# Time-dependent finite-dimensional dynamical system representation of
breather solutions

Yoritaka Iwata<EMAIL_ADDRESS>Faculty of International
Studies, Osaka University of Economics and Law  Yasuhiro Takei  Mizuho
Research & Technologies

###### Abstract

A concept of finite-dimensional dynamical system representation is introduced.
Since the solution trajectory of partial differential equations are usually
represented within infinite-dimensional dynamical systems, the proposed
finite-dimensional representation provides decomposed snapshots of time
evolution. Here we focus on analyzing the breather solutions of nonlinear
Klein-Gordon equations, and such a solution is shown to form a geometrical
object within finite-dimensional dynamical systems. In this paper, based on
high-precision numerical scheme, we represent the breather solutions of the
nonlinear Klein-Gordon equation as the time evolving trajectory on a finite-
dimensional dynamical system. Consequently, with respect to the evolution of
finite-dimensional dynamical systems, we confirm that the rotational motion
around multiple fixed points plays a role in realizing the breather solutions.
Also, such a specific feature of breather solution provides us to understand
mathematical mechanism of realizing the coexistence of positive and negative
parts in nonlinear systems.

##  I Introduction

Due to the underlying nonlinearity, natural events possibly take place in a
complex way. For example, pattern formation, which can be found in physics,
chemistry, biology, and materials science, typically arises from the
nonlinearity. Also nonlinear wave propagation such as shock waves and solitons
are observed in different scales [1, 2, 3]. In order to understand these
nonlinear phenomena quantitatively, we introduce a mathematical tool to
represent and analyse the time evolution driven by nonlinearity.

In this paper, we propose a method to represent a point (the state of the
system at a certain time) on a solution trajectory inside the infinite-
dimensional dynamical system as a geometrical object in finite-dimensional
dynamical system. We call this method "$t$-dependent finite-dimensional
representation" of originally infinite-dimensional dynamical system (for a
textbook of finite-dimensional dynamical systems see [4], and for a textbook
of infinite-dimensional dynamical systems see [5]). Here note that solution
trajectories of partial differential equations are usually represented within
infinite-dimensional dynamical systems. The method is then applied to clarify
the characteristics of breather mode of the nonlinear wave described by
nonlinear Klein-Gordon equations. As a result, the trajectory of breather
solution provides a coexistence of positive and negative parts, which are
geometrically represented in finite-dimensional dynamical systems by
coexistence of rotational motions around multiple fixed points.

##  II Finite-dimensional representation of infinite-dimensional dynamical
systems

###  II.1 Infinite-dimensional dynamical system

A couple of space and semigroup $(X,S(t))$ is called dynamical system. Here we
limit ourselves to the solvable cases with a well-defined semigroup $S(t)$ on
a Banach space $X$, and concentrate on introducing a tool analyzing nonlinear
dynamics in a geometrical manner. One parameter semigroup of operator
satisfies the semigroup property

$\left\\{\begin{array}[]{ll}S(t+s)=S(t)~{}S(s),\quad t,s\geq 0,\vspace{3mm}\\\
S(0)=I.\end{array}\right.$

We consider a Cauchy problem of nonlinear wave equations. It is formulated as
abstract evolution equations of hyperbolic type

$\begin{array}[]{ll}dU(t)/dt+\left(\begin{array}[]{ll}~{}0\qquad-I\\\
-\partial_{x}^{2}\qquad 0\end{array}\right)U(t)=\left(\begin{array}[]{cc}0\\\
-\mu^{2}u(t)+k(u(t))\end{array}\right),\vspace{5mm}\\\
U(0)=U_{0}=(u_{0},v_{0})^{T},\end{array}$ (1)

in an infinite-dimensional Banach space $X$, where the unknown function $U(t)$
is represented by a vector form $U(t)=(u(t),v(t))^{T}$ with a relation
$v(t)=\partial_{t}u(t)$. For a real positive number $\mu$ standing for a mass,
the non-autonomous term $F(U(t))$ consists of linear part
$(0,-\mu^{2}u(t))^{T}$ and the nonlinear part $(0,k(u(t)))^{T}$, where
$k(u(t))$ is, for example, a polynomial of $u(t)$ and/or a trigonometric
function of $u(t)$. Let the space $X$ consist of square-integrable periodic
functions being defined on one-dimensional interval; $X=L_{\rm
per}^{2}(0,L)\times L_{\rm per}^{2}(0,L)$, and semigroup $S(t):~{}U(s)\mapsto
U(t+s)$ with $t,s\geq 0$ defined on $X$. If it comes to partial differential
equations, the dynamical system $(X,S(t))$ becomes infinite-dimensional, as
$X=L_{\rm per}^{2}(0,L)\times L_{\rm per}^{2}(0,L)$ is infinite-dimensional.

###  II.2 Finite-dimensional dynamical system

The finite-dimensional representation is introduced with a focus on the
spatial variables. Such a viewpoint naturally appears, if we limit ourselves
to the stationary problem of originally non-stationary problem. The stationary
problem of (1) is given by

$\begin{array}[]{ll}\begin{array}[]{ll}\left(\begin{array}[]{ll}~{}0\qquad-I\\\
-\partial_{x}^{2}\qquad
0\end{array}\right)\end{array}U(t)=\left(\begin{array}[]{cc}0\\\
-\mu^{2}u(t)+k(u(t))\end{array}\right)\end{array}$ (2)

in $X$ at a certain fixed $t$, which is equivalent to

$\left\\{\begin{array}[]{ll}\partial_{x}^{2}u(t,x)=\mu^{2}u(t,x)-k(u(t,x)),\vspace{3mm}\\\
u(t,0)=u(t,L),\quad\partial_{x}u(t,0)=\partial_{x}u(t,L),\end{array}\right.$
(3)

in $(0,L)$. Equation (3) is an ordinary differential equation holding a
finite-dimensional dynamical system whose evolution direction is intentionally
taken to be the spatial direction. Here it is practical to introduce two
parameter semigroup

$s(x_{2},x_{1}):~{}u(x_{1})~{}\mapsto~{}u(x_{2}),$

on Euclidean space ${\mathbb{R}}^{2}$. Two-parameter semigroup
$s(x_{2},x_{1})$ satisfies the semigroup property

$\left\\{\begin{array}[]{ll}s(x_{3},x_{1})=s(x_{3},x_{2})~{}s(x_{2},x_{1}),\quad
0\leq x_{1}\leq x_{2}\leq x_{3}\leq L,\vspace{3mm}\\\
s(x,x)=I.\end{array}\right.$

Since the periodic boundary condition is assumed to be satisfied in the
present settings, the finite-dimensional dynamical system shows a closed loop,
where note that “closed" loop appears due to the periodic boundary condition.
The loop is not necessarily a simple closed curve, and some crossings possibly
appear. According to the possible appearance of crossing points, here we
introduce a two-parameter semigroup $s(x_{2},x_{1})$, in which the evolution
can be different depending on both the starting point $x_{1}$ and ending point
$x_{2}$.

