A. Carlotto, M. B. Schulz pageheadfoot

# Minimal hypertori in the four-dimensional sphere

Alessandro Carlotto and Mario B. Schulz

###### Abstract

We prove that the four-dimensional round sphere contains a minimally embedded
hypertorus, as well as infinitely many, pairwise non-isometric, immersed ones.
Our analysis also yields infinitely many, pairwise non-isometric, minimally
embedded hyperspheres and thus provides a self-contained solution to Chern’s
spherical Bernstein conjecture in dimensions four and six.

##  1 Introduction

Lawson [Lawson1970] proved in 1970 that closed surfaces of any orientable
topological type can be minimally embedded in the three-dimensional round
sphere, and that, for non-orientable types, there always exists a minimal
immersion with the sole exception of $\mathbb{R}\mathbb{P}^{2}$. For what
concerns the former case, such minimal embeddings are known to be non-unique
for surfaces of genus $g$ that is not prime, while uniqueness theorems are
only at disposal when $g=0$ (indeed by an Hopf differential type argument due
to Almgren [Alm66], and Calabi [Calabi1967]) and when $g=1$ (by much more
recent work of Brendle, see [Brendle2013]). It is therefore quite surprising
that, if we look at higher-dimensional spheres, remarkably little is known: we
simply lack any sort of characterisation of those topological types of
$m$-dimensional manifolds that can actually be realised as minimally embedded
(or even immersed) hypersurfaces in $\mathbb{S}^{m+1}$ (a notation we shall
henceforth adopt to denote the standard, round sphere in $\mathbb{R}^{m+2}$).

A particularly interesting case is that of the four-dimensional sphere. To the
best of our knowledge, till about a decade ago only _three_ topological types
were known to enjoy such a property: the three-dimensional sphere (which
obviously admits a totally geodesic realisation), the product $S^{2}\times
S^{1}$ (for which one can consider the corresponding Clifford-type
hypersurface) and a quotient of $\operatorname{SO}(3)$ by a suitable action of
the group $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ (the minimal embedding
corresponds to one of the so-called Cartan isoparametric hypersurfaces, cf.
[KiNak87] and [Sol90]). In fact, as we shall explain in the sequel of this
introduction, the first two out of these three topological examples are
actually known to be minimally embeddable in $\mathbb{S}^{4}$ in infinitely
many, pairwise non-isometric, different ways (see [Hsiang1983, Hsiang1987]).

In recent years, we have witnessed the impressive development of the min-max
theory for the area functional, which has led (among other things) to the
positive solution of Yau’s conjecture asserting the existence of infinitely
many closed minimal hypersurfaces in any compact manifold without boundary,
possibly accounting for a singular set in ambient dimension larger than seven,
see in particular [MarNev17], [IriMarNev18], [MarNevSon19], [LioMarNev18] and
[Son18]. One should note that it follows from work of Zhou [Zhou20] on the so-
called multiplicity one conjecture, see Theorem B therein, that every
Riemannian manifold, of dimension less than eight and positive Ricci
curvature, contains a sequence of closed minimal hypersurfaces with areas
diverging as prescribed by the Weyl law for the corresponding min-max
spectrum. This conclusion obviously applies, for instance, to the case of
$\mathbb{S}^{4}$, however it is well-known that these variational methods _do
not_ , except for very few special cases, provide topological control of the
minimal hypersurfaces they produce, and thus it is still unknown whether such
hypersurfaces do indeed exhibit increasing topological complexity.

All that said, we wish to present here some advances on the question we stated
above, specified to the most fundamental case of higher-dimensional tori
(henceforth simply referred to as _hypertori_), i. e. products of the form
$T^{m}=S^{1}\times\ldots\times S^{1}$ of $m\geq 2$ copies of $S^{1}$. In
particular, we shall prove that there exists a minimal embedding of $T^{3}$
into the round four-dimensional sphere, as well as infinitely many, pairwise
non-isometric, immersed ones. As we will better explain below, when
contextualising our results and methodology, this seems to be the first
construction of a minimal embedding of $T^{m}$ into $\mathbb{S}^{m+1}$ for any
$m\geq 3$: in other words, prior to our work the only known minimal embedding
of an $m$-dimensional torus into a round $(m+1)$-dimensional sphere was the
standard Clifford torus in $\mathbb{S}^{3}$. In particular, we hereby provide
an affirmative answer to the (somewhat more general) question posed by Choe
and Fraser in [ChoeFra18]. To avoid misunderstandings, we wish to stress that
all immersions we study in the present article happen in codimension equal to
one: if one allows for higher codimension, then on the one hand cheap examples
exist in abundance and, on the other hand, one faces a much richer theory (we
refer the reader to e. g. the seminal paper [Bryant1982] by Bryant).

Actually, the results claimed above follow, as special cases (for $n=2$) from
the combination of two somewhat more general statements. The first one reads
as follows:

######  Theorem 1.1.

For any $2\leq n\in\mathbb{N}$ there exists a minimal embedding of
$S^{n-1}\times S^{n-1}\times S^{1}$ in the round sphere $\mathbb{S}^{2n}$.

We will later elaborate about why we expect the minimal embedding of $T^{3}$
in the round four-dimensional sphere (the case $n=2$ in the statement) to be
unique up to ambient isometries, see Conjecture 3.7; however it is a matter of
fact that if one allows for _immersions_ then the landscape changes quite a
lot, as witnessed by this second main result:

######  Theorem 1.2.

Let $n\in\\{2,3\\}$. Then, there exist infinitely many, pairwise non-
isometric, minimal immersions of $S^{n-1}\times S^{n-1}\times S^{1}$ in
$\mathbb{S}^{2n}$.

Before adding some significant, additional remarks on these two statements we
need to briefly digress on the methodology employed in their proofs. In short,
we employ equivariant techniques (along the lines pioneered in [HsiLaw71]), i.
e. we consider a suitable isometric group action on $\mathbb{R}^{2n+1}$ and
study the reduced equation one obtains in the quotient space
$\mathbb{S}^{n+1}/G$; more specifically, the group $G$ is the product
$\operatorname{O}(n)\times\operatorname{O}(n)$ and we regard
$\mathbb{R}^{2n+1}=\mathbb{R}^{n}\times\mathbb{R}^{n}\times\mathbb{R}$
(namely: we consider the representation
$\rho_{n}\oplus\rho^{\prime}_{n}\oplus\textbf{1}$ that is the outer direct sum
of the standard representations of $\operatorname{O}(n)$ as a group acting on
$\mathbb{R}^{n}$). The idea of considering this group action, and the
corresponding reduced minimal surface equation, is very natural and was indeed
successfully employed in 1983 by Hsiang [Hsiang1983] to disprove (for spheres
of dimension $4$ and $6$) Chern’s spherical Bernstein conjecture, namely to
answer the question whether a minimal embedding of the $m$-dimensional sphere
into $\mathbb{S}^{m+1}$ should necessarily be totally geodesic (hence a
standard equator). As recalled at the beginning of this introduction, a
classical result obtained independently in [Alm66] and [Calabi1967] provides
an affirmative answer for $m=2$, so it was quite notable, at least in certain
respects, that _infinitely many counterexamples_ (i. e. infinitely many non-
equatorial embeddings) can be obtained in higher-dimension.

At a technical level, Hsiang’s work amounts to proving the existence of
certain trajectories for a planar dynamical system of second order (which we
will describe in Section 2). One of the reasons why this is not a trivial task
is that the ODE in question becomes singular at the boundary of the quotient
space, and indeed Hsiang’s construction relies on earlier work [HsiHsi82]
where the authors prove a local existence and uniqueness theorem which applies
to the problem in question, thereby allowing for a _singular shooting method_.
Secondly, Hsiang crucially exploits a blow-up trick, which allows to bypass
some analytic challenges of the problem, by ultimately appealing to a pre-
existing, important result by Bombieri–De Giorgi–Giusti [BdGG69] concerning
the corresponding limit problem in $\mathbb{R}^{m}$ which happens to be
reducible to a first-order planar ODE due to its scaling invariance.

Now, the conclusion of Theorem 1.1 comes from proving the existence of _closed
periodic orbits_ for the reduced equation and, as such, is patently _not
tractable_ with Hsiang’s methods. Such orbits, if any, keep away from the
boundary of the quotient space and thus one cannot shoot from there, and also
our problem cannot be tackled through blow-up shortcuts. That being said, we
took a different path: we reduced the original equation to a $3\times 3$ ODE
system (thus to a non-planar problem) and developed a careful study of various
classes of solutions, which ultimately led to the result by means of a
suitable continuity argument. Such an analysis requires a priori estimates
that turn out to be rather subtle, and yet the resulting outcome is a rather
direct, self-contained proof. In fact, the proof of Theorem 1.2 requires a
significant amount of additional work, but forced us to derive some ancillary
lemmata that immediately allow to gain, as a byproduct, a somewhat simpler
proof of Hsiang’s result:

######  Theorem 1.3.

Let $n\in\\{2,3\\}$. Then, there exist infinitely many, pairwise non-
isometric, minimal, nonequatorial embeddings of $S^{2n-1}$ in
$\mathbb{S}^{2n}$.

Concerning both Theorem 1.2 and 1.3 we wish to add a comment about the
dimensional assumptions: ample numerical evidence seems to suggest that the
trajectories corresponding to such minimal immersions of hypertori and minimal
embeddings of hyperspheres, respectively, cease to exist when $n\geq 4$, and
thus lead us to believe that our results should be sharp in that respect as
well. These phenomena are described at the end of Section 4, see in particular
Conjecture 4.14 therein.

Getting back to hypertori, we would like to mention what follows: in the
concluding remarks at page 550 of [HsiHsi80] it is claimed that the methods
developed there would also allow to prove the existence of minimal embeddings
of $S^{1}\times S^{n-1}\times S^{n-1}\times S^{n-1}$ in the round sphere
$\mathbb{S}^{3n-1}$ in $\mathbb{R}^{3n}$ for any $n\geq 2$ and, thus, that
such an argument would also provide (in the sole case $n=2$) a minimal
embedding of $T^{4}$ in $\mathbb{S}^{5}$. Although there are indeed some
formal similarities between such a problem and those explicitly studied in
[HsiHsi80] (i. e. in the symmetric spaces
$\operatorname{SU}(3)/\operatorname{SO}(3)$, $\operatorname{SU}(3)$,
$\operatorname{SU}(6)/\operatorname{Sp}(3)$ and $E_{6}/F_{4}$), and the
structure of the orbit space is analogous (the quotient space is a ‘triangular
domain’ which, by virtue of symmetry arguments, can be split into six pairwise
isometric subdomains, where one needs to construct a free boundary arc of
prescribed geodesic curvature) we note that the ODE analysis is actually _not_
carried through there, nor in any later article we are aware of. For this
reason, and for the sake of completeness, we show in Appendix A how some of
the arguments we present in the first part of Section 3 can be modified, at
modest additional cost, so to confirm the existence of a minimal embedding of
$T^{4}$ in $\mathbb{S}^{5}$; as the reader will see, this result is arguably
much simpler to prove than Theorem 1.1. In any event, the authors of
[HsiHsi80] also wrote ‘ _One does not know whether there are minimal
imbeddings of codimension-one torus in $\mathbb{S}^{m}$ for $m\neq 3,5$_’ and,
to our knowledge, our result is the first advance on this matter since 1980.

#### Acknowledgements.

The authors would like to thank Lucas Ambrozio, Renato Bettiol and Ben Sharp
for a number of stimulating conversations.
This project has received funding from the European Research Council (ERC)
under the European Union’s Horizon 2020 research and innovation programme
(grant agreement No. 947923). The research of M. S. was partly funded by the
EPSRC grant EP/S012907/1 and by the Deutsche Forschungsgemeinschaft (DFG,
German Research Foundation) under Germany’s Excellence Strategy EXC 2044 –
390685587, Mathematics Münster: Dynamics–Geometry–Structure, and the
Collaborative Research Centre CRC 1442, Geometry: Deformations and Rigidity.

##  2 Setup and preliminaries

Let $2\leq n\in\mathbb{N}$. The group
$G=\operatorname{O}(n)\times\operatorname{O}(n)$ acts on the sphere
$\mathbb{S}^{2n}$ via the representation
$\rho_{n}\oplus\rho_{n}^{\prime}\oplus\textbf{1}$. The quotient
$\mathbb{S}^{2n}/G$ is equipped with spherical coordinates
$(r,\theta)\in[0,\pi]\times[0,\frac{\pi}{2}]$, and the orbital distance metric

$\displaystyle g=dr^{2}+(\sin r)^{2}d\theta^{2}.$

We recall e. g. from [HsiLaw71] that a curve
$s\mapsto\gamma(s)=(r(s),\theta(s))$ in $\mathbb{S}^{2n}/G$ parametrised by
arc-length lifts to a minimal hypersurface in $\mathbb{S}^{2n}$ if and only if
it has vanishing geodesic curvature in the conformal metric $V^{2}g$ where
$V=V(r,\theta)$ stands for the volume of the pre-image of the point
$(r,\theta)$, so in our case
$V(r,\theta)=\omega^{2}_{n-1}\sin^{2n-2}(r)\sin^{n-1}(2\theta)/2^{n-1}$ where
$\omega_{n-1}$ denotes the volume of the unit sphere in $\mathbb{R}^{n}$. As a
result, since with respect to the unit normal

$\eta=-\sin(r)\frac{d\theta}{ds}\frac{\partial}{\partial
r}+\frac{1}{\sin(r)}\frac{dr}{ds}\frac{\partial}{\partial\theta}$

the geodesic curvature in metric $g$ of the curve in question equals

$\kappa_{g}=\frac{d\alpha}{ds}+\cos(r)\frac{d\theta}{ds}$

the well-known equation $\kappa_{V^{2}g}=\kappa_{g}-\nabla_{\eta}\ln V$
implies that the lift is a $G$-equivariant minimal hypersurface if and only if

$\displaystyle\frac{d\alpha}{ds}-\frac{(2n-2)\cot(2\theta)}{\sin(r)}\frac{dr}{ds}+(2n-1)\cos(r)\frac{d\theta}{ds}=0,$
(1)

where $\alpha$ is the (signed) angle between the vectors $d\gamma/ds$ and
$\partial/\partial r$; to avoid ambiguities we agree that $\alpha=\pi/2$
(respectively $\alpha=-\pi/2$) for curves of the form $s\mapsto(r_{0},s)$
(respectively $s\mapsto(r_{0},-s)$) for any fixed $r_{0}\in]0,\pi[$.

Parametrisation by arc-length implies

$\displaystyle 1$
$\displaystyle=g\biggl{(}\frac{d\gamma}{ds},\frac{d\gamma}{ds}\biggr{)}=\biggl{(}\frac{dr}{ds}\biggr{)}^{2}+(\sin
r)^{2}\biggl{(}\frac{d\theta}{ds}\biggr{)}^{2},$ $\displaystyle\cos(\alpha)$
$\displaystyle=\frac{dr}{ds},$ $\displaystyle\sin(\alpha)$
$\displaystyle=(\sin r)\frac{d\theta}{ds}.$ (2)

We wish to rewrite the differential equation above as an autonomous $3\times
3$ ODE system, i. e. to have it in the form $dU/ds=f(U)$ where $U$ varies in a
suitable subset of $\mathbb{R}^{3}$. Indeed, we can take $U=(r,\theta,\alpha)$
and thus consider the system of equations

$\displaystyle\frac{dr}{ds}$ $\displaystyle=\cos(\alpha),$ (3)
$\displaystyle\smash{\left\\{\vphantom{\begin{aligned}
\frac{dr}{ds}&=\cos(\alpha),\\\\[4.30554pt]
\frac{d\theta}{ds}&=\frac{\sin(\alpha)}{\sin(r)},\\\\[4.30554pt]
\frac{d\alpha}{ds}&=\frac{(2n-2)\cot(2\theta)}{\sin(r)}\cos(\alpha)-(2n-1)\cot(r)\sin(\alpha).\end{aligned}}\right.}\,\frac{d\theta}{ds}$
$\displaystyle=\frac{\sin(\alpha)}{\sin(r)},$ (4)
$\displaystyle\frac{d\alpha}{ds}$
$\displaystyle=\frac{(2n-2)\cot(2\theta)}{\sin(r)}\cos(\alpha)-(2n-1)\cot(r)\sin(\alpha).$
(5)

In Section 3 we will design a suitable shooting method to construct solutions
of the system (3),(4),(5) such that the $(r,\theta)$-trajectory is simple and
periodic. Instead, in Section 4 we will obtain on the one hand periodic orbits
(with suitable symmetries) and any odd number of self-intersections and on the
other hand central-symmetric trajectories, without self-intersections,
reaching the boundary of the quotient space orthogonally. When lifted back to
$\mathbb{S}^{2n}$ these curves provide the minimally immersed hypersurfaces
and the minimally embedded hyperspheres claimed in the statements of Theorem
1.2 and 1.3, respectively.

