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\begin{document}
\title{A Real-Analytic Jordan Curve Cannot Bound Infinitely Many Relative Minima of Area}
\author{Michael Beeson\\
Department of Mathematics\\
San Jose State University\\
San Jose, California 95192\\
USA\\
email: beesonpublic@gmail.com\\
}
\maketitle
\medskip
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem{corollary}{Corollary}
\newtheorem{definition}{Definition}
\newtheorem{conjecture}{Conjecture}
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\begin{abstract}
Let $\Gamma$ be a real-analytic Jordan curve in $R^3$. Then $\Gamma$ cannot bound
infinitely many disk-type minimal surfaces that provide relative minima of
area.
\end{abstract}
\section{Introduction}
Minimal surfaces are mathematical objects that are intimately related to the physical surfaces
formed by thin soap films. Aside from the idealization to zero thickness, there are some
other differences between physical soap films and minimal surfaces. First, physical soap films
are {\em stable}, in the sense that if they are disturbed slightly, they regain their shape.
Secondly, physical soap films sometimes have internal edges where several films meet, or
other kinds of more complicated topology. Here we are concerned with the classical
{\em problem of Plateau}, which requires, given a Jordan curve $\Gamma$, to find
(or at least prove the existence of) a surface of the topological type of the disk bounded
by $\Gamma$ and minimizing area among such surfaces. Surfaces forming a {\em relative minimum}
of area correspond to soap films stable in the physical sense just described.
Plateau's problem was solved independently by Douglas and Rado in the early 1930s. Their
solution methods produced minimal surfaces that might have certain
singularities known as {\em branch points}.
It was not until the seventies that the regularity (lack of
branch points) of solutions providing an absolute minimum of area with given real-analytic boundary was proved.
Readers not already intimately familiar with branch points may want to view the animated pictures posted to the
Web at \cite{pictures} for a visual introduction to branch points. One picture of a boundary branch point
is given here in Figure 1.
The ``finiteness problem'' addressed in this paper is to prove that for a given Jordan curve $\Gamma$,
only finitely many disk-type soap films are bounded by $\Gamma$. The reason that minimal surfaces
are called ``minimal'' is that a soap film furnishes a relative minimum of area; that is,
perturbing the surfaces slightly will not decrease the area. Thus the mathematical form of
the finiteness problem is to prove that a given Jordan curve $\Gamma$ cannot bound
infinitely many relative minima of area. We prove that theorem in this paper (for real-analytic
boundaries and disk-type surfaces).
Under those same assumptions,
Tomi proved \cite{tomi2} that there cannot exist infinitely many {\em absolute minima} of area.
But generally there will exist many relative minima of area that are not absolute minima,
so Tomi's theorem does not rule out infinitely many physical soap films.
Tromba proved \cite{tromba} {\em generic finiteness}; that is, the set of boundary
curves for which finiteness holds is open and dense in a suitable topology. Here we prove
a theorem about {\em every} real-analytic Jordan boundary.
The finiteness problem for relative minima that we solve here
is a long-standing open problem.%
\footnote{See e.g. \cite{nitsche}, pages~252 and 379, for discussion and many references.
Although the date of that book is 1989, there has been no progress on finiteness since then.
See also \cite{bohme2} for a somewhat more detailed discussion.
}
Branch points of minimal surfaces not furnishing a relative minimum of area are also of considerable
interest, because they are intimately connected with the finiteness problem. It follows (as shown
in \cite{part1}, p. 121) from the
work of B\"ohme \cite{bohme} that if a real-analytic Jordan curve $\Gamma$ bounds infinitely many
disk-type relative minima of area, then it bounds an analytic one-parameter family of such
minimal surfaces, all with the same Dirichlet's integral, which can be continued until it either loops
or terminates in a branched minimal surface.
In \cite{tomi2}, it is proved that a loop of disk-type minimal surfaces contains one which is not a relative
minimum of area.
Therefore, to prove that $\Gamma$
does not bound infinitely many relative minima of area, we only have to rule out the possibility that $\Gamma$
bounds a
one-parameter family of minimal surfaces $u(t,z)$, terminating in a
branched minimal surface when the parameter $t$ is zero, and furnishing a relative minimum of area
for each positive $t$. Tomi solved the finiteness problem
for absolute minima of area and real-analytic boundaries \cite{tomi2} in this way, since the limit of absolute
minima of area is again an absolute minimum, and hence has no branch points. But the problem for relative minima
has remained open. (A discussion of other open problems in this area is given at the end of this paper.)
\begin{figure}[htp]
\centering
\includegraphics[width=0.9\textwidth]{BoundaryOrder2Index1.eps}
\caption{\small This is a boundary branch point of order 2 and index 1. A boundary branch point of order $2m$
goes ``around'' $(2m+1)/2$ times while going ``up and down'' $(2m+k+1)/2$ times. Could a one-parameter family
of surfaces, each of which is a relative minima of area, and all with the same Jordan boundary, end in a surface like this?
}
\end{figure}
Suppose we have a boundary branch point; orient the boundary so it is tangent to the $X$-axis with the branch point
at origin. Consider the unit normal $N$ restricted to the boundary. As the parameter $t$ varies, we do not know
{\em a priori} how $N$ behaves, but at least it must remain perpendicular to the boundary. Close to the branch
point it must almost lie in the $YZ$ plane. Some of our paper contains
calculations concerning the possible behavior of $N$ as the parameter $t$ goes to zero.
The finiteness problem for relative minima of area was first attacked in \cite{part1},
where interior branch points arising as the limit of one-parameter families of relative minima were
shown not to exist, and in \cite{part2}, such boundary branch points were eliminated under special
conditions on $\Gamma$, but not in general; so the finiteness problem for relative minima remained open
until now, except for special boundaries.
It is well known that there is a natural eigenvalue problem associated to the second variation
of the area functional. That problem, and its first eigenvalue and the associated eigenfunction,
play an important role in our work.
The main line of attack in \cite{part1} is an eigenvalue argument, that in the case of an interior branch point,
the Gaussian image of $u$ for small positive $t$ must include more than a hemisphere, and hence $u$ cannot be
a relative minimum of area. The reason why it must include more than a hemisphere is the
Gauss-Bonnet-Sasaki-Nitsche formula, which expresses the total boundary curvature as a sum of contributions
from the Gaussian area and the branch points. This formula implies that as soon as $t > 0$, the
contribution of the branch point term to the total curvature when $t=0$ must be made up by ``extra'' Gaussian area
totaling $2m\pi$, where $2m$ is the order of the boundary branch point. For boundary branch points, this argument
implies that for small positive $t$, there cannot be a neighborhood in the parameter domain on which the Gauss map
covers more than the upper hemisphere.
Near the boundary the normal is confined to within $O(t)$ of the $YZ$ plane,
and it seems plausible that the extra Gaussian area is contributed in the form of $m$ hemispheres. However, that is
difficult to prove. The behavior of $N$ could, {\em a priori}, be quite complicated.
The main line of attack in \cite{part2} is the calculation of the eigenfunction corresponding to the first
eigenvalue. The starting point of this calculation is the relation between the second variations of area and of Dirichlet's integral;
namely, the condition for $\phi$ to belong to the
kernel of the second variation of area is given by $\phi = k \cdot N$, where $k$ belongs to the kernel of
the second variation of area.%
\footnote{Note that $\phi$ is a scalar, because with area it is enough to consider normal variations given
by $u + \phi N$. On the other hand $k$ is a vector since the second variation of Dirichlet's integral is defined
as a Frechet derivative in the space of (vectors defining) surfaces $u$. The theorem that $\phi = k \cdot N$
is proved in \cite{part1}.
}
The ``tangent vector'' $k = u_t$ can be shown to be a ``forced Jacobi direction'' associated
with the branch point; these are purely tangential, so as a function of the parameter $t$, the eigenfunction
goes to zero. We calculate how the eigenfunction behaves on the $w$-plane, that is, on a region near the origin
that shrinks to zero as $t^\gamma$. The leading term in $t$ of this function is a harmonic function in the
$w$-plane which inherits from the eigenfunction the property that it must have only one sign in the upper half plane.
Using this property, we are able to derive additional information about the behavior of $N$ near the origin.
In this paper, we go over the eigenfunction calculations again, for two reasons: First, in \cite{part2} we failed to include
a term allowing for the possibility that there might be branch points in the lower half plane for $t>0$ that might converge
to the origin as $t$ goes to zero. Second, we now simplify and improve the eigenfunction analysis.
Let $g(z)$ be the stereographic projection of the unit normal $N$ to $u^t$. Then the poles of $g$, say $a_i(t)$,
are the places where $N(z)$ is the ``north pole'' $(0,0,1)$.
We analyze the way in which the $a_i(t)$ approach
zero as $t$ goes to zero. These go to zero asymptotically in this way:
$a_i(t) = \alpha_i t^\gamma$, for some $\gamma$. Possibly some of the $a_i$ go to zero faster than others;
we group the $a_i$ into ``rings'', each ring containing the roots that go to zero with the same power of $t$.
In the vicinity of (some of) these $a_i$
the Gauss map of $u^t$ for $t > 0$ must contribute enough Gaussian area to make about $2m\pi$ ``extra'' Gaussian area,
which will disappear in the limit to account for the branch point term in the Gauss-Bonnet-Sasaki-Nitsche formula.
Consider the behavior of the normal $N$ on the boundary near the point where the branch point is when $t=0$.
The normal must be perpendicular to the boundary, so it is confined to the vicinity of the $YZ$-plane if we take
$\Gamma$ to be tangent to the $X$-axis. But its behavior there might be quite uncontrollable, and moreover, several
of the $a_i$ can be associated with the same $\alpha_i$.
In this argument there is another difficulty: the terms calculated for the eigenfunction might all
cancel out, leaving the true eigenfunction hidden in the error terms. This is the case we call ``$\HH$ constant'', since
the eigenfunction is a power of $t$ times the imaginary part of a complex-valued function we call $\HH$.
It turns out that the main difficulty is to rule out this case.
To do so, we first calculate exactly what the eigenfunction must look like in case $\HH$ is constant. Then, we are able to
show that if $\HH$ is constant, no Gaussian area is contributed on the outermost ``ring'' of roots. We then ``chase the Gaussian area''
from one ring of roots down to the next, each time moving to a higher power of $t$ and showing that if the terms in the eigenfunction
cancel out to that power of $t$, then no Gaussian area can be contributed by the roots in that ring. But since {\em somewhere}
we must pick up $2m\pi$ in extra Gaussian area, $\HH$ cannot be constant ``all the way down''. Once that is proved,
we can show that in fact it must not be constant on the outermost ring, and all the roots go to zero with the same power of $t$.
At that point, the proof is nearly complete. In order to finish the proof, we
have to analyze the behavior of $\HH$ at infinity. Since its imaginary part $\Im\, H$
is the limit of the first eigenfunction,
it has only one sign in the upper half-plane. That means that many terms in the
eigenfunction formula must cancel out. These cancellations are very difficult to analyze
directly, which is why we could not get a better result in \cite{part2}. Here (in the last
section of the paper)
we exhibit a differential equation that is satisfied by the function stereographic
projection of the unit normal, involving the eigenfunction $\HH$. This equation is
the key to the analysis of $\HH$ near infinity (in the $w$-plane, where $z = t^\gamma w$,
so the $w$-plane ``blows up'' a portion of the $z$-plane containing the roots $a_i$ and $b_i$,
that shrinks to zero with $t$).
There is an appendix giving a ``Dictionary of Notation'', in which we list the symbols that are used in
more than one section of the paper, and give a reminder of their definitions, as an aid to the reader.
We note in passing that as of yet, nobody has produced an example of
a real-analytic Jordan curve bounding a minimal surface with a boundary branch point, although Gulliver
has exhibited in \cite{gulliver} a surface with a $C^\infty$ boundary and a boundary branch point. The
experts I have asked all believe that they can exist, but an example is still missing. Also, it is not
known whether Gulliver's example is area-minimizing or not.%
\footnote{Wienholtz has used the solution of Bj\"orling's problem (\cite{hildebrandt1}, page 124)
with boundary values $\Gamma(t^3)$ where
$\Gamma$ is a regular parametrization of a real-analytic arc, and a suitably specified normal, to produce
examples of a minimal surface partially bounded by a real-analytic arc which is not a straight line
segment, with a branch point on the arc.}
I am grateful to Fritz Tomi for his careful reading of several early versions of this paper.
\section{Preliminaries}
\subsection{ Analyticity at the boundary}
One of the reasons we need to work with a real-analytic boundary is that we need to know that the
minimal surface can be analytically extended across the boundary, so that it is defined in some neighborhood
of each boundary point and given by a power series there. This well-known result is due to Lewy, and a proof can
be found in \cite{tromba-hildebrandt}, p.~107, Theorem 3. It will be taken for granted in the rest of
this paper and not cited explicitly.
\subsection{ Boundary parametrization}
By a real-analytic arc, or real-analytic Jordan curve, we mean an arc or curve that can be parametrized as a real-analytic function
of arc length. We will consider surfaces bounded by a real-analytic Jordan curve $\Gamma$.
We suppose that $\Gamma$ passes through the origin tangent to the $X$-axis, and that $u$ takes the
portion of the real axis near origin onto $\Gamma$, with $u(0) = 0$.
We still are free to orient the $Y$ and $Z$ axes. We do this in such
a way that the normal at the branch point (which is well-defined) points in the positive
$Z$-direction. With $\Gamma$ oriented in this way, there will be two positive integers
$p$ and $q$ such that $\Gamma$ has a parametrization in the form
$$ \Gamma(\tau) = \vector
{ \tau + O(\tau^2) }
{ \frac{C_1 \tau^{q+1}}{q+1} + O(\tau^{q+2})}
{ \frac{C_2 \tau^{p+1}}{p+1} + O(\tau^{p+2})}
$$
for some nonzero real constants $C_1$ and $C_2$. But it is possible to choose $\tau$ more
carefully so that we do not have the $O(\tau^2)$ term in the first coordinate.
Let $\tau(z) = X(z) = \Re \onehalf \int f - fg^2 dz $. Then on the boundary we have
\begin{eqnarray*}
u(z) &=& \Gamma(\tau(z))
\end{eqnarray*}
This parametrization and function $\tau(z)$ were inspired by Lewy's equation (see \cite{tromba-hildebrandt}, p.~108).
We therefore have
\begin{eqnarray}
\Gamma^{\prime}(\tau) &=& \vector
{1}
{C_1 \tau^q + O(\tau^{q+1})}
{C_2 \tau^p + O(\tau^{p+1})} \label{eq:taudef}
\end{eqnarray}
\subsection{ Order and index of a branch point}
We write $u(z) = (X(z),Y(z),Z(z))$.
We make use of the Enneper-Weierstrass representation of $u$ (see e.g. \cite{hildebrandt1}, p. 112)
$$ u(z) =
Re \vector
{ \onehalf \int f - fg^2 \,dz}
{ \iover2 \int f + fg^2 \,dz}
{ \int fg \,dz}
$$
where $f$ is analytic and $g$ is meromorphic in the upper half-disk.
\begin{definition} The {\em order} of the branch point is the order of the zero of $f$.
The {\em index} of the branch point is the order of the zero of $g$.
\end{definition}
\noindent{\em Remark}. We follow \cite{nitsche}, p.~315 in this definition of ``index''.
Tromba \cite{tromba-hildebrandt} defines ``index'' to mean the sum of the orders of $f$ and $g$,
i.e. Tromba's index is Nitsche's index plus the order.
The order of a boundary branch point of a solution of Plateau's problem must be even
(since the boundary is taken on monotonically).
It is customary to write it as $2m$, and to use the letter $k$ for the index. Thus
$f(z) = z^{2m} + O(z^{2m+1})$ and $g(z) = cz^k + O(z^{k+1})$ for some constant $c$.
\subsection{Gauss-Bonnet theorem for branched minimal surfaces} \label{section:Gauss-Bonnet}
The ``geodesic curvature'' $\kappa_g$ of a boundary curve $\Gamma$ bounding a surface $u$
is the component of the curvature vector of $\Gamma$ in the plane of $u$. It is therefore
bounded by the magnitude of the curvature vector of $\Gamma$, which is defined independently
of any surface.
The Gauss-Bonnet formula says that for regular surfaces (minimal or not)
$$ \int_\Gamma \kappa_g = 2\pi - \int KW \, dx\, dy $$
Note that for minimal surfaces, $KW$ is negative, so both terms on the right are positive.
For minimal surfaces with branch points, there is another term in the Gauss-Bonnet formula:
$$ \int_\Gamma \kappa_g = 2\pi - \int KW \, dx\, dy + 2\pi M$$
where $M$ is the sum of the orders of the interior branch points and half the
orders of the boundary branch points. See \cite{tromba-hildebrandt}, p. 195,
or \cite{nitsche}, p. 331, equation (155).
The ``total curvature'' of a minimal surface is $-\int KW\, dx\,dy$. This is the
``Gaussian area'', or the area of the image on the Riemann sphere of the unit normal $N$ to the
surface. The expression $\vert KW \vert$ is the Jacobian of $N$.
One way to remember the Gauss-Bonnet formula is that each interior branch of order $2m$
point contributes as much $m$ spheres of Gaussian area, and
each boundary branch point of order $2m$
contributes as much as $m$ hemispheres of Gaussian area. This way of looking at the
formula will be helpful when we consider a family of minimal surfaces without branch
points for positive values of the parameter $t$, but having a branch point when $t$ is zero.
So for positive values of $t$, there must be compensating ``bubbles'' of Gaussian area.
Somewhat confusingly, the phrase ``total curvature'' is applied both to surfaces and boundary
curves. Applied to a boundary curve, it means the integral of the magnitude of the curvature
vector over the whole curve. The total curvature of $\Gamma$ is thus greater than the
left side of the Gauss-Bonnet theorem above, stated using the geodesic curvature.
\subsection{On the zero set of real-analytic functions}
\label{section:analyticity}
A subset $U$ of $R^n$ is a called an {\em analytic set} if locally it is the set of simultaneous zeroes of a finite number of
real-analytic functions from $R^n$ to some $R^m$. There is a classical theorem about the structure of analytic sets:
\begin{lemma} \label{lemma:analyticity} Analytic sets are analytically triangulable. That means that each
analytic set is the union of a finite number of real-analytic homeomorphic images of closed simplexes,
meeting only at their boundaries.
\end{lemma}
Six references for this theorem are given in \cite{part1}, where the theorem is discussed on page 117.
The theorem refers to {\em closed} simplexes, which means that the homeomorphisms
in the conclusion are analytic even at the endpoints of intervals or boundary points of 2-simplexes.
Several times in this paper we have reason to consider the zero set of a function that is analytic
in two variables (one real and one complex). When we refer to a function of a real variable $t$
and a complex variable $z$ as ``analytic'', we mean it is given by a power series in $z$ and $t$,
so in particular it is real-analytic in $t$ for fixed $z$ and complex-analytic in $z$ for fixed $t$.
The following lemma describes the zero sets of such functions.
\begin{corollary} \label{corollary:analyticity}
Let $A$ be an open subset of $\R \times \C$ containing
the point $(0,p)$. Let $f$ be a complex-valued analytic function defined on $A$.
Suppose $f(0,p) = 0$ and that for each sufficiently small $t > 0$, $f(t,\cdot)$ is not constant.
Then there exists a neighborhood $V$ of $(0,p)$ and finitely many
functions $c_i$ such that $f(t,c_i(t)) = 0$, every zero of $f$ in $V$ for $t > 0$
has the form $(t,c_i(t))$,
and the $c_i(t)$ are analytic in some rational power of $t$, with $c_i(0) = p$.
\end{corollary}
\noindent{\em Proof}.
If $f$ does not depend on $t$ at all, the theorem is trivial. Otherwise, if
$f$ is divisible by some power of $t$, we can divide that power out without changing the zero set of $f$ for $t > 0$,
so we can assume that $f(t, \cdot)$ is not constant for any sufficiently small $t$, including $t=0$.
Let $S$ be the zero set of $f$.
Then the dimension of $S$ is one, and by Lemma \ref{lemma:analyticity},
$S$ is composed of finitely many analytic 1-simplexes, i.e.
analytic paths given by $t = d_i(\tau)$, $w =c_i(\tau)$,
where $\tau$ is the parameter in which $c_i$ is analytic.
If any of these simplexes do not pass through the $(0,p)$, decrease $V$ to exclude them. The
ones that do pass through the $(0,p)$ should be divided into two simplexes, each ending at $(0,p)$.
Then we can assume that $\tau = 0$ corresponds to $(0,p)$, i.e. $d_i(0) = 0$ and $c_i(0) = p$.
We have
$$f(d_i(\tau),c_i(\tau)) = 0.$$
Since $d_i$ is analytic and not constant, it has only finitely many critical points.
Decrease the size of $V$ if necessary to exclude all nonzero critical points of $d_i$. Then
either $d_i$ is increasing in some interval $[0,b]$ or decreasing in some interval $[0,b]$.
In either case it has an inverse function $\phi = d_i^{-1}$, so $d_i(\phi(t)) = t$.
If the leading term of $d_i$ is $\tau^n$, then $\phi$ is real-analytic in $t^{1/n}$. We
can parametrize the zero set of $f$ by $\tilde c_i(t) = c_i(\phi(t))$, which is real-analytic
in $t^{1/n}$. That completes the proof.
{\em Remark:} In applications it will usually be possible to replace the original parameter $t$ by
$t^{1/n}$, enabling us to assume that the zero set is analytic in $t$.
\subsection{Regularity results for relative minima of area} \label{section:regularity}
We need the following theorem:
\begin{theorem} \label{theorem:regularity} Let $u$ be a minimal surface bounded by a real-analytic Jordan curve $\Gamma$ and
furnishing a relative minimum of area in the $C^0$ metric on the closed disk. Then $u$ has neither
interior nor boundary branch points.
\end{theorem}
It is an open problem to replace $C^0$ by $C^n$ in this theorem. We now summarize the relevant results
from the literature.
In \cite{tromba-hildebrandt}, pp. 554-560, there is an extensive {\em Scholia} reviewing the various
results and proofs of regularity for interior and boundary branch points as matters stood in 2010.
This {\em Scholia} is reprinted and updated to 2012 in \cite{tromba-book}, pp.~169--175.
We are here concerned only with the case of real-analytic boundary, but we need the regularity
of {\em relative} minimizers (of area or Dirichlet energy), not just absolute minimizers, and
both interior and boundary regularity.
In the theory of branch points, one distinguishes between true branch points and false
branch points (see p.~58 of \cite{tromba-hildebrandt} for the definition). False branch points
cannot exist for any solution of Plateau's problem, minimizing or not; see the discussion
in the {\em Scholia} ({\em op.~cit.}) for references. In the rest of this discussion, we
consider true branch points.
First we take up interior regularity. It seems that \cite{beeson-regularity}
was the first to claim interior regularity for relative minima, as opposed to absolute minima,
but Osserman's proof also
yields such a result. Osserman's proof works for $C^0$ relative minima, but not for $C^1$ relative
minima, because the proof involves ``smoothing off'' a wedge.
The argument in \cite{beeson-regularity} is (also) a local
argument, but does not involve cut-and-paste: it shows how to decrease the area (in a $C^n$ smooth way) in a neighborhood of a branch point. To obtain a global result
one must supplement that argument by a smoothing argument to remove the ``crease'' introduced by decreasing
area locally. Wienholtz has correctly pointed out that smoothing argument given in \cite{beeson-regularity}
works only for $C^1$, although
$C^n$ is claimed. Nevertheless the result is correct for $C^1$, as confirmed in the {\em Scholia} just mentioned.
So as far as interior branch points go, we could replace $C^0$ by $C^1$ in the theorem.
In \cite{tromba-book}, Tromba made direct calculations attempting to decrease the Dirichlet integral (globally) for a
minimal surface with an interior branch point. While his method was successful in most cases, in certain cases he was still forced to make
a local argument, so that $C^1$ can still not be replaced with $C^n$ in the theorem; but Tromba did provide
an independent proof of the interior regularity for $C^1$ relative minima.
Now we consider boundary regularity. In the case of real-analytic boundary curves, area and energy
can be locally decreased near a true boundary branch point, as proved independently in
\cite{gulliver-lesley} and \cite{white}. Both proofs rely
on an Osserman-style cut-and-paste.
Hence they work for $C^0$ relative minima, although the papers claim the theorem only for
absolute minima, and the {\em Scholia} cited above also does not state that the theorem works
for $C^0$ relative minima.
However, we do not have to rely on my claim that these two proofs work for $C^0$ relative minima.
Instead we can rely on the work of
Tromba \cite{tromba-book}, who also made calculations for a minimal surface with a boundary
branch point. While serious difficulties remain in the case of boundaries that are only $C^n$,
Tromba's method does work
in the case of a real-analytic boundary curve. But, as mentioned
above, Tromba also needed a local argument in some cases, so his proof also works only for
$C^0$ relative minima. For Tromba's explicit statement that his method works in the
real-analytic boundary case, see p.~168 of \cite{tromba-book}. It is the last sentence in the book, not counting
the {\em Scholia}.
\subsection{ Dirichlet integral and area}
Let $D$ be a plane domain; for our purposes, we may suppose $D$ is a disk or a half-plane, so we will not
worry about the conditions on the boundary $\partial D$ of $D$.
The Dirichlet integral, or ``energy'', of a surface $u$ defined in $D$ is defined by
$$ E(u) = \onehalf \int_D u_x^2 + u_y^2 \, dx\,dy $$
The area functional is defined by
$$ A(u) = \int_D \vert u_x \times u_y \vert \, dx\, dy$$
These functionals are defined on various spaces of functions on $D$, for example $C^{n,\alpha}$ or the
Sobolev spaces $W^{k,p}$. Since we will be restricting attention to surfaces bounded by a fixed real-analytic Jordan
curve $\Gamma$, we are interested in a space of functions defining surfaces bounded by $\Gamma$.
These function spaces can be considered as Hilbert manifolds, and the tangent space at a surface $u$
is a vector space of functions defined on $D$ whose values are tangent to $\Gamma$ on the boundary of $D$.
These functionals have Frechet derivatives, which we denote
by $DE(u)$ and $DA(u)$. These are linear mappings on the tangent space.
Minimal surfaces are defined as critical points of $E$, but it is well known that they
are also critical points of $A$. In connection with $A$, it is common to consider only ``normal variations'',
i.e. to restrict $DA(u)$ to the subspace of the tangent space consisting of those $k$ such that $k(x,y)$
is normal to $u$. It can be shown that if $DA[u]$ vanishes on this subspace then $u$ is minimal.
(Details can be found in \cite{nitsche}, p. 94). In connection with $E$, it is common to consider
only harmonic surfaces; or equivalently, spaces of functions defined on the boundary of the parameter domain
and mapping it to $\Gamma$ in a way homotopic to the identity (since such functions have a unique harmonic extension
to the interior).