The finite-dimensional representation in this paper is obtained based on the
stationary problem; $\partial_{t}u=0$, so that it is regarded as a snapshot at
a certain time. That is, the solution at a certain time is shown in the $u-v$
plane (i.e., phase space). If those snapshots are connected with respect to
the variable $t$, then a $t$-dependent moving loop is obtained. It is called
here the time-dependent finite-dimensional representation. Although the
finite-dimensional representation is proposed here particularly for nonlinear
hyperbolic evolution equations, it can be applied to any partial differential
equations including fluid dynamics.

##  III Visualization of finite-dimensional dynamical systems

High-precision numerical scheme is indispensable in the numerical treatment of
nonlinear problems. In order to obtain the reliable time-evolving geometric
shapes, we employ high-precision numerical scheme satisfying the hyperbolic
type conservation laws; for benchmark tests, see [7, 9, 8]. In the high-
precision numerical scheme, the spatial discretization is made by the Fourier
spectral method (for example, see [10, 11, 12]), and the time-discretization
by the implicit Runge-Kutta method (for example, see [13, 14, 15, 16]).

Figure 1: Breather solution, $(A=0.04)$ : In each panel, the solid line shows
the function $u$ at each time $t$, and the dotted line shows the function $u$
at time $(t-16)$. A part $u(t,x)$ with $x\leq 4$ is confined in the positive
region ($u(t,x)>0$) and a part $u(t,x)$ with $x>4$ is in the negative region
($u(t,x)<0$).

###  III.1 Nonlinear Klein-Gordon equations with cubic nonlinearity

Let $x\in[0,L]$ be a finite domain of space. The time evolution problem is
considered for $t\geq 0$. We consider the nonlinear Klein-Gordon equation with
the double-well type interaction.

$\begin{array}[]{ll}\ \partial_{t}^{2}u+\alpha\partial_{x}^{2}u+(\beta
u^{2}-\mu)u=0,\vspace{3mm}\\\ \ u(0,x)=A\sin(\pi x/4),\quad
u(t,0)=u(t,L),\vspace{3mm}\\\ \ \partial_{t}u(0,x)=0,\
\partial_{t}u(t,0)=\partial_{t}u(t,L).\end{array}\qquad{\rm(KG)}$

where $\alpha=2^{-8}$, $\beta=1$, $\mu=0.00305$, and $A>0$ are real constants.
This equation is known to hold the symmetry breaking being known as the Higgs
mechanism; $\phi^{4}$-theory in the context of quantum field theory (for a
textbook, see [17]).

###  III.2 Ordinary oscillatory solution and the breather solution

Two different solutions (sometimes a terminology “modes" is used for
“solutions") appears for the Cauchy problem (KG) only by changing an amplitude
$A$ of the initial function. One is the ordinary oscillatory mode and the
other is the breather mode, where it is practical to imagine harmonic
oscillation for the ordinary oscillatory solution. The ordinary oscillatory
mode and the breather mode are the periodic solutions for both $t$ and $x$, so
that they are a kind of doubly periodic and therefore closed curves as for
both space and time.

The breather solution can be distinguished from an ordinary oscillatory
solution by the appearance of certain kinds of collectivity, where activated
modes result in the resonance. For the collectivity, the breather solution
includes the resonating large amplitude oscillation, which is localized only
in the positive or negative side of $u=0$ (Fig. 1). Such a localization is not
seen in the ordinary oscillatory solution. The breather mode is achieved by
the collective resonance of many activated modes, as such curved cannot be
made of a sum of few sine or cosine curves.

Figure 2: The finite-dimensional representation for a breather solution
$(A=0.04)$ : The solid line represents the finite-dimensional representation
for the breather solution at each time $t$, represented by a curve on the
$u-v$ plane. The points on the graph indicate the trajectories of the
endpoints of the curve up to time $t$. The left and right halves of the curve
evolve in time to rotational motion with the fixed point
$(\pm(\mu/\beta)^{1/2},0)$ as the focus, respectively.

Figure 2 shows a $t$-dependent finite-dimensional representation of the
breather solution. At $t=0$, the solution is represented as a line segment
along the $u$-axis on the $u-v$ plane, as can be seen from the initial value
setting. For the ordinary oscillatory solution, the curve rotates around the
origin $(0,0)$. Unlike the ordinary oscillatory solution, the curve does not
rotate around the origin $(0,0)$ in case of breather solution, and
furthermore, the length of segments becomes shorter and larger. Instead
additional two fixed point $(\pm(\mu/\beta)^{1/2},0)$ play roles.

##  IV Summary

We propose $t$-dependent finite-dimensional representation of originally
infinite-dimensional dynamical systems. The proposed method is applied to a
breather solution for nonlinear Klein-Gordon evolution equation with cubic
nonlinearity. By comparing Figs. 1 and 2, more details of breather solution
can be seen by having a finite-dimensional representation (Fig. 2). Although
ordinary oscillatory solution is a rotation centered at the origin $(0,0)$ on
the $u-v$ plane, the breather solution is more complex. Indeed, for the
breather solution, two local rotational motions coexisting at the fixed point
$(\pm(\mu/\beta)^{1/2},0)$.

