# Finiteness of Stationary Configurations of the Planar Five-vortex Problem
††thanks: Supported by NSFC(No.11701464, No.11801537)

Xiang<EMAIL_ADDRESS><EMAIL_ADDRESS>Shuqiang
<EMAIL_ADDRESS>
School of Economic and Mathematics, Southwestern University of Finance and
Economics,
Chengdu 611130, China

###### Abstract

For the planar five-vortex problem, we show that there are finitely many
relative equilibria (uniformly rotating configurations) and collapse
configurations, except perhaps if the vorticities satisfy some explicitly
given algebraic constraints. In particular, if the vorticities are of the same
sign, we show the number of relative equilibria and collapse configurations
are finite.

Keywords: Point vortices; Relative equilibrium; Finiteness.

2020AMS Subject Classification 76B47 70F10 70F15 37Nxx .

##  1 Introduction

The planar _ $N$-vortex problem_, dated back to Helmholtz’s work on
hydrodynamics in 1858 [6], consider the dynamical system given by

$\Gamma_{n}\dot{\mathbf{r}}_{n}=J\nabla_{k}H=J\sum_{1\leq j\leq N,j\neq
n}\frac{\Gamma_{n}\Gamma_{j}(\mathbf{r}_{j}-\mathbf{r}_{n})}{|\mathbf{r}_{j}-\mathbf{r}_{n}|^{2}},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}n=1,2,\cdots,N.$
(1.1)

where $\Gamma_{n}\in\mathbb{R}\backslash\\{0\\}$ is the vortex strength (or
vorticity) and $\mathbf{r}_{n}\in{\mathbb{R}}^{2}$ is the position for
$n=1,...,N$, $J=\left(\begin{array}[]{cc}0&1\\\ -1&0\\\ \end{array}\right)$,
$H=-\sum_{1\leq j<k\leq N}\ln|\mathbf{r}_{j}-\mathbf{r}_{k}|$, and $|\cdot|$
denotes the Euclidean norm in ${\mathbb{R}}^{2}$.

It is of physical interest to study stationary configurations which lead to
self-similar solutions of the $N$-vortex problem[10, 14]. O’Neil [10] has
shown that the only stationary configurations are equilibria, rigidly
translating configurations, relative equilibria (uniformly rotating
configurations) and collapse configurations.

For two and three vorticities, stationary configurations are known to be
finite, [4, 13, 8, 9, 2, 3]. For four vorticities, O’Neil [11], Hampton and
Moeckel [5] independently proved that the numbers of equilibria, rigidly
translating configurations and relative equilibria are generically finite.
However, their methods do not work for collapse configurations.

Motivated by the novel method of Albouy and Kaloshin for celestial mechanics
[1], the first author introduced a system of equations (2.7), whose solution
set contains relative equilibria and collapse configurations. The system would
admit a continuum of complex solutions if there are infinitely many relative
equilibria and collapse configurations. Then the finiteness of all stationary
configurations for the 4-vortex problem is established by analysis of the
singularities of the system [14].

In this study, we shall apply the method introduced in [14] to study the
finiteness problem of relative equilibria and collapse configurations for the
five-vortex problem. Because of the well-known continuum of relative
equilibria [12], (see Section 2), the finiteness of relative equilibria and
collapse configurations would be expected only for generic vorticities. We
have proven the following results:

###### Theorem 1.1

If the vorticities $\Gamma_{n}$ $(n\in\\{1,2,3,4,5\\})$ are nonzero, the five-
vortex problem has finitely many relative equilibria and collapse
configurations, with the possible exception of some vorticity parameter values
on which an explicitly given set of polynomials vanish (see Sect. 5).

###### Theorem 1.2

For any choice of five vorticities, if $\sum_{j\in J}\Gamma_{j}\neq 0$ and
$\sum_{j,k\in J,j\neq k}\Gamma_{j}\Gamma_{k}\neq 0$, where $J$ is a nonempty
subset of $\\{1,2,3,4,5\\}$, there are finitely many normalized central
configurations of the five-vortex problem.

###### Corollary 1.3

If the vorticity $\Gamma_{n}$ $(n\in\\{1,2,3,4,5\\})$ are all positive (or
negative), then the five-vortex problem has finitely many relative equilibria
and collapse configurations.

Recall that in the Newtonian five-body problem, even the masses are all
positive, there exists a continuum of central configurations in the complex
domain [1]. Theorem 1.2 points out one of the many differences between central
configurations of the Newtonian five-body problem and that of the five-vortex
problem when finiteness of central configurations is concerned.

The paper is structured as follows. In Section 2, we give some notations and
definitions. In particular, following Albouy and Kaloshin [1], we introduce
singular sequences of normalized central configurations. In Section 3, we
discuss some tools to classify the singular sequences. In Section 4, we study
all possible singular sequences and reduce the problem to twenty-nine
diagrams. In Section 5, we obtain the constraints on the vorticities
corresponding to each of the twenty-nine diagrams. In Section 6, we prove
Theorem 1.1 and Theorem 1.2.

##  2 Preliminaries

In this section we recall some notations and definitions that will be needed
later.

###  2.1 Stationary configurations

First it is more convenient to consider the vortex positions
$\mathbf{r}_{n}\in\mathbb{R}^{2}$ as complex numbers $z_{n}\in\mathbb{C}$, in
which case the dynamics are given by $\dot{z}_{n}=-\textbf{i}V_{n}$ where

$V_{n}=\sum_{1\leq j\leq N,j\neq
n}\frac{\Gamma_{j}z_{jn}}{r_{jn}^{2}}=\sum_{j\neq
n}\frac{\Gamma_{j}}{{\overline{z}_{jn}}},$ (2.2)

$z_{jn}=z_{n}-z_{j}$, $r_{jn}=|z_{jn}|=\sqrt{z_{jn}{\overline{z}_{jn}}}$,
$\textbf{i}=\sqrt{-1}$ and the over bar denotes complex conjugation.

Let
$\mathbb{C}^{N}=\\{z=(z_{1},\cdots,z_{N}):z_{j}\in\mathbb{C},j=1,\cdots,N\\}$
denote the space of configurations for $N$ point vortex. Let
$\Delta=\\{z\in\mathbb{C}^{N}:z_{j}=z_{k}~{}~{}\emph{for some}~{}~{}j\neq
k\\}$ be the collision set in $\mathbb{C}^{N}$. Then the set
$\mathbb{C}^{N}\backslash\Delta$ is the space of collision-free
configurations.

###### Definition 2.1

The following quantities are defined:

$\begin{array}[]{cc}\text{Total vorticity}&\Gamma=\sum_{j=1}^{N}\Gamma_{j}\\\
\text{Total vortex angular momentum}&L=\sum_{1\leq j<k\leq
N}\Gamma_{j}\Gamma_{k}\\\ \text{ Moment of vorticity
}&M=\sum_{j=1}^{N}\Gamma_{j}z_{j}\\\ \text{ Angular impulse
}&I=\sum_{j=1}^{N}\Gamma_{j}|z_{j}|^{2}=\sum_{j=1}^{N}\Gamma_{j}z_{j}{\overline{z}_{j}}\end{array}$

Then it is easy to see that

$\Gamma I=\sum_{1\leq j<k\leq
N}\Gamma_{j}\Gamma_{k}z_{jk}{\overline{z}_{jk}}=\sum_{1\leq j<k\leq
N}\Gamma_{j}\Gamma_{k}r_{jk}^{2}\triangleq S,$ (2.3)

and

$\Gamma^{2}-2L>0.$ (2.4)

Following O’Neil [10] we will call a configuration stationary if the relative
shape remains constant, i.e., if the ratios of mutual distances
$r_{jk}/r_{lm}$ remain constant (such solutions are often called homographic).
More precisely,

###### Definition 2.2

A configuration $z\in\mathbb{C}^{N}\backslash\Delta$ is stationary if there
exists a constant $\Lambda\in{\mathbb{C}}$ such that

$V_{j}-V_{k}=\Lambda(z_{j}-z_{k}),~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}1\leq j,k\leq
N.$ (2.5)

It is shown that the only stationary configurations of vortices are
equilibria, rigidly translating configurations, relative equilibria (uniformly
rotating configurations) and collapse configurations [10].

###### Definition 2.3

  * i.

$z\in\mathbb{C}^{N}\backslash\Delta$ is an _equilibrium_ if
$V_{1}=\cdots=V_{N}=0$.

  * ii.

$z\in\mathbb{C}^{N}\backslash\Delta$ is _rigidly translating_ if
$V_{1}=\cdots=V_{N}=V$ for some $V\in\mathbb{C}\backslash\\{0\\}$. (The
vortices are said to move with common velocity $V$.)

  * iii.

$z\in\mathbb{C}^{N}\backslash\Delta$ is a _relative equilibrium_ if there
exist constants $\lambda\in\mathbb{R}\backslash\\{0\\},z_{0}\in\mathbb{C}$
such that $V_{n}=\lambda(z_{n}-z_{0}),~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}1\leq
n\leq N$.

  * iv.

$z\in\mathbb{C}^{N}\backslash\Delta$ is a _collapse configuration_ if there
exist constants $\Lambda,z_{0}\in\mathbb{C}$ with $\emph{Im}(\Lambda)\neq 0$
such that $V_{n}=\Lambda(z_{n}-z_{0}),~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}1\leq
n\leq N$.

It is easy to see that $L=0$ is a necessary condition for the existence of
equilibria, and $\Gamma=0$ is a necessary condition for the existence of
configurations [10].

###### Proposition 2.1

Every equilibrium has vorticities satisfying $L=0$; every rigidly translating
configuration has vorticities satisfying $\Gamma=0$

If $z^{\prime}$ differs from $z$ only by translation, rotation, and change of
scale (dilation) in the plane, then it is easy to see that $z$ is stationary
if and only if $z^{\prime}$ is stationary.

###### Definition 2.4

A configuration $z$ is equivalent to a configuration $z^{\prime}$ if for some
$a,b\in\mathbb{C}$ with $b\neq 0$, $z^{\prime}_{n}=b(z_{n}+a),~{}~{}~{}1\leq
n\leq N$.

$z\in\mathbb{C}^{N}\backslash\Delta$ is a translation-normalized configuration
if $M=0$; $z\in\mathbb{C}^{N}\backslash\Delta$ is a rotation-normalized
configuration if $z_{12}\in\mathbb{R}$. Fixing the scale of a configuration,
we can give the definition of dilation-normalized configuration, however, we
do not specify the scale here.

