text:
\section{Introduction}

Multiple-valued functions were first introduced by F. Almgren in \cite{bigreg} to analyze the regularity of mass-minimizing rectifiable currents in arbitrary codimension. Such functions have recently seen a resurgence in the works of C. De Lellis and E. N. Spadaro \cite{DelSeries} (and its continuations), \cite{revisit, Delint}, S. Chang \cite{Schang}, W. Zhu \cite{WZhu2dim, reducing}, C. C. Lin \cite{cclin1}, P. Bouafia \cite{selection}, P. Bouafia and T. De Pauw \cite{Bouafia}, and P. Mattila \cite{Mattila}, to name just a few.

A $Q$-valued function to $R^n$ is essentially a single-valued function taking values in the set $A_Q(R^n)$ of all unordered $Q$ tuples of points in $\mathbb{R}^n$.  Equivalently $A_Q(\mathbb{R}^n)$ may be considered as the space of all sums $\sum_{i=1}^Q [[ a_i ]] $ of $Q$ Dirac measures of points in $R^n$.  Thus, for $\Omega\subset \mathbb{R}^m$, a function $f:\Omega\to A_Q(\mathbb{R}^n)$ corresponds to a $Q$- valued function from $\Omega$ to $\mathbb{R}^n$. We will occasionally refer to such functions as     \emph{multiple-valued functions}.

One can metrize $A_Q(\mathbb{R}^n)$ via a standard `translation' invariant metric. With this metric, Almgren showed that $A_Q(\mathbb{R}^n)$ admits a bi-Lipschitz embedding onto a Lipschitz retract of a high dimensional Euclidean space, and used this to define Sobolev multiple-valued maps. Further, he gave a well defined notion of differentiability and Dirichlet energy, and solved the Dirichlet problem for such functions.

In this paper we seek to formulate and prove a result analogous to the classical Plateau problem in the setting of multiple-valued functions. Our main result is the following:

\begin{thm}
\label{thm:main}
 Let $\Gamma_1, \ \Gamma_2, \ ...\ , \Gamma_K$ be disjoint, simple closed curves of finite length in $\mathbb{R}^n$ so that $\Gamma_i \cap \Gamma_K = \phi$ for $i\neq j$ and $k_1,...,k_K$ positive integers so that $\sum k_i = Q$. Further, suppose that each $\Gamma_i$ is a Lipschitz Neighborhood Retract of an open neighborhood $U_i$. Denote by $\mbb D$ the open unit disk in $\mathbb{R}^2$. Then, if

 \begin{equation*}
 \mathcal A = \{ F\in W^{1,2}(\mbb D, A_Q(\mathbb{R}^n) )\cap C^0(\bar\mbb D,A_Q(\mathbb{R}^n)) \ | \ F|_{S^1}(z)=\sum_i \sum_{\zeta = z^{k_i}}\llbracket f_i(\zeta) \rrbracket \
 \end{equation*}
\begin{equation*}
  \mbox{with $f_i:S^1\rightarrow \Gamma_i$ weakly monotonic} \}
  \end{equation*}

 \noindent then there is a map $f\in \mathcal A$ so that:

 \begin{center}
 $\text{\textnormal{Dir}}(f)=\displaystyle\min_{G\in \mathcal A} \text{\textnormal{Dir}}(G)$.
 \end{center} \end{thm}

Recall that a map $f_i: S^1 \rightarrow \Gamma_i$ is said to be \emph{weakly monotonic} if it may be written as $\varphi_i ( e^{i\tau_i(\theta)})$, where $\varphi_i:S^1\rightarrow \Gamma_i$ is a homeomorphism, and $\tau_i : [0,2\pi] \rightarrow \mathbb{R}$ is a nondecreasing continuous, $2\pi$ periodic function.

Unfortunately, there is a subtlety in the existence of `wrapped' solutions--that is, those solutions with some $k_i > 1$. For example, if the optimal $f_i$ is constant on an arc of length greater than $\frac{ 2\pi }{k_i}$, there will be a branch point on the boundary circle $S^1$. Simple planar examples show that this can indeed occur--and so we must introduce a condition to guarantee such branching does not occur:

\begin{dfn}
\label{dfn:wrapped}
In the setting of Theorem \ref{thm:main}, the boundary data  $(\Gamma_1,...,\Gamma_K,k_1,...,k_K)$ is said to \emph{admit a wrapped solution} if there exists $f$ a minimizing element of $\mathcal A$ so that:

\begin{equation*}
f|_{S^1}(z)=\sum_i \sum_{\zeta = z^{k_i}}\llbracket f_i(\zeta) \rrbracket
\end{equation*}

\noindent for $f_i : S^1 \rightarrow \Gamma_i$ weakly monotonic so that no $f_i$ is constant on an arc of length greater than or equal to $\frac{2\pi}{k_i}$.

\end{dfn}

One should note that, while Theorem \ref{thm:main} guarantees a minimizing element of $\mathcal A$ under fairly general conditions, this element can fail to satisfy the condition of Definition \ref{dfn:wrapped} without contradicting its admissibility.

Definition \ref{dfn:wrapped} is an analogous condition to the Douglas condition for planar domain Plateau solutions, as in \cite[8.6]{Hildmin} where similar degenerations must be ruled out. Interestingly, though, in contrast to the classical case, we are able to establish existence of a solution without this condition--but regularity fails.

As in the classical case, we begin with an admissible sequence $\{f^n\}$ whose energy tends towards the infimum and we extract a subsequence which converges ($W^{1,2}$ weakly) to some function $f: \mbb D \rightarrow A_Q(\mathbb{R}^n)$. Then the issue is showing the admissibility of the function $f$--a fact which ultimately comes down to proving that the boundary data of the subsequence cannot collapse. In fact, a posteriori, a selection for the optimal boundary data will consist of homeomorphisms of the boundary; a stronger result than that $f|_{S^1}$ is a homeomorphism onto a subset of $A_Q(\mathbb{R}^n)$.

To prevent the boundary data from collapsing, one first precomposes with a disk automorphism to guarantee that, if $f^n|_{S^1}$ has boundary data $f^n_j$ for $f^n_j : S^1 \rightarrow \Gamma_j$ as in the definition of $\mathcal A$, then $f^n_1$ satisfies a three point condition. We will then show that (in the case that $f$ is branched) this normalization forces a normalization for each $f^n_j$--on a first attempt, one might try, supposing towards a contradiction, to analyze Dirichlet minimizing functions $G: \mbb D \rightarrow A_Q(\mathbb{R}^n)$ for which $0\in \text{\textnormal{spt}}(G(s))$ for all $s\in S^1$. Unfortunately, such functions can exist with rather elaborate branching behavior, as exhibited in Section \ref{section:degen}, so a more complicated argument is required.

Finally, using recent results regarding the Plateau problem in abstract metric spaces (in particular, \cite{Disks}), in conjunction with Theorem \ref{thm:main}, we are able to prove the following.

\begin{thm}
\label{thm:conformal}
Suppose that $\Gamma_1,...,\Gamma_K$ and $k_1,...,k_K$ are given as in Theorem \ref{thm:main}, and further suppose that  the boundary data  $(\Gamma_1,...,\Gamma_K,k_1,...,k_K)$ admits a wrapped solution. Then there exists  $F : \mbb D \rightarrow A_Q(\mathbb{R}^m)$, an area-minimizing Plateau solution for some collection of disjoint Jordan curves $\Gamma_1,...,\Gamma_K$, so that if $f_1,...,f_K$ is the associated boundary data the following holds:
\begin{enumerate}
\item around each regular point there exists a conformal selection for $F$
\item $\text{\textnormal{Dir}}(F) = \mbox{MV-Area}(F)$ and
\item Each $f_i$ is a homeomorphism.
\end{enumerate}

\noindent Here $\mbox{MV-Area}(F)$ is the two dimensional area of the set $\{\text{\textnormal{spt}}(F(x) \ | \ x\in \mbb D\}$, counting multiplicity.

\end{thm}

This multi-valued Plateau problem may, as described in Section 6, lead to minimal surfaces necessarily having branch points, even in $\mathbb R^3$.   Here the location or number of the branch points is not fixed. Near a single prescribed branch point, a minimal surface has been studied by considering corresponding solution of a ``multi-valued minimal surface equation'' in  \cite{add3},  \cite{add1} and \cite{add2}.

In Section \ref{section:topology},  we also present the sketch of a proof of Theorem \ref{thm:conformal} in the special case where $K=Q$ and $k_i=1$ using topological methods and solutions to the classical Douglas problem.

\section{Preliminaries}

A detailed introduction to the theory of multiple-valued functions may be found in \cite{revisit}. Here we present only the details essential to the current paper. We will mostly follow the notations found therein, with the exception that our definition of ``affinely approximatable'' is slightly weaker than the definition of ``differentiable'' given therein. Throughout the paper, $m$, $n$, and $Q$ will denote natural numbers, $\Omega$ is an open subset of $\mathbb{R}^m$ with sufficiently regular boundary, and, for $x$ an element of any metric space $X$, we will use $B_r(x)$ to denote the open ball of radius $r$ centered at $x$. We also denote by $\mathcal H^\alpha$ the $\alpha$-dimensional Hausdorff measure--in particular, $\mathcal H^0$ is counting measure.

We define the metric space $A_Q(\mathbb{R}^n)$ as the set of positive integer sums of Dirac measures on $\mathbb{R}^n$ with total mass $Q$. If, for $P \in \mathbb{R}^n$, $\llbracket P \rrbracket$ denotes the Dirac mass at $P$, then we may define $A_Q(\mathbb{R}^n)$ in symbols as follows:

\begin{equation*}
A_Q(\mathbb{R}^n) = \{\displaystyle \sum_{i=1}^Q \llbracket P_i \rrbracket \ | \ P_i \in \mathbb{R}^n\}.
\end{equation*}

\noindent We note that we do not require the $P_i$'s to be distinct, and that we will occasionally omit $Q$ and $n$ when they are clear from the context. Following \cite{bigreg}, we metrize $A_Q(\mathbb{R}^n)$ by a metric $\mathcal G$ defined as:

\begin{equation*}
\mathcal G(\sum \llbracket P_i \rrbracket , \sum \llbracket Q_i \rrbracket ) = \min_{\sigma \in S_Q} \sqrt{\sum ||P_i - Q_{\sigma(i)}||^2}.
\end{equation*}

\noindent Where $S_Q$ denotes the permutation group on $\{1,...,Q\}$. It is easy to see that this gives a well-defined metric under which $A_Q(\mathbb{R}^n)$ is a complete metric space.

For an open $\Omega\subset \mathbb{R}^m$, we say a function $f: \Omega \rightarrow A_Q(\mathbb{R}^n)$ is a multiple-valued function. If $g : \Omega \rightarrow A_{Q'}(\mathbb{R}^n)$ is another multiple-valued function, we denote by $\llbracket f \rrbracket + \llbracket g \rrbracket$ the function into $A_{Q+Q'}(\mathbb{R}^n)$ whose value at the point $x\in \Omega$ is the sum of the measures $f(x)$ and $g(x)$.

 If $f_i:\Omega \rightarrow \mathbb{R}^n$ are such that $f=\sum \llbracket f_i\rrbracket$, we say $\{f_i\}$ is a selection for $f$. A major problem in the theory of multiple-valued maps is attempting to derive selections with some degree of regularity from multiple-valued functions. However, we always have the following result:

\begin{thm}
If $f: \Omega \rightarrow A_Q(\mathbb{R}^n)$ is measurable, then there exists a selection for $f$ consisting of measurable functions.
\end{thm}

Almgren developed an extrinsic theory through the help of a bi-Lipschitz embedding of $A_Q(\mathbb{R}^n)$ onto a Lipschitz retract of higher dimensional Euclidean space. A corollary attributed to B. White strengthens this embedding to locally preserve distances, and we present this version below.

\begin{thm}
\label{thm:embedding}

There exists $N=N(Q,n)$, an injective map $\zeta : A_Q(\mathbb{R}^n) \rightarrow \mathbb{R}^N$ and a Lipschitz retraction $\rho : \mathbb{R}^N \rightarrow \zeta(A_Q(\mathbb{R}^n))$ so that:

\begin{enumerate}
\item $\text{\textnormal{Lip}}(\zeta) \leq 1$.
\item If $\mathcal Q = \zeta(A_Q(\mathbb{R}^n))$, then $\text{\textnormal{Lip}}(\zeta^{-1}|_{Q}) \leq C(Q,n)$.
\item For any $T\in A_Q(\mathbb{R}^n)$, there exists $\delta$ so that $\zeta|_{B_\delta(T)}$ preserves distances.
\end{enumerate}

\end{thm}

This embedding allows us to quickly define multiple-valued Sobolev functions. For $\Omega \subset \mathbb{R}^m$, $k\in \mathbb N$ and $1 \leq p \leq \infty$, we define $W^{k,p}(\Omega, A_Q(\mathbb{R}^n))$ to be those functions $f: \Omega \rightarrow A_Q(\mathbb{R}^n)$ so that $\zeta \circ f \in W^{k,p}(\Omega, \mathbb{R}^N)$. While this definition may seem rather contrived, it has many computational benefits to the more natural equivalent definitions discussed in \cite{revisit}. In particular, we note that classical results regarding Lipschitz approximation of Sobolev functions, interpolation lemmas, traces and one-dimensional restrictions of Sobolev functions are seen to immediately carry over to the multiple-valued case.