\subsection{The second variation of Dirichlet's integral and the forced Jacobi fields }
At a minimal surface $u$, we can consider the
second variation $D^2 E(u)$. This is a bilinear mapping on the tangent space at $u$. The members of this
tangent space are ``tangent vectors'' $k$, i.e. functions defined on $\partial D$ such that $k(\xi)$ is tangent
to $u(\xi)$ for $\xi$ on $\partial D$.
The kernel of $D^2 E(u)$ consists of tangent vectors $k$ such that $D^2 E(u)[k,j] = 0$ for all
tangent vectors $k$. We write $D^2 E(u)[k]$ for $D^2 E(u)[k,k]$. By
diagonalizing the bilinear form, one can prove that the kernel of $D^2 E(u)$ consists of those $k$ for which
$D^2 E(u)[k] = 0$.
\begin{theorem}[Tromba] The second variation of Dirichlet's integral is given by
$$ D^2E[u](h,k) = \int k(h_r - \tilde h_\theta)\, d\theta$$
where $k = \lambda u_\theta$ and $h = \eta u_\theta$ and $\tilde h = \eta u_r$ and $\tilde k = \lambda u_r$.
The tangent vector $k$ to the minimal surface $u$ belongs
to Ker $D^2 E[u]$ if and only if
$$ u_\theta(k_r - \tilde k_\theta) = 0$$
or equivalently
$$ k_z u_z = 0$$
\end{theorem}
\noindent{\em Proof}. The first formula is equivalent to the one
given in \cite{tromba} (bottom of p.~53, with $h$ and $k$
interchanged); for a stand-alone one-page proof of it by direct calculation, and the easy derivation
of the second two formulas from the first, see \cite{beeson-notes}.
Consider the kernel equation $u_\theta(k_r - \tilde k_\theta) = 0$. One way in which this could
be satisfied is if $k_r -\tilde k_\theta = 0$; vectors $k$ satisfying this condition and not
induced by the conformal group are called ``forced Jacobi fields'' or ``forced Jacobi directions''.
Tromba proved that they do not occur in the absence of branch points, and that in the presence of branch
points there are two for each interior branch point (counting multiplicities) and one for each
boundary branch point, so that the space of forced Jacobi fields is finite dimensional. (There can
be at most finitely many branch points, even if the boundary is not real-analytic, as long as the
total curvature of the surface is finite, thanks to the Gauss-Bonnet formula for branched minimal
surfaces.) The forced Jacobi directions are just the directions $k$ such that the function
$K = k + i \tilde k$ is complex analytic, i.e. such that $\tilde k$ is the conjugate harmonic
function of $k$.
Another important characterization of the forced Jacobi fields is this: they are exactly the
tangent vectors of the form
$$ k = \Re\, (i\omega z \, u_z)$$
where $i \omega z$ is a function meromorphic in the parameter domain, and having a pole of
order at most $m$ at each branch point of order $m$. Any function $\omega$ with suitable behavior
on the boundary, and poles of the right orders at the branch poitns, will produce a tangent vector
by this equation. The reason for writing the equation with $\omega z$ instead of with $\omega$
is that in case the parameter domain is the unit disk, the appropriate boundary condition is that
$\omega$ be real on $S^1$. In case the parameter domain is the upper half plane, the condition
is that $i \omega z$ be real on the $x$-axis. The Appendix of \cite{bohme-tromba} contains Tromba's
treatment of the forced Jacobi fields.
\begin{lemma}[Tromba \cite{tromba}] \label{lemma:Tromba} Suppose $u$ is a minimal surface, and $k$ is a tangent vector
belong to $Ker D^2E[u]$ whose harmonic extension is everywhere tangent to $u$. Then $k$ is
a forced Jacobi direction or a direction induced by the conformal group.
\end{lemma}
\subsection{The second variation of area}
\begin{lemma} [Laplacian of the Gauss map]\label{lemma:laplacianN} Let $u$ be a minimal surface.
Then the Laplacian of its unit normal is given by
$\Delta N = 2KWN$.
\end{lemma}
\noindent{\em Proof}.
To prove this elegantly, we make use of the general fact that the Laplace-Beltrami operator of any
surface $S$, applied to the position vector of $S$, is exactly twice the mean curvature of $S$.
(For a proof of that fact, see p. 45 of \cite{hildebrandt1}, formula (30), with $f$ the position vector of $X$
so that $H_f$ in formula (27) is the first fundamental form of $X$.)
Apply this fact to the Riemann sphere, whose position vector $h(w)$ coincides with its
unit normal. Thus $\Delta h = -2h$. Next, note that the map from the disk $D$ to the Riemann
sphere induced by the Gauss map of $u$ is a conformal map with Jacobian $-KW$. Under a conformal
map, the Laplace-Beltrami operator changes to the Laplace-Beltrami operator on the range
surface, multiplied by the Jacobian of the mapping. Hence $\triangle N = 2KW N$, and the
lemma is proved.
One can define the second variation of area, $D^2 A(u)$, on the original tangent space, or on the
subspace of the tangent space corresponding to normal variations. Nitsche tells us (\cite{nitsche}, p. 95) that
we get the same result, but that the calculation is too long to put in his 562 page book. It is customary
to consider the second variation of area on normal variations only. The second
variation $D^2A(u)\phi$ is a bilinear functional on normal variations; technically
normal variations are ``tangent vectors'' to a Hilbert manifold of surfaces, but of course
they are not tangent to the surfaces themselves.
Abusing notation slightly, we
write $D^2 A(u)[\phi,\psi]$ to stand for
$D^2 A(u)[k,j]$ where $k$ is the tangent vector $\phi \cdot N$ and $j=\psi\cdot N$ (and $N$ is the unit normal to $u$).
Thus $D^2 A(u)[\phi,\psi]$
is, in classical terms,
$$ \frac {\partial^2}{\partial s \partial t} A(u + s\phi + t\psi) $$
evaluated at $t=0$ and $s=0$. This derivative is only used when $\phi$ and $\psi$
are in the kernel of the first variation $DA$ (which of course holds when $u$ is minimal).
In case $\phi = \psi$ we recover the classical second variation of area:
$$ D^2 A(u)[\phi,\phi] = \frac {\partial^2}{\partial t^2} A(u + t\phi),$$
which by a further abuse of notation is usually written $$D^2A(u)[\phi]$$.
One can calculate that
\begin{equation}
D^2A(u)[\phi,\psi] = \int_D \psi (-\Delta \phi +2KW\phi) \, dx\, dy,
\end{equation}
where $K$ is the Gauss curvature and $W = \vert u_x \times u_y \vert$. See for example
\cite{nitsche}, p. 96, where the case $\psi = \phi$ is calculated, but the calculation easily adapts to the
bilinear second variation.
\begin{lemma} \label{lemma:KerD2A}
$\phi$ belongs to the kernel of $D^2A(u)$ (for normal variations) just in case $\phi$ is zero on the
boundary of the parameter domain and satisfies
$$ \Delta \phi = 2KW\phi$$
in the interior.
\end{lemma}
\noindent{\em Proof.} Apply the fundamental lemma of the calculus of variations to the preceding formula.
\subsection{Connections between $D^2A$ and $D^2E$}
\begin{theorem} \label{theorem:EtoA}
Let $u$ be a minimal surface in $R^3$ with $C^n$ boundary, and unit normal $N$.
Let $k$ be in $Ker D^2E[u]$. Then $\phi = k \cdot N$ belongs to $Ker D^2 A[u]$.
\end{theorem}
\begin{corollary} If $Ker D^2 A[u]$ has no kernel among normal variations, $Ker D^2 E[u]$
contains only the conformal and forced Jacobi directions.
\end{corollary}
\noindent{\em Proof}. The Corollary follows immediately from the theorem and Tromba's
lemma (Lemma \ref{lemma:Tromba}). We now prove the theorem. Suppose $k$ is in $Ker D^2 E[u]$. By
Lemma \ref{lemma:KerD2A}, it will suffice to show that
$\phi = k\cdot N$ satisfies $\triangle \phi - 2KW \phi = 0$. We have
$$\triangle \phi = \triangle(k\cdot N) = (\triangle k) \cdot N + 2\nabla k \nabla N + k \triangle N.$$
The first term vanishes because $k$ is harmonic. We claim the second term vanishes also.
To prove this, fix a point $z_0$ in the unit disk, and choose coordinates $a$ and $b$
in a neighborhood of $z_0$ that diagonalize the first fundamental form at $z_0$, so that
$N_a = \kappa_1 u_a$ and $N_b = \kappa_2 u_b$, where $\kappa_1$ and $\kappa_2$ are the
principal curvatures of $u$ at $z_0$. If these equations hold in a whole
neighborhood, then $a$ and $b$ are called ``local curvature coordinates''; it costs some
trouble to prove they exist, and we do not need them; we need the first fundamental form
to be diagonalized at one point $z_0$ only. To do this,
we take $a$ and $b$ to be a
certain linear combination of $x$ and $y$.
Let $\nu$ be the angle between the positive $x$-direction and the positive $a$-direction
(so $\nu$ is a function of $z_0$ but not of $z$). Then we define $w = e^{i\nu}(z-z_0)$,
and define $a$ and $b$ by $w = a+ib$, so $a$ and $b$ are coordinates rotated by $\nu$
from $(x,y)$.
Because $u$ is a minimal surface, we have
$\kappa_1 = - \kappa_2$. Then at $z_0$ we have
\begin{eqnarray*}
\nabla k \cdot \nabla N &=& k_a N_a + k_b N_b \\
&=& k_a \kappa_1 u_a + k_b \kappa_2 u_b \\
&=& \kappa_1 (k_au_a -k_bu_b)
\end{eqnarray*}
We have $u_w = u_a-iu_b$ and $k_w = k_a -ik_b$, so
\begin{eqnarray*}
\nabla k \cdot \nabla N &=& \kappa_1 \Re\,(k_w \cdot u_w) \\
&=& \kappa_1 \Re\,((k_z z_w) \cdot (u_z z_w)) \\
&=& \kappa_1 \Re\,(z_w (k_z \cdot u_z)) \\
&=& \kappa_1 \Re\,(e^{-i\nu} (k_z \cdot u_z))
\end{eqnarray*}
since $z_w = e^{-i\nu}$.
Since $k$ is assumed to be in $Ker D^2 E[u]$, we have
$k_z u_z = 0$. Hence the term $\nabla k \nabla N$ vanishes at $z_0$.
But $z_0$ was arbitrary; hence $\nabla k \nabla N$ vanishes everywhere,
and we have proved $\triangle \phi = k \cdot N$.
The proof of the theorem is thus reduced to proving $\triangle N = 2KWN$. But this is
Lemma \ref{lemma:laplacianN}. That completes the proof.
The converse of this theorem is also true (but more difficult):
every $\phi$ in the kernel of $D^2 A(u)$ arises as
$k \cdot N$ for some $k$ in the kernel of $D^2 E(u)$. We do not need this result in this paper, but the curious
can find a proof in \cite{beeson-notes} or \cite{beeson-Bonn}.
\subsection{Stereographic projection}
We take the ``Riemann sphere'' to be the unit sphere $\{(x,y,z): x^2 +y^2 + z^2 = 1\}$.
Stereographic projection ${\bf St}$ is defined by
$$ {\bf St}((x_1,x_2,x_3)) = \frac{ x_1+ix_2}{1-x_3}$$
and its inverse is given by
$$ {\bf St}^{-1}(z) = \frac 1 {1 + \vert z\vert^2} \vector { 2 \,\Re\ z} {2 \,\Im\ z}{\vert z\vert^2 -1}.$$
Thus the equator projects onto the unit circle, and the ``north pole'' $(0,0,1)$ projects onto $\infty$,
while the southern hemisphere projects onto the unit disk, with the south pole $(0,0,-1)$ going to origin.
The picture then has the plane passing through the equator of the sphere.
We follow \cite{osserman}, p. 46, in these details, and we mention them because some other authors use
a sphere of radius $1/2$, with the picture having the sphere entirely above the plane, tangent to
the plane where the south pole of the sphere touches the origin of the plane.
\subsection{The eigenvalue problem associated with the second variation of area}
Associated with a conformal map $N$ defined on a region $\Omega$ in the plane and taking values in the Riemann sphere,
there is a natural eigenvalue problem:
\begin{eqnarray*}
\Delta \phi - \onehalf \lambda \vert \nabla N \vert^2 \phi &=& 0 \mbox{\qquad in $\Omega$ } \\
\phi &=& 0 \mbox{\qquad on $\partial \Omega$}
\end{eqnarray*}
The Jacobian of $N$ is $\onehalf \vert \nabla N \vert^2$.
In case $N$ is the Gauss map of a minimal surface, the Jacobian is also $-KW$, so
$ \onehalf \vert \nabla N \vert^2 = - KW$ and
the eigenvalue equation becomes
$$ \Delta \phi + \lambda KW \phi = 0$$
As proved above in Lemma \ref{lemma:KerD2A}, if $\phi$ is in the kernel of $D^2A(u)$,
then $\phi$ is an eigenfunction of this equation for $\lambda = 2$. But for purposes of
this section, $N$ can be any map from the disk to the Riemann sphere.%
\footnote{Some readers may be familiar with another
form of the eigenvalue equation for which the critical eigenvalue is zero rather than 2, or with this form
but with a factor of 2 inserted so that the critical eigenvalue is 1 instead of 2; both
forms are discussed in \cite{nitsche}, p. 103, cf. equations (62) and (62${}^\prime$). }
We denote the least eigenvalue $\lambda$ of this problem by $\lambda_{N,\Omega}$, or simply
by $\lambda_\Omega$ when $N$ is clear from the context. Sometimes we use the notation $\lambda_{\min}$.
Sometimes, for a minimal surface $u$, we speak of the ``least eigenvalue of $u$'' rather than the
``least eigenvalue of the second variation of $u$'' or ``the least eigenvalue of the eigenvalue problem
associated with the second variation of $u$.''
The least eigenvalue is well-known to be equal to the infimum of the
Rayleigh quotient
$$ R[\phi] = \frac { \int \int_\Omega \vert \nabla \phi \vert ^2 \,dx\, dy}
{ \int \int_\Omega \onehalf \vert \nabla N \vert^2 \phi^2 \,dx\,dy}.
$$
When we speak of the least eigenvalue $\lambda_\Omega$ of a region $\Omega$ on the Riemann sphere,
we mean the following: Let $\Delta$ be the stereographic projection of $\Omega$ and $N$ the
inverse of stereographic projection. Then $\lambda_\Omega := \lambda_{N,\Delta}$. The eigenvalue
problem $\Delta \phi - \frac 1 2 \lambda \vert \nabla N \vert^2 \phi$ on $\Delta$ is equivalent
to the problem $\Delta \phi = \lambda \phi$ on $\Omega$, where now $\Delta$ is the Laplace-Beltrami
operator on the sphere. If $\Omega$ contains the north pole, we should use stereographic projection
from some point not contained in $\Omega$. We do not need to discuss the case when $\Omega$ is the
entire sphere.
{\em Example}.
We compute the least eigenvalue when $N(\Omega)$ is a hemisphere.
In this case the eigenfunction in the lower hemisphere is minus the
$Z$-component of $N$. For example with $\Omega$ equal
to the unit disk and $g(z) = z$, we have $N(z)$ the inverse of stereographic
projection. With $\vert z \vert = r$ and $z = x + iy$ we have
$$N(z)= \frac 1 {1+r^2} \vector{2 x} {2y }{r^2 -1 }.$$
The eigenfunction $\phi$ is given, with $\vert z \vert = r$, by
$$ \phi(z) = \frac{ 1-r^2}{1+r^2}.$$
A few lines of elementary computations (or a couple of commands to a computer algebra program) show that
\begin{eqnarray*}
\triangle \phi &=& \phi_{rr} + \frac 1 {r} \phi_r \\
&=& \frac {8(r^2-1)} {(1+r^2)^3} \\
\end{eqnarray*}
and
\begin{eqnarray*}
\vert \nabla N \vert^2 &=& N_x^2 + N_y^2 = \frac{ 8r}{(1+r^2)^2} \\
\vert \nabla N \vert^2 \phi &=&\frac {8(1-r^2)} {(1+r^2)^3}
\end{eqnarray*}
Hence, $\Delta \phi - \vert \nabla N \vert^2 \phi = 0$, which means the eigenvalue of a hemisphere is 2.
\begin{lemma} \label{lemma:lambda-monotonicity} [Passing to the Riemann sphere does not decrease the eigenvalue] Let $\Omega$ be a connected open set on the sphere
with least eigenvalue $\lambda_\Omega$. Suppose $\Delta$ is a region (open set) in the
plane and $\Omega \subseteq N(\Delta)$ and $N(\partial \Delta) \cap \Omega = \phi$. Suppose that
boundary of $\Omega$ is $C^2$ and the boundary of $\Delta$ is piecewise $C^2$ with the pieces meeting
at positive angles.
Then $\lambda_{N,\Delta} \le \lambda_\Omega$,
and strict inequality holds if $N^{-1}\partial\Omega$ contains an interior point of $\Delta$.
\end{lemma}
\noindent{\em Remark}. Regarding the assumptions on the boundaries, the proof requires that
the least eigenvalue be the minimum of the Rayleigh quotient, and that the gradient of the
least eigenfunction of $\Omega$ not vanish at any boundary point. The hypotheses given
imply these conditions but still allow $\Delta$ to be a half-disk. See \cite{gilbarg}.
\noindent {\em Proof}.
Let $\phi$ be the least eigenfunction of $\Omega$ and define $\psi$ on $\Delta$ by
setting $\psi(z) = \phi(N(z))$ if $N(z) \in \Omega$ else $\psi(z) = 0$. Then
$\psi$ is admissible in the Rayleigh quotient for $\Delta$, since $\phi$ is zero outside
$\Omega$ and $N(\partial \Delta) \cap \Omega = \phi$.
The Rayleigh quotient in question is
$$ \frac { \int \int_\Delta \vert \nabla \psi \vert ^2 dx\, dy}
{ \int \int_\Delta \frac 1 2 \vert \nabla N \vert ^2 \psi^2}
$$
(The factor $1/2$ was explained above.)
Since $\Omega \subseteq N(\Delta)$, on the support of $\psi$, $N$ is a covering map (i.e., locally a
homeomorphism), except
at the points of ramification of $N$, which are isolated. Since $\Omega$ is connected
and $N(\partial \Delta) \cap \Omega = \phi$,
the number of sheets over (cardinality of the pre-image of) $N$ of each non-ramification point is the same.
Hence each of the two
integrals in the (numerator and denominator of the) Rayleigh quotient is the number of sheets times the corresponding integral
on the Riemann sphere, with $\phi$ in place of $\psi$. That is, the Rayleigh quotient for
$\psi$ on $\Delta$ equals the Rayleigh quotient for $\phi$ on $\Omega$, which is $\lambda_\Omega$.
Since $\lambda_\Delta$ is the minimum of such Rayleigh quotients, $\lambda_\Delta \le \lambda_\Omega$.
Now suppose there is an interior point $p$ of $\Delta$ in $N^{-1} (\partial\Omega)$.
By Hopf's lemma, $\nabla \phi$ is never zero at a
point on $\partial \Omega$ where the boundary is $C^2$.
By analyticity, $\nabla N$ is zero only at isolated points; so
there is a point $q$ near $p$ which is still on $N^{-1} (\partial\Omega)$ at which $\nabla N$
is not zero and $\nabla \psi$ is not zero. Hence near $q$, the set $N^{-1}(\partial\Omega)$ is a smooth arc,
and $\nabla \psi$ is zero on one side of it and bounded away from zero on the other side.
Hence we can ``smooth out the edge'' near $q$ to obtain a function $\psi^\prime$ which is
admissible for the Rayleigh quotient and has smaller Rayleigh quotient than $\psi$. Hence
$\lambda_\Delta < \lambda_\Omega$. This completes the proof of the lemma.
\begin{corollary}
If the Gaussian image of a minimal surface defined in (the open set)
$\Delta$ contains a hemisphere, and
at least one boundary point of the hemisphere,
then the eigenvalue $\lambda_{N,\Delta}$ is less than 2.
\end{corollary}
\noindent{\em Proof}. The eigenvalue of a hemisphere is 2.
\subsection{ Dependence of the Gauss map on a parameter}
In this section we present a result which will be applied to the Gauss map of a
one-parameter family of minimal surfaces. However, we present it here as a
result about conformal mappings from a plane domain to the Riemann sphere.
Let $D^+$ be the upper half of the unit disk. We will consider a one-parameter
family of analytic mappings from $D^+$ to the Riemann sphere. (These arise
in our work as the Gauss maps of a one-parameter family of minimal surfaces.)
We write $t$ for the parameter and
$N(t,z)$ for the value of the mapping. For brevity we often omit the explicit
$t$-dependence and write $N(z)$, or $N^t(z)$. Thus $N^t$ is the map $N(t,\cdot)$
from $D^+$ to the Riemann sphere, for a fixed $t$. We define $g = {\bf St} \circ N$. The map
$N$ is thus conformal except at the zeros of $g^{\prime}$ and poles of $g$ of order
more than one. (These are sometimes known as {\em ramification points} of $g$.)
We will suppose that for sufficiently small $t$, the map $g$ is a quotient of
functions $B/A$, where $B$ and $A$ are analytic jointly in $t$ and $z$ (in the sense
explained in section~\ref{section:analyticity}), and that when
$t=0$, $A$ has a zero of order $m$, and $B$ has a zero of order $m+k$, for some positive integers
$m$ and $k$. (In our work, these arise from a boundary branch point of order $2m$ and
index $k$, but here it is not necessary to be so specific.)
By Corollary \ref{corollary:analyticity},
near the origin there
exist paths $a_i(t)$ and $b_i(t)$ describing the poles and zeroes of $g$, respectively,
so that each $a_i$ and $b_i$ is analytic in a rational power of $t$. Let $s$ be the
greatest common divisor of the denominators of these rational powers, and replace
$t$ by $t^{1/s}$. Then $a_i$ and $b_i$ will be analytic in $t$. Thus we may suppose that
the parameter $t$ has been chosen such that $a_i$ and $b_i$ are analytic in $t$. We
also may suppose without loss of generality that for all sufficiently small positive $t$,
we have $a_i(t) \neq b_j(t)$
for all $i$, $j$, since otherwise by analyticity we would have,
for some $i$ and $j$, $a_i(t) = b_j(t)$
for all sufficiently small $t$, and then the factor $z-a_i(t)$ could be cancelled out of both
$A$ and $B$. The partial derivatives of a quotient of real analytic functions are
again quotients of real-analytic functions, so the structure of the set of critical points
of the real and imaginary parts of such a function is also known: it is the union of a
finite set of real-analytic arcs.
%Let $A(t)$ be the area of the spherical image of $D^+$ under $N$, counting
%multiple coverings. Specifically, we have
%\begin{eqnarray*}
% A(t) &=& \int \int_{D^+} \frac 1 2 \vert \nabla N \vert ^2 \,dx\,dy\\
%&=& \int \int_{D^+} \left[ \frac { 4\vert g^{\prime} \vert }
% { (1 + \vert g \vert ^2 )^2 } \right]^2 \, dx\, dy
%\end{eqnarray*}
%
\subsection{Dependence of eigenvalues on a parameter}
Let $D$ be the unit disk and let $0 < \alpha < 1$. Let $F:[0,1] \times D \rightarrow \R$, and suppose that
$F(t,x,y) \ge 0$ for all $t$, $x$, and $y$. For each fixed $t$ let $F^t$ be the function of $x,y$ given by $F^t(x,y) = F(t,x,y)$.
Then the eigenvalue problem $\phi = \lambda F^t \phi$ (with $\phi = 0$ on the boundary) has eigenvalues $\lambda_1^t,\lambda_2^t, \cdots \lambda_n^t$, given
in non-decreasing order. For $n$ such that the eigenspace of $\lambda_n$ has dimension greater than 1, that leaves some
ambiguity about the indexing of the eigenvalues. Suppose that $F$ depends in some smooth way on $t$, perhaps even
real-analytically. Then what can be said about the dependence of $\lambda_n$ on $t$? At least this much:
\begin{lemma}[Continuous dependence of eigenvalues] \label{lemma:continuouseigenvalues}
Consider the eigenvalue problem $\Delta \phi = \lambda F_t \phi$, with $\phi = 0$ on the boundary.
For each fixed
value $t_0$ of $t$, there are continuous functions $\lambda_n(t)$ defined in a neighborhood of $t_0$ giving the eigenvalues.
\end{lemma}
\noindent{\em Proof}. The proof can be found on page 419 of \cite{courant-hilbert}.
In fact more is true. If $F$ depends real-analytically on $t$, then it is possible to number the eigenvalues in such a way that
they are given by real-analytic functions of $t$, as shown in \cite{kato}, p. 370, p. 387; in general this numbering will not be in order
of size, since two eigenvalues, for example $\lambda_2$
and $\lambda_3$, might ``cross'' at a certain value of $t$. Of course no other eigenvalue can cross or even touch $\lambda_1$ since
its eigenspace is always one-dimensional. For our purposes simple continuity of $\lambda_1$ and $\lambda_2$ suffices.
\subsection{Tomi's no-immersed-loops theorem}
We state and sketch the proof of a theorem due to Tomi \cite{tomi1}, or maybe it is due to Tomi and B\"ohme \cite{bohme-tomi}. It is difficult to give an exact reference for this
theorem as the paper where it is stated \cite{tomi1} contains deeper theorems about the structure of the solution set of Plateau's problem,
and the calculation needed for this proof is referenced to \cite{bohme-tomi} where, at the crucial point, the paper says
``Durch eine elementare aber etwas m\"uhsame Rechnung finder man$\ldots$'' (by an elementary but somewhat tiresome computation one finds),
and does not give the computation. Also, they required the boundary to
be $C^{4,\alpha}$, but that was for other reasons in their paper; $C^2$ is enough for the theorem stated here (although in our paper,
the boundary is always real-analytic anyway).
\begin{theorem} [Tomi] Let $\Gamma$ be a $C^{2}$ Jordan curve and suppose $u = u(t)$ is a periodic one-parameter family of
minimal surfaces, $C^2$ as a function of $z$ and $C^1$ in $t$, bounded by $\Gamma$ for each $t$, and satisfying a three-point condition. Suppose that $u_t$ is not identically zero as a function of $z$ for any $t$, and that each $u(t)$ has $\lambda_{\min} = 2$.
Then some $u(t)$ has a branch point, either in the
interior or on the boundary.
\end{theorem}
\noindent{\em Remarks}. By a ``periodic family'', we mean that $u(t + 2\pi)= u(t)$; the exact period is not relevant. The condition
$\lambda_{\min} = 2$ will be fulfilled if $u(t)$ is a relative minimum of area, but it is a more general condition. The condition that
the $u(t)$ satisfy a three point condition is only needed to guarantee that $u_t$ is not a conformal direction.
\smallskip
\noindent{\em Proof}. Suppose, for proof by contradiction, that $u(t)$ has no branch point.
Since each $u(t)$ is a minimal surface, the first variation of Dirichlet's integral $E$
is zero, so $E(t) = E[u(t)]$ is
constant. Hence the second derivative $\partial^2 E/\partial t^2 = 0$. Hence the second variation of $E$ is zero
in the direction $u_t$. That is,
$$ D^2 E[u](u_t) = 0$$
Define
$$ \phi := u_t \cdot N$$
where $N$ is the unit normal to $u(t)$.