A configuration, which is translation-normalized, rotation-normalized and
dilation-normalized, is called a normalized configuration.

Note that for a given configuration $z$, there always is certain rotation-
normalized configuration and dilation-normalized configuration being
equivalent to $z$; if $\Gamma\neq 0$, then there is always certain
translation-normalized configuration being equivalent to $z$ too. Hence, if
$\Gamma\neq 0$, there is exactly one normalized configuration being equivalent
to $z$.

###  2.2 Central configurations

In this paper we consider only relative equilibria and collapse
configurations, then it is easy to see that there always is certain
translation-normalized configuration being equivalent to a given
configuration.

The equations of relative equilibria and collapse configurations can be
unified into the following formula

$V_{n}=\Lambda(z_{n}-z_{0}),~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}1\leq n\leq N,$
(2.6)

$\Lambda\in\mathbb{R}\backslash\\{0\\}$ corresponds to relative equilibria and
$\Lambda\in\mathbb{C}\backslash\mathbb{R}$ corresponds to collapse
configurations.

###### Definition 2.5

Relative equilibria and collapse configurations are both called central
configurations.

The equations (2.6) can be reducible to

$\Lambda z_{n}=V_{n},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}1\leq n\leq N,$ (2.7)

if we replace $z_{n}$ by $z_{n}+z_{0}$, i.e., the solutions of equations (2.7)
have been removed the translation freedoms. In fact, it is easy to see that
the solutions of equations (2.7) satisfy

$M=0,$ (2.8) $\Lambda I=L.$ (2.9)

To remove the dilation freedoms, we set $|\Lambda|=1$. Following Albouy and
Kaloshin [1] we introduce

###### Definition 2.6

A real normalized central configuration of the planar $N$-vortex problem is a
solution of (2.7) satisfying $z_{12}\in\mathbb{R}$ and $|\Lambda|=1$.

We remark that we shall study system (2.7) in the complex domain and the word
“real” refers to the reality hypothesis. It is omitted in a real context.

###### Proposition 2.2

[14] Collapse configurations satisfy $\Gamma\neq 0$ and

$S=I=L=0.$ (2.10)

For relative equilibria we have
$S=0~{}~{}~{}\Longleftrightarrow~{}~{}~{}\left\\{\begin{array}[]{lr}\Gamma\neq
0&\\\
I=L=0&\end{array}\right.\text{or}~{}~{}~{}\left\\{\begin{array}[]{lr}\Gamma=0&\\\
I\neq 0,L\neq 0&\end{array}\right.$

###  2.3 Complex central configurations

To eliminate complex conjugation in (2.7) we introduce a new set of variables
$w_{n}$ and a “conjugate” relation:

$\begin{array}[]{c}\Lambda z_{n}=\sum_{j\neq
n}\frac{\Gamma_{j}}{{w_{jn}}},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}1\leq n\leq N,\\\
\overline{\Lambda}w_{n}=\sum_{j\neq
n}\frac{\Gamma_{j}}{{z_{jn}}},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}1\leq n\leq
N,\end{array}$ (2.11)

where $z_{jn}=z_{n}-z_{j}$ and $w_{jn}=w_{n}-w_{j}$.

The rotation freedom is expressed in $z_{n},w_{n}$ variables as the invariance
of (2.11) by the map $R_{a}:(z_{n},w_{n})\mapsto(az_{n},a^{-1}w_{n})$ for any
$a\in\mathbb{C}\backslash\\{0\\}$ and any $n=1,2,\cdots,N$. The condition
$z_{12}\in\mathbb{R}$ we proposed to remove this rotation freedom becomes
$z_{12}=w_{12}$.

To the variables $z_{n},w_{n}\in\mathbb{C}$ we add the variables
$Z_{jk},W_{jk}\in\mathbb{C}$ $(1\leq j<k\leq N)$ such that
$Z_{jk}=1/w_{jk},W_{jk}=1/z_{jk}$. For $1\leq k<j\leq N$ we set
$Z_{jk}=-Z_{kj},W_{jk}=-W_{kj}$. Then equations (2.7) together with the
condition $z_{12}\in\mathbb{R}$ and $|\Lambda|=1$ becomes

$\begin{array}[]{cc}\Lambda z_{n}=\sum_{j\neq n}\Gamma_{j}Z_{jn},&1\leq n\leq
N,\\\ \overline{\Lambda}w_{n}=\Lambda^{-1}w_{n}=\sum_{j\neq
n}\Gamma_{j}W_{jn},&1\leq n\leq N,\\\ Z_{jk}w_{jk}=1,&1\leq j<k\leq N,\\\
W_{jk}z_{jk}=1,&1\leq j<k\leq N,\\\
z_{jk}=z_{k}-z_{j},~{}~{}~{}w_{jk}=w_{k}-w_{j},&1\leq j,k\leq N,\\\
Z_{jk}=-Z_{kj},~{}~{}~{}W_{jk}=-W_{kj},&1\leq k<j\leq N,\\\
z_{12}=w_{12}.\end{array}$ (2.12)

This is a polynomial system in the variables
$\mathcal{Q}=(\mathcal{Z},\mathcal{W})\in(\mathbb{C}^{N}\times\mathbb{C}^{N(N-1)/2})^{2}$,
here

$\mathcal{Z}=(z_{1},z_{2},\cdots,z_{N},Z_{12},Z_{13},\cdots,Z_{(N-1)N})$,
$\mathcal{W}=(w_{1},w_{2},\cdots,w_{N},W_{12},W_{13},\cdots,W_{(N-1)N}).$

It is easy to see that a real normalized central configuration of (2.7) is a
solution $\mathcal{Q}=(\mathcal{Z},\mathcal{W})$ of (2.12) such that
$z_{n}={\overline{w}}_{n}$ and vice versa.

######  Definition 2.7 (Normalized central configuration)

A normalized central configuration is a solution
$\mathcal{Q}=(\mathcal{Z},\mathcal{W})$ of (2.12). A real normalized central
configuration is a normalized central configuration such that
$z_{n}={\overline{w}}_{n}$ for any $n=1,2,\cdots,N$.

Definition 2.7 of a real normalized central configuration coincides with
Definition 2.6.

###### Definition 2.8

We will use the name “distance” for the $r_{jk}=\sqrt{z_{jk}{w_{jk}}}$. We
will use the name $z$-separation (respectively $w$-separation) for the
$z_{jk}$’s (respectively the $w_{jk}$’s) in the complex plane.

Strictly speaking, the distances $r_{jk}=\sqrt{z_{jk}{w_{jk}}}$ are now bi-
valued. However, only the squared distances appear in the system, so we shall
understand $r^{2}_{jk}$ as $z_{jk}w_{jk}$ from now on.

###  2.4 Elimination theory

Let $m$ be a positive integer. Following [7], we define a closed algebraic
subset of the affine space $\mathbb{C}^{m}$ as the set of common zeros of a
system of polynomials on $\mathbb{C}^{m}$.

The polynomial system (2.12) defines a closed algebraic subset
$\mathcal{A}\subset(\mathbb{C}^{N}\times\mathbb{C}^{N(N-1)/2})^{2}$. For the
planar five-vortex problem, we will prove that this subset is finite, then
real normalized central configurations is finite. To distinguish the two
possibilities, finitely many or infinitely many points, we will only use the
following result (see [1]) from elimination theory.

###### Lemma 2.1

Let $\mathcal{X}$ be a closed algebraic subset of $\mathbb{C}^{m}$ and
$f:\mathbb{C}^{m}\rightarrow\mathbb{C}$ be a polynomial. Either the image
$F(\mathcal{X})\subset\mathbb{C}$ is a finite set, or it is the complement of
a finite set. In the second case one says that f is dominating.

###  2.5 Singular sequences of normalized central configurations

We consider a solution $\mathcal{Q}=(\mathcal{Z},\mathcal{W})$ of (2.12), this
is a normalized central configuration. Let $\mathfrak{N}=N(N+1)/2$. Set

$\mathcal{Z}=(\mathcal{Z}_{1},\mathcal{Z}_{2},\cdots,\mathcal{Z}_{\mathfrak{N}})=(z_{1},z_{2},\cdots,z_{N},Z_{12},Z_{13},\cdots,Z_{(N-1)N})$,

$\mathcal{W}=(\mathcal{W}_{1},\mathcal{W}_{2},\cdots,\mathcal{W}_{\mathfrak{N}})=(w_{1},w_{2},\cdots,w_{N},W_{12},W_{13},\cdots,W_{(N-1)N})$.

Let $\|\mathcal{Z}\|=\max_{j=1,2,\cdots,\mathfrak{N}}|\mathcal{Z}_{j}|$ be the
modulus of the maximal component of the vector
$\mathcal{Z}\in\mathbb{C}^{\mathfrak{N}}$. Similarly, set
$\|\mathcal{W}\|=\max_{k=1,2,\cdots,\mathfrak{N}}|\mathcal{W}_{k}|$.

Consider a sequence $\mathcal{Q}^{(n)}$, $n=1,2,\cdots$, of normalized central
configurations. Extract a sub-sequence such that the maximal component of
$\mathcal{Z}^{(n)}$ is always the same, i.e.,
$\|\mathcal{Z}^{(n)}\|=|\mathcal{Z}^{(n)}_{j}|$ for a
$j\in\\{1,2,\cdots,\mathfrak{N}\\}$ that does not depend on $n$. Extract again
in such a way that the vector sequence
$\mathcal{Z}^{(n)}/\|\mathcal{Z}^{(n)}\|$ converges. Extract again in such a
way that there is similarly an integer $k\in\\{1,2,\cdots,\mathfrak{N}\\}$
such that $\|\mathcal{W}^{(n)}\|=|\mathcal{W}^{(n)}_{k}|$ for all $n$. Extract
a last time in such a way that the vector sequence
$\mathcal{W}^{(n)}/\|\mathcal{W}^{(n)}\|$ converges.

If the initial sequence is such that $\mathcal{Z}^{(n)}$ or
$\mathcal{W}^{(n)}$ is unbounded, so is the extracted sequence. Note that
$\|\mathcal{Z}^{(n)}\|$ and $\|\mathcal{W}^{(n)}\|$ are bounded away from
zero: if the first $N$ components of the vector $\mathcal{Z}^{(n)}$ or
$\|\mathcal{W}^{(n)}\|$ all go to zero, then the denominator $z_{12}=w_{12}$
of the component $Z_{12}=W_{12}$ go to zero and $\mathcal{Z}^{(n)}$ and
$\mathcal{W}^{(n)}$ are unbounded. There are two possibilities for the
extracted sub-sequences above:

  * •

$\mathcal{Z}^{(n)}$ and $\mathcal{W}^{(n)}$ are bounded,

  * •

at least one of $\mathcal{Z}^{(n)}$ and $\mathcal{W}^{(n)}$ is unbounded.