##  3 Existence of a minimally embedded hypertorus

This section is devoted to the proof of Theorem 1.1, but also serves as a
preparation for some of the results we will present in the following sections.

######  Lemma 3.1 (Well-posedness).

Given any $r_{0}\in]0,\pi/2[$ there exists $s_{*}>0$ such that the system
(3),(4),(5) has a unique solution

$(r,\theta,\alpha)\colon[0,s_{*}]\to
B\vcentcolon=]0,\tfrac{\pi}{2}]\times]0,\tfrac{\pi}{4}]\times[-\tfrac{\pi}{2},0]$

with initial data $(r,\theta,\alpha)(0)=(r_{0},\pi/4,-\pi/2)$ on which the
solution depends continuously such that $r(s_{*})=\pi/2$ or $\alpha(s_{*})=0$;
in either case $\theta(s_{*})>0$. Moreover, $dr/ds\geq 0$, $d\theta/ds\leq 0$
and $d\alpha/ds\geq 0$.

###### Proof.

As long as $(r,\theta,\alpha)\in B$, equations (3)–(5) imply $dr/ds\geq 0$,
$d\theta/ds\leq 0$ and $d\alpha/ds\geq 0$ (i. e. the motion is weakly monotone
in each of the three coordinates; note also that there are no stationary
points). Therefore, solutions with initial data
$(r,\theta,\alpha)(0)=(r_{0},\pi/4,-\pi/2)$ stay in $B$ at least for a short
time and can leave $B$ only via $r=\frac{\pi}{2}$, $\theta=0$ or $\alpha=0$
(see Figure 1).

Equation (5) implies $d\alpha/ds(0)=(2n-1)\cot(r_{0})>0$ since we assume
$r_{0}\in]0,\pi/2[$. Then, by monotonicity of $\alpha$ along the motion,
$\cot(\alpha)<0$ for any $s>0$. If the trajectory in question intersects
$\\{\theta=\pi/8\\}$ at some $s_{0}\in]0,s_{*}[$ we can find $\delta_{0}>0$
such that $\cot(\alpha)\leq-\delta_{0}$ for all $s\in[s_{0},s_{*}[$. Equations
(4) and (5) then imply

$\displaystyle\frac{d\alpha}{ds}$
$\displaystyle=\Bigl{(}(2n-2)\cot(2\theta)\cot(\alpha)-(2n-1)\cos(r)\Bigr{)}\frac{d\theta}{ds}\geq-(2n-2)\cot(2\theta)\delta_{0}\frac{d\theta}{ds}$

for all $s\in[s_{0},s_{*}[$. Integrating this differential inequality over an
interval $[s_{0},s]$ gives

$\frac{\pi}{2\delta_{0}}\geq\frac{\alpha(s)-\alpha(s_{0})}{\delta_{0}}\geq(n-1)\log\left(\frac{\sin(\pi/4)}{\sin(2\theta(s))}\right).$

This rough bound shows that the trajectory in question must stay at positive
distance away from $\\{\theta=0\\}$. Hence, again appealing to its
monotonicity, the trajectory will either exit $B$ through the roof (namely:
reaching $\alpha=0$) or through the side (namely: reaching $r=\pi/2$). Of
course, if the trajectory does not even reach $\theta=\pi/8$ then the same
conclusions still hold true.

In particular, for any $r_{0}\in]0,\pi/2[$ and any sufficiently small
$\eta>0$, the image through the flow map of $[r_{0}-\eta,r_{0}+\eta]$ will
stay at positive distance from the plane $\\{\theta=0\\}$: in that region, the
vector field generating the flow (cf. right-hand side of (3),(4),(5)) is
smooth and uniformly bounded. Hence given any $\varepsilon>0$ one can find
$\delta\in(0,\eta)$ such that trajectories emanating from initial conditions
less than $\delta$ apart will leave $B$ at times less than $\varepsilon$ apart
and will be $\varepsilon$-close in $C^{\infty}$ for all times of their motion
(in $B$). Thereby the proof is complete. ∎

######  Lemma 3.2.

Given $r_{0}\in]0,\pi/8[$ let $(r,\theta,\alpha)\colon[0,s_{*}]\to B$ be the
solution constructed in Lemma 3.1 and let $s_{1}\in]0,s_{*}]$ be arbitrary. If
$r(s_{1})\geq 2r_{0}$ then

$\displaystyle\theta(s_{1})<\frac{\pi}{4}-\frac{1}{6n}.$

###### Proof.

Let $\delta=1/(6n)$ be fixed and assume there exists $s_{1}\in]0,s_{*}]$ with
$r(s_{1})\geq 2r_{0}$. (Otherwise there is nothing to prove.) Towards a
contradiction, suppose $\theta(s_{1})\geq\pi/4-\delta$. By monotonicity of
$\theta$, we then have $\theta(s)\geq\pi/4-\delta$ for all $s\in[0,s_{1}]$.
Hence,

$\displaystyle 0\leq\cot(2\theta)$ $\displaystyle\leq\tan(2\delta)=\vcentcolon
b,$ $\displaystyle\cot(\alpha)$ $\displaystyle\leq 0,$ $\displaystyle 0$
$\displaystyle\leq\cos(r)\leq 1,$ $\displaystyle\frac{d\theta}{ds}$
$\displaystyle\leq 0$

for all $s\in[0,s_{1}]$. Equations (4) and (5) then imply

$\displaystyle\frac{d\alpha}{ds}$
$\displaystyle=\Bigl{(}(2n-2)\cot(2\theta)\cot(\alpha)-(2n-1)\cos(r)\Bigr{)}\frac{d\theta}{ds}\leq(2n-1)\Bigl{(}b\cot(\alpha)-1\Bigr{)}\frac{d\theta}{ds}.$
(6)

We observe that the function $f\colon]-\pi/2,0[\to\mathbb{R}$ given by
$f(x)=(b\cot(x)-1)^{-1}$ is the derivative of

$\displaystyle F(x)$
$\displaystyle=-\frac{b\log\bigl{(}b\cos(x)-\sin(x)\bigr{)}+x}{b^{2}+1}.$

Multiplying estimate (6) by $f(\alpha)<0$ and integrating from $s=0$ to
$s_{1}$ we obtain

$\displaystyle
F\bigl{(}\alpha(s_{1})\bigr{)}-F\bigl{(}-\tfrac{\pi}{2}\bigr{)}\geq(2n-1)\bigl{(}\theta(s_{1})-\tfrac{\pi}{4}\bigr{)}\geq-(2n-1)\delta.$
(7)

Now, the function $[-\pi/2,0]\ni x\mapsto b\cos(x)-\sin(x)$ is concave and is
checked to attain its minimum $b$ at $x=0$. Hence,

$\displaystyle F(\alpha)$ $\displaystyle\leq\frac{-b\log(b)-\alpha}{b^{2}+1},$
$\displaystyle F(-\tfrac{\pi}{2})$ $\displaystyle=\frac{\pi}{2(b^{2}+1)}$

so estimate (7) implies
$-b\log(b)-\alpha(s_{1})-\frac{\pi}{2}\geq-(b^{2}+1)(2n-1)\delta$ which is
equivalent to

$\displaystyle\alpha(s_{1})$
$\displaystyle\leq-b\log(b)+(b^{2}+1)(2n-1)\delta-\frac{\pi}{2}.$ (8)

Recalling $b=\tan(2\delta)$, it is evident that if $\delta>0$ is sufficiently
small, then (8) implies $\alpha(s_{1})<-\pi/4$; in fact, one can check
$\delta=1/(6n)$ is sufficient to that aim. By monotonicity of $\alpha$, we
have $\alpha(s)<-\pi/4$ for all $s\in[0,s_{1}]$ and equations (3) and (4) then
imply

$\displaystyle\frac{d\theta}{ds}$
$\displaystyle=\frac{\tan(\alpha)}{\sin(r)}\frac{dr}{ds}\leq\frac{-1}{\sin(r)}\frac{dr}{ds}.$
(9)

A primitive of $]0,\pi/2[\ni r\mapsto-1/\sin(r)$ is given by the function
$-\log(\tan(r/2))$. Since by assumption $r(s_{1})\geq 2r_{0}=2r(0)$,
integrating estimate (9) from $s=0$ to $s_{1}$ yields

$\displaystyle\theta(s_{1})-\frac{\pi}{4}\leq\log\biggl{(}\frac{\tan(r_{0}/2)}{\tan(r_{0})}\biggr{)}\leq\log\biggl{(}\frac{1}{2}\biggl{)}<-\delta$
(10)

which gives a contradiction to our initial assumption, thereby proving the
claim. ∎

$r$$\theta$$0$$\frac{\pi}{2}$$\frac{\pi}{4}$(1)(2)(3) Figure 1: Implementing
the shooting method: Trajectory (1) exists $B$ through the “roof”
$\\{\alpha=0\\}$. Trajectory (2) exits $B$ through the “side” $\\{r=\pi/2\\}$.
Trajectory (3) is the desired curve.

######  Lemma 3.3 (Exit through the roof, see Figure 1 (1)).

There exists a constant $c_{n}>0$ depending only on $2\leq n\in\mathbb{N}$
such that if the initial data $r_{0}\in]0,\pi/2[$ is sufficiently small, then
the solution constructed in Lemma 3.1 satisfies $\alpha(s_{*})=0$ and
$r(s_{*})\leq c_{n}r_{0}$.

###### Proof.

Let $r_{0}\in]0,\pi/8[$ be arbitrary. We may assume $r(s_{*})>2r_{0}$,
otherwise the claim follows directly. Let $0<s_{1}<s_{*}$ such that
$r(s_{1})=2r_{0}$. Then,

$\displaystyle\theta(s)<\frac{\pi}{4}-\frac{1}{6n}$

for all $s\in[s_{1},s_{*}]$ by Lemma 3.2 and by monotonicity of $\theta$. This
implies $\cot(2\theta)>\tan(1/(3n))$ for all $s\in[s_{1},s_{*}]$ and we obtain

$\displaystyle\frac{d\alpha}{ds}$
$\displaystyle=\biggl{(}\frac{(2n-2)\cot(2\theta)}{\sin(r)}-(2n-1)\cot(r)\tan(\alpha)\biggr{)}\frac{dr}{ds}\geq\frac{(2n-2)}{r}\tan\biggl{(}\frac{1}{3n}\biggr{)}\frac{dr}{ds}$
(11)

for all $s\in[s_{1},s_{*}]$. Integrating (11) yields

$\displaystyle\frac{\pi}{2}\geq\alpha(s_{*})-\alpha(s_{1})$
$\displaystyle\geq(2n-2)\tan\biggl{(}\frac{1}{3n}\biggr{)}\log\biggl{(}\frac{r(s_{*})}{2r_{0}}\biggr{)}$
(12)

which implies

$r(s_{*})\leq
2r_{0}\exp\biggl{(}\frac{\pi}{4n-4}\cot\biggl{(}\frac{1}{3n}\biggr{)}\biggr{)}=\vcentcolon
c_{n}r_{0}$

and thus proves the second claim. If the initial value $r_{0}>0$ is chosen
such that $c_{n}r_{0}<\pi/2$ then $r(s_{*})<\pi/2$ and we obtain
$\alpha(s_{*})=0$ from Lemma 3.1. ∎

######  Lemma 3.4 (Exit through the side, see Figure 1 (2)).

If the initial data $r_{0}\in]0,\pi/2[$ is sufficiently close to $\pi/2$, then
the solution constructed in Lemma 3.1 satisfies $r(s_{*})=\pi/2$.

###### Proof.

Our task amounts to finding initial data such that $\alpha(s_{*})<0$. We
perform the following change of variables, which turns out to be especially
useful in analysing the motion for initial data $r_{0}\in]0,\pi/2[$ close to
$\pi/2$.

$\displaystyle x$ $\displaystyle=\tan(r),$ $\displaystyle y$
$\displaystyle=\cot(2\theta),$ $\displaystyle z$
$\displaystyle=-\cot(\alpha),$ $\displaystyle\cos(r)$
$\displaystyle=\frac{1}{\sqrt{x^{2}+1}},$ $\displaystyle\sin(2\theta)$
$\displaystyle=\frac{1}{\sqrt{y^{2}+1}},$ $\displaystyle\sin(\alpha)$
$\displaystyle=-\frac{1}{\sqrt{z^{2}+1}},$ $\displaystyle\sin(r)$
$\displaystyle=\frac{x}{\sqrt{x^{2}+1}},$ $\displaystyle\cos(2\theta)$
$\displaystyle=\frac{y}{\sqrt{y^{2}+1}},$ $\displaystyle\cos(\alpha)$
$\displaystyle=\frac{z}{\sqrt{z^{2}+1}}.$

For $s\in[0,s_{*}[$ the system (3),(4),(5) takes the equivalent form

$\displaystyle\frac{dx}{ds}$
$\displaystyle=\frac{(x^{2}+1)z}{\sqrt{z^{2}+1}},$ (13)
$\displaystyle\smash{\left\\{\vphantom{\begin{aligned}
\frac{dx}{ds}&=\frac{(x^{2}+1)z}{\sqrt{z^{2}+1}},\\\\[4.30554pt]
\frac{dy}{ds}&=\frac{2(y^{2}+1)\sqrt{x^{2}+1}}{x\sqrt{z^{2}+1}},\\\\[4.30554pt]
\frac{dz}{ds}&=\frac{(2n-2)yz\sqrt{z^{2}+1}\sqrt{x^{2}+1}}{x}+\frac{(2n-1)}{x}\sqrt{z^{2}+1}.\end{aligned}}\right.}\,\frac{dy}{ds}$
$\displaystyle=\frac{2(y^{2}+1)\sqrt{x^{2}+1}}{x\sqrt{z^{2}+1}},$ (14)
$\displaystyle\frac{dz}{ds}$
$\displaystyle=\frac{(2n-2)yz\sqrt{z^{2}+1}\sqrt{x^{2}+1}}{x}+\frac{(2n-1)}{x}\sqrt{z^{2}+1}.$
(15)

Note that the domain for the ODE above is $(x,y,z)\in
Q=]0,\infty[\times[0,\infty[\times[0,\infty[$. The variables $x,y,z$ are all
(weakly) increasing along the motion, with initial values
$x(0)=x_{0}\vcentcolon=\tan(r_{0})>0$ and $y(0)=0=z(0)$.