We further define the \emph{Dirichlet energy} of a $W^{1,2}(\Omega, A_Q(\mathbb{R}^n))$ function as:

\begin{equation*}
\text{\textnormal{Dir}}(f,\Omega) = \int_\Omega |D(\zeta \circ f)|^2.
\end{equation*}

\noindent Again, this is simply a pragmatic definition--more natural equivalent definitions are available in the literature.

\begin{rmk}
\label{rmk:invar}
From this definition it immediately follows that if $\Omega$ is two dimensional, the Dirichlet energy is invariant under precomposition by conformal maps.
\end{rmk}
We say that a multiple-valued function $f\in W^{1,2}(\Omega, A_Q(\mathbb{R}^n))$ is \emph{Dirichlet minimizing} if

\begin{equation*}
\text{\textnormal{Dir}}(f,\Omega) \leq \text{\textnormal{Dir}}(g,\Omega)
\end{equation*}

\noindent for any $g\in W^{1,2}(\Omega, A_Q(\mathbb{R}^n))$ such that $f|_{\partial \Omega} = g|_{\partial \Omega}$. Here, the last condition means that the $W^{1,2}$ traces on $\partial\Omega$  of $\zeta\circ f$ and $\zeta\circ g$ coincide. In Almgren's original work, he proved that if $g\in W^{1,2}(\Omega, A_Q(\mathbb{R}^n))$, then there exists a Dirichlet minimizing $f\in W^{1,2}(\Omega, A_Q(\mathbb{R}^n))$ with $f|_{\partial \Omega} = g|_{\partial \Omega}$. He further showed that there is a constant $\alpha=\alpha(m,Q) >0$ so that such a minimizing function is equal almost everywhere to a function which is $\alpha$ H\"older continuous on any $\Omega'\subset\subset\Omega$. In the following, we assume that this identification has been made, and in particular that Dirichlet minimizers are continuous functions ``strictly defined'' at every point.

In case $m=1$, Sobolev functions automatically admit an absolutely continuous selection. We may define the space of absolutely continuous multiple-valued functions on an interval $I$, denoted by $AC(I,A_Q(\mathbb{R}^n)$, to be those functions $f:I\rightarrow A_Q(\mathbb{R}^n))$ so that $\zeta \circ f$ is absolutely continuous. We may now state a selection theorem for one dimensional multiple-valued Sobolev functions--see Proposition 1.2 of \cite{revisit}:

\begin{thm}
If $f\in W^{1,p}(I,A_Q(\mathbb{R}^n))$, then,
\begin{enumerate}
\item $f\in AC(I,A_Q(\mathbb{R}^n))$.
\item There exists a selection $\{f_i\} \subset W^{1,p}(I,\mathbb{R}^n)$ and so that $|Df_i| \leq |D(\zeta\circ f)|$.
\end{enumerate}
\end{thm}

Two more definitions will be useful in the following.

\begin{dfn} We say a multiple-valued function $f:\Omega \subset \mathbb{R}^m \rightarrow A_Q(\mathbb{R}^n)$ is \emph{affinely approximatable} at $x_0\in \Omega$ if there exist affine maps $T_i: \mathbb{R}^m \rightarrow \mathbb{R}^n$ for $i=1,...,Q$ so that:

\begin{enumerate}

\item $\mathcal G (f(x_0), \sum \llbracket T_i(x)\rrbracket ) =o(|x-x_0|)$

\item If $f_i(x_0) = f_j(x_0)$, then $L_i=L_j$.

\end{enumerate}
\end{dfn}

\begin{dfn}
A \emph{branch point} for a multiple-valued function $f: \Omega \rightarrow A_Q(\mathbb{R}^n)$ is a point of discontinuity for the function $\sigma(x)=\mathcal{H}^0(\text{\textnormal{spt}}(f(x)))$.
\end{dfn}

\begin{rmk} For a Dirichlet minimizer the set of branch points is at most of Hausdorff dimension $m-2$, and, in case $m=2$, it is locally finite. If $x\in \Omega$ is not a branch point for a Dirichlet minimizing $f$, then in some neighborhood around $x$, there exists a harmonic selection for $f$, and such a point is called a \emph{regular point} for $f$. Note this definition allows the possibility that at a regular point one of the selection functions for $f$ may still have a critical point where the rank of its differential is less than $\min\{ m,n\}$.
\end{rmk}

\section{A Multiple-Valued Plateau Problem}

In this section, we will prove Theorem \ref{thm:main}. Notice that for $K=1$, $k_1=1$, this theorem reduces to the classical case proven by Douglas and Rado, see \cite{Douglas} and \cite{Rado}.

The naive idea is the following: take a minimizing sequence and apply a version of the Courant-Lebesgue Lemma to an appropriate subsequence to get a set of maps which are weakly monotonic on the boundary and then appeal to lower semi-continuity of Dirichlet energy under weak convergence. Unfortunately, the Courant-Lebesgue Lemma requires the so called ``three point condition.'' In the classical case, this condition is obtained by ``normalizing" the functions by precomposing with an automorphism of the disk. In our context, the problem with this approach is that we cannot a priori simultaneously normalize all of the boundary functions--the simplest explanation for our method around this is when $K=2$. We will show that, if there are certain kinds of branch points within the disk for each element of a minimizing set, then normalization of one of the boundary functions automatically normalizes the other. In the absence of these branch points, we may normalize separately.

To do this, we require two lemmas. The first lemma relies upon the following (slight) generalization of two Theorems of W. Zhu, found in \cite{WZhu2dim}.

\begin{thm}
\label{thm:Zhu}
Let $\epsilon >0$ and suppose an energy minimizing $f\in W^{1,2} ((1+\epsilon)\mbb D, A_Q(\mathbb{R}^n))$ is strictly defined and has precisely one branch point in $\mbb D$ at the origin. Then there exists $j \leq Q$ and $k_1, k_2, ..., k_j \in \mathbb N$ so that $\kappa := \sum k_i \leq Q$ and harmonic functions $f_1,...f_{k_j},f_{Q-\kappa},f_{Q-\kappa+1},...,f_Q:\mbb D \rightarrow \mathbb{R}^n$ so that, for all $s\in \mbb D$.

\begin{equation*}
f(s) = \sum_{i=1}^Q \left(    \sum_{\zeta=s^{k_i}} \llbracket f_i(\zeta)\rrbracket\right)
\end{equation*}

\noindent where, to suppress notation, we set $k_{\ell} = 1$ for $\ell \geq j$.

\end{thm}

The next lemma is elementary, but, to the authors's knowledge, isn't explicitly stated in the available literature.

\begin{comment}
There is a much, much easier proof of this fact, at least when $A$ is an annulus (which suffices for the remainder of the paper) involving harmonic extensions and what you used to call the SBV selection--think polygonal paths. (OR just make a cut in the disk and go from there to get a selection--the boundary data requirements mean that there can't be any of what Mike calls monodromy.
\end{comment}

\begin{lemma}
\label{lemma:sepcont}
Suppose $\Omega\subset \mathbb{R}^m$ is path connected and locally path connected and that $f: \Omega \rightarrow A_Q(\mathbb{R}^n)$ is continuous. Additionally, suppose that $f$ satisfies the following for some $\alpha >0$:

\begin{enumerate}
\item $\mathcal H^0(\mbox{\emph{spt}}(f(x)))=Q$ for all $x\in A$
\item $\{ (x,y) \in \Omega \times \mathbb{R}^n \ | \ y\in \text{\textnormal{spt}}\{f(x)\} \}$ has $K$ connected components $E_1,...,E_K$ so that $d(E_i,E_j) > \alpha$ for $i\neq j$.
\end{enumerate}

\noindent  Then $f$ admits a continuous decomposition, that is, there are integers $1', 2',...K'$ so that $\sum i'=Q$, continuous functions $f_j : \Omega \rightarrow A_{j'}(\mathbb{R}^n)$ so that $f = \sum \llbracket f_i \rrbracket$. Moreover, if $\Omega$ is a region in $\mathbb{R}^n$ and $f\in W^{1,p}(\Omega,A_Q(\mathbb{R}^n))$ for some $1<p<\infty$, then each $f_i \in W^{1,p}(\Omega, A_{i'}(\mathbb{R}^n))$.

\end{lemma}

\begin{proof}

 Let

 \begin{equation*}
 Gf(z) = \displaystyle\sum_{y\in \text{\textnormal{spt}}(f(z))} \llbracket (z, y) \rrbracket
 \end{equation*}

 \noindent be the graph map associated to $f$. Note that if $f$ is continuous (or Sobolev), so too is $Gf$. Let $i'$ be the number of elements of $E_i \cap\text{\textnormal{spt}}(Gf(z))$. We will show that $i'$ is indendent of $z \in \Omega$.

Let $x, y\in \Omega$ be given and let $c:[0,1] \rightarrow \Omega$ be a curve with $c(0)=x$ and $c(1) = y$. Since assumptions (1) and (2) guarantee that $f$ has no branch branch points on $c([0,1])$, well known selection theory (see, for example the beginning chapters of \cite{revisit}) gives a unique selection for $Gf\circ c$ as

\begin{equation*}
Gf\circ c = \sum \llbracket g_i \rrbracket.
\end{equation*}

We see that, if $g_i(0) \in E_\ell$, then $d(g_i(t),E_k) > \alpha$ for all $k\neq \ell$ by assumption two. Therefore, $g_i(1) \in E_\ell$, proving $i'$ is well defined.

Thus for $y\in \Omega$, we may put:

\begin{equation*}
f_i(y) = \sum_{(y,p)\in E_i} \llbracket p \rrbracket.
\end{equation*}

To see that each $f_i$ is continuous, given $x\in A$ and  $\epsilon > 0$, we choose $\delta$ so that $|x - y| < \delta$ implies $\mathcal G (Gf(x),Gf(y)) < \min(\epsilon, \tfrac{\alpha}{3} )$.  Let $Gf(x) = \sum \llbracket (x,p_i)\rrbracket$ and $Gf(y) \sum \llbracket (y,q_i) \rrbracket$ be labeled so that:

\begin{equation*}
\mathcal G(Gf(x), Gf(y) )^2 =\sum_i ||x-y||^2 + ||p_i - q_i ||^2.
\end{equation*}

\noindent Then, in particular, for each $i$, $||x-y||^2 + ||p_i - q_i ||^2  < \alpha^2$. So, $(x,p_i) \in E_j$ implies that $(y,q_i) \in E_j$, which implies that $\mathcal G(f_j(x) , f_j(y) ) < \epsilon$, so each $f_j$ is continuous.

Finally we show that each $f_j$ must be Sobolev as follows: for a.e. point $x$ in the interior of $\Omega$, if $\ell(x)$ is a line parallel to the unit axis of $\mathbb{R}^n$ passing through $x\in\mathbb{R}^n$, then $Gf|_{\ell(x)}$ is Sobolev on some neighborhood $U\subset \ell(x)$ of $x$. This implies, by Proposition 1.2 of \cite{revisit} that there exists a Sobolev selection on $U$. However, since each $f_i$ is continuous  we must have that the Sobolev selection corresponds to a Sobolev selection for each $f_i|_U$.  Hence by Theorem 2  of \cite[\S 4.9.2]{EandG} each $f_j \in W^{1,2}_{loc}(\Omega, A_{j'}(\mathbb{R}^n))$, and $f\in W^{1,2}(\Omega, A_Q(\mathbb{R}^n))$ allows us to conclude that $f_j \in W^{1,2}(\Omega,A_{j'}(\mathbb{R}^n))$ for all $j$.

\end{proof}

Our ultimate goal is to apply the following generalization of the Courant-Lebesgue Lemma to the boundary data of a minimizing sequence of multiple-valued functions. For the result in the classical setting, see \cite[\S 4.3 Theorem 3]{Hildmin}.

\begin{thm}
\label{thm:courant}
Suppose $\Gamma_1,...\Gamma_K$ are  closed, oriented Jordan curves as in the statement of Theorem \ref{thm:main} and that $\{f^i_k\}_{k\in \mathbb N}$ are sequences of monotone continuous mappings $f^i_k: S^1\rightarrow \Gamma^i$. Define $F_k : S^1 \rightarrow A_Q(\mathbb{R}^n)$ by:

\begin{equation*}
F_k(s)=\sum_{i} \sum_{\zeta = s^{k_i}}\llbracket f^i_k(\zeta) \rrbracket
\end{equation*}

\noindent  and suppose that the Dirichlet minimizing extensions of the maps $F_k$, denoted by $\tilde F_k: \mbb D \mapsto A_Q(\mathbb{R}^n)$ are such that

\begin{center}
\text{\textnormal{Dir}}$(\tilde F_k;\mbb D) \leq M$
\end{center}

\noindent for some $M\in\mathbb{R}$ independent of $k$, then, for each $i$, the family $\{f^i_k\}_{k\in \mathbb N}$ is equicontinuous if they satisfy a uniform three point condition:

\begin{center}
$f^i_k(\alpha^i_j) = \beta^i_{k,j} $ for $j=1,2,3,$, $i=1,...,Q$
\end{center}

\noindent for some distinct points $\alpha^i_j\in S^1$ and $\beta^i_{k,j} \in \Gamma^i$ so that $\beta^i_{k,j} \rightarrow \beta^i_j$ holds, for $\beta^i_j$ three distinct points of $\Gamma_i$.