######  Definition 2.9 (Singular sequence)

Consider a sequence of normalized central configurations. A sub-sequence
extracted by the above process, in the unbounded case, is called a _singular
sequence_.

_Robert’s continuum_. Roberts found the following continuum of relative
equilibria for five vortices. The masses are
$m_{1}=m_{2}=m_{3}=m_{4}=2,m_{5}=-1$. The fifth vortex is at the origin
$z_{5}=w_{5}=0$, while the first four vortices form a rhombus

$(z_{1},z_{2},z_{3},z_{4})=(a,b,-a,-b),\
(w_{1},w_{2},w_{3},w_{4})=(a,-b,-a,b),\ $

where $a\in\mathbb{R}$ and $\textbf{i}b\in\mathbb{R}$ for the real
configurations. One can check that the configuration defined by the
coordinates above and the restriction
$r^{2}_{12}=r^{2}_{23}=r^{2}_{34}=r^{2}_{14}=a^{2}-b^{2}=1$ satisfies system
(2.12).

For the real configurations, when $a\to 0$, vertices 1, 3, and 5 collide; when
$b\to 0$, vertices 2, 4, and 5 collide. We get a real triple contact
“singularity”. For the complex configurations, when $a\to\infty$, we have two
choices: $b\sim a$ or $b\sim-a$. We get two other singularities at imaginary
infinity.

##  3 Tools to classify the singular sequences

######  Definition 3.1 (Notation of asymptotic estimates)

$a\sim b$

means $a/b\rightarrow 1$

$a\prec b$

means $a/b\rightarrow 0$

$a\preceq b$

means $a/b$ is bounded

$a\approx b$

means $a\preceq b$ and $a\succeq b$

######  Definition 3.2 (Strokes and circles.)

We pick a singular sequence. We write the indices of the bodies in a figure
and use two colors for edges and vertices.

The first color, the $z$-color, is used to mark the maximal order components
of

$\mathcal{Z}=(z_{1},z_{2},\cdots,z_{N},Z_{12},Z_{13},\cdots,Z_{(N-1)N})$.

They correspond to the components of the converging vector sequence
$\mathcal{Z}^{(n)}/\|\mathcal{Z}^{(n)}\|$ that do not tend to zero. We draw a
circle around the name of vertex j if the term $z^{(n)}_{j}$ is of maximal
order among all the components of $\mathcal{Z}^{(n)}$. We draw a stroke
between the names k and l if the term $Z^{(n)}_{kl}$ is of maximal order among
all the components of $\mathcal{Z}^{(n)}$.

###  3.1 Rules of colored diagram

The following rules mainly concern $z$-diagram, but they apply as well to the
$w$-diagram.

If there is a maximal order term in an equation, there should be another one.
This gives immediately the following Rule I.

Rule I

There is something at each end of any $z$-stroke: another $z$-stroke or/and a
$z$-circle drawn around the name of the vertex. A $z$-circle cannot be
isolated; there must be a $z$-stroke emanating from it. There is at least one
$z$-stroke in the $z$-diagram.

######  Definition 3.3 ($z$-close)

Consider a singular sequence. We say that bodies k and l are close in
$z$-coordinate, or $z$-close, or that $z_{k}$ and $z_{l}$ are close, if
$z^{(n)}_{kl}\prec\|\mathcal{Z}^{(n)}\|$.

The following statement is obvious.

Rule II

If bodies k and l are $z$-close, they are both $z$-circled or both not
$z$-circled.

######  Definition 3.4 (Isolated component)

An isolated component of the $z$-diagram is a subset of vertices such that no
$z$-stroke is joining a vertex of this subset to a vertex of the complement.

Rule III

The moment of vorticity of a set of bodies forming an isolated component of
the $z$-diagram is $z$-close to the origin.

Rule IV

Consider the $z$-diagram or an isolated component of it. If there is a
$z$-circled vertex, there is another one. The $z$-circled vertices can all be
$z$-close together only if the total vorticity of these vertices is zero.

######  Definition 3.5 (Maximal $z$-stroke)

Consider a $z$-stroke from vertex k to vertex l. We say it is a maximal
$z$-stroke if k and l are not $z$-close.

Rule V

There is at least one $z$-circle at certain end of any maximal $z$-stroke. As
a result, if an isolated component of the $z$-diagram has no $z$-circled
vertex, then it has no maximal $z$-stroke.

On the same diagram we also draw $w$-strokes and $w$-circles. Graphically we
use another color. The previous rules and definitions apply to $w$-strokes and
$w$-circles. What we will call simply the diagram is the superposition of the
$z$-diagram and the $w$-diagram. We will, for example, adapt Definition 3.4 of
an isolated component: a subset of bodies forms an isolated component of the
diagram if and only if it forms an isolated component of the $z$-diagram and
an isolated component of the $w$-diagram.

######  Definition 3.6 (Edges and strokes)

There is an edge between vertex k and vertex l if there is either a
$z$-stroke, or a $w$-stroke, or both. There are three types of edges,
$z$-edges, $w$-edges and $zw$-edges, and only two types of strokes,
represented with two different colors.

$1$$2$$1$$2$$1$$2$ Figure 1: A $z$-stroke, a $z$-stroke plus a $z$-stroke, a
$w$-stroke, forming respectively a $z$-edge, a $zw$-edge, a $w$-edge.

####  3.1.1 New normalization. Main estimates.

One does not change a central configuration by multiplying the $z$ coordinates
by $a\in\mathbb{C}\backslash\\{0\\}$ and the $w$ coordinates by $a^{-1}$. Our
diagram is invariant by such an operation, as it considers the $z$-coordinates
and the $z$-coordinates separately.

We used the normalization $z_{12}=w_{12}$ in the previous considerations. In
the following we will normalize instead with
$\|\mathcal{Z}\|=\|\mathcal{W}\|$. We start with a central configuration
normalized with the condition $z_{12}=w_{12}$, then multiply the
$z$-coordinates by $a>0$, the $w$-coordinates by $a^{-1}$, in such a way that
the maximal component of $\mathcal{Z}$ and the maximal component of
$\mathcal{W}$ have the same modulus, i.e., $\|\mathcal{Z}\|=\|\mathcal{W}\|$.

A singular sequence was defined by the condition either
$\|\mathcal{Z}^{(n)}\|\rightarrow\infty$ or
$\|\mathcal{W}^{(n)}\|\rightarrow\infty$. We also remarked that both
$\|\mathcal{Z}^{(n)}\|$ and $\|\mathcal{W}^{(n)}\|$ were bounded away from
zero. With the new normalization, a singular sequence is simply characterized
by $\|\mathcal{Z}^{(n)}\|=\|\mathcal{W}^{(n)}\|\rightarrow\infty$. From now on
we only discuss singular sequences.

Set $\|\mathcal{Z}^{(n)}\|=\|\mathcal{W}^{(n)}\|=1/\epsilon^{2}$, then
$\epsilon\rightarrow 0$.

######  Proposition 3.1 (Estimate)

For any $(k,l)$, $1\leq k<l\leq N$, we have $\epsilon^{2}\preceq
z_{kl}\preceq\epsilon^{-2}$, $\epsilon^{2}\preceq w_{kl}\preceq\epsilon^{-2}$
and $\epsilon^{2}\preceq r_{kl}\preceq\epsilon^{-2}$.

There is a $z$-stroke between k and l if and only if
$w_{kl}\approx\epsilon^{2}$, then $r_{kl}\preceq 1$.

There is a maximal $z$-stroke between k and l if and only if
$z_{kl}\approx\epsilon^{-2},w_{kl}\approx\epsilon^{2}$, then $r_{kl}\preceq
1$.

There is a $z$-edge between k and l if and only if
$z_{kl}\succ\epsilon^{2},w_{kl}\approx\epsilon^{2}$, then $\epsilon^{2}\prec
r_{kl}\preceq 1$.

There is a maximal $z$-edge between k and l if and only if
$z_{kl}\approx\epsilon^{-2},w_{kl}\approx\epsilon^{2}$, then $r_{kl}\approx
1$.

There is a $zw$-edge between k and l if and only if
$z_{kl},w_{kl}\approx\epsilon^{2}$, this can be characterized as
$r_{kl}\approx\epsilon^{2}$.

###### Remark 3.1

By the estimates above, the strokes in a $zw$-edge are not maximal. A maximal
$z$-stroke is exactly a maximal $z$-edge.

_Circling method_. Estimate show that in all the cases where there is a
$z$-stroke between vertices k and l, there bodies are close in $w$-coordinate.
Then Rule II applies to the $w$-diagram. Vertices k and l are both $w$-circled
or both not $w$-circled. One vertex isolated in $w$-diagram can not be
$w$-circled. If there is one $w$-circle, there is another $w$-circle.

Rule VI

If there are two consecutive $z$-stroke, there is a third $z$-stroke closing
the triangle.

The Roberts’ continuum example produces the following collection of diagrams
from left to right: $a\to 0$ and 1, 3, 5 collide; $b\to 0$ and 2, 4, 5
collide; $a\to\infty$ and $b\sim a$; $a\to\infty$ and $b\sim-a$, (see Figure
2).

Figure 2: Roberts’ continuum at real triple contact and infinity

##  4 The possible diagrams

In this section, under the assumption that the total vorticity is nonvanishing
and that each vorticity is nonzero, we derive a list of problematic diagrams.
This list of twenty-nine diagrams is in Figure 16.

As in the case of the 5-body problem [1] in celestial mechanics, we divide
possible diagrams into groups according to the maximal number of strokes from
a bicolored vertex. During the analysis of all the possibilities we rule some
of them out immediately, some with further arguments. The ones we cannot
exclude without further hypotheses on the vorticities are incorporated into
the list of twenty-nine in Figure 16.

We list some useful results.

###### Proposition 4.1

[14] Suppose a diagram has two $z$-circled vertices (say 1 and 2) which are
also $z$-close, if none of all the other vertices is $z$-close with them, then
$\Gamma_{1}+\Gamma_{2}\neq 0$ and
$\overline{\Lambda}z_{12}w_{12}\sim\frac{1}{\Gamma_{1}+\Gamma_{2}}$. In
particular, vertices 1 and 2 cannot form a $z$-stroke.