_Step 1: Preliminary bound on $y$ for the first part of the motion._ Equations
(14) and (15) imply

$\displaystyle\frac{dz}{ds}$
$\displaystyle=\biggl{(}\frac{(n-1)yz(z^{2}+1)}{(y^{2}+1)}+\frac{(2n-1)(z^{2}+1)}{2(y^{2}+1)\sqrt{x^{2}+1}}\biggr{)}\frac{dy}{ds}.$
(16)

Let $s_{1}\in]0,s_{*}[$ be fixed, to be chosen. Let $x_{1}=x(s_{1})$,
$y_{1}=y(s_{1})$, $z_{1}=z(s_{1})$ and then set

$\displaystyle\varepsilon=\frac{(2n-1)}{2\sqrt{x_{1}^{2}+1}}.$

Since $\frac{dz}{ds}(0)>\varepsilon\frac{dy}{ds}(0)$ there exists
$s_{\varepsilon}\in]0,s_{1}]$ such that $z\geq\varepsilon y$ for all
$s\in[0,s_{\varepsilon}]$. In particular,

$\displaystyle\frac{dz}{ds}$
$\displaystyle\geq\biggl{(}\frac{(n-1)y^{2}\varepsilon(z^{2}+1)}{(y^{2}+1)}+\frac{(z^{2}+1)\varepsilon}{(y^{2}+1)}\biggr{)}\frac{dy}{ds}\geq(z^{2}+1)\varepsilon\frac{dy}{ds}$

for all $s\in[0,s_{\varepsilon}]$, which implies $\arctan(z)\geq\varepsilon y$
or, equivalently, $z\geq\tan(\varepsilon y)$ for all
$s\in[0,s_{\varepsilon}]$. Thus, in fact, $s_{\varepsilon}=s_{1}$ since
$\tan(\varepsilon y)\geq\varepsilon y$. To conclude, we obtain the preliminary
bound:

$\displaystyle y_{1}\leq\frac{z_{1}}{\varepsilon}\leq
z_{1}\sqrt{x_{1}^{2}+1}.$ (17)

######  Remark 3.5.

The bound (17) is global in time in the sense that one can take any value
$s_{1}=s$ in the interval $]0,s_{\ast}[$, but we will introduce a stopping
condition at a suitably chosen intermediate time $s_{1}$ and use the estimate
above for the _first part of the motion_ only.

_Step 2: Preliminary bound on $y$ for the second part of the motion._ Equation
(16) also implies (recalling that $n\geq 2$ throughout our discussion)

$\displaystyle\frac{d}{ds}\log\biggl{(}\frac{z^{2}}{z^{2}+1}\biggr{)}=\frac{2}{z(z^{2}+1)}\frac{dz}{ds}$
$\displaystyle\geq\frac{2y}{y^{2}+1}\frac{dy}{ds}=\frac{d}{ds}\log(y^{2}+1)$
(18)

for all $s\in]0,s_{*}[$. Integrating (18) from $s_{1}$ to $s\in[s_{1},s_{*}[$
we obtain

$\displaystyle\log\biggl{(}\frac{z^{2}(z_{1}^{2}+1)}{(z^{2}+1)z_{1}^{2}}\biggr{)}$
$\displaystyle\geq\log\biggl{(}\frac{y^{2}+1}{y_{1}^{2}+1}\biggr{)}.$ (19)

Estimates (17) and (19) imply for any $s\in[s_{1},s_{*}[$

$\displaystyle y^{2}+1$
$\displaystyle\leq(y_{1}^{2}+1)\frac{z^{2}(z_{1}^{2}+1)}{(z^{2}+1)z_{1}^{2}}\leq\Bigl{(}x_{1}^{2}+1+z_{1}^{-2}\Bigr{)}\frac{z^{2}(z_{1}^{2}+1)}{(z^{2}+1)}.$
In particular, $\displaystyle\frac{y\sqrt{z^{2}+1}}{z}$
$\displaystyle\leq\sqrt{x_{1}^{2}+1+z_{1}^{-2}}\sqrt{z_{1}^{2}+1}.$ (20)

_Step 3: Global bound on $z$._ Equations (13) and (15) in combination with
(20) imply

$\displaystyle\frac{dz}{ds}$
$\displaystyle=\biggl{(}\frac{(2n-2)y(z^{2}+1)}{x\sqrt{x^{2}+1}}+\frac{(2n-1)(z^{2}+1)}{x(x^{2}+1)z}\biggr{)}\frac{dx}{ds}$
$\displaystyle=\biggl{(}\frac{(2n-2)y\sqrt{z^{2}+1}}{\sqrt{1+x^{-2}}\leavevmode\nobreak\
z}+\frac{(2n-1)\sqrt{z^{2}+1}}{x(1+x^{-2})z^{2}}\biggr{)}\frac{z\sqrt{z^{2}+1}}{x^{2}}\frac{dx}{ds}$
(21) $\displaystyle\leq C_{1}\frac{z\sqrt{z^{2}+1}}{x^{2}}\frac{dx}{ds}$ (22)

for all $s\in[s_{1},s_{*}[$, where we have set

$\displaystyle
C_{1}\vcentcolon=(2n-1)\sqrt{z_{1}^{2}+1}\biggl{(}\sqrt{x_{1}^{2}+1+z_{1}^{-2}}\sqrt{z_{1}^{2}+1}+\frac{1}{x_{1}z_{1}^{2}}\biggr{)}$

and used that the function $0<z\mapsto z^{-2}\sqrt{z^{2}+1}$ in the second
summand of (21) is decreasing. Dividing (22) by $z\sqrt{z^{2}+1}$ and
integrating, we obtain

$\displaystyle\int^{s_{*}}_{s_{1}}\frac{1}{z\sqrt{z^{2}+1}}\frac{dz}{ds}\,ds$
$\displaystyle\leq\frac{C_{1}}{x_{1}}\leq(2n-1)\sqrt{z_{1}^{2}+1}\Biggl{(}\frac{\sqrt{x_{0}^{2}+1+z_{1}^{-2}}}{x_{0}}\sqrt{z_{1}^{2}+1}+\frac{1}{z_{1}^{2}x_{0}^{2}}\Biggr{)},$
(23)

Towards a contradiction, suppose that $s\mapsto z(s)$ is unbounded for any
choice of initial value $x_{0}>1$. Then, in particular, we may choose
$s_{1}\in]0,s_{*}[$ such that $z_{1}=z(s_{1})=1/x_{0}$. With this choice, the
right-hand side of (23) is bounded from above by $12n$ uniformly in $x_{0}>1$,
while the left-hand side of (23) is bounded from below by

$\displaystyle\int^{1}_{1/x_{0}}\frac{1}{z\sqrt{z^{2}+1}}\,dz$

which diverges as $x_{0}\to\infty$. This contradiction yields a global upper
bound on $z$ provided that $x_{0}$ is chosen sufficiently large. Rephrasing
this outcome back in the original variables, the third component $\alpha$ of
trajectories with $r_{0}$ close enough to $\pi/2$ remains negative, which
means that such trajectories exit the domain $B$ from the side $r=\pi/2$, as
claimed. ∎

We can now proceed and wrap things together to prove our main theorem.

###### Proof of Theorem 1.1.

Appealing to Lemma 3.3 and 3.4 respectively, we can find
$r^{\prime}_{0},r^{\prime\prime}_{0}\in]0,\pi/2[$ with
$r^{\prime}_{0}<r^{\prime\prime}_{0}$ such that the trajectories emanating
from $(r^{\prime}_{0},\pi/4,-\pi/2)$ and $(r^{\prime\prime}_{0},\pi/4,-\pi/2)$
leave the domain $B$ from the roof ($\alpha=0$) and from the side ($r=\pi/2$)
respectively. Now, consider the image through the flow map of the relatively
compact interval $[r^{\prime}_{0},r^{\prime\prime}_{0}]\subset]0,\pi/2[$. It
is understood that each trajectory is followed till $\partial B$, where there
is a transverse crossing, and not beyond. In particular, by considering the
exit time along each trajectory the flow defines a $C^{0}$ curve, whose image
will be henceforth denoted by $\Gamma$. Note that (by continuous dependence on
initial data) $\Gamma$ is a connected set and by definition (keeping in mind
Lemma 3.1) its image will be contained in the dihedron

$\Delta\vcentcolon=(B\cap\left\\{\alpha=0\right\\})\cup(B\cap\left\\{r=\pi/2\right\\}).$

Furthermore, $\Gamma$ contains points both on $B\cap\left\\{\alpha=0\right\\}$
and on $B\cap\left\\{r=\pi/2\right\\}$. Hence, by connectedness $\Gamma$ shall
also contain point on the intersection

$(B\cap\left\\{\alpha=0\right\\})\cap(B\cap\left\\{r=\pi/2\right\\}).$

In other words, if we get back to the picture in the planar domain with
coordinates $(r,\theta)$ only, we have singled out a monotone arc that starts
at a point $(r_{0},\pi/4)$ and reaches the segment $r=\pi/2$ orthogonally.
Hence, by simply reflecting such a trajectory along the axes $r=\pi/2$ and
$\theta=\pi/4$ of our planar domain $(r,\theta)\in]0,\pi[\times]0,\pi/2[$ we
get a smooth closed orbit solving our ODE system. Thereby, by virtue of the
equivariant reduction we presented at the beginning of the present section,
the conclusion follows. ∎

######  Remark 3.6.

Concerning a different geometric problem, we recall here that Angenent
[Angenent1992] used the shooting method to construct an embedded, rotationally
symmetric, self-shrinking $S^{1}\times S^{n-1}$ in $\mathbb{R}^{n+1}$.
Concerning, instead, the contributions we will present in the next section,
note that, later, more examples of immersed, rotationally symmetric self-
shrinkers were found in [Drugan2018].

Numerical simulations seem to clearly indicate that the closed periodic orbits
constructed above are unique, and thus the corresponding minimal embedding of
$S^{n-1}\times S^{n-1}\times S^{1}$ in $\mathbb{S}^{2n}$ should be unique in
the class of $G$-equivariant maps for any $n\geq 2$. In fact, in the case when
$n=2$ we wish to formulate here a stronger conjecture:

######  Conjecture 3.7.

There exists, up to ambient isometry, a unique minimal embedding of the three-
dimensional torus in the round four-dimensional sphere.

This can be regarded as a higher-dimensional counterpart of Lawson’s
conjecture, confirmed by Brendle in [Brendle2013]. It is not clear at the
moment whether we have compelling evidence to envision a similar landscape in
$\mathbb{S}^{m+1}$ when $m>3$ (and especially for $m>4$, cf. Appendix A), i.
e. whether minimal embeddings of $T^{m}$ in $\mathbb{S}^{m+1}$ should actually
exist and, if so, whether such embeddings should be unique. These questions
stand as fascinating, yet challenging open problems in the field.

##  4 Existence of infinitely many minimally immersed hypertori and minimally
embedded hyperspheres

To ease the notation in this section, we introduce the shifted variable
$\vartheta(s)=\theta(s)-\pi/4$. Then, the system (3),(4),(5) takes the form

$\displaystyle\frac{dr}{ds}$ $\displaystyle=\cos(\alpha),$ (24)
$\displaystyle\smash{\left\\{\vphantom{\begin{aligned}
\frac{dr}{ds}&=\cos(\alpha),\\\\[4.30554pt]
\frac{d\vartheta}{ds}&=\frac{\sin(\alpha)}{\sin(r)},\\\\[4.30554pt]
\frac{d\alpha}{ds}&=-\frac{(2n-2)\tan(2\vartheta)}{\sin(r)}\cos(\alpha)-(2n-1)\cot(r)\sin(\alpha).\end{aligned}}\right.}\,\frac{d\vartheta}{ds}$
$\displaystyle=\frac{\sin(\alpha)}{\sin(r)},$ (25)
$\displaystyle\frac{d\alpha}{ds}$
$\displaystyle=-\frac{(2n-2)\tan(2\vartheta)}{\sin(r)}\cos(\alpha)-(2n-1)\cot(r)\sin(\alpha).$
(26)

We will study solutions of the system (24),(25),(26) with initial data
$(r,\vartheta,\alpha)(0)=(r_{0},0,\alpha_{0})$, where $r_{0}\in[\pi/2,\pi[$
and $\alpha_{0}\in]0,\pi/2[$. We will then show that if $n\in\\{2,3\\}$ and if
$\alpha_{0}>0$ is sufficiently small, the trajectory will intersect
$\\{\vartheta=0\\}$ again for some $s>0$. Moreover, $\alpha$ will be
arbitrarily close to zero at the point of intersection if $\alpha_{0}>0$ is
sufficiently small. Since the problem is symmetric with respect to the
reflection $(r,\vartheta,\alpha)\mapsto(r,-\vartheta,-\alpha)$ we may iterate
our estimates, and will prove the existence of trajectories whose second
component $\vartheta$ has arbitrarily many zeros.

######  Lemma 4.1.

Given any $r_{0}\in[\pi/2,\pi[$ and any $\alpha_{0}\in]0,\pi/2[$ there exists
$0<s_{*}<\infty$ such that the system (24),(25),(26) has a unique solution

$(r,\vartheta,\alpha)\colon[0,s_{*}[\to
D\vcentcolon=[\tfrac{\pi}{2},\pi[\times]-\tfrac{\pi}{4},\tfrac{\pi}{4}[\times]-\tfrac{\pi}{2},\tfrac{\pi}{2}[$

with initial data $(r,\vartheta,\alpha)(0)=(r_{0},0,\alpha_{0})$, on which the
solution depends continuously; furthermore $dr/ds>0$ and one of the following
alternatives holds:

$\displaystyle\lim_{s\to s_{*}}r(s)$ $\displaystyle=\pi,$
$\displaystyle\limsup_{s\to s_{*}}\lvert\vartheta(s)\rvert$
$\displaystyle=\frac{\pi}{4},$ $\displaystyle\limsup_{s\to
s_{*}}\lvert\alpha(s)\rvert$ $\displaystyle=\frac{\pi}{2}.$ (27)

Moreover, if either $\alpha(s)$ or $\vartheta(s)$ converges to a nonzero limit
as $s\to s_{*}$, then

$\displaystyle\lim_{s\to s_{*}}r(s)<\pi.$ (28)

###### Proof.

By our choice of initial data, there exists a solution to system
(24),(25),(26) which stays in $D$ at least for a short time. Since
$-\pi/2<\alpha<\pi/2$, equation (24) implies $dr/ds=\cos(\alpha)>0$ as
claimed. The domain $D$ is bounded, $r$ is increasing and
$(dr/ds)^{2}+(d\vartheta/ds)^{2}\geq 1$ by (2). Hence, there exists
$s_{*}<\infty$ such that one of the alternatives in (27) holds.

_Case 1:_ $\lim_{s\to s_{*}}\alpha(s)=\alpha_{*}\neq 0$ exists. Since the
problem is invariant under the reflection
$(r,\vartheta,\alpha)\mapsto(r,-\vartheta,-\alpha)$ we may assume
$\alpha_{*}>0$ without loss of generality. By continuity of $\alpha$ there
exist $\varepsilon\in]0,\pi/2[$ and $0<s_{\varepsilon}<s_{*}$ such that
$\alpha(s)\geq\varepsilon$ for all $s\in[s_{\varepsilon},s_{*}[$. Equations
(24) and (25) then imply

$\displaystyle\frac{d\vartheta}{ds}$
$\displaystyle=\frac{\tan(\alpha)}{\sin(r)}\frac{dr}{ds}\geq\frac{\varepsilon}{\sin(r)}\frac{dr}{ds}.$
(29)

Integrating (29) from $s=s_{\varepsilon}$ to any $s\in]s_{\varepsilon},s_{*}[$
yields

$\displaystyle\frac{\pi}{2}\geq\vartheta(s)-\vartheta(s_{\varepsilon})\geq\varepsilon\log\biggl{(}\tan\biggl{(}\frac{r(s)}{2}\biggr{)}\biggr{)}-\varepsilon\log\biggl{(}\tan\biggl{(}\frac{r(s_{\varepsilon})}{2}\biggr{)}\biggr{)}.$
(30)

Passing to the limit $s\to s_{*}$ in (30), keeping in mind the monotonicity of
$s\mapsto r(s)$, proves (28).

