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% K. Grosse-Brauckmann, R. B. Kusner, J.M. Sullivan
% Triunduloids: Embedded constant mean curvature surfaces
% with three ends and genus zero
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\begin{document}
\title[Triunduloids]
{Triunduloids:\\ Embedded constant mean curvature surfaces\\
with three ends and genus zero}
\date{2001 January 12}
\author[Grosse-Brauckmann]{Karsten Gro\ss e-Brauckmann}
\address{Universit\"at Bonn, Mathematisches Institut, Beringstr.~1,
53115 Bonn, Germany}
\email{kgb@math.uni-bonn.de}
\author[Kusner]{Robert B. Kusner}
\address{Mathematics Department, University of Massachusetts,
Amherst MA 01003, USA}
\email{kusner@math.umass.edu}
\author[Sullivan]{John M. Sullivan}
\address{Mathematics Department, University of Illinois, Urbana IL 61801, USA}
\email{jms@uiuc.edu}
\begin{abstract}
In 1841, Delaunay constructed the embedded surfaces of revolution with
constant mean curvature ({\CMC/}); these unduloids have genus zero
and are now known to be the only embedded \cmc surfaces
with two ends and finite genus.
Here, we construct the complete family of embedded \cmc surfaces with three
ends and genus zero; they are classified using their asymptotic necksizes.
We work in a class slightly more general than embedded surfaces,
namely immersed surfaces which bound an immersed three-manifold,
as introduced by Alexandrov.
\end{abstract}
\maketitle
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Introduction}
Surfaces which minimize area under a volume constraint have constant mean
curvature~$H$. This condition can be expressed as a nonlinear elliptic
partial differential equation:
in terms of a local conformal immersion $f(x,y)$ of the surface into $\R^3$,
we have
$$f_{xx} + f_{yy} = 2 H \, f_x \times f_y.$$
For us, a \emph{\CMC/ surface} will mean
a complete surface of nonzero constant mean curvature,
properly immersed in $\R^3$, and rescaled to have $H\ident1$.
We are interested in embedded \CMC/ surfaces.
The reflection technique of Alexandrov~\cite{al1} shows that
the unit sphere~$\S^2$ is the only compact embedded \cmc surface.
Thus the simplest nontrivial case to consider is surfaces
of \emph{finite topology}.
These are homeomorphic to a compact surface~$\Sigma$ of
genus~$g$ with a finite number~$k$ of points removed. A neighborhood of each
of these punctures is called an \emph{end} of the surface.
The \emph{unduloids}, embedded \cmc surfaces of revolution described by
Delaunay~\cite{del}, are genus-zero examples with two ends.
The periodic generating curve for each unduloid solves an
ordinary differential equation.
Up to rigid motion, the entire family is parametrized by the
length of the shortest closed geodesics: this \emph{necksize}
ranges from $0$ at the singular chain of spheres, to~$\pi$ at the cylinder.
Unlike the case of minimal surfaces,
continuous families of embedded \cmc surfaces may develop self-intersections
not only at the ends, but also on compact sets.
Therefore, to study \CMC/ moduli spaces,
it is natural to consider a more general class of
surfaces, as introduced by Alexandrov~\cite{al2}.
A compact immersed surface is \emph{\alex-emb/}
if it bounds a compact immersed three-manifold.
This is the class of compact surfaces to which
the Alexandrov reflection technique applies,
but the technique can also apply to certain noncompact surfaces.
%
\begin{definition}
A \cmc surface $M$ of finite topology is \emph{\alex-emb/}
if $M$ is properly immersed, if each end of $M$ is embedded, and if
there exists a compact three-manifold~$W$ with boundary $\d W =: \Sigma$
and a proper immersion $F\takes W\setm\{q_1,\ldots,q_k\}\to\R^3$ whose
boundary restriction $f\takes \Sigma\setm\{q_1,\ldots,q_k\} \to\R^3$
parametrizes $M$.
Moreover, we require that the mean-curvature normal of $M$ points into $W$.
\end{definition}
We define a \emph{triunduloid} to be an \alex-emb/ \cmc surface with
genus zero and three ends.
These are the most basic \alex-emb/ \cmc surfaces,
since there are no examples with one end~\cite{mee},
and the unduloids are the only examples with two ends~\cite{kks}.
Kapouleas~\cite{kap}, Grosse-Brauckmann~\cite{kgb},
and Mazzeo and Pacard~\cite{mp} established the existence of certain
triunduloids, with small necksizes or high symmetry.
We now note that triunduloids always have a mirror symmetry,
unlike general \cmc surfaces with more than three ends.
By a theorem of Meeks~\cite{mee}, any triunduloid is contained in a half-space.
Korevaar, Kusner and Solomon~\cite{kks} extended Alexandrov's reflection
technique to \alex-emb/ \cmc surfaces contained in a half-space.
It follows that any such surface $M$ is \emph{Alexandrov-symmetric}
in the following sense:
$M$ has a reflection plane~$P$ such that $M\setm P$ consists of two connected
components $M^{\pm}$ whose Gauss images~$\nu(M^{\pm})$ are
contained in the open hemispheres~$\S^2_{\pm}$; moreover $\nu(M\cap P)$
is contained in the equator.
Together with the asymptotics result
of~\cite{kks} this gives for triunduloids (of any genus):
\begin{theorem}[\cite{kks}]\label{thkks} % see pp.~472, 476, 477
A triunduloid is Alexandrov-symmetric.
Each of its ends is exponentially asymptotic to a Delaunay unduloid,
and thus has an asymptotic axis line and necksize $n\in(0,\pi]$.
%\qed
\end{theorem}
Our main theorem below classifies all triunduloids, up to rigid motion.
To avoid orbifold points in the moduli space, we label
the ends of each triunduloid before passing to the quotient space:
\begin{definition}
The \emph{moduli space $\M$ of triunduloids} consists of all proper,
\alex-emb/ \cmc immersions of $\S^2\setm \{q_1,q_2,q_3\}$ in $\R^3$,
modulo diffeomorphisms of the domain fixing each $q_i$,
and modulo orientation-preserving isometries of $\R^3$.
We define the topology on $\M$ as follows: A sequence in $\M$
converges if there are representative surfaces
which converge in Hausdorff distance on every compact subset of $\R^3$.
\end{definition}
The classifying space for triunduloids is quite explicit:
\begin{definition}
The space of all ordered \emph{triples}
of distinct points in $\S^2$, up to rotation, is denoted by
$$\T:=\{(p_1,p_2,p_3)\in \S^2\times\S^2\times\S^2 \colon p_1\not=p_2\not=p_3\not=p_1\}
\,\big/\, \SO(3).$$
\end{definition}
%
\noindent
We observe that $\T$ is an open three-ball:
Simply rotate a triple until it lies on a circle of latitude, with $p_1$ on a
fixed longitude, followed by $p_2$ then $p_3$ proceeding eastward;
the common latitude and the longitudes of $p_2$ and $p_3$ give coordinates.
Alternatively, we may view $\T$ as the group of conformal motions
modulo rotations, i.e., as hyperbolic space.
% written as $PSL_2(\C)/PSU_2$,
Our theorem establishes a homeomorphism between the spaces $\M$ and $\T$:
\begin{maintheorem}
There is a homeomorphism $\Psi$ from the moduli space $\M$ of triunduloids
onto the open three-ball $\,\T$ of triples in the sphere $\S^2$.
The asymptotic necksizes of a triunduloid $M\in\M$ are
the spherical distances of the triple~$\Psi(M)\in\T$.