\end{thm}

\noindent Once we guarantee that we can find an $A$ (independent of $k$) for which the conditions of Lemma \ref{lemma:sepcont} hold for each $\tilde F_k$, the proof of Theorem \ref{thm:courant} follows from identical methods of \cite[\S 4.3 Theorem 3]{Hildmin} after applying Lemma \ref{lemma:sepcont} and analyzing the embeddings $\zeta \circ f_i$ for the decomposition guaranteed by the lemma. Finding such a neighborhood is the content of the next proposition.

\begin{lemma}
\label{lemma:nbd}
Suppose that $\{F_k\}_{k\in \mathbb N} \subset W^{1,2}(\mbb D, A_Q(\mathbb{R}^n))\cap C(\overline{\mbb D}, A_Q(\mathbb{R}^n))$ is a sequence of Dirichlet minimizing functions with a selection for the trace of $F_k$ given, for $s\in S^1$ by

\begin{equation*}
F_k(s)= \sum_{i} \sum_{\zeta = s^{k_i}}\llbracket f^i_k(\zeta) \rrbracket
\end{equation*}

\noindent  for $f_k^i : S^1 \rightarrow \Gamma_i$ continuous, weakly monotonic maps onto curves as in Theorem \ref{thm:main}. Further suppose that each $F_k$ is affinely approximatable a.e. on the interior of $\mbb D$ and that

\begin{equation*}
\sup_k \text{\textnormal{Dir}}(F_k;\mbb D)=M < \infty.
\end{equation*}

\noindent  Then there exists a neighborhood $A\subset \mbb D$ of $S^1$ so that the conditions of Lemma \ref{lemma:sepcont} are satisfied for each $F_k|_{A}$.

Further, if each $\Gamma_i$ is a LNR of some open neighborhood $U_i$ (where we assume that $U_i \cap U_j = \phi$ for $i\neq j$), we may guarantee that: $F_k|_A \in \sum_1^K k_i \llbracket U_i \rrbracket$, where $\sum_1^K k_i \llbracket U_i \rrbracket$ is the set of all $y\in A_Q(\mathbb{R}^n)$ so that $\mathcal H^0 (\text{\textnormal{spt}}(y) \cap U_i )=k_i$ for all $i$.

\end{lemma}

\begin{rmk}
Note that the differentiability condition is immediately satisfied by any sequence of functions which are Dirichlet minimizing for their boundary data, as in \cite{bigreg} or \cite{Delint}.
\end{rmk}

\begin{proof}

Let $\epsilon < \displaystyle \min_{i\neq j} \text{dist}(\Gamma_i,\Gamma_j)$. We will show that for $\delta <1$ close enough to one and any $x\in S^1$,  $\delta \leq \alpha \leq 1$, $\mathcal G(F_k(\alpha x),F_k( \gamma )) < \frac{\epsilon}{3}$, where $\gamma$ is some arc of $S^1$, independent of $k$, and $x\in\gamma$.  By our assumptions on the boundary data of each $F_k$, this will guarantee the assumptions of Lemma \ref{lemma:sepcont}. We will assume that $M=1$, which is possible simply by dividing the functions appropriately.

 Let $1 \geq  \alpha \geq \delta \geq \frac{1}{2}$ where $\delta$ will be chosen shortly, and $0 < \Delta \theta \leq \frac{\pi}{2}$ be given. Let $a \in \alpha\cdot S^1$ be given, and let $\hat a=\frac a {||a||}$. Viewing $\mathbb{R}^2$ as $\mathbb{C}$, we lose no generality in assuming that $\hat a =1$. Consider the region $R_\delta$ defined by:

\begin{equation*}
R_\delta= \{ (x,y) | \ y\in [-\delta\sin(\frac{\Delta \theta}{2}),\delta\sin(\frac{\Delta\theta}{2})], \mbox{and} \ \delta^2 \leq x^2+y^2 \leq 1 \}.
\end{equation*}

\noindent For each $y_0\in [-\delta\sin(\frac{\Delta \theta}{2}),\delta\sin(\frac{\Delta \theta}{2})]$, let $L(y_0)$ be the horizontal line segment $(\mathbb{R}\times \{y_0\}) \cap R_\delta$. Denote by $x_{\delta}(y_0)$ and $x_1(y_0)$ the $x$ coordinates of where this line intersects $\delta S^1$ and $S^1$, respectively.

Since each $F_k$ is Sobolev, for a.e.  line $\ell$ in $\mathbb{C}$, we have that $F_k|_{(\ell \cap \mbb D)}$ is absolutely continuous for all $k$ and that $F_k$ is affinely approximatable at $\mathcal{H}^1$ a.e. point on $\ell$. This implies, by combining Proposition 1.2 of \cite{revisit} and Theorem 6.4 of \cite{selection}\footnote{While the statement of the theorem in \cite{revisit} requires that $F_k$ be Lipschitz, the proof is identical provided that $F_k$ admits an absolutely continuous selection, which is guaranteed by Proposition 1.2 of \cite{revisit}. } that for a.e. $y_0\in [-\delta \sin(\frac{\Delta \theta}{2}),\delta\sin(\frac{\Delta \theta}{2})]$
\begin{equation*}
\mathcal G(F_k(x_{\alpha}(y_0),y_0), F_k(x_1(y_0),y_0))\leq \int_{L(y_0)\cap \{ x \ | \ \delta \leq ||x|| \leq 1\} } ||\nabla F_k || \, d\mathcal{H}^1 \leq \int_{L(y_0)} ||\nabla F_k || \, d\mathcal{H}^1.
\end{equation*}

\noindent Integrating this expression with respect to the variable $y$ gives:

\begin{equation*}
\int_{-\delta\sin(\frac{\Delta \theta}{2})}^{\delta\sin(\frac{\Delta \theta}{2})}\mathcal G(F_k(x_{\\delta}(y),y), F_k(x_1(y),y)) \, dy\leq \int_{-\delta\sin(\frac{\Delta \theta}{2})}^{\delta\sin(\frac{\Delta \theta}{2})}\left(\int_{L(y)} ||\nabla F_k || \, d\mathcal{H}^1\right) \, dy.
\end{equation*}

\noindent The Fubini Theorem and the H\"older Inequality implies:
\begin{equation}
\label{eq:rhsbound}
\int_{-\delta\sin(\frac{\Delta \theta}{2})}^{\delta\sin(\frac{\Delta \theta}{2})}\left(\int_{L(y)\cap R_\delta} ||\nabla F_k || d\mathcal{H}^1\right) \, dy\leq M^{\frac{1}{2}} \left(\mathcal{H}^2(R_\delta)\right)^{\frac1 2}.
\end{equation}

\noindent On the other hand, if $\gamma_1$ and $\gamma_2$ denote the portions of $\alpha\cdot S^1$ and $S^1$ contained in $R_\delta$, we see that:

\begin{equation*}
{2\delta\sin(\frac{\Delta \theta}{2})} \mathcal G(F_k(\gamma_1),F_k(\gamma_2)) \leq \int_{-\delta\sin(\frac{\Delta \theta}{2})}^{\delta\sin(\frac{\Delta \theta}{2})}\mathcal G(F_k(x_{\alpha}(y),y), F_k(x_1(y),y)) \, dy
\end{equation*}

\noindent which, combined with Equations (\ref{eq:rhsbound}) and (\ref{eq:choice}) gives the bound:

\begin{equation}
\label{eq:neccest}
\mathcal G(F_k(\gamma_1),F_k(\gamma_2)) \leq \frac{\mathcal{H}^2(R_\delta)^{\frac1 2} }{2\sin(\frac{\Delta \theta}{2})\delta}
\end{equation}

 (Recall that we've assumed $M=1$.) By rotation, this bound may be established for any arc of length equal to $\Delta \theta$.

Next, one easily checks that $\mathcal{H}^2(R_\delta)\rightarrow 0$ as $\delta \rightarrow 1$ so we may choose (for fixed $\Delta \theta$) $\delta$ close enough to one so that:

\begin{equation}
\label{eq:choice}
\frac{\mathcal{H}^2(R_\delta)^{\frac1 2} }{2\sin(\frac{\Delta \theta}{2})\delta} \leq \frac{\epsilon}{6}.
\end{equation}

We now observe that, if the arc length were chosen to be small enough so that the variation of all of the functions $F_k$ on $\delta \gamma$ was smaller than $\frac \epsilon 6$, we could guarantee that, for each $e^{i\psi} \in \gamma$:

\begin{equation*}
\mathcal G(F_k(\delta e^{i\psi}, F_k(\gamma) ) < \frac \epsilon 3.
\end{equation*}

Obtaining such a bound for smaller arc length is slightly more complex than it seems at first, since $\delta$ must be chosen \emph{after} the arc-length is already fixed, but it can be done via the following method of choosing $\delta$ and the arc length $\ell$.

Recall that, in Theorem 9 of \cite{revisit}, the following inequality is shown for Dirichlet minimizing functions on the disk:

\begin{equation}
\label{eq:crucial}
\int_{B_r(x)} ||DF_k||^2 \, d\mathcal{H}^2 \leq r^{\frac 2 Q} \int_{\mbb D} ||Df||^2 \, d\mathcal{H}^2
\end{equation}

\noindent Whenever $x\in \mbb D$ and $r < 1-||x||$.

Following the arguments of Section 3.2 of \cite{LinPDE}, we see that this implies, some constant $C$ which depends only on $n$ and $Q$ so that:

\begin{equation*}
x,y \in \delta \mbb D \mbox{ with } ||x-y||< \frac{1-\delta}{2} \Rightarrow \mathcal G (F_k(x),F_k(y))\leq C ||x-y||^{\frac 1 Q}.
\end{equation*}

Let $\epsilon$ be as above and choose $n\in \mathbb N$ so that $\frac{2\cdot7^2}{ \epsilon^2 } \leq n $.  For $\delta$ sufficiently close to one,  $\ell = n\left(\frac{1-\delta}2\right)$ defines an arc within the ranges given in the preceding arguments.

Further, we may also choose $\delta$ close enough to one to guarantee the additional inequality:

\begin{equation}
\label{eq:choice2}
n C \left( \frac{1-\delta}{2} \right) ^{\frac 1 Q} \leq \frac \epsilon 6.
\end{equation}

 Then, using the asymptotics:

\begin{equation*}
 \mathcal{H}^{ 2}(R_\delta)  \approx \ell(1-\delta)
 \end{equation*}
 \begin{equation*}
2\sin(\frac{\Delta \theta}{2} ) \approx \ell
 \end{equation*}

 \noindent and our choices of $n$, $\delta$ and $\ell$, one checks that the above integral estimates imply (possibly upon choosing $\delta$ even closer to one):

\begin{equation*}
\mathcal G(F_k(\gamma),F_k(\delta \gamma)) \leq \frac{\epsilon}{6}.
\end{equation*}

\noindent for any arc $\gamma \subset S^1$ with length $\ell$.

Let $\gamma$ be such an arc. For any two points $x, y\in \delta \gamma$, by choice of $\ell$, we may find a chain :

\begin{center}$\{x=x_1,x_2,...,x_n=y\}\subset \delta \gamma$ for which $||x_i - x_{i+1}|| \leq \frac{1-\delta}{2}$.
\end{center}

\noindent Then:

\begin{equation*}
\mathcal G(F_k(x),F_k(y) ) < \sum \mathcal G(F_k(x_i),F_k(x_{i+1})) \leq n C \left(\frac{1-\delta}{2}\right)^{\frac 1 Q} \leq \frac{\epsilon}{6}
\end{equation*}

\noindent where the last inequality follows from Inequality (\ref{eq:choice2}).

To summarize, we have shown the following statement:

\begin{center}
{\it If $\gamma$ is an arc of $S^1$ of length $\ell$ and $x \in \gamma$, $k\in \mathbb N$, there exists $y_k\in \gamma$ so that $\mathcal G(F_k(\delta x) , F_k( y_k) ) \leq \frac{\epsilon}{3}$.}
\end{center}

One checks that this statement proves the initial goal, since choosing $\delta \leq \alpha$ only improves the bounds used above. By further constraining $\epsilon$ to also satisfy $\epsilon < \displaystyle \min_i \text{dist}(\Gamma_i, \partial U_i)$, we may obtain the final statement in the Lemma, completing the proof.

\end{proof}

\begin{remark}
\label{rmk:addit}
Note that Equation (\ref{eq:neccest}) a posteriori gives the additional property, where $\{f_i\}$ is the selection from Lemma \ref{lemma:sepcont}:  There is an $\alpha >0$ such that for all $0 < \alpha \leq \beta \leq 1$, $\beta S^1 \subset A$ and , and all $i$, $\mathcal G (f_i( \beta e^{i\theta}) ,f_i(\gamma)) < \epsilon$ where $\gamma$ is an arc of $S^1$ with length bounded by $C(1-\beta)$ with $e^{i\theta} \in \gamma$.
\end{remark}

Finally, we recall the following interpolation lemma, Lemma 2.15 of \cite{revisit}.

\begin{lemma}[Interpolation Lemma]
\label{lemma:interpolation}
There is a constant $C=C(m,n,Q)$ with the following property. Let $r, \ \epsilon > 0, \ g\in W^{1,2}(\partial B_r, A_Q(\mathbb{R}^n))$ and $f\in W^{1,2}(\partial B_{r(1-\epsilon}), A_Q(\mathbb{R}^n))$. Then, there exists $h\in W^{1,2}(B_r\setminus B_{r(1-\epsilon)}, A_Q(\mathbb{R}^n))$ so that $h|_{\partial B_{r(1-\epsilon)}} = f$, $h|_{\partial B_{r}} = g$ and so that $\text{\textnormal{Dir}}(h,B_r\setminus B_{r(1-\epsilon)}) $ is less than or equal to:

\begin{equation}
C\epsilon r [\text{\textnormal{Dir}}(g,\partial B_r) + \text{\textnormal{Dir}}(f,\partial B_{r(1-\epsilon)})] + \frac{C}{\epsilon r} \int_{\partial B_r} \mathcal G(g(x), f((1-\epsilon)x))^2 \, dx
\end{equation}

\end{lemma}

\begin{proof}[Proof of Theorem \ref{thm:main}] We present a proof in the case where $K=2$, and $k_1$ and $k_2$ are given so that $k_1 + k_2 = Q$. The more general case follows by identical arguments, there are simply more cases to analyze. For more details, see Remark \ref{rmk:higherq}.