## References

  * [1] M. C. Cross, P. C. Hohenberg, “Pattern formation outside of equilibrium”, Rev. Mod. Phys. 65, 851, 1993.
  * [2] H. Nabika, Ma. Itatani, I. Lagzi, “Pattern Formation in Precipitation Reactions: The Liesegang Phenomenon”, Langmuir, 36, 2, 481-497, 2020.
  * [3] P. Ball, “Forging patterns and making waves from biology to geology: a commentary on Turing (1952) ‘The chemical basis of morphogenesis”’, Phil. Trans. R. Soc. B 370, 2015: 20140218.
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  * [6] Y. Takei, Y. Iwata, "Space-time breather solution for nonlinear Klein-Gordon equations", J. Phys.: Conf. Ser. 1730, 2021.
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  * [16] G. A. Staff, K. A. Mardal and T. K. Nilssen, “Preconditioning of fully implicit Runge-Kutta schemes for parabolic PDEs", Modeling, Identification and Control, 27, pp.109-123, 2006.
  * [17] J. D. Bjorken and S. D. Drell, “Relativistic quantum fields”, MacGraw-Hill, 1965.


# A Direct Proof of the 2nd Atiyah-Sutcliffe Conjecture for Convex
Quadrilaterals

Mazen Bou Khuzam
Dept. of Mathematics and Natural Sciences
The American University of Iraq at Sulaimani
Email<EMAIL_ADDRESS>
###### Abstract

We present a direct proof of the second conjecture made by M. Atiyah and P.
Sutcliffe for the case of convex quadrilaterals. Unlike previous work on this
conjecture, our proof does not require any computer aided computations. The
new proof relies on a new geometric inequality proved recently by the author.

$\bf{Keywords}$: Atiyah determinant ; Atiyah-Sutcliffe conjectures

##  1 Introduction

While studying the spin statistics theorem using classical quantum theory, M.
V. Berry and J. M. Robbins [3] came across the following purely geometric
question in $\mathbb{R}^{3}$ : Does there exist a continuous function from the
set of $n$ distinct points in $\mathbb{R}^{3}$ to the flag manifold
$U(n)/T^{n}$ which is compatible with the action of the symmetric group? In
his attempt to answering their question, M. Atiyah gave a very natural and
elegant solution whose validity depends on the nonvanishing of a certain
determinant function. Although another construction was given by Atiyah in [1]
answering the question of Berry and Robbins in the positive, the nonvanishing
of the determinant given in the original construction remained an unresolved
conjecture. Numerical evidence of the validity of the conjecture were provided
by Atiyah and Sutcliffe in their paper [7] where they added two new
conjectures which imply the first one. All three conjectures have remained
open since then except for a few verifications on special configurations of
points [4] , [5], [6].

The construction of the determinant begins with $n$ distinct points in
$\mathbb{R}^{3}$. By choosing a prefered axis in $\mathbb{R}^{3}$ we identify
it with $\mathbb{R}\times\mathbb{C}$ and denote our $n$ points by
$P_{j}=(a_{j},z_{j})$ for $j=1,...,n$. We consider the vector
$V_{jk}=(a_{k}-a_{j},z_{k}-z_{j})$ to be the vector from $P_{j}$ to $P_{k}$
and use the Hopf map $h:\mathbb{C}^{2}\rightarrow\mathbb{R}^{3}$ given by
$h(z,w)=((|z|^{2}-|w|^{2})/2,z\overline{w})$ to lift $V_{jk}$ from
$\mathbb{R}^{3}$ to $\mathbb{C}^{2}$. If we denote also $a_{k}-a_{j}$ by
$a_{jk}$, $z_{k}-z_{j}$ by $z_{jk}$ and set
$d_{jk}=a_{jk}+\sqrt{a_{jk}^{2}+|z_{jk}|^{2}}$ we see that the lift of
$V_{jk}$ is given by

$h^{-1}(V_{jk})=e^{i\theta}\frac{1}{\sqrt{d_{jk}}}(d_{jk},\overline{z}_{jk})\hskip
42.67912pt;\,\,\theta\in\mathbb{R},\,i=\sqrt{-1}$

Since $h^{-1}(V_{jk})$ does not uniquely depend on $V_{jk}$ we follow the
normalization of Atiyah by choosing $(-\overline{w},\overline{z})$ as a lift
(under $h$) for $V_{kj}$ whenever $(z,w)$ is a lift for $V_{jk}$ (see [2]). We
let $C_{k}$ be the symmetric tensor product of $h^{-1}(V_{kj})$ for all $j\neq
k$. The determinant of the $n\times n$ matrix whose $k$th column is $C_{k}$
will be called the Atiyah determinant and will be denoted by
$At(P_{1},...,P_{n})$ or just $At$ if the points are already specified. The
determinant function $At$ is invariant under rotations and translations and
gets conjugated under plane reflections. The first conjecture (C1) of Atiyah
is $At\neq 0$ while the second Atiyah-Sutcliffe conjecture (C2) is
$\displaystyle{At\geq\prod_{j<k}(2||\overrightarrow{P_{j}P_{k}}||)}$, and the
third one (C3) is

$At^{n-2}(P_{1},...,P_{n})\geq\prod_{j=1}^{n}At(P_{1},...,P_{j-1},P_{j+1},P_{n})$

for all distinct points $P_{1}$, …, $P_{n}$ of $\mathbb{R}^{3}$. It is clear
from the statement of these conjectures that $(C3)\implies(C2)\implies(C1)$.

Most of the methods applied in previous work on this determinant considered
special cases of configurations of points (see for example [4], [6]) or used
some sort of computer aid in their computations. The main work on the general
four-point case was done by M. Eastwood and P. Norbury in [5] where they
proved the first conjecture of Atiyah to be true for $n=4$ points in space.
Others followed their formula using Maple (see [8]) and proved the three
conjectures with computer aid. Our paper provides the first direct proof of
the second Atiyah-Sutcliffe conjecture for four points forming a convex
quadrilateral. The proof we present here does not rely on any computer
calculations, but rather on a geometric inequality proved by the author in a
previous paper [9]. We hope that this new approach will pave the way for
generalizations that may solve the full conjecture.