###### Proposition 4.2

[14] Consider an isolated component of the $z$-diagram. If the $z$-circled
vertices are all $z$-close (for example, connected by $w$-strokes), then the
total vorticity of these vertices is zero.

###### Proposition 4.3

[14] Suppose a diagram has an isolated $z$-color triangle, and none of
vertices (say 1,2,3) of it are $z$-circled, then
$L_{123}=\Gamma_{1}\Gamma_{2}+\Gamma_{2}\Gamma_{3}+\Gamma_{3}\Gamma_{1}=0$.

###### Proposition 4.4

[14] Suppose a fully $z$-stroked sub-diagram with four vertices exists in
isolation in a diagram, and none of vertices (say 1,2,3,4) of it are
$z$-circled, then

$L_{1234}=\Gamma_{1}\Gamma_{2}+\Gamma_{2}\Gamma_{3}+\Gamma_{3}\Gamma_{1}+\Gamma_{4}(\Gamma_{1}+\Gamma_{2}+\Gamma_{3})=0.$
(4.13)

###### Proposition 4.5

Suppose a diagram has one $z$-stroke that is isolated in the $z$-diagram and
the two ends are $z$-circled. Call 1 and 2 the ends of this $z$-stoke. Suppose
that there is no other $z$-circle in the diagram. Then
$\Gamma_{1}+\Gamma_{2}\neq 0$. The diagram forces $\Lambda=\pm 1$ or
$\pm\textbf{i}$. Furthermore,

  * •

If $\Lambda=\pm 1$, we have $\sum_{j=3}^{N}\Gamma_{j}=0$;

  * •

If $\Lambda=\pm\textbf{i}$, we have $L=0$ and
$\Gamma_{1}\Gamma_{2}=L_{3...N}$.

###### Proof 1

First, if $\Gamma_{1}+\Gamma_{2}=0$, then two $z$-circled vertices 1 and 2 are
$z$-close since
$\Gamma_{1}z_{1}+\Gamma_{2}z_{2}=(\Gamma_{1}+\Gamma_{2})z_{1}+\Gamma_{2}(z_{2}-z_{1})<\epsilon^{-2}$.
However, this contradicts Proposition 4.1. Thus $\Gamma_{1}+\Gamma_{2}\neq 0$.

Without loss of generality, assume $z_{1}\sim-\Gamma_{2}a\epsilon^{-2}$ and
$z_{2}\sim\Gamma_{1}a\epsilon^{-2}$, then

$z_{12}\sim(\Gamma_{1}+\Gamma_{2})a\epsilon^{-2},\
\frac{1}{z_{2}}-\frac{1}{z_{1}}\sim(\frac{1}{\Gamma_{1}}+\frac{1}{\Gamma_{2}})\frac{\epsilon^{2}}{a}.$

The equations (2.12) gives

$\begin{array}[]{c}\overline{\Lambda}w_{12}\sim(\Gamma_{1}+\Gamma_{2})W_{12}+\sum_{j=3}^{N}\Gamma_{j}(\frac{1}{z_{j2}}-\frac{1}{z_{j1}})\cr\Lambda
z_{2}\sim\Gamma_{1}Z_{12}.\end{array}{}$ (4.14)

The second equation of (4.14) gives $w_{12}\sim\frac{\epsilon^{2}}{a\Lambda}$.
Note that
$\frac{1}{z_{j2}}-\frac{1}{z_{j1}}\sim\frac{1}{z_{2}}-\frac{1}{z_{1}}$ for any
$j>2$ and that $W_{12}=\frac{1}{z_{12}}$. The first equation of (4.14) leads
to

${\overline{\Lambda}}/{\Lambda}=1+\sum_{j=3}^{N}\Gamma_{j}(\frac{1}{\Gamma_{1}}+\frac{1}{\Gamma_{2}}).$
(4.15)

It follows that $\Lambda=\pm 1$ or $\pm\textbf{i}$.

If $\Lambda=\pm 1$, we have

$0=\sum_{j=3}^{N}\Gamma_{j}(\frac{1}{\Gamma_{1}}+\frac{1}{\Gamma_{2}}),\
\Rightarrow\ \sum_{j=3}^{N}\Gamma_{j}=0.$

If $\Lambda=\pm\textbf{i}$, we have

$-2=\sum_{j=3}^{N}\Gamma_{j}(\frac{1}{\Gamma_{1}}+\frac{1}{\Gamma_{2}}),\
L=0,\ \Rightarrow L=0,\ \Gamma_{1}\Gamma_{2}=L_{3...N}.$

Similarly, we have the following result.

###### Proposition 4.6

Suppose a diagram has a triangle of $z$-strokes that is isolated in the
$z$-diagram and the two of the vertices of the triangle are $z$-circled. Call
2, and 3 the two $z$-circled vertices and 1 the other vertex. Suppose that
there is no other $z$-circle in the diagram. Then $\Gamma_{2}+\Gamma_{3}\neq
0$. The diagram forces $\Lambda=\pm 1$ or $\pm\textbf{i}$. Furthermore,

  * •

If $\Lambda=\pm 1$, we have $\Gamma_{1}=\sum_{j=4}^{N}\Gamma_{j}$;

  * •

If $\Lambda=\pm\textbf{i}$, we have $L=0$ and
$L_{123}=L_{4...N}+\Gamma_{1}(\sum_{j=4}^{N}\Gamma_{j})$.

###### Proof 2

Similarly as in the above case, we have $\Gamma_{2}+\Gamma_{3}\neq 0$. Note
that

$\Gamma_{2}z_{2}+\Gamma_{3}z_{3}\prec\epsilon^{-2},\ \Lambda
z_{1}\sim\Gamma_{2}Z_{21}+\Gamma_{3}Z_{31}\prec\epsilon^{-2}.$

Without loss of generality, assume

$z_{2}\sim-\Gamma_{3}a\epsilon^{-2},\ z_{3}\sim\Gamma_{2}a\epsilon^{-2},\
Z_{21}\sim-\Gamma_{3}b\epsilon^{-2},\ Z_{31}\sim\Gamma_{2}b\epsilon^{-2}.$

then

$\displaystyle z_{23}\sim(\Gamma_{2}+\Gamma_{3})a\epsilon^{-2},\
\frac{1}{z_{3}}-\frac{1}{z_{2}}\sim(\frac{1}{\Gamma_{2}}+\frac{1}{\Gamma_{3}})\frac{\epsilon^{2}}{a},$
$\displaystyle
Z_{23}=\frac{1}{w_{23}}=\frac{1}{1/Z_{21}+1/Z_{13}}\sim-b\frac{\Gamma_{2}\Gamma_{3}}{(\Gamma_{2}+\Gamma_{3})}\epsilon^{-2}.$

Then similar to the above case, we have

$\displaystyle\overline{\Lambda}w_{23}\sim(\Gamma_{2}+\Gamma_{3})W_{23}+\sum_{j\neq
2,3}^{N}\Gamma_{j}(\frac{1}{z_{3}}-\frac{1}{z_{2}})$ $\displaystyle\Lambda
z_{23}\sim(\Gamma_{2}+\Gamma_{3})Z_{23}+\Gamma_{1}(Z_{13}-Z_{12}).$

Short computation reduces the two equations to

$\displaystyle-\overline{\Lambda}\frac{a}{b}=\frac{\Gamma_{2}\Gamma_{3}}{\Gamma_{2}+\Gamma_{3}}(1+\sum_{j\neq
2,3}\Gamma_{j}\frac{\Gamma_{2}+\Gamma_{3}}{\Gamma_{2}\Gamma_{3}}),$
$\displaystyle-\Lambda\frac{a}{b}=\frac{L_{123}}{\Gamma_{2}+\Gamma_{3}}.$

Then we obtain

$\frac{\overline{\Lambda}}{\Lambda}L_{123}=\Gamma_{2}\Gamma_{3}+(\Gamma_{2}+\Gamma_{3})\sum_{j\neq
2,3}\Gamma_{j}.$

It follows that $\Lambda=\pm 1$ or $\pm\textbf{i}$.

If $\Lambda=\pm 1$, we have $\Gamma_{1}=\sum_{j=4}^{N}\Gamma_{j}$. If
$\Lambda=\pm\textbf{i}$, we have $L=0$ and

$L=0,\ -L_{123}=\Gamma_{2}\Gamma_{3}+(\Gamma_{2}+\Gamma_{3})\sum_{j\neq
2,3}\Gamma_{j},$

which is equivalent to
$L_{123}=L_{4...N}+\Gamma_{1}(\sum_{j=4}^{N}\Gamma_{j}),\ L=0.$

We call a _bicolored vertex_ of the diagram a vertex which connects at least a
stroke of $z$-color with at least a stroke of $w$-color. The number of edges
from a bicolored vertex is at least 1 and at most $N-1$. The number of strokes
from a bicolored vertex is at least 2 and at most $2N-2$. Given a diagram, we
define $C$ as the maximal number of strokes from a bicolored vertex. We use
this number to classify all possible diagrams.

Recall that the $z$-diagram indicates the maximal terms among a finite set of
terms. It is nonempty. If there is a circle, there is an edge of the same
color emanating from it. So there is at least a $z$-stroke, and similarly, at
least a $w$-stroke.

###  4.1 No bicolored vertex

Figure 3: No bicolored vertex

There is at least one isolated edge, which is not a $zw$-edge. Let us say it
is a $z$-edge. The complement has 3 bodies. There three can have one or three
$w$-edges according to Rule VI.

For one $w$-edge, the attached bodies have to be $w$-circled by Rule I. This
is the first diagram in Figure 3.

For three $w$-edges, the three edges form a triangle. There are three
possibilities for the number of $w$-circled vertices: it is either zero, or
two or three (one is not possible by Rule III.) They constitute the last three
diagrams in Figure 3.

The case with zero $w$-circles is impossible. Applying Proposition 4.5 to
vertex 1 and vertex 2, we see $\Lambda=\pm 1$ or $\pm\textbf{i}$. The mass
constraints are

  * •

$\Gamma_{3}+\Gamma_{4}+\Gamma_{5}=0,\Gamma_{1}+\Gamma_{2}\neq 0$, if
$\Lambda=\pm 1$;

  * •

$\Gamma_{1}\Gamma_{2}=L_{345},L=0,\Gamma_{1}+\Gamma_{2}\neq 0$, if
$\Lambda=\pm\textbf{i}$.