\end{maintheorem}
%
\noindent
Thus spherical trigonometry can be used to give necessary conditions
on triunduloids:
\begin{cornn}
Three numbers $n_1$, $n_2$, $n_3$ in $(0,\pi]$ can be
the asymptotic necksizes of a triunduloid
if and only if they satisfy the spherical triangle inequalities
\begin{equation*} \begin{split}
n_1 \le n_2+n_3, & \quad
n_2 \le n_3+n_1, \quad
n_3 \le n_1+n_2, \\
& n_1+n_2+n_3 \le 2\pi.
% Another form is: 2n_k \le n_1+n_2+n_3\le 2\pi.
\end{split} \end{equation*}
In particular, at most one end of a triunduloid can be cylindrical,
that is, have necksize~$\pi$.
\end{cornn}
%
\noindent
When all the inequalities are strict, the triple in~$\T$ has a distinct
mirror image; thus there are exactly two triunduloids with the
same three necksizes (see \figr{triunds}).
%
\figsix{Three triunduloids and their classifying triples.}
{Three typical triunduloids (top),
and their classifying spherical triples (bottom).
The triunduloid in the middle column
has necksizes $\tfrac\pi2$, $\tfrac{2\pi}3$ and $\tfrac{5\pi}6$; since these
sum to the maximal $2\pi$, its classifying triple is equatorial.
In all three columns, the points of the triples
lie on these same meridians, but on the left and right,
the common latitude is $\pm\tfrac\pi4$ instead of $0$.
These two mirror-image triples correspond to a pair of triunduloids
with a common set of three necksizes.
%the smallest of which is $\tfrac\pi3$.
}
The necksize bounds, in the special case $n_1=n_2$,
were described earlier in~\cite{gk}.
The fact that the forces of the ends (see Section~\ref{se:prop}) must balance
leads to explicit trigonometric formulas relating the necksizes to
the angles made by the asymptotic axes of the ends (see~\cite{gks1}).
However, we do not have equally good control on the phase of each end,
meaning the position of the asymptotic necks along the axis.
\subsection*{Outline of the proof}
%
The main steps in the proof of our theorem
(see also~\cite{pnas})
are to define the classifying map $\Psi$ from triunduloids to spherical
triples, and then to prove it is injective, proper, and surjective.
In Section~\ref{se:class} we define $\Psi$.
We make use of Alexandrov symmetry to decompose a triunduloid in~$\M$
into two halves.
Each half is simply connected, so Lawson's conjugate-cousin
construction~\cite{law} gives a minimal cousin in the three-sphere~$\S^3$.
The three boundary components of the cousin are fibers in a single
Hopf fibration~\cite{kar},
and hence they project under the Hopf map to the three points of a
spherical triple in~$\T$.
Our approach to conjugate cousins, explained in Section~\ref{se:coco},
is different from Lawson's. It allows us to show that the minimal cousin itself
Hopf-projects to an immersed disk in~$\S^2$; the spherical metric
induced on this disk depends only on the triple in~$\T$.
Pulling back the Hopf bundle, we get a circle-bundle over this disk,
and we regard the minimal cousin as a section.
The cousin of any other triunduloid with the same triple
gives another minimal section of the same bundle,
and we use the maximum principle
to show these sections are congruent. This proves the
injectivity of~$\Psi$, the main result of Section~\ref{se:inj}.
It remains to show $\Psi$ is surjective, that is, to
prove a triunduloid exists for each triple.
We employ a novel continuity method in the noncompact space
$\T\supset\Psi(\M)$.
In Section~\ref{se:prop} we prove $\Psi$ is continuous and proper;
we use the area estimate from~\cite{kk} and extend their curvature estimate.
In Section~\ref{se:surj} we prove the surjectivity of $\Psi$.
We depend on the fact~\cite{kmp} that the
moduli space~$\M$ of triunduloids is locally a real analytic variety.
We show that this variety has dimension three, by using a nondegenerate
minimal trinoid~\cite{mr} in a gluing construction~\cite{mp}.
A topological argument then shows that
the proper injective map $\Psi$ from the three-variety~$\M$
to the connected three-manifold~$\T$ must be surjective and,
in fact, a homeomorphism, completing the proof.
\subsection*{Open problems and related work}
%
Our classification theorem suggests that all triunduloids may be
nondegenerate, a prerequisite to apply known gluing constructions.
We expect that triunduloids can be glued end-to-end
to construct \CMC/ surfaces of any finite genus $g$ with $k\ge3$ ends.
Our results generalize to the case of \emph{coplanar $k$-unduloids} of
genus~$0$, meaning those \alex-emb/ \cmc surfaces with $k$~ends whose
asymptotic axes lie in a common plane. Since this case is technically
more involved, we devote a forthcoming paper to it.
Cos{\'\dotlessi}n and Ros~\cite{cr} adapted our approach to classify
coplanar minimal $k$-noids in~$\R^3$, and more recently, Rosenberg has
announced a result like ours for mean-curvature-1 trinoids in
hyperbolic space.
The Weierstrass-type representation of~\cite{dpw} might be used to give
explicit formulas for triunduloids as well as nonembedded \CMC/ trinoids;
this is explored in work in progress by Dorfmeister and Wu, and by Schmitt
(see~\cite{kms}).
\subsection*{Acknowledgements}
%
We wish to thank Claus Hertling for his advice on real analytic
varieties, Frank Pacard for many useful suggestions,
and Hermann Karcher for helpful discussions.
We also gratefully acknowledge the hospitality of
the Institute for Advanced Study
and the Center for GANG (through NSF grant DMS~96-26804)
at various times during this project, as well as that
of the CMLA at \'Ecole Normale Sup\'erieure Cachan (K.G.B.)
and the NCTS in Hsinchu, Taiwan (R.B.K.).
This work was partly supported by the Deutsche Forschungsgemeinschaft
through SFB 256 (K.G.B.), and by the National Science Foundation
under grants DMS~97-04949 and DMS~00-76085 (R.B.K.)
and DMS~97-27859 and DMS~00-71520 (J.M.S.).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Conjugate cousins}\label{se:coco}
\subsection{Conjugate minimal surfaces}
A simply connected minimal surface $M$ in $\R^3$ has,
as is well known, a conjugate minimal surface $\Mconj$.
These surfaces are isometric, and have a common Gauss map~$\nu$:
at corresponding points they have the same tangent plane.
However, tangent vectors are rotated by $90^\circ$ under the isometry.
In terms of a parametrization $f\colon\Omega\to\R^3$ of $M$
(where $\Omega\subset\R^2$) and the corresponding parametrization
$\fconj$ of~$\Mconj$, conjugation can be described as follows.
Since $f$ and $\fconj$ are isometric, they induce the same pullback
metric on $\Omega$. We let $J$ denote rotation by $90^\circ$
with respect to this metric.
Then the parametrizations are related by the first-order
differential equation
\begin{equation}\label{eq:first}
\dfconj= df\circ J.
\end{equation}
Equivalent formulations of this equation include
$\dfconj=-\star df$, using the Hodge star induced on $\Omega$,
and $\dfconj = \nu\times df$,
using the cross product in $\R^3$ with the unit normal vector $\nu$.
If $f$ is conformal, $J$ is the ordinary rotation in $\R^2$, and
\eqn{first} becomes the Cauchy-Riemann system
$$
\fconj_x = f_y, \qquad \fconj_y = -f_x.
$$
That is, when a minimal surface $M$ is given a conformal parametrization,
its coordinate functions are harmonic, and those of $\Mconj$ are the
conjugate harmonic functions.
Symmetry curves on conjugate surfaces have a striking relationship.
If $f\circ\gamma$ is a curve in $M$ (parametrized by arclength)
then~\eqn{first} implies that its
\emph{conormal} vector $\eta:=df(J\gamma')$ equals the tangent
vector $\dfconj(\gamma')$ of the conjugate curve in~$\Mconj$.