(We suppress the $t$-dependence in our notation, writing $u$ instead of $u(t)$ and not indicating the $t$-dependence of $\phi$ and $u_t$.)
Because there is no branch point, there are no forced Jacobi directions. Because of the three-point condition, $u_t$ is not a
conformal direction.
Then $D^2[A](\phi) = 0$ as shown in the previous section. Since $u$ has no branch points, $\phi$ is not identically zero.
By Theorem \ref{theorem:EtoA}, $\phi$ is an eigenfunction of
$$ \Delta \phi = 2 KW\phi$$
over the parameter domain $D$, with $\phi = 0$ on the boundary $\partial D$.
We define the ``volume integral'' to be
$$ V(t) := \int_D u \cdot ( u_x \times u_y) \, dx dy \ = \ \int_D u \cdot N\, W\, dx dy$$
where $N$ is the unit normal to $u$.
The key to the proof is the ``m\"uhsame Rechnung'' that
$$ \frac {\partial V} {\partial T} = V_t = \int \phi W \, dx dy $$
That computation, not found in either \cite{tomi1} or \cite{bohme-tomi}, is written out in detail in \cite{beeson-notes}. Since by hypothesis $\lambda_{\min} = 2$, for each $t$ the function $\phi = \phi(t)$ has only
one sign in the interior of the parameter domain. Since $\phi$ is continuous in $t$ (because $u$ is $C^1$ in $t$),
that sign is the same for all $t$ in
$[0,2\pi]$. Now choose $t_0$ at which $V(t)$ has its minimum value. Then we have
\begin{eqnarray*}
0 &=& V_t(t_0) \\
&=& \int u_t \cdot NW\, dx dy \\
&=& \int \phi W\, dx dy
\end{eqnarray*}
But since $\phi$ is not identically zero and has one sign, this is a contradiction.
\subsection{B\"ohme and Tomi's structure theorem}
Since the 1930s it has been known that, if $\Gamma$ bounds
infinitely many minimal surfaces, there is a sequence $u_n$
of them converging to a minimal surface $u$ bounded by $\Gamma$.
We need an ``analytic compactness'' result that will give us
a one-parameter family of minimal surfaces instead of just a sequence.
B\"ohme and Tomi \cite{bohme-tomi, bohme} and Tromba \cite{tromba} applied
nonlinear global analysis to the theory of minimal surfaces.
We will state and use
the structure theorem of B\"ohme and Tomi. This theorem is Satz 2.4, page 15 in \cite{bohme-tomi}.
That theorem is stated more generally, to cover the case of constant mean curvature as well as
the minimal-surface case; here we state it only for minimal surfaces.
\def\F{\mathcal F}
\begin{theorem} [B\"ohme-Tomi structure theorem] Let $u$ be a minimal surface bounded by a real-analytic
Jordan curve in $R^3$. Let $0 < \alpha < 1$.
Let $\F$ be the space of $C^{2+\alpha}$ maps from $D$ to $\R^3$ satisfying a three-point condition
and whose restrictions to the boundary map $S^1$ onto the range of $\Gamma$ and
are homotopic to $\Gamma$. Let ${{\mathcal M^\prime}(\Gamma)}$ be the set of minimal surfaces in $\F$.
Then there is a neighborhood $W$ of $u$ in the space $C^{2,\alpha}$, such that
$W \cap {\mathcal M^\prime}(\Gamma)$
is an analytic subset of a finite-dimensional analytic submanifold of $\F$.
\end{theorem}
\noindent{\em Warning}. The surfaces in $W$ are {\em not} required to map the boundary of the parameter domain
onto $\Gamma$ monotonically. That is why we do not simply say that surfaces in $\F$ are ``bounded by $\Gamma$.''
The notation ${\mathcal M^\prime}$ instead of ${\mathcal M}$ means that monotonicity is not required.
\smallskip
\noindent{\em Remarks}. The proof goes by introducing a map ${\mathcal L}$ from $W$ to $C^{2+\alpha}(D) \times C^{2+\alpha}{S^1}$,
defined by
$${\mathcal L}(u) = (\Delta u, u_r \cdot u_\theta)$$
Since a harmonic function conformal on the boundary is also conformal in the interior, the zeroes of ${\mathcal L}$ are
exactly the minimal surfaces. B\"ohme and Tomi show that ${\mathcal L}$ is a Fredholm map, which means that
the Frechet derivative $D{\mathcal L}(u)$ has finite-dimensional kernel and cokernel. The finite-dimensional
manifold in the theorem is parametrized by the kernel of ${\mathcal L}$.
The theorem essentially boils down to an application of the implicit function theorem;
if $D{\mathcal L}$ is zero in the $n$ directions $t = (t_1,\ldots,t_n)$, and nonzero in other directions,
then the non-isolated zeroes of ${\mathcal L}$ are locally analytic functions of $t$. That remark, of course,
is not a proof--it is intended only to convey the idea of the proof in \cite{bohme-tomi}.
The kernel of $D{\mathcal L}$ is essentially the kernel of $D^2 E$, the second variation
of Dirichlet's integral. We now know that kernel consists of the forced Jacobi directions (in case $u$ has branch points),
plus directions with non-zero normal components.
In \cite{bohme}, Satz 6 and Satz 11, B\"ohme spells out the consequences of the structure theorem more explicitly.
With $\Gamma$, $u$, and $\F$ be as in the previous theorem, let ${\mathcal M}^\prime (\Gamma)$ be the set of minimal surfaces in $\F$.
The notation ${\mathcal M}^\prime(\Gamma)$ is B\"ohme's. It contrasts with ${\mathcal M}(\Gamma)$, in that
members of ${\mathcal M}(\Gamma)$ must take on $\Gamma$ monotonically (that is, topologically); both sets impose
a three-point condition.
Let ${\mathcal L}$ be as defined above, so ${\mathcal L}(u) $ is zero just when $u$ is a minimal surface in $\F$. Then
the kernel of $D{\mathcal L}(u)$ is a finite-dimensional vector space $V$. B\"ohme's Satz 11 uses terminology from his
Satz 6; putting in that terminology, Satz 11 becomes:
\begin{theorem}[B\"ohme analytic structure theorem]
There is a neighborhood W of $u$ and an analytic map $j$ defined on the unit ball $U$ in $R^n$, where $n$ is the
dimension of the kernel of $D^2 E$, such that
$ {\mathcal A} := j^{-1} {\mathcal M}^\prime(\Gamma)\cap W)$ is an analytic subset of $U$.
\end{theorem}
B\"ohme does not define ``analytic subset'', and we wish to make sure there is no ambiguity about it. It means,
the set of simultaneous zeroes of a finite number of real-analytic functions from $U$ to some $R^m$.
Explicitly, we have
$$ {\mathcal M}(\Gamma)\cap W) = \{ v \in j(U) : {\mathcal L}(v) = 0$$
Then $j(t)$ is a function of $z$, and $j(t)(x,y)$, which we can write as $j(t,x,y)$, is real-analytic in
all three variables, as B\"ohme points out. We have
$$ {\mathcal A} = \{ t \in V \ : \ ({\mathcal L} \circ j)(t)) = 0 \} $$
This ``analytic set'' is the set of zeroes of a real-analytic function from a finite-dimensional space to
an infinite-dimensional space, but since $U$ is finite-dimensional and ${\mathcal L} \circ j$ is analytic,
${\mathcal L} \circ j(U)$ is also a finite-dimensional
manifold.
The structure of the set of minimal surfaces ${\mathcal M}^\prime(\Gamma)$ bounded by $\Gamma$ in a neighborhood of $u$ is thus
reduced to the study of the local structure of analytic subsets of $R^n$. If we want to also require monotonicity
on the boundary, the set ${\mathcal M}(\Gamma)$ becomes {\em semianalytic}; that is, defined by a Boolean combination
of analytic sets. For our purposes, it is enough to consider ${\mathcal M}^\prime(\Gamma)$.
\begin{corollary} Let $\Gamma$ be a real-analytic Jordan curve
in $\R^3$. Then the set of minimal surfaces bounded by $\Gamma$ is locally analytically triangulable.
That is, if $u$ is a minimal surface bounded by $\Gamma$,
and $\F$ is the space of minimal surfaces bounded by $\Gamma$ and satisfying a three-point condition,
then there is a neighborhood $W$ of $u$ in $\F$ such that the set of minimal
surfaces in $W$ is a union of analytic images of simplexes in $R^n$.
\end{corollary}
\noindent{\em Proof}. By B\"ohme's structure theorem and Lemma \ref{lemma:analyticity}.
\smallskip
B\"ohme's theorem leaves open the possibility that the analytic simplexes have their vertices at $u$.
If $u$ is a relative minimum of area, more can be said: the space of minimal surfaces is one-dimensional
and a 1-simplex passes right {\em through} $u$, not just up {\em to} it:
\begin{lemma}[Tomi] \label{lemma:TomiStructureLemma} Let $u$ be a relative minimum of area bounded by a real-analytic Jordan curve $\Gamma$.
If $u$ is not isolated, then the set of minimal surfaces bounded by $\Gamma$ in some neighborhood of $u$
is exactly a one-parameter family $u^t$, with $u^0 = u$, defined for $t$ in some open interval containing 0.
\end{lemma}
\noindent{\em Proof}. See \cite{tomi2}.
\section{ Basic setup and Weierstrass representation}
\subsection{A one-parameter family terminating in a branched surface}
The starting point of our work is the following theorem. The statement and proof can be sketched as follows:
if $\Gamma$ bounds infinitely many relative minima of area, then there is a one-parameter family of minimal surfaces
$u^t$ bounded by $\Gamma$. If they are all without branch points, then they all have $\lambda_{\min} = 2$, and
by compactness the family must loop, but Tomi's theorem prevents that, so they must run into a branched minimal surface.
To make this ``run into'' precise we need B\"ohme's structure theorem; and to guarantee that they remain relative
minima until they run into the branched surface, we need the continuity of the first two eigenvalues. Here are
the details:
\begin{theorem} \label{theorem:one-parameter}
Let $\Gamma$ be a real-analytic Jordan curve bounding infinitely many minimal surfaces without branch
points (satisfying a three-point condition) and with $\lambda_{\min} \ge 2$.
Then there exists a one-parameter family of minimal
surfaces $u^t$, bounded by $\Gamma$, defined for $t$ in some closed interval containing 0,
such that for $t > 0$, $u^t$ is a minimal surface without branch points and with $\lambda_{\min} \ge 2$, and $u^0$ has
an (interior or boundary) branch point.
\end{theorem}
\noindent{\em Proof}. Suppose that $\Gamma$ bounds infinitely many minimal surfaces with $\lambda_{\min} \ge 2$ and
without branch points.
Then by compactness, there is a surface $u$ that is a limit of a sequence $u_n$ of such surfaces, satisfying the same
three-point condition as the $u_n$.
Since the limit of minimal surfaces is minimal, this surface is also minimal. Let $\F$ be the set of minimal surfaces
satisfying the same three-point condition and taking $S^1$ to $\Gamma$ with winding number 1, but not necessarily monotonically.
By (the corollary to) B\"ohme's structure theorem, the set of minimal surfaces in $\F$ near $u$
is analytically triangulable. Pick $n$ large enough that $u_n$ lies on the triangulation,
i.e., $u_n$ is the image under some analytic map $j$ defined on a $k$-simplex $S$ in $R^n$ taking
values in $\F$. Since $u_n$ has no branch points, it takes the boundary monotonically.
By Lemma \ref{lemma:TomiStructureLemma}, the set of minimal surfaces bounded by $\Gamma$ near $u_n$ is one-dimensional,
so the simplex $S$ is a $1$-simplex, and there is a one-parameter family of minimal surfaces $u^t$ defined for
$t$ in $[0,1]$, connecting
$u^0 = u$ to $u^1 = u_n$.%
\footnote{Actually, B\"ohme's structure theorem suffices for this proof, since according to that theorem there
is {\em some} simplex containing $u_n$, possibly of dimension more than 1, that extends analytically to $u^0$.
In that simplex we can find some analytic path.}
Let $q$ be the infimum of the set of $t \le 1$ such that $u^t$ has no boundary or interior branch points. This set
is an open set, since it is the set where $ \vert \nabla u \vert^2 > 0$ on the closed unit disk;
hence $q < 1$.
Now, we just forget about the $u^t$ for $t < q$. That is,
replacing $t$ by $t + q(1-t)$, we can assume without loss of generality that $q=0$; that is, for $t > 0$ the surface
$u^t$ has no branch points and, in view of the fact that there are no boundary branch points, each $u^t$
takes the boundary monotonically.
Let $\phi = u_t \cdot N$. Since for $t > 0$, $u^t$ has no branch points,
Theorem \ref{theorem:EtoA} implies that
for $t > 0$, $\phi$ is an eigenfunction of
$\Delta \phi + 2KW\phi$. Hence 2 is an eigenvalue of $\Delta \phi + \lambda KW \phi$ for each $t > 0$.
For $t = 1$ we have $\lambda_1 = 2 $ and $\lambda_2 > \lambda_1$, since the least eigenvalue
has a one-dimensional eigenspace. By Lemma \ref{lemma:continuouseigenvalues}, $\lambda_1$ and $\lambda_2$ depend continuously on $t$ for
$t >0$, since for $t>0$ we have $KW >0$. Because the eigenspace of $\lambda_1$ is one-dimensional, we cannot
have $\lambda_2 = \lambda_1$ for any value of $t$; hence the two continuous functions $\lambda_1(t)$ and
$\lambda_2(t)$ do not cross (or even touch), and we have $\lambda_1 = 2$ for all $t$ in $(0,1]$. That completes the proof.
\begin{corollary} Let $\Gamma$ be a real-analytic Jordan curve bounding infinitely many relative minima
of area in the $C^0$ metric.
relative minima of area. There there exists a one-parameter family of minimal
surfaces $u^t$, bounded by $\Gamma$, defined for $t$ in some closed interval containing 0,
such that for $t > 0$, $u^t$ is a minimal surface without branch points and with $\lambda_{\min} \ge 2$, and $u^0$ has
an (interior or boundary) branch point.
\end{corollary}
\noindent{\em Proof}. This would be true with $C^n$ in place of $C^0$, if we knew that $C^n$ relative minima have no
branch points. See the discussion in Section \ref{section:regularity} above.
\smallskip
\noindent{\em Remark}. The conclusion of the corollary can be strengthened to claim that each $u^t$ for $t > 0$ not
only has $\lambda_{\min} = 2$ but is also a relative minimum of area. We do not need this result, and it is more
complicated to prove than the lemma above; in \cite{part1} it is proved by reference to a normal form theorem for
the Dirichlet integral due to Tromba.
\subsection{ Weierstrass Representation}
We suppose we are in the following situation: a real-analytic Jordan curve, not lying in a plane, bounds a
one-parameter family of minimal surfaces $u(t,z)$. The surfaces are parametrized by $z$ in the upper half plane,
so that on the real line each $u(t,\cdot)$ is a reparametrization of $\Gamma$. The surface $u(0,\cdot)$ has
a boundary branch point of order $2m$ and index $k$ at the origin. The surfaces $u(t,\cdot)$ for $t > 0$
have no branch points, and they have $\lambda_{\min} = 2$.
All our arguments will be local; we shall only be concerned
with what happens near the origin.\footnote{
We do not need to assume that the branch point is a true branch point. We can rule out the existence of a
one-parameter family of the type considered without that assumption. This does not, however, provide a new proof
of the non-existence of false branch points which are not the terminus of a one-parameter family of relative
minima of area.}
We suppose that $\Gamma$ passes through the origin tangent to the $X$-axis. Then since $\Gamma$ is
a Jordan curve, it is not contained in a line. We still are free to orient the $Y$ and $Z$ axes. We do this in such
a way that the normal to $u(0, \cdot)$ at the branch point (which is well-defined) points in the positive
$Z$-direction.
Finally, we can assume that for all $t$ we have $u(0,0) = 0$. If this is not already the case,
then we can make it so by applying (for each $t > 0$) a conformal transformation that moves $u(0,0)$ to the origin.
If we do this while using the upper half-plane for a parameter domain, then the transformation in question is
simply the map $z \mapsto z- \xi(t)$, where $u(\xi(t)) = 0$. $\xi(t)$ can be found analytically in a rational power of $t$,
and we can replace $t$ by a new parameter in which the family will be analytic. This assumption will be used (for example)
to know that $u$ can be found on the boundary by integrating $u_x$ starting from 0.
The surfaces $u(t,z)$ have an Enneper-Weierstrass representation
$$ u(t,z) =
\Re \vector
{ \onehalf \int f - fg^2 \,dz}
{ \iover2 \int f + fg^2 \,dz}
{ \int fg \,dz}
$$
The branch points of $u(t,\cdot)$ are the places where $f$ and $fg^2$ vanish simultaneously.
In \cite{part1} we were concerned with interior branch points only. In that case we can
argue that since there are no branch points for $t > 0$, the zeroes of $f$ are of even
order and coincide with the poles of $g$, so that $f$ and $g$ have the forms stated in
Lemma 3.3 of \cite{part1}, namely:
\begin{eqnarray*}
f(z) &=& f_0(z) \prod_{i=1}^m(z-a_i)^2\\
g(z) &=& \frac{\displaystyle h(z) \prod_{i=1}^{m+k}(z-b_i)}{\displaystyle \prod_{i=1}^m(z-a_i)}
\end{eqnarray*}
However, this lemma was also used in \cite{part2}, where boundary branch points are considered,
and the formula was not established in that case. In this paper we correctly consider
also the possibility that $f$ and $fg^2$ may have some common zeroes $s_i(t)$, corresponding
to a branch point (or branch points) in the lower half plane that converge to the origin as $t$
goes to zero.
In general we might, {\em a priori}, have some common
zeroes $s_i(t)$ of $f$ and $fg^2$. These common zeroes of $f$ and $fg^2$
could occur with different multiplicities in $f$ and $fg^2$. Say that $s_i$ occurs
with multiplicity $m_i$ in $f$ and $n_i$ in $fg^2$.
We must have $\Im(s_i(t)) < 0$ for $t > 0$, since there will be branch points at the $s_i$,
and $s_i(t)$, like $a_i(t)$ and $b_i(t)$, depends analytically on some power of $t$
called $t^\gamma$ and converging to 0 as $t$ approaches 0.
We define
$$S(z) = \prod_{i=1}(z-s_i)^{\min(n_i,m_i)} $$
Then all zeroes of $S$ are zeroes of $f$ and$fg^2$ with at least the multiplicity they have in $S$.
If $s_i$ occurs with a greater multiplicity in $fg^2$ than in $f$, the extra occurrences are zeroes
of $g^2$, and hence of $g$, so $n_i - m_i$ is even. Similarly, if $s_i$ occurs with a greater
multiplicity in $f$ than in $fg^2$, the extra occurrences are poles of $g^2$, and hence poles of $g$,
so $m_i-n_i$ is even. We have
$$ \frac {fg^2} f = g^2 $$
so the remaining zeroes of $f$ and $fg^2$ are double. We can therefore define functions $A$ and
$B$, real-analytic in $t$ and complex-analytic in $z$, such that
\begin{eqnarray*}
f(z) &=& A^2 S \\
f(z)g^2(z) &=& - B^2 S\\
g(z) &=& \frac{fg^2}{f} = \frac{i B}{A} \\
f(z)g(z) &=& i A B S
\end{eqnarray*}
We already defined the $a_i(t)$ and $b_i(t)$ to be the poles and zeroes of $g$, respectively.
Let $2N$ be the number of zeros of $S$. (It must be even since the number of zeroes of $f$
is $2m$ and all the zeroes of $f$ besides those of $S$ are double.)
Then we can find functions $A_0$ and $B_0$, real-analytic in $t$ and complex-analytic in $z$,
not vanishing at the origin, such that
\begin{eqnarray*}
A(t,z) = A_0(t,z) \prod_{i=1}^{m-N}(z-a_i) \\
B(t,z) = i \lambda B_0(t,z) \prod_{i=1}^{m+k-N}(z-b_i) \\
\end{eqnarray*}
and $A_0(0,0) = 1$, $B_0(0,0)$ is real, and $\vert \lambda \vert = 1$ with $\lambda \neq -1$.
Note that the common zeroes $s_i$ do not enter into the formula for $g$. In \cite{part2}, in
effect we assumed the $s_i$, and hence $S$, do not occur.
We can assume $A_0(0,0) = 1$,
since we can always change $z$ to a constant times $z$. The leading term of $B(z)$, namely
$i\lambda B_0(0,0)$ is some nonzero complex constant, but because of the presence of $\lambda$
in the formula, we can assume $B_0(0,0)$ is real.
Note that with this definition of $A$ and $B$, some of the $a_i$ and/or some of the $b_i$ may
have been defined to be equal to some of the $s_i$, in case $s_i$ occurs with a greater multiplicity
in $f$ than in $fg^2$ or vice-versa. This is possible since these extra occurrences are double,
since they occur as roots or poles of $g$. Each branch point $s_i$ will either occur as an $a_i$,
or as a $b_i$, or neither one (in case its multiplicities in $f$ and $fg^2$ are equal); so it will
still be true that $a_i(t)$ is not identically equal to $b_j(t)$ as $t$ varies, even if the $s_i$
occur and are treated this way.
As we have already remarked, it follows from Corollary \ref{corollary:analyticity} that the $a_i$, $b_i$,
and $s_i$ depend analytically on $t$.
Although they all approach zero as $t$ goes to zero,
some may go to zero faster than others. The roots $a_i$, $b_i$, and
$s_i$ occur in ``rings'', going to zero at different speeds. There is a (finite) sequence of powers $\gamma_1 < \gamma_2 < \ldots $
such that every root goes to zero as $t^{\gamma_n}$ for some $n$; we say those roots
belong to the $n$-th ring. Specifically, the roots in the $n$-th ring are those for which
there exist nonzero complex constants $\alpha_i$, $\beta_i$, and $\zeta_i$ such that
\begin{eqnarray*}
a_i &=& \alpha_i t^{\gamma_n} + O(t^{\gamma_n + 1}) \\
b_i &=& \beta_i t^{\gamma_n} + O(t^{\gamma_n + 1}) \\
s_i &=& \zeta_i t^{\gamma_n} + O(t^{\gamma_n + 1})
\end{eqnarray*}
The roots in the first ring, i.e. those that go to
zero as $t^{\gamma_1}$, are called the {\em principal roots}. These are the ones that
go to zero the slowest; the roots in the higher rings go to zero faster.
\section{The Gauss map and the $\w$-plane} \label{section:wplane}
Suppose that $\gamma$ is one of the
numbers $\gamma_n$; that is, $\gamma$ is a number
such that some of the roots $a_i$, $b_i$, or $s_i$ go to zero as $t^\gamma$, but not necessarily the least such number.
Roots that go to zero as some higher
power of $t$ are called ``fast roots'', and ones that go to zero as a lower power of $t$ are called ``slow roots''.
\begin{eqnarray}
&& \mbox{$Q$ is the number of $a_i$ that go to zero as $t^\gamma$ or faster.} \label{eq:Qdef} \\
&& \mbox{$S$ is the number of $s_i$ that go to zero as $t^\gamma$ or faster.} \label{eq:Sdef} \\
&& \mbox{$R$ is the number of $b_i$ that go to zero as $t^\gamma$ or faster.}\label{eq:Rdef} \\
&& \mbox{$P$ is the total number of fast roots.}\label{eq:Pdef}
\end{eqnarray}
Thus if $\gamma = \gamma_1$, we have $Q = m$, $S = 2N$, and $R = m+k$.
Recall that $z = a_i(t)$ are the zeroes of $A$ and $z = b_i(t)$ are the zeroes of $B$.
We will generally be considering only one ``ring'' of roots at a time, and when we speak of
$\alpha_i$, $\beta_i$, and $\zeta_i$, we have in mind those roots that go to zero as $t^\gamma$:
\begin{eqnarray*}
a_i &=& \alpha_i t^\gamma(1 + O(t)) \\
b_i &=& \beta_i t^\gamma(1 + O(t)) \\
a_i &=& \zeta_i t^\gamma(1 + O(t)) \\
\end{eqnarray*}
We have
$$A = A_0(t,z) \prod_{i=1}^{m-N}(z-a_i(t))$$
We introduce a new variable $\w$ by $z = t^\gamma \w$. Then a factor
$z-\alpha_i t^\gamma$ becomes $t^\gamma (\w - \alpha_i) $, and a factor $z - \alpha_i t^{\gamma_i}$ for $\gamma_i < \gamma$
becomes $$t^{\gamma_i}(z t^{\gamma - \gamma_i} - \alpha_i) = t^{\gamma_i}(-\alpha_i + O(t)).$$
Thus the slow roots introduce constants and powers of $t$, and the roots that go to zero as $t^\gamma$
introduce factors containing $w-\alpha_i$, and the roots that go to zero faster than $t^\gamma$ introduce powers of $w$ times $(1+O(t))$.
All products without explicit indices will be understood to be over those roots which are $O(t^\gamma)$,
and in such products (but not elsewhere) we will assume that the $\alpha_i$ corresponding to the fast roots are zero. Thus
$\prod (w-\alpha_i)$ contains a factor of $w$ for each fast root, and the total number of factors is the number of $a_i$
that go to zero as $t^\gamma$ or faster. Define, for the roots that go to zero as $t^\gamma$ or faster,
\begin{eqnarray*}
\alpha &:=& \mbox{the product of all the $-\alpha_i$ over the slow $a_i$ } \\
\beta &:=& \mbox{the product of the $-\beta_i$ over the slow $b_i$ times $B_0(0,0)$ }\\
\zeta &:=& \mbox{the product of the $-\zeta_i$ over the slow $\zeta_i$} \\
\a_i &:=& a_i/t^\gamma \\
\b_i &:=& b_i/t^\gamma \\
\s_i &:=& s_i/t^\gamma \\
\tAA &:=& \alpha\prod (\w - \a_i) \qquad\mbox{(the product is over $i$ such that $a_i = O(t^\gamma)$)}\\
\tBB &:=& i\lambda B_0\beta\prod (\w - \b_i) \\
\tSS &:=& \zeta\prod (\w - \s_i)
\end{eqnarray*}
Thus $\a_i$ is zero for the fast roots $a_i$, but it could also be zero for some roots $a_i$ that
go to zero as $t^\gamma$, and $\a_i$ is not defined for the slow roots. When $t=0$, $\a_i$
becomes $\alpha_i$. Over $\AA$, $\BB$, or $\SS$, a tilde
indicates $t$-dependence. When $t = 0$ we drop the tilde:
\begin{eqnarray*}
\AA &=& \alpha \, \prod(\w-\alpha_i) \\
\BB &=& i \lambda \beta \, \prod(\w-\beta_i) \\
\SS &=& \zeta\prod(\w-\zeta_i)
\end{eqnarray*}
Now we consider the relation between $A(t,z)$ and $\tAA(t,w)$. As discussed above, the roots that go to zero
as $t^\gamma$ or faster contribute factors of $t^\gamma w$ or $t^\gamma(w-\alpha_i)$ to $\tAA$, and the slow
roots contribute factors $-\alpha_i t^{\gamma_i}$, with $\gamma_i < \gamma$. The product of the factors
contributed by all the slow roots is thus a constant $\alpha$ (defined above) times a power $t^d$ of $t$,
where $d$ is the sum of the $\gamma_i$ over all the slow roots $a_i$. Since the constant $\alpha$ is
part of the definition of $\AA$, it does not show explicitly in the relation between $A$ and $\AA$:
\begin{eqnarray}
A = t^d t^{Q\gamma} \AA (1 + O(t)) \label{eq:ddef}
\end{eqnarray}
Similarly, we have numbers $e$ and $f$ such that
\begin{eqnarray}
B &=& t^e t^{R\gamma} \BB (1 + O(t)) \label{eq:edef} \\
S &=& t^f t^{S\gamma} \SS (1 + O(t))\label{eq:fdef}
\end{eqnarray}
No ambiguity will result from using $f$ for this power of $t$ since in this meaning, it will always occur in an exponent,
and the Weierstrass function $f$ never occurs in an exponent. Similarly the letter ``$S\,$'' has
to do double duty, in exponents as the number of roots $s_i$ that go to zero as $t^\gamma$ or faster,
and not in exponents as a function of $z$, as can be seen in (\ref{eq:fdef}). (We are short of letters in the alphabet, and the function $S$ will eventually disappear.) We define
\begin{equation} \label{eq:Kdef}
\K := (R-Q)\gamma + (e-d).
