# Cutting a cake for infinitely many guests

Zsuzsanna Jankó  Zsuzsanna Jankó, Department of Operations Research and
Actuarial Sciences, Corvinus University of Budapest, Budapest, Hungary;
Institute of Economics, Centre for Economic and Regional Studies, Budapest,
Hungary<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Attila
Joó  Attila Joó, Department of Mathematics, University of Hamburg, Hamburg,
Germany; Set theory, logic and topology research division, Alfréd Rényi
Institute of Mathematics, Budapest, Hungary<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

Fair division with unequal shares is an intensively studied recourse
allocation problem. For $i\in[n]$, let $\mu_{i}$ be an atomless probability
measure on the measurable space $(C,\mathcal{S})$ and let $t_{i}$ be positive
numbers (entitlements) with $\sum_{i=1}^{n}t_{i}=1$. A fair division is a
partition of $C$ into sets $S_{i}\in\mathcal{S}$ with $\mu_{i}(S_{i})\geq
t_{i}$ for every $i\in[n]$.

We introduce new algorithms to solve the fair division problem with irrational
entitlements. They are based on the classical Last diminisher technique and we
believe that they are simpler than the known methods. Then we show that a fair
division always exists even for infinitely many players.

###### Key words and phrases:

fair division, cake cutting, last diminisher

###### 2020 Mathematics Subject Classification:

Primary: 91B32, 91A07 Secondary: 68W30

Jankó is grateful for the support of NKFIH OTKA-K128611

Joó would like to thank the generous support of the Alexander von Humboldt
Foundation and NKFIH OTKA-129211

##  1\. Introduction

Cake cutting is a metaphor of the distribution of some inhomogeneous continuos
goods and is intensively investigated by not just mathematicians but
economists and political scientists as well. The preferences of the players
$P_{i}$ involved in the sharing are usually represented as atomless
probability measures $\mu_{i}$ defined on a common $\sigma$-algebra
$\mathcal{S}\subseteq\mathcal{P}(C)$ of the possible ‘slices’ of the ‘cake’
$C$. One option of how a division can be “good” is _proportionality_. This
means that each of the $n$ players gets at least one $n$th of the cake
according to their own measurement, i.e. $C=\bigsqcup_{i=1}^{n}S_{i}$ with
$\mu_{i}(S_{i})\geq\frac{1}{n}$. The division is called _strongly
proportional_ if all these inequalities are strict. For $n=2$ a proportional
division can be found by the so called “Cut and choose” procedure. This was
used by Abraham and Lot in the Bible to share Canaan. Abraham divided Canaan
into two parts which have equal value for him and then Lot chose his favourite
among these two parts leaving Abraham the other one. For a general $n$,
Steinhaus challenged his students Banach and Knaster to find a solution that
they successfully accomplished by developing the so called ”Last Diminisher”
procedure (see [steihaus1948problem]). In this method $P_{1}$ picks a slice
$T_{1}$ with $\mu_{1}(T_{1})=\frac{1}{n}$. If $\mu_{2}(T_{1})>\frac{1}{n}$,
then $P_{2}$ diminishes $T_{1}$ in the sense that he takes an $T_{2}\subseteq
T_{1}$ with $\mu_{2}(T_{2})=\frac{1}{n}$, otherwise he lets $T_{2}:=T_{1}$.
They proceed similarly and slice $T_{n}$ is allocated to the player who lastly
diminished or to $P_{1}$ if nobody did so. Then the remaining cake worth at
least $\frac{n-1}{n}$ for each of the remaining $n-1$ players and they can
continue using the same protocol.

A natural extension of the concept of proportional division is the so called
“fair division with unequal shares”. In this variant there are entitlements
$t_{i}>0$ associated to the players satisfying $\sum_{i}t_{i}=1$. A division
is called (strongly) _fair_ if the slice $S_{i}$ given to player $P_{i}$
worths for him at least (more than) $t_{i}$, i.e. $\mu_{i}(S_{i})\geq t_{i}$
($\mu_{i}(S_{i})>t_{i}$) holds for each $i$. If all of these entitlements are
rational numbers, say $\frac{p_{1}}{q},\dots,\frac{p_{n}}{q}$, then a fair
division according to them can be reduced to a proportional division problem
for $\sum_{i=1}^{n}p_{i}$ players where measure $\mu_{i}$ is “cloned” to
$p_{i}$ copies. In the presence of irrational entitlements such a “player-
cloning“ argument is no more applicable.

Several finite procedures were developed to find a (strongly) fair division
allowing irrational entitlements (see [barbanel1995game, shishido1999mark,
cseh2020complexity]). Our first contribution (Section 2) is two such
proceeders which we believe are simpler than the known methods. Both of them
are based on Last Diminisher-type of principles. In the first one we use
rational approximation of irrationals while in the second one not even that is
necessary.

It was shown in [dubins1961cut] based on Lyapunov’s theorem that if not all
the measures are identical, then a strongly proportional division exists. A
constructive proof was obtained later in [woodall1986note] which was then
further developed for the case of unequal shares (i.e. strong fairness) in
[barbanel1995game]. We point out in Section 3 that the strongly fair division
problem (for potentially infinitely many players) can be actually reduced to
the fair division problem in a completely elementary way, which reduction we
need later.

In the last section (Section 4) we consider the (strongly) fair division
problem for infinitely many players. Rational entitlements do not make this
problem easier since representing them with a common denominator is impossible
in general. Since the entitlements sum up to $1$, they must converge to $0$,
thus extension of protocols in which one need to start with the smallest
positive entitlement (like the one given in [cseh2020complexity]) is
problematic. By Last Diminisher-type of methods we are facing in addition the
difficulty that diminishing infinitely often might be necessary in which case
no “last diminisher” exists, moreover, we may end up with the empty set as a
limit of the iterated trimmings. Eliminating one player and using induction
for the rest is also not applicable for obvious reasons. Although the so
called Fink protocol (see [fink1964note]) can be considered as such a player-
eliminating recursive algorithm, it inspired our procedure that finds a fair
division for infinitely many players.

thmcake Assume that $(C,\mathcal{S})$ is a measurable space and for
$i\in\mathbb{N}$, $\mu_{i}$ is an atomless probability measure defined on
$\mathcal{S}$ and $t_{i}$ is a positive number such that
$\sum_{i=0}^{\infty}t_{i}=1$. Then there is a partition
$C=\bigsqcup_{i=0}^{\infty}S_{i}$ such that $S_{i}\in\mathcal{S}$ with
$\mu_{i}(S_{i})\geq t_{i}$ for each $i\in\mathbb{N}$. Furthermore, if not all
the $\mu_{i}$ are identical, then ‘$\mu_{i}(S_{i})\geq t_{i}$’ can be
strengthened to ‘$\mu_{i}(S_{i})>t_{i}$’ for every $i\in\mathbb{N}$.

Let us mention that cake cutting problems have a huge literature and this
particular model and notion of fairness that we consider is only a tiny
fragment of it. About the so called exact, envy-free and equitable divisions
(none of which are extendable to infinitely many players for obvious reasons)
and the corresponding existence results a brief but informative survey can be
found in [cheze2017existence]. For a more general picture about this field,
including completely different mathematical models of the problem, we refer to
[barbanel2005geometry], [brams1996fair], [procaccia2015cake] and
[robertson1998cake].

##  2\. ‘Last Diminisher’-type of procedures for fair division with irrational
entitlements

Our aim is to find a fair division $S_{1},\dots,S_{n}$ for players
$P_{1},\dots,P_{n}$ with respective atomless probability measures
$\mu_{1},\dots,\mu_{n}$ and (potentially irrational) entitlements $0<t_{1}\leq
t_{2}\leq\dots\leq t_{n}<1$ where $\sum_{i=1}^{n}t_{i}=1$. An algorithm was
given in Section 7 of [cseh2020complexity] that reduces this problem to two
sub-problems in one of which the number of players is smaller by one while in
the other all the entitlements are rational and the number of players remains
the same. We introduce two algorithms both of which solves the problem in
finitely many steps and based on ‘Last diminisher’-type of ideas. The first
one reduces the problem to another one in which either the number of players
is smaller by one or all the entitlements are rationals and the number of
players is the same. In this algorithm we need to pick rational numbers from
non-degenerate intervals (which was used in the algorithms given in
[woodall1986note, cseh2020complexity]). In the second algorithm no such
rational approximation is needed.

As it is standard in the cake cutting literature, algorithms use certain
queries. We allow the following operations.

  * •

The four basic arithmetical operations and comparison on $\mathbb{R}$.

  * •

The set operations on $\mathcal{S}$.

  * •

Computing $\mu_{i}(S)$ for some $i\in[n]$ where slice $S$ is obtained in a
previous step.

  * •

Cutting a slice $S^{\prime}\subseteq S$ with $\mu_{i}(S)=\alpha$ for an
$i\in[n]$ and $\alpha\in[0,\mu_{i}(S)]$ where either $S=C$ or $S$ is obtained
in a previous step.111It is well-defined because $\mu_{i}$ is atomless (see
[lorenc2013darboux]*Theorem 5).

###  2.1. Algorithm I

Player $P_{1}$ picks some $T_{1}$ with $\mu_{1}(T_{1})=t_{1}$. If $T_{i}$ is
already defined for some $i<n$, we let $T_{i+1}:=T_{i}$ if
$\mu_{i+1}(T_{i})\leq t_{1}$ and we define $T_{i+1}$ to be a subset of $T_{i}$
with $\mu_{i+1}(T_{i+1})=t_{1}$ if $\mu_{i+1}(T_{i})>t_{1}$. After the
recursion is done, $\mu_{i}(T_{n})\leq t_{1}$ holds for each $i$ and there is
equality for at least one index.

If $\mu_{1}(T_{n})=t_{1}$, then we let $S_{1}:=T_{n}$ and remove player
$P_{1}$ from the process. Since the rest of the cake worth at least $1-t_{1}$
for all the players, dividing it fairly with respect to the entitlements
$\frac{t_{i}}{1-t_{1}}$ for $1<i\leq n$ leads to a fair division. Thus we
invoke the algorithm for this sub-problem with less players.

If $\mu_{1}(T_{n})<t_{1}$, then there must be a player who diminished the
slice during the recursion. Let $k$ be the largest index for which $P_{k}$ is
such a player. We allocate $T_{n}$ to $P_{k}$ but we do not remove $P_{k}$
from the process unless $t_{1}=t_{k}$. In order to satisfy $P_{k}$, he needs
to get at least the $t^{\prime}_{k}:=\frac{t_{k}-t_{1}}{\mu_{k}(C\setminus
T_{n})}$ fraction of the rest of the cake $C\setminus T_{n}$ according to his
measure $\mu_{k}$, while for $i\neq k$ player $P_{i}$ should get at least the
fraction $t^{\prime}_{i}:=\frac{t_{i}}{\mu_{i}(C\setminus T_{n})}$ of
$C\setminus T_{n}$ w.r.t. $\mu_{i}$. As we already noticed $\mu_{i}(T_{n})\leq
t_{1}$ and hence $\mu_{i}(C\setminus T_{n})\geq 1-t_{1}$ for every $i$,
furthermore, the inequality is strict for $i=1$ in this branch of the case
distinction. Therefore

$\sum_{i=1}^{n}t^{\prime}_{i}<\frac{t_{k}-t_{1}}{1-t_{1}}+\sum_{i\neq
k}\frac{t_{i}}{1-t_{1}}=\frac{(\sum_{i=1}^{n}t_{i})-t_{1}}{1-t_{1}}=1.$

Thus we can pick rational numbers $t^{\prime\prime}_{i}>t_{i}^{\prime}$ with
$\sum_{i\leq n}t^{\prime\prime}_{i}=1$. Finally, we use a subroutine to divide
$C\setminus T_{n}$ fairly among the players w.r.t. the rational entitlements
$t^{\prime\prime}_{i}$ to obtain a strongly fair division for the original
problem.

###  2.2. Algorithm II

In this algorithm no ‘rounding up to rationals’ is necessary. We shall make
several rounds and in each of them allocate a slice chosen in a ‘Last
diminisher’ manner. The satisfied players are dropping out of the process. The
algorithm itself is quite simple in this case as well but the proof of the
correctness is somewhat more involved.