##  2 The Planar Case

In this section, we explore the Atiyah determinant for four planar points and
present it in standard form. First note that $At$ is a real number when all
points lie in a plane. This is because a reflection in their plane leaves the
points fixed since $\overline{At}=At$. This was already mentioned in [2] and
will also be seen here concretely from the expansion of its standard form.
Moreover, after applying a solid motion, we may assume our points are in
$\\{0\\}\times\mathbb{C}$ so that all $a_{jk}$ values become zeros and our
points can be written as $(0,z_{1})$, …, $(0,z_{4})$. Based on that, the
construction of the Atiyah determinant is done as follows:

When the first point is considered as an observer of the three other points we
obtain $(0,z_{12})$, $(0,z_{13})$, $(0,z_{14})$ and these lift under the Hopf
map $h$ to

$\frac{1}{\sqrt{r_{12}}}(r_{12},\overline{z}_{12}),\,\,\,\,\,\,\,\frac{1}{\sqrt{r_{13}}}(r_{13},\overline{z}_{13}),\,\,\,\,\,\,\,\frac{1}{\sqrt{r_{14}}}(r_{14},\overline{z}_{14})$

where $r_{jk}$ denotes $|z_{jk}|$.

We do the same thing when $(0,z_{2})$ is a vision point and get the vectors

$(0,z_{21})$, $(0,z_{23})$, $(0,z_{24})$ whose lifts are

$\frac{1}{\sqrt{r_{21}}}(z_{21},r_{21}),\,\,\,\,\,\,\,\frac{1}{\sqrt{r_{23}}}(r_{23},\overline{z}_{23}),\,\,\,\,\,\,\,\frac{1}{\sqrt{r_{24}}}(r_{24},\overline{z}_{24}).$

Similarly, the lifts corresponding for the vision points $(0,z_{3})$ and
$(0,z_{4})$ are

$\frac{1}{\sqrt{r_{31}}}(z_{31},r_{31}),\,\,\,\,\,\,\,\frac{1}{\sqrt{r_{32}}}(z_{32},r_{32}),\,\,\,\,\,\,\,\frac{1}{\sqrt{r_{34}}}(r_{34},\overline{z}_{34}).$

and

$\frac{1}{\sqrt{r_{41}}}(z_{41},r_{41}),\,\,\,\,\,\,\,\frac{1}{\sqrt{r_{42}}}(z_{42},r_{42}),\,\,\,\,\,\,\,\frac{1}{\sqrt{r_{43}}}(z_{43},r_{43}).$

respectively.

Using $u_{jk}=\frac{z_{jk}}{r_{jk}}$ to be the direction of $z_{jk}$ we see
that the lifts can be written as:

$\sqrt{r_{12}}(1,\overline{u}_{12}),\,\,\,\,\,\,\sqrt{r_{13}}(1,\overline{u}_{13}),\,\,\,\,\,\,\sqrt{r_{14}}(1,\overline{u}_{14})$

$\sqrt{r_{21}}u_{21}(1,\overline{u}_{21}),\,\,\,\,\,\,\sqrt{r_{23}}(1,\overline{u}_{23}),\,\,\,\,\,\,\sqrt{r_{24}}(1,\overline{u}_{24})$

$\sqrt{r_{31}}u_{31}(1,\overline{u}_{31}),\,\,\,\,\,\,\sqrt{r_{32}}u_{32}(1,\overline{u}_{32}),\,\,\,\,\,\,\sqrt{r_{34}}(1,\overline{u}_{34})$

$\sqrt{r_{41}}u_{41}(1,\overline{u}_{41}),\,\,\,\,\,\,\sqrt{r_{42}}u_{42}(1,\overline{u}_{42}),\,\,\,\,\,\,\sqrt{r_{43}}u_{43}(1,\overline{u}_{43})$

The symmetric tensor product of the vectors in each of the lines above gives
us:

$\sqrt{r_{12}r_{13}r_{14}}(1,\overline{u}_{12}+\overline{u}_{13}+\overline{u}_{14},\,\,\,\,\,\,\overline{u}_{12}\overline{u}_{13}+\overline{u}_{12}\overline{u}_{14}+\overline{u}_{13}\overline{u}_{14},\,\,\,\,\,\,\overline{u}_{12}\overline{u}_{13}\overline{u}_{14})$

$\sqrt{r_{21}r_{23}r_{24}}u_{21}(1,\overline{u}_{21}+\overline{u}_{23}+\overline{u}_{24},\,\,\,\,\,\,\overline{u}_{21}\overline{u}_{23}+\overline{u}_{21}\overline{u}_{24}+\overline{u}_{23}\overline{u}_{24},\,\,\,\,\,\,\overline{u}_{21}\overline{u}_{23}\overline{u}_{24})$

$\sqrt{r_{31}r_{32}r_{34}}u_{31}u_{32}(1,\overline{u}_{31}+\overline{u}_{32}+\overline{u}_{34},\,\,\,\,\,\,\overline{u}_{31}\overline{u}_{32}+\overline{u}_{31}\overline{u}_{34}+\overline{u}_{32}\overline{u}_{34},\,\,\,\,\,\,\overline{u}_{31}\overline{u}_{32}\overline{u}_{34})$

$\sqrt{r_{41}r_{42}r_{43}}u_{41}u_{42}u_{43}(1,\overline{u}_{41}+\overline{u}_{42}+\overline{u}_{43},\,\,\,\,\,\,\overline{u}_{41}\overline{u}_{42}+\overline{u}_{41}\overline{u}_{43}+\overline{u}_{42}\overline{u}_{43},\,\,\,\,\,\,\overline{u}_{41}\overline{u}_{42}\overline{u}_{43}).$

Consequently, if we define the angular part of $At$ to be $At_{ang}$ given by
the product $u_{12}u_{13}u_{14}u_{23}u_{24}u_{34}$ times the determinant

$\left|\begin{array}[]{cccc}1&1&1&1\\\ \\\
\overline{u}_{12}+\overline{u}_{13}\par&\overline{u}_{21}+\overline{u}_{23}&\overline{u}_{31}+\overline{u}_{32}&\overline{u}_{41}+\overline{u}_{42}\\\
+\overline{u}_{14}&+\overline{u}_{24}&+\overline{u}_{34}&+\overline{u}_{43}\\\
\\\
\overline{u}_{12}\overline{u}_{13}+\overline{u}_{12}\overline{u}_{14}&\overline{u}_{21}\overline{u}_{23}+\overline{u}_{21}\overline{u}_{24}&\overline{u}_{31}\overline{u}_{32}+\overline{u}_{31}\overline{u}_{34}&\overline{u}_{41}\overline{u}_{42}+\overline{u}_{41}\overline{u}_{43}\\\
+\overline{u}_{13}\overline{u}_{14}&+\overline{u}_{23}\overline{u}_{24}&+\overline{u}_{32}\overline{u}_{34}&+\overline{u}_{42}\overline{u}_{43}\\\
\\\
\overline{u}_{12}\overline{u}_{13}\overline{u}_{14}&\overline{u}_{21}\overline{u}_{23}\overline{u}_{24}&\overline{u}_{31}\overline{u}_{32}\overline{u}_{34}&\overline{u}_{41}\overline{u}_{42}\overline{u}_{43}\\\
\end{array}\right|$

then the Atiyah determinant will be $At=PAt_{ang}$ where
$P=r_{12}r_{13}r_{14}r_{23}r_{24}r_{34}$ is the product of the lengths of all
sides. The decomposition $At=PAt_{ang}$ will be called the standard form of
the Atiyah determinant for four planar points. It is also natural to call $P$
the scalar part of $At$ since the rest of the product $At_{ang}$ is the
angular part of $At$.

