle
w^{*}_{+}|w_{+}\rangle=\langle
u^{*}\oplus\boldsymbol{0}_{u}|u\oplus\boldsymbol{0}_{u^{*}}\rangle=\langle
u^{*}|u\rangle+\langle\boldsymbol{0}_{u}|\boldsymbol{0}_{u^{*}}\rangle=0\,.$
$\displaystyle\langle
w^{*}_{-}|w_{-}\rangle=\langle\boldsymbol{0}_{u^{*}}\oplus
u|\boldsymbol{0}_{u}\oplus
u^{*}\rangle=\langle\boldsymbol{0}_{u^{*}}|\boldsymbol{0}_{u}\rangle+\langle
u|u^{*}\rangle=0\,.$ $\displaystyle\langle
w^{*}_{\mathrm{null}}|w_{\mathrm{null}}\rangle=\langle\boldsymbol{0}|\boldsymbol{0}\rangle=0\,.$

Next, we would like to construct a larger basis for the
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ group from the 4-fundamental tableaux
representation. We construct another 4 representations from the 4-fundamental
tableaux representation by reflections along the horizontal and vertical axes,
and define the abelian $\mathbb{Z}_{4}$ group (which is also the cyclic group
$C_{4}$) with elements $\\{I,\sigma_{L},\sigma_{D},\sigma_{d}\\}$(where
$\sigma_{d}=\sigma_{L}\sigma_{D}$) over the 4 representations as follow. (Here
$L$ means reflect left/right-wise and $U$ means reflect up/down-wise, $d$
means reflect diagonal-wise. It is illustrated as follow.

Figure 34: Extensive 4-network

Such construction allows us to define all other larger interesting objects.
The joining of the four individual tableau glued by the original $O$ defines
the repeating unity of the 4-duality network, and we suppose the network is
defined infinitely.

Figure 35: Extensive 4-network formed by the 34. Note that the white gaps in
between do not exist but is left for clear demonstration.

Note that the choice of repeating unit is not unique at least locally, but it
has to be a 16-unit square. We can see that the extended version of the
4-fundamental tableau reappears as a 16-unit repeating unit of the network,
shown in the upper left corner of 35.

We can have different sub-diagrams in the network, and the sub-diagram can
have different units of red, blue, magenta and white boxes. We defined the
standard notation as $(\mathrm{C_{1}m_{1}:C_{2}m_{2}:C_{3}m_{3}:C_{4}m_{4}})$,
where $\mathrm{C_{i}}$ refers to the colour and $\mathrm{m_{i}}$ refers to the
number of units possessed by that colour. If $m_{i}=0$ we can choose to omit
the $\mathrm{C_{i}m_{i}}$ term.

Now we need to introduce some formal definitions in a rigorous manner for the
repeating units.

######  Definition 3.1.

In a 4-network, the standard repeating unit is a repeating unit of 16 area-
units which is the extension of the 4-fundamental tableau with no reflectional
symmetry.

######  Definition 3.2.

The regular repeating unit is a 16 area-unit that is formed by the horizontal,
vertical and diagonal reflections of the 4-fundamental tableau by 35 with 4
reflectional symmetries (1 horizontal, 1 vertical and 2 diagonals).

There are 4 possible regular repeating units in total. Starting from the
lower-left one in 35, translation in the horizontal direction by 2 units,
translation in the vertical direction by 2 units, and their composition would
give the remaining 3 repeating units. The total 4 repeating units form the
basis representation of $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ 4-dual group.

######  Definition 3.3.

The diamond representation a shrinked or extended representation with a
rotation by $\pi/4$, which is defined as the dual representation of the square
representation, which defines the deficit colour representation.

The reason for why the diamonds are dual to the square would be apparent in
the moment. It is noted that the diamond representation is not a repeating
unit of the network, ad we require two set of different diamond
representations in order to cover the whole network.

Next we would introduce the concept of level $n$.

######  Definition 3.4.

The level $n\in\mathbb{Z}$ of the box or diamond representation defines the
index of the layers. The $n=0$ is the lowest level and defined as the ground
level in which the box cannot be further split into other colour except for
itself.