Let $\{F_k\}_{k=1}^\infty\subset W^{1,2}(\mbb D, A_2(\mathbb{R}^n))$ be a minimizing sequence so that:

\begin{enumerate}
\item $F_k |_{S^1} (s)= \sum_{\zeta=s^{k_1}} \llbracket f_k(\zeta) \rrbracket +  \sum_{\zeta=s^{k_1}}\llbracket g_k (\zeta)\rrbracket $ for $f_k : S^1 \rightarrow \Gamma_1$ and $g_k : S^1 \rightarrow \Gamma_2$ monotone.
\item $\text{\textnormal{Dir}}(F_k, \mbb D ) \rightarrow \displaystyle \inf_{G\in \mathcal A} \text{\textnormal{Dir}}(G,\mbb D)$
\end{enumerate}

We may additionally assume that each $F_k$ is Dirichlet minimizing for its boundary data--since replacing each $F_k$ with a Dirichlet minimizer will only decrease the energy.

Standard compactness arguments allow us to extract a subsequence (which we won't relabel) which weakly converges in $W^{1,2}$ to a Dirichlet minimizing function $F$. Furthermore, by Arzela-Ascoli, uniform Holder continuity on any compact $\Omega \subset \subset \mbb D$ and a diagonal argument applied to a compact exhaustion of $\mbb D$,  we may guarantee that the $f_k$ converge uniformly on compact subsets of the disk.

Let $A$ be the neighborhood of $S^1$ provided by Lemma \ref{lemma:nbd}. For simplicity, we assume that $A$ is a disk of inner radius $\delta$. For reasons that will be elucidated shortly, we will mostly be interested in the branch points of $F$ within $\mbb D \setminus A$, so  denote by $\Sigma_k$ the branch set of $F_k$ within the compact set $\mbb D \setminus A$ and $\Sigma_0$ the branch set of $F$ within $\mbb D \setminus A$. Since interior branch points for 2-dimensional multiple-valued functions are isolated, we know that $F$ may have only finitely many branch points in $\mbb D \setminus A$. Further, since the $F_k$'s converge to $F$ uniformely on $\mbb D \setminus A$, we may, by throwing out finitely many elements of the sequence $\{F_k\}$, also assume that $\mathcal H^0(\Sigma_k) \geq \mathcal H^0 (\Sigma_0)=\ell$. Put $\Sigma_0 =\{x_1,...,x_\ell\}$.

For the remainder of the proof it will be crucial to have more control on the location and quantity of points in $\Sigma_k$--the following claim shows that we can do this without changing our sequence's properties substantially.

\begin{claim}
\label{claim:modification}
We may modify the sequence $\{F_k\}$ to obtain a new sequence, $\{\tilde F_n \}$ still converging uniformly on compact sets to $F$ so that if $\tilde \Sigma_n$ is the branch set of $\tilde F_n$ in $\mbb D \setminus A$, then $\mathcal H^0 (\tilde \Sigma_n) = \ell$ and so that $\text{\textnormal{Dir}}(\tilde F_n) = \text{\textnormal{Dir}}( F_{k(n)}) +o(1)$ as $n\rightarrow \infty$ for some increasing sequence $k(n)$.
\end{claim}

\begin{proof}[Proof of Claim]

Using the uniform convergence of the sequence $\{F_k\}$ on $\mbb D \setminus A$, we may find $r_1,...,r_\ell$ so that the following holds for all large $k$:

\begin{enumerate}
\item $B_{r_i}(x_i) \cap B_{r_j}(x_j) = \phi$ for $i\neq j$.
\item $\Sigma_k \subset \cup B_{r_i}(x_i)$
\item The image of the graph map of $F_k |_{\partial B_{r_i}(x_i)}$ has the same number of connected components for all $k$.
\end{enumerate}

\noindent For each $k\in \{1,...,\ell\}$, we consider $B_{r_i}(x_i)$ independently. For ease of notation, assume that $x_i=0$, and, as before, let $B_s$ be the open ball of radius $s$ about the origin.

Let the image of the graph map of $F_k |_{\partial B_{r_i}}$ have $\ell$ connected components.

Then there are positive integers $\ell_{i,j}$ for $j=1,...,\ell$ so that $\sum_j\ell_{i,j}=Q$   and there are $\ell$ continuous functions $f_{k,i,j}$ defined on $\partial B_{r_i}$ mapping to $A_{\ell_{i,j}}(\mathbb{R}^n)$  so that:
\begin{equation}
F_k|_{\partial B_{r_i}} =\sum_j  \llbracket f_{k,i,j} \rrbracket
\end{equation}

\noindent and so that $\mathcal G (f_{k,i,j} ,f_{k,i,m})$ is bounded away from zero independent of $k$ for $j\neq m$. Since $F$ is Dirichlet minimizing, has only one branch point within $B_{r_i}(x_i)$, and has $\ell$ connected components on $\partial B_{r_i}$,  we conclude via Theorem \ref{thm:Zhu} that there are $\ell$ harmonic functions $h_j : B_{1} \rightarrow \mathbb{R}^n$ so that:

\begin{equation*}
 F|_{B_{r_i}}(z) = \sum_j \sum_{\zeta^{\ell_{i,j}}=r_i z} \llbracket h_j(\zeta) \rrbracket.
 \end{equation*}

\noindent Uniform convergence implies that (up to relabeling),

  \begin{equation*}
  f_{k,i,j}(z) \rightarrow \text{Trace}\left(\sum_{\zeta^{\ell_{i,j}}=r_i z} \llbracket h_j(\zeta) \rrbracket\right )
  \end{equation*}

  \noindent uniformly as $k\rightarrow \infty$. For arbitrary $0 < \epsilon < r_i$, we may apply Lemma \ref{lemma:interpolation} to the following sequence of functions:

\begin{enumerate}
\item $F_k|_{\partial B_{r_i}} =\sum_j \llbracket f_{k,i,j} \rrbracket$
\item $F_\epsilon(s)=F(\frac{s}{1-\epsilon})$ defined on $\partial B_{r_i(1-\epsilon)}$.
\end{enumerate}

\noindent We obtain a sequence $h_{k,\epsilon} \in W^{1,2} (B_{r_i}\setminus B_{r_i(1-\epsilon)}, A_Q(\mathbb{R}^n))$ with bounds on $\text{\textnormal{Dir}}(h_{k,\epsilon},B_{r_i}\setminus B_{r_i(1-\epsilon)}) $; in particular:

\begin{equation}
\label{eq:bounds}
\text{\textnormal{Dir}}(h_{k,\epsilon},B_{r_i}\setminus B_{r_i(1-\epsilon)})  \leq C\epsilon r [\text{\textnormal{Dir}}(F_k,\partial B_{r_i}) + \text{\textnormal{Dir}}(F_{\epsilon},\partial B_{r_i(1-\epsilon)})] + \frac{C}{\epsilon r} \int_{\partial B_{r_i}} \mathcal G(F_k(x), F_\epsilon((1-\epsilon)x))^2 \, dx.
\end{equation}

Notice that in the above equation, as $k \rightarrow \infty$, the right hand side tends to:

\begin{equation}
2C\epsilon r \text{\textnormal{Dir}}(F_\epsilon,\partial B_{r_i}) \leq M \epsilon
\end{equation}

\noindent where $M$ is independent of $\epsilon$.

Let $\epsilon_n$ be any sequence tending to zero, and further suppose that $0 < \epsilon_n < \min(r_i,\epsilon)$. For each $n$, choose an increasing sequence $k(n)$ so that:

\begin{equation*}
C\epsilon_n r [\text{\textnormal{Dir}}(f_{k(n)},\partial B_{r_i}) + \text{\textnormal{Dir}}(h_{\epsilon_n},\partial B_{r_i(1-\epsilon_n)})] + \frac{C}{\epsilon_n r} \int_{\partial B_{r_i}} \mathcal G(f_{k(n)}(x), h_{\epsilon_n}((1-\epsilon)x))^2 \, dx < 2M\epsilon.
\end{equation*}

\noindent Now, define $\tilde F_n$ by:

 \begin{displaymath}
   \tilde F_n(x) = \left\{
     \begin{array}{llr}
       F_{k(n)} (x) & : & x \in \mbb D \setminus B_{r_i}\\
       h_{k(n),\epsilon_n} (x) & : & x\in B_{r_i} \setminus B_{r_i(1-\epsilon)}\\
       F (\frac{x}{1-\epsilon_n}) & : & x\in B_{r_i(1-\epsilon)}\\
     \end{array}
   \right.
\end{displaymath}

\noindent One may check that $\text{\textnormal{Dir}}(\tilde F_n, B_{r_i}) = \text{\textnormal{Dir}}( F_k(n),B_{r_i}) +o(1)$ holds, and further that $\tilde F_n$ has only one branch point within $B_{r_i}$ for sufficiently large $n$--namely $0$.

By applying the above construction to each $B_{r_i}(x_i)$, we obtain a new sequence $\{\tilde F_k\}$ satisfying the conclusions of the Claim.

\end{proof}

\begin{remark}
\label{rmk:local}
Note that the new functions $\tilde F_k$ are no longer Dirichlet minimizing for their given boundary data, however, they are \emph{locally} Dirichlet minimizing in some neighborhood of their branch points which is independent of $k$.
\end{remark}

There is a sort of dichotomy between the two types of branch points that may occur in $\Sigma_k$, which we now analyze.

\begin{dfn}
Call $\{x_i^k\}_{k\in \mathbb N}$ \emph{connecting} if there is a smoothly bounded neighborhood $U$ so that:

\begin{enumerate}
\item for all large $k$, $U\cap \Sigma_k = \{x_1^k\}$ and,
\item $U$ is smoothly contractible to $x_1^k$ for all $k$ as in (1),
\item $\partial U$ is a smooth curve whose intersection with $A$ has a smooth arc $\Gamma$ of positive $\mathcal{H}^1$ measure
\item  If $GF_k$ denotes the graph map (as in the proof of Lemma \ref{lemma:sepcont}), then $\text{\textnormal{spt}}(GF_k(\partial U)$ has a connected component $\Lambda$ so that $(\Lambda \cap \text{\textnormal{spt}}(GF_k(\Gamma)))\cap B_i\neq \phi$ for $i=1,2$.
\end{enumerate}
\end{dfn}

The proof now reduces to two cases.

\emph{Case I:} Suppose first that for every $j$, $\{x_j^k\}_{k\in \mathbb N}$ are not connecting. We will show that the decomposition of $F_k$ on $A$ extends to a decomposition on all of $\mbb D$, which then allows us to independently normalize the boundary data to satisfy the three point condition of Theorem \ref{thm:courant}.

To see this, choose $x_0 \in \mbb D \setminus (A \cup \Sigma_0)$, and let $\gamma: [0,1]\rightarrow A$ be a path so that $\gamma(0) = x_0$ and $\gamma(1)=z \in A$ so that $\gamma([0,1]) \subset \mbb D \setminus (A \cup \Sigma_0)$. Standard selection theory (see, for example the beginning chapters of \cite{revisit}) gives a unique selection

\begin{equation*}
F_k\circ \gamma = \sum \llbracket f^i_k \rrbracket
\end{equation*}

\noindent Partition $F_k(x_0) = \sum \llbracket f^i_k(0) \rrbracket$ into two multiple-valued functions by:
\begin{equation*}
F_k(x_0) = \sum_{(z,f^i_k(1)) \in B_1} \llbracket f^i_k(0) \rrbracket + \sum_{(z,f^i_k(1)) \in B_2} \llbracket f^i_k(0) \rrbracket =:\llbracket F_k^1(x_0)\rrbracket +\llbracket F_k^2(x_0)\rrbracket
\end{equation*}

One checks that the fact that no element of $\Sigma_0$ is connecting implies that this choice is independent of path and therefore well-defined. Since each $F_k$ is continuous, this decomposition may be extended to the finite set $\Sigma_0$ by continuity. It readily follows that each $F_k^i$ is Sobolev. Further, the decomposition on $A$ gives (up to possible relabeling)

\begin{equation*}
F_k^1 |_{S^1} (s)= \sum_{\zeta=s^{k_1}} \llbracket f_k(\zeta) \rrbracket
\end{equation*}
\begin{equation*}
F_k^2 |_{S^1} (s)= \sum_{\zeta=s^{k_2}} \llbracket g_k(\zeta) \rrbracket
\end{equation*}

\noindent as in (1) at the beginning of the proof. Therefore, by picking two sequences of conformal disk automorphisms, $\{\sigma_k\}$ and $\{\psi_k\}$ so that $f_k \circ \sigma_k$ and $g_k \circ \psi_k$ satisfies a three point condition, we see that $\llbracket F_k^1 \circ \sigma_k \rrbracket+ \llbracket F_k^2 \circ \psi_k\rrbracket$ converges to an admissible map $\tilde F$ as $k\rightarrow \infty$--but since this normalization doesn't affect the energy, lower semicontinuity of the Dirichlet energy gives that $\tilde F$ is a minimizer, as desired.