_Case 2:_ $\lim_{s\to s_{*}}\vartheta(s)=\vartheta_{*}\neq 0$ exists. Without
loss of generality we may assume $\vartheta_{*}>0$ by symmetry of the problem.
Therefore, there exist $\varepsilon>0$ and $0<s_{\varepsilon}<s_{*}$ such that
$\tan(2\vartheta)\geq\varepsilon$ for all $s\in[s_{\varepsilon},s_{*}[$.
Equations (24)–(26) imply

$\displaystyle\frac{d\alpha}{ds}$
$\displaystyle=-(2n-2)\frac{\tan(2\vartheta)}{\sin(r)}\frac{dr}{ds}-(2n-1)\cos(r)\frac{d\vartheta}{ds}.$
(31)

Integrating (31), we obtain

$\displaystyle\int^{s_{*}}_{s_{\varepsilon}}\frac{(2n-2)\varepsilon}{\sin(r)}\frac{dr}{ds}\,ds$
$\displaystyle\leq-\int^{s_{*}}_{s_{\varepsilon}}\frac{d\alpha}{ds}\,ds-(2n-1)\int^{s_{*}}_{s_{\varepsilon}}\cos(r)\frac{d\vartheta}{ds}\,ds$
$\displaystyle=\int^{s_{*}}_{s_{\varepsilon}}\biggl{(}(2n-1)\frac{d\vartheta}{ds}-\frac{d\alpha}{ds}\biggr{)}\,ds-(2n-1)\int^{s_{*}}_{s_{\varepsilon}}\frac{1+\cos(r)}{\sin(r)}\sin(\alpha)\,ds.$
(32)

Noting that $\lvert(1+\cos(r))/\sin(r)\rvert\leq 1$ for $r\in[\pi/2,\pi[$ we
conclude that the right-hand side of (32) is bounded from above by
$(n+1)\pi+(2n-1)(s_{*}-s_{\varepsilon})$. Since $r(s)$ is increasing in $s$,
its limit as $s\to s_{*}$ exists, and (28) readily follows. ∎

###  4.1 Preparation: Estimates for small initial angles

######  Lemma 4.2 (see Figure 2).

Given initial data $r_{0}\in[\pi/2,\pi[$ and $\alpha_{0}\in]0,\pi/2[$ let
$(r,\vartheta,\alpha)\colon[0,s_{*}[\to D$ be the solution constructed in
Lemma 4.1. Then there exists $s_{1}\in]0,s_{*}[$ such that

$\displaystyle\vartheta(s_{1})$
$\displaystyle=\frac{1}{4n}\Bigl{(}\frac{\pi}{2}-\alpha_{0}\Bigr{)}\alpha_{0}.$

Moreover, $0<\vartheta(s)\leq\vartheta(s_{1})$ and
$\tan\bigl{(}\alpha(s)\bigr{)}\geq\frac{1}{2}\tan(\alpha_{0})$ for all
$s\in]0,s_{1}]$.

###### Proof.

Let $s_{1}\in[0,s_{*}]$ be largest with the property that

$\displaystyle
0\leq\vartheta(s)\leq\frac{1}{4n}\Bigl{(}\frac{\pi}{2}-\alpha_{0}\Bigr{)}\alpha_{0}$
(33)

for all $s\in[0,s_{1}[$. Then, $s_{1}>0$ because $\vartheta(0)=0$ and
$\alpha_{0}\in]0,\pi/2[$. Similarly and independently we consider the largest
$s_{\alpha}\in]0,s_{*}]$ with the property that $\alpha(s)>0$ for all
$s\in[0,s_{\alpha}[$. By equation (25) we have $d\vartheta/ds\geq 0$ in the
interval $[0,s_{\alpha}]$ and $\cos(r)\leq 0$ since $r\in[\pi/2,\pi]$. Thus,
(25) and (26) imply

$\displaystyle\frac{d\alpha}{ds}$
$\displaystyle=\Bigl{(}-(2n-2)\tan(2\vartheta)\cot(\alpha)-(2n-1)\cos(r)\Bigr{)}\frac{d\vartheta}{ds}\geq-(2n-2)\tan(2\vartheta)\cot(\alpha)\frac{d\vartheta}{ds}$
(34)

for all $s\in[0,s_{\alpha}]$. Multiplying estimate (34) by $\tan(\alpha)>0$,
we obtain

$\displaystyle-\frac{d}{ds}\log\bigl{(}\cos(\alpha)\bigr{)}=\tan(\alpha)\frac{d\alpha}{ds}$
$\displaystyle\geq-(2n-2)\tan(2\vartheta)\frac{d\vartheta}{ds}=(n-1)\frac{d}{ds}\log\bigl{(}\cos(2\vartheta)\bigr{)}.$
(35)

Integrating (35) from $s=0$ to any $s\in[0,s_{\alpha}[$ implies

$\displaystyle\frac{\cos(\alpha_{0})}{\cos(\alpha)}$
$\displaystyle\geq\bigl{(}\cos(2\vartheta)\bigr{)}^{n-1}.$ (36)

Estimate (36) is equivalent to
$\cos(\alpha)\leq\bigl{(}\cos(2\vartheta)\bigr{)}^{1-n}\cos(\alpha_{0})$.
Since $\tan\bigl{(}\arccos(x)\bigr{)}=\sqrt{x^{-2}-1}$ for $0<x\leq 1$, we
obtain for any $s\in[0,s_{\alpha}[$

$\displaystyle\tan(\alpha)$
$\displaystyle\geq\sqrt{\frac{\cos^{2n-2}(2\vartheta)}{\cos^{2}(\alpha_{0})}-1}=\tan(\alpha_{0})\sqrt{\frac{\cos^{2n-2}(2\vartheta)-\cos^{2}(\alpha_{0})}{\sin^{2}(\alpha_{0})}}.$
(37)

On the other hand, in $[0,s_{1}[$ the assumption $0\leq
2\vartheta\leq(\pi/2-\alpha_{0})\alpha_{0}/(2n)\leq\alpha_{0}/n$ implies
$\cos(2\vartheta)\geq\cos(\alpha_{0}/n)$; the map
$f\colon]0,\pi/2]\to\mathbb{R}$ given by

$\displaystyle f(a)=\frac{\cos^{2n-2}(a/n)-\cos^{2}(a)}{\sin^{2}(a)}$

is decreasing with minimum value $f(\pi/2)>1/4$ (since $n\geq 2$), hence we
obtain

$\displaystyle\tan(\alpha)\geq\tan(\alpha_{0})\sqrt{f(\alpha_{0})}\geq\frac{1}{2}\tan(\alpha_{0})$
(38)

for all $s\in[0,s_{\alpha}[\cap[0,s_{1}[$. In particular, (38) implies that
$\alpha$ is strictly positive in the intersection of both intervals. Hence,
$s_{1}\leq s_{\alpha}$ because otherwise, we would have $s_{\alpha}<s_{1}\leq
s_{*}$ and thus $\alpha(s_{\alpha})=0$ in contradiction to (38). In
particular, $\vartheta$ is increasing in $[0,s_{1}[$. It remains to show
$s_{1}<s_{*}$ because then the upper bound in (33) must be attained and the
claim follows.

Since equations (25) and (26) imply $d\alpha/ds\leq(2n-1)d\vartheta/ds$ in
$[0,s_{1}[$ we have

$\displaystyle\alpha(s)$
$\displaystyle\leq\alpha_{0}+(2n-1)\vartheta(s)\leq\alpha_{0}+\frac{1}{2}\Bigl{(}\frac{\pi}{2}-\alpha_{0}\Bigr{)}\alpha_{0}<\frac{\pi}{2}$
(39)

for all $s\in[0,s_{1}[$. Indeed, the function $h\colon[0,\pi/2]\to\mathbb{R}$
given by $h(a)=a+(\pi/2-a)a/2$ satisfies $h^{\prime}(a)=1+\pi/4-a>0$.
Therefore $h(\alpha_{0})<h(\pi/2)=\pi/2$ for any choice of
$\alpha_{0}\in]0,\pi/2[$.

If $s_{1}=s_{*}$ then (39) and (33) together with Lemma 4.1 (27) imply
$r(s)\to\pi$ as $s\to s_{*}$. However, since $\vartheta$ is bounded and
increasing in $[0,s_{1}[$, the limit $\lim_{s\to s_{*}}\vartheta(s)>0$ exists.
Hence, Lemma 4.1 (28) applies, which yields a contradiction. As a result,
$s_{1}<s_{*}$ and the claim follows. ∎

######  Lemma 4.3 (see Figure 2).

Let $n\in\\{2,3\\}$ and let $\delta>0$ be arbitrary. Then there exists
$0<a_{\delta}<\pi/2$ such that for any $\alpha_{0}\in]0,a_{\delta}]$ and any
$r_{0}\in[\pi/2,\pi[$ the solution $(r,\vartheta,\alpha)\colon[0,s_{*}[\to D$
with $(r,\vartheta,\alpha)(0)=(r_{0},0,\alpha_{0})$ as constructed in Lemma
4.1 allows for $s_{2}\in]0,s_{*}[$ such that

$\alpha(s_{2})=0$

and, in addition, $\alpha(s)>0$ and $\tan\bigl{(}2\vartheta(s)\bigr{)}<\delta$
for all $s\in[0,s_{2}[$.

$r$$\vartheta$$0$$-\frac{\pi}{4}$$\frac{\pi}{2}$$\pi$$0$$s_{2}$$s_{*}$$r_{0}$$\vartheta(s_{2})$$s_{1}$
Figure 2: Visualisation of Lemmata 4.2 and 4.3 for $n=2$, $r_{0}=2\pi/3$ and
$\alpha_{0}=\pi/36$.

###### Proof.

Let $s_{1}\in]0,s_{*}[$ be as in Lemma 4.2. Let $s_{2}\in]0,s_{*}]$ be largest
with the property that

$\displaystyle 0<\alpha(s)<\tfrac{1}{3}$ (40)

for all $s\in[0,s_{2}[$. We claim that if $\alpha_{0}>0$ is sufficiently
small, then $s_{2}>s_{1}$. Since Lemma 4.2 implies that $\alpha$ is positive
in $[0,s_{1}]$ it suffices to prove $\alpha(s)<1/3$ for all $s\in[0,s_{1}]$.
As in estimate (39) we obtain for all $s\in[0,s_{1}]$

$\displaystyle\alpha(s)\leq\alpha_{0}+(2n-1)\vartheta(s)\leq\alpha_{0}+\frac{2n-1}{4n}\Bigl{(}\frac{\pi}{2}-\alpha_{0}\Bigr{)}\alpha_{0}\leq
2\alpha_{0}.$ (41)

In particular, we have $\alpha(s)<1/3$ for all $s\in[0,s_{1}]$ provided that
$\alpha_{0}<1/6$. Thus, $s_{2}>s_{1}$ for any choice of $0<\alpha_{0}<1/6$.
Since $\vartheta(s)>0$ for all $s\in[s_{1},s_{2}[$ there exists
$\beta\colon[s_{1},s_{2}[\to\mathbb{R}$ such that

$\displaystyle\alpha(s)=\vartheta(s)\,\beta(s)$

for all $s\in[s_{1},s_{2}[$. Equations (25) and (26) then imply

$\displaystyle\vartheta\frac{d\beta}{ds}$
$\displaystyle=\Bigl{(}-(2n-2)\tan(2\vartheta)\cot(\alpha)-\beta-(2n-1)\cos(r)\Bigr{)}\frac{d\vartheta}{ds}$
$\displaystyle\leq\Bigl{(}-4(n-1)\vartheta\cot(\alpha)-\frac{\alpha}{\vartheta}+(2n-1)\Bigr{)}\frac{d\vartheta}{ds}$
$\displaystyle\leq\Bigl{(}-4\sqrt{(n-1)\alpha\cot(\alpha)}+(2n-1)\Bigr{)}\frac{d\vartheta}{ds}$
(42)

in $[s_{1},s_{2}[$, where the last inequality above relies on the standard AM-
GM inequality. The assumptions $n\in\\{2,3\\}$ and $0<\alpha<1/3$ in
$[s_{1},s_{2}[$ imply

$\displaystyle-4\sqrt{(n-1)\alpha\cot(\alpha)}+(2n-1)<-\frac{1}{2}.$ (43)

The negativity of the left-hand side of (43) is crucial here and in fact fails
for $n\geq 4$ even if $\alpha$ is arbitrarily small. Dividing estimate (42) by
$\vartheta>0$ and integrating from $s=s_{1}$ to any $s\in[s_{1},s_{2}[$, we
obtain

$\displaystyle\beta(s)-\beta(s_{1})$
$\displaystyle\leq\frac{1}{2}\log\bigl{(}\vartheta(s_{1})\bigr{)}-\frac{1}{2}\log\bigl{(}\vartheta(s)\bigr{)}.$
(44)

Moreover, since
$\vartheta(s_{1})=(\pi/2-\alpha_{0})\alpha_{0}/(4n)\geq\alpha_{0}/9$ for
$\alpha_{0}<1/6$ and $n\in\\{2,3\\}$ we have

$\displaystyle\beta(s_{1})=\frac{\alpha(s_{1})}{\vartheta(s_{1})}\leq\frac{\alpha_{0}}{\vartheta(s_{1})}+(2n-1)\leq
9+5=14.$

We also have $\vartheta(s_{1})\leq\pi\alpha_{0}/(8n)<\alpha_{0}/5$. Estimate
(44) then implies for any $s\in[s_{1},s_{2}[$

$\displaystyle\alpha$
$\displaystyle\leq\biggl{(}14+\frac{1}{2}\log\Bigl{(}\frac{\alpha_{0}}{5}\Bigr{)}\biggr{)}\vartheta-\frac{\vartheta}{2}\log(\vartheta).$
(45)

$\vartheta$$0$$1$$\vartheta_{0}$$\frac{1}{10}$ Figure 3: Plot of the function
$\vartheta\mapsto\Bigl{(}14+\frac{1}{2}\log\bigl{(}\alpha_{0}/5\bigr{)}\Bigr{)}\vartheta-\frac{1}{2}\vartheta\log(\vartheta)$
for $\alpha_{0}=e^{-27}$.

The function on the right-hand side of (45) can be made arbitrarily small by
choosing $\alpha_{0}>0$ sufficiently small. In particular, we can achieve
(say) $\alpha<1/10$ for all $s\in[s_{1},s_{2}[$. Moreover, as illustrated in
Figure 3, the right-hand side of (45) as a function of $\vartheta>0$ vanishes
for some $\vartheta_{0}>0$ which again is arbitrarily small if $\alpha_{0}>0$
is chosen sufficiently small. Since $\vartheta$ is increasing in $[0,s_{2}[$,
we obtain $\vartheta(s)<\vartheta_{0}$ for all $s\in[0,s_{2}[$.

It remains to prove $s_{2}<s_{*}$ because then the lower bound in (40) must be
attained and $\alpha(s_{2})=0$ would follow. Towards a contradiction we assume
$s_{2}=s_{*}$. Then, the bounds on $\vartheta$ and $\alpha$ imply $r(s)\to\pi$
as $s\to s_{*}$ by Lemma 4.1 (27). However, since $\vartheta$ is increasing it
has a nonzero limit as $s\to s_{*}$. Consequently, Lemma 4.1 (28) applies,
which yields a contradiction. ∎

######  Lemma 4.4.

Let $n\in\\{2,3\\}$ and let $0<\varepsilon<\pi/2$ be arbitrary. Then there
exists $a_{\varepsilon}\in]0,\pi/2[$ such that for any $r_{0}\in[\pi/2,\pi[$
and any $\alpha_{0}\in]0,a_{\varepsilon}]$ the solution
$(r,\vartheta,\alpha)\colon[0,s_{*}[\to D$ with initial data
$(r,\vartheta,\alpha)(0)=(r_{0},0,\alpha_{0})$ as constructed in Lemma 4.1
allows for $s_{3}\in]0,s_{*}[$ such that

$\displaystyle\vartheta(s_{3})$ $\displaystyle=0,$
$\displaystyle\lvert\alpha(s_{3})\rvert$ $\displaystyle<\varepsilon$

and such that $\vartheta(s)>0$ for all $s\in]0,s_{3}[$.