Applying Proposition 4.3 to vertices 3, 4 and 5 we have $L_{345}=0$. This
constraint is not compatible with neither one of the above two constraints.
Hence, this diagram is impossible.

Hence, we could not exclude the first, the third and the last diagrams in
Figure 3. There are three possible diagrams.

###  4.2 $C=2$

There are two cases: a $zw$-edge exists or not.

Figure 4: $C=2$

If it is present, it is isolated. Let us say, vertex 1 and vertex 2 are
connected by one $zw$-edge. Among vertices 3, 4, and 5, there exist both $z$
and $w$-circle. If none of the three vertices is $z$-circled, we have
$\Gamma_{1}+\Gamma_{2}=0$ by Proposition 4.2. On the other hand, since
vertices 3, 4, and 5 are not $z$-circled, they are not $z$-close to vertex 1
and vertex 2 . Then Proposition 4.1 implies that $\Gamma_{1}+\Gamma_{2}\neq
0$. This is a contradiction.

Then Rule I implies that there is at least one $z$-stroke and one $w$-stroke
among the cluster of vertices 3, 4, and 5. There are two possibilities:
whether there is another $zw$-edge or not.

If another $zw$-edge is present, then it is again isolated. This is the first
diagram in Figure 4.

If another $zw$-edge is not present, there is at least one edge in both color.
By the circling method, the adjacent vertex is $z$\- and $w$-circled. By the
Estimate, the $z$-edge implies the two attached vertices are $w$-close. Then
the two ends are both $w$-circled by Rule II. Thus, all three vertices are
$z$\- and $w$-circled. Then there is more strokes in the diagram. This is a
contradiction.

If there is no $zw$-edge, there are adjacent $z$-edges and $w$-edges. From any
such adjacency there is no other edge. Suppose that vertex 1 connects with
vertex 4 by $w$-edges and connects with 2 by $z$-edges. The circling method
implies that 1 is $z$\- and $w$-circled, 2 is $w$-circled and 4 is
$z$-circled. The color of 2 and 4 forces the color of edges from the circle.
If the two edgers go to the same vertex, we get the diagram corresponding to
Roberts’ continuum at infinity, shown as the second in Figure 4.

If the two edgers go to the different vertices, the circling method demands a
cycle with alternating colors, which is impossible since we have only five
edges.

Hence, there are two possible diagrams, see Figure 4.

###  4.3 $C=3$

Consider a bicolored vertex with three strokes. It is Y-shaped or connects a
single stroke to a $zw$-edge.

Figure 5: $C=3$

We start with the Y-shaped case. Let us say vertex 1 connects with vertex 2
and vertex 3 by $z$-edges, and connects with vertex 4 by a $w$-edge. There is
a $z_{23}$-stroke by Rule VI. The circling method implies that the vertices 1,
2 and vertex 3 are all $w$-circled, see Figure 5. Then there is $w$-stroke
emanating from 2 and vertex 3. The $w$-stroke may go to vertex 4, vertex 5, or
it is a $w_{23}$-stroke.

If one $w$-stroke goes from vertex 2 to vertex 4, then there is extra
$w_{12}$-stroke by Rule VI, which contradict with $C=3$. If all two
$w$-strokes go to vertex 5, then Rule VI implies the existence of
$w_{23}$-stroke. This is again a contradiction with $C=3$.

If the $w$-strokes emanating from 2 and vertex 3 is just $w_{23}$-stroke. Then
we have an $zw$-edge between 2, and 3. Then it is not necessary to discuss the
$zw$-edge case. Then we consider the vertex 5 . It is connected with the
previous four vertices or isolated.

If the diagram is connected, vertex 5 can only connects with vertex 4 by a
$z$-edge (other cases is not possible by Rule VI). Then the circling method
implies that vertex 5 is $w$-circled. Then there is $w$-stroke emanating from
5. This is a contradiction.

If vertex 5 is isolated. Then the circling method implies that all vertices
except vertex 5 are $w$-circled. Only 2 and vertex 3 can be $z$-circled, and
they are both $z$-circled or both not $z$-circled by Rule IV, see Figure 5.

If both vertex 2 and vertex 3 are not $z$-circled, then we have
$\Gamma_{2}+\Gamma_{3}=0$ and $L_{123}=0$ by Proposition 4.2 and Proposition
4.3. This is a contradiction since

$L_{123}=\Gamma_{1}(\Gamma_{2}+\Gamma_{3})+\Gamma_{2}\Gamma_{3}.$

If both vertex 2 and vertex 3 are $z$-circled, then Proposition 4.2 implies
that $\Gamma_{2}+\Gamma_{3}=0$. On the other hand, Proposition 4.1 implies
that $\Gamma_{2}+\Gamma_{3}\neq 0$. This is a contradiction.

Thus, the two diagrams in Figure 5 are both excluded. we have no possible
diagram.

###  4.4 $C=4$

There are five cases: two $zw$-edges, only one $zw$-edge and one edge of each
color, only one $zw$-edge and two edge of the same color, one $z$-edge and
three $w$-edges, or two edges of each color emanating from the same vertex.

Figure 6: $C=4$, two $zw$-edges

#### 4.4.1

Suppose that there are two $zw$-edges emanating from, e.g., vertex 1 as on the
first diagram in Figure 6.

We get a fully $zw$-edged triangle by Rule VI. This triangle is isolated since
$C=4$. Since vertices 1, 2, and 3 are $z$-close and $w$-close, if one of them
is circled in some color, all of them will be circled in the same color. Thus,
the first three vertices may be all $z$-circled, all $z$-and $w$-circled, or
all not circled.

If vertices 1, 2, and 3 are $z$-circled but not $w$-circled, we have
$\Gamma_{1}+\Gamma_{2}+\Gamma_{3}=0$ and $L_{123}=0$ by Proposition 4.2 and
Proposition 4.3. This is a contradiction since
$(\Gamma_{1}+\Gamma_{2}+\Gamma_{3})^{2}-2L_{123}\neq 0$. Then the first three
vertices can only be all $z$-and $w$-circled, or all not circled.

The other two vertices can be disconnected, connected by one , e.g., $z$-edge,
or by one $zw$-edge.

If vertex 4 and vertex 5 are connected by one $zw$-edge, then there are two
possibilities: whether the first three vertices are all $z$-and $w$-circled,
or all not circled. In the first case, we have
$\Gamma_{1}+\Gamma_{2}+\Gamma_{3}=0$ and $\Gamma_{4}+\Gamma_{5}=0$ by
Proposition 4.2. Then the total vorticity is zero, which contradicts with our
assumption on the vorticity. In the second case, we have
$\Gamma_{4}+\Gamma_{5}=0$ by Proposition 4.2. On the other hand, since
vertices 1, 2, and 3 are not $z$-circled, they are not $z$-close to vertex 4
and vertex 5 . Then Proposition 4.1 implies that $\Gamma_{4}+\Gamma_{5}\neq
0$. This is a contradiction.

Then we have the four possibilities, see Figure 6. The third diagram is
impossible. The argument is similar to that corresponding to the second
diagram in Figure 3.

#### 4.4.2

Suppose that there are one $zw$-edges and one edge of each color emanating
from vertex 1, as in the first digram of Figure 7. We have a “butterfly
diagram” by Rule VI. Note that no more strokes can emanating from vertex 1 and
vertex 4 since $C=4$. If there are more strokes from vertex 2, it can not goes
to vertex 3, since it implies one more stroke emanating from vertex 1.
Similarly, there is no $z_{35}$-stroke or $w_{25}$-stroke. Then between vertex
5 and the first four vertices, there can have no edge, one $z_{25}$-stroke, or
one $z_{25}$-stroke and one $w_{35}$-stroke.

For the disconnected diagram, vertex 1 and vertex 4 can not be circled.
Otherwise, the circling method implies either vertex 2 is be $z$-circled or 3
is $w$-circled, there should be stroke emanating from vertex 2 or vertex 3 .
This is a contradiction. Then vertex 2 can not be circled, otherwise, vertex 3
is of the same color by Rule IV. This is a contradiction. Hence, there is no
circle in the diagram and this is the first diagram in Figure 7.

If there is only $z_{25}$-stroke, then the circling method implies that
vertices 1, 2, 4, and 5 are $z$-circled. Note that vertex 3 and vertex 5 can
not be $w$-circled, otherwise there are extra $w$-strokes. Then vertices 1, 2,
and 4 are not $w$-circled by the circling method. Note that vertex 3 must be
$z$-circled, otherwise, we have $L_{124}=0$ and $\Gamma_{1}+\Gamma_{4}=0$.
This is a contradiction. Then we have the second diagram in Figure 7.

If there are $z_{25}$-stroke and $w_{35}$-stroke. Then vertex 5 is $z$-and
$w$-circled. By the circling method, all vertices are $z$-and $w$-circled.
This the third diagram in Figure 7.

Figure 7: $C=4$, one $zw$-edges

#### 4.4.3

Suppose that there are one $zw$-edges and two $z$-edges emanating from vertex
1, as in the fourth diagram in Figure 7. Then there are $z_{23}$-stroke,
$z_{24}$-stroke and $z_{34}$-stroke by Rule VI. Note that vertex 1 is
$z$-circled, then the circling method implies all vertices except possibly
vertex 5 are $w$-circled. Then there is $w$-stroke emanating from 3 and vertex
4. The $w$-stroke may go to vertex 5, or it is a $w_{34}$-stroke.

If all two $w$-strokes go to vertex 5, then Rule VI implies the existence of
$w_{34}$-stroke, which contradicts with $C=4$. Then the $w$-strokes from 3 and
vertex 4 is $w_{34}$-stroke, and vertex 5 is disconnected.

Consider the circling the the diagram. We have three different cases: all
vertices are not $z$-circled, only two of the first four vertices, say, vertex
1 and vertex 2 are $z$-circled, or all the first four vertices are
$z$-circled.

If none of the first four vertices is $z$-circled, then we have
$\Gamma_{1234}=0$ and $L_{1234}=0$ by Proposition 4.2 and Proposition 4.4.
This is a contradiction.

If only vertex 1 and vertex 2 are $z$-circled, we have
$\Gamma_{1}+\Gamma_{2}=0$ by Proposition 4.2. On the other hand, since
vertices 3, 4, and 5 are not $z$-circled, they are not $z$-close to vertex 1
and vertex 2 . Then Proposition 4.1 implies that $\Gamma_{1}+\Gamma_{2}\neq
0$. This is a contradiction.