\end{equation}
When $\gamma = \gamma_1$, so we are working with the principal, or first-ring, roots, then $\K = k\gamma_1$, since no roots
go to zero slower than $\gamma_1$, so $e=d=0$, and $R-Q = k$.
Our next aim is transform our formulas for the Weierstrass representation from their
expressions in $z$ and $t$ to express them in terms of $w$ and $t$. Starting with the
expressions in terms of $z$ and $t$, we have
\begin{eqnarray*}
u &=& \Re \vector{ \onehalf \int f - fg^2 \, dz }
{ \frac i 2 \int f + fg^2\, dz}
{ \int fg \, dz} \\
&=&\Re \vector { \onehalf \int A^2 S + B^2 S \, dz}
{ \frac i 2 \int A^2 S - B^2 S \, dz}
{ i \int ABS \, dz}
\end{eqnarray*}
Now we express $A$, $B$, and $S$ as appropriate powers of $t$ times $\tAA$, $\tBB$, and $\tSS$, as given above; and
we note that since $z = t^\gamma w$, we have $dz = t^\gamma dw$, so when we change the variable of integration to $w$,
we pick up an extra power of $t^\gamma$. Replacing $\tAA$ by $(1+O(t))\AA$, and similarly for $\tBB$ and $\tSS$, we have
\begin{equation}
u = (1 + O(t))\, \Re \vector {\onehalf \int t^{2d + f} t^{(2Q+2S+1)\gamma} \AA^2 \SS + t^{2e+f} t^{2R+2S+1)\gamma} \BB^2 \SS \,dw }
{\frac i 2 \int t^{2d + f} t^{(2Q+2S+1)\gamma} \AA^2\SS - t^{2e+f} t^{(2R+2S+1)\gamma} \BB^2 \SS \,dw }
{ \int t^{d +e + f} t^{(Q+R+S + 1)\gamma} \AA\BB\SS \, dw} \label{eq:WeierstrassAA}
\end{equation}
We next review the formula for the unit normal.
The function $g$ in the Weierstrass representation, which is the stereographic projection
of the unit normal $N$, is given by $ g(z) = i B/A$, and the normal itself is given by
\begin{equation}
N = \frac 1 {\vert g\vert^2 + 1} \vector{ 2\Re\,(g)}{ 2\,\Im(g)}{ \vert g \vert^2 -1}.
\end{equation}
We write $\bar A$ for the complex conjugate of $A$.
Substituting $g = i B/A$, we have
$$ N = \frac {1}{1 + \vert B/A\vert^2} \vector{ -2\Im\,(B/A)} { 2\Re(B/A)} {-1 + \vert B/A\vert^2}$$
Expressing this in terms of $\tBB/\tAA$, we have $B/A = t^\K \tBB/\tAA$, and hence
\begin{equation} \label{eq:Nofwexact}
N = \frac {1}{1 + t^{2\K}\vert \tBB/\tAA\vert^2} \vector{ -2t^{\K}\Im\,(\tBB/\tAA)}
{2 t^{\K} \Re\,(\tBB/\tAA)}
{-1 + t^{2\K}\vert \tBB/\tAA\vert^2}
\end{equation}
We remind the reader that $\None$ is the first component of $N$, and $\Ntwo$ the second component. We have
\begin{equation} \label{eq:N1OverN2}
\frac \None \Ntwo = - \frac {\Im\,(\tBB/\tAA)} {\Re\,(\tBB/\tAA)}
\end{equation}
We record in a lemma the geometric significance of $\K$:
\begin{lemma} \label{lemma:Kpositive} $\K > 0$ if and only if the unit normal $N$ converges to $(0,0,-1)$ on compact subsets of the $w$-plane away from the zeroes of $\AA$, as $t$ goes to zero.
\end{lemma}
\noindent{\em Proof}. The stereographic projection of the unit normal $N$ is $t^\K \tBB / \tAA$. This goes to
zero on compact subsets of the $w$ plane if and only if $\K > 0$; but 0 is the stereographic projection of $(0,0,-1)$.
Alternately the lemma can be proved by direct examination of (\ref{eq:Nofwexact}).
\begin{lemma} \label{lemma:realAA}
Suppose that on compact subsets of the $\w$-plane away from the zeroes of $\AA$, $N$ converges to $(0,0,-1)$ as $t$ goes to zero. Then
\smallskip
(i) $\AA$, $\SS$, and $\BB$ are real when $w$ is real;
\smallskip
(ii) the $\alpha_i$ and $\zeta_i$ are real and the $\beta_i$ are
real and/or occur in complex-conjugate pairs.
\end{lemma}
\noindent{\em Remark}. The hypothesis is certainly true for the first ring of $a_i$, the principal roots. We will later show inductively
that it is true for all the rings of $a_i$, i.e. when each $\gamma_n$ is used to define $\w$.
\smallskip
\noindent{\em Proof}.
First we prove that $\alpha_i$ does not have positive imaginary part.
Suppose, for proof by contradiction, that it does. Then consider a circle of
radius $\rho$ about $\alpha_i$, where $\rho$ small enough that the entire
circle lies in the upper half plane and each other $\alpha_i$ is either
inside or outside the circle. Then by hypothesis, on the boundary of the circle,
$N$ goes to $(0,0,-1)$ as $t$ goes to zero.
Hence the Gauss map covers the upper hemisphere inside the circle, so
the least eigenvalue of $u^t$ is less than 2, for $t>0$ sufficiently small; but that is a contradiction. That contradiction proves that
$\alpha_i$ does not have positive imaginary part.
Let $Y(\w) := {}^2u(t^\gamma \w)$, where again, ${}^2u$ is the second component of $u$.
Then for real $w$
$$Y(\w) = \frac {C_2}{q+1} \tau^{q+1}(1 + O(s))$$
where $\tau$, $q$, and $C_2$ are as in (\ref{eq:taudef}), and $q > 0$.
\begin{eqnarray*}
\tau &=& X(w) = \int_0^z A^2S + B^2S\, dz \\
&=& t^{2d + f + (2Q + S + 1) \gamma} \int_0^\w \AA^2\SS + t^{2\K} \BB^2 \, d\w \,( 1 + O(t))
\end{eqnarray*}
by (\ref{eq:ddef}),(\ref{eq:fdef}), (\ref{eq:edef}), and (\ref{eq:Kdef}). The hypothesis that $N$ converges to $(0,0,-1))$
away from the zeroes of $\AA$ implies $\K > 0$, by Lemma \ref{lemma:Kpositive}.
Hence the term in $t^\K$ can be absorbed into the error term, and we have for real $w$
$$ \tau = t^{2d + f + (2Q + S + 1) \gamma} \int_0^\w \AA^2\SS \, d\w \,( 1 + O(t)) $$
and hence
$$ Y(\w) = \frac {C_2}{q+1} t^{(q+1)(2d + f + (2Q + S + 1) \gamma)} \int_0^\w \AA^2\SS \, d\w \,(1 + O(t))$$
On the other hand from the Weierstrass representation (\ref{eq:WeierstrassAA}), we have
$$Y(\w) = t^{2d + f + (2Q +S + 1)\gamma }\, \Im \int_0^\w \tAA^2\tSS\, d\w + O(t^{2d + 2Q\gamma + f + S\gamma + 2 }) $$
Note that, since $q > 0$, the power of $t$ shown explicitly in this equation is greater than the power of $t$ in the preceding equation.
It follows that
$$\Im \int_0^\w \AA^2\SS\, dw = 0$$
since it is independent of $t$ but from the two previous equations must be $O(t)$.
Therefore the polynomial $\int_0^\w \AA^2\SS \,d\w$ is real on the real axis. It follows that
its coefficients are real. Hence the coefficients of its derivative $\AA^2\SS$ are real. Hence the
roots of $\AA^2\SS$, which are the $\alpha_i$ and $\zeta_i$, come in
complex-conjugate pairs. But we proved above that no $\alpha_i$ has positive imaginary part.
Hence they are all real when $t=0$. Similarly, the $\zeta_i$ do not have positive imaginary part, since
$u^t$ has no branch points for $t > 0$. Hence the $\zeta_i$ are real when $t=0$.
Similarly, with $ Z(\w) = {}^3u(t^{\gamma}) \w $ in place of $Y$, we find
$$\Im \int_0^\w \AA\BB\SS\, dw = 0$$
That is, $\int_0^\w \AA\BB\SS\, d\w = 0$ is real on the real axis.
Therefore also its derivative $\AA\BB\SS$ is real on the
real axis. But $\AA\BB\SS$ is a polynomial; since it is real on
the real axis, its coefficients are real. We have already proved that
$\AA\SS$ is real on the real axis; hence $\BB$ is also real on the real
axis. Hence its roots, which are the $\hat \b_i$ when $t=0$, are real or occur in complex-conjugate
pairs. That completes the proof of the lemma.
\section{The eigenfunction}
In this section, we compute the eigenfunction of the
variational problem associated with the second variation of area. This is shown in \cite{part1} to be
$\phi = u_t \cdot N$, where $ u_t = du/dt$ is the
``tangent vector'' to the one-parameter family, and $N$ is the unit normal. The calculations in this section
are similar to those in \cite{part2}, but apply to the whole $\w$-plane rather than only to compact subsets
away from the $\alpha_i$, and apply not only in the vicinity of the roots going to zero at the slowest rate, but
also to the inner ``rings'' of roots.
Only one assumption will be made for our eigenfunction computation: we assume that $\K > 0$.
(See (\ref{eq:ddef}), (\ref{eq:edef}), (\ref{eq:fdef}), and (\ref{eq:Kdef}) for the definition of $\K$ and the quantities
used to define $\K$.)
This assumption implies that
the power of $t$ that goes with $\BB$ is larger than the power of $t$ that goes with $\AA$. In the case $\gamma = \gamma_1$,
we have $b=d=0$ and $R = m+k-N$ and $Q = m-N$, so the inequality in question is $m+k-N > m-N$. That
is equivalent to $k > 0$, which is certainly true. It turns out that the calculations we would
make for the principal (slowest) roots work for every ring, if we replace $k$ by $\K$ and
assume $\K > 0$. This assumption ensures that
the unit normal, whose stereographic projection is $B/A$, is equal to $t$ to a positive power times $\BB/\AA$, so that
on compact subsets of the $\w$-plane away from zeroes of $\AA$, the unit normal tends to $(-1,0,0)$. Specifically we have
\begin{eqnarray*}
\frac B A &=& t^{(R-Q)\gamma + (e-d)} \frac \tBB \tAA \\
&=& t^{\K\gamma} \frac \tBB \tAA \qquad \mbox{by (\ref{eq:Kdef}) }
\end{eqnarray*}
and the $\K > 0$, by our assumptions.
Differentiating the Weierstrass representation for $u$, we have
$$2 u_t = \frac{d}{dt} \Re \vector{ \int_0^z (A^2S+B^2S) \,dz}
{ i \int_0^z (A^2S-B^2S) \,dz}
{ 2i \int_0^z ABS \,dz}
$$
Expressing this in terms of $\tAA$ and $\tBB$ we have
\begin{equation}
2 u_t = \frac{d}{dt} \Re \vector{ t^{2d+f} t^{(2Q+S+1)\gamma}\int_0^\w (\tAA^2\tSS+ t^{2\K}\tBB^2\tSS) \,d\w }
{ i t^{2d+f} t^{(2Q+S+1)\gamma}\int_0^\w (\tAA^2\tSS-t^{2\K}\tBB^2\tSS) \,d\w}
{ 2i t^{2d + f}t^{(2Q+S+1)\gamma + \K} \int_0^\w \tAA\tBB\tSS \,d\w }
\label{eq:1409}
\end{equation}
where the power of $t$ in the last term is calculated thus:
\begin{eqnarray*}
e+d+f + (R +Q + S + 1)\gamma &=& 2d + f + (2Q+S+1)\gamma + \K \qquad\mbox{by (\ref{eq:Kdef})}
\end{eqnarray*}
Since the expressions for these exponents are a bit unwieldy, we shorten them by introducing
\begin{eqnarray} \label{eq:Mdef}
\mm &:=& 2d + f + (2Q+S+1)\gamma
\end{eqnarray}
To get an intuitive grasp of these formulas, consider the special case in which we are
working with the principal roots. Then $2Q+S+1 = 2m+1$, and if there are also no slow
roots then $e=d=f=0$, so $\mm = (2m+1)\gamma$. If there are some fast roots they
contribute extra powers of $t$, namely $t^{e+d+f}$. In general, we will find that $\mm$
plays the role that $(2m+1)\gamma$ would play in the simplest case.
We have
\begin{eqnarray}
\int_0^\w \tAA^2\, d\w &=& t^\mm \int_0^z A^2 \, dz + O(t^{\mm + 1}) \label{eq:Mprop}\\
\int_0^\w \tBB^2\, d\w &=& t^{\mm +\K} \int_0^z B^2 \, dz + O(t^{\mm + \K + 1})\label{eq:Kprop}
\end{eqnarray}
When we put in the abbreviation $\mm$, the equation (\ref{eq:1409}) for $u_t$ becomes
\begin{equation}
2 u_t = \frac{d}{dt} \Re \vector{ t^{\mm}\int_0^\w (\tAA^2\tSS+ t^{2\K}\tBB^2\tSS) \,d\w }
{ i t^{\mm}\int_0^\w (\tAA^2\tSS-t^{2\K}\tBB^2\tSS) \,d\w }
{ 2i t^{\mm + \K} \int_0^\w \tAA\tBB\tSS \,d\w }
\label{eq:1434}
\end{equation}
The rule for differentiating with respect to $t$ in this situation is found by applying the chain rule as follows,
where $d/dt$ means the partial derivative with respect to $t$, holding $z$ fixed, and $\partial/\partial t$
means the partial derivative with respect to $t$, holding $\w$ fixed.
\begin{eqnarray*}
dH/dt &=& \frac{\partial H}{\partial t} + \frac{dH}{d\w} \frac{d\w}{dt} \nonumber \\
&=& \frac{\partial H}{\partial t} + \frac{ dt^{-\gamma} z}{dt} \frac{dH}{d\w} \nonumber \\
&=& \frac{\partial H}{\partial t} - \gamma t^{-\gamma-1}z \frac{dH}{d\w}\nonumber \\
&=& \frac{\partial H}{\partial t} - \gamma t^{-\gamma-1}t^\gamma w \frac{dH}{d\w}\nonumber \\
\end{eqnarray*}
The final result for differentiating with respect to $t$ is then
\begin{equation}
\label{eq:10}
\frac{dH}{dt} = \frac{\partial H}{\partial t} - t^{-1}\gamma w \frac{dH}{d\w}
\end{equation}
Applying this rule on the right side of (\ref{eq:1434}), we find
\begin{eqnarray*}
&& 2u_t = \\
&& \Re \vector
{ t^{\mm-1}\mm\int_0^\w\tAA^2\tSS \,d\w + t^{\mm + 2\K -1}(\mm+2\K)\int_0^w \tBB^2\tSS \,d\w + O(t^{\mm})}
{ it^{\mm-1}\mm\int_0^\w \tAA^2\tSS \,dw + it^{\mm + 2\K-1}(\mm + 2\K )\int_0^\w \tBB^2\tSS \,d\w + O(t^{\mm})}
{ 2it^{\mm + \K -1} (\mm + \K) \int_0^\w \tAA\tBB\tSS \,d\w + O(t^{\mm + \K})} \\
&&- \Re \vector
{ t^{\mm-1}\gamma \tAA^2\tSS\, \w + t^{\mm + 2\K -1} \tBB^2\tSS\, \w }
{ it^{\mm-1}\gamma \tAA^2\tSS\, \w + it^{\mm + 2\K -1} \tBB^2\tSS\, \w }
{ 2it^{\mm + \K -1} \gamma \tAA\tBB\tSS \, \w} \\
&& + \Re \vector
{t^{\mm} \mm \frac \partial {\partial t} \int_0^\w \tAA^2\tSS \,d\w +
t^{\mm + 2\K}(\mm + 2\K) \frac \partial {\partial t} \int_0^w \tBB^2\tSS \,\w
}
{ it^{\mm} \mm \frac \partial {\partial t} \int_0^\w \tAA^2\tSS \,d\w +
it^{\mm + 2\K} (\mm + 2\K)\frac \partial {\partial t} \int_0^\w \tBB^2\tSS \,d\w
}
{ 2it^{\mm + \K }(\mm + \K) \frac \partial {\partial t} \int_0^\w \tAA\tBB\tSS \,d\w
}
\end{eqnarray*}
The ``error terms'' $O(t^{\mm})$ in the first two components (that come from differentiating
with respect to $t$ under the integral) are uncontrollable, and swamp the $t^{\mm+2\K-1}$ terms;
also the terms in the third vector, involving $\partial/\partial t$, have one higher power of $t$ and are $O(t^{\mm})$.
Eliminating the swamped terms we have
\begin{eqnarray*}
&& 2u_t = \\
&& \Re \vector
{ t^{\mm-1}\mm\int_0^\w\tAA^2\tSS \,d\w - t^{\mm-1}\gamma \tAA^2\tSS\, \w + O(t^{\mm})}
{ it^{\mm-1}\mm\int_0^\w \tAA^2\tSS \,dw -it^{\mm-1}\gamma \tAA^2\tSS\, \w + O(t^{\mm}) }
{ 2it^{\mm + \K -1} (\mm + \K) \int_0^\w \tAA\tBB\tSS \,d\w-2it^{\mm + \K -1} \gamma \tAA\tBB\tSS \, \w + O(t^{\mm+\K})}
\end{eqnarray*}
We may also drop the tildes over $\AA$, $\BB$, and $\SS$, since
the terms with nonzero powers of $t$ can be absorbed in the error terms. Dropping the tildes
and dividing by the leading power of $t$, we have
\begin{eqnarray*}
\frac {2u_t} { t^{\mm-1}}
&=& \Re \vector
{ \mm \int_0^\w \AA^2 \SS \,d\w - \gamma\AA^2\SS\, \w + O(t)}
{ \mm i\int_0^\w \AA^2\SS \,d\w - i\gamma\tA^2\SS\, \w + O(t)}
{ 2it^{\K}(\mm + \K) \int_0^\w \AA\BB\SS \,d\w - \gamma\AA\BB\SS \, \w + O(t^{\K + 1})}
\end{eqnarray*}
Taking the dot product with $N$, we have the following formula for the eigenfunction, valid in the whole $\w$-plane,
including subsets that contain the zeroes of $\AA$:
\begin{eqnarray} \label{eq:ExactEigenFunction1}
&&\frac 2 { t^{\mm-1}} \phi = \\
&& [ \None]\, \Re\, \bigg(\mm\int_0^\w \AA^2\SS\,d\w - \gamma\AA^2\SS\, \w + O(t)\bigg)\nonumber \\
&&- [ \Ntwo]\, \Im\, \bigg(\mm\int_0^\w \AA^2\SS\,d\w - \gamma\AA^2\SS\, \w + O(t) \bigg) \nonumber \\
&& - t^{\K}(\mm + \K)[\Nthree]2\,\Im\,\bigg(\int_0^\w \AA\BB\SS \,d\w - \gamma \AA\BB\SS \, \w +
O(t) \bigg) \nonumber
\end{eqnarray}
We can now substitute for $\None$, $\Ntwo$, and $\Nthree$ the values obtained in (\ref{eq:Nofwexact}),
to express the formula entirely in terms of $\tBB$ and $\tAA$.
We get (after multiplying by the denominator and changing the signs of both sides)
\begin{eqnarray*}
&& \frac {-(1 + t^{2\K} \vert \tBB/\tAA\vert^2)} {t^{\mm-1}} \phi = \\
&& \bigg(t^{\K\gamma} \Im\,(\tBB/\tAA)\bigg)\, \Re\, \bigg(\mm\int_0^\w \AA^2\SS\,d\w - \gamma\AA^2\SS\, \w + O(t)\bigg) \\
&&+ \bigg( t^{\K\gamma} \Re\,(\tBB/\tAA)\bigg)\, \Im\, \bigg(\mm \int_0^\w \AA^2\SS\,d\w - \gamma\AA^2\SS\, \w + O(t)\bigg) \\
&& + t^{\K\gamma}(\mm + 2\K)\bigg(-1+t^{2\K\gamma}\vert \tBB/\tAA\vert^2\bigg)2\,\Im\, \bigg(\int_0^\w \AA\BB\SS \,d\w - \gamma\AA\BB\SS \, \w +O(t)\bigg)
\end{eqnarray*}
\begin{lemma} \label{lemma:eigenfunctionNearAlpha} The following formula for the least eigenfunction of $u^t$
in terms of $N$ is valid everywhere, even near the zeroes of $\AA$.
\begin{eqnarray*}
&& \frac {-(1 + t^{2\K\gamma} \vert \tBB/\tAA\vert^2)} { t^{(\mm + \K)\gamma-1}} \phi \ = \
\Im\,\bigg\{(\tBB/\tAA)\, \bigg(\mm\int_0^\w \AA^2\SS\,d\w -\gamma \AA^2\SS\, \w \bigg)\bigg\} \\
&& - \bigg(1-t^{2\K\gamma}\vert \tBB/\tAA\vert^2\bigg) 2\, \Im\,\bigg\{(\mm + \K)\int_0^\w \AA\BB\SS \,d\w - \gamma\AA \BB \SS \, \w \bigg\} \\
&& + \Re\,(\tBB/\tAA) O(t)+ \Im\,(\tBB/\tAA) O(t) + \vert \tBB/\tAA \vert^2 O(t^{2\K})
\end{eqnarray*}
\end{lemma}
\noindent{\em Remark}. The meaning of ``valid everywhere'' is that the big-$O$ terms are analytic in $t$ and $w$.
\smallskip
\noindent{\em Proof}.
Starting with the formula given just before the lemma, we separate out the error terms:
\begin{eqnarray*}
&& \frac {-(1 + t^{2\K} \vert \tBB/\tAA\vert^2)} {t^{\mm-1}} \phi = \\
&& \bigg(t^{\K\gamma} \Im\,(\tBB/\tAA)\bigg)\, \Re\, \bigg(\mm\int_0^\w \AA^2\SS\,d\w - \gamma\AA^2\SS\, \w \bigg) \\
&&+ \bigg( t^{\K\gamma} \Re\,(\tBB/\tAA)\bigg)\, \Im\, \bigg(\mm \int_0^\w \AA^2\SS\,d\w - \gamma\AA^2\SS\, \w \bigg) \\
&& + t^{\K\gamma}(\mm + 2\K)\bigg(-1+t^{2\K\gamma}\vert \tBB/\tAA\vert^2\bigg)2\,\Im\, \bigg(\int_0^\w \AA\BB\SS \,d\w - \gamma\AA\BB\SS \, \w \bigg) \\
&& + t^{\K \gamma} \Im(\tBB/\tAA)\, O(t) + t^{\K \gamma} (\BB/\AA)\, O(t) + t^{\K\gamma}\, O(t)
+ \vert \BB/\AA\vert^2 \,O(t^{3\K \gamma} )\\
\end{eqnarray*}
Now we combine real and imaginary
parts according to the rule $\Re(u)\Im(v) + \Re(v)\Im(u) = \Im(uv)$, applied to the first two terms on the
right, and divide both sides by $t^{\K \gamma}$. That yields the formula of the lemma.
\smallskip
\begin{lemma} \label{lemma:eigenfunctionCompact} The following two formulas for the least eigenfunction of $u^t$
in terms of $N$ are valid on each compact subset of the $\w$ plane away from the zeroes of $\AA$,
provided $\K > 0$.
\begin{eqnarray*}
\frac {-1 } { t^{(\mm + \K)\gamma-1}} \phi &=&
\Im\,\bigg\{ \frac \BB \AA \, \bigg(\mm\int_0^\w \AA^2\SS\,d\w - \gamma\AA^2\SS\, \w \bigg) \\
&& - (\mm + \K)\int_0^\w \AA\BB\SS \,d\w - \gamma\AA \BB \SS \, \w \bigg\} + O(t) \\
&=& \Im\,\bigg\{\frac \BB \AA \, \bigg(\mm\int_0^\w \AA^2\SS\,d\w \bigg)
- (\mm + \K)\int_0^\w \AA\BB\SS \,d\w \bigg\} + O(t)\\
\end{eqnarray*}
\end{lemma}
\noindent{\em Remark}. The provision that $\K > 0$ is the same as providing that $N$ approach $(0,0,-1)$ on compact subsets
away from the zeros of $\AA$.
It is certainly true when $\gamma = \gamma_1$, and later we will inductively show it is true on the inner rings of roots as well.
\smallskip
\noindent{\em Proof}. The first formula of the lemma follows from the previous lemma by replacing $t^\K \tBB/\tAA$ with $O(t)$
on the left, and noting that $\tBB = \BB + O(t)$, and $\tAA = \AA + O(t)$, and on a given compact subset away from zeros of $\AA$, we have $\tBB/\tAA = \BB/\AA + O(t)$.
The second formula follows by canceling two terms in the first formula. That completes the proof of the lemma.
\smallskip
We now define the function $\HH$ by
\begin{eqnarray} \label{eq:HHdef}
\HH(\w) &=& \frac \BB \AA\, \mm\int_0^\w \AA^2\SS\,d\w - (\mm + \K)\int_0^\w \AA\BB\SS \,d\w \,
\end{eqnarray}
Then by the previous lemma we have
\begin{eqnarray*}
\Im\, \HH &=& \lim_{t \to 0} \frac {-1} { t^{(\mm + \K)\gamma-1}} \phi
\end{eqnarray*}
on compact subsets of the $\w$-plane away from the zeroes of $\AA$.