For $1\leq i\leq n$, we denote by $S^{m}_{i}$ the portion allocated to player
$P_{i}$ at the beginning of round $m$. We set $S^{0}_{i}=\emptyset$ for every
$i$. The rest of the cake is $C_{m}:=C\setminus\bigcup_{i=1}^{n}S_{i}^{m}$. We
also have improved entitlements $t_{i}^{m}$ where $t_{i}^{0}:=t_{i}$. Let us
define the set of indices of the players that are unsatisfied at the beginning
of round $m$ as

$I_{m}:=\\{i\in[n]:\ t_{i}>\mu_{i}(S_{i}^{m})\\}.$

If $I_{m}$ is a singleton, $I_{m}=\\{i\\}$ say, then we allocate $C_{m}$ to
player $P_{i}$ and the algorithm terminates. As long as $I_{m}$ is not a
singleton, the algorithm does the following. It considers the smallest
$i_{m}\in I_{m}$ that minimizes
$\frac{t_{i}^{m}-\mu_{i}(S_{i}^{m})}{\mu_{i}(C_{m})}$. Then player $P_{i_{m}}$
takes a $T_{1}^{m}\subseteq C_{m}$ with
$\mu_{i_{m}}(T_{1}^{m})=t_{i_{m}}^{m}-\mu_{i_{m}}(S_{i_{m}}^{m})$. After that
players $P_{i}$ for $i\in I_{m}\setminus\\{i_{m}\\}$ diminish or keep
unchanged the actual slice depending on if the value of their normed measures
$\frac{\mu_{i}}{\mu_{i}(C_{m})}$ exceed the constant
$\frac{t_{i_{m}}^{m}-\mu_{i_{m}}(S_{i_{m}}^{m})}{\mu_{i_{m}}(C_{m})}$ on it or
not. Eventually they obtain a $T^{m}_{\left|I_{m}\right|}=:R_{m}$ such that

(1)
$\frac{\mu_{i}(R_{m})}{\mu_{i}(C_{m})}\leq\frac{t_{i_{m}}^{m}-\mu_{i_{m}}(S_{i_{m}}^{m})}{\mu_{i_{m}}(C_{m})}$

for every $i\in I_{m}$ and there is equality for at least one index. Let
$j_{m}:=i_{m}$ if there is equality at (1) for $i_{m}$ and let $j_{m}$ be the
smallest index in $I_{m}$ for which we have equality if the inequality is
strict for $i_{m}$. We allocate $R_{m}$ to player $P_{j_{m}}$, formally
$S_{j_{m}}^{m+1}:=S_{j_{m}}^{m}\cup R_{m}$ and $S_{i}^{m+1}:=S_{i}^{m}$ for
$i\in\\{1,\dots,n\\}\setminus\\{j_{m}\\}$. For $i\in I_{m+1}$ let

$t_{i}^{m+1}:=\mu_{i}(S_{i}^{m+1})+\frac{t_{i}^{m}-\mu_{i}(S_{i}^{m+1})}{\sum_{j\in
I_{m+1}}\frac{t_{j}^{m}-\mu_{j}(S_{j}^{m+1})}{\mu_{j}(C_{m+1})}},$

which completes the description of the general step of the algorithm.

We turn to the proof of the correctness. First, we show by induction that the
steps described above can be done, the algorithm maintains the equation

(2) $\sum_{i\in I_{m}}\frac{t_{i}^{m}-\mu_{i}(S_{i}^{m})}{\mu_{i}(C_{m})}=1$

and $t_{i}^{m}$ is an increasing function of $m$ for every $i$. For $m=0$, (2)
says $\sum_{i=1}^{n}t_{i}=1$ which we assumed. Suppose we know the statement
up to some $m$. If $I_{m}$ is a singleton, then the algorithm terminates after
round $m$ and there is nothing to prove. Suppose that $\left|I_{m}\right|>1$.
Since the summands at (2) are all positive, we have
$\frac{t_{i_{m}}^{m}-\mu_{i_{m}}(S_{i_{m}}^{m})}{\mu_{i_{m}}(C_{m})}<1$.
Therefore $t_{i_{m}}^{m}-\mu_{i_{m}}(S_{i_{m}}^{m})<\mu_{i_{m}}(C_{m})$, thus
there is indeed a $T_{1}^{m}\subseteq C_{m}$ with
$\mu_{i_{m}}(T_{1}^{m})=t_{i_{m}}^{m}-\mu_{i_{m}}(S_{i_{m}}^{m})$. By
subtracting both sides of (1) from $1$ and taking the reciprocates we obtain

(3)
$\frac{\mu_{i}(C_{m})}{\mu_{i}(C_{m+1})}\leq\frac{1}{1-\frac{t_{i_{m}}^{m}-\mu_{i_{m}}(S_{i_{m}}^{m})}{\mu_{i_{m}}(C_{m})}}.$

Note that player $P_{i_{m}}$ will be satisfied after round $m$ if
$j_{m}=i_{m}$ because in that case we have equality at (1) for $i_{m}$. We
claim that

$\sum_{i\in
I_{m+1}}\frac{t_{i}^{m}-\mu_{i}(S_{i}^{m+1})}{\mu_{i}(C_{m+1})}\leq 1$

and therefore $t_{i}^{m+1}\geq t_{i}^{m}$ by the definition of $t_{i}^{m+1}$,
moreover, both of these inequalities are strict if $j_{m}\neq i_{m}$. Indeed

$\displaystyle\sum_{i\in
I_{m+1}}\frac{t_{i}^{m}-\mu_{i}(S_{i}^{m+1})}{\mu_{i}(C_{m+1})}\leq\sum_{i\in
I_{m}}\frac{t_{i}^{m}-\mu_{i}(S_{i}^{m+1})}{\mu_{i}(C_{m+1})}=\sum_{i\in
I_{m}}\frac{t_{i}^{m}-\mu_{i}(S_{i}^{m+1})}{\mu_{i}(C_{m})}\cdot\frac{\mu_{i}(C_{m})}{\mu_{i}(C_{m+1})}\overset{(3)}{\leq}$
$\displaystyle\sum_{i\in
I_{m}}\frac{t_{i}^{m}-\mu_{i}(S_{i}^{m+1})}{\mu_{i}(C_{m})}\cdot\frac{1}{1-\frac{t_{i_{m}}^{m}-\mu_{i_{m}}(S_{i_{m}}^{m})}{\mu_{i_{m}}(C_{m})}}\leq\left[\left(\sum_{i\in
I_{m}}\frac{t_{i}^{m}-\mu_{i}(S_{i}^{m})}{\mu_{i}(C_{m})}\right)-\frac{\mu_{j_{m}}(R_{m})}{\mu_{j_{m}}(C_{m})}\right]\cdot\frac{1}{1-\frac{t_{i_{m}}^{m}-\mu_{i_{m}}(S_{i_{m}}^{m})}{\mu_{i_{m}}(C_{m})}}$
$\displaystyle\overset{(2)}{=}\left[1-\frac{t_{i_{m}}^{m}-\mu_{i_{m}}(S_{i_{m}}^{m})}{\mu_{i_{m}}(C_{m})}\right]\frac{1}{1-\frac{t_{i_{m}}^{m}-\mu_{i_{m}}(S_{i_{m}}^{m})}{\mu_{i_{m}}(C_{m})}}=1,$

where the overestimation of $\frac{\mu_{i_{m}}(C_{m})}{\mu_{i_{m}}(C_{m+1})}$
via (3) was strict if $i_{m}\neq j_{m}$.

Suppose for a contradiction that the algorithm does not terminate for
$\mu_{1},\dots,\mu_{n}$ and $t_{1},\dots,t_{n}$. Let $k$ be the smallest
number for which $I_{k}=I_{m}$ for every $m>k$. Then $j_{k}\neq i_{k}$ since
otherwise we had $I_{k+1}=I_{k}\setminus\\{i_{k}\\}\subsetneq I_{k}$. As we
have already seen, this implies $t_{i}^{k+1}>t_{i}^{k}\geq t_{i}$ for every
$i\in I_{k}$. Let $(m_{\ell})_{\ell\in\mathbb{N}}$ be a strictly increasing
sequence of natural numbers with $m_{0}>k$ such that there are $i^{*},j^{*}\in
I_{k}$ with $i_{m_{\ell}}=i^{*}$ and $j_{m_{\ell}}=j^{*}$ for every $\ell$.
There cannot be a $\varepsilon>0$ such that
$\mu_{j^{*}}(R_{m_{\ell}})\geq\varepsilon$ for infinitely many $\ell$ because
then $P_{j^{*}}$ would be eventually satisfied and removed from the process,
contradicting the definition of $k$. Thus
$\lim_{\ell\rightarrow\infty}\mu_{j^{*}}(R_{m_{\ell}})=0$. Since there is
equality for $j^{*}$ at (1) for each $m_{\ell}$, we know that

$\mu_{j^{*}}(R_{m_{\ell}})=\left[t_{i^{*}}^{m_{\ell}}-\mu_{i^{*}}(S_{i^{*}}^{m_{\ell}})\right]\frac{\mu_{j^{*}}(C_{m_{\ell}})}{\mu_{i^{*}}(C_{m_{\ell}})}.$

If
$\lim_{\ell\rightarrow\infty}t_{i^{*}}^{m_{\ell}}-\mu_{i^{*}}(S_{i^{*}}^{m_{\ell}})=0$,
then $\mu_{i^{*}}(S_{i^{*}}^{m_{\ell}})\geq t_{i^{*}}$ for a large enough
$\ell$ because $t_{i^{*}}^{m_{0}}>t_{i^{*}}$ and $t_{i^{*}}^{m_{\ell}}$ is
increasing in $\ell$, a contradiction. Therefore we must have
$\lim_{\ell\rightarrow\infty}\frac{\mu_{j^{*}}(C_{m_{\ell}})}{\mu_{i^{*}}(C_{m_{\ell}})}=0$.
Since $\mu_{i^{*}}(C_{m_{\ell}})\leq\mu_{i^{*}}(C_{m_{0}})$, this implies
$\lim_{\ell\rightarrow\infty}\mu_{j^{*}}(C_{m_{\ell}})=0$. But then it follows
from (2) that
$\lim_{\ell\rightarrow\infty}t_{j^{*}}^{m_{\ell}}-\mu_{j^{*}}(S_{j^{*}}^{m_{\ell}})=0$.
As earlier with $i^{*}$, this implies that player $P_{j^{*}}$ will be
eventually satisfied, which is a contradiction.

Finally, if $I_{m}=\\{i\\}$ for some $m\in\mathbb{N}$ and $i\in[n]$, then (2)
ensures $\mu_{i}(C_{m})=t_{i}^{m}-\mu_{i}(S_{i}^{m})$ and therefore the
inequality $t_{i}^{m}\geq t_{i}$ combined with the definition of $I_{m}$
guarantee that all the players are satisfied when the algorithm terminates
after round $m$.

##  3\. From fairness to strong fairness, an elementary approach

######  Lemma 3.1.

Assume that $(C,\mathcal{S})$ is a measurable space, $I$ is a countable index
set, and for $i\in I$, $\mu_{i}$ is an atomless probability measure defined on
$\mathcal{S}$ and $t_{i}$ is a positive number such not all the $\mu_{i}$ are
identical and $\sum_{i\in I}t_{i}=1$. Then there is a partition
$C=C^{\prime}\sqcup C^{\prime\prime}$ and
$t_{i}^{\prime},t_{i}^{\prime\prime}>0$ with $\sum_{i\in
I}t^{\prime}_{i}=\sum_{i\in I}t^{\prime\prime}_{i}=1$ such that
$t^{\prime}_{i}\cdot\mu_{i}(C^{\prime})+t^{\prime\prime}_{i}\cdot\mu_{i}(C^{\prime\prime})>t_{i}$
for each $i\in I$.

###### Proof.

Suppose that $j,k\in I$ and $C^{\prime}\in\mathcal{S}$ such that
$\mu_{j}(C^{\prime})<\mu_{k}(C^{\prime})$. It is enough to find
$s^{\prime}_{i},s^{\prime\prime}_{i}>0$ with $\sum_{i\in
I}s^{\prime}_{i},\sum_{i\in I}s^{\prime\prime}_{i}<1$ and
$s^{\prime}_{i}\cdot\mu_{i}(C^{\prime})+s^{\prime\prime}_{i}\cdot\mu_{i}(C^{\prime\prime})=t_{i}$
for every $i\in I$ because then

$t^{\prime}_{i}:=\frac{s^{\prime}_{i}}{\sum_{\ell\in
I}s^{\prime}_{\ell}}\text{ and
}t^{\prime\prime}_{i}:=\frac{s^{\prime\prime}_{i}}{\sum_{\ell\in
I}s^{\prime\prime}_{\ell}}$

are as desired. We are looking for $\varepsilon,\delta>0$ for which the
definitions

  * •

$s^{\prime}_{j}:=t_{j}-\varepsilon$

  * •

$s^{\prime\prime}_{j}:=t_{j}+\varepsilon\cdot\frac{\mu_{j}(C^{\prime})}{\mu_{j}(C^{\prime\prime})}$

  * •

$s^{\prime}_{k}:=t_{k}+\delta\cdot\frac{\mu_{k}(C^{\prime\prime})}{\mu_{k}(C^{\prime})}$

  * •

$s^{\prime\prime}_{k}:=t_{k}-\delta$

  * •

$s_{i}^{\prime\prime}:=s_{i}^{\prime}:=t_{i}$ for
$i\in\mathbb{N}\setminus\\{j,k\\}$

are suitable. Note that whatever $\varepsilon$ and $\delta$ we choose,
$s^{\prime}_{i}\cdot\mu_{i}(C^{\prime})+s^{\prime\prime}_{i}\cdot\mu_{i}(C^{\prime\prime})=t_{i}$
will hold for each $i\in\mathbb{N}$. Thus the requirements
$s_{i}^{\prime},s_{i}^{\prime\prime}>0$ and $\sum_{i\in
I}s^{\prime}_{i},\sum_{i\in I}s^{\prime\prime}_{i}<1$ mean for $\varepsilon$
and $\delta$ that they satisfy

$\displaystyle\varepsilon$ $\displaystyle\in(0,t_{j})$ $\displaystyle\delta$
$\displaystyle\in(0,t_{k})$ $\displaystyle\varepsilon$
$\displaystyle>\delta\cdot\frac{\mu_{k}(C^{\prime\prime})}{\mu_{k}(C^{\prime})}$
$\displaystyle\delta$
$\displaystyle>\varepsilon\cdot\frac{\mu_{j}(C^{\prime})}{\mu_{j}(C^{\prime\prime})}$

If $\mu_{j}(C^{\prime})=0$, then the last inequality is redundant and the
existence of a solution is straightforward. Otherwise the last two
inequalities demand

$\frac{\mu_{k}(C^{\prime\prime})}{\mu_{k}(C^{\prime})}<\frac{\varepsilon}{\delta}<\frac{\mu_{j}(C^{\prime\prime})}{\mu_{j}(C^{\prime})}.$

Since
$\frac{\mu_{k}(C^{\prime\prime})}{\mu_{k}(C^{\prime})}<\frac{\mu_{j}(C^{\prime\prime})}{\mu_{j}(C^{\prime})}$
follows from $\mu_{j}(C^{\prime})<\mu_{k}(C^{\prime})$, the desired
$\varepsilon$ and $\delta$ exist in this case as well. ∎

Let $\mu_{i}^{\prime}$ be the restriction of
$\frac{\mu_{i}}{\mu_{i}(C^{\prime})}$ to
$\mathcal{S}\cap\mathcal{P}(C^{\prime})$ if $\mu_{i}(C^{\prime})\neq 0$ and an
arbitrary atomless probability measure on
$\mathcal{S}\cap\mathcal{P}(C^{\prime})$ if $\mu_{i}(C^{\prime})=0$. We define
$\mu_{i}^{\prime\prime}$ analogously with respect to $C^{\prime\prime}$.