Hence, all the first four vertices are $z$-circled. This gives the fourth
diagram in Figure 7.

#### 4.4.4

Suppose that from vertex 1 there are three $w$-edges go to vertex 2, vertex 3
and vertex 4 respectively and one $z$-edges goes to vertex 5 . There is a
$w$-stroke between any pair of $\\{\textbf{1, 2, 3, 4}\\}$ by Rule VI, and the
four vertices are all $z$-circled.

Then there are $z$-strokes emanating from vertex 2, vertex 3 and vertex 4 .
None of them can go to vertex 5 by Rule VI. Then there are $z_{23}$-,
$z_{24}$\- and $z_{34}$-strokes. This contradict with $C=4$. Hence, there is
no possible diagram in this case.

Figure 8: $C=4$, no $zw$-edges

#### 4.4.5

Suppose that there are two $w$-edges and two $z$-edges emanating from vertex
1, with numeration as in the first digram of Figure 8. By Rule VI, there is
$w_{23}$\- and $z_{45}$-stroke. Thus, we have two attached triangles. By Rule
VI, if there is more stroke, it must be $z_{23}$\- and/or $w_{45}$-stroke.

Case I: If there is no more stroke, vertex 2 and vertex 3 can only be
$w$-circled, and vertex 4 and vertex 5 can only be $z$-circled. Then vertex 1
can not be circled, otherwise, the circling method would lead to
contradiction. Then we have the first three diagrams in Figure 8.

Case II: If there is only $z_{23}$-stroke, the circling method implies vertex
1, vertex 2 and vertex 3 are $z$-circled. Note that only vertex 2 and vertex 3
can be $w$-circled. There are two possibilities: whether vertex 2 and vertex 3
are both $w$-circled or both not $w$-circled.

If vertex 2 and vertex 3 are both $w$-circled, we have
$\Gamma_{2}+\Gamma_{3}=0$ by Proposition 4.2. On the other hand, since
vertices 1, 4, and 5 are not $w$-circled, they are not $w$-close to vertex 2
and vertex 3 . Then Proposition 4.1 implies that $\Gamma_{2}+\Gamma_{3}\neq
0$. This is a contradiction.

If vertex 2 and vertex 3 are both not $w$-circled, we have
$\Gamma_{2}+\Gamma_{3}=0$ and $L_{123}=0$ by Proposition 4.2 and Proposition
4.3. This is a contradiction. Hence there is no possible diagram in this case.

Case III: If there are $z_{23}$\- and $w_{45}$-strokes, the circling method
implies vertex 2 and vertex 3 are $z$-circled, vertex 4 and vertex 5 are
$w$-circled, and vertex 1 is $z$\- and $w$-circled. Consider the circling of
vertex 2 and vertex 3, there are three possibilities: there are zero, one, or
two $w$-circles.

If both vertex 2 and vertex 3 are not $w$-circled, then the upper triangle,
which is isolated in $w$-diagram, has only one $w$-circle, namely, vertex 1.
This is a contradiction by Rule IV.

If both vertex 2 and vertex 3 are $w$-circled, then we have
$\Gamma_{2}+\Gamma_{3}=0$ by Proposition 4.2. Consider the upper triangle in
$w$-diagram. We have

$\Gamma_{1}w_{1}+\Gamma_{2}w_{2}+\Gamma_{3}w_{3}=\Gamma_{1}w_{1}+(\Gamma_{2}+\Gamma_{3})w_{2}\approx\Gamma_{1}w_{1}<\epsilon^{-2}.$

Then we have $\Gamma_{1}=0$, which is a contradiction.

Repeat the argument for vertex 4 and vertex 5. Hence, the only possible
diagram in this case is that only one of vertex 2 and vertex 3 is $w$-circled
and only one of vertex 4 and vertex 5 is $z$-circled. This is the last diagram
in Figure 8.

However, this diagram is impossible. Vertex 2 and vertex 3 are $w$-close, so
they should be both $w$-circled by Rule II.

In summary, among the twelve diagrams in Figure 6, 7 and 8, we have excluded
the third one in Figure 6 and the fourth one in Figure 8. We have ten possible
diagrams.

###  4.5 $C=5$

There are three cases: two $zw$-edges, one $zw$-edge with one $z$-edge and two
$w$-edges, one $zw$-edge with three $z$-edges.

Figure 9: $C=5$, two $zw$-edges

#### 4.5.1

Suppose that there are two $zw$-edges and one $z$-edges emanating from vertex
1, with numeration as in the first digram of Figure 9. Rule VI implies the
existence of $zw_{23}$-, $z_{24}$\- and $z_{34}$-edges. Then vertex 5 can be
either disconnected or connects with vertex 4 by one $w$-edge, otherwise, it
would contradicts with $C=5$.

For the disconnected diagram, note that any of the connected four vertices can
not be $w$-circled. Otherwise the circling method implies vertex 4 is
$w$-circled, which is a contradiction. There are three cases: none of the
vertices are circled, all four vertices except vertex 4 are $z$-circled, or
all four vertices are $z$-circled.

If none of the vertices are circled, by Proposition 4.3 we have $L_{123}$
since vertices 1, 2, and 3 form a triangle with no $w$-circled attached.
Similarly, we have $L_{1234}=0$ by Proposition 4.4. This is a contradiction
since

$L_{1234}=L_{123}+\Gamma_{4}(\Gamma_{1}+\Gamma_{2}+\Gamma_{3}),\
(\Gamma_{1}+\Gamma_{2}+\Gamma_{3})^{2}-2L_{123}\neq 0.$

If only vertices 1, 2, and 3 are $z$-circled, we also have $L_{123}$. By
Proposition 4.2, we have $\Gamma_{1}+\Gamma_{2}+\Gamma_{3}=0$. This is a
contradiction.

Hence, we have only one possible diagram for the disconnected diagram, and it
is the first in Figure 9.

For the connected diagram with $w_{45}$-edge, the circling method implies that
all vertices are $w$-circled. Note that vertex 4 and vertex 5 can not be
$z$-circled. Then we have two cases: all of the five vertices are not
$z$-circled, or vertices 1, 2, and 3 are $z$-circled.

If all of the five vertices are not $z$-circled, we have
$\Gamma_{1}+\Gamma_{2}+\Gamma_{3}=0$ by Proposition 4.2 and $L_{1234}=0$ by
Proposition 4.4. This is a contradiction.

Hence, we have only one possible diagram for the connected diagram, and it is
the second in Figure 9.

#### 4.5.2

Suppose that there are one $zw$-edge, one $z$-edge, and two $w$-edges
emanating from vertex 1, with numeration as in the first digram of Figure 10.
Rule VI implies that vertices 1, 2, 3, and 4 are fully $w$-stroked, and that
vertices 1, 3, and 5 are fully $z$-stroked. There is no more stroke emanating
from vertices 1, 3, and 5, since that would contradicts with $C=5$. There are
two cases, $z_{24}$-stroke is present or not.

If it is not present, vertex 2 and vertex 4 can only be $w$-circled, and
vertex 5 can only be $z$-circled. Then vertex 1 and vertex 3 can not be
circled, otherwise, the circling method would lead to contradiction. Then we
have the first two diagrams in Figure 10 by Rule IV.

If $z_{24}$-stroke is present, then vertices 1, 2, 3, and 4 are all
$z$-circled. Vertex 1 and vertex 3 can not be $w$-circled, which would lead to
vertex 5 also $w$-circled and a contradiction. Vertex 2 and vertex 4 are
$w$-close, so they are both $w$-circled or both not $w$-circled. If they are
both $w$-circled, then Proposition 4.2 implies that $\Gamma_{2}+\Gamma_{4}=0$.
On the other hand, Proposition 4.1 implies that $\Gamma_{2}+\Gamma_{4}\neq 0$.
This is a contradiction.

Then according to whether vertex 5 is $z$-circled or not, we have two cases,
which are the last two diagrams in Figure 10.

The third diagram in Figure 10 is impossible. We would have $L_{1234}=0$ and
$\sum_{j=1}^{4}\Gamma_{j}=0$ by Proposition 4.4 and Proposition 4.2. Hence, we
have only one possible diagram if $z_{24}$-stroke is present, and it is the
last diagram in Figure 10.

Figure 10: $C=5$, one $zw$-edges

#### 4.5.3

Suppose that there are one $zw$-edge and three $w$-edges emanating from vertex
1 . Let us say, the $zw$-edge goes to vertex 2 and the other edges go to the
other three vertices. Rule VI implies that vertices 1, 2, 3, 4, 5 are fully
$w$-stroked. The circling method implies that all vertices are $z$-circled.
Then there are $z$-strokes emanating from vertices 3, 4, and 5. Since $C=5$ at
vertex 1 and vertex 2, the $z$-strokes from vertices 3, 4, and 5 must go to
vertices 3, 4, and 5. By Rule VI, they form a triangle of $z$-strokes. This is
a contradiction with $C=5$.

In summary, among the six diagrams in Figure 9 and 10, we have excluded the
third one in Figure 10. We have five possible diagrams.

###  4.6 $C=6$

There are three cases: three $zw$-edges, two $zw$-edge with one edge in each
color, two $zw$-edge with two $z$-edge.

Figure 11: $C=6$, three $zw$-edges

#### 4.6.1

Suppose that there are three $zw$-edges emanating from vertex 1, with
numeration as in the first digram of Figure 11. Rule VI implies that the
vertices 1, 2, 3, and 4 are fully $zw$-edged. Then vertex 5 must be
disconnected since $C=6$. The first four vertices can be circled in the same
way by the circling method.

If all of the first four vertices are $z$-circled but not $w$-circled, then we
would have we would have $L_{1234}=0$ and $\sum_{j=1}^{4}\Gamma_{j}=0$ by
Proposition 4.4 and Proposition 4.2. This is one contradiction.

Hence, we have two possible diagrams if there are three $zw$-edges, which are
the two diagrams in Figure 11.

#### 4.6.2

Suppose that there are two $zw$-edges and two $z$-edges emanating from vertex
1, with numeration as in the first digram of Figure 12. Rule VI implies the
existence of $z_{25}$-, $z_{35}$-, $z_{24}$-, $z_{45}$\- and $z_{34}$-strokes,
and that the vertices 1, 2, and 3 are fully $zw$-edged. If there is more
stroke, it must be $w_{45}$-stroke since $C=6$.

If there is no more stroke, then the vertices can only be $z$-circled. There
are two cases, either vertices 1, 2, and 3 are $z$-circled or not.