Thus $f\circ\gamma$ has constant conormal $u\in\S^2$ if and only if
the curve $\fconj\circ\gamma$ is contained in a straight line in
the direction $u$;
that is, an \emph{arc of planar reflection} in~$M$ corresponds to an
\emph{arc of rotational symmetry} in~$\Mconj$.
According to Bonnet's fundamental theorem,
a surface in space is determined up to rigid motion by second-order
data: its metric and its shape operator.
This leads to an equivalent, second-order description of conjugation:
$M$ and~$\Mconj$ are isometric, and their shape operators are related by
$J\circ\Sconj = S$.
The Gauss and Codazzi equations, which are the integrability conditions in
Bonnet's theorem, confirm that the conjugate surface $\Mconj$ exists exactly
when $M$ is minimal.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Conjugate cousins and quaternions}
Lawson~\cite{law} described a similar notion of \emph{conjugate cousin}
surfaces, where $M$ is a \cmc surface in~$\R^3$ and $\Mconj$
is an isometric minimal surface in~$\S^3$.
He gave a second-order description, in which the shape operators are
related by $J\circ\Sconj=S-\id$.
Again, the integrability conditions in Bonnet's theorem
prove existence of~$\Mconj$ exactly when $M$ is \cmc and, conversely, existence
of~$M$ exactly when $\Mconj$ is minimal~\cite{law,kar,kgb}.
It is the Gauss equation which shifts the unit mean curvature of
$M\subset\R^3$ to ambient sectional curvature of $\S^3\supset\Mconj$.
(With the terminology \emph{cousin} we follow Bryant~\cite{bry}, who similarly
relates \cmc surfaces in hyperbolic space to minimal surfaces in $\R^3$.)
Here we will give a description of conjugate cousins via
a first-order system of differential equations.
When properly interpreted, it says again that
the surfaces share a common normal,
and that tangent vectors rotate by a right angle.
This description seems to have been mentioned first in~\cite{gbp}
(see also~\cite[p.66]{hel}),
while the first proof (in a different context) appears in~\cite{obe}.
Karcher's earlier work~\cite{kar} can be regarded as a first-order
description of the symmetry curves alone.
System~\eqn{first} for minimal surfaces in~$\R^3$ equates tangent vectors at
two different points of~$\R^3$, implicitly making use of parallel translation.
In the case of conjugate cousins, to compare tangent vectors to~$\R^3$
and~$\S^3$, we use the Lie group structure of~$\S^3$.
If $L_p$ denotes left translation of $\S^3$ by $p\in\S^3$, then
we use the left translation of tangent vectors,
$dL_p\colon T_1\S^3\to T_p\S^3$, to generalize~\eqn{first} to
\begin{equation}\label{eq:cousin}
\dfconj=dL_\fconj \, (df\circ J).
\end{equation}
That is, given a tangent vector to the \cmc surface $f\takes \Omega\to\R^3$,
we first rotate it by $90^\circ$ within the tangent plane.
This rotated vector is an element of $T_{f(z)}\R^3 \isom \R^3$,
which we identify with $T_1\S^3$. Finally,
we left-translate by~$dL_{\fconj(z)}$ to obtain a vector in~$T_{\fconj(z)}\S^3$.
It is convenient to view the Lie group~$\S^3$ as the set of unit quaternions.
Then tangent vectors to~$\S^3$ are also represented by quaternions,
and left translation (of points or tangent vectors) is simply multiplication.
We identify $\R^3$ with $T_1\S^3=\Im\H$, the imaginary quaternions.
This way, \eqn{cousin} becomes the quaternion-valued equation
$\dfconj=\fconj\,df\after J$.
Again, if $f$ (or equivalently $\fconj$) is conformal, we can write this as
\begin{equation}\label{eq:fstHcnf}
\fconj_x = \fconj\,f_y, \qquad \fconj_y = -\fconj\,f_x.
\end{equation}
In order to investigate the integrability of this system, let us now
briefly review some of the geometry of surfaces in $\R^3=\Im\H$ and in
$\S^3\subset\R^4=\H$ using quaternionic notation.
For imaginary quaternions, $p,q\in\Im\H$, we can express the quaternionic
product in terms of the inner product and cross product:
$\Re(pq)=-\langle p, q\rangle\in\Re\H=\R$
and $\Im(pq)= p\times q\in\Im\H=\R^3$, giving
$pq=-\langle p, q\rangle + p\times q$.
In particular, imaginary $p$ and $q$ commute if and only if they are
parallel, and anticommute if and only if they are orthogonal.
Therefore, conformality for a map $f\takes\Omega\to\Im\H$ can be expressed as
\begin{equation}\label{eq:conf}
|f_x|^2=|f_y|^2, \qquad f_xf_y=-f_yf_x.
\end{equation}
Given a surface in an oriented three-manifold,
a choice of unit normal $\nu$ is equivalent to
a choice of orientation for the surface.
We always orient a \cmc surface by choosing the
(inward) \emph{mean curvature normal}.
Once we orient a surface, we will always use oriented parametrizations
such that $f_x\times f_y$ is a positive multiple of~$\nu$.
The $90^\circ$ rotation~$J$ on the domain is then oriented in
the usual counterclockwise manner, so that
$\big(df(X),df(JX),\nu\big)$ is positively oriented for any $X\in T\Omega$.
With the above conventions, the mean curvature~$H$
of a conformally parametrized surface $f\colon\Omega\to\R^3$
can be measured by $\Delta f = 2 H\, f_x\times f_y$, where
$\Delta=\tfrac{\d^2}{\d x^2}+\tfrac{\d^2}{\d y^2}$
is the usual Laplacian on~$\Omega$.
Expressing the cross product quaternionically, we see that
$f$ is \CMC/ (meaning that $H\ident1$) if and only if
\begin{equation}\label{eq:cmc}
\Delta f = 2f_xf_y .
\end{equation}
Similarly, a conformal $\fconj\takes\Omega\to\S^3$ is minimal
if and only if it is harmonic:
\begin{equation}\label{eq:minimal}
\Delta \fconj = -\fconj\,|\dfconj|^2,
\end{equation}
where $|\dfconj|^2 = |\fconj_x|^2 + |\fconj_y|^2$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Existence of cousins}
We are now ready to prove the theorem giving the existence of
isometric conjugate cousin surfaces $M\subset\R^3$ and $\Mconj\subset\S^3$.
We will find the curvature equation~\eqn{cmc} for~$f$ is the integrability
condition for~$\fconj$, and similarly~\eqn{minimal} is the one for~$f$.
Thus cousins exist exactly when $f$ is \cmc or $\fconj$ is minimal.
\begin{theorem}\label{thcous}
Let $\Omega$ be a simply connected domain in $\R^2$.
Then, for each immersion $f\takes\Omega\to\R^3=\Im\H$ of
constant mean curvature~$1$,
there is an isometric minimal immersion $\fconj\takes\Omega\to\S^3$,
determined uniquely up to left translation, satisfying
\begin{equation}\label{eq:firstH}
\dfconj = \fconj \,df\circ J.
\end{equation}
Conversely, given an oriented minimal~$\fconj$, there is a
\cmc immersion~$f$, unique up to translation, satisfying~\eqn{firstH}.
In either case, the surface normals agree, in the sense that
$\nuconj=\fconj \nu$.
\end{theorem}
\begin{proof}
The results are invariant under
orientation-preserving diffeomorphisms of~$\Omega$.
Thus it is sufficient to prove the theorem
assuming a conformal parametrization.
Because \eqn{firstH} implies that tangent vectors to the two surfaces
have the same lengths, and lie within $T\S^3$ or $T\R^3$
as appropriate, it is clear that solutions are isometric surfaces
in $\S^3$ and $\R^3$.