###### Proof.

Given any $\delta>0$ let $0<\alpha_{0}<a_{\delta}$ and $s_{2}\in]0,s_{*}[$ be
as in Lemma 4.3. Let $s_{3}\in[0,s_{*}]$ be largest with the property that
$\vartheta(s)\geq 0$ for all $s\in[0,s_{3}[$. By Lemma 4.3 we have $\alpha>0$
and hence $d\vartheta/ds>0$ in the interval $[0,s_{2}[$. Therefore,
$s_{2}<s_{3}$. Concerning the subsequent part of the motion, equation (26)
implies

$\displaystyle\frac{d\alpha}{ds}\leq-\frac{(2n-2)\tan(2\vartheta)}{\sin(r)}\cos(\alpha)<0$
(46)

as long as $\vartheta>0$ and $\alpha\leq 0$. In the interval $]s_{2},s_{3}[$,
we thus have $\alpha<0$ and $d\vartheta/ds<0$. Therefore, the bound
$\tan(2\vartheta)\leq\delta$ from Lemma 4.3 extends to the interval
$[s_{2},s_{3}[$. Equations (25) and (26) then yield

$\displaystyle\frac{d\alpha}{ds}$
$\displaystyle=\Bigl{(}-(2n-2)\tan(2\vartheta)\cot(\alpha)-(2n-1)\cos(r)\Bigr{)}\frac{d\vartheta}{ds}\geq\Bigl{(}-\delta\cot(\alpha)+1\Bigr{)}(2n-1)\frac{d\vartheta}{ds}$
(47)

in $[s_{2},s_{3}[$. Dividing inequality (47) by $(-\delta\cot(\alpha)+1)>0$
and integrating from $s=s_{2}$ to any $s\in[s_{2},s_{3}[$ we obtain

$\displaystyle\frac{\alpha+\delta\log\bigl{(}\delta\cos(\alpha)-\sin(\alpha)\bigr{)}}{\delta^{2}+1}-\frac{\delta\log(\delta)}{\delta^{2}+1}$
$\displaystyle\geq(2n-1)\bigl{(}\vartheta-\vartheta(s_{2})\bigr{)}\geq-(2n-1)\delta$

which implies

$\displaystyle\frac{\alpha(s)}{\delta^{2}+1}$
$\displaystyle\geq-(2n-1)\delta-\frac{\delta\log(\delta+1)}{\delta^{2}+1}+\frac{\delta\log(\delta)}{\delta^{2}+1}$
(48)

for any $s\in[s_{2},s_{3}[$. The right-hand side of (48) converges to $0$ as
$\delta\searrow 0$. Therefore, we may choose $\delta>0$ such that (48) implies
$\alpha(s)>-\varepsilon$ for all $s\in[s_{2},s_{3}[$.

We must have $s_{3}<s_{*}$ because otherwise, Lemma 4.1 (27) implies
$r(s)\to\pi$ as $s\to s_{*}$ which by Lemma 4.1 (28) contradicts the fact that
$\alpha(s)$ must have a nonzero limit as $s\to s_{*}$ due to (46).
Consequently, $\vartheta(s_{3})=0$ and the claim follows. ∎

######  Lemma 4.5.

Let $n\in\\{2,3\\}$ and let $0<\varepsilon<\pi/2$ and $1\leq k\in\mathbb{N}$
be arbitrary. Then there exists $a_{\varepsilon,k}\in]0,\pi/2[$ such that for
any $\alpha_{0}\in]0,a_{\varepsilon,k}]$ the solution
$(r,\vartheta,\alpha)\colon[0,s_{*}]\to D$ with initial data
$(r,\vartheta,\alpha)(0)=(\pi/2,0,\alpha_{0})$ as constructed in Lemma 4.1
allows $0<\sigma_{1}<\sigma_{2}<\ldots<\sigma_{k}<s_{*}$ such that for all
$j\in\\{1,\ldots,k\\}$

$\displaystyle\vartheta(\sigma_{j})$ $\displaystyle=0,$
$\displaystyle\lvert\alpha(\sigma_{j})\rvert$ $\displaystyle\leq\varepsilon.$

###### Proof.

We prove the claim by induction in $k$. The case $k=1$ follows by Lemma 4.4
with $\sigma_{1}=s_{3}$. Given any fixed $\varepsilon>0$ let
$a_{\varepsilon}\in]0,\pi/2[$ be as in Lemma 4.4. If the statement is true for
a certain $k\in\mathbb{N}$, we may apply it with
$\min\\{\varepsilon,a_{\varepsilon}\\}$ in place of $\varepsilon$ to further
obtain $\lvert\alpha(\sigma_{k})\rvert\leq a_{\varepsilon}$. Since the problem
is invariant under the reflection
$(r,\vartheta,\alpha)\mapsto(r,-\vartheta,-\alpha)$ we may reapply Lemmata
4.1–4.4 with initial data
$\bigl{(}r(\sigma_{k}),0,\lvert\alpha(\sigma_{k})\rvert\bigr{)}$ to conclude
that the original solution allows $\sigma_{k+1}\in]\sigma_{k},s_{*}[$ with
$\vartheta(\sigma_{k+1})=0$ and
$\lvert\alpha(\sigma_{k+1})\rvert\leq\varepsilon$. ∎

The following estimate can be interpreted as a complement to Lemma 4.4, which
we will crucially employ at the end of the next section.

######  Lemma 4.6.

Given $r_{0}\in]\pi/2,\pi[$ and $\alpha_{0}\in]0,\pi/2[$ let
$(r,\vartheta,\alpha)\colon[0,s_{*}[\to D$ be the solution to system
(24),(25),(26) with initial data
$(r,\vartheta,\alpha)(0)=(r_{0},0,\alpha_{0})$. If there exists
$\sigma\in]0,s_{*}[$ such that $\vartheta(\sigma)=0$ and $\vartheta(s)>0$ for
all $s\in]0,\sigma[$, then

$\displaystyle\bigg{\lvert}\frac{\alpha(\sigma)}{\alpha_{0}}\bigg{\rvert}$
$\displaystyle\geq\max\left\\{1,\leavevmode\nobreak\
\frac{2n-1}{2n-2}\big{\lvert}\cos(r_{0})\big{\rvert}\right\\}.$

Moreover, there exists $s_{2}\in]0,\sigma[$ such that

$\displaystyle\max_{s\in[0,\sigma]}\lvert\vartheta(s)\rvert$
$\displaystyle=\vartheta(s_{2})\geq\frac{\alpha_{0}}{2n-2}.$

###### Proof.

By assumption, $\vartheta$ must attain an interior maximum in $[0,\sigma]$. In
particular, by equations (25) and (26), there exists $s_{2}\in]0,\sigma[$ such
that

$\displaystyle\vartheta(s_{2})$ $\displaystyle>0,$
$\displaystyle\alpha(s_{2})$ $\displaystyle=0.$ (49)

By equation (26), we have $d\alpha/ds(s_{2})<0$, and in fact $d\alpha/ds(s)<0$
as long as $\vartheta(s)>0$ and $\alpha(s)\leq 0$. Therefore, $\alpha(s)$ is
decreasing in $[s_{2},\sigma[$ and $s_{2}\in]0,\sigma[$ is uniquely determined
by property (49). As in (36) above, we can show

$\displaystyle\bigl{(}\cos\bigl{(}2\vartheta(s_{2})\bigr{)}\bigr{)}^{n-1}$
$\displaystyle\leq\cos(\alpha_{0}).$ (50) This implies
$\displaystyle\cos\bigl{(}2\vartheta(s_{2})\bigr{)}$
$\displaystyle\leq\bigl{(}\cos(\alpha_{0})\bigr{)}^{\frac{1}{n-1}}\leq\cos\Bigl{(}\frac{\alpha_{0}}{n-1}\Bigr{)},$
(51)

where the last inequality follows for all $\alpha_{0}\in]0,\pi/2[$ and all
$2\leq n\in\mathbb{N}$ simply by computing

$\displaystyle\frac{d}{d\alpha}\bigl{(}\cos(\alpha)\bigr{)}^{\frac{1}{n-1}}$
$\displaystyle=\frac{-\sin(\alpha)}{n-1}\bigl{(}\cos(\alpha)\bigr{)}^{-\frac{n-2}{n-1}}\leq\frac{-\sin(\alpha)}{n-1}\leq\frac{-1}{n-1}\sin\Bigl{(}\frac{\alpha}{n-1}\Bigr{)}=\frac{d}{d\alpha}\cos\Bigl{(}\frac{\alpha}{n-1}\Bigr{)}.$

In particular, (51) implies

$\displaystyle\vartheta(s_{2})$ $\displaystyle\geq\frac{\alpha_{0}}{2n-2}.$
(52)

For any $s\in[s_{2},\sigma]$ we have $d\vartheta/ds(s)\leq 0$ and

$\displaystyle\frac{d\alpha}{ds}$
$\displaystyle=\Bigl{(}-(2n-2)\tan(2\vartheta)\cot(\alpha)-(2n-1)\cos(r)\Bigr{)}\frac{d\vartheta}{ds}\leq-(2n-2)\tan(2\vartheta)\cot(\alpha)\frac{d\vartheta}{ds}.$
(53)

Multiplying by $\tan(\alpha)<0$ we obtain exactly the same estimate as in
(35), here valid for $s\in[s_{2},\sigma]$. Integration implies
$-\log\bigl{(}\cos\bigl{(}\alpha(\sigma)\bigr{)}\bigr{)}\geq-(n-1)\log\bigl{(}\cos\bigl{(}2\vartheta(s_{2})\bigr{)}\bigr{)}$
which is equivalent to

$\displaystyle\cos\bigl{(}\alpha(\sigma)\bigr{)}\leq\bigl{(}\cos\bigl{(}2\vartheta(s_{2})\bigr{)}\bigr{)}^{n-1}.$
(54)

Combining (50) and (54) we obtain $\cos(\alpha(\sigma))\leq\cos(\alpha_{0})$
which implies $\lvert\alpha(\sigma)\rvert\geq\lvert\alpha_{0}\rvert$. In
analogy to estimate to (53) we have
$d\alpha/ds\leq-(2n-1)\cos(r_{0})d\vartheta/ds$ for $s\in[s_{2},\sigma]$.
Integration then yields

$\displaystyle\alpha(\sigma)\leq(2n-1)\cos(r_{0})\,\vartheta(s_{2})\leq\frac{2n-1}{2n-2}\cos(r_{0})\,\alpha_{0}$

where we used $\cos(r_{0})\leq 0$ and estimate (52). Since
$\alpha(\sigma)<0<\alpha_{0}$ the claim follows. ∎

######  Remark 4.7.

For later reference, we explicitly note that it follows from the previous
lemma that

$-\frac{d\vartheta}{ds}(\sigma)\geq\frac{d\vartheta}{ds}(0)$

and that, furthermore, $\frac{d\vartheta}{ds}(s)$ is monotonically decreasing
on $[s_{2},\sigma]$.

###  4.2 Start of the Induction: The $\boldsymbol{\infty}$-figure

The construction of an immersed minimal hypertorus marks the start of an
iteration process which will ultimately provide a family of infinitely many
examples. We apply a continuity argument twice: it first yields an auxiliary
trajectory which we then suitably reflect before we use the argument a second
time. The entire approach, which we are about to explain in detail, is
summarised visually in Figure 4.

######  Lemma 4.8 (see Figure 4 (1)).

For every $\rho\in]\pi/2,\pi[$ there exists $a\in]0,\pi/2[$ such that for all
$r_{0}\in[\rho,\pi[$ and all $\alpha_{0}\in[a,\pi/2[$ the solution
$(r,\vartheta,\alpha)\colon[0,s_{*}[\to D$ with
$(r,\vartheta,\alpha)(0)=(r_{0},0,\alpha_{0})$ as constructed in Lemma 4.1
satisfies $\vartheta(s)>0$ for all $s\in]0,s_{*}[$ and $\alpha(s_{*})\to\pi/2$
as $s\nearrow s_{*}$.

###### Proof.

Let $\rho\in]\pi/2,\pi[$ be fixed and $r_{0}\in[\rho,\pi[$ arbitrary. We
recall from equations (24) and (25) that $r$ is increasing, and that
$\vartheta$ is also increasing as long as $\alpha>0$. Equation (26) implies
$d\alpha/ds(0)=-(2n-1)\cot(r_{0})\sin(\alpha_{0})>0$. Hence there exists
$s_{\alpha}\in]0,s_{*}]$ which is largest with the property that $\alpha$ is
(strictly) increasing in $s\in[0,s_{\alpha}[$. Let $s_{\vartheta}\in]0,s_{*}]$
instead be largest with the property that

$\displaystyle\tan\bigl{(}2\vartheta(s)\bigr{)}\leq-\cos(\rho)\tan(\alpha_{0}).$
(55)

for all $s\in[0,s_{\vartheta}[$. We claim $s_{\alpha}\geq s_{\vartheta}$.
Equations (25) and (26) imply

$\displaystyle\frac{d\alpha}{ds}$
$\displaystyle=\Bigl{(}-(2n-2)\tan(2\vartheta)\cot(\alpha)-(2n-1)\cos(r)\Bigr{)}\frac{d\vartheta}{ds}.$

If $s_{\alpha}<s_{\vartheta}$, then $d\alpha/ds(s_{\alpha})=0$. Thus, since
$-\cos(r)$ and $\tan(\alpha)$ are both increasing in $[0,s_{\alpha}]$,

$\displaystyle\tan\bigl{(}2\vartheta(s_{\alpha})\bigr{)}$
$\displaystyle=-\frac{(2n-1)}{(2n-2)}\cos\bigl{(}r(s_{\alpha})\bigr{)}\tan\bigl{(}\alpha(s_{\alpha})\bigr{)}>-\cos(\rho)\tan(\alpha_{0})$

in contradiction to (55). Hence, the claim holds and we have indeed
$s_{\alpha}\geq s_{\vartheta}$. In the interval $[0,s_{\alpha}[$, the estimate

$\displaystyle\frac{d\alpha}{ds}$
$\displaystyle\geq\Bigl{(}-(2n-2)\tan(2\vartheta)\cot(\alpha_{0})-(2n-1)\cos(\rho)\Bigr{)}\frac{d\vartheta}{ds}$
(56)

is true. Integrating (56) from $s=0$ to any $s\in]0,s_{\alpha}[$, we obtain

$\displaystyle\frac{\pi}{2}-\alpha_{0}\geq\alpha-\alpha_{0}$
$\displaystyle\geq(n-1)\cot(\alpha_{0})\log\bigl{(}\cos(2\vartheta)\bigr{)}-(2n-1)\cos(\rho)\vartheta.$
(57)

Now, towards a contradiction, suppose that $s_{\vartheta}<s_{*}$. Then, by
definition of $s_{\vartheta}$,

$\displaystyle\tan\bigl{(}2\vartheta(s_{\vartheta})\bigr{)}$
$\displaystyle=-\cos(\rho)\tan(\alpha_{0})=\vcentcolon c_{0}$
$\displaystyle\Rightarrow\leavevmode\nobreak\
\cos\bigl{(}2\vartheta(s_{\vartheta})\bigr{)}$
$\displaystyle=\bigl{(}c_{0}^{2}+1\bigr{)}^{-\frac{1}{2}}.$

Multiplying estimate (57) by $-2/\cos(\rho)>0$, we obtain

$\displaystyle\frac{\pi-2\alpha_{0}}{-\cos(\rho)}+2(n-1)\frac{\log\bigl{(}\cos(2\vartheta)\bigr{)}}{\cos(\rho)\tan(\alpha_{0})}$
$\displaystyle\geq(2n-1)2\vartheta$ which, specified to $s=s_{\vartheta}$,
yields
$\displaystyle\frac{\pi-2\alpha_{0}}{-\cos(\rho)}+(n-1)\frac{\log\bigl{(}c_{0}^{2}+1\bigr{)}}{c_{0}}$
$\displaystyle\geq(2n-1)\arctan(c_{0}).$ (58)

Since $c_{0}\to\infty$ as $\alpha_{0}\nearrow\pi/2$, the left-hand side of
(58) converges to zero as $\alpha_{0}\nearrow\pi/2$ while the right-hand side
of (58) converges to $(2n-1)\pi/2$. This contradiction proves
$s_{\vartheta}=s_{*}$ provided that $\alpha_{0}$ is sufficiently close to
$\pi/2$. Since $s_{\alpha}\geq s_{\vartheta}$ we also have $s_{\alpha}=s_{*}$
in this case. In particular, all three variables are increasing for
$s\in[0,s_{*}[$. Hence, $\vartheta(s)\to\vartheta_{*}>0$ and
$\alpha(s)\to\alpha_{*}>0$ as $s\nearrow s_{*}$. Thus, by Lemma 4.1 (28), we
have $\lim_{s\to s_{*}}r(s)<\pi$. Moreover, estimate (55) implies
$\vartheta_{*}<\pi/4$. Therefore, $\alpha_{*}=\pi/2$ follows by (27). ∎

######  Remark 4.9.

The assumption $r_{0}\neq\pi/2$ is crucial in the proof of Lemma 4.8. In fact,
(e. g. by analysing the second derivative of $\alpha$) one can show for
$r_{0}=\pi/2$ and any choice of $\alpha_{0}\in]0,\pi/2[$ that $\alpha(s)$ is
actually _decreasing_ while $\vartheta\geq 0$. In what follows, we specialise
to the case $r_{0}=2\pi/3$.