A crucial question is whether $\HH$ is constant or not. If it is constant, then all the visible terms cancel out, and
we can thus get no information from them. If it is not constant, we can get some information by the asymptotic behavior of
$\HH$ near the origin or near $\alpha_i$.
\section{Gaussian area and the eigenfunction}
We next draw some consequences from the calculations of the eigenfunction and the properties of $\HH$.
\begin{lemma} \label{lemma:hemisphericcovering}
Suppose that on compact subsets of the upper half of the $\w$-plane away from the zeroes of $\AA$,
$N$ converges to $(0,0,-1,)$ as $t$ goes to zero. Then
exactly one of the points $(1,0,0)$ and $(-1,0,0)$ (the ``east pole'' and ``west pole'') can be taken by $N$ at points
in the upper half plane in the vicinity of one $\alpha_i$, and that point is taken on at most $J$ times, where $J$ is the number of
$\a_j$ that converge to $\alpha_i$.
\end{lemma}
\noindent{\em Proof}. Fix $i$, and fix a disk $U$ centered at $\alpha_i$ and excluding all other $\alpha_i$ and zero.
Consider a point $\xi$ in $U$ at which the normal takes on $(1,0,0)$ or $(-1,0,0)$. Let $U^+$ be the intersection
of $U$ with the upper half plane and let $U^-$ be the intersection of $U$ with the lower half plane.
Since $N = t^\K \tBB/\tAA$, the hypothesis about the convergence of $N$ is equivalent to
$\K > 0$. Hence $N$ is $O(t)$ also on the boundary of $U$.
There are $J$ values of $j$ for which $\a_j$ lies in $U$ (for sufficiently small $t$), and at these points $N$ takes
on the north pole. Hence in the whole neighborhood $U$, the map $N$ is a $J$-sheeted covering of the portion of the
Riemann sphere lying north of a small circle of latitude within $O(t)$ of the south pole. Hence each of the east pole
and the west pole can be taken on at most $J$ times over the whole neighborhood $U$.
We will show that if one of the east or west poles is taken on in $U^+$, the other cannot be. At $\xi$ we have $\None = \pm 1$,
and $\Ntwo = \Nthree = 0$. The formula (\ref{eq:ExactEigenFunction1}) calculated above for the eigenfunction $\phi$ in terms of the components of $N$ simplifies greatly when we substitute $\Ntwo = \Nthree = 0$. According to that formula, at $\xi$ we have
\begin{eqnarray*}
\frac 2 {t^{\mm -1} } \phi &=& \None\,\Re\,\bigg( \mm \int_0^\xi \AA^2\,\SS\,d\w - \AA^2(\xi)\,\SS(\xi)\,\xi + O(t)\bigg)
\end{eqnarray*}
Note that
$$ \int_0^\xi \AA^2\SS\, dw = \int_0^{\alpha_i} \AA^2\SS\, dw + \int_{\alpha_i}^\xi \AA^2\SS\, dw $$
The first term is a constant and the second is $O(\xi-\alpha_i)$. Hence the previous equation can be written as
\begin{eqnarray*}
\frac 2 {t^{\mm -1} } \phi &=& \None \, c(1+O(\xi-\alpha_i)) - \tAA^2(\xi)\,\tSS(\xi)\,\xi + O(t)
\end{eqnarray*}
where $c$ is the constant $\mm \int_0^{\alpha_i} \AA^2\,\SS\,d\xi$. The constant $c$ is nonzero since (i) $\alpha_i$ is real,
by Lemma \ref{lemma:realAA}, and
and (ii) the integrand $\AA^2\,\SS$ is nonnegative on the real axis,
because $u$ takes the boundary monotonically and $u_x = t^{\mm + 2d + f} \AA^2\,\SS(1 + O(t))$, and (iii)
$\alpha_i \neq 0$, since $a_i$ goes to zero as $t^\gamma$.
As $t$ goes to zero, $\xi$ will converge to $\alpha_i$, since by assumption
outside every disk centered at $\alpha_i$, $N$ converges to $(0,0,-1)$.
We do not require that $\xi$ depend smoothly on $t$; a sequence of values of $t$ going to zero will suffice, or even
a single choice of $t$ depending appropriately on the constants hidden in the $\O(t)$ and $O(w)$ terms.
Therefore the $O(\xi-\alpha_i)$ term goes to zero too; that is, $\xi -\alpha_i(t) = O(t)$.
We have
\begin{eqnarray}
\frac 2 {t^{\mm -1} } \phi(\xi) &=& \None(\xi) \, c + O(t) \mbox{\qquad where $c = \mm \int_0^{\alpha_i} \AA^2 \SS$} \label{eq:phieastwest}
\end{eqnarray}
For small enough $t$, we see that the sign of $\phi(w)$ is equal to the sign of $\None$ if $\alpha > 0$, or is opposite to the sign
of $\None$ if $\alpha < 0$. In particular, for all such points in the vicinity of a fixed $\alpha_i$, the
sign of $\None$ is the same, since the sign of $\phi$ is the same in the entire upper half plane, because $\phi$ is
the eigenfunction of the least eigenvalue.
That proves that not both the east and the west pole can be taken on over $U^+$ near $\alpha_i$.
But the west (or east) pole can be covered at most $J$ times, since $N$ is a $J$-sheeted covering over $U$.
That completes the proof.
\smallskip
{\em Discussion}. The significance of this lemma is that the Gauss map, considered as a covering map, covers at most one hemisphere
(plus $O(t)$ in a narrow band near the great circle dividing the two hemispheres, whose plane is perpendicular to the $X$-axis), since if it could be extended across the narrow strip near the $YZ$ plane where the image of the boundary
must lie, it would reach the other of the two points in question. Of course that hemisphere could be covered multiple times;
or possibly not at all, as it is conceivable that $\tilde a_i$ has a slightly negative imaginary part for $t>0$ and that $N$ then
covers the entire sphere in the lower half plane near $a_i$, even though $\tilde a_i$ tends to the real number $\alpha_i$.
At this point we have not ruled
out the possibility that one or more sheets of the covering might lie entirely over the lower half $U_-$ of $U$.
The argument does not work for $U_-$, because although $\phi$ continues analytically into $U_-$,
we do not know that it has the same sign throughout $U_-$.
However, the proof does permit us to extract a bit more information than is stated in the above lemma. For each $\alpha_i$,
we have shown that there is one ``permitted hemisphere'', either the west hemisphere or the east hemisphere, according as
the east pole or the west pole may possibly be a value of $N$ near that $\alpha_i$. We can say more about which hemisphere
is permitted:
\begin{lemma} \label{lemma:permittedhemisphere}
All positive $\alpha_i$ have the same permitted hemisphere, and all negative $\alpha_i$ have the
opposite permitted hemisphere.
\end {lemma}
\noindent{\em Proof}. Let $\xi$ be a point near $\alpha_i$ where the east or west pole is taken on.
The formula for $\phi(\xi)$ given in (\ref{eq:phieastwest}) shows that $\phi(\xi)$ has the same sign as $\None(\xi)$
when $\alpha_i > 0$, since then $\int_0^{\alpha_i} \AA^2\SS\, dw > 0$, and the opposite sign with $\alpha_i < 0$,
since then the integral is negative. But $\phi$, since it is the eigenfunction for the least eigenvalue, has the
same sign throughout the upper half plane. Hence the permitted sign of $\None$ is the same for all positive
$\alpha_i$, and the opposite sign for all negative $\alpha_i$. That completes the proof.
\begin{definition}
We say that $\alpha_i$ {\em contributes Gaussian area} if, in the vicinity of $\alpha_i$, the
normal restricted to a small upper half-disk covers an area at least $2\pi - O(t)$ on the sphere.
To make this definition precise, let $D_\rho$ be a disk of radius $\rho$ about $\alpha_i$ in the $w$-plane,
excluding all other $\alpha_j$ and zero,
and let $D_\rho^+$ be its intersection with the upper half plane. Then $\alpha_i$ contributes Gaussian
area if, for every $\rho > 0$, $N(D_\rho^+)$ has area at least $2\pi - O(t)$ for sufficiently small $t$.
In case $N(D_\rho^+)$ has area $J\pi - O(t)$ then we say $\alpha_i$ contributes Gaussian area $J\pi$.
\end{definition}
Since the normal $N$ is a projection mapping (away from its ramification points), and since the
image of the boundary is confined to within $O(t)$ of the $YZ$-plane, as soon as $N$ takes on a value
more than $O(t)$ from the $YZ$-plane (in $D_t^+$) then $\alpha_i$ contributes Gaussian area, and in that
case it must contribute Gaussian area $2J\pi$ for some $J$, i.e. area equal to an integer number of hemispheres.
In particular, if $N$ takes on the ``east pole'' $(1,0,0)$ or the ``west pole'' $(-1,0,0)$ in $D^+_t$
for every $t > 0$, then $\alpha_i$ contributes Gaussian area.
We next turn our attention to the points $(0,1,0)$ and $(0,-1,0)$.
These two points are not the north, south, east, or west poles of the sphere, but they are the remaining
two ``poles''; perhaps we could call them the ``front pole'' and ``back pole.'' Since the
normal $N$, restricted to the boundary near origin,
is confined to the vicinity of the $YZ$ plane, the question is obviously relevant whether it
stays near the south pole (in which case nearly a whole sphere of Gaussian area would be contributed)
or not (in which case, it must come near the front and back poles). To prove this
rigorously is the point of the following lemma.
\smallskip
\begin{lemma}
\label{lemma:eqt}
Suppose that on compact subsets of the $\w$-plane away from the zeroes of $\AA$, $N$ converges to $(0,0,-1,)$ as $t$ goes to zero.
If $\alpha_i$ contributes Gaussian area, then
\smallskip
(i) There exist two points on the real $w$-axis, say $\eq_1(t)$ and $\eq_2(t)$, both converging to $\alpha_i$ as $t$ goes to zero,
such that $\Nthree(\eq_i(t)) = 0 $ and $\Ntwo(\eq_1(t)) = 1 + \O(t)$
and $\Ntwo(\eq_2(t)) = -1 + \O(t)$.
\smallskip
(ii) $\eq_i(t)$ can be found in such a way that
$\eq_i(t)$ has a limit as $t$ goes to zero, and even depends analytically on a rational power of $t$.
Therefore such expressions as $\tAA[\xi:=eq_i(t)]$ are analytic in a rational power of $t$, and
have limits as $t \to 0$.
\end{lemma}
\noindent{\em Remark}.
Condition (i) says that $N(\eq_i(t))$ lies exactly on the equator and within $O(t)$ of the $YZ$ plane.
Hence it is close to either the ``back pole'' or the ``front pole''. Although the points $\eq_i(t)$
are not exact pre-images of the front and back poles, they do lie exactly on the boundary, and
their images under $N$ are close to the front and back poles.
\smallskip
\noindent{\em Proof}. The stereographic projection of the unit normal $N$ is given by $t^{\K} \tilde \BB/ \tilde \AA$. The first
hypothesis is equivalent to $\K > 0$.
Pick a small number $\rho$.
Let $D$ be the disk of radius $\rho$ about $\alpha_i$; let $D^+$ be the part $D$ lying in the upper half-plane.
By hypothesis, $\alpha_i$ contributes Gaussian area, so
it is not the case that $N(D^+)$ lies within $O(t)$
of the $YZ$ plane. Then $N$ does cover one of the points $(-1,0,0)$ and $(1,0,0)$ (the west and east poles).
$\Nthree$ is $-1 + O(t^{2k})$ at the endpoints of the interval $[\alpha-\rho,\alpha+\rho]$, and
indeed on the entire circle $\partial D$, by the hypothesis of the lemma.
Ad (i): In Lemma \ref{lemma:hemisphericcovering} we proved that the Gauss map in the vicinity of $\alpha_i$ covers at most one of the two possible
hemispheres. By hypothesis it covers at least one of them, since it covers either $(-1,0,0)$ or $(1,0,0)$ and can be continued from there
over either the half-space $X \ge O(t)$ or the half-space $X \le O(t)$.
The pre-image of the equator over $D^+$, that is, the set of points $w$ in $D^+$ such that $\Nthree(w) = 0$,
must be mapped either into the half-space $X \le O(t)$ or into the half-space $X \ge O(t)$; otherwise both points $(1,0,0)$ and $(-1,0,0)$
would be covered, contrary to Lemma \ref{lemma:hemisphericcovering}. According to that lemma, at most one of those points is covered, and
as remarked above, at least one of them is covered. Suppose that $(1,0,0)$ is covered (the argument is similar if $(-1,0,0)$ is covered).
Since away from the boundary, $N$ is a covering projection, we can find a path $\pi$ in $D^+$ such that $N(\pi(t))$ lies on the equator,
and $N(\pi(0)) = (1,0,0)$. This path can be continued in both directions
(even through possible ramification points, i.e. points where $\nabla N = 0$)
until the image reaches $(-1,0,0)$, or until for some value of $\tau$, $\pi(\tau)$ lies on the boundary of $D^+$.
The boundary of $D^+$ consists of a circular part $C^+$ (the upper half of the circle of radius $\rho$ about $\alpha_i$)
and a diameter of that circle on the real axis.
On $C^+$, we have $\Nthree(t) = -1 + O(t)$,
since $C^+$ lies in a compact subset of the upper half plane
away from the $\alpha_i$, namely the intersection of the upper half plane with the annulus about $\alpha_i$ of inner radius $\rho/2$
and outer radius $3\rho/2$. Since $(-1,0,0)$ is far from the equator of the sphere, and $N(\pi(\tau))$ lies on the equator,
if $N(\pi(\tau))$ does reach the boundary of $D^+$ for some $\tau$, then $\pi(\tau)$ has to be real. But for real $w$,
$N(w)$ lies within $O(t)$ of the $XZ$ plane, since it must be perpendicular to the boundary curve $\Gamma$ at the point $z = t^{\gamma} w$.
Hence, when $\pi(\tau)$ reaches the boundary of $D^+$, then $N(\pi(\tau))$ lies within $O(t)$ of the $XZ$ plane.
If $N(\pi(\tau))$ reaches the $XZ$ plane and passes more than $O(t)$ beyond the $XZ$ plane before $\pi(\tau)$ reaches the boundary of $D^+$, then
the path can be continued all the way to $(-1,0,0)$, since $\pi(\tau)$ cannot lie on the boundary except when $N(\pi(\tau))$
is within $O(t)$ of the
$XZ$ plane. But by Lemma \ref{lemma:hemisphericcovering}, the path cannot be continued to $(-1,0,0)$. Hence $\pi(\tau)$ must reach the boundary
of $D^+$ while $N(\pi(\tau))$ is within $O(t)$ of the $XZ$-plane. Then that value of $\pi(\tau)$ can be taken as $eq_i(t)$. The
path on the sphere that goes along the equator can be continued in both directions; on the
sphere one direction eventually reaches the front pole, and the other reaches the back pole.
Continuing the path $\pi$ in the parameter domain correspondingly,
one end of $\pi$ terminates in $eq_1(t)$ and
the other end in $eq_2(t)$. As shown above, when $\pi(\tau)$ meets the boundary, $\tau$ must be real, so $eq_i(t)$ as specified
here is real.
Ad (ii): The zero set of $\Nthree$ is defined by a quotient of functions analytic jointly in $\xi$ and $t$.
Its zero set therefore consists of a finite number of curves parametrizable as analytic functions of a rational
power of $t$, by Corollary \ref{corollary:analyticity}.
That completes the proof.
\smallskip
{\em Discussion}. How much Gaussian area is contributed by each $a_i$? We hope to show just
one hemisphere, rather than one sphere. We know that no $a_i$ can contribute a whole sphere
in the most obvious way, as then the Gaussian image would contain more than a hemisphere and the
eigenvalue would be less than 2. Moreover in the previous lemma we showed that the image of the
boundary sticks up at least to the equator, hinting that it divides the sphere into hemispheres.
But this still does not rule out the possibility that somehow a whole sphere of Gaussian area
might be contributed. In the end, we are going to count the Gaussian area, and it will be
vital that each $a_i$ contributes only a hemisphere. The next lemma addresses this crucial issue.
The key to the proof is the fact that near $a_i$, and in the upper half-plane, $N$ can take on the
east or the west pole, but not both. That means it is confined to a hemisphere. Here are the details:
\smallskip
\begin{lemma} \label{lemma:GaussAreaPrincipal}
Assume that $N$ goes to $(0,0,-1)$ on compact subsets of the $w$-plane away from the zeroes of $\AA$.
Then every $\alpha_i$ contributes Gaussian area at most equal to one hemisphere for each $\a_j$
that converges to $\alpha_i$.
\end{lemma}
\noindent{\em Proof.} According to the Gauss-Bonnet theorem (discussed in Section \ref{section:Gauss-Bonnet}), we
have when $t=0$
$$ \int KW \, dx\, dy + \int_\Gamma \kappa_g = 2\pi + 2M \pi$$
where $M$ is the sum of the orders of the boundary branch points and twice the sum of the
orders of the interior branch points. When $t>0$, there are no branch points in the closed upper half plane,
so we have
$$ \int KW \, dx\, dy + \int_\Gamma \kappa_g = 2\pi $$
We let $t$ go to zero in this formula.
Since $\kappa_g$ is bounded by the curvature of $\Gamma$, the integral containing $\kappa_g$ converges as $t$ goes
to zero to the corresponding term in the first formula. Hence the first term, which represents the Gaussian area,
must jump upwards by $2M\pi$. That is, each interior branch point contributes one sphere (times the order)
and each boundary branch point contributes a hemisphere (times the order).
Near each $a_i$, disregarding the boundary, one whole sphere of Gaussian area (times the order) will exist in a small
neighborhood of $a_i$ for small positive $t$, within $O(t)$. Since the Gaussian image of the boundary is confined
to lie near the $YZ$ plane, and since (in the case of several nearby $a_i$'s) the Gauss map may be a multi-sheeted
covering of the sphere, there are ({\em a priori}) several possibilities. For example, there may be no Gaussian area contributed
(in the upper half plane), if the whole ``bubble'' is taken on in the lower half plane. Or, the ``bubble'' may
be divided by the image of the boundary, so that one hemisphere is taken on in the upper half plane, and one in the
lower half plane. Or, if there is more than one sheet of the covering in a small neighborhood of $\alpha_i$, which
is possible if several $\a_j$ converge to the same $\alpha_i$, then some sheets may be divided by the boundary
and others not. {\em A priori} it is also possible that the image of the boundary may rise part way up the sphere and
then reverse direction, so as to only partly divide the sphere.
However, we can now rule out some of these possibilities. Let $p$ be one of the $\alpha_i$ or zero.
Fix a neighborhood $U$ of $p$ in the $w$-plane
small enough to exclude all other $\alpha_i$, or all $\alpha_i$ if $p$ is zero,
and let $U^+$ be the intersection of $U$ with the upper half-plane.
We know that each $\alpha_i$ is real, so $U^+$ is half of $U$. Let $J$ be the number of $\a_j$ that converge to $p$.
Then the total extra Gaussian area over $U$ is $4\pi J$, and over $U$, the Gauss map is a $J$-sheeted
branched covering of the part of the Riemann sphere north of any fixed circle of latitude (say in the southern
hemisphere). That is, fixing such a circle of latitude, for $t$ sufficiently small $N$ will be a branched covering
of the upper part of the Riemann sphere, north of the fixed circle of latitude.
If this $\alpha_i$ contributes any Gaussian area at all (in the upper
half plane), then $N$ takes on either the east or west pole somewhere in $U^+$. Let us suppose that $N$ takes on
the west pole $W$ (the other case is symmetric). It may do so on more than one sheet.
On each sheet (that is, starting from each different pre-image of $W$, we can continue a path $\pi$ such
that $N(\pi(\tau))$ lies on the equator (even if the path must path through ramification points of $N$),
and as in the proof of the previous lemma, $\pi(\tau)$ must run into
real axis, and cannot be continued into the east hemisphere. Over the whole neighborhood $U$ there are $J$
sheets covering the sphere $J$ times, for a total area of $4\pi J$. But now we have proved that each
of these $J$ sheets has at most the west hemisphere covered by $N(U^+)$.
Hence the total Gaussian area contributed by $\alpha_i$
is at most $2\pi J$; that is, one hemisphere per $a_i$ at the most.
That completes the proof of the lemma.
%Since we have already ruled out the
%existence of an interior branch point when $t=0$, all the $\alpha_i$ correspond to boundary branch points,
%so the Gauss-Bonnet theorem requires the total Gaussian area to be $2\pi$ times the sum of orders of the
%branch points. But the sum of orders of the branch points is the $m-2N$, where $2N$ is the number of the $s_i$.
% Now assume, for proof by contradiction, that $N > 0$, i.e., that some $s_i$ exist. Then,
%since each $a_i$ contributes at most $2\pi$ of Gaussian area, the total Gaussian area is only $2\pi(m-2N) < 2\pi m$.
%But the Gauss-Bonnet theorem requires it to be $2\pi m$, so that is a contradiction. Hence the $s_i$ do not occur.
%The Gauss-Bonnet theorem requires the extra Gaussian area to be $2\pi m$, the equivalent of one hemisphere per $a_i$.
%Since we have proved that no $\alpha_i$ can contribute more than its share (i.e. more than the number of $\a_i$
%that converge to that $\alpha_i$, it follows that each $\alpha_i$ must actually contribute its full share,
%the maximum possible: one hemisphere per $\a_i$.
\section{What is true if $H$ is not constant}
In this section, we suppose that $\gamma$ is one of the $\gamma_n$, that $z = t^\gamma \w$,
and that $N$ goes to $(0,0,-1)$ on compact subsets away from the zeroes of $\AA$. This condition
we call the ``$N$-condition''. Since
$$ \frac B A = t^{\K} \frac \BB \AA$$
the $N$-condition can be succinctly expressed as $\K > 0$.
The $\alpha_i$ are the coefficients of the $a_i$ in this ring of roots,
i.e., $a_i = \alpha_i t^\gamma ( 1 + O(t))$, and we define $\alpha_i = 0$ for
roots that go to zero faster than $t^\gamma$. Similarly for the $\beta_i$ and $\zeta_i$.
We refer to the roots that go to zero slower than $t^\gamma$ as ``slow roots'', and
the roots that go to zero faster as ``fast roots''.
\begin{lemma} \label{lemma:H}
If\, $\HH$ is not constant, then there are no fast roots; that is, no $a_i$ or $b_i$ or $s_i$ goes to zero faster than $t^\gamma$.
\end{lemma}
\noindent{\em Proof}. Recall that $Q$, $S$, and $R$, are the numbers of $a_i$, $s_i$, and $b_i$
going to zero as fast or faster than $t^\gamma$. But now we need notation for the numbers of
fast roots.
\def\Qf{Q_f}
\def\Sf{S_f}
\def\Rf{R_f}
Let $\Qf$, $\Sf$, and $\Rf$ be respectively the number of fast $a_i$, fast $s_i$, and fast $b_i$.
The total number of roots going to zero faster than $t^\gamma$, is $$ P := \Qf + \Rf + \Sf.$$
Let $\alphat$ be the product of the nonzero $\alpha_i$ (all of which are real) over the $a_i$
that go to zero as $t^\gamma$. Let
$\betat$ be the product of the nonzero $\beta_i$.
If the $\beta_i$ are not real, they come in complex conjugate pairs, by Lemma \ref{lemma:realAA},
so the product $\betat$ is real. Let $\zetat$ be the product of the nonzero $\zeta_i$, which is
real since all the $\zeta_i$ are real. (Do not confuse $\alphat$ with
$\alpha$, which is the product of the $\alpha_i$ over the slow roots.)
\
We will analyze the asymptotic form of $\HH$ near the origin (without yet assuming $\HH$ is not constant.)
We will show that, if
$\HH$ is not constant, its
leading term is a nonzero real constant times $ \beta \w^{P+1}$.
We analyze the first formula of Lemma \ref{lemma:eigenfunctionCompact}, rather than the
second one. Letting $t$ go to zero in that formula and using the definition of $\HH$ we have
\begin{eqnarray}
&& \HH = \frac \BB \AA \, \bigg(\mm\int_0^\w \AA^2\SS\,d\w - \gamma \AA^2\SS\, \w \bigg) \label{eq:25} \\
&&
- (\mm + \K)\int_0^\w \AA\BB\SS \,d\w - \gamma\AA \BB \SS \, \w \nonumber
\end{eqnarray}
Consider the second term in that equation, namely
\begin{equation} \label{eq:secondterm23}
(\mm + \K)\int_0^\w \AA\BB\SS\, d\w - \gamma \AA\BB\SS \,\w.
\end{equation}
Asymptotically for small $w$, $\AA\BB\SS$ is a
constant times $w^P$, since each fast root contributes a factor of $w$ and there are $P$ fast roots in all.
Hence, asymptotically for small $\w$, (\ref{eq:secondterm23}) has the form
$c \w^{P+1}$ for some real constant $c$.
The value of $c$ is
\begin{equation}\label{eq:cdef}
c = c_0 \bigg(\frac{\mm + K }{P+1} -\gamma \bigg).
\end{equation}
where
\begin{equation} \label{eq:cdef2}
c_0 = \alpha\alphat\beta\betat\zeta\zetat
\end{equation}
The constants $\alphat$, etc., are defined in the first paragraph of this proof, and are all products of nonzero quantities,
so $c_0 \neq 0$.
To see whether $c$ is zero, we recall the definitions of $\mm$ and $\K$:
\begin{eqnarray*}
\mm &=& 2d + f + (2Q+ S+1)\gamma \qquad \mbox{by (\ref{eq:Mdef})} \\
\K &=& (R-Q)\gamma + (e-d) \qquad \mbox{by (\ref{eq:Kdef})}\\
\mm+ \K &=& d+e+f +(R+Q+S+1)\gamma
\end{eqnarray*}
Now $P \le R+Q+S$, and if equality held, that would mean every root that goes to zero at least as fast
as $t^\gamma$ actually goes faster; but there are by assumption {\em some} roots that go to zero as $t^\gamma$.
Therefore, strict inequality holds: $P < R+Q+S$. Adding 1 to both sides and multiplying by $\gamma$,
we have $$(P+1)\gamma < (R+Q+S+1)\gamma \le \mm +\K.$$ It follows from (\ref{eq:cdef}) and the fact that
$c_0 \neq 0$ that $c \neq 0$.
The second term (\ref{eq:secondterm23}) then has the form
\begin{eqnarray}
(\mm + \K)\int_0^\w \AA\BB\SS\, d\w - \gamma \AA\BB\SS \,\w &=& cw^{P+1} + O(w^{P+2}) \label{eq:secondterm}
\end{eqnarray}
Now consider the behavior of the first term in (\ref{eq:25}) near the origin, namely
\begin{equation}
\label{eq:27}
\frac{\BB}{\AA} \bigg(\mm\int_0^w \AA^2\SS \, d\w - \gamma\AA^2\SS\, \w\bigg).