\emph{Case II:} Suppose now that $\{x_1^k\}_k$ is a sequence of connecting branch points. We will prove that the boundary data $f_k$ and $g_k$ cannot degenerate in the limit.  Let $\alpha_1, \ \alpha_2,$ and $\alpha_3$ be three distinct points of $S^1$, and let $s_i^k = f_k(\alpha_i)$, $t_i^k = g_k(\alpha_i)$ for $i=1,2,3$. By compactness of $\Gamma_1$ and $\Gamma_2$, we may, upon extracting to a subsequence (which we again will not relabel) suppose that $s_i^k \rightarrow s_i$ and $t_i^k\rightarrow t_i$ for each $i$ as $k\rightarrow \infty$.

By precomposing with a disk diffeomorphism, we may assume that the set $\{t_1,t_2,t_3\}$  consists of three distinct elements , and hence  that the boundary data $g_k$ converges uniformly to some $g$ by Theorem \ref{thm:courant}. We want to show that the set  $\{s_1,s_2,s_3\}$ also consists of three distinct elements, since then we may apply Theorem \ref{thm:courant} to obtain the result immediately.

We now assume towards a contradiction that $\{s_1, s_2, s_3\}$ are not distinct, say that $s_1 = s_3$. Under these assumptions, by the monotonicity of the $f_k$, and once again, possibly extracting a (still un-relabeled) subsequence  we may select a sequence $\beta^k$ with the following properties:

\begin{enumerate}
\item $\beta^k \rightarrow \beta \notin \{\alpha_1, \alpha_2\}$ as $k\rightarrow \infty$.
\item $f_k(\beta^k) = \tilde s_3\notin \{s_1, s_2\}$.
\end{enumerate}

Note that $g$ must be non-constant on one of the arcs $(\beta, \alpha_1)$, or $(\beta, \alpha_2)$, where $(\beta,\alpha_1)$ is the arc that doesn't pass through $\alpha_2$ and vice versa.  We suppose that $g$ is non-constant on $(\beta, \alpha_1)$. Since for each $k$ there are only finitely many  branch points to avoid and their locations are constrained to balls of fixed radii, we may find a smoothly bounded, simply connected $U \subset \mbb D$ satisfying the properties in the definition of connecting branch point, with the additional property that:

\begin{itemize}
\item $\partial U$ is a smooth curve whose intersection with $A$ has an arc $L$ so that $\text{dist}(L \cap (\beta, \alpha_1)) < \frac{\delta}{2}$, where $(\beta, \alpha_1)$ is the arc of $S^1$ connecting $\beta$ and $\alpha_1$ which does not pass through $\alpha_2$.
\end{itemize}

Using the estimate given by Inequality (\ref{eq:neccest}) in the proof of Lemma \ref{lemma:nbd}, we may further assert that if $r_f : U_f \rightarrow \Gamma_i$ is a Lipschitz retract and $\epsilon >0$ is given, we may choose $U$ so that $(\tilde s^3 \pm \epsilon , s_1 \pm \epsilon ) \subset r_f( F_k(L)\cap U_f)$, where the $\pm$'s are independent of one another.

 Let $\phi : \mbb D \rightarrow U$ be the conformal map guaranteed by the Riemann Mapping Theorem. Further, by Caratheadory's Theorem (see, for example Theorem 5.5 of \cite{Conway} and the preceeding sections), we see that $\phi$ extends to a homeomorphism  $\varphi: S^1 \rightarrow \partial U$. Then since $\{x_1^k\}$ is a sequence of connecting branch points, we may put

\begin{equation*}
F_k|_{\partial U}(z) = \sum_i \sum_{\zeta^{\ell_i} = \phi^{-1}(z) } \llbracket h_{k,i} (\zeta \rrbracket) .
\end{equation*}

\noindent For some integers $\ell_i$ with $\sum \ell_i = Q$ and some continuous functions $h_{k,i} : S^1 \rightarrow \mathbb{R}^n$.  Without loss of generality, we may assume that $h_{k,1}$ is a parameterizing map for $\Lambda$ as in the definition of connecting branch points. Set $h_k = h_{k,i}$.

 Consider now the maps $\psi_k(z)= (z, h_k(z) )$ for $z\in \mbb D$. Since $\partial U \Subset \mbb D$, uniform convergence on compact sets of the maps $F_k$ guarantees that the curves $\psi_k(S^1)$ converge in the sense of Fr\' echet (see, for example, \cite[\S 4.2]{Hildmin} ) to a curve $\Psi$. Further, since $g_k$ converges uniformly to a non-constant function (by our choice of arc $(\beta, \alpha_1)$), by the comments in Remark \ref{rmk:addit}, the traces of $\psi_k$ are monotone and satisfy the three point condition of Theorem \ref{thm:courant} and thus $\text{Trace}(\psi_k)$ converges uniformly to some $\psi: S^1 \rightarrow \mathbb{R}^n$, which is monotonic onto the curve $\Psi$. Put $\psi(z) = (z, h(z))$. Then, for each $k$, there is a segment of the image curve of $h$ contained in $((\tilde s_3, s_1))_\delta$ (where this is the interval inside $\Gamma_1$), which follows from Remark \ref{rmk:addit} and so, by applying the retraction $r_f : (\Gamma_1)_\delta \rightarrow \Gamma_1$, we see that this property must persist in the limit, and therefore the $f_k$ restricted to the interval $(\beta, \alpha_1)$ cannot degenerate, and hence we may choose three points and apply Theorem \ref{thm:courant} to obtain convergence to a monotone map.

Finally, notice that we may (possibly upon extracting to another subsequence and utilyzing another diagonal argument) assume that the sequence of renormalized functions weakly converges to some $\tilde F \in W^{1,2}(\mbb D, A_Q(\mathbb{R}^n))$. Further, Remark \ref{rmk:addit} applied to smaller and smaller neighborhoods paired with uniform convergence on compact sets and uniform convergence on $S^1$, one obtains using a standard $\frac{\epsilon}{3}$ argument that this subsequence converges uniformly to $\tilde F$ on $\mbb D$, which implies that $\tilde F$ is admissible, and lower semi-continuity of the Dirichlet Energy then gives the result.

\end{proof}

\begin{rmk}[Generalization to $K >2$]
\label{rmk:higherq}
One notices that there is nothing unique for the case where $K=2$, there are simply fewer cases of the branching behavior. Indeed, one first checks that Claim \ref{claim:modification} holds for arbitrary $K$, and then, to guarantee the boundary convergence, one uses an identical argument as presented above, there are simply more cases to be concerned with since there are more possibilities for branching behavior.
\end{rmk}

\section{Proof of the Regularity Theorem}
\label{section:topology}
In this section we  prove Theorem \ref{thm:conformal} via  methods gleaned from the classical situation for Douglas Minimizers. The proof involves using topological methods and complex analysis to produce a single-valued map from a planar domain with the same minimization properties and boundary data. It relies heavily on the classical results regarding the Douglas problem--see Chapter 8 of \cite{Hildmin}.

Before we begin, we need a few results from complex analysis. We start by stating a weakened version of a theorem of Bieberbach \cite{bieber}--see also \cite{complex}.

\begin{thm}
\label{thm:bieber}
Let $\Omega\subset \mathbb{C}$ be a bounded domain in the plane bounded by $n$ non-intersecting Jordan curves $\gamma_1,\gamma_2,...,\gamma_n$, and, for each $i$, choose a point $b_i \in \gamma_i$. Then there exists a proper holomorphic mapping $f: \Omega \rightarrow \mbb D$ which is an $n$-to-one branched covering map. Further, since $f$ is proper, $f$ extends continuously to the boundary of $\Omega$, and this extension (which we will also denote by $f$) maps each boundary curve monotonically onto the unit circle and has $f(b_i) =1$ for all $i$.
\end{thm}

The above Theorem leads, in a fairly straightforward way, to the following. The main idea is to translate the following Theorem into a problem on the upper half-plane, and then add functions produced via the above with clever selections of points--for full details, see \cite[\S 5]{complex}.

\begin{thm}
\label{thm:genahlfors}
Suppose that $\Omega \subset \mathbb{C}$ is as in Theorem \ref{thm:bieber} and $d_1,...,k=d_n$ are positive integers. Then there exists a proper holomorphic map $f: \Omega \rightarrow \mbb D$ such that:

\begin{enumerate}
\item $f$ is a $d_1+d_2+...+d_n$-to-one branched cover of $\mbb D$
\item $f$ admits a continuous extension to $\overline \Omega$ (which we will also denote by $f$) so that, for each $i$, $f|_{\gamma_i}$ has degree $k_i$.
\end{enumerate}

\end{thm}

\begin{proof}[Proof of Theorem \ref{thm:conformal}]   Let $F$, $\Gamma_1,...,\Gamma_K$ and $k_1,...,k_K$ be as in Definition \ref{dfn:wrapped}. Consider the set $\Gamma(F) = \{ (x,y) \ | \ x\in \mbb D \ y\in\text{\textnormal{spt}}(F(x))\}$, the \emph{graph} of $F$ (note that $\Gamma(F)$ is also the multiple-valued image of the map $GF$ from the above sections).

Notice that, since the boundary data admits a wrapped solution, $F$ can have no branch points on the boundary $S^1$, and thus by Remark \ref{rmk:addit} $F$ has no branch points on some annulus $A$, and hence only finitely many branch points in $\mbb D$.

Next let $\pi_1$ denotes the projection of $\mathbb{R}^2 \times \mathbb{R}^n$ onto the $\mathbb{R}^2$ coordinates, and $\pi_2$ denote the projection onto $\mathbb{R}^n$ coordinates. Comments in the above paragraph show that $\pi_1 : \Gamma(F) \rightarrow \mbb D$  is a branched cover.

Further, one readily checks that $\Gamma(F)$ is a Riemann surface with boundary for which any loop is either null-homotopic or homotopic to a concatenation of boundary curves--and so $\Gamma(F)$ is therefore topologically a punctured Riemann Sphere. By Koebe's General Uniformization Theorem (see, for example, \cite{foundit}) that there is a planar domain $\Omega$ with $K$ boundary components and a conformal diffeomorphism $\varphi : \Omega \rightarrow \Gamma(F)$.

By conformal invariance of energy and construction, $\kappa :=\pi_2 \circ \varphi : \Omega \rightarrow \mathbb{R}^n$ has the same energy as (the multiple-valued function) $F$.

Our goal will be to show that that $\kappa$ is conformal, as this will imply (2) of Theorem \ref{thm:conformal}. (2) paired with the energy/area inequality will then imply (1).

Supposing towards a contradiction, suppose that $\kappa$ is not conformal. Following \cite[\S 8.2 and \S 4.5]{Hildmin}, we find a domain $\tilde \Omega$ and a diffeomorphism $\sigma : \tilde \Omega \rightarrow \Omega$ so that $\text{\textnormal{Dir}}(\kappa \circ \sigma) < \text{\textnormal{Dir}}(\kappa)$. Further, we may assume that $\sigma$ is arbitrarily close to the identity in the uniform topology, and thus that $\tilde \Omega$ has the same number of boundary components as $\Omega$.

Our next goal is to use $\tilde \Omega$ and $\sigma$ to produce a competitor for $F$ with less energy, which will contradict $F$'s minimizing property.

Let $\gamma_1,...,\gamma_k$ be the boundary curves of $\tilde \Omega$ ordered so that $\kappa \circ \sigma (\gamma_i) \subset \Gamma_i$. Then, apply Theorem \ref{thm:genahlfors} to $\tilde \Omega$ with $d_i = k_i$ to obtain a $Q$-to-one branched cover $\Phi :\tilde  \Omega \rightarrow \mbb D$ which extends to a map on the closure of $\tilde \Omega$ for which the maps $\Phi|_{\gamma_i}$ are monotonic and of degree $k_i$ onto the circle $S^1$.

We then define a multiple-valued map $G : \mbb D \rightarrow A_Q$ as follows: for $z$ a regular point of the branched cover $\Phi$,

\begin{equation}
G(z) = \displaystyle \sum_{y\in \Phi^{-1}(z) } \llbracket \kappa \circ \sigma (y) \rrbracket
\end{equation}

\noindent and extended by continuity to the finitely many branch points of $\Phi$. One readily checks that the energy of $G$ is (since $\Phi$ is holomorphic and extends to a monotonic map on the boundary, and hence conformal except at finitely many interior points) equal to the energy of $\kappa \circ \sigma$, a contradiction, since $G$ is also admissible. Therefore $\kappa$ is conformal, proving (2) and hence (1).

\noindent

To prove (3), we follow an argument similar to the one presented in \cite[Theorem 3, \S 4.5]{Hildmin}. If one of the $f_i's$ were not a homeomorphism, since it is monotonic it must map some arc $\gamma$ along $S^1$ to a constant $\alpha \in \mathbb{R}^n$. Let $x_0^{k_i}$ be some point on the interior of this arc. We then find a harmonic selection on some small ball $B_r(x_0) \cap \mbb D$ by Lemma \ref{lemma:nbd}. By (1), and since this selection is unique (up to reordering), it must be conformal. Let $F_i$ be the function on $B_r(x_0)\cap \mbb D$ which maps $B_r(x_0) \cap \gamma$ to a constant. By Schwarz reflection and conformality, we obtain that $F_i \equiv \alpha$ and hence, since $\mbb D \setminus \sigma(F)$ is path connected (where $\sigma(F)$ is the set of branch points of $F$), that $\alpha\in \text{\textnormal{spt}}(F(z))$ for all $z\in \mbb D$, an obvious impossibility given the separation assumptions on the curves $\Gamma_i$.