Equivalently, it is straightforward to check algebraically
that any quaternionic solutions to \eqn{firstH} have
the following two properties:
first, $|\fconj|$ is constant if and only if $\Re f$ is constant;
second, the metrics induced by $f$ and $\fconj$ are conformal,
with lengths scaling by the factor $|\fconj|$.
Now let $f\takes\Omega\to\R^3=\Im\H$ be conformal,
and set $\alpha=df\after J=f_y\,dx-f_x\,dy$, so that~\eqn{firstH}
becomes $d\fconj=\fconj\alpha$.
The integrability condition is then
$$
0=d(\fconj\alpha)
= \dfconj \wedge\alpha+\fconj d\alpha
= \fconj(\alpha\wedge\alpha+d\alpha).
$$
But we calculate $d\alpha=-\Delta\! f\,dx\wedge dy$, and
$\alpha\wedge\alpha = f_xf_y\,dx\wedge dy + f_yf_x\,dy\wedge dx
= 2f_xf_y\,dx\wedge dy$, using~\eqn{conf}.
Thus the integrability condition is precisely~\eqn{cmc},
so a solution $\fconj$ exists exactly when $f$ is \CMC/,
and it is determined up to left translation.
Conversely, given a conformal $\fconj\takes\Omega\to\S^3$,
and setting $\beta=-\dfconj\after J$,
we can write~\eqn{firstH} as $df=\fconj^{-1}\beta$.
Thus the integrability condition is
$$
0=d(\fconj^{-1}\beta)
= d(\fconj^{-1})\wedge\beta + \fconj^{-1}d\beta
= -\fconj^{-1}\dfconj\fconj^{-1}\wedge\beta+\fconj^{-1}d\beta.
$$
From \eqn{fstHcnf} we have $(\fconj^{-1}\fconj_x)^2=f_y^2$.
Moreover, $f_y$ is pure imaginary, so $f_y^2=-|f_y|^2$,
which in turn equals $-|\fconj_y|^2$ since the cousins are isometric.
Similarly, $(\fconj^{-1}\fconj_y)^2=-|\fconj_x|^2$, so we get
$(\fconj^{-1}\fconj_x)^2+(\fconj^{-1}\fconj_y)^2=-|\dfconj|^2$.
Thus the first term in the integrability condition
becomes $|\dfconj|^2\,dx\wedge dy$.
The second term is $\fconj^{-1}\,\Delta\fconj\,\,dx\wedge dy$.
Comparing with~\eqn{minimal}, we see that our system is integrable for $f$
exactly when $\fconj$ is minimal.
In either case, $f$ and $\fconj$ are conjugate cousins; either
could be derived from the other via~\eqn{firstH},
and both integrability conditions are satisfied.
This verifies that $\fconj$ is minimal and $f$ is \CMC/,
no matter which we started with.
\end{proof}
The simplest example of cousins is the unit sphere~$\S^2$ contained
both in $\S^3$ and $\Im\H$ (as their intersection).
It is its own conjugate cousin via the identity map.
Indeed, if $f$ parametrizes $\S^2$ then
$f\,df\circ J=f\times df\circ J=df$ and so $\fconj=f$.
In this case, the point $1\in\S^3$ and its left translate
$f\in\S^2\subset\S^3$ always have spherical distance~$\tfrac\pi 2$;
therefore, on the level of tangent vectors a $90^\circ$ rotation is induced
which cancels the effect of~$J$.
For the record, we show that our first-order description of conjugate
cousins agrees with Lawson's original second-order description~\cite{law}:
\begin{proposition}
The shape operators $S$ of~$f$ and $\Sconj$ of~$\fconj$
are related by $J\after\Sconj=S-\id$.
\end{proposition}
\begin{proof}
By definition of the shape operator,
$\dfconj\after \Sconj=-d\nuconj = -\dfconj\,\nu - \fconj\, d\nu$,
using $\nuconj=\fconj\nu$.
Substituting $\dfconj=\fconj\,df\after J$ and $d\nu=-df\after S$, we get
$$
df\after J \after \Sconj(\cdot) =
-\big(df\after J(\cdot) \big)\,\nu + \,df\after S(\cdot).
$$
The first term on the right can be simplified:
$df\after J$ takes values in the tangent plane to $M$;
since multiplication with the normal $\nu$ simply rotates
by $90^\circ$, we get
$$\big(df\after J(\cdot)\big)\nu = \big(df\after J(\cdot)\big)\times\nu
= df(\cdot).$$
Therefore, $J\after\Sconj=-\id + S$ as desired.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Hopf fields and spherical cousins}
Left-translating a unit vector $u\in\S^2\subset\Im\H=T_1\S^3$
gives us a left-invariant
vector field $p\mapsto pu$ on $\S^3$, called the \emph{$u$-Hopf field}.
Its integral curve through $p\in\S^3$ is a \emph{$u$-Hopf circle},
the great circle through~$p$ and~$pu$ parametrized by
$pe^{tu} := p(\cos t+u\sin t)$.
The \emph{$u$-Hopf projection} $\Pi_u\takes\S^3\to\S^2$ is
defined by $\Pi_u(p) = pu\bar p$. This projection is a fibration;
its fibers are the $u$-Hopf circles, since
$$\Pi_u\big(p(\cos t+u\sin t)\big)
= p(\cos t+u\sin t)u(\cos t-u\sin t)\bar p
= \Pi_u(p).$$
Later, we will also consider the $u$-Hopf projection of
a $v$-Hopf circle~$\gamma$ with $\langle u,v \rangle =\cos\theta$.
It is a round circle $\Pi_u(\gamma)$ of spherical
radius~$\theta$, covered twice at constant speed $2\sin\theta$.
Indeed, the $u$-Hopf circles through the different points of $\gamma$
foliate a distance torus~$T$ in~$\S^3$.
They are curves of homology class $(1,1)$ on~$T$, while $\gamma$ is $(1,-1)$.
The two polar core circles of~$T$ are $u$-Hopf, and project under~$\Pi_u$
to the two antipodal centers of $\Pi_u(\gamma)$ in~$\S^2$.
As on conjugate minimal surfaces,
symmetry curves on cousin surfaces are related.
This time, using \eqn{firstH}, we find that
a curve $f\circ \gamma$ in~$M$ with constant conormal $u\in\S^2$
corresponds to a curve $\fconj\circ \gamma$ in~$\Mconj$
with tangent vector $\fconj u$ in the $u$-Hopf field:
\begin{proposition}\label{pr:bdycurves}
Let $M$ and $\Mconj$ be conjugate cousins.
A curve $f\circ\gamma$ in~$M$ is a curve of planar reflection for~$M$,
contained in a plane with unit normal $u\in\S^2$,
if and only if the curve $\fconj\circ\gamma$ in $\Mconj$ is contained
in a $u$-Hopf circle in~$\S^3$.
%\qed
\end{proposition}
Since left translation is an isometry for
the metric on $\S^3$ we conclude from $\nuconj=\fconj\nu$ that
$\langle \nuconj,\fconj u\rangle_{\S^3} = \langle \nu,u \rangle_{\R^3}$
on~$\Omega$.
When this is nonzero, we get:
\begin{proposition}\label{pr:trvs}
A \cmc surface $M$ is transverse to the foliation of~$\R^3$
by lines in the $u$~direction if and only if its minimal cousin
$\Mconj$ is transverse to the foliation of~$\S^3$ by $u$-Hopf circles.
In this case, the Hopf projection $\Pi_u$ immerses $\Mconj$ to $\S^2$.