\end{equation}
We have
\begin{eqnarray*}
\AA &=& \alpha \alphat w^{\Qf} + O(w^{\Qf+1}) \\
\SS &=& \zeta\zetat w^{\Sf} + O(w^{\Sf+1}) \\
\AA^2 \SS &=& \alpha^2\alphat^2\zeta\zetat w^{2\Qf + \Sf} + O(w^{2\Qf+\Sf+1})
\end{eqnarray*}
To simplify this equation we introduce a new constant:
$$ c_1 = \alpha^2 \alphat^2\zeta\zetat.$$
Then we have
\begin{eqnarray*}
\AA^2 \SS &=& c_1 w^{2\Qf + \Sf} + O(w^{2\Qf+\Sf+1}) \\
\mm \int_0^w \AA^2 \SS \, d\w &=& \frac {\mm c_1} {2\Qf+\Sf+1} w^{2\Qf+\Sf+1} + O(w^{2\Qf+\Sf+2})\\
\gamma\AA^2 \SS \w &=& \gamma c_1 \w^{2\Qf+\Sf+1} + O(w^{2\Qf+\Sf+2})
\end{eqnarray*}
Subtracting the last two equations,
\begin{eqnarray*}
\mm \int_0^w \AA^2 \SS \, d\w -\gamma\AA^2 \SS \w &=& c_1\bigg( \frac \mm {2Q+S+1} - \gamma\bigg) w^{2\Qf+\Sf+1} + O(w^{2\Qf+\Sf+2})
\end{eqnarray*}
We have
\begin{eqnarray*}
\frac {\BB}{\AA} &=& \frac { \BB} {\alpha\alphat w^{\Qf} + O(w^{\Qf+1})} \\
&=& \frac {\beta\betat}{\alpha\alphat} w^{\Rf-\Qf} (1 + O(w))\\
&=& \frac {\beta\betat}{\alpha\alphat} w^{\Rf-\Qf} + O(w^{\Rf-\Qf+1})
\end{eqnarray*}
so the first term of (\ref{eq:25}), that is (\ref{eq:27}), comes to
\begin{eqnarray}
&&c_0\bigg( \frac \mm {2\Qf+\Sf+1} - \gamma\bigg) w^{\Rf+\Qf+\Sf+1} + O(w^{\Rf+\Qf+\Sf+2}) \nonumber\\
&=& c_0\bigg( \frac \mm {2\Qf+\Sf+1} - \gamma\bigg) w^{P+1} + O(w^{P+2}) \label{eq:firstterm}
\end{eqnarray}
Subtracting the contributions of the two terms as given in (\ref{eq:firstterm}) and (\ref{eq:secondterm}), we have\begin{eqnarray*}
\HH &=& w^{P+1}(c_0 \bigg( \frac \mm {2\Qf + \Sf+1} - \gamma\bigg) - c) +O(w^{P+2})
\end{eqnarray*}
Recalling that $c = c_0((\mm + K)/(P+1)-\gamma)$, we see that the two $c_0\gamma$ terms cancel, leaving
\begin{eqnarray}
\HH &=& c_0w^{P+1} \bigg( \frac \mm {2\Qf+\Sf+1} - \frac{\mm+\K}{P+1}\bigg) + O(w^{P+2} )\label{eq:ImH57}
\end{eqnarray}
Since by hypothesis, $\HH$ is not constant, we must have $P=0$ because
$\Im\, \HH$ has to have only one sign in the upper half plane. It does not matter whether the coefficient is or
is not zero. But $P=0$ means there are no fast roots. That completes the proof of the lemma.
\begin{lemma} \label{lemma:H2}
If $H$ is not constant, and the $N$-condition holds, then at each $\alpha_i$, either $\BB/\AA$ has a simple zero or a simple pole, or is
analytic non-vanishing at $\alpha_i$.
\end{lemma}
\noindent{\em Remark}. This means that the number of $\b_j$ converging to $\alpha_i$ and the number of $\a_i$ converging to $\alpha_j$
differ by at most 1.
\smallskip
\noindent{\em Proof}. Let $q$ be one of the $\alpha_i$.
Then by Lemma \ref{lemma:realAA}, $q$ is real.
We note that $\tAA^2 \tSS$ is, up to multiplication by a power of $t$, the derivative $u_x$
of the minimal surface $u$ along the $x$-axis, which is nonnegative since $u$
takes the boundary monotonically, and can be zero only at the origin. Dividing by a power of $t$
and taking the limit as $t$ goes to zero, we find that $ \AA^2\SS > 0$ when $w > 0$.
Hence $\int_0^q \AA^2\SS \, d\w > 0$ as well.
We will make an analysis near $q$ similar to the analysis
we made above near zero. The role played in that analysis by roots that go to zero as $t^\gamma$
or faster will be played now by the roots that
converge to $q$ when $t$ goes to zero. We re-use the letters ``$Q$'' , ``$R$'', and ``$S$'' with new
meanings: Let $Q$ be the number of $\alpha_j$
that are equal to $q$, let
$R$ be the number of $\beta_j$ that are equal to $q$, and let $S$ be the number of $s_i$ that are equal to $q$.
Let $\betat$ be the product of $\alpha-\beta_j$ over the $j$ such that $\beta_j \neq q$,
(so $\betat \neq 0$); let $\alphat$ be the product of $q- \alpha_j$ over the $j$ such that $\alpha_j \neq q$; and
let $\zetat$ be the product of $q-\zeta_j$ over the $j$ such that $\zeta_j \neq q$.
We consider the behavior of the terms in (\ref{eq:25}) for $w$ near $q$. Let $w = q + \xi$. We have
\begin{eqnarray*}
\int_0^w \AA^2 \SS \, dw &=& \int_0^{q} \AA^2\SS\, dw + \alpha^2\alphat^2 \zeta \zetat \frac{\xi^{2Q+S+1}}{2Q+S+1} + O(\xi^{2Q+S+2})\\
\frac{\BB}{\AA} &=& \frac{ \beta \betat} {\alpha\alphat} \xi^{R-Q} ( 1 + O(\xi))
\end{eqnarray*}
Multiplying these expressions we have
\begin{eqnarray*}
\frac {\BB}{\AA} \int_0^w \AA^2\SS \, dw &=& \frac{ \beta\betat} {\alpha\alphat}\xi^{R-Q}
\bigg\{\int_0^{q} \AA^2 \SS \, d\w \\
&& + \alpha^2\alphat^2 \zeta\zetat \frac{\xi^{2Q+1}}{2Q+S+1} + O(\xi^{2Q+S+2}) \bigg\}(1+O(\xi))
\end{eqnarray*}
\begin{eqnarray*}
&=& \frac{ \beta \betat} {\alpha\alphat} \int_0^{q} \AA^2\SS \, dw \ \xi^{R-Q}
+ \frac {\alpha\alphat\beta\betat\zeta\zetat}{2Q+1} \xi^{R+Q+S+1} + O(\xi^{R+Q+S+2}) \\
&=& \frac{ \beta \betat} {\alpha\alphat} \int_0^{q} \AA^2 \SS\, dw \ \xi^{R-Q}
+ O(\xi^{R+Q+S+1})
\end{eqnarray*}
Define the constant
$c_2 = \frac{ \beta \betat} {\alpha\alphat} \int_0^q \AA^2\SS\, dw.$
We proved in the first paragraph of this proof that the integral appearing in $c_2$ is not zero, and $\beta\betat/(\alpha\alphat)$ is
not zero either, since it is a quotient of products of nonzero numbers.
The last equation can be written
\begin{eqnarray*}
\frac {\BB}{\AA} \int_0^w \AA^2\SS \, dw &=& c_2\xi^{R-Q} + O(\xi^{R+Q+S+1})
\end{eqnarray*}
Similarly
\begin{eqnarray*}
\int_0^w \AA\BB\SS \, dw &=& \int_0^{q} \AA \BB\SS \, dw + \alpha\beta\alphat\betat\zeta\zetat\, \frac{ \xi^{R+Q+S+1}}{R+Q+S+1} + O(\xi^{R+Q+S+2})
\end{eqnarray*}
The last two lines can be combined using (\ref{eq:25}) to yield
\begin{eqnarray*}
H(w) &=& c_1 + c_2 \xi^{R-Q} + O(\xi^{R+Q+S+1})
\end{eqnarray*}
where the constant $c_1 = \int_0^q \AA\BB\SS$ is real because $\AA$, $\BB$, and $\SS$ are real on the real axis, and $q$ is real
and nonzero.
Thus we have
$$ \Im\, \HH(w) = c_2 \Im\, \xi^{R-Q} + O(\xi^{R-Q+1})$$
Unless $\vert R-Q \vert \le 1$, i.e. $R = Q \pm 1$ or $R=Q$, the imaginary part of the leading term $c_2 \xi^{R-Q}$
will not have one sign in the upper half-plane. Therefore, $R =Q \pm 1$ or $R=Q$.
That completes the proof of the lemma.
\section{What is true if $\HH$ is constant}
In this section we draw out some consequences of the assumption that $\HH$ is constant.
We start with a formula for the derivative of $\HH$:
\begin{lemma} \label{lemma:Hdif}
$$\frac d {dw} \HH = \frac d {dw} \bigg( \frac \BB \AA \bigg) \int_0^\w \AA^2\SS\,d\w
-\K \AA\BB\SS $$
\end{lemma}
\noindent{\em Proof}. By the definition in (\ref{eq:HHdef}) we have
\begin{eqnarray*}
\HH(\w) &=& \frac \BB \AA\, \mm\int_0^\w \AA^2\SS\,d\w - (\mm + \K)\int_0^\w \AA\BB\SS \,d\w \,
\end{eqnarray*}
Differentiating we have
\begin{eqnarray*}
\frac d {dw} \HH &=& \frac d {dw} \bigg( \frac \BB \AA \bigg) \int_0^\w \AA^2\SS\,d\w
+ \bigg(\frac \BB \AA\bigg) \AA^2 \SS - (\mm +\K) \AA \BB \SS
\end{eqnarray*}
Simplifying, we have the formula of the lemma. That completes the proof.
\begin{lemma} \label{lemma:Hconstant}
If $H$ is constant, then
$$ \frac \BB \AA = C \bigg(\int\,\AA^2\SS\,d\w\bigg)^{ \K/\mm}$$
for some constant $C$,
and $\BB/\AA$ is a polynomial.
\end{lemma}
\noindent{\em Proof}.
Let $\Q$ be the rational function $\BB/\AA$.
By Lemma~\ref{lemma:Hdif}, we have
\begin{eqnarray*}
\HH^\prime = 0
&=& \mm \Q^\prime \int\, \AA^2\SS\,d\w - \K\AA^2\SS\Q
\end{eqnarray*}
The equation has one solution
$$\Q = \bigg(\int\,\AA^2\SS\,d\w\bigg)^{ \K/\mm}$$
Since it is a linear equation, every solution is a multiple of that one, so the formula in the lemma
is proved.
Whenever $C$ occurs in the context of this integral, it will mean this constant.
Multiplying by $\AA$ we have
$$\BB = C \AA \bigg(\int\,\AA^2\SS\,d\w\bigg)^{ \K/\mm}.$$
This equation shows that $\BB$ vanishes wherever $\AA$ does, and to at least
as great an order. Hence the rational function $\BB/\AA$ is actually a polynomial.
That completes the proof of the lemma.
The converse is also true:
\begin{lemma} \label{lemma:impliesHconstant}
If for some constant $C$, we have
$$ \frac \BB \AA = C \bigg(\int\,\AA^2\SS\,d\w\bigg)^{ \K/\mm}$$
then $\HH$ is constant.
\end{lemma}
\noindent{\em Proof}. By Lemma~\ref{lemma:Hdif},
\begin{eqnarray*}
\frac d {dw} \HH &=& \mm \frac d {dw} \bigg( \frac \BB \AA \bigg) \int\, \AA^2\SS\, d\w - \K\AA\BB\SS \\
\end{eqnarray*}
Substituting the expression assumed in the lemma for $\BB/\AA$ we have
\begin{eqnarray*}
\frac d {dw}\HH &=& \mm C \frac d {dw}\bigg( \int\,\AA^2\SS\,dw \bigg)^{\K/\mm}\int\, \AA^2\SS\, d\w - \K\AA\BB\SS \\
&=& \K C\bigg( \int\,\AA^2\SS\,dw \bigg)^{\K/\mm-1}\AA^2 \SS \int\, \AA^2\SS\, d\w - \K\AA\BB\SS \\
&=&\K C\bigg( \int\,\AA^2\SS\,dw \bigg)^{\K/\mm} \AA^2\SS - \K\AA\BB\SS \\
&=& \K \bigg(\frac \BB \AA \bigg) \AA^2 \SS - \K\AA\BB\SS \\
&=& 0
\end{eqnarray*}
Since $\HH$ has derivative 0, $\HH$ is constant. That completes the proof of the lemma.
\begin{lemma} \label{lemma:H-constant}
Let $R$, $Q$, and $S$ again be the numbers of $a_i$, $b_i$, and $s_i$ going to zero as $t^\gamma$ or faster. Let
$R_f$, $Q_f$, and $S_f$ be the numbers of $b_i$, $a_i$, and $s_i$ going to zero strictly faster than $t^\gamma$ (the ``fast roots'').
If \,$\HH(w)$ is constant then
\smallskip
(i) $ R-Q = (2Q+S+1) \K/\mm$.
\smallskip
(ii) $R_f - Q_f = (2Q_f+S_f+1) \K/\mm$.
\smallskip
(iii) if $\K > 0$ then $R_f > Q_f$.
\end{lemma}
\noindent{\em Proof}.
By the previous lemma we have
$$ \frac \BB \AA = C\bigg(\int\,\AA^2\SS\,d\w\bigg)^{ \K/\mm}$$
Ad (i). We look at the asymptotic expansion for large $w$.
We note that as $\w$ goes to infinity, $\Q = \BB/\AA$ must be asymptotic to $ (\beta/\alpha)\w^{R-Q}$, where
$R$ and $Q$ are degrees of $\BB$ and $\AA$ respectively. The right hand side gives us
\begin{eqnarray*}
\bigg(\int\,\AA^2\SS\,d\w\bigg)^{ \K/\mm} &=&
\bigg( \frac {\alpha^2\zeta}{2Q+S+1}( w^{2Q+S+1} + O(w^{2Q+S}))\bigg)^{\K/\mm}
\end{eqnarray*}
\begin{eqnarray*}
&=& \bigg(\frac {\alpha^2\zeta}{2Q+S+1}\bigg)^{\K/\mm} w^{(2Q+S+1)\K/\mm} + O(w^{(2Q+S+1)\K/\mm -1})
\end{eqnarray*}
The exponents of the leading power of $w$ must be equal; that yields the equation in (i). Comparing the coefficients
also enables us to determine the constant $C$; it turns out to be
$$ C = \frac { \beta}{\alpha} \bigg(\frac{2Q+S+1}{\alpha^2\zeta}\bigg)^{\K/\mm} $$
Ad (ii). Now we look at the behavior for small $w$.
Asymptotically near the origin we have
$$ \int\,\AA^2\SS\,d\w = \frac {\alpha^2 \zeta} {2Q_f+S_f+1} \w^{2Q_f + S_f + 1} (1 + O(\w))$$
and hence
$$\bigg(\int\,\AA^2\SS\,d\w\bigg)^{ \K/\mm} = \bigg( \frac {\alpha^2\zeta} {2Q_f+S_f+1}\bigg)^{\K/\mm} \w^{(2Q_f + S_f + 1)\K/\mm} (1 + O(\w))$$
and hence
\begin{eqnarray*}
\BB &=& C \bigg( \frac {\alpha^2\zeta} {2Q_f+S_f+1}\bigg)^{\K/\mm} \AA\, \w^{(2Q_f + S_f + 1)\K/\mm} (1 + O(\w)) \\
&=& c \w^{(2Q_f + S_f + 1)\K/\mm + Q_f} (1 + O(\w)) \qquad\mbox{for some constant $c \neq 0$} \\
\beta \w^{R_f} &=& c \w^{(2Q_f + S_f + 1)\K/\mm + Q_f} (1 + O(\w)) \\
\end{eqnarray*}
Equating the exponents of the leading terms on left and right we have
$$ R_f = (2Q_f+S_f+1)\K/\mm + Q_f$$
as required in (ii). Assertion (iii) follows immediately from (ii).
That completes the proof of the lemma.
\smallskip
%The following lemma reverses the above calculations, giving a sufficient
%condition for $\HH$ to be constant.
%\begin{lemma} \label{lemma:ConditionsForHconstant}
%Suppose $\K= \mm(p-q)$ and $\BB/\AA = c(\int_0^\w \AA^2\, d\w)^{p-q}$ for some
%constant $c$, and $\SS = 1$. Then $\HH$ is constant.
%\end{lemma}
%\noindent{\em Proof}. We have
%\begin{eqnarray*}
%\HH &=& \mm \frac \BB \AA \int_0^\w \AA^2 d\w - (\mm + \K ) \int_0^\w \AA \BB \,d\w \\
% &=& \mm c \left(\int_0^w \AA^2\, dw\right)^{p-q} \int_0^w \AA^2 \, d\w - (\mm + \K) \int_0^\w \AA \BB \,d\w \\
% &=& \mm c\left(\int_0^w \AA^2 d\w\right)^{p-q+1} - (\mm + \K) \int_0^\w \AA \BB \, d\w
%\end{eqnarray*}
%Differentiating with respect to $\w$, we have
%\begin{eqnarray*}
%\frac {d\HH} {d\w} &=& \mm c(p-q+1) \left(\int \AA^2\, d\w\right)^{p-q} \A^2 - (\mm + \K )\AA\BB \\
%&=& \mm (p-q+1) \left(\frac \BB \AA\right) \AA^2 - (\mm + \K )\AA\BB \\
%&=& \mm (p-q+1) \AA\BB - (\mm + \K )\AA\BB \\
%&=& (\mm(p-q) + \mm) \AA \BB - (\mm + \K) \AA\BB \\
%&=& (\mm(p-q) - \K) \AA \BB \\
%&=& 0 \qquad\mbox{since $\K= \mm(p-q)$ by hypothesis}
%\end{eqnarray*}
%We have proved $d\HH/d\w = 0$; hence $\HH$ is constant. That completes the proof
%of the lemma.
\section{Descending through the rings of roots}
Recall that the roots $a_i$, $b_i$, and $s_i$ fall into ``rings'', where the $n$-th ring of roots
goes to zero as $t^{\gamma_n}$, and $\gamma_1 > \gamma_2 \ldots$. For each ring we can
consider the corresponding $\w$-plane, where $z = t^{\gamma_i} \w$. In each ring we
can consider the $\HH$ for that ring. As above, we abbreviate by the phrase ``$N$-condition''
the proposition that $N$ goes to $(0,0,-1)$ on compact subsets of the $w$-plane apart from
the zeros of $\AA$. The $N$-condition depends on $n$, since the power of $t$ used to
define the $w$-plane for the $n$-th ring of roots is $\gamma_n$. If the $n$-th ring satisfies
the $N$-condition, and $\HH$ for ring $n$ is constant, then by
Lemma \ref{lemma:H-constant}, there are more fast $b_i$'s than there are fast $a_i$'s, so the $n+1$-st ring
also will satisfy the $N$-condition. This is the fundamental observation behind the
following lemma:
\begin{lemma} \label{lemma:descendingrings}
Suppose that the $N$-condition holds on all rings from the first up to and including the $n$-th ring,
and that $H$ is constant on all those rings. Then
the $N$-condition holds on ring $n+1$.
\end{lemma}
The $N$-condition is equivalent to $\tBB /\tAA = O(t)$ on compact subsets of the $w$-plane
away from the $\alpha_i$, and also (by Lemma~\ref{lemma:Kpositive}) to $\K > 0$.
Let $R_n$ and $S_n$ be the number of roots that go to zero as or faster than $\gamma_n$, and let
$e_n$ be the sum of $n_i \gamma_i$ for $i < n$, where $n_i$ is the number of $b_i$ that go to zero as $t^\gamma_i$;
and let $d_n$ be the sum of $n_i \gamma_i$ for $i < n$, where $n_i$ is the number of $a_i$ that go to zero as $t^\gamma_i$.
Then in the $n$-th ring we have
\begin{eqnarray*}
\frac \tBB \tAA &=& t^{\K} \frac B A \\
&=& t^{e_n-d_n + \gamma(R_n-Q_n)} \frac B A
\end{eqnarray*}
Since the $N$-condition holds for ring $n$, we have
$$ e_n - d_n + \gamma(R_n-Q_n) > 0.$$
When we pass to ring $n+1$, the roots that go to zero slower than $t^\gamma_{n+1}$ include those that
contributed $t^{e_n - d_n}$ at ring $n$, but also those that go to zero as $t^{\gamma_n}$. Those now
contribute to $t^{e_{n+1}-d_{n+1}}$ the powers of $t$ that they were contributing to $\BB/\AA$, making
no net change in the power of $t$ in $\BB/\AA$. The remaining roots, the roots that go to zero
faster than $t^{\gamma_n}$, now contribute a factor $t^{\gamma_{n+1}}$ instead of the former
$t^{\gamma_n}$. That is because if $\alpha_i$ is a fast root, when we replace $z$ by $t^\gamma w$,
the factor $z-\alpha_i$
becomes $t^\gamma w ( 1 + O(t))$.
Therefore the $N$-condition will hold at ring $n+1$ if there are more of the
$b_i$ going to zero faster than $t^{\gamma_n} $ than there are $a_i$ going to zero faster than
$t^{\gamma_n}$. But that has been proved in Lemma~\ref{lemma:H-constant}, part (ii),
which is applicable since
$\HH$ is constant on ring $n$, and the $N$-condition is equivalent to $\K > 0$.
That completes the proof of the lemma. (Actually $\K \ge 0$ in Lemma~\ref{lemma:H-constant}, part (ii),
would have been enough.)
\begin{lemma} \label{lemma:Hnotconstant} On the ``innermost'' ring of roots, i.e. the roots that go to zero as the
highest power of $t$, $\HH$ is not constant, and the $N$-condition holds (i.e. $\K > 0$) on that ring, and
indeed on all the rings. On all but the innermost ring $\HH$ is constant. On all the rings we have
$$ \frac {\K} \mm = \frac k {2m+1} $$
\end{lemma}
\noindent{\em Proof.} There are only finitely many rings, because there are only finitely many of the $a_i$, $b_i$,
and $s_i$. Consider the first one on which $\HH$ is not constant, if any.
By Lemma \ref{lemma:descendingrings},
each ring of roots down to and including that ring satisfies the $N$-condition, i.e. $\K > 0$.
On a ring where $\HH$ is constant and the $N$-condition is satisfied, there are some fast roots, by Lemma \ref{lemma:H-constant}, part (iii).
Assume, for proof by contradiction, that $\HH$ is constant on all the rings. Then the $N$-condition holds on the
innermost ring. But by Lemma \ref{lemma:H-constant}, part (iii) there are some fast roots on that ring, which is impossible
for the innermost ring, as the fast roots would lie on another ring of roots. That contradiction shows that $\HH$
cannot be constant on all the rings, so there is a ring on which $\HH$ is not constant. By Lemma \ref{lemma:descendingrings},
the $N$-condition holds
on the slowest (outermost) ring for which $\HH$ is not constant. Then by Lemma \ref{lemma:H}, there are no
fast roots on that ring. Hence, it is the innermost ring. It now remains only to prove
the last equation of the lemma.
On the outermost ring we have $\K = k$ and $\mm = 2m+1$, so $\K/\mm = k/(2m+1)$ on the outermost ring. Now look at
the formulas in Lemma \ref{lemma:H-constant}. These involve the numbers $Q$, $R$, and $S$ (numbers of roots that
go to zero as $t^\gamma$ or faster) and $Q_f$, $R_f$, and $S_f$ (numbers of roots that go to zero faster than $t^\gamma$).
When we
pass from a ring on which $\HH$ is constant to the next ring, the $Q_f$, $R_f$, and $S_f$ of the first ring become
the $Q$, $R$, and $S$ on the next ring in. According to Lemma \ref{lemma:H-constant}, we have
$\K/\mm$ on the first ring equal to $(R_f -Q_f)/(2Q_f+S_f+1)$. Now passing to the second ring, the values on the right
become $(R-Q)/(2Q+S+1)$, which, according to the other formula in Lemma \ref{lemma:H-constant}, is equal to
$\K/\mm$ on the next ring. Hence, the value of $\K/\mm$ never changes as we pass from one ring to the next,
all the way down to the innermost ring. Hence, it retains the value $k/(2m+1)$ that it has on the outermost ring.
That completes the proof of the lemma.
\begin{lemma} \label{lemma:NoS} The $s_i$ do not occur.
\end{lemma}
\noindent{\em Proof}. According to the Gauss-Bonnet-Sasaki-Nitsche formula, a total of $2\pi m$ in extra
Gaussian area must be contributed by the $a_i$, since the branch point has order $2m$,
and each boundary branch point contributes an amount equal to a hemisphere of Gaussian
area, times half the order. If any $s_i$ exist, there are
fewer than $2m$ of the $a_i$.
We shall show that each $a_i$ can contribute at most a hemisphere, so the total will be insufficient unless there are actually $m$ of the $a_i$, in which case, no $s_i$ exist. By Lemma \ref{lemma:descendingrings},
on the innermost ring $\HH$ is not constant, and the $N$-condition holds on each ring.
By Lemma \ref{lemma:hemisphericcovering}, all the $a_i$ that belong to that ring or to slower rings
contribute only one hemisphere of Gaussian area per $a_i$. Thus every $a_i$ contributes only one hemisphere.
That completes the proof of the lemma.
\smallskip
\noindent{\em Remark.} Tony Tromba asked whether the proof (of our main theorem about finiteness)
requires an appeal to external references to know that
the branch point is not a false branch point. It does not, and the key reason why not is the application of the Gauss-Bonnet
formula. The surfaces $u^t$ for $t > 0$ must contribute Gaussian area. Therefore $\HH$ is not constant, as we have shown.
That means that the surfaces $u^t$ are not simply ``sliding'' over the same geometrical surface as $t$ becomes positive; there is
a normal component, with {\em some} power of $t$, to the motion.
\begin{corollary} \label{lemma:HwithoutS}
We have
$$ \HH = \frac \BB \AA \, \bigg(\mm\int_0^\w \AA^2 \,d\w \bigg)
- (\mm + \K)\int_0^\w \AA\BB \,d\w
$$
and $\HH$ cannot take two different signs in the upper half plane.
\end{corollary}
\noindent{\em Proof}. By the previous lemma we can now put $\SS = 1$ in the formula
\ref{eq:HHdef}.
Since $\HH$ is (according to Lemma~\ref{lemma:eigenfunctionCompact}) the limit of a power of $t$ times
the first eigenfunction, and the first eigenfunction has just one sign in the upper half plane,
either $\HH$ takes just one sign or possibly is identically zero. That completes the proof.
\begin{lemma} \label{lemma:crucialidentity}
Suppose that on the outermost ring of roots, $\HH$ is constant. Then on the innermost ring, we have
$ R-Q = (2Q+1) \K/\mm$, where $Q$ and $R$ are respectively the degrees of $\AA$ and $\BB$; that is, the number
of $a_i$ and $b_i$ belonging to the innermost ring.