######  Corollary 3.2.

Assume the settings of Lemma 3.1. If $\\{S^{\prime}_{i}:\ i\in I\\}$ is a fair
division with respect to $\mu_{i}^{\prime},t_{i}^{\prime}\ (i\in I)$ and
$\\{S^{\prime\prime}_{i}:\ i\in I\\}$ is a fair divisions with respect to
$\mu_{i}^{\prime\prime},t_{i}^{\prime\prime}\ (i\in I)$, then for
$S_{i}:=S_{i}^{\prime}\cup S^{\prime\prime}_{i}$, $\\{S_{i}:\ i\in I\\}$ is a
strongly fair division with respect to $\mu_{i},t_{i}\ (i\in I)$.

###### Proof.

We have $\mu_{i}(S^{\prime}_{i})\geq t_{i}^{\prime}\cdot\mu_{i}(C^{\prime})$
and $\mu_{i}(S^{\prime\prime}_{i})\geq
t_{i}^{\prime\prime}\cdot\mu_{i}(C^{\prime\prime})$ by fairness, thus by Lemma
3.1

$\mu_{i}(S_{i})=\mu_{i}(S_{i}^{\prime}\sqcup
S_{i}^{\prime\prime})=\mu_{i}(S_{i}^{\prime})+\mu_{i}(S_{i}^{\prime\prime})\geq
t_{i}^{\prime}\cdot\mu_{i}(C^{\prime})+t_{i}^{\prime\prime}\cdot\mu_{i}(C^{\prime\prime})>t_{i}.$

∎

##  4\. Existence of a fair division for infinitely many players

We repeat the theorem here for convenience. *

###### Proof.

Without loss of generality we may looking for a sub-partition instead of a
partition, i.e. we can relax ‘$C=\bigsqcup_{i=0}^{\infty}S_{i}$’ to
‘$C\supseteq\bigsqcup_{i=0}^{\infty}S_{i}$’ since the remaining surplus part
of the cake can be given to anybody. The last sentence of Theorem 1 follows
from the rest of it via Corollary 3.2.

For $n\in\mathbb{N}$, we let $t_{0}^{n},t_{1}^{n},\dots,t_{n}^{n}$ to be the
first $n+1$ entitlements scaled to sum up to $1$, i.e.

$t_{i}^{n}:=\frac{t_{i}}{\sum_{j=0}^{n}t_{j}}.$

######  Observation 4.1.

$(1-t_{n+1}^{n+1})t_{i}^{n}=t^{n+1}_{i}$ and
$\lim_{n\rightarrow\infty}t_{i}^{n}=t_{i}$.

###### Proof.

$\displaystyle\frac{t^{n+1}_{i}}{t_{i}^{n}}$
$\displaystyle=\frac{\sum_{j=0}^{n}t_{j}}{\sum_{j=0}^{n+1}t_{j}}=\frac{\sum_{j=0}^{n+1}t_{j}-t_{n+1}}{\sum_{j=0}^{n+1}t_{j}}=1-t_{n+1}^{n+1},$
$\displaystyle\lim_{n\rightarrow\infty}t_{i}^{n}$
$\displaystyle=\lim_{n\rightarrow\infty}\frac{t_{i}}{\sum_{j=0}^{n}t_{j}}=\frac{t_{i}}{\lim_{n\rightarrow\infty}\sum_{j=0}^{n}t_{j}}=t_{i}.$

∎

We shall define recursively $S_{i}^{n}\in\mathcal{S}$ for $i,n\in\mathbb{N}$
with $i\leq n$ in such a way that

  1. (i)

$C=\bigsqcup_{i\leq n}S_{i}^{n}$ for every $n$;

  2. (ii)

$\mu_{i}(S_{i}^{n})\geq t_{i}^{n}$;

  3. (iii)

For every fixed $i\in\mathbb{N}$ the sequence $(S_{i}^{n})_{n\geq i}$ is
$\subseteq$-decreasing.

Observe that conditions (i) and (ii) say that for each fixed $n$ the sets
$S_{0}^{n},S_{1}^{n},\dots,S_{n}^{n}$ form a fair division with respect to the
measures $\mu_{i}$ and entitlements $t_{i}^{n}$. Although such a fair division
can be found for every particular $n$, it cannot be guaranteed without
condition (iii) that they have a meaningful “limit” which provides a fair
division in the original settings.

We let $S_{0}^{0}:=C$ which obviously satisfies the conditions. Suppose that
$S_{0}^{n},S_{1}^{n}\dots,S_{n}^{n}$ are already defined for some
$n\in\mathbb{N}$. We need to find for each $i\leq n$ an $S_{i}^{n+1}\subseteq
S_{i}^{n}$ with $\mu_{i}(S_{i}^{n+1})\geq t_{i}^{n+1}$ in such a way that for

$S_{n+1}^{n+1}:=C\setminus\bigcup_{i\leq n}S_{i}^{n+1}$

we have $\mu_{n+1}(S_{n+1}^{n+1})\geq t_{n+1}^{n+1}$. For the last inequality
it is enough to ensure that

(4) $\mu_{n+1}(S_{i}^{n}\setminus S_{i}^{n+1})\geq\mu_{n+1}(S_{i}^{n})\cdot
t_{n+1}^{n+1}\text{ for }i\leq n.$

Indeed, since

$S_{n+1}^{n+1}=\bigsqcup_{i\leq n}S_{i}^{n}\setminus S_{i}^{n+1},$

the inequalities (4) imply

$\displaystyle\mu_{n+1}(S_{n+1}^{n+1})$
$\displaystyle=\mu_{n+1}\left(\bigsqcup_{i\leq n}S_{i}^{n}\setminus
S_{i}^{n+1}\right)=\sum_{i=0}^{n}\mu_{n+1}(S_{i}^{n}\setminus
S_{i}^{n+1})\geq\sum_{i=0}^{n}\mu_{n+1}(S_{i}^{n})\cdot t_{n+1}^{n+1}$
$\displaystyle=t_{n+1}^{n+1}\cdot\sum_{i=0}^{n}\mu_{n+1}(S_{i}^{n})=t_{n+1}^{n+1}\cdot\mu_{n+1}(C)=t_{n+1}^{n+1}\cdot
1=t_{n+1}^{n+1},$

where we used (i) combined with the fact that $\mu_{n+1}$ is a probability
measure. Therefore it is enough to find for every $i\leq n$ an
$S_{i}^{n+1}\subseteq S_{i}^{n}$ such that

(5) $\displaystyle\mu_{i}(S_{i}^{n+1})$ $\displaystyle\geq t_{i}^{n+1}$ (6)
$\displaystyle\mu_{n+1}(S_{i}^{n}\setminus S_{i}^{n+1})$
$\displaystyle\geq\mu_{n+1}(S_{i}^{n})\cdot t_{n+1}^{n+1}.$

Let $i\leq n$ be fixed. If $\mu_{n+1}(S_{i}^{n})=0$, then we let
$S_{i}^{n+1}:=S_{i}^{n}$ which is clearly appropriate since $t_{i}^{n}\geq
t_{i}^{n+1}$ (see Observation 4.1). Suppose that $\mu_{n+1}(S_{i}^{n})>0$ and
note that $\mu_{i}(S_{i}^{n})\geq t_{i}^{n}>0$ by assumption. We claim that
choosing $S_{i}^{n+1}$ to be the slice corresponding to $i$ in a fair division
of $S_{i}^{n}$ between $P_{i}$ and $P_{n+1}$ with respect to the restrictions
of $\frac{\mu_{i}}{\mu_{i}(S_{i}^{n})}$ and
$\frac{\mu_{n+1}}{\mu_{n+1}(S_{i}^{n})}$ to
$\mathcal{S}\cap\mathcal{P}(S_{i}^{n})$ and respective entitlements
$1-t_{n+1}^{n+1}$ and $t_{n+1}^{n+1}$ is suitable. Indeed, by the fairness of
the obtained bipartition $\\{S_{i}^{n+1},\ S_{i}^{n}\setminus S_{i}^{n+1}\\}$
of $S_{i}^{n}$ we have

$\displaystyle\frac{\mu_{i}(S_{i}^{n+1})}{\mu_{i}(S_{i}^{n})}$
$\displaystyle\geq 1-t_{n+1}^{n+1},$
$\displaystyle\frac{\mu_{n+1}(S_{i}^{n}\setminus
S_{i}^{n+1})}{\mu_{n+1}(S_{i}^{n})}$ $\displaystyle\geq t_{n+1}^{n+1}.$

Here the second inequality is equivalent with (6) and the first one implies
(5) since

$\mu_{i}(S_{i}^{n+1})\geq(1-t_{n+1}^{n+1})\mu_{i}(S_{i}^{n})\geq(1-t_{n+1}^{n+1})t_{i}^{n}=t^{n+1}_{i},$

where we used $\mu_{i}(S_{i}^{n})\geq t_{i}^{n}$ and Observation 4.1. The
recursion is done.

We define $S_{i}:=\bigcap_{n\geq i}S_{i}^{n}$ for $i\in\mathbb{N}$. Then for
$i<j$ we have $S_{i}\cap S_{j}=\emptyset$ because $S_{i}\subseteq S_{i}^{j},\
S_{j}\subseteq S_{j}^{j}$ and $S_{i}^{j}\cap S_{j}^{j}=\emptyset$ by (i),
furthermore,

$\mu_{i}(S_{i})=\mu_{i}\left(\bigcap_{n\geq
i}S_{i}^{n}\right)=\lim_{n\rightarrow\infty}\mu_{i}(S_{i}^{n})\geq\lim_{n\rightarrow\infty}t_{i}^{n}=t_{i}$

by (iii), (ii) and Observation 4.1. This completes the proof of Theorem 1. ∎

## References


To find the other critical points, let us consider the function $h$ defined in
(7.1). It takes the form

$\displaystyle h(\nu,\zeta)={}$ $\displaystyle
2^{-1}(\alpha_{1}-\alpha_{2})\sin^{2}2\nu\cos
2\zeta-2^{-1}(\alpha_{1}+3\alpha_{2})\sin^{2}2\nu+2\alpha_{3}\cos
2\nu+3\alpha_{1}+\alpha_{2}+r,$

where $\nu\in(0,\pi/2)\cup(\pi/2,\pi)$ and $\zeta\in[0,\pi)$. Note that

$\partial_{\nu}h(\nu,\zeta)=2\sin 2\nu(\cos 2\nu((\alpha_{1}-\alpha_{2})\cos
2\zeta-(\alpha_{1}+3\alpha_{2}))-2\alpha_{3})$

and

$\partial_{\zeta}h(\nu,\zeta)=-(\alpha_{1}-\alpha_{2})\sin^{2}2\nu\sin
2\zeta.$

Since $\sin 2\nu\neq 0$, $\partial_{\zeta}h(\nu,\zeta)=0$ implies $\sin
2\zeta=0$ and hence $\cos 2\zeta=\pm 1$.

If $\cos 2\zeta=1$ then $\partial_{\nu}h(\nu,\zeta)=0$ implies $\alpha_{2}>0$
and $\cos 2\nu=-\frac{\alpha_{3}}{2\alpha_{2}}$. We remark that such $\nu$
exists in $(0,\pi/2)\cup(\pi/2,\pi)$ if $\alpha_{3}<2\alpha_{2}$. The critical
value in this case is $g_{3}$ and the set of the corresponding critical points
is $T_{0,3}$.