%\qed
\end{proposition}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Unduloids and their helicoid cousins}
The \emph{spherical helicoid} is parametrized by $h\colon\R^2\to\S^3$,
\begin{align*}
h(x,y)
% :=& \cos x \, e^{-ny\,\i} + \j\,\sin x\,e^{-(2\pi-n)y\,\i} \\
:=& \cos x\,\cos ny - \i\,\cos x\,\sin ny
+ \j\,\sin x\,\cos(2\pi-n)y + \k\,\sin x\,\sin(2\pi-n)y,
\end{align*}
where $\{1, \i, \j, \k\}$ is the standard basis of $\H=\R^4$
and $n$ is a real parameter.
For $n\not=0,2\pi$ the helicoid $h$ is an immersion;
moreover it is minimal (see \cite[\paragraph 2]{kgb}
and \cite[\paragraph 7]{law}). % Lawson skips the calculation
To analyse the symmetries of the cousin, we rewrite $h$ as
\begin{equation}\label{eq:helic}
h(x,y)
= p(y)\big(\cos x+u(y)\sin x\big),
% := p(y)e^{xu(y)} = p(y)\big(\cos x+u(y)\sin x\big),
\end{equation}
where $p(y) := \cos ny-\i\,\sin ny \in\S^3$
and $u(y) := \j\,\cos 2\pi y + \k\,\sin2\pi y \in\S^2\subset\Im\H$.
The $x$-coordinate lines $x\mapsto h(x,y)$
trace out the $u(y)$-Hopf circles through $p(y)$.
These great-circle rulings are orthogonal to the \emph{axis} of the helicoid,
$y\mapsto h(0,y)=p(y)$, along the $(-\i)$-Hopf circle through~$1$.
In fact, the helicoid has another axis,
$$
y\mapsto h(\tfrac\pi 2,y)
= p(y)u(y)
% = e^{(2\pi-n)y\,\i}\,\j
= \j\,\cos(2\pi-n)y + \k\,\sin(2\pi-n)y
$$
along the $(-\i)$-Hopf circle through~$\j$;
this is the polar circle of the original axis.
Each ruling intersects both axes, and its tangent vector at the first axis
coincides with its position on the polar axis:
$\tfrac{\d h}{\d x}(0,y)=h(\tfrac\pi 2,y)$.
Since the polar axis is parametrized with constant speed,
the tangent vectors rotate with constant speed along the first axis,
thus justifying the name spherical helicoid.
Our parametrization $h(x,y)$ is along orthogonal asymptotic lines,
confirming that~$h$ is a minimal surface.
Indeed, the $x$-coordinate lines are the great-circle rulings.
The $y$-coordinate lines are helices contained in
distance tori around the polar axes of~$h$.
Within the torus, the helix is geodesic and so
its curvature vector in~$\S^3$ is normal to the torus, and hence
tangent to the helicoid~$h$.
The next lemma will show that the cousins of these helicoids
are the \cmc surfaces of revolution characterized by Delaunay~\cite{del}:
for $n\in(0,\pi]$, the helicoid~$h$ is the cousin of
an embedded unduloid of necksize~$n$;
for $n<0$, it is the cousin of an immersed nodoid.
Indeed, by Proposition~\ref{pr:bdycurves} the cousin \cmc surface
is foliated by arcs of planar reflection; the normals to the mirror planes are
the tangent Hopf fields $u(y)$ of the rulings.
Since $u(y)$ rotates through all vectors perpendicular to~$\i$, the cousin is
a surface of revolution around the $\i$-axis.
\begin{lemma}\label{le:unduloids}
If we fix $n\in(0,\pi]$, the spherical helicoid $h(x,y)$,
given by~\eqn{helic} for $x\in\R$ and $y\in(-\tfrac 14,\tfrac 14)$,
is a minimal surface, the cousin $\plustilde U$
of a half-unduloid $U^+$ with necksize~$n$, whose boundary $\d U^+$
is contained in the $\i\j$-plane.
Moreover, the Hopf projection $\Pik$ immerses $\plustilde U$ into the
two-sphere $\S^2$ in such a way that the boundary
$\Pik(\d\plustilde U)$
consists of two points at spherical distance~$n$.
\end{lemma}
\begin{proof}
The cousin of $h$ is a \cmc surface of revolution with meridians contained
in mirror planes with normal $u(y)$.
Since $y\in(-\tfrac14,\tfrac14)$ is a maximal interval
for which these planes are nonhorizontal, the upper half $U^+$ is
indeed conjugate to~$h$ on the claimed domain.
The axes of the helicoid give circles of planar reflection
on the \cmc surface, contained in planes with normal~$\i$.
In particular, the first axis is parametrized with speed $n$
and so gives a neck circle with circumference~$n$, while the polar axis
is parametrized with speed $2\pi-n$ and so gives a circle with larger
circumference $2\pi-n$.
Since $n>0$, this latter bulge circumference is less than $2\pi$, showing
that the \CMC/ surface is an unduloid, not a nodoid.
Note that this description shows easily that the meridians of any unduloid
have length $\tfrac\pi2$ between neck and bulge.
Since unduloids are embedded, $U^+$ is transverse to the vertical lines,
and so $\Pik$ immerses $\plustilde U$ by Proposition~\ref{pr:trvs}.
The bounding rulings
$h(x,\pm\tfrac14) =p(\pm\tfrac14)(\cos x \pm \k\,\sin x)$
are $(\pm\k)$-Hopf circles through $h(0,\pm\tfrac14)=p(\pm \tfrac14)$,
and hence \kHopf/ project to single points
$\k\,\cos\tfrac n2 \mp \j\,\sin\tfrac n2$,
which are at spherical distance~$n$.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The classifying map}\label{se:class}
Our triunduloid moduli space $\M$ consists of
equivalence classes of triunduloids up to rigid motion.
For convenience, we assume the triunduloid $M$ representing a class
has been rotated as follows: The symmetry plane $P$,
guaranteed by Theorem~\ref{thkks}, is the $\i\j$-plane in~$\R^3$;
moreover, the asymptotic axis of the first end points in the $\i$-direction,
and successively labeled ends occur in counterclockwise order in~$P$.
For each class in~$\M$, this determines the representative~$M$
uniquely up to horizontal translation.
The intersection of $M$ with the open half-space above the symmetry plane
will be denoted $M^+$, the \emph{upper half} of $M$.
Since $M$ has genus zero, $M^+$ is simply connected.
Hence Theorem~\ref{thcous} gives a conjugate cousin $\plustilde M\subset\S^3$,
which is determined uniquely up to left translation.
Since $\d M^+$ consists of three arcs of reflection, all contained in
the $\i\j$-plane~$P$ with normal $\k\in\S^2$,
it follows from Proposition~\ref{pr:bdycurves} that
$\plustilde M$ is bounded by three arcs of \kHopf/ circles.
Thus $\Pik(\d\plustilde M)$ consists of three points in~$\S^2$.
We want to use this triple to define our map $\Psi\colon\M\to\T$,
so we need to check that the triple is well defined up to rotation,
and that the three points are distinct.
But~$\plustilde M$, and hence its boundary, is well defined up
to left translation by some $a\in \S^3$,
and $\Pik(ap)= ap\,\k\,\bar p\,\bar a = a\Pik(p)\bar a$.
This is indeed a rotation of~$\S^2$ applied to $\Pik(p)$.
%
We show the triple is distinct by proving the second part of our Main Theorem.
\begin{theorem}\label{th:necksizes}
The spherical distances between the points of the triple
$\Pik(\d\plustilde M)$ are the necksizes of the triunduloid $M$.
\end{theorem}
\begin{proof}
Let $E_i$ be the $i^{\text{th}}$ end of $M$, and $E_i^+=E_i\cap M^+$.
Theorem~\ref{thkks} shows that $E_i$ is exponentially
asymptotic to a Delaunay unduloid $U$ of necksize~$n_i$.