If vertices 1, 2, and 3 are $z$-circled, then either vertex 4 or vertex 5 must
also be $z$-circled. Otherwise, we have $\Gamma_{1}+\Gamma_{2}+\Gamma_{3}=0$
and $L_{123}=0$ by Proposition 4.4 and Proposition 4.2. This is one
contradiction. Then we have the first two diagrams in Figure 12.

If vertices 1, 2, and 3 are not $z$-circled, then vertex 4 and vertex 5 are
both $z$-circled or both not $z$-circled. Then we have the second two diagrams
in Figure 12.

If $w_{45}$-stroke is present, then the circling method implies that all
vertices are $w$-circled. Then we have $\Gamma_{4}+\Gamma_{5}=0$ and
$\Gamma_{1}+\Gamma_{2}+\Gamma_{3}=0$ by Proposition 4.2, which contradicts
with our assumption that the total vorticity is not zero.

#### 4.6.3

Suppose that there are two $zw$-edges and one edge of each color emanating
from vertex 1, with numeration as in the first digram of Figure 13. Rule VI
implies the existence of $w_{25}$-, $w_{35}$-, $z_{24}$-, and
$z_{34}$-strokes, and that the vertices 1, 2, and 3 are fully $zw$-edged. If
there is more stroke, it must be $w_{45}$-stroke, but it would lead to extra
stroke emanating from vertex 1, which contradicts with $C=6$.

Vertex 4 can only be $w$-circled, and vertex 4 can only be $z$-circled. Then
vertices 1, 2 and 3 can not be circled, otherwise, the circling method would
lead to contradiction. Then we have the diagram in Figure 13 by Rule IV.

In summary, we have seven possible diagrams, see Figure 11, 12, and 13.

Figure 12: $C=6$, two $zw$-edges, case 1 Figure 13: $C=6$, two $zw$-edges,
case 2

###  4.7 $C=7$

Suppose that there are three $zw$-edges and one $z$-edge emanating from vertex
1, with numeration as in the digram of Figure 14. Rule VI implies that the
vertices 1, 2, 3, and 4 are fully $zw$-edged and that vertex 5 connects with
the first four vertices by one $z$-edge. There is no more stroke since $C=7$.
If any vertex is $w$-circled, all are $w$-circled, which would lead to
$w$-stroke emanating from vertex 5 . This is a contradiction. There are two
cases, either vertex 1 is $z$-circled or not.

If vertex 1 is not $z$-circled, then none of the vertices is $z$-colored by
the circling method. Then $L_{1234}=0$ by Proposition 4.4. By the estimate, we
have

$r_{ij}\approx\epsilon^{2},\ 1\leq i<j\leq 4,\
r_{5k}^{4}=w_{5k}z_{5k}<\epsilon^{2}\epsilon^{-2}<1.$

Then $S=0$ on the continuum corresponding to this diagram. Then we have
$\Gamma=0$ or $L=0$ by Proposition 2.2. The first case contradicts with our
assumption of the total vorticity. The second case leads to $\Gamma_{5}=0$
since

$L=L_{1234}+\Gamma_{5}(\sum_{j=1}^{4}\Gamma_{j}),\
(\sum_{j=1}^{4}\Gamma_{j})^{2}-2L_{1234}\neq 0.$

If vertex 1 is $z$-circled, then vertices 1, 2, 3, and 4 are all $z$-circled.
Then vertex 5 can be $z$-circled or not.

If vertex 5 is not $z$-circled, we have $L_{1234}=0$ and
$\sum_{j=1}^{4}\Gamma_{j}=0$ by Proposition 4.4 and Proposition 4.2. This is a
contradiction.

Hence, we only have one diagram in the case of $C=7$, as in Figure 14.

Figure 14: $C=7$ Figure 15: $C=8$

###  4.8 $C=8$

Suppose that there are four $zw$-edges emanating from vertex 1 . Rule VI
implies that the vertices 1, 2, 3, 4, 5 are fully $zw$-edged. Since all
vertices are both $z$-close and $w$-close, any circle would implies that the
total vorticity is zero, which contradicts with our assumption on the total
vorticity.

Hence, we only have one diagram in the case of $C=8$, as in Figure 15.

To summarize, we have excluded the second diagram in Figure 3, the two
diagrams in Figure 5, the third diagram in Figure 6, the fourth diagram in
Figure 8, the third diagram in Figure 10. The conclusion of the section is
that _any singular sequence should converge to one of the twenty-nine diagrams
in Figure 16_.

Figure 16: The twenty-nine 5-vortex diagrams

##  5 The twenty-nine remaining 5-vortex diagrams

In the previous process of eliminating 5-body diagrams, we could not eliminate
the twenty-nine diagrams in Figure 16. Some singular sequence could still
exist and approach any of these diagrams. These diagrams will be excluded
except if the vorticities satisfy a polynomial relation. We discuss the
constraints on the vorticities corresponding to each of the twenty-nine
diagrams. We number 5.1 to 5.29 the discussions of the constraints on the
vorticities corresponding to each of the twenty-nine diagrams from Figure 16,
ordered horizontally from top left to bottom right.

$5.1$. Applying Proposition 4.5 to vertex 1 and vertex 2, then to vertex 3 and
vertex 4, we see $\Lambda=\pm 1$ or $\pm\textbf{i}$. The mass constraints are

  * •

$\Gamma_{1}+\Gamma_{2}+\Gamma_{5}=0,\Gamma_{1}+\Gamma_{2}=\Gamma_{3}+\Gamma_{4}\neq
0$, if $\Lambda=\pm 1$;

  * •

$\Gamma_{1}\Gamma_{2}=L_{345},\Gamma_{3}\Gamma_{4}=L_{125},L=0,\Gamma_{1}+\Gamma_{2}\neq
0,\Gamma_{3}+\Gamma_{4}\neq 0$, if $\Lambda=\pm\textbf{i}$.

$5.2$. Applying Proposition 4.5 to vertex 1 and vertex 2, we see $\Lambda=\pm
1$ or $\pm\textbf{i}$. The mass constraints are

  * •

$\Gamma_{3}+\Gamma_{4}+\Gamma_{5}=0,\Gamma_{1}+\Gamma_{2}\neq 0$, if
$\Lambda=\pm 1$;

  * •

$\Gamma_{1}\Gamma_{2}=L_{345},L=0,\Gamma_{1}+\Gamma_{2}\neq 0$, if
$\Lambda=\pm\textbf{i}$.

Applying Proposition 4.6 to vertices 3, 4 and 5, we see $\Lambda=\pm 1$ or
$\pm\textbf{i}$. The mass constraints are

  * •

$\Gamma_{1}+\Gamma_{2}=\Gamma_{5},\Gamma_{3}+\Gamma_{4}\neq 0$, if
$\Lambda=\pm 1$;

  * •

$L_{345}=L_{125},L=0,\Gamma_{3}+\Gamma_{4}\neq 0$, if $\Lambda=\pm\textbf{i}$.

Hence, the mass constraints are

  * •

$\Gamma_{3}+\Gamma_{4}+\Gamma_{5}=0,\Gamma_{1}+\Gamma_{2}=\Gamma_{5},\Gamma_{1}+\Gamma_{2}\neq
0,\Gamma_{3}+\Gamma_{4}\neq 0$, if $\Lambda=\pm 1$;

  * •

$\Gamma_{1}\Gamma_{2}=L_{345},L=0,L_{345}=L_{125},\Gamma_{1}+\Gamma_{2}\neq
0,\Gamma_{3}+\Gamma_{4}\neq 0$, if $\Lambda=\pm\textbf{i}$.

Note the case $\Lambda=\pm\textbf{i}$ implies $\Gamma_{5}=0$, which
contradicts our assumption on the masses. Thus, we have

$\Gamma_{3}+\Gamma_{4}+\Gamma_{5}=0,\Gamma_{1}+\Gamma_{2}=\Gamma_{5},\Gamma_{1}+\Gamma_{2}\neq
0,\Gamma_{3}+\Gamma_{4}\neq 0,\ \Lambda=\pm 1$

$5.3$. Applying Proposition 4.5 to vertex 1 and vertex 2, we see $\Lambda=\pm
1$ or $\pm\textbf{i}$. The mass constraints are

  * •

$\Gamma_{3}+\Gamma_{4}+\Gamma_{5}=0,\Gamma_{1}+\Gamma_{2}\neq 0$, if
$\Lambda=\pm 1$;

  * •

$\Gamma_{1}\Gamma_{2}=L_{345},L=0,\Gamma_{1}+\Gamma_{2}\neq 0$, if
$\Lambda=\pm\textbf{i}$.

$5.4$. Applying Proposition 4.2 to vertex 1 and vertex 2, then to vertex 3 and
vertex 4, we obtain

$\Gamma_{1}+\Gamma_{2}=0,\ \Gamma_{3}+\Gamma_{4}=0.$

$5.5$. Consider the isolated component $z_{12}$-edge. By Rule III,
$\Gamma_{1}z_{1}+\Gamma_{2}z_{2}\prec\epsilon^{-2}$. Without loss of
generality, assume $z_{1}\sim-\Gamma_{2}a\epsilon^{-2}$ and
$z_{2}\sim\Gamma_{1}a\epsilon^{-2}$.Similarly, we assume
$z_{3}\sim-\Gamma_{4}b\epsilon^{-2}$ and $z_{4}\sim\Gamma_{3}b\epsilon^{-2}$.
Since $z_{14}\approx z_{23}\approx\epsilon^{2}$ by the estimate, we obtain

$-\Gamma_{2}a=\Gamma_{3}b,\ \Gamma_{1}a=-\Gamma_{4}b.$

Hence we obtain

$\Gamma_{1}\Gamma_{3}=\Gamma_{2}\Gamma_{4}.$

$5.6$. Applying Proposition 4.3 to vertices 1, 2, and 3, we obtain

$L_{123}=0.$

$5.7,5.8$. Applying Proposition 4.2 to vertices 1, 2, and 3, we obtain

$\Gamma_{1}+\Gamma_{2}+\Gamma_{3}=0.$

$5.9$. Applying Proposition 4.3 to vertices 1, 2, and 4, then to vertices 1,
3, and 4, we obtain