$r$$\vartheta$$0$$-\frac{\pi}{4}$$\frac{\pi}{2}$$\pi$$0$$\frac{2\pi}{3}$$r_{1}$$r_{\infty}$$\pi-
r_{1}$$\frac{\pi}{3}$(1)(2)(3)(4)(5)(6)

Figure 4: Trajectories used to construct the “$\infty$-figure”: (1)–(3) start
from fixed $r_{0}=2\pi/3$ with varying $\alpha_{0}\in]0,\pi/2[$ while (4)–(6)
start with fixed $\alpha_{0}=-\pi/2$ and $r_{0}\in]0,\pi/2[$ varies.

  1. (1)

see Lemma 4.8: If $\alpha_{0}$ is close to $\pi/2$ and $r_{0}>\pi/2$ then
$\alpha(s_{*})=\pi/2$.

  2. (2)

see Lemma 4.4: If $\alpha_{0}$ is close to $0$ then the trajectory intersects
$\\{\vartheta=0\\}$ for some $s>0$.

  3. (3)

see Lemma 4.10: Given $r_{0}\in]\pi/2,\pi[$ there exists
$\alpha_{0}\in]0,\pi/2[$ such that $(\vartheta,\alpha)(s_{*})=(0,-\pi/2)$.

  4. (4)

reflection of trajectory (3): Intersection with
$\\{\vartheta=0\\}\cap\\{r<\pi/2\\}$ for $r_{0}=\pi-r_{1}$.

  5. (5)

see Lemma 3.4: Intersection with $\\{r=\pi/2\\}$ if $r_{0}\in]0,\pi/2[$ is
close to $\pi/2$ and $\alpha_{0}=-\pi/2$.

  6. (6)

see Lemma 4.11: Intersection with $\\{r=\pi/2\\}\cap\\{\vartheta=0\\}$ for
some $r_{0}\in]0,\pi/2[$ and $\alpha_{0}=-\pi/2$.

######  Lemma 4.10 (see Figure 4 (3)).

Let $n\in\\{2,3\\}$ and $r_{0}=2\pi/3$. There exists $\alpha_{0}\in]0,\pi/2[$
such that the solution $(r,\vartheta,\alpha)\colon[0,s_{*}[\to D$ with initial
data $(r_{0},0,\alpha_{0})$ satisfies $\vartheta(s)>0$ for all $s\in]0,s_{*}[$
and

$\displaystyle\lim_{s\to
s_{*}}(r,\vartheta,\alpha)(s)=(r_{1},0,-\tfrac{\pi}{2})$

for some $r_{1}\in]r_{0},\pi[$.

###### Proof.

On the one hand, Lemma 4.4 implies that if $\alpha_{0}\in]0,\pi/2[$ is chosen
sufficiently small, then there exists $\sigma_{1}\in]0,s_{*}[$ such that
$\vartheta(\sigma_{1})=0$ and $\vartheta(s)>0$ for all $s\in]0,\sigma_{1}[$.
On the other hand, for any choice of $\alpha_{0}\in]0,\pi/2[$ which is
sufficiently close to $\pi/2$ Lemma 4.8 implies $\vartheta(s)>0$ for all
$s\in]0,s_{*}[$. Therefore, there exists $\tilde{\alpha}_{0}\in]0,\pi/2[$
which is largest with the property that for all
$\alpha_{0}\in]0,\tilde{\alpha}_{0}[$ the corresponding trajectory intersects
$\\{\vartheta=0\\}$ for some $s=\sigma_{1}>0$.

Let $(\tilde{r},\tilde{\vartheta},\tilde{\alpha})\colon[0,\tilde{s}_{*}[\to D$
be the solution to system (24),(25),(26) with initial data
$(\tilde{r},\tilde{\vartheta},\tilde{\alpha})(0)=(2\pi/3,0,\tilde{\alpha}_{0})$
as defined above. If there exists $\sigma_{1}\in]0,\tilde{s}_{*}[$ such that
$\tilde{\vartheta}(\sigma_{1})=0$, then $\tilde{\alpha}(\sigma_{1})\neq 0$
because otherwise $\tilde{\vartheta}(s)$ and $\tilde{\alpha}(s)$ vanish
identically. Therefore, $d\tilde{\vartheta}/ds(\sigma_{1})\neq 0$ and
${\sigma}_{1}$ is a nondegenerate zero of the map
$s\mapsto\tilde{\vartheta}(s)$. By continuous dependence on the initial data
there exists some $\tilde{\alpha}^{\prime}_{0}>\tilde{\alpha}_{0}$ such that
all solutions with $\alpha_{0}<\tilde{\alpha}^{\prime}_{0}$ still intersect
$\\{\vartheta=0\\}$. This contradicts our choice of $\tilde{\alpha}_{0}$.
Therefore, $\tilde{\vartheta}(s)>0$ for all $s\in]0,\tilde{s}_{*}[$. Moreover,
again by continuous dependence on the initial data, we have

$\displaystyle\lim_{s\to s_{*}}\tilde{\vartheta}(s)=0.$

Hence, $\tilde{\vartheta}$ attains an interior, positive local maximum at some
$s_{2}\in]0,\tilde{s}_{*}[$, where $\tilde{\alpha}(s_{2})=0$. Moreover, as
long as $\tilde{\vartheta}>0$ and $\tilde{\alpha}\leq 0$ equation (26) implies
$d\tilde{\alpha}/ds<0$. Therefore, $\tilde{\alpha}(s)$ is decreasing in
$]s_{2},\tilde{s}_{*}[$ and the limit

$\lim_{s\to s_{*}}\tilde{\alpha}(s)=\tilde{\alpha}_{*}<0$

exists. Lemma 4.1 (28) then implies $\lim_{s\to s_{*}}\tilde{r}(s)<\pi$ and
thus necessarily $\tilde{\alpha}_{*}=-\pi/2$ by (27). ∎

The following lemma differs from the previous one only in the fact that we
assume $r_{0}=\pi/2$ rather than $r_{0}=2\pi/3$. However, the proof is very
different because (as mentioned in Remark 4.9) we cannot rely on Lemma 4.8 in
the case $r_{0}=\pi/2$. Instead we aim for the reflection of the desired
trajectory using the results from Section 3.

######  Lemma 4.11 (see Figure 4 (6)).

Let $n\in\\{2,3\\}$. There exists $\alpha_{\infty}\in]0,\pi/2[$ such that the
solution $(r,\vartheta,\alpha)\colon[0,s_{\infty}[\to D$ with initial data
$(r,\vartheta,\alpha)(0)=(\pi/2,0,\alpha_{\infty})$ satisfies $\vartheta(s)>0$
for all $s\in]0,s_{\infty}[$ and

$\displaystyle\lim_{s\to
s_{\infty}}(r,\vartheta,\alpha)(s)=(r_{\infty},0,-\tfrac{\pi}{2})$

for some $r_{\infty}\in]\pi/2,\pi[$.

###### Proof.

As in Section 3, we consider solutions of the system (24),(25),(26) with
initial data

$\displaystyle r(0)$ $\displaystyle=r_{0}\in]0,\tfrac{\pi}{2}[,$
$\displaystyle\vartheta(0)$ $\displaystyle=0,$ $\displaystyle\alpha(0)$
$\displaystyle=-\tfrac{\pi}{2}$ (59)

(recalling $\vartheta=\theta-\pi/4$) but now we extend the domain
$]0,\tfrac{\pi}{2}]\times]-\tfrac{\pi}{4},0]\times[-\tfrac{\pi}{2},0]$ we had
employed there to

$\displaystyle\tilde{B}\vcentcolon=]0,\tfrac{\pi}{2}]\times]-\tfrac{\pi}{4},0]\times[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$

We recall that the motion is weakly monotone in each of the three coordinates
as long as $\alpha\leq 0$. If $\alpha$ becomes positive then we might loose
its monotonicity but we still have $dr/ds\geq 0$. As long as $\vartheta<0$ and
$r<\pi/2$, equation (26) ensures the implications

$\displaystyle\alpha=0$ $\displaystyle\leavevmode\nobreak\
\Rightarrow\leavevmode\nobreak\ \frac{d\alpha}{ds}>0,$
$\displaystyle\alpha=\frac{\pi}{2}$ $\displaystyle\leavevmode\nobreak\
\Rightarrow\leavevmode\nobreak\ \frac{d\alpha}{ds}<0.$

This implies that if $\alpha(s^{\prime})\geq 0$ for some
$s^{\prime}\in]0,s_{*}[$ then $\alpha(s)\in]0,\pi/2[$ for all
$s\in]s^{\prime},s_{*}]$. By Lemma 3.1 we also have $\vartheta>-\pi/4$ as long
as $\alpha\leq 0$. If $\alpha$ becomes positive, then the inequality
$d\vartheta/ds\geq 0$ ensures $\vartheta>-\pi/4$. To summarize, the trajectory
can exit $\tilde{B}$ only at $r=\pi/2$ or at $\vartheta=0$.

On the one hand, Lemma 3.4 states that if $r_{0}\in]0,\pi/2[$ is chosen
sufficiently close to $\pi/2$ then the trajectory exits the domain $\tilde{B}$
at $r(s_{*})=\pi/2$. On the other hand, we may consider a central reflection
of the trajectory constructed in Lemma 4.10 with respect to
$r=\pi/2,\vartheta=0,\alpha=0$ and suitably reparametrise to obtain a solution
$(\tilde{r},\tilde{\vartheta},\tilde{\alpha})\colon[0,s_{*}[\to\tilde{B}$ with
initial data

$\displaystyle(\tilde{r},\tilde{\vartheta},\tilde{\alpha})(0)=(\pi-
r_{1},0,-\tfrac{\pi}{2})$

where $r_{1}\in]\pi/2,\pi[$ is as in Lemma 4.10 (see also Figure 4) which
exits the domain $\tilde{B}$ through $\\{\vartheta=0\\}$. Hence, arguing
exactly as in the proof of Theorem 1.1 there exists some $r_{0}\in]0,\pi/2[$
such that the corresponding solution with initial data (59) satisfies
$r(s_{*})=\pi/2$ and $\vartheta(s_{*})=0$.

Reflecting back and setting $r_{\infty}=\pi-r_{0}$ proves the claim. ∎

###  4.3 The Iteration process

######  Lemma 4.12.

Let $n\in\\{2,3\\}$. For all $1\leq k\in\mathbb{N}$ there exist
$0<\hat{\alpha}_{0}^{(k)}<\check{\alpha}_{0}^{(k)}<\pi/2$ such that
$\check{\alpha}_{0}^{(k+1)}<\hat{\alpha}_{0}^{(k)}$ for all $k$ and such that
the solutions

$\displaystyle\bigl{(}\hat{r}^{(k)},\hat{\vartheta}^{(k)},\hat{\alpha}^{(k)}\bigr{)}\colon[0,\hat{s}_{*}^{(k)}[$
$\displaystyle\to D,$
$\displaystyle\bigl{(}\check{r}^{(k)},\check{\vartheta}^{(k)},\check{\alpha}^{(k)}\bigr{)}\colon[0,\check{s}_{*}^{(k)}[$
$\displaystyle\to D$ with initial data
$\bigl{(}\pi/2,0,\hat{\alpha}_{0}^{(k)}\bigr{)}$ respectively
$\bigl{(}\pi/2,0,\check{\alpha}_{0}^{(k)}\bigr{)}$ as constructed in Lemma 4.1
satisfy
$\displaystyle\lim_{s\to\hat{s}_{*}^{(k)}}\bigl{(}\hat{\vartheta}^{(k)},\hat{\alpha}^{(k)}\bigr{)}(s)$
$\displaystyle=(-1)^{k}\Bigl{(}\frac{\pi}{4},\frac{\pi}{2}\Bigr{)},$
$\displaystyle\lim_{s\to\check{s}_{*}^{(k)}}\bigl{(}\check{\vartheta}^{(k)},\check{\alpha}^{(k)}\bigr{)}(s)$
$\displaystyle=(-1)^{k}\Bigl{(}0,\frac{\pi}{2}\Bigr{)}$

and such that the closures of their images intersect the plane
$\\{\vartheta=0\\}$ exactly $k+1$ times.

$r$$\theta$$0$$\frac{\pi}{2}$$\frac{\pi}{4}$ Figure 5: The trajectories of
$\bigl{(}\hat{r}^{(1)},\hat{\vartheta}^{(1)}\bigr{)}$ (orange curve) and
$\bigl{(}\check{r}^{(2)},\check{\vartheta}^{(2)}\bigr{)}$ (blue curve) and
their respective reflections (dotted curves), as described within the proof of
Theorem 1.3 and Theorem 1.2, in the case $n=2$.

As an example, we visualise the trajectories of
$\bigl{(}\hat{r}^{(1)},\hat{\vartheta}^{(1)}\bigr{)}$ and
$\bigl{(}\check{r}^{(2)},\check{\vartheta}^{(2)}\bigr{)}$ for $n=2$ in Figure
5.