By \eqn{firstH}, conjugation preserves exponential convergence,
and so the bounding rays of $\plustilde{E_i}$ are asymptotic to the
great circles bounding the helicoid~$\plustilde U$.
However, by Proposition~\ref{pr:bdycurves} the two rays
in the boundary of $\plustilde{E_i}$ are themselves great-circle rays.
Thus they must cover the great circles bounding the cousin of the asymptotic
unduloid.
Lemma~\ref{le:unduloids} shows that these Hopf-project to points
at spherical distance~$n_i$.
\end{proof}
\begin{definition}
The \emph{classifying map} for triunduloids is given by
$$\Psi\colon\M\to\T, \qquad
\Psi(M):=\Pik(\d \plustilde M )\in \T.$$
\end{definition}
\noindent
By Theorem~\ref{th:necksizes} and the discussion above, $\Psi$ is
well defined with values in~$\T$.
In Section~\ref{se:prop}, we will use curvature bounds to
check that $\Psi$ is continuous and proper.
\begin{remark}
Equally well, we could define $\Psi$ on a triunduloid $M$ with arbitrary
mirror plane~$P$, by Hopf-projecting in the direction $u$ normal to $P$.
If $M$ is rotated by $A\in\SO(3)$, then its cousin is also rotated by~$A$,
extended to act on $\S^3$ by fixing~$1$ and preserving
$\S^2\subset\R^3=\Im\H$.
Hopf projection is equivariant with respect to such rotations:
if $A$ is realized through conjugation by a unit quaternion $a$, then
$$
\Pi_{Au}(Ap)
=(ap\bar a)(au\bar a)\bar{(ap\bar a)}
=a(pu\bar p)\bar a
=A\big(\Pi_u(p)\big).
$$
Thus surfaces representing the same class in $\M$ have Hopf images
related by a rotation, and indeed give the same triple in~$\T$.
\end{remark}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Injectivity}\label{se:inj}
The injectivity of our classifying map $\Psi$ is a uniqueness
theorem for triunduloids with the same classifying triple.
By definition of the classifying map, their cousin disks
are bounded by the same three Hopf fibers;
we will apply the maximum principle to show
the cousins are uniquely determined by these bounding fibers.
It is well known that two minimal graphs over the same domain
in $\R^2$ coincide, provided they have the same boundary data.
Indeed, by a vertical translation, the graphs can be assumed to have a
one-sided contact, in which case they agree by the maximum principle.
We can consider the two graphs as minimal sections of the
trivial (vertical) line bundle over the domain;
this suggests how we will generalize the argument
to certain minimal surfaces in $\S^3$.
As in the previous section,
we assume $M$ is rotated to have horizontal Alexandrov symmetry plane.
Since the cousin minimal disk $\plustilde M\subset\S^3$ is then
transverse to the \kHopf/ circles,
we can use the \kHopf/ projection of $\plustilde M$ to define the
domain disk for a generalized graph.
\begin{proposition}\label{pr:phiimmerses}
Define $\Phi(M):=\Pik(\plustilde M)$. This is an immersed open disk in~$\S^2$,
well defined up to rotation of~$\S^2$.
\end{proposition}
\begin{proof}
By Theorem~\ref{thkks} the immersed open disk $M^+$ is transverse to the
vertical lines in~$\R^3$.
It follows from Proposition~\ref{pr:trvs} that
$\Pik$ immerses $\plustilde M$ into~$\S^2$.
Finally, $\plustilde M$ is well defined up to left translation,
so as in Section~\ref{se:class}, $\Phi$ is well defined up to rotation.
\end{proof}
%
\noindent
Note that, by our conventions, this immersion preserves the orientation
given by the inward normals on $M$ and $\S^2$.
The disk $\Phi(M)$ carries a spherical metric which, as we will show in
section~\ref{ss:sphtri} below, depends on the classifying triple alone.
Therefore the pullback of the Hopf bundle to this disk defines a bundle
whose geometry depends on the classifying triple only.
This allows us, in section~\ref{ss:unique}, to apply the maximum principle
in the bundle to prove the injectivity of~$\Psi$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Spherical metrics on half-unduloids}
A metric on the oriented open two-disk~$D$ is called
\emph{spherical} if it is locally isometric to $\S^2$.
We will also consider the metric completion $\hat D$,
and refer to $\hat D \setm D$ as the \emph{completion boundary} of $D$.
A spherical metric defines an oriented \emph{developing map}
$\phi\colon D \to\S^2$, unique up to rotation,
which extends continuously and isometrically to~$\hat D$.
A particular case we consider is metrics whose completion boundary
includes an arc developing onto a great-circle arc in~$\S^2$ of some
length $n\in(0,2\pi)$.
We can \emph{join} two such metrics if the lengths $n$ are the same,
by developing the two disks so that the images in~$\S^2$ extend one
another across the common boundary arc. The union of the disks,
together with the open arc, is again a disk with a spherical metric.
Suppose we are given a \emph{minimizing great-circle arc} in~$\S^2$,
that is, a closed arc~$e$ of length $n\in(0,\pi]$.
Its complement is a disk with a spherical metric; we
call this a \emph{slit sphere} of \emph{slit length} $n$.
The completion boundary of a slit sphere
is a loop consisting of two arcs of length $n$.
We can join two, or more generally $k\ge 2$,
slit spheres of the same slit length.
Their join is a chain of $k$ slit spheres,
whose developing map has degree $k$ almost everywhere.
The completion boundary is again a loop of two arcs,
each developing onto the original arc~$e$.
A \emph{ray of slit spheres} is obtained by
successively joining infinitely many slit spheres of common slit length.
The developing map of such a ray has infinite degree.
Its completion boundary is a single closed arc of length $n$.
Joining two rays of slit spheres gives a \emph{line of slit spheres},
whose developing map is the universal covering
of~$\S^2\setm\{p,q\}$, where $p$ and $q$ are at distance~$n$.
The completion boundary can be identified with $\{p,q\}$.
\begin{lemma}\label{le:slitray}
If the completion boundary of a spherical metric on the disk
is a closed arc developing onto a minimizing great-circle arc,
then the metric is a ray of slit spheres.
\end{lemma}
\begin{proof}
By joining two copies of the metric, we obtain a spherical metric
on the disk~$D$ whose completion boundary develops
to the two endpoints~$\{p,q\}$ of the original arc.
Any path in $\S^2\setm\{p,q\}$ then lifts to~$D$,
so $D$~is the universal isometric covering space of
$\S^2\setm\{p,q\}$, that is, a line of slit spheres.
The arc across which we joined the two copies of
the original metric develops to~$\overline{pq}$,
and thus decomposes~$D$ into two rays of slit spheres.
\end{proof}
Given an unduloid~$U$ of necksize~$n$, by Lemma~\ref{le:unduloids}
the Hopf projection of the cousin induces a spherical metric
on~$U^+$.
A~\emph{half-bubble} of an unduloid $U$ is the open subset of the
upper half $U^+$ enclosed by the semicircles around two consecutive
necks. On a cylinder, we let a half-bubble be the region
enclosed by any two semicircles at distance~$\pi$.
%
\begin{lemma}\label{le:hopfund}
Let $U^+$ be the upper half of an unduloid of necksize $n$
(with boundary in a horizontal plane as in Lemma~\ref{le:unduloids}),
and let $B^+\subset U^+$ be a half-bubble.
Then $\Pik(\plustilde B)$ is a slit sphere of slit length~$n$.
Consequently, $\Pik(\plustilde U)$ is a line of slit spheres.
\end{lemma}
%
\begin{proof}
The unduloid $U$ is the cousin of the
spherical helicoid $h$ from Lemma~\ref{le:unduloids}.