\end{lemma}
\noindent{\em Proof}. If on the outermost ring $\HH$ is constant, then the innermost ring (where $\HH$ is not constant)
is not the outermost ring. Consider the next-to-innermost ring. On that ring, the ``fast roots'' are the
roots that belong to the innermost ring. We apply Lemma \ref{lemma:H-constant} on the next-to-innermost ring,
where $\HH$ is constant. Formula (ii) of that lemma tells us that
$$ R_f - Q_f = (2Q_f + S_f + 1) \frac \K \mm $$
where the numbers $Q_f$, $R_f$, and $S_f$ are the numbers of $a_i$, $b_i$, and $s_i$ that go to zero faster
than the roots of the next-to-innermost ring. But those are exactly the roots of the innermost ring. By Lemma \ref{lemma:NoS},
we have $S_f = 0$. Hence
$$ R - Q = (2Q + 1) \frac \K \mm$$
That completes the proof of the lemma.
\begin{lemma} \label{lemma:someA}
On the innermost ring of roots, there exists some $a_i$; that is,
some $a_i$ goes to zero as $t^\gamma$, where $\gamma$ is the largest number such that
some $a_i$, $b_i$, or $s_i$ goes to zero as $t^\gamma$.
\end{lemma}
\noindent{\em Proof}. On the innermost ring, $\HH$ is not constant, and $K>0$, as shown in Lemma \ref{lemma:Hnotconstant}.
Suppose, for proof by contradiction, that there is no principal $a_i$ on this ring. Then $\AA$ is constant, $\AA = \alpha$.
By Lemma \ref{lemma:NoS} we have $\SS = 1$. Putting $\SS = 1$ into (\ref{eq:HHdef}) we have
\begin{eqnarray*}
\HH(\w) &=& \frac \BB \AA\, \mm\int_0^\w \AA^2\,d\w - (\mm + \K)\int_0^\w \AA\BB\,d\w \,
\end{eqnarray*}
Substituting $\AA= \alpha$ we have
\begin{eqnarray*}
\HH(\w) &=& \BB \alpha \mm w - \alpha (\mm + \K)\int_0^\w \BB\,d\w
\end{eqnarray*}
Looking at the asymptotic behavior for large $w$, we have $\BB = w^R + O(w^{R-1})$,
where $R$ is the degree of $\BB$. Then
\begin{eqnarray*}
\HH(\w) &=& w^R \alpha \mm w - \alpha (\mm + \K) \frac {w^{R+1}}{R+1} + O(w^R)\\
&=&\alpha w^{R+1}\bigg( \mm - \frac {\mm + \K}{R+1} \bigg)
\end{eqnarray*}
We have $R > 0$ since there must be {\em some} root on the innermost ring, and we have supposed there are no $a_i$
and proved in Lemma \ref{lemma:NoS} that there are no $s_i$. Hence, unless the coefficient is zero,
$\HH$ will have a leading term in a power at least 2, and hence the eigenfunction, which is a multiple of $\Im\, \HH$,
will not have one sign in the upper half plane. But it must have one sign. Hence the coefficient is zero:
$$\mm = \frac {\mm + \K}{R+1} $$
Clearing the denominator we have
\begin{eqnarray*}
\mm(R+1) &=& \mm + \K
\end{eqnarray*}
Subtracting $\mm$ from both sides we have
\begin{eqnarray}
\mm R &=& \K \label{eq:2197}
\end{eqnarray}
Now we recall the definitions of $\mm$ and $\K$ from (\ref{eq:Mdef}) and $\ref{eq:Kdef}$. Since there
no $a_i$ on this ring, we have $Q = 0$ in those definitions; also since there no $s_i$ at all, we have
$S = 0$ and $f = 0$. The equations for $\K$ and $\mm$ become
\begin{eqnarray*}
\K &=& R\gamma + (e-d) \\
\mm &=& 2d +\gamma
\end{eqnarray*}
Putting those expressions into \ref{eq:2197} we get
\begin{eqnarray*}
(2d + \gamma)R &=& R\gamma + e-d
\end{eqnarray*}
Subtracting $R\gamma$ from both sides we have
\begin{eqnarray}
2d R &=& e-d \nonumber \\
R &=& \frac {e-d}{2d} \label{eq:31}
\end{eqnarray}
Now recall what $e$ and $d$ are. Let $\gamma$ be the exponent associated with the innermost ring,
i.e., the largest number such that some root goes to zero as $t^\gamma$.
The number $d$ is, according to its definition in (\ref{eq:ddef}),
the product of one factor $t^{\gamma}$ for each $a_i$ that
goes to zero on a slower ring. Since we are working on the innermost ring and assuming that there are no $a_i$
on that ring, that is one factor of $t^{\gamma}$ for {\em every} $a_i$. Similarly, according to (\ref{eq:edef}),
$e$ is a product of
one factor $t^{\gamma}$ for each root $b_i$ on any but the innermost ring. Now, on all the rings but the
innermost, $\HH$ is constant, so on those rings, each $a_i$ is matched by a $b_i$ converging to
the same $\alpha_i$. That means that for each factor $t^{\gamma_i}$ in the product defining $d$, the same
factor occurs in the product defining $e$, and with at least as great a multiplicity,
since when $\HH$ is constant, $\BB$ is a polynomial multiple of $\AA$.
Since all factors are powers of $t$, the more factors,
the smaller the number. Hence $e \le d$. Therefore $e-d \le 0$ and since $d > 0$, (\ref{eq:31}) gives us
$R \le 0$. But as remarked above, $R >0$, so this is a contradiction. That completes the proof of the lemma.
\begin{lemma}\label{lemma:onering}
There is just one ring of roots, and on that ring, $\HH$ is is not constant.
\end{lemma}
\noindent{\em Proof}.
We count the number of roots on each ring and add them up. Specifically, we count
the excess of roots of $\BB$ over roots of $\AA$, i.e. the number of $b_i$'s minus the
number of $a_i$'s. The total over all rings is $(m+k)-m = k$. On the rings where
$\HH$ is constant (that is, by Lemma~\ref{lemma:Hnotconstant}, all but the innermost ring),
we have by Lemma~\ref{lemma:Hconstant}
$$\BB/\AA = C\sigma^{\K/\mm}.$$
By that same
lemma $\BB/\AA$ is a polynomial; its degree is the number of $b$'s minus the number of $a$'s
on that ring of roots. Note that $\K$ and $\mm$ may be different for each ring;
before we analyzed only one ring at a time, but now we need to consider all the rings
at once. So let $Q_i$ be the number of $a_i$ on the $i$-th ring, i.e. the degree of
$\AA$ on the $i$-th ring. On the right side, for large $w$ $\sigma$ is
asymptotic to $w^{2Q_i+1}$, so right side is asymptotic to $w^{(2Q_i+1)\K/\mm}$.
Hence the excess of $b$'s over $a$'s on that ring is $(2Q_i+1)\K/\mm$.
By Lemma~\ref{lemma:Hnotconstant}, on all the rings $\K/\mm = k/(2m+1)$,
so the excess of $b$'s over $a$'s on rings where $\HH$ is constant is
$(2Q_i+1) k/(2m+1)$. Since $\HH$ is constant on all but the $n$-th ring,
adding up the excess of $b$'s over $a$'s on all the rings we get
$$k = R_n - Q_n + \sum_{i=1}^{n-1} \frac{(2Q_i+1) k}{2m+1} $$
Multiplying both sides by $(2m+1)/k$ we have
\begin{eqnarray*}
2m+1 &=& \frac{2m+1} k (R_n-Q_n) + \sum_{i=1}^{n-1} (2Q_i + 1) \\
&=& \frac{2m+1} k (R_n-Q_n) + \sum_{i=1}^{n-1} 2Q_i + \sum_{i=1}^{n-1} 1\\
&=& \frac{2m+1} k (R_n-Q_n)+ 2\sum_{i=1}^{n-1} Q_i + n-1
\end{eqnarray*}
The sum of all the $Q_i$ is $m$, the total number of $a_i$. Therefore the
sum of the first $n-1$ of them is $m-Q_n$. Putting that in we have
\begin{eqnarray*}
2m+1 &=& \frac{2m+1} k (R_n-Q_n) + 2(m-Q_n) + n-1
\end{eqnarray*}
Adding $1-2m$ to both sides we have
\begin{eqnarray}
2 &=& \frac{2m+1} k (R_n-Q_n) -2Q_n + n \label{eq:2437}
\end{eqnarray}
By Lemma~\ref{lemma:crucialidentity}, we have
$$ \frac {2m+1} k (R_n-Q_n) = \frac \mm \K (R_n-Q_n) = 2Q_n + 1$$
Putting that into (\ref{eq:2437}), we have
\begin{eqnarray*}
2 &=& (2Q_n+1)-2Q_n + n \\
&=& 1 + n \\
n &=& 1
\end{eqnarray*}
Since $n$ is the number of rings, that completes the proof of the lemma.
\section{Finiteness}
Now we know several important things:
\begin{itemize}
\item There is just one ring of roots; all the $a_i$ and $b_i$ go to zero
with the same power $t^\gamma$.
\item All the roots are principal, so $\AA$ and $\BB$ do not vanish at the origin.
\item The possible branch points $s_i$ in the lower half plane do not occur, so $\SS = 1$.
\item $\HH$ is not constant.
\end{itemize}
With these results in hand,
our formulas become simpler. We have $\K = k$ and $\mm = 2m+1$. In this section,
we will use $k$ instead of $\K$, but we retain the letter $\mm$ because it makes
the formulas shorter. We define
\begin{equation}
\sigma := \int_0^w \AA^2 \, dw \label{eq:sigmadef}
\end{equation}
As an example of the use of $\sigma$,
the formula for $\HH$ in Corollary~\ref{lemma:HwithoutS} becomes
\begin{equation} \label{eq:HHdef2}
\HH(\w) = \frac \BB \AA\, \mm\sigma - (\mm + k)\int_0^\w \AA\BB \,d\w
\end{equation}
To assist the intuition: $\sigma$ is, except for a factor of a power of $t$,
the first term in the arc length along the boundary $\Gamma$ from the origin.
A second example of the use of $\sigma$: we can express the condition shown in
in Lemma~\ref{lemma:Hconstant} and Lemma~\ref{lemma:impliesHconstant} to be
equivalent to ``$\HH$ is constant'' as
\begin{equation}
\frac \BB \AA = c \sigma^{k/\mm} \label{eq:2528}
\end{equation}
We have dropped the factor $\SS$ occurring in the cited lemmas, because it is
now known to be 1.
\subsection{A differential equation for $\BB/\AA$ and its solution}
Starting with the formula for $\HH$, we will differentiate, change variables,
and integrate. The result is a formula that connects the (stereographic projection of the)
normal $\BB/\AA$ with the derivative of the
eigenfunction. The formula involves a complex path integral. The basic idea
of this final section of the paper is to analyze the behavior of the terms in
this formula for $\BB/\AA$, especially the complex integral, at zero, at the complex
zeroes of $\sigma$, and finally at infinity. Using the facts that $\HH$ must have
just one sign in the upper half plane, and is not constant, we are able to reach
a contradiction. Here is the key formula on which these results are based:
\begin{lemma} \label{lemma:exactB/A}
Where both sides are defined we have for some constant $c$
\begin{eqnarray*}
\frac {\BB}{\AA} &=& c\sigma^{k/\mm} \exp \bigg(\int \frac {\HH_w}{\mm \sigma \BB /\AA} \, d\w\bigg)
\end{eqnarray*}
The path of integration begins somewhere on the positive real axis, at a point
greater than all the $\alpha_i$ and $\beta_i$.
For real $w$ we assume the path
of integration is on the real axis except for small semicircles in the upper half plane
around the poles of $\BB/\AA$.
\end{lemma}
\medskip
\noindent{\em Remarks}. If $\HH$ is constant, then $\HH_w = 0$, so the exponential factor is 1,
and the formula simplifies to the formula given in (\ref{eq:2528})
for $\BB/\AA$ when $\HH$ is constant. In some
sense, this new formula measures the deviation from what is true when $\HH$ is constant.
The lemma mentions the zeroes of $\sigma$; of course 0 is the only real zero of $\sigma$, but
it may have complex zeroes. Even for real $w$, we must use a
complex path of integration, that deviates from the real axis to make small detours in the
upper half-plane around the poles of $\BB/\AA$. Then the constant $c$ depends only
on the starting point of the path of integration.
\medskip
\noindent{\em Proof}.
By Lemma~\ref{lemma:Hdif} (with $\SS = 1$)
we have
\begin{eqnarray}
\HH_w &=&\mm \int_0^w \AA^2 \, d\w \ \frac d { d\w} \bigg( \frac \BB \AA \bigg) - k\, \AA\BB
\label{eq:Hw}
\end{eqnarray}
Recall that $\sigma = \int_0^w \AA^2 \, dw$, so $\sigma_w = \AA^2$. Introducing $\sigma$, the
previous equation becomes
\begin{eqnarray*}
\HH_w &=& \mm \sigma \frac d { d\w} \bigg( \frac \BB \AA \bigg) -k \sigma_w \frac {\BB}{\AA}
\end{eqnarray*}
This can be considered as a differential equation satisfied by $f = (\BB/\AA)$:
\begin{eqnarray*}
\HH_w = \mm \sigma f_w - k\sigma_w f
\end{eqnarray*}
with $w$ as the independent variable. We now want to regard $\sigma$ as the independent
variable. By definition $\sigma = \int_0^w \AA^2\, dw$. Hence
$\sigma_w = \AA^2$. Then $\sigma$ is a one-one function of $w$ at least locally,
away from the $\alpha_i$ (which are zeroes of $\AA$), so $f(w)$ can be regarded
(at least locally) as a function of $\sigma$. We write $f_\sigma$ for the derivative
of this function.
By the chain rule $f_w = f_\sigma \sigma_w = \AA^2 f_\sigma$.
Then the previous equation becomes
\begin{eqnarray*}
\mm \sigma f_\sigma \sigma_w&=& k \sigma_w f + \HH_w \\
\mm \AA^2 \sigma f_\sigma &=& k \AA^2 f + \HH_w \mbox{\qquad since $\sigma_w = \AA^2$}
\end{eqnarray*}
Dividing both sides by $\mm \AA^2 \sigma f$ we have
\begin{eqnarray*}
\frac{f_\sigma} f &=& \frac k { \mm \sigma} + \frac{\HH_w} { \mm\AA^2 \sigma f}
\end{eqnarray*}
Replacing $f$ by $\BB/\AA$ in the last term we have
\begin{eqnarray*}
\frac{f_\sigma} f &=& \frac k { \mm \sigma} + \frac{\HH_w} { \mm \sigma \AA \BB }
\end{eqnarray*}
We are going to integrate; this will be a complex path integral and we want to
specify the path. Fix a starting point $q$ on the positive real $w$-axis away from 0 and the $\alpha_i$. (We take $q > 0$ so that $\sigma(q)$ will be positive.)
Let $P$ be any path from $q$ to $w$, avoiding 0, any complex zeroes of $\sigma$, and
the zeroes and poles of $\BB/\AA$; to avoid those zeroes and poles it suffices to
avoid the $\alpha_i$ and $\beta_i$. By $P\sigma$ we mean the path in the $\sigma$-plane defined by
composing $P$ with $\sigma$.
We integrate with respect to $\sigma$ along the path $P\sigma$
(introducing $\log c$ as a constant of integration):
\begin{eqnarray*}
\log f &=& \frac k \mm \log \sigma + \int \frac {\HH_w}{\mm \sigma \AA \BB} \, d\sigma + \log c
\end{eqnarray*}
where the constant $c$ is explicitly given by
$$ \log c = \log f(q) -\log \sigma(q) = \log \frac {\BB}{\AA \sigma} \mbox{\qquad evaluated at $w=q$}
$$
Since $\BB/\AA$ is finite and non-zero at $q$, and $\sigma$ is non-zero at $q$, the
constant $\log c$ is well-defined; in particular $c \neq 0$. (If $f(q) < 0$ then both
$\log c$ and $\log f$ have an imaginary part $i\pi$.) We have
\begin{eqnarray*}
\log f &=& \log \bigg( c \sigma^{k/\mm}\bigg) + \int \frac {\HH_w}{\mm \sigma \AA \BB} \, d\sigma \\
&=& \log \bigg( c \sigma^{k/\mm}\bigg) + \int \frac {\HH_w}{\mm \sigma \AA \BB} \, \AA^2 \, d\w
\mbox{\qquad since $d\sigma = \AA^2\, d\w$} \\
&=& \log \bigg( c \sigma^{k/\mm}\bigg) + \int \frac {\HH_w}{\mm \sigma \BB /\AA} \, d\w
\end{eqnarray*}
Here the path of integration must avoid the singularities
of the integrand, which occur at $\sigma = 0$ (the only real zero of $\sigma$ is $w=0$, but there may be
complex zeroes), at zeroes of $\BB/\AA$, and possibly at singularities of $\HH_w$.
The branch of the logarithm is chosen to be continuous along the path of integration.
Applying the function $\exp(x) = e^x$ to both
sides of the equation, we have
\begin{eqnarray*}
f &=& c\sigma^{k/\mm} \exp \bigg(\int \frac {\HH_w}{\mm \sigma \BB/\AA} \, dw \bigg)
\end{eqnarray*}
Recalling that $f = \BB/\AA$ we have proved the lemma.
\medskip
\subsection{$k/\mm$ is an integer}
This is an important step in the proof. Among other things,
it implies that $\sigma^{k/\mm}$ is
a polynomial. We prove it by analyzing the behavior of the path integral
in Lemma~\ref{lemma:exactB/A} at the origin.
\begin{lemma} $k$ is a multiple of $\mm = 2m+1$; in other words $k/\mm$ is an integer.
\end{lemma}
\noindent{\em Proof}. First we calculate $\HH_w$ at the origin.
We have by (\ref{eq:Hw})
\begin{eqnarray*}
\HH_w &=&\mm \sigma \ \frac d { d\w} \bigg( \frac \BB \AA \bigg) - k\, \AA\BB
\end{eqnarray*}
At the origin, $\sigma$ is zero. Let $\alpha$ be the product of
all the $-\alpha_i$ and let $\delta$ be the product of the $-\beta_i$.
Since $\AA = \prod(w-\alpha_i)$ and $\BB = \beta \prod(w-\beta_i)$
the value of $\AA\BB$ at the origin is $\alpha\beta\delta$. Hence
the value of $\HH_w$ at the origin is $-k\alpha\beta \delta \neq 0$.
By Lemma~\ref{lemma:exactB/A},
\begin{eqnarray*}
\frac \BB \AA &=& c\sigma^{k/\mm} \exp \bigg(\int \frac {\HH_w}{\mm \sigma \BB/\AA} \, dw \bigg)
\end{eqnarray*}
Since $\AA$ is not zero at the origin, the left side does not change
sign as $w$ moves from small positive values to small negative ones.
The factor $\sigma^{k/\mm}$ changes sign if and only if $k$ is odd.
Hence the exponential factor changes sign if and only if $k$ is odd.
For the exponential factor to change sign, the complex path integral in
the exponent must pick up an odd multiple of $i\pi$ on a small
semicircle in the upper half plane around the origin. That is possible,
since the integrand has a simple pole there. The
denominator is
$$ \mm\sigma\BB/\AA = \mm w\beta\delta/\alpha + O(w^2).$$
Since the value of
the numerator at the origin is $-k\alpha\beta\delta$,
the residue is $-k \alpha^2/\mm$.
But $\alpha^2 = 1$, since $\AA^2 = \sigma_w = 1$ because we
normalized the parametrization of $\Gamma$ so that $\Gamma^\prime(0) = 1$.
Hence the residue is $-k/\mm$. The exponential factor will change
sign if and only if this is an odd integer. Since the exponential factor
must be real on both sides of the origin, $k$ must be a multiple of $\mm$.
That completes
the proof of the lemma.
\medskip
{\em Remark}.
Just to check that the sign is right:
Since $\mm = 2m+1$ is
odd, $k/\mm$ is odd if and only if $k$ is odd, which means that
the exponential factor changes sign just when the $\sigma^{k/\mm}$
factor changes sign.
\medskip
\subsection{The non-real zeroes of $\sigma$ are also zeroes of $\BB$}
In this section, we analyze the behavior of the formula in Lemma~\ref{lemma:exactB/A}
at the non-real zeroes of $\sigma$. The conclusion is that these zeroes are
also zeroes of $\BB$, and moreover, they have the same multiplicity as zeroes of $\BB$
as they do as zeroes of $\sigma^{k/\mm}$ (which is a polynomial because $k/\mm$ is an
integer).
\smallskip
We start with the following fact from complex analysis:
\begin{lemma} \label{lemma:complexpole}
Let $f(w)$ be a rational function with a pole at $q$.
Let $\gamma:[0,1] \rightarrow \C$
be a path in the complex plane avoiding the poles of $w$, except that
at the last point of $\gamma$, $\gamma$ reaches $q$, i.e. $\gamma(1) = q$.
Let $p = \gamma(0)$. Then
$$\lim_{\tau \to 1} \int_p^{\gamma(\tau)} f(w)\, dw$$
does not have a value independent of $\gamma$; indeed, for appropriate choices
of $\gamma$ it can be made $-\infty$, $\infty$, or 0.
\end{lemma}
\noindent{\em Proof}. By considering the new rational function $f(w-q)$,
we can assume without loss of generality that the pole $q$ is
at $w=0$ and the function $f$ has the asymptotic form
$$ f(w) = Cw^{-j} + O(w^{-j+1}) = Cw^{-j} (1 + O(w))$$
for some constant $C \neq 0$. Also without loss of generality, we can assume that
$C$ is real. We seek a change of variable $w \mapsto \zeta$ such that as a
function of $\zeta$ we have
$$ f(w) = C\zeta^{-j} \mbox{\qquad exactly} $$
Once we have such a variable $\zeta$, the desired paths of integration in the $\zeta$ plane
are simply lines given by $ g(\tau) =(1-\tau)e^{i\theta}$ for various $\theta$. Along
such a path we have
\begin{eqnarray*}
\int f(w)\, dw &=&\int_0^{g(1-\tau)}\zeta^{-j} d\zeta \\
&=& \int_0^{1-\tau} g(t)^{-j} g^\prime(t)\, dt \\
&=& \int_0^{1-\tau} (1-t)^{-j} e^{-ij\theta} g^\prime(t)\, dt\\
&=& e^{-ij\theta} \int_0^{1-\tau} (1-t)^{-j} g^\prime(t)\, dt\\
&=& e^{-ij\theta} \int_0^{1-\tau} (1-t)^{-j} (-e^{i\theta})\, dt\\
&=& -e^{-(j+1)i\theta} \int_0^{1-\tau} (1-t)^{-j}\, dt \\
&=& e^{-(j+1)i\theta} \int_1^\tau x^{-j}\, dx
\end{eqnarray*}
Since the pole at $q$ is at least a double pole, we have $j \neq 1$, and
\begin{eqnarray*}
\int f(w)\, dw &=& e^{(j-1)i\theta} \bigg(1 - \frac{ \tau^{-j+1}}{-j+1}\bigg)
\end{eqnarray*}
and choosing $\theta$ appropriately we can make the right side approach
$\pm \infty$ or 0 as desired.
If $j=1$, we have
\begin{eqnarray*}
\int f(w)\, dw &=& e^{-2i\theta}\log \tau
\end{eqnarray*}
which also can be made to approach $\pm \infty$ or 0 by an appropriate choice of $\theta$.
It remains to show how to find $\zeta$ as a function of $w$. We have
$$f(w)= Cw^j( 1+ g(w))$$
for some analytic function $g(w)$ with $g(w) = 0$. Then by the binomial series,
$(1+g(w))^{1/j}$ is analytic in some neighborhood of the origin. That is,
there is a function $h$ analytic in some neighborhood of the origin such that
$$(h(w))^j = 1 + g(w)$$
Now define
$\zeta = w h(w).$
Then $\zeta^j = w^j (h(w))^j = w^j(1+g(w)) = f(w)$, so we have found the required
change of variables. That completes the proof of the lemma.
\medskip
Next we extract information by looking at the formula of Lemma~\ref{lemma:exactB/A}
at the non-real zeroes of $\sigma$. Those zeroes occur in complex-conjugate pairs,
but our argument does not depend on which half-plane contains the zero in question.
\smallskip
\begin{lemma} \label{lemma:nonrealzeroes}
Every non-real zero of $\sigma$ is also a zero of
$\BB$, and as a zero of $\BB$ it has multiplicity $k/\mm$ times greater
than its multiplicity as a zero of $\sigma$.
\end{lemma}
\noindent{\em Proof}.
By (\ref{eq:Hw}) we have
\begin{eqnarray*}
\HH_w &=&\mm \sigma \ \frac d { d\w} \bigg( \frac \BB \AA \bigg) - k \AA\BB.
\end{eqnarray*}
Suppose that $q$ is a non-real zero of $\sigma$. Since the zeroes of
$\AA$ are all real, $\BB/\AA$ is analytic at $q$. Then at $q$ we have
$$ \HH_w = -k\AA\BB $$
Therefore
$$\frac {\HH_w} {\BB/\AA} = -k\AA^2 \neq 0.$$
Therefore
$$ \frac {\HH_w \,dw}{\mm \sigma \BB/\AA} \mbox{\qquad has a pole at $q$} $$
By Lemma~\ref{lemma:exactB/A}, we have
\begin{eqnarray*}
\frac \BB \AA &=& c\sigma^{k/\mm} \exp \bigg(\int \frac {\HH_w \,dw}{\mm \sigma \BB/\AA} \bigg).
\end{eqnarray*}
We consider this equation for $w$ near $q$. As $w$ approaches $q$, the left side
goes to a nonzero finite value, since neither $\AA$ nor $\BB$ is zero at $q$.
On the right side, $\sigma^{k/\mm}$ is zero at $q$. Therefore the exponential factor
has to tend to infinity as $w$ approaches $q$. That means that the integral
also tends to infinity as $w$ approaches $q$. But by Lemma~\ref{lemma:complexpole},
the integral does not
have a limit as $w$ approaches $q$; choosing an appropriate path approaching
$q$ the real part of the (absolute value of the) integrand will remain bounded. Hence the
exponential factor remains bounded away from both zero and infinity.
Hence the right side of the equation goes to zero.
Now assume, for proof by contradiction, that $\BB$ is not zero at $q$. Then
the left side, $\BB/\AA$, is not zero at $q$ and continuous there, since
$q$ is not real but the zeroes of $\AA$ are real. That is a contradiction.
Hence in fact $\BB$ is zero at $q$.
It remains to prove that $q$ has the same multiplicity as a zero of $\B$ as
it does as a zero of $\sigma^{k/\mm}$. Suppose, for proof by contradiction,
that it has a different multiplicity. Because the exponential factor has
a non-zero finite limit (for a suitable approach to $q$), along that
approach $\BB/\AA$ and $\sigma^{k/\mm}$ must approach zero at the same rate.