If $\cos 2\zeta=-1$ then $\partial_{\nu}h(\nu,\zeta)=0$ implies
$\alpha_{1}+\alpha_{2}\neq 0$ and $\cos
2\nu=-\frac{\alpha_{3}}{\alpha_{1}+\alpha_{2}}$. We remark that such $\nu$
exists in $(0,\pi/2)\cup(\pi/2,\pi)$ if the modulus of the right hand side is
less than one, that is, if $\alpha_{3}<|\alpha_{1}+\alpha_{2}|$. The critical
value in this case is $g_{4}$ and the set of the corresponding critical points
is $T_{0,4}$.

Let us find $g_{\min}$ and $T_{0}$. One has $g_{1}>g_{2}$. Hence, if
$\max(2\alpha_{2},|\alpha_{1}+\alpha_{2}|)\leqslant\alpha_{3}$ then
$g_{\min}=g_{2}$ and $T_{0}=T_{0,2}$. Further, it follows that $g_{2}>g_{3}$
for $\alpha_{3}<2\alpha_{2}$ since
$g_{2}>g_{3}\Leftrightarrow(2\alpha_{2}-\alpha_{3})^{2}>0$. Similarly,
$g_{2}>g_{4}$ holds if $\alpha_{3}<\alpha_{1}+\alpha_{2}$ and $g_{4}>g_{2}$
holds if $\alpha_{3}<-(\alpha_{1}+\alpha_{2})$. Finally, under the assumption
$\alpha_{3}<\min(2\alpha_{2},\alpha_{1}+\alpha_{2})$, $g_{3}>g_{4}$ holds if
and only if $\alpha_{1}>\alpha_{2}$. One can summarize as in the statement. ∎

###  7.5. System (NLS5)

In this case, we have

$\displaystyle g(u_{1},u_{2})=$
$\displaystyle\alpha_{1}|u_{1}^{2}+u_{2}^{2}|^{2}+\alpha_{2}|u_{1}^{2}-u_{2}^{2}|^{2}-4\alpha_{2}|u_{1}|^{2}|u_{2}|^{2}+2\alpha_{3}\cos\eta(|u_{1}|^{4}-|u_{2}|^{4})$
$\displaystyle+4\alpha_{3}\sin\eta(|u_{1}|^{2}+|u_{2}|^{2})\operatorname{Re}(\overline{u_{1}}u_{2})+(2\alpha_{1}+r)(|u_{1}|^{2}+|u_{2}|^{2})^{2},$

where $\alpha_{2}>0$, $\alpha_{3}>0$,
$\alpha_{1}^{2}+\alpha_{2}^{2}+\alpha_{3}^{2}=1$, $\eta\in(0,\pi)$, and
$r\in\mathbb{R}$. We put an additional assumption $\alpha_{1}>0$ if
$\eta\in(\pi/2,\pi)$. For $\alpha_{2},\alpha_{3}>0$ and $k\in(0,\pi/2]$,
recall that $\theta_{0}=\theta_{0}(k)\in[k,\pi/2]$ is a unique solution to
$\alpha_{2}\sin 2\theta=\alpha_{3}\sin(\theta-k)$ in $[k,\pi/2]$. Corollary
1.19 follows from the following proposition.

######  Proposition 7.5.

  1. (1)

Let $\theta_{j}$ ($j=0,1,2,3$) be the solution to (1.20) with
$\rho=\alpha_{3}/\alpha_{2}$ and $\tau=\eta$ and define

$\tilde{T}_{c,j}:=\bigcup_{\theta\in\mathbb{R}/2\pi\mathbb{Z}}(e^{i\theta}\cos\tfrac{\theta_{j}}{2},e^{i\theta}\sin\tfrac{\theta_{j}}{2}).$

The function $g|_{\partial B}$ has the critical points
$\cup_{j=0,1,2,3}\tilde{T}_{c,j}$ if
$\alpha_{3}/\alpha_{2}\leqslant\rho_{*}(\eta)$ and
$\cup_{j\in\mathcal{J}(\eta)}\tilde{T}_{c,j}$ if
$\alpha_{3}/\alpha_{2}>\rho_{*}(\eta)$, where

$\mathcal{J}(\eta):=\begin{cases}\\{0,3\\}&\text{if }\eta<\pi/2,\\\
\\{1,3\\}&\text{if }\eta=\pi/2,\\\ \\{2,3\\}&\text{if }\eta>\pi/2.\end{cases}$

If

(7.2)
$\alpha_{1}\neq\alpha_{2}\quad\text{and}\quad\alpha_{3}^{2}\leqslant\tfrac{(\alpha_{1}+\alpha_{2})^{2}(\alpha_{1}-\alpha_{2})^{2}}{\alpha_{1}^{2}+\alpha_{2}^{2}-2\alpha_{1}\alpha_{2}\cos
2\eta}$

then the set

$\displaystyle{T}_{c}:=\bigcup_{\theta\in\mathbb{R}/2\pi\mathbb{Z},\,\sigma\in\\{\pm
1\\}}\Big{\\{}\Big{(}$ $\displaystyle\sigma
e^{i\theta}\sqrt{\tfrac{\alpha_{1}+\alpha_{2}-\alpha_{3}\cos\eta}{2(\alpha_{1}+\alpha_{2})}},$
$\displaystyle
e^{i\theta}\sqrt{\tfrac{\alpha_{1}+\alpha_{2}}{2(\alpha_{1}+\alpha_{2}-\alpha_{3}\cos\eta)}}\Big{(}-\tfrac{\sigma\alpha_{3}\sin\eta}{\alpha_{1}-\alpha_{2}}+i\sqrt{1-\tfrac{\alpha_{3}^{2}(\alpha_{1}^{2}+\alpha_{2}^{2}-2\alpha_{1}\alpha_{2}\cos
2\eta)}{(\alpha_{1}+\alpha_{2})^{2}(\alpha_{1}-\alpha_{2})^{2}}}\Big{)}\Big{)}\Big{\\}}$

also gives critical points of $g|_{\partial B}$. These are all the critical
points of $g|_{\partial B}$.

  2. (2)

$g_{\min}$ and $T_{0}$ is given as follows:

     * •

If the condition (1.21) is fulfilled then

$g_{\min}=-\tfrac{\alpha_{3}^{2}(\alpha_{1}-\alpha_{2}\cos
2\eta)}{\alpha_{1}^{2}-\alpha_{2}^{2}}+2\alpha_{1}+r$

and $T_{0}=T_{c}$.

     * •

If (1.21) is not satisfied then

$g_{\min}=\alpha_{2}\cos
2\theta_{3}+2\alpha_{3}\cos(\theta_{3}-\eta)+3\alpha_{1}+\alpha_{2}+r$

and $T_{0}=\tilde{T}_{c,3}$.

######  Remark 7.6.

In (NLS4) case, $\eta=0$, the function $g$ has the symmetry
$g(-u_{1},u_{2})=g(u_{1},-u_{2})=g(u_{1},u_{2})$. This shows that the set
$T_{0}$ of minimum points of $g|_{\partial B}$ is symmetric with respect to
$\\{z_{1}=0\\}$ and $\\{z_{2}=0\\}$. Indeed, as is observed in Proposition
7.4, the set $T_{0}$ actually has this symmetry. However, this symmetry is
broken in (NLS5) case, $\eta>0$. We see that
$g(-u_{1},u_{2})=g(u_{1},-u_{2})\neq g(u_{1},u_{2})$ unless $u_{1}=0$ or
$u_{2}=0$. Consequently, the set $T_{0}$ does not have such symmetry any more.

####  7.5.1. Preliminaries

Let us first prove Lemma 1.18.

###### Proof of Lemma 1.18.

Recall that $f_{\rho,\tau}(\theta):=\sin 2\theta+\rho\sin(\theta-\tau)$.

(1) Fix $\tau\in(0,\pi/2)$. It is obvious that $f_{0,\tau}(\theta)=0$ has four
solutions $\theta=\pi/2,\pi,3\pi/2,2\pi$. Hence, we let $\rho>0$ in the rest
of the proof. Since $f_{\rho,\tau}$ solves the ODE $y^{\prime\prime}=-y-3\sin
2\theta$, it follows from the Duhamel principle that

(7.3)
$f_{\rho,\tau}(\theta)=\cos(\theta-\tilde{\theta})f_{\rho,\tau}(\tilde{\theta})+\sin(\theta-\tilde{\theta})f_{\rho,\tau}^{\prime}(\tilde{\theta})-3\int_{\tilde{\theta}}^{\theta}\sin(\theta-\tau)\sin
2\tau d\tau$

for any $\theta,\tilde{\theta}\in\mathbb{R}$.

Let us begin with the study of solutions in the range $[0,\pi/2)$. Since
$f_{\rho,\tau}(0)=-\rho\sin\tau<0$ and $f_{\rho,\tau}(\pi/2)=\rho\cos\tau>0$,
there exists a solution $\theta_{0}\in(0,\pi/2)$ to $f_{\rho,\tau}(\theta)=0$.
Without loss of generality, we may suppose that $\theta_{0}$ is the largest
solution in this interval. Then, $f_{\rho,\tau}>0$ on $(\theta_{0},\pi/2)$.
Further, one sees from (7.3) with $\tilde{\theta}=\theta_{0}$ that

$f_{\rho,\tau}(\theta)=\sin(\theta-\theta_{0})f^{\prime}_{\rho,\tau}(\theta_{0})-3\int_{\theta_{0}}^{\theta}\sin(\theta-\tau)\sin
2\tau d\tau.$

By putting $\theta=\pi/2$ in this formula and using the fact that
$f_{\rho,\tau}(\pi/2)>0$, one obtains $f^{\prime}_{\rho,\tau}(\theta_{0})>0$.
Then, the above formula yields $f_{\rho,\tau}<0$ on $[0,\theta_{0})$. Hence,
$\theta_{0}$ is the only solution in this interval. Finally, since
$f_{\rho,\tau}(\tau)=\sin 2\tau>0$, one has the bound $\theta_{0}<\tau$.

Let us move on to the solutions in $(\pi,\frac{3\pi}{2}]$. Since
$f_{\rho,\tau}(\pi)=\rho\sin\tau>0$ and
$f_{\rho,\tau}(3\pi/2)=-\rho\cos\tau<0$, the equation
$f_{\rho,\tau}(\theta)=0$ has a solution in $(\pi,3\pi/2)$. We let
$\theta_{3}\in(\pi,3\pi/2)$ be the smallest solution in this range. Then, one
has $f_{\rho,\tau}(\theta_{3})=0$ and $f_{\rho,\tau}>0$ on $(\pi,\theta_{3})$.
One sees from (7.3) with $\tilde{\theta}=\theta_{3}$ that

$f_{\rho,\tau}(\theta)=\sin(\theta-\theta_{3})f_{\rho,\tau}^{\prime}(\theta_{3})-3\int_{\theta_{3}}^{\theta}\sin(\theta-\tau)\sin
2\tau d\tau.$

By substituting $\theta=\pi$ to this formula and using the fact that
$f_{\rho,\tau}(\pi)>0$, one sees that $f_{\rho,\tau}^{\prime}(\theta_{3})<0$.
Together with the above formula, this shows $f_{\rho,\tau}<0$ on
$(\theta_{3},3\pi/2)$. Hence, $\theta_{3}$ is the unique solution in this
interval. Noting that $f_{\rho,\tau}(\pi+\tau)=\sin 2\tau>0$, we have
$\pi+\tau<\theta_{3}$.

Let us next consider solutions in $(\frac{3\pi}{2},2\pi)$. One has $\sin
2\theta<0$. Since $\tau\in(0,\pi/2)$, we see that $\theta-\tau\in(\pi,2\pi)$
and hence that $\rho\sin(\theta-\tau)<0$. Therefore, we have
$f_{\rho,\tau}(\theta)<0$ for all $\theta\in(3\pi/2,2\pi)$, which means that
$f_{\rho,\tau}(\theta)=0$ has no solution in this range.

We finally consider the interval $[\pi/2,\pi]$. Define

$\rho_{*}:=\inf\left\\{\rho>0\ \middle|\
\inf_{\theta\in[{\pi}/2,\pi]}f_{\rho,\tau}(\theta)>0\right\\}.$

As
$\inf_{\theta\in[\frac{\pi}{2},\pi]}\sin(\theta-\tau)=\min(\sin\tau,\cos\tau)>0$,
$\rho_{*}$ is finite. Since $f_{\rho,\tau}(\tau+\pi/2)=-\sin
2\tau+\rho\leqslant 0$ for $\rho\leqslant\sin 2\tau$, one has the lower bound
$\rho_{*}\geqslant\sin 2\tau>0$.

By definition, if $\rho>\rho_{*}$ then there is no solution to
$f_{\rho,\tau}(\theta)=0$ in $[{\pi}/2,\pi]$.