The half-bubble $\plustilde B$ is the image under $h$
of the rectangle $00$) and for unduloids
($H=1$, $00$, there exists a constant $C$ such that
any triunduloid~$M$ with necksizes $n_1,n_2,n_3>\epsilon$
has uniformly bounded curvature, $|A|\le C$.
\end{proposition}
\begin{proof}
Suppose we have $\sup_{M_k}|A|\to\infty$
for some sequence of triunduloids~$M_k$
with necksizes greater than~$\epsilon$.
Then the rescaled surfaces $M'_k$, as in Lemma~\ref{le:minimallimit},
converge to an Alexandrov-symmetric minimal surface~$N$.
The~$M_k$ have genus~$0$, so there are no closed loops in $M_k\cap P$.
Since the $M'_k$ converge to $N$ with finite multiplicity,
there are no closed loops in~$N\cap P$ either.
But $N$ meets $P$, and hence does so in an unbounded curve.
Thus, we can apply Lemma~\ref{lemsce} to get a
closed geodesic $\gamma$ in $N$ with nonzero force.
This is the limit of closed curves $\gamma'_k$ on $M'_k$,
whose lengths and forces converge to those of $\gamma$.
So on the unrescaled $M_k$, the corresponding curves $\gamma_k$
have nonzero forces, which converge to zero.
Because $\gamma_k$ has nonzero force, it is nontrivial; a simple nontrivial
curve on a triunduloid is homologous to the boundary of one of the ends.
By the necksize bound, the force of \eqn{weight} is bounded away from zero,
a contradiction.
\end{proof}
\begin{theorem}\label{th:proper}
The classifying map $\Psi\colon\M\to\T$ is continuous and proper.
\end{theorem}
\begin{proof}
Consider a compact subset $\mathcal K\subset\T$.
By Theorem~\ref{th:necksizes} there is an $\epsilon>0$ such that each surface
$M\in\Psi^{-1}(\mathcal K)$ has necksizes bounded below by~$\epsilon$,
and thus uniformly bounded curvature by Proposition~\ref{pr:curvbound}.
Suppose two triunduloids in $\Psi^{-1}(\mathcal K)$ are sufficiently close
in Hausdorff distance on an open domain in $\R^3$ with compact closure.
Because of the uniform curvature bounds,
either one can then be written as a normal graph over the other
in any closed subset of the domain. By elliptic regularity,
these subsets of the triunduloids are then also $C^1$ (or~$C^\infty$) close.
Thus, up to left translation in~$\S^3$,
the corresponding subsets of the cousins are again close,
and so are the Hopf projections of their boundaries.
We conclude that $\Psi$ is continuous.
To show that $\Psi^{-1}(\mathcal K)$ is compact,
let us now prove that any sequence $M_k\in \Psi^{-1}(\mathcal K)$
subconverges with respect to the topology on~$\M$.
The~$M_k$ are given only up to horizontal translation; to
get convergence, we choose representatives such that
the enclosing balls $B(M_k)$ from Theorem~\ref{th:kk}
are centered at the origin.
Due to the area and curvature estimates we find a subsequence of the~$M_k$
that converges on each compact subset to some \cmc surface~$M$.
By smooth convergence, it is clear that $M$ is \alex-emb/.
Because a sequence of contractible loops has a contractible limit,
$M$ has genus zero and at most three ends.
On the other hand, by the enclosure theorem, on each $M_k$
there are three closed curves in $B_{12}(0) \setm B_{11}(0)$
which bound the three ends. These three sequences of
curves subconverge to three closed curves which are disjoint in
the limit surface $M$, and thus bound three different ends.
Thus $M$ is a triunduloid, as desired.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Surjectivity}\label{se:surj}
\subsection{The real-analytic variety structure of $\M$}
The moduli space of \cmc surfaces of any fixed finite topology
is locally a real-analytic variety, by the local structure theory
of Kusner, Mazzeo, and Pollack~\cite{kmp}. In order to discuss the regular
points of this variety, we recall the notion of nondegeneracy:
\begin{definition}\label{dedege}
A (minimal or \textsc{cmc}) surface $M$ is called \emph{degenerate}
if the Jacobi equation $\Delta_{{}_M} u+|A|^2 u=0$
has a nonzero square-integrable solution $u\in L^2(M)$.
Otherwise $M$ is called \emph{nondegenerate}.
\end{definition}
At a nondegenerate \cmc surface of finite topology with $k$~embedded ends,
\cite{kmp} used the implicit function theorem to show the $3k$~asymptotic
parameters given by the $k$~force vectors of the ends locally determine the
surface completely. Because the sum of the forces on all ends must be zero,
and because we mod out by the action of $\SO(3)$,
the number of parameters is reduced to~$3k-6$.
Thus for triunduloids, we expect a three-dimensional moduli space.
\begin{theorem}[\cite{kmp}]\label{th:kmp}
The moduli space $\M$ is locally
a real-analytic variety of finite dimension.
In a neighborhood of any nondegenerate triunduloid,
$\M$ is a manifold of dimension three.
%\qed
\end{theorem}
We remark that the converse is not necessarily true: a smooth
three-manifold point of~$\M$ may still be a degenerate triunduloid.
Therefore, while our main theorem says that $\M$ is a three-manifold
everywhere, it remains an open question whether all triunduloids are
nondegenerate.
A \CMC/ moduli space can have dimension different from
the expected~$3k-6$, and thus consist entirely of degenerate surfaces.
In fact, the moduli space of \CMC/ surfaces with a fixed finite topology
can have components of arbitrarily large dimension; examples include the
so-called multibubbletons with two ends, and the \CMC/ tori.
On the other hand, it seems likely that most \alex-emb/ surfaces,
and perhaps all the Alexandrov-symmetric ones, are nondegenerate.
While the methods of~\cite{kgb} give certain degenerate
embedded triply periodic \cmc surfaces (considered as
embeddings of a compact surface into a flat three-torus), no similar example
is known with finite topology.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Nondegenerate triunduloids}
The existence of the injective map $\Psi\takes \M\to\T$
shows that the variety $\M$ cannot have dimension greater than three.
However, to apply topological methods to get surjectivity, we need
to know the dimension is exactly three. To get this, we want to
prove the existence of a nondegenerate triunduloid, near which $\M$
will be a three-manifold.
Montiel and Ros~\cite{mr} have a theorem about nondegeneracy of minimal
surfaces. Here we specialize this result to
\emph{minimal $k$-noids}, minimal surfaces of finite total curvature and
genus zero with only catenoidal ends.
\begin{theorem}[{\cite[Cor.~15]{mr}}]
Suppose $M$ is a minimal $k$-noid
such that the Gauss image of its umbilic points
is contained in a great circle in~$\S^2$.
Then the only bounded Jacobi fields on~$M$ are those induced by
translations in~$\R^3$.
%\qed
\end{theorem}
%
\noindent
In our reformulation we have used the fact that the Gauss map of
a $k$-noid does not branch at the ends.
\begin{corollary}\label{co:trinoids}
All minimal trinoids are nondegenerate.
\end{corollary}
\begin{proof}
Note that a trinoid $M$ has total curvature $-8\pi$.
Hence the Gauss map has degree two and so by the Riemann-Hurwitz formula
at most two branch points, which are the umbilic points of $M$.
Therefore the theorem applies. If $M$ were degenerate, there
would be an $L^2$~Jacobi field, which would be bounded, and thus
induced by a translation. But a translation of a catenoidal end
(even if orthogonal to the axis) has normal component which fails to be~$L^2$.