It is easy to see that for $n\in\mathbb{Z}^{-}$ being negative, the results
would be the same $n=0$ case, as we just continue to split within a same
coloured box. Therefore we would only have non-trivial results when $n>0$.

The following shows the illustration,

Figure 36: Levels of square or diamond representation

Next we will introduce the concept of observation frame. We will define two
frames, one the interior frame and its dual, exterior frame.

######  Definition 3.5.

Let $R$ be the vector space that the box is situated in, while $S$ be the
vector space that the diamond is situated in, where
$R,S\subset\mathbb{R}^{2}$. The interior frame is defined by the space
enclosed by $R$ or $S$. Denote $A(R)$ and $A(S)$ the area of the box and
diamond respectively defined by $A:R,S\rightarrow\mathbb{R}$. The interior
space $I$ is defined by

$\mathrm{I}=\begin{cases}R&\text{if}\quad A(R)<A(S)\\\ S&\text{if}\quad
A(S)<A(R)\end{cases}\,.$ (259)

The exterior space $E$ is defined by

$\mathrm{E}=\begin{cases}R\setminus S&\text{if}\quad A(R)>A(S)\\\ S\setminus
R&\text{if}\quad A(S)>A(R)\end{cases}\,.$ (260)

Define the interior frame ${\mathrm{I}}=F$ and the exterior frame as its dual
${\mathrm{E}}=F^{*}$. If we have an isomorphism between $R$ and $S$, $R\cong
S$, then $A(R)=A(S)$. if follows that $A(\mathrm{E})=A(\mathrm{I})$. Then we
have $\mathrm{E}\cong\mathrm{I}$. We say $\mathrm{E}$ and $\mathrm{I}$ is dual
invariant $F=F^{*}$, writing $F\equiv F^{*}$, in which the condition is the
same area of both representations.

We would also call the interior frame normal (existing) frame and the exterior
frame the null frame. The reason for this terminology is because, the interior
is considered as an ownership while the exterior is considered as things
belonging to ”outside”.

Finally we have to define the translational operation on the network.

######  Definition 3.6.

Let $k\in\mathbb{Z}$ be the units to be translated across the 4-network.
Define the map $\rho^{(k)}$ as the translation of $k$ units along the
horizontal direction $LR$ or vertical direction $UD$, with $\rho^{(k)}_{LR}$
and $\rho^{(k)}_{UD}$. The $\rho^{(k)}$ is independent of layer $n$, in which
$\rho^{(k)}=\rho^{(k)}_{n}$ for all $n\in\mathbb{Z}$.

It is easy to see that the map has a periodicity of $4$,

$\rho^{(k)}=\rho^{(k+4)}\,.$ (261)

######  Definition 3.7.

The map of translation $\rho^{k}$ for $k=0,1,2,3$ forms an abelian group which
is isomorphic to the cyclic group $C_{4}$ (or $\mathbb{Z}_{4}$), explicitly
$C_{4}=\\{I,\rho^{1},\rho^{2},\rho^{3}\\}$ with $I=\rho^{0}$ the identity
element.

We have naturally three of these groups $C_{4\,LR}$ , $C_{4\,UD}$ and
$C_{4\,d}$ for the horizontal, vertical and diagonal translations
respectively.

Now let’s first study the ground level $n=0$. It can be illustrated by the
following diagram.

Figure 37: The study of ground level. The quandrant number is labelled by the
binary number.

For the $n=0$ case, both the box and diamond representations share the same
quadrant number. It is noted that in general for $n>0$, the quadrant number
will not be invariant for both representations as we will see. We can see that
the only difference between the box the diamond representation is the number
of coloured area units it enclose, for which is halved for the diamond case.

The most important feature for the $n=0$ ground level is that it is interior
and exterior dual invariant by definition 2.0.5 . It is a very special
property for the ground level. It is easy to see that cases for $n<0$ is also
dual invariant. Thus for all $n\leq 0$, the box representation and the diamond
representation is dual invariant under the condition of same area.

Thus in summary, the $\mathrm{E}-\mathrm{I}$ dual invariant for the $n=0$
ground level case, which means $F^{*}\equiv F^{*\star}$ implies

  1. 1.