Let us prove if $\rho=\rho_{*}$ the equation $f_{\rho_{*},\tau}(\theta)=0$ has
a unique solution in $[{\pi}/2,\pi]$. Since $f_{\rho,\tau}$ is continuous in
$L^{\infty}([{\pi}/2,\pi])$ with respect to $\rho$, one has

$\inf_{\theta\in[{\pi}/2,\pi]}f_{\rho_{*},\tau}(\theta)=0.$

Since $f_{\rho_{*},\tau}$ is continuous in $\theta$ and since
$f_{\rho_{*},\tau}(\pi/2)>0$ and $f_{\rho_{*},\tau}(\pi)>0$, there exists
$\theta_{*}\in(\pi/2,\pi)$ such that $f_{\rho_{*},\tau}(\theta_{*})=0$. It
follows that $f_{\rho_{*},\tau}^{\prime}(\theta_{*})=0$. Then, we see from
(7.3) with $\tilde{\theta}=\theta_{*}$ that

$f_{\rho_{*},\tau}(\theta)=-3\int_{\theta_{*}}^{\theta}\sin(\theta-\tau)\sin
2\tau d\tau.$

Hence, we see that $f_{\rho_{*},\tau}\geqslant 0$ on $[\pi/2,\pi]$. Further,
in this interval, the equality holds if and only if $\theta=\theta_{*}$. Thus,
$f_{\rho_{*},\tau}(\theta)=0$ has the solution $\theta_{*}\in(\pi/2,\pi)$
which is unique in $[{\pi}/2,\pi]$.

We now prove that if $0\leqslant\rho<\rho_{*}$ then there exist exactly two
solutions to $f_{\rho,\tau}(\theta)=0$ in $[{\pi}/2,\pi]$. Fix
$\rho\in(0,\rho_{*})$. Since we have $f_{\rho,\tau}(\pi/2)>0$,
$f_{\rho,\tau}(\pi)>0$, and

$f_{\rho,\tau}(\theta_{*})=(\rho_{*}-\rho)\sin(\theta_{*}-\tau)<0,$

there exist two solutions $\theta_{1}\in(\pi/2,\theta_{*})$ and
$\theta_{2}\in(\theta_{*},\pi)$. By Rolle’s theorem, there exists
$\theta_{**}\in(\theta_{2},\theta_{3})$ such that
$f_{\rho,\tau}^{\prime}(\theta_{**})=0$. Then, (7.3) with
$\tilde{\theta}=\theta_{**}$ yields

$f_{\rho,\tau}(\theta)=\cos(\theta-\theta_{**})f_{\rho,\tau}(\theta_{**})-3\int_{\theta_{**}}^{\theta}\sin(\theta-\tau)\sin
2\tau d\tau.$

If $f(\theta_{**})\geqslant 0$ then $f_{\rho,\tau}(\theta)=0$ has at most one
solution in $[{\pi}/2,\pi]$, a contradiction. Hence,
$f_{\rho,\tau}(\theta_{**})<0$. Then, the above formula shows that
$f_{\rho,\tau}$ is strictly decreasing in $(\pi/2,\theta_{**})$ and strictly
decreasing in $(\theta_{**},\pi)$. In particular, $\theta_{1}$ and
$\theta_{2}$ are only solutions to $f_{\rho,\tau}(\theta)=0$ in
$[{\pi}/2,\pi]$. Combining the above, we obtain the first assertion.

(2) is obvious.

(3) follows from (1) by symmetry. If we define $\zeta=\pi-\theta$,
$f_{\rho,\tau}(\theta)=0$ is equivalent to $\sin
2\zeta+\rho\sin(\zeta-(\pi-\tau))=0$. Thus, we have
$\rho_{*}(\tau)=\rho_{*}(\pi-\tau)$ and
$\theta_{0}(\tau)=\pi-\theta_{2}(\pi-\tau)$,
$\theta_{1}(\tau)=\pi-\theta_{1}(\pi-\tau)$,
$\theta_{2}(\tau)=\pi-\theta_{0}(\pi-\tau)$, and
$\theta_{3}(\tau)=3\pi-\theta_{3}(\pi-\tau)$. ∎

######  Lemma 7.7.

For $\rho>0$ and $\tau\in(0,\pi)$, the function

$p_{\rho,\tau}(\theta):=\cos 2\theta+2\rho\cos(\theta-\tau)$

takes its minimum at $\theta=\theta_{3}$. There is no other minimum point in
$[0,2\pi)$.

###### Proof.

We only consider the case $\tau\neq\pi/2$ since $\tau=\pi/2$ is obvious. Note
that $p(\theta)=\cos 2\theta+2\rho\cos\tau\cos\theta+2\rho\sin\tau\sin\theta$.

If $\tau\in(0,\pi/2)$ then one sees that minimum is attained only in
$[\pi,3\pi/2]$. Indeed, if $\theta\in[0,\pi/2]$ then one has
$p_{\rho,\tau}(\theta)>p_{\rho,\tau}(\theta+\pi)$. If $\theta\in[\pi/2,\pi)$
then $p_{\rho,\tau}(\theta)>p_{\rho,\tau}(2\pi-\theta)$ holds. Similarly, if
$\theta\in(3\pi/2,2\pi]$ then
$p_{\rho,\tau}(\theta)>p_{\rho,\tau}(3\pi-\theta)$. These imply that minimum
of $p_{\rho,\tau}$ is not achieved in $[\pi,3\pi/2]^{c}$. Since a minimum
point is a critical point and since $\theta_{3}$ is the only critical point in
this interval for $\rho>0$, one sees that the minimum point must be this
point.

If $\tau\in(\pi/2,\pi)$ then one sees that minimum is attained only in
$[3\pi/2,2\pi]$ by a similar symmetry argument. By the previous lemma,
$\theta_{3}$ is the only critical point in this range and hence it must be the
minimum point. ∎

####  7.5.2. Proof of Proposition 7.5

###### Proof.

Regarding $g$ as a function $\mathbb{R}^{4}\to\mathbb{R}$ by
$z_{1}=x_{1}+iy_{1}$ and $z_{2}=x_{2}+iy_{2}$, we have

$\nabla_{x_{1}.y_{1},x_{2},y_{2}}g=4(\operatorname{Re}F_{1}(z_{1},z_{2}),\operatorname{Im}F_{1}(z_{1},z_{2}),\operatorname{Re}F_{2}(z_{1},z_{2}),\operatorname{Im}F_{2}(z_{1},z_{2})).$

If $|z_{1}|=1$ and $z_{2}=0$ then

$F_{1}(z_{1},0)=(3\alpha_{1}+\alpha_{2}+2\alpha_{3}\cos\eta+r)z_{1},\quad
F_{2}(z_{1},0)=\alpha_{3}\sin\eta z_{1}.$

Since $\alpha_{3}\sin\eta\neq 0$, $(F_{1}(z_{1},0),F_{2}(z_{1},0))$ is not a
constant multiple of $(z_{1},z_{2})$. Hence, $(z_{1},0)$ is not a critical
point of $g|_{\partial B}$.

If $|z_{2}|=1$ and $z_{1}=0$ then

$F_{1}(0,z_{2})=\alpha_{3}\sin\eta z_{2},\quad
F_{2}(0,z_{2})=(3\alpha_{1}+\alpha_{2}-2\alpha_{3}\cos\eta+r)z_{2}.$

Since $\alpha_{3}\sin\eta\neq 0$, $(0,z_{2})$ is not a critical point of
$g|_{\partial B}$.

To find other critical points, let us consider the function $h$ defined in
(7.1). It takes the form

$\displaystyle h(\nu,\zeta)={}$ $\displaystyle
2^{-1}(\alpha_{1}-\alpha_{2})\sin^{2}2\nu\cos
2\zeta-2^{-1}(\alpha_{1}+3\alpha_{2})\sin^{2}2\nu$
$\displaystyle+2\alpha_{3}\cos\eta\cos 2\nu+2\alpha_{3}\sin\eta\sin
2\nu\cos\zeta+3\alpha_{1}+\alpha_{2}+r,$

where $\nu\in(0,\pi/2)\cup(\pi/2,\pi)$ and $\zeta\in[0,\pi)$. One has

$\partial_{\nu}h(\nu,\zeta)=2\sin 2\nu(\cos 2\nu((\alpha_{1}-\alpha_{2})\cos
2\zeta-(\alpha_{1}+3\alpha_{2}))-2\alpha_{3}\cos\eta)+4\alpha_{3}\sin\eta\cos
2\nu\cos\zeta$

and

$\displaystyle\partial_{\zeta}h(\nu,\zeta)={}$ $\displaystyle-2\sin\zeta\sin
2\nu((\alpha_{1}-\alpha_{2})\sin 2\nu\cos\zeta+\alpha_{3}\sin\eta).$

Since $\sin 2\nu\neq 0$, $\partial_{\zeta}h=0$ implies that $\zeta=0$ or

(7.4) $(\alpha_{1}-\alpha_{2})\sin 2\nu\cos\zeta+\alpha_{3}\sin\eta=0.$

Let us first consider the case $\zeta=0$. One has

$h(\nu,0)=\alpha_{2}\cos
4\nu+2\alpha_{3}\cos(2\nu-\eta)+3\alpha_{1}+r=\alpha_{2}p_{{\alpha_{3}}/{\alpha_{2}},\eta}(2\nu)+3\alpha_{1}+r.$

Hence, by means of Lemma 1.18, the solutions $\partial_{\nu}h(\nu,0)$ in
$(0,\pi/2)\cup(\pi/2,\pi)$ are
$\nu=\tfrac{1}{2}\theta_{j}(\alpha_{3}/\alpha_{2})$ for $j=0,1,2,3$ if
$\alpha_{3}/\alpha_{2}\leqslant\rho_{*}(\eta)$ and
$\nu=\tfrac{1}{2}\theta_{j}(\alpha_{3}/\alpha_{2})$ for
$j\in\mathcal{J}(\eta)$ if $\alpha_{3}/\alpha_{2}>\rho_{*}(\eta)$. They are
critical points of $h$ and correspond to the points given by
$\tilde{T}_{c,j}$.

If $\alpha_{1}=\alpha_{2}$ there is no more critical points since (7.4) has no
solution. Suppose $\alpha_{1}\neq\alpha_{2}$. Then, (7.4) is equivalent to

$\cos\zeta=-\tfrac{\alpha_{3}\sin\eta}{(\alpha_{1}-\alpha_{2})\sin 2\nu}.$

Using this relation, we see that $\partial_{\zeta}h=0$ is equivalent to

$\cos 2\nu=-\tfrac{\alpha_{3}\cos\eta}{\alpha_{1}+\alpha_{2}}.$

To obtain a solution $(\nu,\zeta)$ such that $\sin 2\nu\neq 0$ to these
identities, we need

$\tfrac{\alpha_{3}^{2}\sin^{2}\eta}{(\alpha_{1}-\alpha_{2})^{2}}\leqslant
1-\tfrac{\alpha_{3}^{2}\cos^{2}\eta}{(\alpha_{1}+\alpha_{2})^{2}}.$

Let us denote the set of the critical points
$\mathcal{C}\subset((0.\pi/2)\cup(\pi/2,\pi))\times[0,\pi)$ of $h$. The
condition for the existence of this critical point is summarized as (7.2).
Further, for any $(\nu,\zeta)\in\mathcal{C}$, we have

$h(\nu,\zeta)=-\tfrac{\alpha_{3}^{2}(\alpha_{1}-\alpha_{2}\cos
2\eta)}{\alpha_{1}^{2}-\alpha_{2}^{2}}+2\alpha_{1}+r.$

We denote this value $h_{c}$. The set of the corresponding critical points is
$T_{c}$.

Let us find $g_{\min}$. We compare critical values. By virtue of Lemma 7.7,

$\min_{j=0,1,2,3}h(\tfrac{1}{2}\theta_{j}(\alpha_{3}/\alpha_{2}),0)=h(\tfrac{1}{2}\theta_{3}(\alpha_{3}/\alpha_{2}),0).$

One sees that this value is equal to

$\tilde{h}_{c}:=\alpha_{2}\cos(2\theta_{3}(\tfrac{\alpha_{3}}{\alpha_{2}}))+2\alpha_{3}\cos(\theta_{3}(\tfrac{\alpha_{3}}{\alpha_{2}})-\eta)+3\alpha_{1}+\alpha_{2}+r.$

If (7.2) fails then there is no other critical point and hence
$g_{\min}=\tilde{h}_{c}$ and $T_{0}=\tilde{T}_{c,3}$.

We consider the case (7.2) holds. Let us first consider the subcase
$\alpha_{1}<\alpha_{2}$. If $(\nu,\zeta)\in\mathcal{C}$ then

$\partial_{\zeta}^{2}h=2(\alpha_{1}-\alpha_{2})(1-\tfrac{4\alpha_{3}^{2}\sin^{2}\eta\alpha_{1}\alpha_{2}}{(\alpha_{1}+\alpha_{2})^{2}(\alpha_{1}-\alpha_{2})^{2}}).$

This has the same sign as $\alpha_{1}-\alpha_{2}$ since (7.2) yields

$1-\tfrac{4\alpha_{3}^{2}\sin^{2}\eta\alpha_{1}\alpha_{2}}{(\alpha_{1}+\alpha_{2})^{2}(\alpha_{1}-\alpha_{2})^{2}}\geqslant\tfrac{(\alpha_{1}-\alpha_{2})^{2}}{\alpha_{1}^{2}+\alpha_{2}^{2}-2\alpha_{1}\alpha_{2}\cos
2\eta}>0.$

Hence, if $\alpha_{1}<\alpha_{2}$ then $\partial_{\zeta}^{2}h<0$ holds and
hence $h_{c}$ is not the minimum value. Thus, $g_{\min}=\tilde{h}_{c}$ and
$T_{0}=\tilde{T}_{c,3}$.