###### Proof.

_Construction of $\check{\alpha}_{0}^{(1)}$._ In Lemma 4.11 we obtained the
“$\infty$-figure” $(r,\vartheta,\alpha)\colon[0,s_{\infty}[\to D$, a solution
of system (24),(25),(26) with initial data
$(r,\vartheta,\alpha)(0)=(\pi/2,0,\alpha_{\infty})$ for some
$\alpha_{\infty}\in]0,\pi/2[$. The trajectory of this solution exits the
domain
$D=[\tfrac{\pi}{2},\pi[\times]-\tfrac{\pi}{4},\tfrac{\pi}{4}[\times]-\tfrac{\pi}{2},\tfrac{\pi}{2}[$
via

$\lim_{s\to s_{\infty}}(r,\vartheta,\alpha)(s)=(r_{\infty},0,-\tfrac{\pi}{2})$

for some $r_{\infty}\in]\pi/2,\pi[$. Actually, we did not show that for any
$\alpha_{0}\in]0,\alpha_{\infty}[$ the corresponding trajectory will also
intersect $\\{\vartheta=0\\}$ for $s>0$, but Lemma 4.4 states that such
intersections actually occur for any sufficiently small $\alpha_{0}$.
Therefore, we may define $\check{\alpha}_{0}\in]0,\alpha_{\infty}]$ which is
largest with the property that for all $\alpha_{0}\in]0,\check{\alpha}_{0}[$
the corresponding trajectory intersects $\\{\vartheta=0\\}$ for some
$s=\sigma_{1}>0$. (Based on our numerical simulations we expect
$\check{\alpha}_{0}=\alpha_{\infty}$ because otherwise we would obtain two
distinct $\infty$-figures.) Let

$\displaystyle(\check{r},\check{\vartheta},\check{\alpha})\colon[0,\check{s}_{*}[\to
D$ (60)

be the solution to system (24),(25),(26) with initial data
$(\check{r},\check{\vartheta},\check{\alpha})(0)=(\pi/2,0,\check{\alpha}_{0})$
as defined above. With the same argument as in the proof of Lemma 4.10, we can
show $\check{\vartheta}(s)>0$ for all $0<s<\check{s}_{*}$ and

$\displaystyle\lim_{s\to\check{s}_{*}}(\check{\vartheta},\check{\alpha})(s)$
$\displaystyle=(0,-\tfrac{\pi}{2}).$ (61)

In particular, the closure of the trajectory of (60) intersects
$\\{\vartheta=0\\}$ exactly twice and we may define
$\check{\alpha}_{0}^{(1)}\vcentcolon=\check{\alpha}_{0}$.

_Construction of $\hat{\alpha}_{0}^{(1)}$._ Let
$(r,\vartheta,\alpha)\colon[0,s_{*}[\to D$ be the solution with
$(r,\vartheta,\alpha)(0)=(\pi/2,0,\alpha_{0})$ for some
$\alpha_{0}\in]0,\check{\alpha}_{0}[$ to be chosen. By definition of
$\check{\alpha}_{0}$, the set $\vartheta^{-1}(\\{0\\})\setminus\\{0\\}$ is
nonempty and Lemma 4.2 implies that $s=0$ is an isolated zero of
$\vartheta(s)$. Therefore, since $\vartheta$ is continuous

$\sigma_{1}\vcentcolon=\min\bigl{(}\vartheta^{-1}(\\{0\\})\setminus\\{0\\}\bigr{)}$

is well-defined. Then, $\alpha(\sigma_{1})\leq 0$ since $\vartheta(s)\geq 0$
in $[0,\sigma_{1}]$. In fact, $\alpha(\sigma_{1})<0$ because otherwise,
$\alpha$ and $\vartheta$ would both vanish identically by ODE uniqueness.
Hence, we may define $s_{\alpha}\in]\sigma_{1},s_{*}]$ being largest with the
property that $-\pi/2<\alpha(s)<0$ for all $s\in[\sigma_{1},s_{\alpha}[$. For
$\alpha_{0}=\check{\alpha}_{0}$ we just set $s_{\alpha}=s_{*}$. In this case,
we have

$\displaystyle\lim_{s\to s_{\alpha}}\alpha(s)=-\frac{\pi}{2}.$ (62)

For any sufficiently small $\alpha_{0}>0$ however, Lemma 4.5 implies
$s_{\alpha}<s_{*}$ and $\alpha(s_{\alpha})=0$. Therefore, there exists
$\hat{\alpha}_{0}\in]0,\check{\alpha}_{0}]$ which is smallest with the
property that for all $\alpha_{0}\in]\hat{\alpha}_{0},\check{\alpha}_{0}]$ the
corresponding solution has property (62). Let

$\displaystyle(\hat{r},\hat{\vartheta},\hat{\alpha})\colon[0,\hat{s}_{*}[$
$\displaystyle\to D$ (63)

be the solution with initial data
$(\hat{r},\hat{\vartheta},\hat{\alpha})(0)=(\pi/2,0,\hat{\alpha}_{0})$. If
$s_{\hat{\alpha}}<\hat{s}_{*}$ then $\hat{\alpha}(s_{\hat{\alpha}})=0$ and
equation (26) implies $d\hat{\alpha}/ds(s_{\hat{\alpha}})\neq 0$, that is,
$s=s_{\hat{\alpha}}$ is a nondegenerate zero of $\hat{\alpha}$. Moreover,
$\hat{\alpha}$ is bounded away from $-\pi/2$ in
$]\hat{\sigma}_{1},s_{\hat{\alpha}}[$. Therefore, there exists
$\alpha_{0}>\hat{\alpha}_{0}$ such that the corresponding trajectory still
satisfies $\alpha(s_{\alpha})=0$ in contradiction to our choice of
$\hat{\alpha}_{0}$. Hence, we must have $s_{\hat{\alpha}}=\hat{s}_{*}$. In
particular, $\hat{\alpha}(s)<0$ for all $s\in]\sigma_{1},\hat{s}_{*}[$. Then,
by equation (25), $\hat{\vartheta}(s)$ is decreasing and negative in
$]\sigma_{1},\hat{s}_{*}[$. This implies on the one hand that
$\hat{\vartheta}(s)$ converges to a limit $\hat{\vartheta}_{*}<0$ as
$s\to\hat{s}_{*}$ and on the other hand that the trajectory of (63) intersects
$\\{\vartheta=0\\}$ exactly twice: at $s=0$ and at $s=\sigma_{1}$. By Lemma
4.1, $\hat{r}(s)\leq\hat{r}_{*}<\pi$ for all $s\in[0,\hat{s}_{*}[$ and we have

$\displaystyle\hat{\vartheta}_{*}=\lim_{s\to\hat{s}_{*}}\hat{\vartheta}(s)$
$\displaystyle=-\frac{\pi}{4}$ (64) or
$\displaystyle\limsup_{s\to\hat{s}_{*}}\bigl{(}-\hat{\alpha}(s)\bigr{)}$
$\displaystyle=\frac{\pi}{2}.$ (65)

We claim that in fact both things are true. If on the contrary
$0>\hat{\vartheta}_{*}>-\pi/4$ then by (65) there is a sequence
$s_{k}\to\hat{s}_{*}$ such that $\hat{\alpha}(s_{k})\to-\pi/2$ as
$k\to\infty$. For sufficiently large $k\in\mathbb{N}$, equation (26) and the
monotonicity of $\hat{r}(s)$ imply

$\displaystyle\frac{d\hat{\alpha}}{ds}(s_{k})$
$\displaystyle\leq-\frac{(2n-2)\tan(2\hat{\vartheta}_{*})}{\sin(\hat{r}_{*})}\cos\bigl{(}\hat{\alpha}(s_{k})\bigr{)}-(2n-1)\cot\bigl{(}\hat{r}(s_{k})\bigr{)}\sin\bigl{(}\hat{\alpha}(s_{k})\bigr{)}$
(66)
$\displaystyle\leq-\cot\bigl{(}\hat{r}(s_{k})\bigr{)}\sin\bigl{(}\hat{\alpha}(s_{k})\bigr{)}<0$
(67)

because as soon as $\cos\bigl{(}\hat{\alpha}(s_{k})\bigr{)}$ is sufficiently
small, the second term in (66) dominates the first. In particular, for any $k$
large enough, there is $\tilde{s}_{k}>s_{k}$ such that $\hat{\alpha}(s)$ is
decreasing in $[s_{k},\tilde{s}_{k}[$. Yet, combining this fact with the
monotonicity of $\hat{r}(s)$ we obtain that actually

$\displaystyle\frac{d\hat{\alpha}}{ds}(s)$
$\displaystyle\leq-\cot\bigl{(}\hat{r}(s_{k})\bigr{)}\sin\bigl{(}\hat{\alpha}(s_{k})\bigr{)}<0$
(68)

for all $s\in[s_{k},\tilde{s}_{k}[$ and, therefore, we may choose
$\tilde{s}_{k}=\hat{s}_{*}$. So, a posteriori, $\hat{\alpha}(s)$ is decreasing
in a left neighbourhood of $\hat{s}_{*}$ and (65) is in fact a limit. Then,
since we assume $\hat{\vartheta}_{*}>-\pi/4$, we can extend the solution to
the interval $[0,\hat{s}_{*}+\epsilon]$ for some small $\epsilon>0$ (allowing
values $\hat{\alpha}<-\pi/2$) and the trajectory will “turn clockwise” in the
sense that

$\displaystyle\frac{d\hat{\alpha}}{ds}(\hat{s}_{*})$
$\displaystyle=(2n-1)\cot(\hat{r}_{*})<0.$

This implies that the function $[0,\hat{s}_{*}+\epsilon]\ni
s\mapsto\hat{\alpha}(s)+\pi/2$ has a nondegenerate zero at $s=\hat{s}_{*}$.
Hence, by continuous dependence on the initial data there exists
$\hat{\alpha}^{\prime}_{0}<\hat{\alpha}_{0}$ such that the (extended)
solutions corresponding to any
$\alpha_{0}\in]\hat{\alpha}^{\prime}_{0},\hat{\alpha}_{0}]$ still intersect
$\\{\alpha=-\pi/2\\}$ in contradiction to our choice of $\hat{\alpha}_{0}$.

Conversely, if (64) was true but (65) was not, then equation (26) would imply
that $d\hat{\alpha}/ds$ is unbounded (with divergent integral) as
$s\to\hat{s}_{*}$ which again yields a contradiction. Therefore, (64) and (65)
both hold. In fact, we can show that (65) is actually a limit. If, on the
contrary, there was a sequence $s_{k}\nearrow s_{*}$ as $k\to\infty$ with
$0>\hat{\alpha}(s_{k})\geq\hat{\alpha}_{*}>-\pi/2$ for all $k\in\mathbb{N}$,
then since $-\tan(2\hat{\vartheta}(s_{k}))\to+\infty$ by (64), equation (26)
implies

$\displaystyle\frac{d\hat{\alpha}}{ds}(s_{k})$
$\displaystyle\geq-\frac{(2n-2)\tan\bigl{(}2\hat{\vartheta}(s_{k})\bigr{)}}{\sin\bigl{(}\hat{r}(s_{k})\bigr{)}}\cos(\hat{\alpha}_{*})-(2n-1)\cot(\hat{r}_{*})\sin\bigl{(}\hat{\alpha}(s_{k})\bigr{)}>0$
(69)

if $k$ is sufficiently large because then the first term in (69) eventually
dominates the second. In particular, there exists $\tilde{s}_{k}>s_{k}$ such
that $\hat{\alpha}(s)$ is increasing in $[s_{k},\tilde{s}_{k}[$. But then,
since $\cos(\hat{\alpha}(s))$ and $\hat{r}(s)$ are increasing, we obtain that
actually $d\hat{\alpha}/ds(s)>0$ for all $s\in[s_{k},\tilde{s}_{k}[$ with a
lower bound only depending on $k$ (and not on $\tilde{s}_{k}$). Hence, we may
choose $\tilde{s}_{k}=\hat{s}_{*}$ and conclude
$\hat{\alpha}(s)\geq\hat{\alpha}(s_{k})$ for all $s\in[s_{k},\hat{s}_{*}[$ in
contradiction to (65).

With $\check{\alpha}_{0}^{(1)}\vcentcolon=\check{\alpha}_{0}$ and
$\hat{\alpha}_{0}^{(1)}\vcentcolon=\hat{\alpha}_{0}$ the statement follows in
the case $k=1$ since we showed that the solutions (63) and (60) satisfy the
desired properties. We now proceed by induction in $k\in\mathbb{N}$ and
suppose that the statement holds for some $1\leq k\in\mathbb{N}$. By the way
we set up the construction, we have that for all
$\alpha_{0}\in]0,\check{\alpha}_{0}^{(k)}[$ the image of the corresponding
solution intersects $\\{\vartheta=0\\}$ at least $k+1$ times for some
$0=\sigma_{0}<\sigma_{1}<\ldots<\sigma_{k}<s_{*}$.

_Construction of $\check{\alpha}_{0}^{(k+1)}$._ On the one hand, the
trajectory corresponding to
$\alpha_{0}=\hat{\alpha}_{0}^{(k)}<\check{\alpha}_{0}^{(k)}$ intersects
$\\{\vartheta=0\\}\cap D$ exactly $k+1$ times ($s=0$ included). On the other
hand, Lemma 4.5 implies that if $\alpha_{0}\in]0,\hat{\alpha}_{0}^{(k)}[$ is
sufficiently small, then the corresponding trajectory intersects
$\\{\vartheta=0\\}\cap D$ at least $k+2$ times. Therefore, there exists
$\check{\alpha}_{0}^{(k+1)}\in]0,\hat{\alpha}_{0}^{(k)}[$ which is largest
with the property that for all $\alpha_{0}\in]0,\check{\alpha}_{0}^{(k+1)}[$
the corresponding trajectory intersects $\\{\vartheta=0\\}\cap D$ at least
$k+2$ times. Let

$\displaystyle\bigl{(}\check{r}^{(k+1)},\check{\vartheta}^{(k+1)},\check{\alpha}^{(k+1)}\bigr{)}\colon[0,\check{s}_{*}^{(k+1)}[\to
D$

be the solution to system (24),(25),(26) with initial data
$\bigl{(}\pi/2,0,\check{\alpha}_{0}^{(k+1)}\bigr{)}$ as defined above. With
the same argument as in the proof of Lemma 4.10 we can show that the
trajectory of this solution intersects $\\{\vartheta=0\\}\cap D$ exactly $k+1$
times for some $0=\sigma_{0}<\sigma_{1}<\ldots<\sigma_{k}<s_{*}$ and,
furthermore,

$\bigl{(}\check{\vartheta}^{(k+1)},\check{\alpha}^{(k+1)}\bigr{)}(s)\to(-1)^{k+1}\bigl{(}0,\tfrac{\pi}{2}\bigr{)}$

as $s\to\check{s}_{*}^{(k+1)}$. In particular, the closure of the trajectory
intersects $\\{\vartheta=0\\}$ exactly $k+2$ times.

_Construction of $\hat{\alpha}_{0}^{(k+1)}$._ The claim follows by exactly the
same arguments as in the construction of $\hat{\alpha}_{0}^{(1)}$ replacing
$\check{\alpha}_{0}$ by $\check{\alpha}_{0}^{(k+1)}$ and $\sigma_{1}$ by
$\sigma_{k+1}\vcentcolon=\min\bigl{(}\vartheta^{-1}(\\{0\\})\setminus\\{0,\sigma_{1},\ldots,\sigma_{k}\\}\bigr{)}$.
∎

######  Remark 4.13.

In the notation of Lemma 4.12 above, we have $\hat{\alpha}_{0}^{(k)}\to 0$ and
$\check{\alpha}_{0}^{(k)}\to 0$ as $k\to\infty$. Indeed, if either of these
sequences was bounded from below by a positive constant, then necessarily (by
virtue of the second clause of Lemma 4.6, and recalling Remark 4.7)

$\bigg{\lvert}\frac{d\vartheta}{ds}(\sigma_{k})\bigg{\rvert}\to\infty$

as $k\to\infty$, which, however, is only possible if $r(\sigma_{k})\to\pi$
(and, in fact, $r(\sigma_{k-1})\to\pi$ as one lets $k\to\infty$); but then, if
this is the case, we employ the first conclusion of Lemma 4.6 to note that,
for $k$ large enough, $\alpha(\sigma_{k-1})$ would, depending on the parity of
$k$, either be above the threshold $a$ provided by Lemma 4.8 or below $-a$,
hence forcing blow-up (before reaching $\sigma_{k}$) and thus a contradiction.