\end{proof}

\section{A Class of Examples}

In this section we produce an example of two curves $\Gamma_1$ and $\Gamma_2$ for which the Plateau solution produced in the above section does not allow for a global selection so that there is necessarily a branch point and the image is connected. Further, we give a method for producing a large number of such examples.

Our last proposition is the following:

\begin{prop}
\label{prop:classex}
Suppose that $V \subset \mathbb{C} \times \mathbb{C}^n$ is a holomorphic variety for which the projection $\pi (z_1,...,z_{n+1} ) = (z_1)$ is a $Q$ to $1$ cover of the disk $\mbb D$ except for possibly finitely many points on the interior of $\mbb D$, and for which $S^1 \times \mathbb{C}^n$ intersects $V$ in $Q$ disjoint curves. Then the multiple-valued function $ z\mapsto \pi^{-1} (z) \cap V$ is the Plateau solution for the given curves.
\end{prop}

The proof of the proposition relies on the following elementary lemma.

\begin{lemma}
\label{lemma:mass}
Suppose $\Omega \subset \mathbb{R}^2$ is compact, and that $F: \Omega \rightarrow \mathbb{R}^n$ is injective and Lipschitz. Then, if $\tilde F (x,y) = (x,y,F(x,y)) \in \mathbb{R}^2 \times \mathbb{R}^n$,and $\llbracket\Omega\rrbracket$ is the standardly-oriented  two-dimensional current in $\mathbb{R}^2$, we have:

\begin{equation*}
\mbb{M} (F_\sharp (\llbracket \Omega\rrbracket) ) +|\Omega| \leq  \mbb{M}(\tilde F_\sharp(\llbracket\Omega)\rrbracket )
\end{equation*}
\end{lemma}

\begin{proof}
One readily checks that, if $a$ and $b$ are non-negative so that:

\[ DF^T \circ DF =\left( \begin{array}{cc}
a & b  \\
c &d  \\
\end{array} \right)\]

\noindent then,

\[ D\tilde F^T \circ D\tilde F =\left( \begin{array}{cc}
1+a & b  \\
c &1+d  \\
\end{array} \right).\]

\noindent Therefore,

\begin{equation*}
1 + \sqrt{ \det (DF^T \circ DF)} \leq \sqrt{\det ( D\tilde F^T \circ D\tilde F) }
\end{equation*}

\noindent which combined with the area formula implies the result.

\end{proof}

To prove the proposition, we recall two results from \cite{higher}, see Lemma 1.8 and Proposition 2.2 and the definitions preceeding them for more details:

\begin{thm}
\label{thm:higherests}

Let $f \in W^{1,2}(\Omega, A_Q(\mathbb{R}^n)$ be so that $\mbb{M} (T_{f,\Omega}) < \infty$. Then:

\begin{equation*}
\mbb{M} (T_{\lambda f, \Omega}) = Q|\Omega| + \frac{\lambda ^2}{2} \text{\textnormal{Dir}}(f,\Omega) + o(\lambda^2) \ \mbox{as $\lambda \rightarrow 0$}
\end{equation*}

\noindent Further, if $F$ is the associated $Q$-valued function of a holomorphic variety over $\Omega \subset \mathbb{C}$, then:

\begin{equation*}
\mbb{M} (\llbracket V \rrbracket \restr \Omega \times \mathbb{C}^n ) = Q |\Omega| + \frac{\text{\textnormal{Dir}}(F,\omega)}{2}
\end{equation*}

\end{thm}

\begin{proof}[Proof of Proposition \ref{prop:classex}]

Note that $V_\lambda = \{ (z,\lambda w) \ | \ (z,w) \in V\}$is also a holomorphic variety, and, if $F_\lambda$ is the corresponding $Q$-valued function, Theorem \ref{thm:higherests} yields:

\begin{equation*}
\mbb{M} (\llbracket V_\lambda \rrbracket \restr \mbb D \times \mathbb{C}^n ) = Q |\mbb D| + \frac{\text{\textnormal{Dir}}(F_\lambda,\mbb D)}{2} = Q |\mbb D| + \frac{\lambda^2}{2} \text{\textnormal{Dir}}(F,\mbb D)
\end{equation*}

Let $G : \mbb D \rightarrow \mathbb{C} \times \mathbb{C}^n$ be any admissible map for $\Gamma_1,...,\Gamma_Q$. Then, note that if $P : \mathbb{C} \times \mathbb{C} \times \mathbb{C}^n$ is the projection onto the last components, $\pi (z, w, v) = (w,v)$ for $z,w\in \mathbb{C}$ and $v\in \mathbb{C}^n$ and $\lambda >0 $ is given,

\begin{equation*}
\mbb{M} (\llbracket V_\lambda \rrbracket \restr \mbb D \times \mathbb{C}^n ) \leq \mbb{M} ( P_\sharp (T_{\lambda G, \mbb D}) )
\end{equation*}

\noindent since complex varieties are absolutely mass minimizing and $G$ is admissible.

However, using the above inequalities, this amounts to:

\begin{equation*}
 Q |\mbb D| + \frac{\lambda^2}{2} \text{\textnormal{Dir}}(F,\mbb D) \leq \mbb{M} ( P_\sharp (T_{\lambda G, \mbb D}) ).
\end{equation*}

\noindent However, by the definition of the current $T_{\lambda G, \mbb D}$ and Lemma \ref{lemma:mass}, we see that the above gives:

 \begin{equation*}
2 Q |\mbb D| + \frac{\lambda^2 \text{\textnormal{Dir}}(F,\mbb D)}{2} \leq \mbb{M} ( P_\sharp (T_{\lambda G, \mbb D}) )  + Q |\mbb D| \leq   \mbb{M} ( T_{\lambda G, \mbb D} ) = Q|\mbb D| + \frac{\lambda ^2}{2} \text{\textnormal{Dir}}(f,\mbb D) + o(\lambda^2)
\end{equation*}

\noindent and therefore,

\begin{equation*}
Q |\mbb D| + \frac{\lambda^2}{2} \text{\textnormal{Dir}}(F,\mbb D)  \leq  \frac{\lambda ^2}{2} \text{\textnormal{Dir}}(G,\mbb D) + o(\lambda^2)
\end{equation*}

\noindent Finally, notice that $\lambda F$ allows for a  local conformal selection, and, if $\tilde F$ denotes the graph map as in Lemma \ref{lemma:mass}, that $1 + \frac{\lambda ^2}{2} |DF|^2  = \sqrt{\det (D\tilde F^T \circ \tilde F )}$, so the area-energy equality for conformal maps, the above gives:

\begin{equation*}
\frac{\lambda^2}{2} \text{\textnormal{Dir}}(\tilde F,\mbb D)  \leq  \frac{\lambda ^2}{2} \text{\textnormal{Dir}}(G,\mbb D) + o(\lambda^2)
\end{equation*}

\noindent which implies $\text{\textnormal{Dir}}(\tilde F,\mbb D)  \leq  \text{\textnormal{Dir}}(G,\mbb D) $, proving the proposition.

\end{proof}

For a specific example, consider the multiple-valued map $F$ defined by $z \mapsto (z,\sqrt{z^2 - \frac{1}{4}})$ for $z\in \mathbb{C}$.  It is clear that $F$ has two connecting branch points located at the points $z=\pm \frac{1}{2}$.

Note that $F|_{S^1}$ has a selection given by:

\begin{itemize}
\item $f_1(\theta) = (\exp (i \theta),\sqrt[4]{ (1+\frac{1}{16} - \frac{\cos(2\theta)}{2}) }\exp(i \cdot\frac{ \text{\textnormal{atan2}} (\sin(2\theta), \cos(2\theta)-\frac{1}{4})}{2} ))$
\item $f_2(\theta)= (\exp (i \theta),  \sqrt[4]{ (1+\frac{1}{16} - \frac{\cos(2\theta)}{2}) }\exp(i\cdot (\frac{  \text{\textnormal{atan2}} (\sin(2\theta), \cos(2\theta)-\frac{1}{4})}{2} )+\pi))$
\end{itemize}

\noindent Let $f_i = (f^1_i,f^2_i,f^3_i)$. Then, even though the above selection is discontinuous it may be used to show that the boundary has two connected components, as is suggested by the following images of the real and imaginary parts of the functions $f_i^3$.

\begin{center}
\includegraphics[scale=.5]{imaginaryboundary} \vspace{5mm}\includegraphics[scale=.5]{realboundary} \vspace{5mm}
\end{center}

\noindent   One may check that $\Re(f^3_1(\theta)) = \Re(f^3_2(\theta) ) $ if and only if $\theta \in \{ \pm \frac{\pi}{2}\}$ and $\Im(f^3_1(\theta)) = \Im(f^3_2(\theta))$ if and only if $\theta \in \{0,\pi\}$.

Further, these images suggest that the following will give a continuous selection:

 \begin{displaymath}
   g_1(\theta) = \left\{
     \begin{array}{llr}
      f_1(\theta)& : & \theta \in [\frac{-\pi}{2},\frac{\pi}{2}]\\
      f_2(\theta) & : & \theta \in [\frac{\pi}{2},\frac{3\pi}{2}]\\
     \end{array}
   \right.
\end{displaymath}

 \begin{displaymath}
   g_2(\theta) = \left\{
     \begin{array}{llr}
      f_2(\theta)& : & \theta \in [\frac{-\pi}{2},\frac{\pi}{2}]\\
      f_1(\theta) & : & \theta \in [\frac{\pi}{2},\frac{3\pi}{2}]\\
     \end{array}
   \right.
\end{displaymath}

\noindent Which one checks provides a continuous selection of the boundary data. Further, by (locally) holomorphically defining $z \mapsto \sqrt{z \pm \frac{1}{2}}$ we see that, for sufficiently small $\rho > 0$, $F|_{\partial B_\rho (\pm \frac{1}{2}) }$ is a connecting branch point, since then it has  structure $\varphi \circ \sqrt{z\mp\frac{1}{2}} $, for $\varphi$ holomorphic.  Therefore, Proposition \ref{prop:classex} guarantees that $F$ is the Plateau solution, and thus that the Plateau solution is branched.

Branch points could also occur in $\mathbb R^3$ for essentially topological reasons.  For $0<\epsilon<1$, consider on the circle $S^1$, a single branch of the 2-valued function
$\phi(z)=(\sqrt{z-\epsilon}, {\mathcal Re}\sqrt{z-\epsilon})$. Note that the image, {\it under this 2 valued map} is a single smoothly embedded circle in $\mathbb R^2\times\mathbb R$ whose projection onto $\mathbb R^2$ circulates the origin twice.  Then $\phi$ does not admit a continuous $2$-valued extension $F:\mathbb D\to A_2(\mathbb R^2\times\mathbb R)$ which is a locally smooth embedding without branch points. In fact otherwise one could start at the origin with two values and smoothly extend to obtain a selection on the whole disk, the image of whose boundary values would, unlike $\phi$, be two embedded curves. In particular, any 2-valued Plateau solution in $\mathbb{R}^3$ starting with this ``doubly-wrapped'' 2-valued boundary $\phi$ must have branch point.  By the same argument we see that:

{\it Any solution of the Plateau problem in Theorem 1.2 starting with boundary parameterization as in Theorem 1.1 with multiple wrapping (i.e. some $k_i > 1$) , whose boundary curves satisfy condition \ref{dfn:wrapped}  must have a branch point. In case $N=3$, the image then must curves of self-intersection. }

It is the subject of current work to determine whether or not one can verify the conditions of Definition \ref{dfn:wrapped} in $\mathbb{R}^3$.

However, with $Q=2$ and $k_1=k_2=1$, for two closed curves lying in different planes in $\mathbb{R}^3$, one can, using similar arguments to those in \cite[\S 8.8]{Hildmin} to guarantee that our Plateau solutions do in fact admit branch points in three dimensional space. Further, constancy along a boundary curve \emph{can} have elaborate branching behavior, as illustrated in the example below--although this example does differ slightly from the type of degeneracy that would occur if given boundary data did not admit a wrapped solution.

\section{An Example of a Degenerate Behavior}
\label{section:degen}

In this section we provide a counterexample to the following question:

\begin{question}
\label{question:main} Let $Q\geq 2$ and suppose that $F: \mbb D \rightarrow A_Q(\mathbb{R}^n)$ is Dirichlet minimizing, and $F|_{S^1} = \sum_{i=1}^{Q-1} \llbracket f_i \rrbracket+\llbracket 0 \rrbracket$, for monotone, continuous maps $f_i: S^1 \rightarrow \Gamma_i$, with $\Gamma_i$ closed Jordan curves with $0\notin \Gamma_i$. Then, is  $0 \in \text{\textnormal{spt}}(F(x))$ for all $x\in \mbb D$?
\end{question}

Note that the answer is \emph{yes} for $Q=1$, by the maximum principle for harmonic functions. Similarly,  for $n=1$, the answer to Question \ref{question:main} is again \emph{yes}, since one may find a continuous (and therefore Sobolev, since $F$ will be affinely approximatable almost everywhere, with $|DF|$ square integrable) selection for $F$, one of whose traces is zero on the boundary.

 These facts leads to the question for higher $Q$--and is an \emph{a priori} type of degeneracy that can occur for certain types of sequences that could have occurred in our existence proof. The answer to this question also illustrates just how weak the maximum principal (as in Section 3.2 of \cite{revisit}) for multiple-valued functions actually is.

Our counter example is in the case $Q=2$ and $n=2$, and our proof is unique to this case. However, simple examples show that this example may be extended to all $Q \geq 2$, provided $n > 1$.