Let us consider the subcase $\alpha_{1}>\alpha_{2}$ and

(7.5)
$\alpha_{3}^{2}<\tfrac{(\alpha_{1}+\alpha_{2})^{2}(\alpha_{1}-\alpha_{2})^{2}}{\alpha_{1}^{2}+\alpha_{2}^{2}-2\alpha_{1}\alpha_{2}\cos
2\eta}.$

One has

$\partial_{\zeta}^{2}h(\nu,0)=-2\sin
2\nu(\alpha_{3}\sin\eta+(\alpha_{1}-\alpha_{2})\sin 2\nu).$

Since $\theta_{3}\in(\pi,2\pi)$,
$\partial_{\zeta}^{2}h(\tfrac{1}{2}\theta_{3},0)<0$ holds if and only if
$\sin\theta_{3}<-\tfrac{\alpha_{3}\sin\eta}{\alpha_{1}-\alpha_{2}}.$ If
$\eta<\pi/2$ then the condition reads as
$f_{{\alpha_{3}}/{\alpha_{2}},\eta}(\arcsin\tfrac{\alpha_{3}\sin\eta}{\alpha_{1}-\alpha_{2}})>0$.
This is equivalent to (7.5). A similar argument shows that the condition for
$\partial_{\zeta}^{2}h(\tfrac{1}{2}\theta_{3},0)<0$ in the case
$\eta\in[\pi/2.\pi)$ is also (7.5). In this case $\tilde{h}_{c}$ is not the
minimum point. Hence, we have $g_{\min}=h_{c}$ and $T_{0}=T_{c}$.

Finally, in the subcase $\alpha_{1}-\alpha_{2}>0$ and

$\alpha_{3}^{2}=\tfrac{(\alpha_{1}+\alpha_{2})^{2}(\alpha_{1}-\alpha_{2})^{2}}{\alpha_{1}^{2}+\alpha_{2}^{2}-2\alpha_{1}\alpha_{2}\cos
2\eta},$

we see that $(\frac{1}{2}\theta_{3},0)\in\mathcal{C}$. Hence, we have
$g_{\min}=h_{c}=\tilde{h}_{c}$ and $T_{0}=T_{c}=\tilde{T}_{c,3}$. ∎

##  8\. Proof of Theorem 1.24

###### Proof.

In view of Theorems 1.2, 1.3, 1.11, and 1.12, it suffices to investigate
critical points of

$g(z_{1},z_{2})=-\kappa|z_{1}|^{3}-|z_{2}|^{3}-\tfrac{3\gamma}{2}\operatorname{Re}(\overline{z_{1}}^{2}z_{2}).$

Regarding $g$ as a function $\mathbb{R}^{4}\to\mathbb{R}$ by
$z_{1}=x_{1}+iy_{1}$ and $z_{2}=x_{2}+iy_{2}$, we have

$\nabla_{x_{1}.y_{1},x_{2},y_{2}}g=4(\operatorname{Re}F_{1}(z_{1},z_{2}),\operatorname{Im}F_{1}(z_{1},z_{2}),\operatorname{Re}F_{2}(z_{1},z_{2}),\operatorname{Im}F_{2}(z_{1},z_{2})).$

If $|z_{1}|=1$ and $z_{2}=0$ then $F_{1}(z_{1},0)=\kappa z_{1}$ and
$F_{2}(z_{1},0)=\tfrac{\gamma}{2}z_{1}^{2}$. Since
$(F_{1}(z_{1},0),F_{2}(z_{1},0))$ is not a constant multiple of $(z_{1},0)$,
one sees from Lagrange’s multiplier theorem that $(z_{1},0)$ is not a critical
point of $g|_{\partial B}$.

If $|z_{2}|=1$ and $z_{1}=0$ then $(F_{1}(0,z_{2}),F_{2}(0,z_{2}))=(0,z_{2})$.
This implies that $(0,z_{2})$ is a critical point of $g|_{\partial B}$. The
critical value is $g(0,z_{2})=-1$. Hence,

$A_{0,\omega}:=\bigcup_{\theta\in\mathbb{R}/2\pi\mathbb{Z}}\mathcal{R}((0,e^{2i\theta}),\omega,1)\subset\mathcal{A}_{\omega}$

for any $\omega>0$.

To find other critical points, let us introduce

$h(\nu,\zeta)=g(\cos\nu,e^{i\zeta}\sin\nu)=-\kappa\cos^{3}\nu-\sin^{3}\nu-\tfrac{3\gamma}{2}\cos^{2}\nu\sin\nu\cos\zeta$

By the symmetry $g(-z_{1},z_{2})=g(z_{1},z_{2})$, we consider $h$ on
$(\nu,\zeta)\in((-\pi/2,0)\cup(0,\pi/2))\times(\mathbb{R}/2\pi\mathbb{Z})$. As
in Lemma 7.2, critical points of $h$ in this domain corresponds to those of
$g|_{\partial B}$. We see that
$\partial_{\zeta}h(\nu,\zeta)=\tfrac{3\gamma}{2}\cos^{2}\nu\sin\nu\sin\zeta=0$
implies $\zeta=0$ or $\pi$.

We first consider the case $\zeta=0$. Since

$\partial_{\nu}h(\nu,0)=\tfrac{3}{2}\cos^{3}\nu(2\kappa\tan\nu+2(\gamma-1)\tan^{2}\nu-\gamma),$

one sees that $\partial_{\nu}h(\nu,0)=0$ holds if and only if
$2\kappa\tan\nu+2(\gamma-1)\tan^{2}\nu-\gamma=0$. If $\gamma\neq 1$ and
$\kappa^{2}+2\gamma(\gamma-1)>0$ then it has two solutions

$\nu_{1}=\arctan\tfrac{\gamma}{\kappa+\sqrt{\kappa^{2}+2\gamma(\gamma-1)}},\quad\nu_{2}=\arctan\tfrac{\gamma}{\kappa-\sqrt{\kappa^{2}+2\gamma(\gamma-1)}}.$

If $\gamma\in(0,1)$ and $\kappa^{2}+2\gamma(\gamma-1)=0$ then the two
solutions are the same. If $\gamma=1$ and $\kappa>0$ then it has a unique
solution $\nu_{1}=\gamma/2\kappa$.

Let us see $h(\nu_{j},0)<0$ if and only if $(\gamma,\kappa)\in{J}_{j}$ for
$j=1,2$. Looking at the sign of $\partial_{\nu}h(\nu,0)$, one sees that
$h(\nu_{1})<h(\pi/2,0)=-1$ holds for $\gamma>1$ or if $\gamma=1$ and
$\kappa>0$. If $\gamma<1$ and $\kappa\geqslant\sqrt{2\gamma(1-\gamma)}$ then
one obtains $h(\nu_{1},0)<h(0,0)=-\kappa<0$. If $\gamma<1$ and
$\kappa\leqslant-\sqrt{2\gamma(1-\gamma)}$ then one obtains
$h(\nu_{1})>h(-\pi/2)=1$. Thus, $h(\nu_{1},0)<0$ if and only if
$(\gamma,\kappa)\in{J}_{1}$. To see the sign of $h(\nu_{2})$, we put
$b_{1}=\kappa-\sqrt{\kappa^{2}+2\gamma(\gamma-1)}\neq 0$. Then,
$\cos\nu_{2}=|b_{1}|/(\gamma^{2}+b_{1}^{2})^{1/2}$,
$\sin\nu_{2}=b_{1}\gamma/|b_{1}|(\gamma^{2}+b_{1}^{2})^{1/2}$, and
$\kappa=(b_{1}^{2}+2\gamma(1-\gamma))/2b_{1}$. Hence,

$h(\nu_{2},0)=-\tfrac{b_{1}}{2|b_{1}|}(\gamma^{2}+b_{1}^{2})^{-\frac{1}{2}}(b_{1}^{2}+2\gamma).$

We see that $h(\nu_{2},0)<0$ if and only if $(\gamma,\kappa)\in{J}_{2}$.

We next consider the case $\zeta=\pi$. Since

$\partial_{\nu}h(\nu,\pi)=\tfrac{3}{2}\cos^{3}\nu(2\kappa\tan\nu-2(\gamma+1)\tan^{2}\nu+\gamma),$

one sees that $\partial_{\nu}h(\nu,\pi)=0$ holds if and only if
$-2\kappa\tan\nu+2(\gamma+1)\tan^{2}\nu-\gamma=0$. It has two solutions

$\nu_{3}=\arctan\tfrac{\gamma}{\sqrt{\kappa^{2}+2\gamma(\gamma+1)}-\kappa},\quad\nu_{4}=-\arctan\tfrac{\gamma}{\sqrt{\kappa^{2}+2\gamma(\gamma+1)}+\kappa}.$

One deduces from the sign of $\partial_{\nu}h(\nu,\pi)$ that
$h(\nu_{4},\pi)>h(-\pi/2,\pi)=1$. Let us observe that $h(\nu_{3},\pi)<0$ if
and only if $(\gamma,\kappa)\in J_{3}$. To this end, out
$b_{2}=\sqrt{\kappa^{2}+2\gamma(\gamma+1)}-\kappa\in(0,\sqrt{2\gamma(\gamma+1)})$.
Then, since $\cos\nu_{2}=b_{2}/(\gamma^{2}+b_{2}^{2})^{1/2}$,
$\sin\nu_{2}=\gamma/(\gamma^{2}+b_{2}^{2})^{1/2}$, and
$\kappa=(-b_{2}^{2}+2\gamma(\gamma+1))/2b_{2}$, one has

$h(\nu_{3},\pi)=\tfrac{1}{2}(b_{2}^{2}+\gamma^{2})^{-\frac{1}{2}}(b_{2}^{2}-2\gamma).$

Hence, $h(\nu_{3},\pi)<0$ if and only if $b_{2}<\sqrt{2\gamma}$. The condition
is equivalent to $(\gamma,\kappa)\in J_{3}$.

Let us find $\mathcal{G}_{\omega}$. Recall that $g(0,e^{i\theta})=-1$. Hence,
$g_{\min}=-1$ and $\mathcal{G}_{\omega}=A_{0,\omega}$ for
$(\gamma,\kappa)\in(\mathbb{R}_{+}\times\mathbb{R})\setminus(J_{1}\cup
J_{2}\cup J_{3})$.

Note that $J_{1}\supset J_{2}\cup J_{3}$. For $(\gamma,\kappa)\in J_{2}$, we
have $h(\nu_{2},0)\geqslant h(\nu_{1},0)$. Further, the equality holds if
$\gamma<1$ and $\kappa=\sqrt{2\gamma(1-\gamma)}$. For $(\gamma,\kappa)\in
J_{2}$, recalling that $\nu_{3}>0$, we have
$h(\nu_{3},\pi)>h(\nu_{3},0)\geqslant h(\nu_{1},0)$. Hence, $(\nu_{3},\pi)$ is
not the minimum point. These imply that

$g_{\min}=\min(-1,h(\nu_{1},0))$

for $(\gamma,\kappa)\in J_{1}\cup J_{2}\cup J_{3}=J_{1}$.

Let us compare this value with $h(\nu_{1},0)$ for $(\gamma,\kappa)\in J_{1}$.
If $\gamma>1$ or if $\gamma=1$ and $\kappa>0$ then we see from the sign of
$\partial_{\nu}h(\nu,0)$ that $h(\nu_{1},0)<h(\pi/2,0)=-1$. Hence,
$g_{\min}=h(\nu_{1},0)$ and hence $\mathcal{G}_{\omega}=A_{1,\omega}$. Let us
consider the region $0<\gamma<1$ and
$\kappa\geqslant\sqrt{2\gamma(1-\gamma)}$. Set
$b_{3}=\kappa+\sqrt{\kappa^{2}+2\gamma(\gamma-1)}\geqslant\sqrt{2\gamma(1-\gamma)}$.
Then, since $\cos\nu_{1}=b_{3}/(\gamma^{2}+b_{3}^{2})^{1/2}$,
$\sin\nu_{1}=\gamma/(\gamma^{2}+b_{3}^{2})^{1/2}$, and
$\kappa=(b_{3}^{2}-2\gamma(\gamma-1))/2b_{3}$, one has

$h(\nu_{1},0)=-\tfrac{1}{2}(\gamma^{2}+b_{3}^{2})^{-1/2}(b_{3}^{2}+2\gamma).$

Hence, $h(\nu_{1},0)<-1$ if and only if $b_{3}>2\sqrt{1-\gamma}$. This
condition reads as $\kappa>\kappa_{c}(\gamma)$. In this case, we see
$g_{\min}=h(\nu_{1},0)$ and $\mathcal{G}_{\omega}=A_{1,\omega}$. A similar
argument shows that if $\kappa=\kappa_{c}(\gamma)$ then $h(\nu_{1},0)=-1$ and
hence $g_{\min}=-1$ and $\mathcal{G}_{\omega}=A_{0,\omega}\cup A_{1,\omega}$.
Similarly, if $\kappa\in[\sqrt{2\gamma(1-\gamma)},\kappa_{c}(\gamma))$ then
$g_{\min}=-1$ and $\mathcal{G}_{\omega}=A_{0,\omega}$. ∎