$r$$\theta$$0$$\frac{\pi}{2}$$\frac{\pi}{4}$$n=2$

$r$$\theta$$0$$\frac{\pi}{2}$$\frac{\pi}{4}$$n=3$

$r$$\theta$$0$$\frac{\pi}{2}$$\frac{\pi}{4}$$n=4$

Figure 6: Shooting from the centre with varying $\alpha_{0}\in]0,\pi/2[$ for
the cases $n\in\\{2,3,4\\}$.

###### Proof of Theorems 1.2 and 1.3.

At this stage, the conclusion of both theorems follow at once by simply
reflecting the trajectories constructed in Lemma 4.12. Specifically, the
projection to the coordinates $(r,\vartheta)$ of the solutions of the first
type, i. e.

$\bigl{(}\hat{r}^{(k)},\hat{\vartheta}^{(k)}\bigr{)}\colon[0,\hat{s}_{*}^{(k)}[\to[\tfrac{\pi}{2},\pi[\times]-\tfrac{\pi}{4},\tfrac{\pi}{4}[$

when reflected through the origin give rise to curves that lift to minimally
embedded hyperspheres in $\mathbb{S}^{2n}$, while the solutions of the second
type, i. e.

$\bigl{(}\check{r}^{(k)},\check{\vartheta}^{(k)}\bigr{)}\colon[0,\check{s}_{*}^{(k)}[\to[\tfrac{\pi}{2},\pi[\times]-\tfrac{\pi}{4},\tfrac{\pi}{4}[$

when reflected with respect to both axes give rise to closed curves that lift
to minimal immersions of $S^{n-1}\times S^{n-1}\times S^{1}$ in
$\mathbb{S}^{2n}$. Also, it follows from our construction that any pair of
elements of either family are mutually distinct, but we do need to discuss why
we do actually get two infinite families of _pairwise non-isometric_
hypersurfaces. As far as Theorem 1.2 is concerned, it suffices to note that
the intersections of the trajectories we construct with the
$\vartheta$-bisector $Z$ lift to the connected components of the self-
intersections of the lifted hypersurface (and of course the number of such
connected components is invariant under ambient isometries). For Theorem 1.3
we need a somewhat less direct argument, instead. Clearly, if either the
volume or the Morse index of the hyperspheres we construct were unbounded then
we would be done, so let us assume, without loss of generality, that these two
quantities are both uniformly bounded. By Sharp’s compactness theorem, see
[Sha17], there would be a subsequence of minimal hyperspheres converging
smoothly and graphically, with finite multiplicity $m\geq 1$, away from
finitely many points, to a smooth (embedded) minimal hypersurface in
$\mathbb{S}^{2n}$. Such a hypersurface inherits the $G$-equivariance, and thus
it is the lift of a solution curve in the $(r,\vartheta)$ domain. However,
since each curve $(\hat{r}^{(k)},\hat{\vartheta}^{(k)})$ is actually a graph,
and we know that $\hat{\alpha}^{(k)}_{0}\to 0$ as $k\to\infty$, we conclude
that the convergence above actually happens with unit multiplicity (i. e.
$m=1$), hence smoothly at all points, and the curve in question must be
tangent to $Z$ at the origin. Hence, by ODE uniqueness, we conclude that the
hypersurface in question must be the lift of $Z$ itself, which contradicts the
smoothness of the limit hypersurface we obtained. ∎

In Figure 6 we display multiple trajectories of solutions to system
(3),(4),(5) with initial data $(\pi/2,\pi/4,\alpha_{0})$. Note that here we
are working with respect to the original variable $\theta\in]0,\pi/2[$ and do
not restrict the trajectories to any specific subdomain. In each of the cases
$n\in\\{2,3,4\\}$ we make different choices of $\alpha_{0}\in]0,\pi/2[$. When
$n\in\\{2,3\\}$, we observe that depending on the choice of $\alpha_{0}$ the
trajectory either turns clockwise ($d\alpha/ds\leq 0$) or anticlockwise
($d\alpha/ds\geq 0$) at the first interior local minimum of $\theta$. In fact,
we either have $\alpha=-\pi$ or $\alpha=0$ at these points. In between there
is a trajectory with $\theta\to 0$. In the case $n=4$ however, we observe that
trajectories always turn clockwise at the first interior local minimum of
$\theta$ even if $\alpha_{0}>0$ is chosen extremely small. This suggests that
no trajectory through the centre may exit at $\theta=0$. This phenomenon has
been observed for all values of $n\geq 4$ we ran the numerical simulations
for, and leads to formulate what follows:

######  Conjecture 4.14.

For $n\geq 4$ any $G$-equivariant, minimal embedding of $S^{2n-1}$ in
$\mathbb{S}^{2n}$ is equatorial.

######  Remark 4.15.

Conjecture 4.14 (where $G=\operatorname{O}(n)\times\operatorname{O}(n)$
consistently with the notation we have employed throughout the paper) should
be compared with Hsiang’s [Hsiang1983] Conjecture 1 stating that for each
$m\geq 7$ there exist infinitely many distinct minimal embeddings of $S^{m-1}$
into $\mathbb{S}^{m}$.

##  Appendix A A minimal embedding of $T^{4}$ in $\mathbb{S}^{5}$

In this appendix, we describe how the analysis presented in Section 3 allows,
with simple modifications, to also prove the following statement:

######  Theorem A.1.

There exists a minimal embedding of $S^{1}\times S^{1}\times S^{1}\times
S^{1}$ in the round sphere $\mathbb{S}^{5}$.

To that aim, one considers the action of
$H=\operatorname{O}(n)\times\operatorname{O}(n)\times\operatorname{O}(n)$ on
$\mathbb{S}^{3n-1}\subset\mathbb{R}^{3n}=\mathbb{R}^{n}\times\mathbb{R}^{n}\times\mathbb{R}^{n}$
given by the outer direct sum of the standard representations of the
orthogonal group. The quotient $\mathbb{S}^{3n-1}/H$, endowed with the orbital
metric, is (isometric to) the first octant of the unit round sphere, i. e.
$\bigl{\\{}(X,Y,Z)\in\mathbb{R}^{3}\,:\ 0\leq X,Y,Z\leq 1,\leavevmode\nobreak\
X^{2}+Y^{2}+Z^{2}=1\bigr{\\}}$ and we note, following [HsiHsi80], that the
geodesic arcs emanating from the vertices to the midpoints of the opposite
sides divide such a domain into six subdomains, so that the problem amounts to
constructing, in any one of those, an arc solving the appropriate geodesic
curvature prescribed equation and meeting the boundary orthogonally (see
Figure 7, right image). That being said, diverting from that source, we prefer
to setup the problem in polar coordinates and construct such a free boundary
arc relying on arguments similar to those in Section 3 above.

Thus, we work in coordinates $(r,\theta)\in[0,\pi/2]\times[0,\pi/2]$ and with
respect to the spherical metric $g=dr^{2}+\sin^{2}(r)d\theta^{2}$; one needs
to prove the existence of certain geodesic arcs in the conformal metric
$W^{2}g$ where
$W(r,\theta)=\omega^{3}_{n-1}\lvert\sin(r)\cos(\theta)\rvert^{n-1}\lvert\sin(r)\sin(\theta)\rvert^{n-1}\lvert\cos(r)\rvert^{n-1}$
(if we take, from now onwards, $n=2$ then simply
$W(r,\theta)=4\pi^{3}\sin^{2}(r)\cos(r)\sin(2\theta)$; this suffices for our
purposes). Thus, proceeding exactly as in Section 2, we need to find a closed
smooth orbit for the $3\times 3$ system given by (3), (4) and

$\displaystyle\frac{d\alpha}{ds}$
$\displaystyle=\frac{2\cot(2\theta)}{\sin(r)}\cos(\alpha)-\frac{2+4\cos(2r)}{\sin(2r)}\sin(\alpha).$
(70)

By virtue of the aforementioned symmetry arguments we work in the ‘fundamental
domain’ (see Figure 7 left image) defined in these coordinates by letting

$\displaystyle
B^{\prime}\vcentcolon=\left\\{(r,\theta)\in]0,\pi/2]\times]0,\pi/4]\,:\
\tan(r)\cos(\theta)\leq 1\right\\}$ (71)

and will further consider $B\vcentcolon=B^{\prime}\times[-\pi/2,0]$. We will
consider solutions of the system above with initial data of the form
$(r_{0},\pi/4,-\pi/2)$, exactly as in Section 3. For the sake of convenience,
in the setting above let us define $u(r,\theta)=\tan(r)\cos(\theta)$ and,
furthermore

$v(r,\theta)=-\arctan\bigl{(}\cos(r)\tan(\theta)\bigr{)},$

that is precisely the value of $\alpha$ corresponding to an outward-pointing
vector normal to the regular curve $\left\\{u(r,\theta)=1\right\\}$ at the
point $(r,\theta)\in B^{\prime}$.

We note that, set $\varrho\vcentcolon=\arctan(\sqrt{2})<\pi/3$, as long as
$(r,\theta,\alpha)\in]0,\varrho]\times]0,\pi/4]\times[-\pi/2,0]$ equations
(3), (4) and (70) imply that $r$ and $\alpha$ are weakly increasing and
$\theta$ is weakly decreasing, and there are no stationary points for the
motion. One can then easily prove a well-posedness lemma analogous to Lemma
3.1, where (in the final clause) one shall read that _either_
$\tan(r(s_{*}))\cos(\theta(s_{*}))=1$ _or_ $\alpha(s_{*})=0$; in either case
$\theta(s_{*})>0$. Indeed, equations (5) and (70) for $\alpha$ have the same
first summand up to a constant factor and their second summand (restricted to
the respective domain) is nonnegative in both cases. Therefore, we may argue
exactly as in the proof of Lemma 3.1, where we simply drop the second summand.
The following estimates are analogous to Lemmata 3.2 and 3.3.

######  Lemma A.2.

Given $r_{0}\in]0,\pi/8[$ let $(r,\theta,\alpha)\colon[0,s_{*}]\to B$ be the
solution to system (3),(4),(70) and let $s_{1}\in]0,s_{*}]$ be arbitrary. If
$r(s_{1})\geq 2r_{0}$ then $\theta(s_{1})<\pi/4-1/12$.

###### Proof.

Let $\delta=1/12$ be fixed and assume there exists $s_{1}\in]0,s_{*}]$ with
$r(s_{1})\geq 2r_{0}$. Towards a contradiction, suppose
$\theta(s_{1})\geq\pi/4-\delta$. By monotonicity of $\theta$, we then have
$\theta(s)\geq\pi/4-\delta$ for all $s\in[0,s_{1}]$. Hence,
$0\leq\cot(2\theta)\leq\tan(2\delta)=\vcentcolon b$ for all $s\in[0,s_{1}]$.
Equations (4) and (5) then imply

$\displaystyle\frac{d\alpha}{ds}$
$\displaystyle=\Bigl{(}2\cot(2\theta)\cot(\alpha)-\frac{2+4\cos(2r)}{\sin(2r)}\sin(r)\Bigr{)}\frac{d\theta}{ds}\leq
3\Bigl{(}b\cot(\alpha)-1\Bigr{)}\frac{d\theta}{ds}.$ (72)

Estimate (72) is identical to (6) for $n=2$ in the proof of Lemma 3.2.
Analogously to (8),

$\displaystyle\alpha(s_{1})$
$\displaystyle\leq-b\log(b)+(b^{2}+1)2\delta-\frac{\pi}{2}<-\frac{\pi}{4}.$
(73)

We then obtain the contradiction $\theta(s_{1})-\pi/4<-\delta$ with the same
argument as for (9)–(10). ∎

######  Lemma A.3 (see Figure 7 (1)).

There exists a constant $c>0$ such that if the initial data
$r_{0}\in]0,\pi/8[$ is sufficiently small, then the solution to system
(3),(4),(70) satisfies $\alpha(s_{*})=0$ and $r(s_{*})\leq cr_{0}$.

###### Proof.

We may argue exactly as in the proof of Lemma 3.3 for $n=2$. By Lemma A.2, we
obtain $d\alpha/ds\geq(2/r)\tan(1/6)dr/ds$ for all $s\in[s_{1},s_{*}]$ as in
(11). The claim then follows by integration. ∎

$r$$\varrho\mathrlap{{}\vcentcolon=\arctan\sqrt{2}}$$\theta$$0$$\frac{\pi}{4}$$\frac{\pi}{4}$(1)(2)(3)$B^{\prime}$

$\theta$$r$

Figure 7: Left image: Implementing the shooting method. Trajectory (1) exists
$B$ through the “roof” $\\{\alpha=0\\}$. Trajectory (2) exits $B$ through the
“side”. Trajectory (3) is the desired curve.
Right image: Visualisation of the reflection symmetries in the quotient
$\mathbb{S}^{3n-1}/H$.

######  Lemma A.4 (see Figure 7 (2)).

For every $\varepsilon>0$ there exists $c>0$ such that if
$r_{0}\in]\varrho-c,\varrho[$ then $\alpha(s)<\varepsilon-\pi/2$ for all
$s\in[0,s_{*}]$ and $\tan(r(s_{*}))\cos(\theta(s_{*}))=1$.

###### Proof.

The monotonicity of $r$ and $\theta$ and the definition (71) imply that for
every $\delta>0$ there exists $c>0$ such that if $r_{0}\geq\varrho-c$ then,
$\vartheta\geq\pi/4-\delta$ for all $s\in[0,s_{*}]$. In fact, solving the
equation $\tan(r)\cos(\theta)=1$ for $r=\arctan\bigl{(}1/\cos(\theta)\bigr{)}$
we obtain a smooth, strictly increasing function of $\theta$ which attains the
value $r=\varrho$ at $\theta=\pi/4$. Hence,
$0\leq\cot(2\theta)\leq\tan(2\delta)$ as in the proof of Lemma A.2 but now the
estimate holds for all $s\in[0,s_{*}]$. Thus, we may argue as we did for (73)
to conclude that $\alpha(s)$ is arbitrarily close to $-\pi/2$ for all
$s\in[0,s_{*}]$ provided that $\delta>0$ is chosen sufficiently small. In
particular, this implies that the trajectory must exit the domain $B$ via
$\tan(r(s_{*}))\cos(\theta(s_{*}))=1$. ∎

Now, given these ancillary lemmata, the proof of Theorem A.1 is easily
obtained. Indeed, we have already shown the existence of trajectories
emanating from initial data of the form $(r_{0},\pi/4,-\pi/2)$ that exit from
the domain $B$ at points in the interior of the sets

$(B\cap\left\\{\alpha=0\right\\})\cup(B\cap\left\\{u(r,\theta)=1,\ \alpha\geq
v(r,\theta)\right\\}),\ \text{and}\ (B\cap\left\\{u(r,\theta)=1,\ \alpha\leq
v(r,\theta)\right\\})$

respectively for $r_{0}$ sufficiently small, and sufficiently close to the
threshold $\varrho$. At this stage, arguing exactly as we did for the proof of
Theorem 1.1 we obtain a trajectory that exits the domain $B$ at the
intersection of the two sets in question, which implies the desired
conclusion.

######  Remark A.5.

We explicitly note that, as can be readily checked, the minimal embeddings of
$T^{3}$ in $\mathbb{S}^{4}$ and of $T^{4}$ in $\mathbb{S}^{5}$ are the only
ones that can be obtained via equivariant methods relying on the outer direct
sum of orthogonal groups determining a cohomogeneity two action.

## References

Alessandro Carlotto
ETH D-Math, Rämistrasse 101, 8092 Zürich, Switzerland
E-mail address<EMAIL_ADDRESS>
Mario B. Schulz
University of Münster, Mathematisches Institut, Einsteinstrasse 62, 48149
Münster, Germany
E-mail address<EMAIL_ADDRESS>