No further colour splitting possible

  2. 2.

Unchanging quadrant number

  3. 3.

Same area $A(\mathrm{E)}=A{\mathrm{(}I)}$

Also, we can define weak dual invariant if only some items of the above is
satisfied. For our $n=0$ case, $F\equiv F^{*}$ is a weak dual invariant as it
just satisfies two items, as we have a change from $\mathrm{C_{i}}4$ to
$\mathrm{C_{i}}2$ when we go from box representation to diamond
representation.

Now for $n=1$ case, things become much more complicated. First remember there
are two levels of duality. First, it is the duality between the box
representation and diamond representation. Next it the duality for the
interior and exterior frame. There are sub-duality inside a duality structure.
In both cases they are not dual invariant. For $n=1$ case, the concept of
lacking colour units for the dual diamond representation is clearly
demonstrated in figure 38.

Figure 38: The study of first level

Note that all of them are basis of irreducible representation of
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. Let’s first consider the general
differences between the box representation and diamond representation. In the
box case, all bases have the same number or red, blue, magenta and white
boxes, which are 4; while in the diamond case, all bases have different
coloured units. However for each of them we can figure out one particular
colour is missing. Thus this is the reason we call it the dual representation.
(Note that in the $F^{*\star}$ the white diamond square do not count as we
only contribute the exterior part by definition. ) Due to the obvious
difference from the box representation, the quadrant indexes have to be
relabelled. Using 244, the new quadrant index in $F^{*}$ frame is given by

$q_{Q}^{*}=({\mathrm{max}}\,q_{Q})-Q=3-q_{Q}\,.$ (262)

For the ease of comparison, we define internal folding for the diagramatic
basis for $F^{*\star}$ basis. The internal folding is defined by joining the
four corners to the center. The result is shown in 39.

Figure 39:  The study of first level with internal folding for $F^{*\star}$
frame and the comparison to $F^{*}$.

Thus we can see that they are apparently different, not just in position for
the colour of the square but the colour in the triangles have swapped.

It is remarked that these colour representations are just denoting the dual
set or dual space, in the end the would translate back the language of colour
back to the elements of dual set or dual space. Let’s define the lacking of an
element in dual space by the action ‘!’ 444Not to confuse with the use of
negation $NOT$ in programming languages.

Next we introduce the concept of perspective. We can have two perspectives,
the perspective of presence and the perspective of absence. For example !$W$
is in the absence perspective, while R,B,M is in the presence perspective. The
full analysis is shown below.

Figure 40:

The concept of lacking colour denoted by the action ‘!’ is in particular
important and requires detailed investigation. Let’s give the formal
definition.

######  Definition 3.8.

Let $(\,\,\,\,|\,\,\,)$ be the formal notation for the description of elements
under perspectives or frames, where the left bracket $(\,\,\,\,|$ holds the
basis element set $\xi$ that are of interest and the right bracket
$|\,\,\,\,)$ holds the frame or perspective $P$, explicitly $(\xi|P)$. Define
the absence perspective as $0$ and the presence perspective as $1$.

The frame and perspective themselves forms a natural basis of irreducible
representation of 4-duality group $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$.

Figure 41:

######  Definition 3.9.

Define in a particular quadrant $Q$ that, the set of elements of lacking $E\in
Q$ in the absence perspective $(E|0)$, where
$E=\\{\text{!}E_{1},\text{!}E_{2},\cdots,\text{!}E_{N}\\}$; and the set of
elements of presence $e\in Q$ in the presence perspective $(e|1)$ where
$e=\\{e_{1},e_{2},\cdots,e_{N}\\}$. $E_{i}$ and $e_{j}$ are elements in dual
set or dual space, which are set or vector space, and
$\mathrm{max}\,M,\mathrm{max}\,N=4$. If there exists more than one $e_{i}$
such that $e_{i}\cup e_{j}=E_{k}$ ; or if $E_{k}\subset e_{i}$ and $E_{k}$ is
not a subset of all elements in $e$ , then !$E_{k}$ is pseudo-lacking of
$E_{k}$. Otherwise, !$E_{k}$ is real-lacking.