\end{proof}
% Riemann-Hurwitz: g(M) = 1 +deg(nu)*(g(S^2)-1) + (total branch order)/2
% For any trinoid: 0 = 1 -2 + (total branch order)/2 ==> total br. or. = 2
Mazzeo and Pacard~\cite{mp} find \cmc surfaces by gluing
unduloids or nodoids of small necksize to
the ends of an appropriately scaled nondegenerate minimal $k$-noid.
%
\begin{theorem}[\cite{mp}]\label{th:mp}
Let $M_0\subset\R^3$ be a nondegenerate minimal $k$-noid that
can be oriented so that each end has normal pointing inwards towards its
asymptotic axis.
Then for some $\epsilon_0$,
there is a family $M_{\epsilon}$, $0<\epsilon<\epsilon_0$,
of nondegenerate \cmc surfaces with
$k$~embedded ends, such that on any compact set in~$\R^3$,
the dilated surfaces
$\tfrac 1{\epsilon} M_{\epsilon}$ converge smoothly and uniformly to~$M_0$.
%\qed
\end{theorem}
%
Gluing an unduloid to each end of a minimal trinoid
yields a surface which, though not embedded for small necksizes,
is \alex-emb/.
Thus we can combine Corollary~\ref{co:trinoids}
with Theorem~\ref{th:mp} to yield:
\begin{corollary}\label{co:nondegtriund}
There exists a nondegenerate triunduloid.
%\qed
\end{corollary}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Topological arguments for surjectivity}
We recall (see~\cite[\paragraph 7-4]{mun}) that a space $Y$ is
\emph{compactly generated} provided that any subset $A$ is open (resp.~closed)
in $Y$ if and only if $A\cap K$ is open (resp.~closed) in $K$
for every compact $K\subset Y$.
Any locally compact space $Y$ is compactly generated, as is any
space $Y$ which has a countable base at every point.
\begin{lemma}\label{le:homeo}
If $f\takes X\to Y$ is a continuous, proper, injective map from
an arbitrary space $X$ to a compactly generated Hausdorff space $Y$,
then $f$ is an embedding, that is,
a homeomorphism onto its image $f(X)$.
Furthermore, this image is closed in $Y$.
\end{lemma}
\begin{proof}
Because $f$ is a continuous injection, it is an embedding
if and only if it is a closed (or open) map onto its image.
Given a closed set $A\subset X$, we claim $f(A)$ is closed in~$Y$,
and hence in $f(X)$; applying the claim to $A=X$ will confirm the
last statement of the lemma.
We check the claim
by showing $f(A)\cap K$ is closed for all compact $K\subset Y$.
Since $f$ is proper, $f^{-1}(K)$ is compact, and so the closed subset
$A\cap f^{-1}(K)$ is also compact. Thus its image $f(A)\cap K$
is compact, and (since $Y$ is Hausdorff) closed.
\end{proof}
\begin{definition}
Let $X$ be a locally finite $d$-dimensional simplicial complex
and $S$ a $(d{-}1)$-simplex in $X$. The \emph{valence} of~$S$ is the
number of $d$-simplices which contain~$S$ as a face.
We say the complex~$X$ is \emph{borderless}
if no $(d{-}1)$-simplex has valence $1$.
\end{definition}
\begin{theorem}\label{th:homeo}
If $f\takes X\to Y$ is a continuous, proper, injective map from
a borderless $d$-complex $X$ to a connected $d$-manifold $Y$,
then $f$ is surjective and thus (by the previous lemma) a homeomorphism.
\end{theorem}
%
\begin{proof}
Consider the union of the closed $d$-simplices in~$X$ and remove its
$(d{-}2)$-skeleton. Let~$X'$ be the closure (in~$X$) of some connected
component of this set. Its $(d{-}1)$-simplices have
valence at least two because $X$ is borderless. Since $X'$~is closed,
the restriction $f|_{X'}$ is still proper.
We will show this restriction is surjective.
(This implies that $X=X'$ and that every valence is exactly two.)
By Lemma~\ref{le:homeo}, $f$ is an embedding and $f(X')$ is closed in~$Y$.
In particular, the image of the $(d{-}2)$-skeleton $X'_{d-2}$ is
a properly embedded $(d{-}2)$-complex in the connected $d$-manifold $Y$;
therefore its complement $B$ is connected.
Letting $Z$ be the complement of $X'_{d-2}$ in $X'$,
we have that $f(Z)=B\cap f(X')$ is a closed subset of $B$.
We claim that $f(Z)$ is also open in~$B$.
It then follows by the connectedness of~$B$ that $f$ maps $Z$ onto~$B$.
Since $\bar Z = X'$ and $f(X')$ is closed,
$f(X') \supset \bar B = Y$, as desired.
To prove the claim, suppose first that $z\in Z$ is interior to a
$d$-simplex. Since the dimension of $Y$ is also~$d$, and $f$
is an embedding, $f(Z)$ contains an open neighborhood of~$f(z)$.
Otherwise $z$ is interior to a $(d{-}1)$-simplex $S$.
Because $X'$ is borderless, $z$~is in the interior of the union of two
$d$-simplices with common face~$S$,
so again $f(Z)$ contains a neighborhood of~$f(z)$.
\end{proof}
To apply this theorem to real-analytic varieties, we recall the notion
of an Euler space (see~\cite[Def.~4.1]{fm}).
\begin{definition}
Let $X$ be a $d$-dimensional locally finite simplicial complex.
We say $X$ is a \emph{$\mod2$--Euler space} if the link of each
simplex in $X$ has even Euler characteristic, or, equivalently, if
the local Euler characteristic $\chi(X,X\setm x)$ is odd for all $x\in X$.
\end{definition}
Recall that the link of a $(d{-}1)$-simplex $S \subset X$ is
obtained as follows: consider the vertices of the $d$-simplices with
$S$ as a face; those which are not in~$S$ itself form the link of~$S$.
The valence of~$S$ gives the number of points in the link. For a
$\mod2$--Euler space the link of any $(d{-}1)$-simplex consists of an
even number of vertices. Therefore, a $\mod2$--Euler space is
borderless.
{\L}ojasiewicz~\cite{loj} showed real-analytic varieties are triangulable
as locally finite simplicial complexes,
and Sullivan discovered that they have the $\mod2$--Euler property
(see \cite[Thm.~4.4]{fm} or~\cite{sul,bv,har1}).
As a consequence of these results, we have:
\begin{proposition}\label{pr:borderless}
A $d$-dimensional real analytic variety, or any space locally
homeomorphic to one, is a borderless $d$-complex.
\end{proposition}
Making use of our previous results on injectivity and properness we can
now prove:
\begin{theorem}
The classifying map $\Psi\colon\M\to \T$ is a homeomorphism.
\end{theorem}
\begin{proof}
To apply Theorem~\ref{th:homeo} to $\Psi\colon\M\to \T$, we need to
verify its assumptions.
The space $\T$ of triples is clearly a connected three-manifold. The
classifying map $\Psi$ is proper and injective by
Theorems~\ref{th:proper} and~\ref{th:uniq}. From the first part of
Theorem~\ref{th:kmp}, the moduli space~$\M$ of triunduloids is
locally a real analytic variety; injectivity of $\Psi$ shows it has
dimension at most three. So Proposition~\ref{pr:borderless}
implies~$\M$ is a borderless $d$-complex for some $d\le3$.
It remains to check that $\M$ has dimension~$d=3$. By
Corollary~\ref{co:nondegtriund} there is a nondegenerate
triunduloid, so by the second part of Theorem~\ref{th:kmp}, $\M$ is
a three-manifold in some neighborhood of this triunduloid.
Theorem~\ref{th:homeo} now applies to conclude $\Psi$ is a homeomorphism.
\end{proof}
Together with Theorem~\ref{th:necksizes} this completes the
proof of our main theorem.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\end{document}