##  Appendix A Derivation of the standard systems

In this appendix, we show that the systems (NLS1)–(NLS5) exhaust all standard
form of the systems which have a conserved energy which has a coercive
kinetic-energy part. The precise statement is as follows:

######  Theorem A.1.

Consider a system of the form

(A.1) $\left\\{\begin{aligned}
(i\partial_{t}+\Delta)u_{1}&=\lambda_{1}|u_{1}|^{2}u_{1}+\lambda_{2}|u_{1}|^{2}u_{2}+\lambda_{3}u_{1}^{2}\overline{u_{2}}+\lambda_{4}|u_{2}|^{2}u_{1}+\lambda_{5}u_{2}^{2}\overline{u_{1}}+\lambda_{6}|u_{2}|^{2}u_{2},\\\
(i\partial_{t}+\Delta)u_{2}&=\lambda_{7}|u_{1}|^{2}u_{1}+\lambda_{8}|u_{1}|^{2}u_{2}+\lambda_{9}u_{1}^{2}\overline{u_{2}}+\lambda_{10}|u_{2}|^{2}u_{1}+\lambda_{11}u_{2}^{2}\overline{u_{1}}+\lambda_{12}|u_{2}|^{2}u_{2},\end{aligned}\right.$

where $(t,x)\in\mathbb{R}\times\mathbb{R}^{3}$ and $\lambda_{j}\in\mathbb{R}$.
Suppose that there exist $a,b,c\in\mathbb{R}$ with $b^{2}-ac<0$ and a real-
valued quartic polynomial $g=g(z_{1},z_{2})$ satisfying
$g(e^{i\theta}z_{1},e^{i\theta}z_{2})=g(z_{1},z_{2})$ for any
$(z_{1},z_{2},\theta)\in\mathbb{C}\times\mathbb{C}\times\mathbb{R}$ such that

(A.2) $\int_{\mathbb{R}^{3}}(a|\nabla
u_{1}|^{2}+2b\operatorname{Re}\overline{\nabla u_{1}}\cdot\nabla
u_{2}+c|\nabla u_{2}|^{2}+g(u_{1},u_{2}))dx$

becomes a time-independent quantity for any $H^{1}(\mathbb{R}^{3})\times
H^{1}(\mathbb{R}^{3})$-solution $(u_{1}(t),u_{2}(t))$ to the system. Then,
there exists $M\in GL_{2}(\mathbb{R})$ such that the system which

$\begin{pmatrix}\tilde{u}_{1}\\\
\tilde{u}_{2}\end{pmatrix}=M\begin{pmatrix}u_{1}\\\ u_{2}\end{pmatrix}$

solves becomes either one of (NLS1)–(NLS5).

######  Remark A.2.

The condition $b^{2}-ac<0$ corresponds to the coercivity of quadratic part the
quantity defined in (A.2). Here, we restrict ourselves to the three
dimensional case. However, this is purely for the sake of simplicity. Note
that the action of $M\in GL(\mathbb{R})$ on the coefficients
$\\{\lambda_{j}\\}_{j=1,\dots,12}$ is independent of the dimensions, and more
generally, of the choice of the linear part of the equations.

###### Proof.

A system of the form (A.1) is identified with a vector
$(\lambda_{j})_{j=1}^{12}\in\mathbb{R}^{12}$. For a vector
$(\lambda_{j})_{j=1}^{12}\in\mathbb{R}^{12}$, we define a matrix $C\in
M_{3}(\mathbb{R})$ and a vector $V\in\mathbb{R}^{3}$ by

$C=\begin{pmatrix}\lambda_{2}-\lambda_{3}&-\lambda_{1}+\lambda_{8}-\lambda_{9}&-\lambda_{7}\\\
\lambda_{5}&-\lambda_{3}+\lambda_{11}&-\lambda_{9}\\\
\lambda_{6}&-\lambda_{4}+\lambda_{5}+\lambda_{12}&-\lambda_{10}+\lambda_{11}\end{pmatrix},\quad
V=\begin{pmatrix}\lambda_{8}-2\lambda_{9}\\\
\tfrac{1}{2}(-\lambda_{2}+2\lambda_{3}-\lambda_{10}+2\lambda_{11})\\\
\lambda_{4}-2\lambda_{5}\end{pmatrix}.$

Note that the inverse map is given as follows: For given
$((c_{jk})_{1\leqslant j,k\leqslant 3},(v_{\ell})_{1\leqslant\ell\leqslant
3})\in M_{3}(\mathbb{R})\times\mathbb{R}^{3}$, the corresponding
nonlinearities are given by

$\displaystyle F_{1}(u_{1},u_{2}):={}$
$\displaystyle-(c_{12}+c_{23})|u_{1}|^{2}u_{1}+c_{11}(2|u_{1}|^{2}u_{2}+u_{1}^{2}\overline{u_{2}})+c_{21}(2u_{1}|u_{2}|^{2}+\overline{u_{1}}u_{2}^{2})+c_{31}|u_{2}|^{2}u_{2}$
$\displaystyle-(\operatorname{tr}{C})\operatorname{Re}(\overline{u_{1}}u_{2})u_{1}+(v_{1}|u_{1}|^{2}+2v_{2}\operatorname{Re}(\overline{u_{1}}u_{2})+v_{3}|u_{2}|^{2})u_{1},$
$\displaystyle F_{2}(u_{1},u_{2}):={}$ $\displaystyle-
c_{13}|u_{1}|^{2}u_{1}-c_{23}(2|u_{1}|^{2}u_{2}+u_{1}^{2}\overline{u_{2}})-c_{33}(2u_{1}|u_{2}|^{2}+\overline{u_{1}}u_{2}^{2})+(c_{21}+c_{32})|u_{2}|^{2}u_{2}$
$\displaystyle+(\operatorname{tr}{C})\operatorname{Re}(\overline{u_{1}}u_{2})u_{2}+(v_{1}|u_{1}|^{2}+2v_{2}\operatorname{Re}(\overline{u_{1}}u_{2})+v_{3}|u_{2}|^{2})u_{2}.$

In the sequel, we will freely use the identification between a system of the
form (A.1) and a pair $(C,V)\in M_{3}(\mathbb{R})\times\mathbb{R}^{3}$.

By [MSU2]*Proposition A.9, there exists $g$ such that the quantity (A.2)
becomes a conserved quantity if and only if the pair $(C,V)$ satisfies

(A.3) $C\begin{pmatrix}a\\\ b\\\
c\end{pmatrix}=0,\quad\begin{pmatrix}\operatorname{tr}C-2v_{2}&2v_{1}&0\\\
-v_{3}&\operatorname{tr}C&v_{1}\\\
0&-2v_{3}&\operatorname{tr}C+2v_{2}\end{pmatrix}\begin{pmatrix}a\\\ b\\\
c\end{pmatrix}=0,$

where $v_{j}$ ($j=1,2,3$) is the $j$-th component of the vector $V$. In
particular, we have $\operatorname{rank}C\leqslant 2$.

Case 1. We first consider the case $\operatorname{rank}C=2$. The assumption
$b^{2}-ac<0$ implies that $C$ belongs to $Z_{\mathrm{e}}$ given in [M]*Section
1.4. Then, thanks to [M]*Theorem 1.10, there exists $M\in GL_{2}(\mathbb{R})$
such that the matrix $C^{\prime}$ corresponding to the transformed system
becomes either one of the element in $T_{\mathrm{e}}$ given as follows:
$T_{\mathrm{e}}:=T_{\mathrm{e},1}\cup T_{\mathrm{e},2}\cup T_{\mathrm{e},3}$,
where

$A_{1}:=\begin{pmatrix}1&0&-1\\\ 0&2&0\\\ -1&0&1\end{pmatrix},\quad
A_{2}:=\begin{pmatrix}0&-2&0\\\ 1&0&-1\\\ 0&2&0\end{pmatrix},\quad
A_{3}:=\begin{pmatrix}0&-2&0\\\ -1&0&1\\\ 0&2&0\end{pmatrix}$

and

$A_{4}(\eta):=\begin{pmatrix}\sin\eta&-2\cos\eta&-\sin\eta\\\ 0&0&0\\\
\sin\eta&-2\cos\eta&-\sin\eta\end{pmatrix}$

and

$\displaystyle T_{\mathrm{e},1}:={}$
$\displaystyle\left\\{p_{1}A_{1}+p_{2}A_{2}+p_{3}A_{3}+p_{4}A_{4}(0)\
\middle|\ (p_{1},p_{2},p_{3},p_{4})\in\mathcal{T}_{1}\right\\},$
$\displaystyle T_{\mathrm{e},2}:={}$
$\displaystyle\bigcup_{\eta\in[0,\frac{\pi}{2}]}\left\\{p_{1}A_{1}+p_{2}A_{2}+p_{3}A_{3}+p_{4}A_{4}(\eta)\
\middle|\ (p_{1},p_{2},p_{3},p_{4})\in\mathcal{T}_{2}(\zeta)\right\\},$
$\displaystyle T_{\mathrm{e},3}:={}$
$\displaystyle\bigcup_{\zeta\in(\frac{\pi}{2},\pi)}\left\\{p_{1}A_{1}+p_{2}A_{2}+p_{3}A_{3}+p_{4}A_{4}(\eta)\
\middle|\ (p_{1},p_{2},p_{3},p_{4})\in\mathcal{T}_{3}\right\\}$

with $S^{3}_{+}:=\\{(w,x,y,z)\in\mathbb{R}^{4}\ |\
w^{2}+x^{2}+y^{2}+z^{2}=1,\,w\geqslant 0,\,y\geqslant 0,\,0\leqslant z<1\\}$
and

$\displaystyle\mathcal{T}_{1}:={}$ $\displaystyle\\{(w,x,y,z)\in S^{3}_{+}\ |\
y=0\\}\cup\\{(w,x,y,z)\in S^{3}_{+}\ |\ y\neq 1/\sqrt{2},\,z=0\\},$
$\displaystyle\mathcal{T}_{2}(\eta):={}$ $\displaystyle\\{(w,x,y,z)\in
S^{3}_{+}\ |\ y>0,\,z>0\\}$ $\displaystyle\setminus\\{(w,x,y,z)\in S^{3}_{+}\
|\ (w,x,y)=k(\sin 2\eta,\cos 2\eta,1),\,\exists k\in\mathbb{R}\\},$
$\displaystyle\mathcal{T}_{3}:={}$ $\displaystyle\\{(w,x,y,z)\in S^{3}_{+}\ |\
x>0,\,y>0,\,z>0\\}.$

Since the original system has a conserved quantity of the form (A.2), so does
the transformed system. Further, since $\ker C^{\prime}$ is spanned by
$\ltrans{(1,0,1)}$, we see from the criterion (A.3) that the conserved energy
of the transformed system takes the form

$\int_{\mathbb{R}^{3}}(|\nabla\tilde{u}_{1}|^{2}+|\nabla\tilde{u}_{2}|^{2}+\tilde{g}(\tilde{u}_{1},\tilde{u}_{2}))dx.$

Let $\tilde{v}_{j}$ ($j=1,2,3$) be the $j$-th component of the transformed
vector $\tilde{V}$. Then we see from the criterion (A.3) that
$\operatorname{tr}C^{\prime}-2\tilde{v}_{2}=\operatorname{tr}C^{\prime}+2\tilde{v}_{2}=0$
and $\tilde{v}_{1}=\tilde{v}_{3}$. In particular,
$\operatorname{tr}C^{\prime}=0$ and $\tilde{v}_{2}=0$. Thus, we see from the
formula $C^{\prime}=p_{1}A_{1}+p_{2}A_{2}+p_{3}A_{3}+p_{4}A_{4}(\eta)$ that
$p_{1}=\frac{1}{4}\operatorname{tr}C^{\prime}=0$.

If $C^{\prime}=\alpha_{1}A_{2}+\alpha_{2}A_{3}\in T_{\mathrm{e},1}$
($(0,\alpha_{1},\alpha_{2},0)\in\mathcal{T}_{1}$) then the transformed system
is (NLS3). In the case $C^{\prime}=\alpha_{1}A_{2}+\alpha_{3}A_{4}(0)\in
T_{\mathrm{e},1}$ ($(0,\alpha_{1},0,\alpha_{3})\in\mathcal{T}_{1}$) and the
case $C^{\prime}=\alpha_{1}A_{2}+\alpha_{2}A_{3}+\alpha_{3}A_{4}(0)\in
T_{\mathrm{e},2}$
($(0,\alpha_{1},\alpha_{2},\alpha_{3})\in\mathcal{T}_{2}(0)$), we have (NLS4).
Finally, (NLS5) is obtained if $C^{\prime}\in T_{\mathrm{e},3}$ or if
$C^{\prime}=\alpha_{1}A_{2}+\alpha_{2}A_{3}+\alpha_{3}A_{4}(\zeta)\in
T_{\mathrm{e},2}$
($(0,\alpha_{1},\alpha_{2},\alpha_{3})\in\mathcal{T}_{2}(\zeta)$)) for some
$\zeta\in(0,\pi/2]$. This completes the proof in the case
$\operatorname{rank}C=2$.