The key is to look instead for zero average multiple-valued functions on $D$ that admit a selection on $S^1$. The following notation will be useful in proving an equivalence between the two concepts.

\begin{notation}
\label{notation:sums}
For $f: \mbb D\rightarrow \mathbb{R}^n$ and $H \in W^{1,2} (\mbb D ,  A_2(\mathbb{R}^n))$, we denote by $H+f$ the function $\llbracket h_1 + f\rrbracket + \llbracket h_2 + f \rrbracket$, where $h_1$ and $h_2$ is a measurable selection for $H$. Note that $H+f$ is independent of the selection chosen.
\end{notation}

Next we recall a result of \cite{higher}:

\begin{thm}
\label{thm:mins}
Let $V\subset \mathbb{C}^\mu \times \mathbb{C}^\nu$ be an irreducible holomorphic variety which is a $Q$-to-one cover of the ball $B_2\subset \mathbb{C}^\mu$ under the orthogonal projection. Then there is a Dirichlet minimizing $f\in W^{1,2}(B_1,  A_Q(\mathbb R^{2\nu} ) )$ so that graph$(f)=V \cap (B_1 \times C^\mu)$.
\end{thm}

Consider the variety $V=\{ (z,w) \ | \ z^2-\frac 1 4  = w^2 \}$. It is easy to see that $p(z,w) = w^2 - z^2 - \tfrac 1 4$ is irreducible. Notice that for $z\neq \pm \frac 1 2$, $V$ is $2:1$ cover under the orthogonal projection. Therefore,  Theorem \ref{thm:mins}\footnote{In the proof of Theorem \ref{thm:mins}, \cite{higher} actually only requires (using standard notation for currents in Euclidean space) that $\pi_\sharp (\llbracket V  \rrbracket ) = 2 \llbracket B_2 \rrbracket$} gives that the multiple-valued function $z\mapsto \sqrt{ z^2-\frac 1 4}$ is Dirichlet Minimizing over the disk $\mbb D$.

Further, as shown in Section 6, $F$ admits a continuous selection on the boundary $S^1$ of $\mbb D$, since this result follows from the fact that the graph map $z\mapsto (z, \sqrt{z^2-1/4})$ maps $S^1$ to two components. We also notice that $F$ is symmetric--i.e. that if $f_1$ and $f_2$ represent a measurable selection for $F$, then, since $f_1+f_2$ is harmonic and zero on the boundary, we must have $f_1 + f_2 =0$.

Therefore, let $F|_{S^1} = \llbracket f \rrbracket + \llbracket - f \rrbracket$. With a slight abuse of notation, let $f$ also denote the harmonic extension of $f$ to the disk $\mbb D$. We finally check that, since $F$ is Dirichlet minimizing, $\tilde F = F+f$ defined as in Notation \ref{notation:sums} is also Dirichlet minimizing for its boundary data.

This follows immediately from the following proposition.

\begin{prop}
Let $H \in W^{1,2} (\mbb D ,  A_2(\mathbb{R}^n))$ be Dirichlet minimizing and suppose that $H|_{S^1} = \llbracket H_1 \rrbracket + \llbracket H_2\rrbracket$ for $H_1, \ H_2 : S^1 \rightarrow \mathbb{R}^n$ continuous. Let $h$ denote the harmonic extension of $H_1$ to $\mbb D$. Then:

\begin{enumerate}
\item if $H_1 = -H_2$, then $\text{\textnormal{Dir}}(H + h, \mbb D) =\text{\textnormal{Dir}}(H, D) +2\text{\textnormal{Dir}}(h, \mbb D)$
\item if $H_2 = 0$, then $\text{\textnormal{Dir}}(H - \frac{h}{2}, \mbb D) =\text{\textnormal{Dir}}(H, D) -2\text{\textnormal{Dir}}( \frac{h}{2}, \mbb D)$
\end{enumerate}
\end{prop}

\begin{proof}[Proof of Proposition]

Denote for two $m\times n$ matricies $(a_{ij})$ and $(b_{ij})$ the matrix inner product:

\begin{equation*}
( (a_{ij} ): (b_{ij}) ) = \sum_{i,j} a_{ij}\overline{b_{ij}}.
\end{equation*}

\noindent Note that the norm induced by this inner product is the Hilbert-Schmidt norm.

Let $h_1$ and $h_2$ denote a measurable selection for $G$. In case (1), properties of the inner product above imply that at any point where $G$ is affinely approximatable (which constitutes a full measure set of $\mbb D$) we have:

\begin{equation*}
||Dh_1 + Dh || ^2 + ||Dh_2 + D h||^2 =  ||DH||^2 + 2||Dh||^2 + 2(Dh_1 + Dh_2 : Dh) .
\end{equation*}

\noindent But, $ 2(Dh_1 + Dh_2 : Df) =0$ since $h_1 + h_2 = 0$, and this gives the result.

For case (2):

\begin{equation*}
||Dh_1 - \frac{Dh} 2  || ^2 + ||Dh_2 -\frac{Dh} 2||^2 =  ||DH||^2 + 2||\frac{Dh} 2||^2 - 2(Dh_1 + Dh_2 :\frac{Dh} 2) .
\end{equation*}

\noindent In this case, though, since $h_1+h_2$ is harmonic and has the same boundary data as $h$ it must be the case that  $h_1+h_2 = h$, so:

\begin{equation*}
 2(Dh_1 + Dh_2 :\frac{Dh} 2) = 2(Dh,\frac{Dh}2) = 4 ||\frac{Dh}{2}||^2.
 \end{equation*}

 \noindent Which implies (2).

\end{proof}

Now we prove $\tilde F$ is Dirichlet minimizing. (1) of the above proposition guarantees:

\begin{equation*}
\text{\textnormal{Dir}}(\tilde F, \mbb D) = \text{\textnormal{Dir}} (F+f, \mbb D) = \text{\textnormal{Dir}}(F,\mbb D) + 2 \text{\textnormal{Dir}}(f, \mbb D)
\end{equation*}

Let $G \in W^{1,2} (\mbb D ,  A_2(\mathbb{R}^n))$ be Dirichlet minimizing and suppose that $G|_{S^1} =\tilde F_{S^1}=\llbracket 2f\rrbracket + \llbracket 0\rrbracket$. Then case (2) of the proposition implies:

\begin{equation*}
 \text{\textnormal{Dir}} (G-f, \mbb D) = \text{\textnormal{Dir}}(G,\mbb D) - 2 \text{\textnormal{Dir}}(f, \mbb D).
\end{equation*}

\noindent So, if

\begin{equation*}
 \text{\textnormal{Dir}}(G,\mbb D) < \text{\textnormal{Dir}}(\tilde F, \mbb D)
 \end{equation*}

 \noindent then:

\begin{equation*}
\text{\textnormal{Dir}}(G,\mbb D) < \text{\textnormal{Dir}}(F, \mbb D) + 2\text{\textnormal{Dir}}(f,\mbb D)
\end{equation*}

\noindent which implies:

\begin{equation*}
\text{\textnormal{Dir}}(G-f,\mbb D)=\text{\textnormal{Dir}}(G,\mbb D)- 2\text{\textnormal{Dir}}(f,\mbb D) < \text{\textnormal{Dir}}(F, \mbb D) .
\end{equation*}

\noindent However, this last inequality is impossible since $F$ is minimizing, so $\tilde F$ is Dirichlet minimizing.

 However, since $F$ admits no global selection (due to its connecting branch points), $\tilde F$ also admits no such selection, and therefore $0\notin \text{\textnormal{spt}}(\tilde F(x))$ for some $x\in \mbb D$, so $\tilde F$ is a counterexample to Question \ref{question:main}.

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


I don't use enough new products nowadays to justify doing monthly favourites, so I thought I'd begin doing seasonal favourites. As Spring ends today and Summer begins, I thought it would be the perfect time to post all about my Spring favourites. It wasn't intentional for everything to be beauty related in my Spring favourites but it just so happens that's the way it's turned out.
I'd heard many great things about the Giorgio Armani Si Eau de Parfum, and when I received the Eau de Toilette as part of a sampling community, I was hoping I would love it. Obviously I did as it is in this post! This smells so incredible, it lasts for almost the whole day and as you can probably tell, the bottle is almost empty. This perfume bottle is also one of the most simplistic yet beautiful bottles I've ever seen.
I have been using the Bondi Sands Self Tan Foam for 3 years now. I first used it in Light/Medium when I wasn't confident with tanning, but now I am I use the Ultra Dark as it is the perfect olive tan shade. I'm not really a regular fake tanner, or rather I haven't been in the past, but I've been using this tan every Thursday since the end of April because I'm obsessed with being the amazing tan shade it produces. It is such a great product, especially when combined with the Bondi Sands Gradual Tanning Milk, which I need to repurchase!
This is the best mascara I've used in a while, maybe ever, and it's only a mini. I need to purchase this as a full size mascara! It adds both length and volume to my lashes, and curls them perfectly. I know many people say drugstore mascara's do the same job but I've always found high end mascara's much better for achieving the results I want.
I've featured this lip product on my blog many times, and that is because I love it so much. It's been perfect to just pop on when I don't want to put lipstick on or I don't have enough time and I need a quick fix. It leaves my lips feeling moisturised, and it smells great too!
I've had the Urban Decay Naked Basics Palette since it was first released, my Mom kindly bought it for me and I've been using it ever since. I'm usually a fan of a shimmer eye but every so often I dip back into this palette and remember how much I love it, especially because it's so easy to pop in your makeup bag! I posted about the look I've been doing using this palette in my everyday spring makeup post, which you can find here.
What have been your favourites this Spring?