Since $\AA$ is not zero at $q$, it follows that $q$ has the same multiplicity
as a zero of $\BB$ and as a zero of $\sigma^{k/\mm}$. That completes the proof
of the lemma.
\subsection{Laurent expansions at infinity}
Looking at the formula of Lemma~\ref{lemma:exactB/A}, we see that the
left side $\BB/\AA$ and the factor $\sigma^{k/\mm}$ on the right both
are asymptotic to $w^k$ for large $k$. The ratio of the two therefore
goes to 1 as $w$ goes to infinity. But what is the next power in its
Laurent expansion? We need to show that it can't be too very far out.
Our answer is that it is no farther out than the term in $w^{-(k-m)}$.
It is important that $\sigma^{k/\mm}$ is a polynomial--otherwise we would
have no hope of answering this question! Therefore it is natural to
begin with a lemma about the Laurent expansions of rational functions.
\smallskip
\begin{lemma} \label{lemma:Laurent} Let $f$ and $g$ be two unequal
rational functions of the same degree $n$
with the same leading coefficient. Let $n-J$ be the highest power of $w$
such that the coefficients of $w^J$ in $f$ and $g$ are different. Then
the Laurent expansion of $f/g$ at infinity is given by
$$ \frac {f(w)}{g(w)} = 1 + \mu w^{-J} + O(w^{-J-1})$$
for some constant $\mu \neq 0$. In particular $J \le n$.
\end{lemma}
\noindent{\em Proof}.
The worst case (largest $J$) occurs when the numerator
and denominator have their corresponding coefficients equal. Here is
an example:
\begin{eqnarray*}
\frac {w^n + 3}{w^n + 5} &=& \frac { 1 + 3w^{-k}}{1 + 5w^{-k}} \\
&=& (1 + 3w^{-k})(1-5w^{-k})(1 + O(w^{-1}))\\
&=& (1-2w^{-k})(1+O(w^{-1}))
\end{eqnarray*}
The general proof is only notationally more complicated: Let
$f(w) = a_0w^n + \ldots +a_n$ and $g(w) = b_0w^n + \ldots + b_n$,
with $a_0 = b_0$.
Let $J$ be
the smallest integer such that $a_J \neq b_J$.
Then with $h(w)$ defined as the sum of the
terms with lower-indexed coefficients we have
\begin{eqnarray*}
\frac {\sum a_i w^{n-i}} {\sum b_i w^{n-i}}
&=& \frac {h(w) + \sum_{i=J}^n a_i w^{n-i}}
{h(w) + \sum_{i=J}^n b_i w^{n-i}}
\end{eqnarray*}
Dividing numerator and denominator by $h(w)$ we have
\begin{eqnarray*}
\frac {\sum a_i w^{n-i} }{\sum b_i w^{n-i}}
&=& \frac {1 + (a_J/a_0) w^{-J} + O(w^{-J-1})}
{ 1 +(b_J/b_0) w^{-J} + O(w^{-J-1})} \\
&=& 1 + \frac {a_J} {a_0} - \frac {b_J}{b_0} w^J + O(w^{-J-1}) \\
&=& 1 + \frac {a_J-b_J}{a_0}w^{-J} + O(w^{-J-1}) \mbox{\qquad since $a_0 = b_0$}
\end{eqnarray*}
Since by hypothesis $a_J \neq b_J$, we have found the first nonzero
term in the Laurent expansion. That completes the proof of the lemma.
\begin{lemma} \label{lemma:Laurent2} For some constant $\mu \neq 0$
and some integer $J$ with $1 \le J \le k-m$ we have the Laurent
expansion at infinity:
$$ \sigma^{-k/\mm} \bigg( \frac \BB \AA \bigg) = 1 + \mu w^{-J} + O(w^{-J-1}).$$
\end{lemma}
\noindent{\em Remark}. Whether or not $k/\mm$ is an integer, we always
have a Laurent expansion. The point is that since $k/mm$ is an
integer, we have a rational function on the left and can therefore
say where the first nonzero term in the Laurent expansion must occur,
while without knowing that $k/\mm$ is an integer, the first nonzero
term could occur very far out in the Laurent series.
\medskip
\noindent{\em Proof}. Since $k/\mm$ is an integer, the function
$\sigma^{k/\mm}$ is a polynomial. Since the degree of $\sigma$ is
$\mm$, the degree of $\sigma^{k/\mm}$ is $k$. The
expression in the lemma can be written as
$$ \frac {\BB} {\AA\sigma^{k/\mm}}.$$
This is a rational function. It is not constant, since
if it were constant, then by Lemma~\ref{lemma:impliesHconstant}, $\HH$
would be constant, but by Lemma~\ref{lemma:onering}, $\HH$ is not constant.
On the face of it, the numerator has
degree $m+k$, and the denominator has degree $m+(2m+1)k\mm$, which is
also $m+k$ since $\mm=2m+1$. But appearances are deceptive:
by Lemma~\ref{lemma:nonrealzeroes}, the non-real zeroes of the denominator
are all zeroes of the numerator with the same multiplicity. Now $\sigma$
is of degree $2m+1$ and has only one real zero (namely zero), so there
are $2mk/\mm$ of these zeroes, counting multiplicities. After these
common zeroes are cancelled out, the degrees of numerator and denominator
will be
\begin{eqnarray*}
n &:=& m+k- \frac {2mk}{2m+1}
\end{eqnarray*}
Since $k/(2m+1)$ is an integer, the last term being subtracted off is at least $2m$. Hence
$$ n \le k-m.$$
Applying Lemma~\ref{lemma:Laurent} to this rational function,
there exists a term $w^{-J}$ in the Laurent expansion with $J \le k-m$.
That completes the proof of the lemma.
\subsection{The final contradiction}
Now we are ready to analyze the behavior of the formula of Lemma~\ref{lemma:exactB/A}
at infinity. The numerator of the integrand is $\HH_w$. On the face of it, that is
a rational function asympotic to $w^{2m+k}$. One can easily calculate that the leading
coefficient vanishes, but since $\HH$ has only one sign in the upper half plane, we must
have $\HH$ asymptotic to $w$ or $w^{-1}$. That means that the integrand goes to zero
as a large negative power of $w$, as those higher powers of $\HH$ are not there to cancel
the denominator. That makes the integral (to infinity) finite; that is the first
key point of the proof. Then we divide the formula of Lemma~\ref{lemma:exactB/A}
by $\sigma^{k/\mm}$, isolating the exponential factor on the right:
\begin{eqnarray*}
\sigma^{-k/\mm} \frac \BB \AA &=& c \exp \bigg(\int \frac {\HH_w \,dw}{\mm \sigma \BB/\AA} \bigg).
\end{eqnarray*}
Now we have worked out the Laurent expansion of the left side, and the order
of the pole in the integrand. Differentiating this equation, we arrive at a comparison
between the first non-zero term of the Laurent expansion, and the order of the pole.
The first term on the left is $w^J$ (or $w^{J-1}$ after differentiating),
where $J$ at most so much,
and the order of the pole on the right is at least so much,
and it turns out that the two estimates are incompatible. The details are below.
\smallskip
\begin{theorem}[Finiteness] \label{theorem:main1}
Let $\Gamma$ be a real-analytic Jordan curve in $R^3$. Then $\Gamma$ cannot bound
infinitely many immersed disk-type minimal surfaces whose least eigenvalue
satisfies $\lambda_{\min} \ge 2$.
\end{theorem}
\noindent{\em Proof}.
By Theorem \ref{theorem:one-parameter},
$\Gamma$ bounds a one-parameter family of minimal surfaces $u^t$ with $u^t$ immersed for $t > 0$, and
each $u^t$ having least eigenvalue 2, so the assumptions that have been in force for all
our calculations are satisfied. By Lemma~\ref{lemma:exactB/A}, we have
\begin{eqnarray*}
\frac \BB \AA &=& c\sigma^{k/\mm} \exp \bigg(\int \frac {\HH_w \,dw}{\mm \sigma \BB/\AA} \bigg).
\end{eqnarray*}
Dividing both sides by $\sigma^{k/\mm}$ we have
\begin{eqnarray*}
\sigma^{-k/\mm} \frac \BB \AA &=& c \exp \bigg(\int \frac {\HH_w \,dw}{\mm \sigma \BB/\AA} \bigg).
\end{eqnarray*}
We consider the behavior
for large $w$. By Lemma~\ref{lemma:Laurent2}, there is an integer $J$ with
$1 \le J \le k-m$ such that the Laurent expansion of the left side has
a term in $w^{-J}$. The leading term is $\beta$, the leading coefficient of $\BB$.
Then we have for some $\mu \neq 0$
\begin{eqnarray}
\beta + \mu w^{-J} + O(w^{-J-1}) &=& c \, \exp \bigg(\int \frac {\HH_w \,dw}{\mm \sigma \BB/\AA} \bigg).
\label{eq:2728}
\end{eqnarray}
Since $\Im\, \HH$ has one sign in the upper half plane, the Laurent expansion
of $\HH$ has to begin with $w$ or $w^{-1}$. Hence the Laurent expansion of
$H_w$ begins with a constant term or with $w^{-2}$.
Therefore the first term of $H_w$ can be written $\nu w^{-1\pm 1}$ for
some constant $\nu \neq 0$. The integrand can then be written like this:
\begin{eqnarray*}
\frac {\HH_w}{\mm\sigma\BB/\AA} &=& \frac{\AA \HH_w} {\mm \sigma \BB} \\
&=& \frac {\nu w^{m-1\pm1}} {w^{\mm} \beta w^{m+k}} (1 + O(w^{-1}))\\
&=& \frac \nu \beta w^{-1\pm 1 - (2m+1 + k)} (1 + O(w^{-1}))\\
&=& \frac \nu \beta w^{-(2+2m+k \pm 1)}(1 + O(w^{-1}))
\end{eqnarray*}
Summarizing that calculation,
\begin{eqnarray}
\frac {\HH_w}{\mm\sigma\BB/\AA} &=&\frac \nu \beta w^{-(2+2m+k \pm 1)} (1 + O(w^{-1}))\label{eq:2751}
\end{eqnarray}
Since $k/(2m+1)$ is an integer, we have $k \ge 2m+1$, so $2+2m+ k \pm 1 \ge 2m+k+ 1$.
Therefore the exponent is negative and has magnitude at least $2m+k+1$. Since $k/(2m+1)$
is an integer, we have $k \ge 2m+1$, so the magnitude of the exponent is at least $4m+2$.
It follows that the integral has a finite limit as $w$ goes to infinity.
Let that finite limit be $L$. Then the integral is $L + O(w^{-1})$.
Differentiating (\ref{eq:2728}), which is legal since the $O(w^{-J-1})$ term stands
for a Laurent series, we have
\begin{eqnarray*}
-J\mu w^{-J-1} + O(w^{-J-2}) &=&
c \, \exp \bigg(\int \frac {\HH_w \,dw}{\mm \sigma \BB/\AA} \bigg)
\frac {\HH_w }{\mm \sigma \BB/\AA} \\
&=& c \, \exp \bigg(L + O(w^{-1}) \bigg)
\frac {\HH_w }{\mm \sigma \BB/\AA} \\
&=& ce^L (1 + O(w^{-1}))\frac {\HH_w }{\mm \sigma \BB/\AA}
\end{eqnarray*}
Putting in the result from (\ref{eq:2751}) we have
\begin{eqnarray*}
-J\mu w^{-J-1} + O(w^{-J-2})
&=& ce^L (1 + O(w^{-1}))\nu w^{-(2+2m+k \pm 1)}
\end{eqnarray*}
On the left, we have a negative exponent $J+1 \le k-m+1$.
On the right, we have a negative exponent of magnitude at least $2m+k+1$.
It is manifestly impossible for these exponents to be equal,
so we have reached a contradiction.
That contradiction depends on no other assumptions than the existence of a one-parameter
family of minimal surfaces $u^t$ bounded by $\Gamma$, such that $\lambda_{\min} = 2$ for $t > 0$.
That completes the proof by contradiction that no such family of minimal surfaces exists.
\section{Relative minima in which topology?}
We are going to prove a theorem limiting the number of relative minima of area with a given boundary.
To state such a theorem precisely, we must fix a topology to use in defining ``relative minimum.''
Once a topology is fixed, a relative minimum is a surface $u$ such that there exists a neighborhood
of $u$ (in the chosen topology) such that no surface in that neighborhood has smaller area than $u$.
As usual in Plateau's problem, we work with harmonic surfaces.
We wish to use the $C^n$ topology. First consider whether it is a stronger theorem for smaller $n$
or larger $n$. {\em A priori} it might be possible to decrease area by a $C^0$ variation but not by
a $C^n$ variation, but if area can be decreased by a $C^n$ variation, that same variation counts as
a $C^0$ variation. Therefore a $C^0$ relative minimum is automatically a $C^n$ relative minimum, but
not vice-versa. Thus if there were infinitely many $C^0$ relative minima, there would automatically
be infinitely many $C^n$ relative minima. Taking the contrapositive, if there cannot be infinitely
many $C^n$ relative minima, there cannot be infinitely many $C^0$ relative minima. Thus the theorem will be stronger,
when we use the $C^n$ topology for as large an $n$ as possible.
Unfortunately, our result depends on the regularity result that a relative minimum cannot have an
interior or boundary branch point, which at present is known only for the $C^0$ topology.%
When and if this regularity result is proved for larger $n$, our proof will immediately apply to the $C^n$
topology; but for now we can prove it for the $C^0$ case.
\begin{theorem} [Main theorem] \label{theorem:main2}
Let $\Gamma$ be a real-analytic Jordan curve in $R^3$.
Suppose that $n$ is a positive integer and every relative minimum of area
in the $C^n$ topology bounded by $\Gamma$ has no (interior or boundary) branch points.
Then $\Gamma$ cannot bound infinitely many disk-type minimal surfaces that are relative
minima of area in the $C^n$ topology.
\end{theorem}
\noindent
{\em Remark\/}: As usual in Plateau's problem, we are counting two surfaces
that differ only by a conformal reparametrization as the same surface.
\noindent
{\em Proof\/}: Suppose, for proof by contradiction, that $\Gamma$ bounds
infinitely many disk-type minimal surfaces furnishing $C^n$ relative minima of
area. Then those surfaces are immersed, by hypothesis, and have $\lambda_{\min} \ge 2$.
Then by Theorem \ref{theorem:one-parameter},
$\Gamma$ bounds a one-parameter family of minimal surfaces $u^t$ with $u^t$ immersed for $t > 0$, and
each $u^t$ having least eigenvalue 2. But that contradicts Theorem~\ref{theorem:main1}.
That completes the proof.
\medskip
\begin{corollary} Let $\Gamma$ be a real-analytic Jordan curve in $R^3$.
Then $\Gamma$ cannot bound infinitely many disk-type minimal surfaces that are relative
minima of area in the $C^0$ topology.
\end{corollary}
\noindent{\em Proof}. By Theorem~\ref{theorem:regularity} and the known regularity
results, which are fully discussed in
section~\ref{section:regularity} above.
\section{Open Problems and Conjectures}
The theorem proved here could possibly be generalized by weakening any of these hypotheses: (i) that the
boundary is real-analytic; (ii) that the surfaces considered are of the topological type of the disk; or (iii) that the surfaces considered
are relative minima of area. All questions formed by weakening one or more of these hypotheses are open.
These questions are all ``equal conclusion'' strengthenings of the theorem, formed by weakening the hypotheses.
We will comment on these questions one by one below.
There is also an interesting ``stronger conclusion'' version of the theorem, in which we ask for an explicit bound on the number of relative minima as a function of the geometry of the boundary curve
$\Gamma$. Our ignorance is scandalous: Not counting curves for which uniqueness is known, there is only one Jordan curve $\Gamma$ known for which anyone can give an explicit bound on the number of solutions of Plateau's problem (of disk type) bounded by $\Gamma$. That curve is ``Enneper's wire'' $\Gamma_r$ (the image of the circle of radius $r$ in the usual parametrization of Enneper's surface), for values of $r$ slightly larger
than $1$, and the bound is three. This bound follows from the theorem of Ruchert \cite{ruchert} that uniqueness holds up to $r=1$ and
the theorem of \cite{beeson-tromba} characterizing the space of minimal surfaces bounded by $\Gamma_r$ in the vicinity of Enneper's surface for $r=1$.
The following
conjecture, for example, seems presently out of reach: if the real-analytic Jordan curve $\Gamma$ has total curvature less than or equal to $6\pi$, then it bounds at most two relative minima of area (of disk type). We do know that there can
be no such estimate for the number of minimal surfaces (minimizing or not), since B\"ohme proved in \cite{bohme3} that
for any $\epsilon > 0$ and any $n$, there is a real-analytic Jordan curve with total curvature less than $4\pi + \epsilon$
bounding more than $n$ minimal surfaces. But these are branched surfaces, not relative minima,
so they do not show the impossibility of a bound on the number of immersed minimal surfaces
bounded by $\Gamma$.
Since Enneper's wire bounds two relative minima when the total curvature exceeds $4\pi$, we could expect to string together
sequences of Enneper's wires with small bridges connecting them, obtaining for each $n$ a Jordan curve with
total curvature $4n\pi + \epsilon$ bounding (at least) $2^n$ relative minima (as well as some unstable minimal surfaces).
Since we seem to have to increase the total boundary curvature to get many solutions of Plateau's problem, we might hope that for some explicitly given
function $F$, we might have, for all real-analytic Jordan curves
$\Gamma$ of total curvature less than $2 \pi n$, there are at most $F(n)$ relative minima of area. As far as we
know, this might be true with $F(n) = 2^{n - 1 }$. Thus: (when $n=2$) curvature less than $4\pi$ implies uniqueness (which is known); when $n=3$, curvature less than $6\pi$ implies at most two relative minima (the conjecture mentioned above);
when $n=4$, curvature less than $8\pi$ implies at most 4 relative minima (also, of course, not known).
Next we take up the possible ``equal conclusion'' versions. First, what about weakening the smoothness of the
boundary to $C^n$? Then we have some troubles about the boundary branch point representation, but perhaps that can
be solved adequately. A more pressing difficulty is that we then cannot control the dependence of the zeroes, poles, and
ramification points of the Gauss map $N$ on the parameter $t$. There is also a third difficulty:
even the proof of Tomi's theorem that there can't be infinitely many absolute minima does not go through for $C^\infty$
boundaries, because the regularity result that a minimum of area cannot have a boundary branch point is still an open problem for $C^\infty$ boundaries.
Second, what about generalizing the theorem to other topological types of surfaces or boundaries? Frank Morgan has
given an example \cite{morgan81} of a boundary consisting of four circles, such that it bounds a continuum of unstable
minimal surfaces of arbitrarily high genus. That still leaves open the finiteness problem for relative minima. I
conjecture that a system of real analytic Jordan curves cannot bound infinitely many relative minima of any fixed orientable topological type (that is, any fixed number of boundary components and Euler characteristic).
For non-orientable surfaces, there is a problem in that the least eigenfunction
no longer has one sign. Instead, it has just one nodal line. Thus even Tomi's argument that there cannot be a loop of
minimal surfaces all of which are absolute minima of area fails to go through for surfaces of the type of the M\"obius
strip. That is the simplest open question for non-orientable surfaces: can there be a one-parameter family of absolute
minima of surfaces of the type of the M\"obius strip?
Incidentally, another paper of Morgan \cite{morgan76} gives an example of a curve in $R^4$ that bounds a continuum of
absolute minima of disk type. Therefore we cannot generalize the theorem to higher dimensional spaces.
Third, what about dropping the requirement that the surfaces be relative minima of area? Then our methods say nothing
at all, since they all depend on analysis of the least eigenvalue and its eigenfunction.
The only known results (other than uniqueness theorems) along these lines are two
``$6 \pi$ theorems'', both of which assume that $\Gamma$ is a real-analytic Jordan curve whose
total curvature is less than or equal to $6 \pi$. In \cite{beeson-sixpi}, it is proved that such a curve
cannot bound infinitely many minimal surfaces (stable or unstable). Earlier, Nitsche proved that same conclusion
(see \textsection A29 of \cite{nitsche}, p. 447) using the additional hypothesis that $\Gamma$
bounds no minimal surface with branch points.
At the time of Nitsche's
proof, the forced Jacobi fields had not yet been discovered. In \cite{beeson-sixpi}, the theory of the forced Jacobi fields
is used.
\section*{Appendix: Dictionary of Notation}
To assist the reader we provide in one place a summary of the notations used in more than one section of this paper.
This summary is provided as a memory aid, not as a list of complete definitions. The symbols are more or less in
order of introduction; there is no natural ``lexicographical'' order for such a variety of symbols.
\begin{tabbing}
principal roots \= \kill \\
$X,Y,Z$ \> coordinates in $R^3$ \\
$\Gamma$ \> a real-analytic Jordan curve in $R^3$, \\
\> tangent to the $X$-axis at the origin \\
$u = u^t$ \> a one-parameter family of minimal surfaces, \\
\> analytic in $t$ in some interval $[0,t_0]$ \\
\> $u^t$ is analytic as a function of $z$ in a closed upper half-disk \\
$t$ \> the parameter defining that one-parameter family \\
${}^2u$ \> the second component of the vector $u = ({}^1u, {}^2u,{}^3u)$. \\
$z = x + iy$ \> the parameter used to define $u^t$ \\
$N$ \> the unit normal to $u^t$ \\
$g$ \> the stereographic projection of the Gauss map $N$ \\
$u_t$ \> subscript denotes differentiation: $\frac {\partial u} {\partial t}$ \\
$\lambda_{\min}$ \> least eigenvalue \\
$\phi$ \> $u_t \cdot N$. If nonzero, this is an eigenfunction for eigenvalue 2. \\
$n$ and $\chi$ \> $\phi = t^n \chi$. \\
$f$ \> ${}^1u_z - {}^2u_z$, occurs in the Weierstrass representation of $u^t$ \\
$2m$ \> order of the boundary branch point of $u$ (when $t=0$). \\
$s_i$ \> branch points in the lower half plane for $t>0$ that \\
\>converge to 0 as $t \to 0$. There are $S$ of these.\\
$N$ \> There are $2N$ of the $s_i$. Hopefully no confusion will arise \\
\>with $N$ the unit normal.\\
$a_i$ \> zeroes of $f$. The $a_i$ depend analytically on $t$. \\
\>There are $Q$ of these. \\
$b_i$ \> zeroes of $fg^2$. The $b_i$ depend analytically on $t$. \\
\>There are $R$ of these. \\
$\gamma_n$ \> The $n$-th ring of roots consists of $a_i$, $b_i$, and $s_i$ that go to zero as $t^{\gamma_n}$. \\
$\gamma$ \> One of the $\gamma_n$. \\
{\em principal roots} \> those $a_i$, $b_i$ and $s_i$ whose leading term is $t^{\gamma_1}$ \\
$\alpha_i$ \> coefficient such that $a_i = \alpha_i t^\gamma + O(t^{\gamma+1})$ \\
$\beta_i$ \> coefficient such that $b_i = \beta_i t^\gamma + O(t^{\gamma+1})$ \\
$\zeta_i$ \> coefficient such that $s_i = \zeta_i t^\gamma + O(t^{\gamma+1})$ \\
\> $\alpha_i$, $\beta_i$, and $\zeta_i$ are used only when non-zero, i.e. \\
\>for the roots that go to zero as $t^\gamma$. \\
$\alpha$\> product of all the $-\alpha_i$ over roots going to zero slower than $t^\gamma$ \\
$\beta$\> product of all the $-\beta_i$ over roots going to zero slower than $t^\gamma$ times $B_0(0,0)$ \\
$\zeta$\> product of all the $-\zeta_i$ over roots going to zero slower than $t^\gamma$ \\
$A $ \> $A = A_0 \prod_{i=1}^{m-N} (z-a_i)$ so $f = A^2$, and $A_0 = 1$ when $t=0$ \\
$B$ \> $B = \beta \prod_{i=1}^{m+k-N} (z-b_i)$ so $fg^2 = B^2$. \\
$C$ \> complex constant, value of $B_0$ when $t=0$ \\
$S$ \> $S = \prod_{i=1}^{2N} (z-s_i(t))$\\
$w = z/t^{\gamma}$ \> a ``blow-up'' of the parameter domain \\
$O(t)$ \> refers to uniform convergence on compact subsets of the \\
\> $w$-plane away from the $\alpha_i$ \\
$\O(t)$ \> refers to uniform convergence on compact subsets of the \\
\> $w$-plane (possibly including the $\alpha_i$) \\
$\AA $ \> $\AA = \alpha \prod_{i=1}^{Q}(w-\alpha_i)$ so $A = t^{(m-N)\gamma} \AA (1+\O(t))$ \\
\> $\AA$ does not depend on $t$ \\
$\BB $ \> $\BB = \beta \prod_{i=1}^{R} (w-\beta_i)$ so $B = t^{(m+k-N)\gamma} \BB (1+\O(t))$\\
\> $\BB$ does not depend on $t$, and $B_0$ turns out to be real. \\
$\tAA $ \> $\alpha\prod (\w - \a_i)$. The product is over $i$ such that $a_i = O(t^\gamma)$\\
$\tBB $ \> $ i\lambda B_0\beta\prod (\w - \b_i)$ \\
$\tSS$ \> $\zeta\prod (\w - \s_i)$ \\
$ \a_i$ \> $\a_i = a_i/t^\gamma$ \\
$ \b_i$ \> $\b_i = b_i/t^\gamma$ \\
$ \s_i$ \> $\s_i = s_i/t^\gamma$ \\
$\tAA_0(t,w)$ \> $\tAA_0(t,w) = A_0(t,t^\gamma w)$ \\
$\tBB_0(t,w)$ \> $\tBB_0(t,w) = B_0(t,t^\gamma w)$ \\
$\chi$ \> $\chi(w) = \phi(t^{\gamma}w)\ /\ \gamma t^{(2m+k+1)\gamma -1} $\\
\> $\chi$ shows what happens to the eigenfunction in the $w$-plane. \\
$\chi^0$ \> limit of $\chi$ when $t$ goes to $0$ \\
$\HH$ \> complex analytic function such that $\chi^0 = -\Im \,\HH$ \\
$\sigma$ \> abbreviation for $\int_0^w \tAA^2 \, dw$ in the last two sections \\
$\tau$ \> the parameter used to parametrize the boundary curve, equal\\
\> to the $X$-coordinate \\
\end{tabbing}
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\medskip
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\end{document}
@incollection{
year={1981},
isbn={978-3-540-10285-4},
booktitle={Global Differential Geometry and Global Analysis},
volume={838},
series={Lecture Notes in Mathematics},
editor={Ferus, Dirk and K�hnel, Wolfgang and Simon, Udo and Wegner, Bernd},
doi={10.1007/BFb0088840},
title={A plateau problem with many solutions for boundary curves in a given knot class},
url={http://dx.doi.org/10.1007/BFb0088840},
publisher={Springer Berlin Heidelberg},
author={B�hme, R.},
pages={36-41},
language={English}
}
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