##  3 The Convex Case

$z_{1}$$z_{2}$$W^{\prime}$$W$$\alpha_{2}$$\beta_{2}$$\alpha_{4}$$\beta_{4}$$\alpha_{3}$$\beta_{3}$$z_{4}$$z_{3}$$X$$Y$$\alpha_{1}$$\beta_{1}$
Figure 1: Convex Quadrilateral

Let us consider the terms of the Atiyah determinant corresponding to the
permutation $(1,2,3,4)$ in the expansion of $At$. This corresponds to the
product of the main diagonal entries:

$(\overline{u}_{21}+\overline{u}_{23}+\overline{u}_{24})(\overline{u}_{31}\overline{u}_{32}+\overline{u}_{31}\overline{u}_{34}+\overline{u}_{32}\overline{u}_{34})\overline{u}_{41}\overline{u}_{42}\overline{u}_{43}$

multiplied with
$r_{12}r_{13}r_{14}r_{23}r_{24}r_{34}u_{12}u_{13}u_{14}u_{23}u_{24}u_{34}$. We
decompose this expression as an angular part
$(1+u_{21}\overline{u}_{23}+u_{21}\overline{u}_{24})(1+u_{31}\overline{u}_{34}+u_{32}\overline{u}_{34})$
multiplied with the scalar part $P=r_{12}r_{13}r_{14}r_{23}r_{24}r_{34}$. The
angular part, expressed with respect to the angles appearing in figure 1, can
be written as

$(1+e^{i\gamma_{2}}+e^{i\beta_{2}})(1+e^{i\alpha_{3}}+e^{i\gamma_{3}})$

This presentation of the terms of the Atiyah determinant can be done in a
similar way for any permutation of the set $\\{1,2,3,4\\}$. For example, the
terms corresponding to the permutation $(1,4,3,2)$ can also be written as the
scalar part $P=r_{12}r_{13}r_{14}r_{23}r_{24}r_{34}$ multiplied with the
angular part
$(1+e^{-i\gamma_{4}}+e^{-i\alpha_{4}})(1+e^{-i\beta_{3}}+e^{-i\gamma_{3}})$.
Taking all terms of the Atiyah determinant into account requires considering
all 24 permutations. Let us first start with the terms of the identity
permutation: Expanding the angular part corresponding for the identity
permutation yields a linear part given by
$\displaystyle{1+e^{i\gamma_{2}}+e^{i\beta_{2}}+e^{i\gamma_{3}}+e^{i\alpha_{3}}}$
and a quadratic part given by
$e^{i(\gamma_{2}+\gamma_{3})}+e^{i(\beta_{2}+\gamma_{3})}+e^{i(\gamma_{2}+\alpha_{3})}+e^{i(\beta_{2}+\alpha_{3})}$.
We need to collect all similar terms from all permutations to get a general
formula for the Atiyah determinant:

The Linear Terms: These terms can be summarized as follows: We have 24 ones,
one from each permutation. We also have all angles of the quadrilaterial
appearing with an equal number of times as $e^{i\theta}$, where $\theta$ is
any one of the $\alpha_{j}$’s, $\beta_{j}$’s or $\gamma_{j}$’s, and
$j=1,2,3,4$. It is easy to see that conjugate terms appear from permutations
of the form $(a,b,c,d)$ and their reverse order $(d,c,b,a)$, which confirms
that the determinant is a real number. This allows us to see that each linear
term $e^{i\theta}$ appears 4 times and each $e^{-i\theta}$ appears 4 times.
Accordingly, the linear terms in the expansion of the Atiyah determinant are

$24+8\sum\cos\theta_{k}$

where $\theta_{k}$ runs over the 12 angles $\alpha_{j}$, $\beta_{j}$, and
$\gamma_{j}$, where $j=1,2,3,4$.

The Quadratic Terms: Listing all quadratic terms based on their corresponding
permutations gives us:

(1,2,3,4):
$\displaystyle{e^{i(\gamma_{2}+\gamma_{3})}+e^{i(\beta_{2}+\gamma_{3})}+e^{i(\gamma_{2}+\alpha_{3})}+e^{i(\beta_{2}+\alpha_{3})}}$

(1,2,4,3):
$\displaystyle{e^{i(\beta_{2}-\beta_{4})}+e^{i(\gamma_{2}-\beta_{4})}+e^{i(\beta_{2}-\gamma_{4})}+e^{i(\gamma_{2}-\gamma_{4})}}$

(1,3,2,4):
$\displaystyle{e^{i(-\beta_{3}-\alpha_{2})}+e^{i(\alpha_{3}-\alpha_{2})}+e^{i(-\beta_{3}+\beta_{2})}+e^{i(\alpha_{3}+\beta_{2})}}$

(1,3,4,2):
$\displaystyle{e^{i(\alpha_{3}+\beta_{4})}+e^{i(-\beta_{3}+\beta_{4})}+e^{i(\alpha_{3}-\alpha_{4})}+e^{i(-\beta_{3}-\alpha_{4})}}$

(1,4,2,3):
$\displaystyle{e^{i(-\alpha_{4}+\alpha_{2})}+e^{i(-\gamma_{4}+\alpha_{2})}+e^{i(-\alpha_{4}+\gamma_{2})}+e^{i(-\gamma_{4}+\gamma_{2})}}$