The concept of pseudo-lacking is introduced because it means it is not really
totally lacking. Although the particular color is lacking in the absence
perspective, it can be formed or hidden in the elements in the presence
perspective. The negation of the two if statements would be real-lacking, as
the particular colour cannot be joined by some other elements in the presence
perspective, nor it is contained in those elements.

######  Definition 3.10.

For a basis of $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ represented by the dual
diamond lacking representation, the 4-lacking is subdivided into one real-
lacking basis and three pseudo-lacking bases and , denoted by $4=1\oplus 3$.

This can be easily checked by using the case for existing frame.

  * •

For !All under $|F^{*},0)$, we have
$({\mathrm{\text{!}All}}|F^{*},0)\equiv(0,U,U^{*}|F^{*},1)$. Yet
$\text{All}=U\cup U^{*}$ thus we can form All in the presence perspective.
Hence this is a pseudo-lacking.

  * •

For !$V^{*}$ under $|F^{*},0)$, we have
$({\text{!}}V^{*}|F^{*},0)\equiv(\text{All},0,V|F^{*},0)$. Yet
$V^{*}\subset\text{All}$, thus $V^{*}$ is hidden in the presence perspective.
This is a pseudo-lacking.

  * •

For !$V$ under $|F^{*},0)$, we have
$({\mathrm{\text{!}V}}|F^{*},0)\equiv(V^{*},\text{All},0|F^{*},0)$. Yet
$V\subset\text{All}$, thus $V$ is hidden in the presence perspective. This is
a pseudo-lacking.

  * •

For $0$,under $|F^{*},0)$, we have
$(0|F^{*},0)\equiv(V,V^{*},\text{All}|F^{*},0)$. These spaces do not contain
$0$, but $0$ is a subset or subspace in all elements $V,V^{*},\text{All}$.
This is a real-lacking.

Thus in the diamond representation, we can diagramatically represent as

Figure 42:

The white box represents the real-lacking basis while the three dark-grey
boxes represent the three pseudo-lacking bases, thus this represents the
structure of $4=1\oplus 3$.

Note that this structure is the same as the basis of irreducible
representations of SO(4), which is an isomorphic to SU(2)$\times$SU(2).
Therefore the diamond representation, which is a dual representation of the
box representation, can be used to represent the non-abelian group SO(4).

##  4 Representation theory for multi-duality

Let the basis of $V$ be $|0\rangle$ and the dual basis of $V^{*}$ be
$|1\rangle$. The tensor product of vector spaces above has the tensor product
states as the basis. For example, $V^{*}\otimes V\otimes V$ has the basis
$|100\rangle$. In this section, we will study multi-dual symmetry through
groups and representations. The dual symmetry here we can naturally refer to
the parity symmetry $\mathbb{Z}_{2}$. The state $|0\rangle$ and $|1\rangle$
form the basis of irreducible representation of the duality (parity) group
$\mathbb{Z}_{2}$.

Next we will use two basic theorems from group theory. For two groups $G_{1}$
and $G_{2}$ with order $|G_{1}|$ and $|G_{2}|$ respectively, the direct
product group $G_{1}\times G_{2}$ has group order $|G_{1}\times
G_{2}|=|G_{1}||G_{2}|$. And if $G_{1}$ and $G_{2}$ are abelian, then
$G_{1}\times G_{2}$ is also abelian. We can apply these two theorems to our
parity group $\mathbb{Z}_{2}$.

######  Definition 4.1.

Let
$\mathbb{Z}_{2}^{N}=\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}$
with $N$ be the multiple parity group for the $n$-level duality symmetry
group, where the group order
$|\mathbb{Z}_{2}^{N}|=|\mathbb{Z}_{2}|^{n}=2^{n}$.