Case 2. Let us proceed to the case $\operatorname{rank}C=1$. In this case, by
means of [MSU2]*Theorem A.16, there exist $M\in GL_{2}(\mathbb{R})$ and
$(k_{1},k_{2},k_{3})\in\mathbb{R}^{3}\setminus\\{(0,0,0)\\}$ such that the
matrix $C^{\prime}$ corresponding to the transformed system becomes either one
of the following three forms:

$\begin{pmatrix}0&k_{1}&0\\\ 0&k_{2}&0\\\
0&k_{3}&0\end{pmatrix},\quad\begin{pmatrix}0&0&k_{1}\\\ 0&0&k_{2}\\\
0&0&k_{3}\end{pmatrix},\quad\begin{pmatrix}k_{1}&0&k_{1}\\\ k_{2}&0&k_{2}\\\
k_{3}&0&k_{3}\end{pmatrix}.$

The assumption $b^{2}-ac<0$ implies that $\ker C^{\prime}$ contains a vector
$\ltrans{(a^{\prime},b^{\prime},c^{\prime})}\in\mathbb{R}^{3}$ such that
$(b^{\prime})^{2}-a^{\prime}c^{\prime}<0$. This implies that $C^{\prime}$
takes the first form and that $\ker C^{\prime}$ is spanned by
$\ltrans{(1,0,0)}$ and $\ltrans{(0,0,1)}$. Hence, by means of (A.3), there
exists a pair of positive numbers $a^{\prime}$ and $c^{\prime}$ such that

$\begin{pmatrix}\operatorname{tr}C^{\prime}-2\tilde{v}_{2}&2\tilde{v}_{1}&0\\\
-\tilde{v}_{3}&\operatorname{tr}C^{\prime}&\tilde{v}_{1}\\\
0&-2\tilde{v}_{3}&\operatorname{tr}C^{\prime}+2\tilde{v}_{2}\end{pmatrix}\begin{pmatrix}a^{\prime}\\\
0\\\ c^{\prime}\end{pmatrix}=0,$

where $\tilde{v}_{j}$ ($j=1,2,3$) is the $j$-th component of the vector part
of the transformed system. One then sees that
$\operatorname{tr}C^{\prime}=\tilde{v}_{2}=0$. Further, either
$(\tilde{v}_{1},\tilde{v}_{3})=(0,0)$ or $\tilde{v}_{1}\tilde{v}_{3}>0$ holds.

Hence, by virtue of [MSU2]*Theorem A.13, the triplet
$(C^{\prime},\tilde{v}_{1},\tilde{v}_{3})$ can be chosen one of the following
three forms:

(A.4) $C^{\prime}=\begin{pmatrix}0&1&0\\\ 0&0&0\\\
0&1&0\end{pmatrix},\quad\tilde{v}_{1},\tilde{v}_{3}\in\mathbb{R},$ (A.5)
$C^{\prime}=\pm\begin{pmatrix}0&1&0\\\ 0&0&0\\\
0&-1&0\end{pmatrix},\quad\tilde{v}_{1}\geqslant\tilde{v}_{3},$

or

(A.6) $C^{\prime}=\pm\begin{pmatrix}0&1&0\\\ 0&0&0\\\
0&0&0\end{pmatrix},\quad\tilde{v}_{1}\in\mathbb{R},\,\tilde{v}_{3}\in\\{-1,0,1\\}.$

If $C^{\prime}$, $\tilde{v}_{1}$, and $\tilde{v}_{3}$ are as in (A.4), the
transformed system takes the form

$\left\\{\begin{aligned}
(i\partial_{t}+\Delta)\tilde{u}_{1}&=-|\tilde{u}_{1}|^{2}\tilde{u}_{1}+(\tilde{v}_{1}|\tilde{u}_{1}|^{2}+\tilde{v}_{3}|\tilde{u}_{2}|^{2})\tilde{u}_{1},\\\
(i\partial_{t}+\Delta)\tilde{u}_{2}&=|\tilde{u}_{2}|^{2}\tilde{u}_{2}+(\tilde{v}_{1}|\tilde{u}_{1}|^{2}+\tilde{v}_{3}|\tilde{u}_{2}|^{2})\tilde{u}_{2}.\end{aligned}\right.$

If $\tilde{v}_{1}=\tilde{v}_{3}=0$ then it is (NLS1). If
$\tilde{v}_{1}\tilde{v}_{3}>0$ then by introducing a new variable
$(u^{\dagger}_{1},u^{\dagger}_{2})=(\sqrt{|\tilde{v}_{3}|}\tilde{u}_{2},\sqrt{|\tilde{v}_{1}|}\tilde{u}_{1})$,
we obtain

$\left\\{\begin{aligned}
(i\partial_{t}+\Delta)u^{\dagger}_{1}&=\tfrac{1}{|\tilde{v}_{3}|}|u^{\dagger}_{1}|^{2}u^{\dagger}_{1}+(\operatorname{sign}\tilde{v}_{1})(|u^{\dagger}_{1}|^{2}+|u^{\dagger}_{2}|^{2})u^{\dagger}_{1},\\\
(i\partial_{t}+\Delta)u^{\dagger}_{2}&=-\tfrac{1}{|\tilde{v}_{1}|}|u^{\dagger}_{2}|^{2}u^{\dagger}_{2}+(\operatorname{sign}\tilde{v}_{1})(|u^{\dagger}_{1}|^{2}+|u^{\dagger}_{2}|^{2})u^{\dagger}_{2},\end{aligned}\right.$

which is (NLS2) with $\alpha\beta<0$.

If $C^{\prime}$, $\tilde{v}_{1}$, and $\tilde{v}_{3}$ are as in (A.5), the
system takes the form

$\left\\{\begin{aligned}
(i\partial_{t}+\Delta)\tilde{u}_{1}&=\tilde{\sigma}|\tilde{u}_{1}|^{2}\tilde{u}_{1}+(\tilde{v}_{1}|\tilde{u}_{1}|^{2}+\tilde{v}_{3}|\tilde{u}_{2}|^{2})\tilde{u}_{1},\\\
(i\partial_{t}+\Delta)\tilde{u}_{2}&=\tilde{\sigma}|\tilde{u}_{2}|^{2}\tilde{u}_{2}+(\tilde{v}_{1}|\tilde{u}_{1}|^{2}+\tilde{v}_{3}|\tilde{u}_{2}|^{2})\tilde{u}_{2}\end{aligned}\right.$

with $\tilde{\sigma}\in\\{\pm 1\\}$. If $\tilde{v}_{1}=\tilde{v}_{3}=0$ then
it is (NLS1). If $\tilde{v}_{1}\tilde{v_{3}}$ then by introducing a new
variable
$(u^{\dagger}_{1},u^{\dagger}_{2})=(\sqrt{|\tilde{v}_{1}|}\tilde{u}_{1},\sqrt{|\tilde{v}_{3}|}\tilde{u}_{2})$,
we obtain

$\left\\{\begin{aligned}
(i\partial_{t}+\Delta)u^{\dagger}_{1}&=\tfrac{\tilde{\sigma}}{|\tilde{v}_{1}|}|u^{\dagger}_{1}|^{2}u^{\dagger}_{1}+(\operatorname{sign}\tilde{v}_{1})(|u^{\dagger}_{1}|^{2}+|u^{\dagger}_{2}|^{2})u^{\dagger}_{1},\\\
(i\partial_{t}+\Delta)u^{\dagger}_{2}&=\tfrac{\tilde{\sigma}}{|\tilde{v}_{3}|}|u^{\dagger}_{2}|^{2}u^{\dagger}_{2}+(\operatorname{sign}\tilde{v}_{1})(|u^{\dagger}_{1}|^{2}+|u^{\dagger}_{2}|^{2})u^{\dagger}_{2}.\end{aligned}\right.$

By swapping $u^{\dagger}_{1}$ and $u^{\dagger}_{2}$ if necessary, we obtain
(NLS2) with $\alpha\beta>0$.

If $C^{\prime}$, $\tilde{v}_{1}$, and $\tilde{v}_{3}$ are as in (A.6), the
system takes the form

$\left\\{\begin{aligned}
(i\partial_{t}+\Delta)\tilde{u}_{1}&=\tilde{\sigma}|\tilde{u}_{1}|^{2}\tilde{u}_{1}+(\tilde{v}_{1}|\tilde{u}_{1}|^{2}+\tilde{v}_{3}|\tilde{u}_{2}|^{2})\tilde{u}_{1},\\\
(i\partial_{t}+\Delta)\tilde{u}_{2}&=(\tilde{v}_{1}|\tilde{u}_{1}|^{2}+\tilde{v}_{3}|\tilde{u}_{2}|^{2})\tilde{u}_{2}\end{aligned}\right.$

with $\tilde{\sigma}\in\\{\pm 1\\}$. If $\tilde{v}_{1}=\tilde{v}_{3}=0$ then
it is (NLS1). If $\tilde{v}_{1}\neq 0$ and
$\tilde{v}_{3}=\operatorname{sign}\tilde{v}_{1}$ then by introducing a new
variable
$(u^{\dagger}_{1},u^{\dagger}_{2})=(\sqrt{|\tilde{v}_{1}|}\tilde{u}_{1},\tilde{u}_{2})$,
we obtain

$\left\\{\begin{aligned}
(i\partial_{t}+\Delta)u^{\dagger}_{1}&=-\tfrac{\tilde{\sigma}}{|\tilde{v}_{1}|}|u^{\dagger}_{1}|^{2}u^{\dagger}_{1}+(\operatorname{sign}\tilde{v}_{1})(|u^{\dagger}_{1}|^{2}+|u^{\dagger}_{2}|^{2})u^{\dagger}_{1},\\\
(i\partial_{t}+\Delta)u^{\dagger}_{2}&=(\operatorname{sign}\tilde{v}_{1})(|u^{\dagger}_{1}|^{2}+|u^{\dagger}_{2}|^{2})u^{\dagger}_{2}.\end{aligned}\right.$

By swapping $u^{\dagger}_{1}$ and $u^{\dagger}_{2}$ if $\tilde{\sigma}=1$, we
obtain (NLS2) with $\alpha\beta=0$ but $(\alpha,\beta)\neq(0,0)$.

Case 3. Finally, we consider the case $\operatorname{rank}C=0$. In this case
$C=0$. By the presence of a conserved energy with a coercive kinetic-energy
part, after a change of variable if necessary, there exists a pair of positive
numbers $a^{\prime}$ and $c^{\prime}$ such that

$\begin{pmatrix}-2\tilde{v}_{2}&2\tilde{v}_{1}&0\\\
-\tilde{v}_{3}&0&\tilde{v}_{1}\\\
0&-2\tilde{v}_{3}&2\tilde{v}_{2}\end{pmatrix}\begin{pmatrix}a^{\prime}\\\ 0\\\
c^{\prime}\end{pmatrix}=0,$

where $\tilde{v}_{j}$ ($j=1,2,3$) is the $j$-th component of the vector part
of the transformed system. One then sees that $\tilde{v}_{2}=0$ and that
either $(\tilde{v}_{1},\tilde{v}_{3})=(0,0)$ or $\tilde{v}_{1}\tilde{v}_{3}>0$
holds, as in the $\operatorname{rank}C=1$ case. Then, the transformed system
takes the form

$\left\\{\begin{aligned}
(i\partial_{t}+\Delta)\tilde{u}_{1}&=(\tilde{v}_{1}|\tilde{u}_{1}|^{2}+\tilde{v}_{3}|\tilde{u}_{2}|^{2})\tilde{u}_{1},\\\
(i\partial_{t}+\Delta)\tilde{u}_{2}&=(\tilde{v}_{1}|\tilde{u}_{1}|^{2}+\tilde{v}_{3}|\tilde{u}_{2}|^{2})\tilde{u}_{2}.\end{aligned}\right.$

If $\tilde{v}_{1}=\tilde{v_{3}}=0$ then it is (NLS1). If $\tilde{v}_{1}\neq 0$
then the system fir
$(u^{\dagger}_{1},u^{\dagger}_{2})=(\sqrt{|\tilde{v}_{1}|}\tilde{u}_{1},\sqrt{|\tilde{v}_{3}|}\tilde{u}_{2})$
becomes

$\left\\{\begin{aligned}
(i\partial_{t}+\Delta)u^{\dagger}_{1}&=(\operatorname{sign}\tilde{v}_{1})(|u^{\dagger}_{1}|^{2}+|u^{\dagger}_{2}|^{2})u^{\dagger}_{1},\\\
(i\partial_{t}+\Delta)u^{\dagger}_{2}&=(\operatorname{sign}\tilde{v}_{1})(|u^{\dagger}_{1}|^{2}+|u^{\dagger}_{2}|^{2})u^{\dagger}_{2}.\end{aligned}\right.$

which is the remaining case $\alpha=\beta=0$ of (NLS2). ∎

### Acknowledgements

The author would like to express the most profound appreciation to professors
Masahito Ohta, Noriyoshi Fukaya, and Yoshinori Nishii for their fruitful
discussions and for giving constructive suggestions on the preliminary version
of the results. The author would like to thank professor Norihisa Ikoma for
the helpful discussions and for the information on references. The author was
supported by JSPS KAKENHI Grant Numbers JP21H00991 and JP21H00993.

## References