(1,4,3,2):
$\displaystyle{e^{i(-\gamma_{4}-\gamma_{3})}+e^{i(-\alpha_{4}-\gamma_{3})}+e^{i(-\gamma_{4}-\beta_{3})}+e^{i(-\alpha_{4}-\beta_{3})}}$

(2,1,3,4):
$\displaystyle{e^{i(-\alpha_{1}+\alpha_{3})}+e^{i(-\gamma_{1}+\alpha_{3})}+e^{i(-\alpha_{1}+\gamma_{3})}+e^{i(-\gamma_{1}+\gamma_{3})}}$

(2,1,4,3):
$\displaystyle{e^{i(-\gamma_{1}-\gamma_{4})}+e^{i(-\alpha_{1}-\gamma_{4})}+e^{i(-\gamma_{1}-\beta_{4})}+e^{i(-\alpha_{1}-\beta_{4})}}$

(3,1,2,4):
$\displaystyle{e^{i(\alpha_{1}+\beta_{2})}+e^{i(-\beta_{1}+\beta_{2})}+e^{i(\alpha_{1}-\alpha_{2})}+e^{i(-\beta_{1}-\alpha_{2})}}$

(3,1,4,2):
$\displaystyle{e^{i(-\beta_{1}-\alpha_{4})}+e^{i(\alpha_{1}-\alpha_{4})}+e^{i(-\beta_{1}+\beta_{4})}+e^{i(\alpha_{1}+\beta_{4})}}$

(4,1,2,3):
$\displaystyle{e^{i(\gamma_{1}+\gamma_{2})}+e^{i(\beta_{1}+\gamma_{2})}+e^{i(\gamma_{1}+\alpha_{2})}+e^{i(\beta_{1}+\alpha_{2})}}$

(4,1,3,2):
$\displaystyle{e^{i(\beta_{1}-\beta_{3})}+e^{i(\gamma_{1}-\beta_{3})}+e^{i(\beta_{1}-\gamma_{3})}+e^{i(\gamma_{1}-\gamma_{3})}}$

The other 12 permutations are nothing but the conjugates of the ones appearing
above. Accordingly, adding all terms would result in a sum of twice the
cosines of the angles appearing in this list. We note that all the terms
appearing in the first column will cancel out. For example,
$\cos(\gamma_{2}+\gamma_{3})+\cos(\alpha_{3}-\alpha_{1})=0$. This can be seen
to apply to all terms of the first column. On the other hand, the terms of the
fourth column appear twice in the list (including the conjugates).

Summarizing all terms can be done as follows: the multiplicity-one terms can
be put in 6 families: $2\cos(\alpha_{j}-\alpha_{j+1})$,
$2\cos(\beta_{j}-\beta_{j+1})$, $2\cos(\gamma_{j}+\alpha_{j+1})$,
$2\cos(\gamma_{j+1}+\beta_{j})$, $2\cos(\gamma_{j}-\alpha_{j+2})$,
$2\cos(\gamma_{j}-\beta_{j+2})$, where $j=1,2,3,4$, modulo 4, and the
multiplicity-two terms can be listed as follows:
$4\cos(\alpha_{1}+\beta_{4})$, $4\cos(\alpha_{2}+\beta_{1})$,
$4\cos(\alpha_{3}+\beta_{2})$, $4\cos(\alpha_{4}+\beta_{3})$,
$4\cos(\gamma_{1}-\gamma_{3})$, and $4\cos(\gamma_{2}-\gamma_{4})$.
Contemplating this list, we realize that these terms can be grouped based on
vertices facing opposite triangles in the following manner:

$\cos(\alpha_{j}-\alpha_{j+1})$: where $\alpha_{j}$ is at vertex $z_{j}$ and
$\alpha_{j+1}$ is in the opposite triangle

$\cos(\beta_{j}-\beta_{j-1})$: where $\beta_{j}$ is at vertex $z_{j}$ and
$\beta_{j-1}$ is in the opposite triangle

$\cos(\gamma_{j}+\alpha_{j+1})$: where $\gamma_{j}$ is at vertex $z_{j}$ and
$\alpha_{j+1}$ is in the opposite triangle

$\cos(\gamma_{j}+\beta_{j-1})$: where $\gamma_{j}$ is at vertex $z_{j}$ and
$\beta_{j-1}$ is in the opposite triangle

$\cos(\alpha_{j}-\gamma_{j+2})$: where $\alpha_{j}$ is at vertex $z_{j}$ and
$\gamma_{j+2}$ is in the opposite triangle

$\cos(\beta_{j}-\gamma_{j+2})$: where $\beta_{j}$ is at vertex $z_{j}$ and
$\gamma_{j+2}$ is in the opposite triangle

in addition to the multiplicity-two terms:

$\cos(\alpha_{1}+\beta_{4})$: where $\alpha_{1}$ is at vertex $z_{1}$ and
$\beta_{4}$ is in the opposite triangle

or $\beta_{4}$ is at vertex $z_{4}$ and $\alpha_{1}$ is in the opposite
triangle

$\cos(\alpha_{2}+\beta_{1})$: where $\alpha_{2}$ is at vertex $z_{2}$ and
$\beta_{1}$ is in the opposite triangle

or $\beta_{1}$ is at vertex $z_{1}$ and $\alpha_{2}$ is in the opposite
triangle

$\cos(\alpha_{3}+\beta_{2})$: where $\alpha_{3}$ is at vertex $z_{3}$ and
$\beta_{2}$ is in the opposite triangle

or $\beta_{2}$ is at vertex $z_{2}$ and $\alpha_{3}$ is in the opposite
triangle

$\cos(\alpha_{4}+\beta_{3})$: where $\alpha_{4}$ is at vertex $z_{4}$ and
$\beta_{3}$ is in the opposite triangle

or $\beta_{3}$ is at vertex $z_{3}$ and $\alpha_{4}$ is in the opposite
triangle

$\cos(\gamma_{1}-\gamma_{3})$: where $\gamma_{1}$ is at vertex $z_{1}$ and
$\gamma_{3}$ is in the opposite triangle

or $\gamma_{3}$ is at vertex $z_{3}$ and $\gamma_{1}$ is in the opposite
triangle

$\cos(\gamma_{2}-\gamma_{4})$: where $\gamma_{2}$ is at vertex $z_{2}$ and
$\gamma_{4}$ is in the opposite triangle

or $\gamma_{4}$ is at vertex $z_{4}$ and $\gamma_{2}$ is in the opposite
triangle