text:
Explore Uttar Pradesh- It has its own beauty and charm
Located in the northern India, Uttar Pradesh is often tagged as 'The heartland of India' due to its geographical location. But if we take a deep look, it is indeed the heart of India. Be it culture, history or tourism, UP is probably the most shining star in the map of India. Without a mention of UP, no chapter of India can be completed.
Holding the pride of being the birthland of Lord Ram, Krishna, and Buddha and housing the most splendid monuments of India, UP is a tourist's hub in the country that offers an unending list of historical places in Uttar Pradesh. After serving as the origination point of Hinduism and Buddhism in ancient history, the state witnessed the Muslim rule at its peak in medieval history.
Thus today, it is an unmatched fusion of different cultures packed with man-made architectural marvels as well as natural wonders. Set in the lap of an awe-inspiring landscape of Himalaya, UP has treasured a rich history within. So, let's explore the heartland of India and the history associated with it.
Historical places in Uttar Pradesh
–source-
Even if you ask an alien what he knows about UP, the first thing he would say Agra. Lucknow is of course the state capital but when it comes to tourism, Agra rules, not only in UP but in whole India. And you obviously know the reason. After all, it is the home to one of Seven Wonders of the World, Taj Mahal. Being the sneak peak of architectural style, Taj Mahal allures the millions of worldwide visitors monthly.
In front of the charm and romantic history of Taj Mahal, the other monuments of Agra get eclipsed sometimes. When you'll see the other monuments, you'll wonder which one to be ranked first and which one to be last. Akbar's Tomb, Agra Fort, and Fatehpur Sikri, all built by Akbar, are the other major tourist attractions in Agra that render the sheer brilliance of architecture.
Being the royal capital of Mughals for more than 100 years, Agra is today entwined with the stellar monuments constructed in Hindu-Muslim architectural style. It boasts many historical places in Uttar Pradesh. Being the home to three UNESCO World Heritage sites, Agra is an absolute delight for the history and architectural lovers. If you want to learn about Mughal history, architecture, culture, and governance, head over to Agra.
Soaked into the vibes of holy love, Mathura has its own definition of tourism. Once you have landed in this holy land of Lord Krishna, you'll feel inner peace and a love for all human beings. The vibrant life of Mathura always entices me the most. Extremely rich in history and even more vivid in lifestyle, Mathura should definitely be in your must-to-visit list.
Located on the bank of sacred Yamuna River and packed with many old age temples, Mathura is one of seven holy cities of India. The city was also a major center of Buddhism and had 20 monasteries but like other Indian cities, it too faced some destructive invasions that left the city in ruins.
However, now the Hindu pilgrimage spots have been renovated. Krishna Janmabhoomi Temple, Mathura Museum, Kusum Sarovar, Dwarkadish Temple, and Goverdhan Hill are a few must-visit places of Mathura. On your visit to Mathura, don't forget to visit Vrindavan, a small town near Vrindavan. It is the gist of Mathura.
If you want to see the innocence and pure devotion of human beings, no place can be better than Varanasi. Glorified by the soaring waves of Ganga and many glorious temples, Varanasi is the oldest city in India according to the Vedas. Not only this, the city is one of very few places of the world that has been continuously inhabited since thousands of thousands years ago.
The abode of Lord Shiva is mainly known for prehistoric Kashi Vishwanath Temple and the majestic Ganga River but it has a lot more to offer you. This is a place where Indian culture, religion, education, art, and history found their root in a real sense.
Varanasi is one of the most popular historical places in Uttar Pradesh. Being a pilgrimage spot, it's all about spirituality and divinity. But if you want to have a thrilling experience in Varanasi, head over to Chunar Fort. The intriguing and abnormal tales associated with the fort would really make your day.
About 10 km from Varanasi, Sarnath is one of the most revered pilgrimage places of Buddhist but is a must-visit place for all beliefs. The best thing about visiting Sarnath is that you can spend some quality time amid the transcendent serenity of the place.
This place is away from the hustle and bustle of urban life. Dotted with many consummate Stupas and temples, Sarnath is believed to have had existed since 2500 years ago. According to legend, Gautam Buddha preached his first sermon here in Sarnath that's why this is held in high reverence in Buddhism.
The great emperor Ashoka was an ardent follower of Buddhism, so the most of Stupas are credited to be built by him. Chaukhandi Stupa, Ashoka Pillar, Sarnath Museum, Dharmarajika Stupa, and Dhamek Stupa are the real beauty of Sarnath.
As we read the word 'Jhansi', the first thing that hits the mind is the name of 'Rani Lakshmi Bai'. This city is predominantly known for its fearless queen Lakshmi Bai and her heroic deeds that can instil a feel of self-respect in any lady.
Earlier, Jhansi was not a big name in Indian history but when Rani Lakshmi Bai came in the regime, she eclipsed the entire history and set her own definition of heroism and governance. Jhansi was the home of Maratha Empire before it was captured by the British during the first war of Independence in 1857. The sublime aura, moving history, and the unmatched architecture of Jhansi are the reasons that entice the millions of visitors.
The rundown forts of Jhansi would surely evoke the feel of patriotism in you and the heroism of Rani Lakshmi Bai will touch your heart. Jhansi Fort, Rani Mahal, Maharaja Gangadhar Rao ki Chhatri, Mahalakshmi Temple, and Government Museum are the major catchment of eyes in Jhansi and historical places in Uttar Pradesh.
Tagged as 'The city of Nawabs', the state capital has really a touch of royalty in everything. Even today, you can feel the grandeur of Nawabi style in the language dialect, attires, culture, cuisines and of course historical monuments of Lucknow.
Located on the bank of Gomati River, Lucknow was also once ruled by the Mughals and the remnants are still seen. Being home to many striking monuments, the city is extremely rich in culture and literature.
Here, culture, architecture, art, music, history, and politics flourished altogether. Maybe, this is the reason that today it holds the honor of being the state capital. Bada Imambara, Chhota Imambara, Hazratganj Market, Lucknow Zoo, Rumi Darwaza, and clock Tower are some of the noticeable places of Lucknow. When you visit Lucknow, don't forget to go on a shopping spree in Chowk Market.
Here comes my favorite city in UP. I'm emotionally attached with this city. Anyways, earlier known as Prayag, Allahabad is yet another pilgrimage spot of Hindus in India. If you have enjoyed Varanasi, the mighty waves of Ganga will surely drag you to Allahabad too because Varanasi and Allahabad are known for the same kind of tourism.
Like Varanasi, Allahabad also gave root to India to flourish. The city has been existed even before the era of Lord Ram and holds a very comprehensive description in Ramayana. The Ganga River of Allahabad is considered to be the most pious in India because here it meets River Yamuna and Saraswati. This confluence is known as 'Triveni Sangam' which is the central reason of Allahabad's tourism.
People from all over the country flock Allahabad to take a dip in Triveni Sangam. Apart from this, Allahabad Fort, Aanand Bhawan, Moti Mahal, Khusro Bagh, and Allahabad Museum are the major tourist attractions of the city. Allahabad is also hugely known for the 'Kumbh Mela' that takes place once in every 12 years.
Here comes one of the most popular historical places in Uttar Pradesh. Being the birthland of Lord Ram, Ayodhya is a revered place for Hindus. Packed with many significant temples, it is a great place to take a plunge into the sea of devoutness. The spellbinding waves of Saryu River and the enigmatic aura of spirituality can urge anybody to admire the beauty of this pleasing place.
But I feel today, the city and its tourism are not as famous as it should be. Today Ayodhya has become just a political agenda over Ram Mandir and Babar Masjid. However, millions of devotees visit the city annually. Except for Ram Janmabhoomi Temple, there are many other tourist spots too.
Best time to visit UP
Uttar Pradesh can be visited all-year round. The state is mainly known for spiritual and historical tourism, so no matter when you visit, it remains almost same in all seasons. UP receives a heating summers, moderate rainfall, and extremely chilly winters.
In winters, the temperature can go up to 4-5°C and in summers, it can rise up to 45°C. So, plan your trip accordingly. Like other states and Indian tourist destinations, visiting Uttar Pradesh between the months of September to November is also considered the ideal time.
During this time, weather remains pleasing and also some of the most popular festivals like Dussehra, Nav Ratra, and Diwali fall in between.So, this festive aura accentuates the allure of the state millions times more.
How to reach UP
Uttar Pradesh is a vital state of India from every aspect. Be it education, art & culture, politics, economics, tourism, or technology, the state has always contributed immensely to the country. So today, it is highly developed and accessible from all part of the country and with all modes of transportation. Having a top-class airways service, UP has four domestic airports namely Lucknow, Kanpur, Varanasi, and Agra that connect it to the other states and cities of India.
If we talk about railways, this is undoubtedly the best way to explore entire UP without any hassle. Uttar Pradesh is well-connected by the rail network throughout the country. From anywhere and anytime, you can get a train going to or going from Uttar Pradesh. Now comes the road network. The all major cities of UP are linked with each other through the national highways. And from a particular city, you can easily get the local vehicles to explore the tourist places of the city.
From numerous magnificent historical monuments to countless holy pilgrimage spots and from colorful culture to the modern-age vibrancy, UP is undoubtedly a jewel in the crown of India. Adorned with some of the most revered and splendid cities of country, it is a place that you can never get enough of. This article is just a glimpse of historical places in Uttar Pradesh, the state has much more to offer you. So, pack your bags.
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redpajama_set_name: RedPajamaCommonCrawl


text:
THE DIGITAL TRAVEL GUIDE FOR FOOTBALL FANS
FA Cup semi-finalists in 2014, League Cup semi-finalists in 2015, Sheffield United haven't held a regular spot in the top flight for decades.
Formed, as rivals Wednesday, by a local cricket club at the city's Adelphi Hotel under the auspices of Sir Charles Clegg, United have been based at Bramall Lane since their foundation year of 1889.
Bramall Lane/Paul Martin
Within a decade, United had been crowned league champions and FA Cup winners, Walter Bennett the prolific, bullish centre-forward who would soon lose his life in a mining tragedy.
Hero of United's second cup win was legendarily large goalkeeper William 'Fatty' Foulke, whose saves kept his club in the replayed final against Southampton in 1902.
Club secretary-cum-manager John Nicholson was still in charge when the Blades won a third trophy in 1915 – the so-called Khaki Cup Final against Chelsea – and still in charge ten years later when United beat Cardiff at Wembley. This loyal club servant of 33 years and four FA Cup wins was killed by a lorry in 1932 when running for a train to see his beloved United play at Aston Villa.
The Blades never recaptured the glory days either side (and during) World War I. After a first relegation in 1934, United only flitted back to the top flight intermittently, before a longer run in the 1960s and 1970s.
Sheaf House/Paul Martin
Stalwarts such as goalkeeper Alan Hodgkinson, defender Len Badger and winger Alan Woodward notched some 500 United games each and became recognised names in this early TV era as the Blades spent 12 seasons in the First Division, taking on Revie's Leeds, Shankly's Liverpool and Busby's Manchester United. Most of all, there was Tony Currie, a mercurial presence in the midfield, a classic 1970s' maverick whose number of England caps barely reflected his ability. Today is he commonly recognised as United's best-ever player.
Currie played at Bramall Lane until 1976, Woodward until 1978, by which time United were adrift of the top flight. They would even sink as low as the fourth. Rare bright moments were provided by Argentine Alejandro ('Alex') Sabella, signed by silver-tongued manager Harry Haslam.
Paying for Bramall Lane's new South Stand forced the sale of flair players Currie and Sabella, and few managers lasted long. The coaching assistant Haslam brought with him from Luton, Montevideo-born Danny Bergara, gave way to Dave 'Harry' Bassett. Picking up strikers Tony Agana from his previous club, Watford, and Brian Deane from Doncaster, Bassett masterminded a two-season jump from third flight to first.
Once back in the First Division, Bassett's men didn't win a game until Christmas but rallied in the new year to avoid relegation. Deane's goals then helped United finish ninth the following season but he was soon sold to Leeds. Blunted, the Blades succumbed to a late turnaround by Chelsea in the decisive last fixture of the 1993-94 campaign.
Relegation consigned United to 12 years in the Championship. Locally born Neil Warnock led the club back to the Premier League in 2006, for one short season – central midfield lynchpin Phil Jagielka duly opted for a move to Everton after heroic efforts to keep the Blades up, including keeping a clean sheet in goal for most of a second half against Arsenal.
With Warnock and Jagielka gone, and owner Kevin McCabe expanding his worldwide property and football empire, United sank from the first to third flights in five seasons.
With Saudi prince Abdullah bin Mosaad Al Saud buying taking over the club's debts in 2013, the Blades made the League One play-offs in 2015, a crazy 5-5 draw meaning that United had lost eight straight times at the sudden-death stage.
A year earlier, Nigel Clough's United had starred in an eight-goal thriller at Wembley, losing 5-3 to Hull in the FA Cup semi-final. In January 2015, 30,000 packed into Bramall Lane to see teenage substitute Che Adams score his first goals for the club, two in three minutes, to push Spurs to the wire for a place in the League Cup final.
Without promotion though, United brought in Nigel Adkins for Clough for the 2015-16 campaign. Despite goals from returning Billy Sharp, a mid-table finish was the end result. Former United defender Chris Wilder then stepped in as manager and, with Sharp still prolific, 2016-17 saw The Blades embark on two 15-game unbeaten runs, culminating in winning the League One crown in front of nearly 27,000 at Bramall Lane.
Bramall Lane is etched into football history. Scene of the game's first floodlit match, it remains the world's oldest stadium still staging professional fixtures.
Originally a cricket ground, Bramall Lane held its first football match in 1862, between Sheffield FC and Hallam. Home to both Sheffield United and Wednesday, Bramall Lane witnessed a friendly game with experimental lighting in 1877 and a handful of England internationals from the 1880s onwards.
Until the development of the larger Hillsborough after Wednesday moved out in 1887, Bramall Lane was also a popular choice of FA Cup semi-final venue. For major tournaments, though, it was always Hillsborough.
The building of the South Stand in 1975 all but bankrupted the club. Forced to sell the likes of Tony Currie, United were relegated and Bramall Lane would not witness top-flight football for 15 years. With the Premier League revenues of the early 1990s, the club built the John Street Stand, holding executive boxes. Its current name is the Duchy Homes Family Stand.
With its corners also developed, one containing an enterprise centre, and a hotel opened, Bramall Lane is modern, impressive and, as opposed to Hillsborough, close to the city centre, an easy stroll from the train station. Capacity is an all-seated 32,600.
The club shop and ticket office are behind the Topspring South Stand. The steep-seated Redbrik Stand houses away fans (turnstiles 13-23), Kennedy's Kop is the home end.
Sheffield United transport/Paul Martin
Bramall Lane is close to the city centre, less than 15 minutes' walk from Sheffield train station. Exit the station left, along the main road. After less than 5min, this road forks – take right-hand Shoreham Street, past BBC Radio Sheffield, and keep going for 10min.
If you're coming from the city centre, four stops from central Cathedral on the blue and purple tramlines, Granville Road by Sheffield College is reasonably close. Once you arrive, walk up to the main junction, away from the college, keeping McDonald's to your left, and keep heading in the same direction. Shoreham Street is about 300 metres ahead.
Sheffield United tickets/Paul Martin
Tickets go on sale four to six weeks before every home game, sold to away fans two to four weeks in advance. With a capacity of 30,000-plus and average gates around 20,000, availability is not a problem – and at present there's no danger of a Steel City derby, not in the league at least.
The main ticket office (Mon-Fri 9am-5pm, non-matchday Sat 9am-2pm, match days 9am-half-time then 30min after match) is behind the Topspring South Stand by the car park on Cherry Street. A booking fee of £1 is charged for telephone (0114 253 7200) and online sales.
Blades Superstore/Paul Martin
Pricing is in two types of category, premium A dearer by £2. On the day for a category B game, you'll pay £20 in the Kop, £24 in the South Stand and John Street family enclosure, £22 in the South Corner and away end. Advance purchases are £2 cheaper. Over 60s pay around £20, students/young adults £10-£17 and under-18s £9-£15.
The Blades Superstore (Mon-Fri 9am-5pm, non-matchday Sat 9am-4pm, matchdays 9am-kick-off, Sat 30min after match) is behind the South Stand. More unusual gifts include masks modelled after bearded defender John Brayford, tape measures and latte mugs.
Sheffield United museum/Paul Martin
Museum & tours
Club historian John Garrett takes tours (£6/£4 reductions) of the stadium and Legends of the Lane museum on occasional match days. See www.sufc.talent-sport.co.uk or call 0871 995 1889 for details.
With Bramall Lane so close to the train station, away fans can drink nearby. Or, in fact, at the station itself: the Sheffield Tap at its two dozen handpull ales/continental lagers can accessed by platform 1b or Sheaf Street alongside the station building. Set in the station's Edwardian dining rooms, the Tap is fiercely independent and proud of its brews – no away colours please.
The Globe/Paul Martin
Close by, just the other side of Sheaf Square, The Globe welcomes both sets of fans if not rowdy, with decent pub grub and live sports on TV.
Another option is Sheaf Island, (not to be confused with Sheaf House, see below), a large Wetherspoons set in an early Victorian brewery. You'll find it on Ecclesall Road, the other side of a large Waitrose, some 7min walk from the ground.
Home fans gather nearer the ground at the Golden Lion on Alderson Road, with TV sports, darts and a small beer garden, and, on Bramall Lane itself, Sheaf House and the partisan Cricketers Arms.
Sheffield Tap/Paul Martin
Sheaf Island/Paul Martin
Golden Lion/Paul Martin
Cricketer's Arms/Paul Martin
Sheffield FC