$L_{124}=L_{134}=0.$

$5.10$. Applying Proposition 4.3 to vertices 1, 2, and 4, we obtain

$L_{124}=0.$

$5.11$. Note that

$\Gamma_{1}z_{1}+\Gamma_{4}z_{4}\sim(\Gamma_{1}+\Gamma_{4})z_{1},\
\Gamma_{1}w_{1}+\Gamma_{4}w_{4}\sim(\Gamma_{1}+\Gamma_{4})w_{1},\ z_{2}\sim
z_{1},\ w_{3}\sim w_{1}.$

Then $\Gamma_{1}+\Gamma_{4}\neq 0$. By Rule III, we have

$\begin{array}[]{c}(\Gamma_{1}+\Gamma_{4})w_{1}+\Gamma_{2}w_{2}\prec\epsilon^{-2},\
(\Gamma_{1}+\Gamma_{4})z_{1}+\Gamma_{3}z_{3}\prec\epsilon^{-2},\cr\Gamma_{3}w_{3}+\Gamma_{5}w_{5}\prec\epsilon^{-2},\
\Gamma_{2}z_{1}+\Gamma_{5}z_{5}\prec\epsilon^{-2}.\end{array}$

These relations imply

$w_{2}\sim-\frac{\Gamma_{1}+\Gamma_{4}}{\Gamma_{2}}w_{1},\
w_{5}\sim-\frac{\Gamma_{5}}{\Gamma_{3}}w_{1},\
z_{3}\sim-\frac{\Gamma_{1}+\Gamma_{4}}{\Gamma_{3}}z_{1},\
z_{5}\sim-\frac{\Gamma_{5}}{\Gamma_{2}}z_{1}.\ $

Then the relations $z_{3}\sim z_{5},w_{2}\sim w_{5}$ lead to
$\Gamma_{2}(\Gamma_{1}+\Gamma_{4})=\Gamma_{3}\Gamma_{5}$ and
$\Gamma_{3}(\Gamma_{1}+\Gamma_{4})=\Gamma_{2}\Gamma_{5}$. Thus, the
constraints are

$\Gamma_{2}=\Gamma_{3},\ \Gamma_{1}+\Gamma_{4}=\Gamma_{5}.$

$5.12$. Applying Proposition 4.2 to vertices 1 and 2, then to vertices 4 and
5, we obtain

$\Gamma_{1}+\Gamma_{2}=0,\ \Gamma_{4}+\Gamma_{5}=0.$

$5.13$. Applying Proposition 4.3 to the two attached triangles respectively,
we obtain

$L_{123}=0,\ L_{145}=0.$

$5.14$. Applying Proposition 4.6 to vertex 4 and vertex 5, Proposition 4.3 to
vertices 1, 2, and 3, we see $\Lambda=\pm 1$ or $\pm\textbf{i}$. The mass
constraints are

  * •

$\Gamma_{1}=\Gamma_{2}+\Gamma_{3},L_{123}=0,\Gamma_{4}+\Gamma_{5}\neq 0$, if
$\Lambda=\pm 1$;

  * •

$L_{145}=0,L=0,L_{123}=0,\Gamma_{4}+\Gamma_{5}\neq 0$, if
$\Lambda=\pm\textbf{i}$.

$5.15$. Applying Proposition 4.6 to the upper and lower triangle respectively,
we see $\Lambda=\pm 1$ or $\pm\textbf{i}$. The mass constraints are

  * •

$\Gamma_{1}=\Gamma_{4}+\Gamma_{5},\Gamma_{1}=\Gamma_{2}+\Gamma_{3},\Gamma_{2}+\Gamma_{3}\neq
0$, if $\Lambda=\pm 1$;

  * •

$L=0,L_{145}=L_{123},\Gamma_{2}+\Gamma_{3}\neq 0,\Gamma_{4}+\Gamma_{5}\neq 0$,
if $\Lambda=\pm\textbf{i}$.

$5.16$. Applying Proposition 4.3 to vertices 1, 2, and 3, we see

$L_{123}=0.$

$5.17$. Applying Proposition 4.2 to vertices 1, 2, and 3, we see

$\Gamma_{1}+\Gamma_{2}+\Gamma_{3}=0.$

$5.18$. Applying Proposition 4.3 to vertices 1, 3 and 5, we see $L_{135}=0$.
Applying Proposition 4.4 to vertices 1, 2, 3, 4, we see $L_{1234}=0$.

$5.19$. Applying Proposition 4.3 to vertices 1, 3 and 5, we see $L_{135}=0$.

$5.20$. Applying Proposition 4.4 to vertices 1, 2, 3, 4, we see $L_{1234}=0$.
Applying Proposition 4.2 to vertex 2 and vertex 4, we see
$\Gamma_{2}+\Gamma_{4}=0$.

$5.21$. Applying Proposition 4.4 to vertices 1, 2, 3, 4, we see $L_{1234}=0$.

$5.22$. Applying Proposition 4.2 to vertices 1, 2, 3, 4, we see
$\Gamma_{1}+\Gamma_{2}+\Gamma_{3}+\Gamma_{4}=0$.

$5.23,5.24,5.25\ {\rm and}\ 5.26$. Applying Proposition 4.3 to vertices 1, 2,
and 3, we see $L_{123}=0$.

$5.27$. Applying Proposition 4.4 to vertices 1, 2, 3, and 4, we see
$L_{1234}=0$. Applying Proposition 4.4 to vertices 1, 2, 3, 5, we see
$L_{1235}=0$.

$5.28$. Applying Proposition 4.4 to vertices 1, 2, 3, and 4, we see
$L_{1234}=0$.

$5.29$. Note that $z_{kl}\approx\epsilon^{2}$ and $w_{kl}\approx\epsilon^{2}$
for $1\leq k<l\leq 5$. Then we have $S=0$ on the continuum corresponding to
this diagram. Then we have $\Gamma=0$ or $L=0$ by Proposition 2.2.

###### Remark 5.1

A 5-tuple of vorticities with some equal masses easily falls in the
exceptional set. This is consistent with Roberts’ example of continuum in
Section 2. The masses are $(2,2,2,2,-1)$, which satisfy the constraints
corresponding to 5.5 and 5.6.

##  6 Proofs

######  Proof 3 (Theorem 1.1)

Suppose that there are infinitely many solutions of system (2.12) in the
complex domain. At least one the squared distance $r^{2}_{kl}$, say
$r_{12}^{2}$, should take infinitely many values. Lemma 2.1 implies that
$r_{12}^{2}=z_{12}w_{12}$ is dominating. There is a sequence of normalized
central configurations such that $r_{12}^{2}\to 0$. Then $\mathcal{Z}$ or
$\mathcal{W}$ is unbounded on this sequence. We extract a singular sequence,
which must correspond to one of the twenty-nine diagrams found in Section 4,
or the total vorticity is zero. In either case, some explicit polynomials on
the five vorticities must be satisfied.

The existence of infinitely many central configurations remains possible for
vorticities satisfying some explicit polynomial relations. However, we confirm
the finiteness under the following restrictions of Theorem 1.2, namely,

$\sum_{j\in J}\Gamma_{j}\neq 0,\ \sum_{j,k\in J,j\neq
k}\Gamma_{j}\Gamma_{k}\neq 0,{\rm\ where}\ J\ {\rm is\ a\ nonempty\ subset\
of\ }\\{1,2,3,4,5\\}.$

Under the above restriction, a singular sequence can only approach one of the
three diagrams in Figure 17, numbered from left to right. Let us first
estimate the distances and the $4$-product
$p_{jn}=r_{jn}^{2}r_{km}^{2}r_{ml}^{2}r_{lk}^{2}$ for each diagram.

Figure 17: The three 5-vortex diagrams

The quadrilateral diagram

Consider $r_{12}^{2}=z_{12}w_{12}$. Since $Z_{12}$ is of maximal order,
$w_{12}\approx\epsilon^{2}$. Since

$\Gamma_{1}z_{1}+\Gamma_{2}z_{2}=(\Gamma_{1}+\Gamma_{2})z_{1}+\Gamma_{2}z_{12}\prec\epsilon^{-2},\
\Gamma_{1}+\Gamma_{2}\neq 0,$

the edge $z_{12}$ is maximal. Then we have $r_{12}^{2}\approx 1$. Similarly,
$w_{23},z_{34},w_{14}$ are maximal, and the side-lengths are of order $1$.

Consider the diagonals. We have $z_{24}=z_{23}+z_{34}\approx\epsilon^{-2}$
since $z_{34}$ is maximal and $z_{23}\approx\epsilon^{2}$. Similarly,
$w_{24}\approx\epsilon^{-2}$. Then we have $r_{24}^{2}\approx\epsilon^{-4}$.
The other diagonal have the same order. It is easy to see that squared
distance from the isolated vertex 5 to other vertices are of order
$\epsilon^{-4}$.

Among the ten mutual squared distances, four of them are of order $1$ and six
of them are of order $\epsilon^{-4}$. We conclude that the $4$-product
$p_{jn}\to\infty$ along corresponding singular sequences, for any $1\leq
j<n\leq 5$.

The kite diagram

Vertex 1 and vertex 4 are close in both $z$ and $w$-coordinates. Similar to
the first diagram, all strokes except $z_{14},w_{14}$ are maximal. Then the
corresponding squared distances are of order $1$, and the pairs of vertices
without edges correspond to squared distances of order $\epsilon^{-4}$,
$r_{14}^{2}\approx\epsilon^{4}$. It is easy to see that $p_{25}$ and $p_{35}$
are of order $\epsilon^{4}$, $p_{14}$ is of order $1$, and the remaining seven
$4$-product $p_{jn}$’s are of order $\epsilon^{-4}$ or $\epsilon^{-8}$, along
corresponding singular sequences, for any $1\leq j<n\leq 5$. Then the product
of any five $4$-products would not go to zero.

The two attached triangles

All strokes are maximal. Then the corresponding squared distances are of order
$1$, and the pairs of vertices without edges correspond to squared distances
of order $\epsilon^{-4}$. We conclude that the $4$-product $p_{jn}\to\infty$
along corresponding singular sequences, for any $1\leq j<n\leq 5$.

######  Proof 4 (Theorem 1.2)

Suppose that there are infinitely many solutions of system (2.12) in the
complex domain. Repeat the argument in the proof of Theorem 1.1. We suppose
that $r_{12}^{2}$ is dominating. Push it to zero. Obviously, we are in the
second diagram. Then the product $p_{41}p_{42}p_{43}p_{45}p_{23}$ goes to
infinity, thus is dominating by Lemma 2.1. Push the product to zero and
extract a singular sequence. However, the singular sequence would correspond
to none of the three diagrams. This is a contradiction.

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