For example, let us consider the vertex $z_{1}$ facing the triangle
$z_{2}z_{3}z_{4}$. The angle $\alpha_{1}$ at the vertex $z_{1}$ is coupled
with each of the angles of the triangle $z_{2}z_{3}z_{4}$ in the terms:
$\cos(\alpha_{1}-\alpha_{2})$, $\cos(\alpha_{1}-\gamma_{3})$, and
$\cos(\alpha_{1}+\beta_{4})$. Also, the angle $\beta_{1}$ at the vertex
$z_{1}$ is coupled with each of the angles of the triangle $z_{2}z_{3}z_{4}$
in the terms: $\cos(\beta_{1}+\alpha_{2})$, $\cos(\beta_{1}-\gamma_{3})$, and
$\cos(\beta_{1}-\beta_{4})$. Finally, the angle $\gamma_{1}$ is coupled with
each of the angles of the triangle $z_{2}z_{3}z_{4}$ in the terms:
$\cos(\gamma_{1}+\alpha_{2})$, $\cos(\gamma_{1}-\gamma_{3})$, and
$\cos(\gamma_{1}+\beta_{4})$.

Collecting the product of cosines $\cos\alpha\cos\beta$ from
$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$ for each of
the quadratic terms we obtain the sum of products

$2\sum_{j=1}^{4}\left(\cos\alpha_{j}+\cos\beta_{j}+\cos\gamma_{j}\right)\left(\cos\alpha_{j+1}+\cos\gamma_{j+2}+\cos\beta_{j+3}\right)$

where all indices are taken modulo four. Here, we see that
$\displaystyle{\cos\alpha_{j}+\cos\beta_{j}}$

$\displaystyle{+\cos\gamma_{j}}$ is the sum of cosines of all angles at the
vertex $z_{j}$ whereas $\cos\alpha_{j+1}+\cos\gamma_{j+2}+\cos\beta_{j+3}$ is
the sum of cosines of all angles of the opposite triangle.

To accommodate for the remaining product of sines $\mp\sin\alpha\sin\beta$
from $\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$, let us
call the sum of all products of two sines (together with their signs) to be
the expression $E$. Based on that we have proved the following theorem:

###### Theorem 1

: The Atiyah determinant can be written as
$r_{12}r_{13}r_{14}r_{23}r_{24}r_{34}$ times the angular part

$24+8\sum_{j=1}^{4}\left(\cos\alpha_{j}+\cos\beta_{j}+\cos\gamma_{j}\right)$

$+\,2\sum_{j=1}^{4}\left(\cos\alpha_{j}+\cos\beta_{j}+\cos\gamma_{j}\right)\left(\cos\alpha_{j+1}+\cos\gamma_{j+2}+\cos\beta_{j+3}\right)$

$+E$

Let us call the first sum
$\displaystyle{S_{1}=\sum_{j=1}^{4}\left(\cos\alpha_{j}+\cos\beta_{j}+\cos\gamma_{j}\right)}$
and the second sum
$\displaystyle{S_{2}=\sum_{j=1}^{4}\left(\cos\alpha_{j}+\cos\beta_{j}+\cos\gamma_{j}\right)\left(\cos\alpha_{j+1}+\cos\gamma_{j+2}+\cos\beta_{j+3}\right)}$

###### Corollary 1

:

$S_{1}=\sum_{j=1}^{4}\left(\cos\alpha_{j}+\cos\beta_{j}+\cos\gamma_{j}\right)\geq
4$

Proof: This proof relies on the identity

$\cos\alpha+\cos\beta+\cos\gamma=1+4\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}\hskip
14.22636pt;\hskip 14.22636pt\alpha+\beta+\gamma=\pi$

(*)

Accordingly, we regroup the terms in our summation $S_{1}$ in such a way that
angles of the same triangle are grouped together to obtain

$S_{1}=\sum_{j=1}^{4}\left(\cos\alpha_{j+1}+\cos\gamma_{j+2}+\cos\beta_{j+3}\right)$

$=\sum_{j=1}^{4}\left(1+4\sin\frac{\alpha_{j+1}}{2}\sin\frac{\gamma_{j+2}}{2}\sin\frac{\beta_{j+3}}{2}\right)$

$=4+4\sum_{j=1}^{4}\sin\frac{\alpha_{j+1}}{2}\sin\frac{\gamma_{j+2}}{2}\sin\frac{\beta_{j+3}}{2}$

which is obviously greater than or equal to 4 since all other terms are non-
negative.

###### Corollary 2

:

$8S_{1}+2S_{2}\geq 40$

Proof: First, we collect from corollary 1 the simplified expression of $S_{1}$
and write it as

$8S_{1}=32+32\sum_{j=1}^{4}\sin\frac{\alpha_{j+1}}{2}\sin\frac{\gamma_{j+2}}{2}\sin\frac{\beta_{j+3}}{2}$

We now turn our attention to $S_{2}$. Here we use the identity (*) again on
the second factors of the sum $S_{2}$ to write it as

$S_{2}=\sum_{j=1}^{4}\left(\cos\alpha_{j}+\cos\beta_{j}+\cos\gamma_{j}\right)(1+4\sin\frac{\alpha_{j+1}}{2}\sin\frac{\gamma_{j+2}}{2}\sin\frac{\beta_{j+3}}{2})$

Consequently,

$2S_{2}=2\sum_{j=1}^{4}\left(\cos\alpha_{j}+\cos\beta_{j}+\cos\gamma_{j}\right)$

$+8\sum_{j=1}^{4}\left(\cos\alpha_{j}+\cos\beta_{j}+\cos\gamma_{j}\right)\sin\frac{\alpha_{j+1}}{2}\sin\frac{\gamma_{j+2}}{2}\sin\frac{\beta_{j+3}}{2}$

The first line of $2S_{2}$ can also be simpified since angles can be regrouped
into four groups, each corresponding to one of the four triangles of the
quadrilateral, so we write as

$2\sum_{j=1}^{4}\left(\cos\alpha_{j}+\cos\gamma_{j+1}+\cos\beta_{j+2}\right)=$

$=8+8\sum_{j=1}^{4}\sin\frac{\alpha_{j+1}}{2}\sin\frac{\gamma_{j+2}}{2}\sin\frac{\beta_{j+3}}{2}$