Note that for the multi direct product for $\mathbb{Z}_{2}$, it is isomorphic
to the multiple tensor product of the $\mathbb{Z}_{2}$, then we can write

$\mathbb{Z}_{2}^{N}=\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}\cong\mathbb{Z}_{2}\otimes\mathbb{Z}_{2}\otimes\cdots\otimes\mathbb{Z}_{2}\,.$
(263)

Using the group homomorphism, we have

$D(\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2})=D(\mathbb{Z}_{2}\otimes\mathbb{Z}_{2}\otimes\cdots\otimes\mathbb{Z}_{2})=D(\mathbb{Z}_{2})\otimes
D(\mathbb{Z}_{2})\otimes\cdots\otimes D(\mathbb{Z}_{2})\,.$ (264)

Next we will use the following property of abelian groups. For an abelian
group $G$, since each group element form the conjugacy class of itself, the
number of classes $N_{c}$ is just the group order $|G|$. Then for our case
$N_{c}$ is just $2^{n}$. And by group theory the number of classes is equal to
the number of irreducible representations $N_{\tau}$, thus the number
$N_{\tau}=2^{n}$ for our case. Let $d_{i}$ be the dimension of the irreducible
representation. Then we will use the theorem of group that

$\sum_{i=1}^{N_{c}}d_{i}^{2}=|G|\,.$ (265)

But since $N_{c}=|G|=2^{n}$ for our case, then this forces the dimension of
each irreducible representation as 1. Therefore, the multiple-duality group of
dimension $2^{n}$ can be decomposed into the direct sum of one-dimensional
irreducible representations. Let
$\mathrm{g}\in\mathbb{Z}_{2}\otimes\mathbb{Z}_{2}\otimes\cdots\otimes\mathbb{Z}_{2}$
, and $D(\mathrm{g})$ be the matrix representation of the group 555We will use
the _non-italic_ $\mathrm{g}$ for the group element of $Z_{2}\otimes
Z_{2}\otimes\cdots\otimes Z_{2}$, while the _italic_ $g$ for group elements in
each $\mathbb{Z}_{2}$.

We will find the representation of the multiple duality group
$\mathbb{Z}_{2}^{N}$ of general $n\geq 2$ levels. Here we will use the
definition by the tensor product
$\mathbb{Z}_{2}\otimes\mathbb{Z}_{2}\otimes\cdots\otimes\mathbb{Z}_{2}$.

First consider the simplest case, which is the 0-th level with $N=1$, this is
just the parity group $\mathbb{Z}_{2}$. The $\mathbb{Z}_{2}$ group only has
two elements $\\{I,P\\}$ and has two classes. Therefore the representation is
reducible to the direct sum of two 1D irreducible representation. Let
$g\in\mathbb{Z}_{2}$, we have

$D(g)=\mathcal{A}_{1}(g)\oplus\mathcal{A}_{2}(g)\,.$ (266)

The $\mathcal{A}_{1}(g)$ is the trivial irreducible representation, with all
characters equal to 1 for all group elements. And we have
$\mathcal{A}_{2}(I)=1$ and $\mathcal{A}_{2}(P)=-1$. This can be easily checked
by the orthogonality theorem in group theory. The basis of reducible
representation is $|0\rangle$ and $|1\rangle$, which can be written as a basis
doublet,

$|v\rangle=\begin{pmatrix}|0\rangle\\\ |1\rangle\end{pmatrix}\,.$ (267)

For any general $n$-levels, the multiple duality group follows the general
duality theorem.

######  Definition 4.2.

Let $|V\rangle$ be the basis of the multiple duality group
$\mathbb{Z}_{2}\otimes\mathbb{Z}_{2}\otimes\cdots\otimes\mathbb{Z}_{2}$,
mathematically

$|V\rangle=|v_{1}v_{2}\cdots
v_{N}\rangle=|v_{1}\rangle\otimes|v_{2}\rangle\otimes\cdots\otimes|v_{N}\rangle\,.$
(268)

The basis is of $N=2^{n}$ dimension, the whole set of

$|\eta_{1}\eta_{2}\cdots\eta_{N}\rangle=|\eta_{1}\rangle\otimes|\eta_{2}\rangle\otimes\cdots\otimes|\eta_{N}\rangle\quad\text{for
all}\,\,\eta_{j}=0,1$ (269)

form the basis of irreducible representation of the multiple duality group.
Hence the representation in $n$-levels is the natural basis of the multiple
duality group.

######  Definition 4.3.

The basis $|V\rangle$ transform under the tensor product representation of the
parity group $\mathbb{Z}_{2}$. Let $g_{i_{j}}$ be the element of the the
$j^{\mathrm{th}}$ parity group, then we have

$|v_{1}\rangle^{\prime}\otimes|v_{2}^{\prime}\rangle\otimes\cdots\otimes|v_{N}^{\prime}\rangle=\Big{(}D(g_{i_{1}})\otimes
D(g_{i_{2}})\otimes\cdots\otimes
D(g_{i_{N}})\Big{)}\,(|v_{1}\rangle\otimes|v_{2}\rangle\otimes\cdots\otimes|v_{N}\rangle)\,.$
(270)

It can be written as

$|V^{\prime}\rangle=D(\mathrm{g})|V\rangle$ (271)

where

$D(\mathrm{g})=D(g_{i_{1}})\otimes D(g_{i_{2}})\otimes\cdots\otimes
D(g_{i_{N}})\,.$ (272)

######  Definition 4.4.

The multiple duality group
$\mathbb{Z}_{2}\otimes\mathbb{Z}_{2}\otimes\cdots\otimes\mathbb{Z}_{2}$ of
order $N=2^{n}$ can be decomposed to the direct sum of 1D irreducible
representations with each of them having the multiplicity as 1,

$D(\mathrm{g}\in\mathbb{Z}_{2}\otimes\mathbb{Z}_{2}\otimes\cdots\otimes\mathbb{Z}_{2})=A_{1}(\mathrm{g})\oplus
A_{2}(\mathrm{g})\oplus\cdots\oplus
A_{2^{n}}(\mathrm{g})=\bigoplus_{i=1}^{N_{\tau}=2^{n}}a_{i}\mathcal{A}_{i}(\mathrm{g})\,.$
(273)

for multiplicity $a_{1}=a_{2}=\cdots a_{2^{n}}=1$.

The proof of the above theorems are as follow,

$\displaystyle\quad|v_{1}\rangle^{\prime}\otimes|v_{2}^{\prime}\rangle\otimes\cdots\otimes|v_{N}^{\prime}\rangle$
(274)
$\displaystyle=\Big{(}D(g_{i_{1}})\,|v_{1}\rangle\Big{)}\otimes\Big{(}D(g_{i_{2}})\,|v_{1}\rangle\Big{)}\otimes\cdots\otimes\Big{(}D(g_{i_{N}})\,|v_{N}\rangle\Big{)}$
$\displaystyle=\Big{(}\,D(g_{i_{1}})\otimes D(g_{i_{2}})\otimes\cdots\otimes
D(g_{i_{N}})\,\Big{)}\,(|v_{1}\rangle\otimes|v_{2}\rangle\otimes\cdots\otimes|v_{N}\rangle)$
$\displaystyle=\Big{[}\Big{(}\,\mathcal{A}_{1}(g_{i_{1}})\oplus\mathcal{A}_{2}(g_{i_{1}})\Big{)}\otimes\Big{(}\mathcal{A}_{1}(g_{i_{2}})\oplus\mathcal{A}_{2}(g_{i_{2}})\Big{)}\otimes\cdots\otimes\Big{(}\mathcal{A}_{1}(g_{i_{N}})\oplus\mathcal{A}_{N}(g_{i_{N}})\Big{)}\,\Big{]}\,(|v_{1}\rangle\otimes\cdots\otimes|v_{N}\rangle)$
$\displaystyle=\bigg{(}\bigoplus_{i_{1},i_{2},\cdots,i_{N}=1,2}\mathcal{A}_{i_{1}}(g_{i_{1}})\,\mathcal{A}_{i_{2}}(g_{i_{2}})\cdots\mathcal{A}_{i_{N}}(g_{i_{N}})\bigg{)}\,(|v_{1}\rangle\otimes|v_{2}\rangle\otimes\cdots\otimes|v_{N}\rangle)\,.$

From the second line to the third line we have used the identity of tensor
product $(A\otimes B)(u\otimes v)=Au\otimes Bv$. In the forth line we have
used (266) for each $D(g_{i_{j}})$. From the forth line to the fifth line,
since all the $A_{i_{j}}$ must be in one dimension, therefore we can apply the
distribution rule. Note that the distribution rule for tensor product cannot