Accordingly, $8S_{1}+2S_{2}$ is equal to

$=40+40\sum_{j=1}^{4}\sin\frac{\alpha_{j+1}}{2}\sin\frac{\gamma_{j+2}}{2}\sin\frac{\beta_{j+3}}{2}$

$+8\sum_{j=1}^{4}\left(\cos\alpha_{j}+\cos\beta_{j}+\cos\gamma_{j}\right)\sin\frac{\alpha_{j+1}}{2}\sin\frac{\gamma_{j+2}}{2}\sin\frac{\beta_{j+3}}{2}$

which finally gives us that

$8S_{1}+2S_{2}=40+8\sum_{j=1}^{4}\left(5+\cos\alpha_{j}+\cos\beta_{j}+\cos\gamma_{j}\right)\sin\frac{\alpha_{j+1}}{2}\sin\frac{\gamma_{j+2}}{2}\sin\frac{\beta_{j+3}}{2}$

Since $5+\cos\alpha_{j}+\cos\beta_{j}+\cos\gamma_{j}\geq 0$, we can easily see
that $8S_{1}+2S_{2}\geq 40$ which proves the second corollary.

Based on these two corollaries, we can prove the second Atiyah-Sutcliffe
conjecture for a convex quadrilateral if we can prove that $E\geq 0$. This
will finish the proof. To do that, we will prove that $E$ is nothing but the
expression $4(E_{12}+E_{23}+E_{34}+E_{14}-E_{13}-E_{24})$ which was proved to
be non-negative in [9]. Before we start comparing our work with [9], note that
the side-lengths in our quadrilateral were denoted in [9] as follows:
$r_{23}=a$, $r_{13}=b$, $r_{12}=c$, $r_{14}=d$, $r_{24}=e$, and $r_{34}=f$.
Let us proceed by regrouping the terms in the following way:

Consider each side of the quadrilaterial separately. For example, let us
consider the products of sines corresponding to the side $z_{3}z_{4}$. This
side belong to two triangles, namely $z_{2}z_{3}z_{4}$ and $z_{1}z_{3}z_{4}$.
When $z_{1}$ is facing triangle $z_{2}z_{3}z_{4}$, the terms corresponding for
$z_{3}z_{4}$ are all the terms corresponding for the angle $\beta_{1}$ at
$z_{1}$ combined with the angles of the triangle $z_{2}z_{3}z_{4}$. These are:
$\cos(\beta_{1}-\beta_{4})$, $\cos(\beta_{1}-\gamma_{3})$, and
$\cos(\beta_{1}+\alpha_{2})$. By doing a similar choice of the terms when
$z_{2}$ is facing triangle $z_{1}z_{3}z_{4}$, we see that the terms
corresponding for the side $z_{3}z_{4}$ are: $\cos(\alpha_{2}-\alpha_{3})$,
$\cos(\alpha_{2}-\gamma_{4})$, and $\cos(\alpha_{2}+\beta_{1})$. When we
collect the product of two sines from these terms as was formed in $E$ we
obtain

$\sin\beta_{1}\sin\beta_{4}+\sin\beta_{1}\sin\gamma_{3}+\sin\alpha_{2}\sin\alpha_{3}+\sin\alpha_{2}\sin\gamma_{4}-2\sin\alpha_{2}\sin\beta_{1}$

Multiplying this with the scalar part $abcdef$, we note that we can write this
expression in terms of the areas of the triangles as follows:

$\displaystyle{(abcdef)\sin\beta_{1}\sin\beta_{4}=4ac(\frac{1}{2}bd\sin\beta_{1})(\frac{1}{2}ef\sin\beta_{4})=4acA_{134}A_{234}}$

$\displaystyle{(abcdef)\sin\beta_{1}\sin\gamma_{3}=4ce(\frac{}{2}bd\sin\beta_{1})(\frac{1}{2}af\sin\gamma_{3})=4ceA_{134}A_{234}}$

$\displaystyle{(abcdef)\sin\alpha_{2}\sin\alpha_{3}=4cd(\frac{}{2}ae\sin\alpha_{2})(\frac{1}{2}bf\sin\alpha_{3})=4cdA_{134}A_{234}}$

$\displaystyle{(abcdef)\sin\alpha_{2}\sin\gamma_{4}=4bc(\frac{}{2}ae\sin\alpha_{2})(\frac{1}{2}df\sin\gamma_{4})=4bcA_{134}A_{234}}$

$\displaystyle{(abcdef)\sin\alpha_{2}\sin\beta_{1}=4cf(\frac{}{2}ae\sin\alpha_{2})(\frac{1}{2}bd\sin\beta_{1})=4cfA_{134}A_{234}}$

Accordingly, when factorizing $4A_{134}A_{234}$, the terms corresponding for
the side $z_{3}z_{4}$ can be written as

$4A_{134}A_{234}c(a+e+d+b-2f)$

This is precisely the quantity $4E_{34}$ as defined in [9]. Repeating this
computation for each of the sides of the quadrilateral, we find that:

The products of sines corresponding to the side $z_{1}z_{2}$ is
$4A_{123}A_{124}f(a+b+e+d-2c)$ which is $4E_{12}$ as defined in [9].

The products of sines corresponding to the side $z_{1}z_{3}$ is
$-4A_{123}A_{134}e(a+c+d+f-2b)$ which is $-4E_{13}$ as defined in [9].

The products of sines corresponding to the side $z_{1}z_{4}$ is
$4A_{124}A_{134}a(c+e+b+f-2d)$ which is $4E_{14}$ as defined in [9].

The products of sines corresponding to the side $z_{2}z_{3}$ is
$4A_{123}A_{234}d(c+b+e+f-2a)$ which is $4E_{23}$ as defined in [9].

The products of sines corresponding to the side $z_{2}z_{4}$ is
$-4A_{124}A_{234}b(c+d+f+a-2e)$ which is $-4E_{24}$ as defined in [9].

Consequently, requiring $E\geq 0$ is the same as requiring that

$E_{12}+E_{23}+E_{34}+E_{14}-E_{13}-E_{24}\geq 0$

This is what we proved in [9]. With that accomplished, we can see that the
proof of the second Atiyah-Sutcliffe conjecture for a convex quadrilateral is
now complete.

## References

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