text:
\section{Main result}\label{sec:main}

\subsection{Configuration}

We consider $L+1$ vertical planes, $L\ge 1$, labeled by integers $l \in [1,
L+1]$.  Up to horizontal rotations, we assume that these planes are all
parallel to the $xz$-plane, which we identify to the complex plane $\mathbb{C}$, with
the $x$-axis (resp.\ $z$-axis) corresponding to the real (resp.\ imaginary) axis.
We use the term ``layer'' for the space between two adjacent parallel planes.
So there are $L$ layers.

We want $n_l \ge 1$ catenoid necks on layer $l$, i.e.\ between the planes $l$
and $l+1$, $1 \le l \le L$.  For convenience, we adopt the convention that $n_l
= 0$ if $l < 1$ or $l > L$, and write $N = \sum n_l$ for the total number of
necks.  Each neck is labeled by a pair $(l,k)$, where $1 \le l \le L$ and $1
\le k \le n_l$.

To each neck is associated a complex number $q_{l,k} \in \mathbb{C}^\times = \mathbb{C}
\setminus\{0\}$, $1 \le l \le L$, $1 \le k \le n_l$.  Then the position of the
necks are prescribed at $\ln q_{l,k} + 2m\pi{\rm i}$, $m \in \mathbb{Z}$.  Recall that the
$z$-axis is identified to the imaginary axis of the complex plane $\mathbb{C}$, so the
necks are periodic with period vector $(0,0,2\pi)$.  Note that, if we multiply
$q_{l,k}$'s by the same complex factor $c$, then the necks are all translated
by $\ln c \pmod{2\pi{\rm i}}$.  So we may quotient out translations by fixing
$q_{1,1} = 1$.

Moreover, each plane has two ends asymptotic to vertical planes.  We
label the end of plane $l$ that expands in the $-x$ (resp.\ $x$) direction by
$0_l$ (resp.\ $\infty_l$).  To be compatible with the language of graph theory
that were used for gluing saddle towers~\cite{saddle1}, we use
\[
	\graph{H} = \{\eta_l \colon 1 \le l \le L+1, \eta \in \{0,\infty\} \}
\]
to denote the set of ends.  When $0_l$ is used as subscript for
parameter $x$, we write $x_{l,0}$ instead of $x_{0_l}$ to ease the notation;
the same applies to $\infty_l$.

To each end is associated a real number $\dot\theta_h$, $h \in \graph{H}$.  They
prescribe infinitesimal changes of the directions of the ends.  More
precisely, for small $\tau$, we want the unit vector in the direction of the
end $h$ to have a $y$-component or the order $\tau \dot\theta_h +
\mathcal{O}(\tau^2)$.

\begin{remark}\label{rmk:normalize}
	Multiplying $\dot\theta$ by a common real constant leads to a
	reparameterization of the family.  Adding a common real constant to
	$\dot\theta_{l,0}$ and subtracting the same constant from
	$\dot\theta_{l,\infty}$ leads to horizontal rotations of the surface.
\end{remark}

In the following, we write
\[
	q = (q_{l,k})_{\substack{1 \leqslant l \leqslant L\\1 \leqslant k \leqslant n_l}}
	\quad\text{and}\quad
	\dot\theta = (\dot\theta_h)_{h \in \graph{H}}.
\]
Then a \emph{configuration} refers to the pair $(q,\dot\theta)$.

\subsection{Force}

Given a configuration $(q,\dot\theta)$, for $1 \le l \le L+1$, let $\psi_l$ be the
meromorphic $1$-form on the Riemann sphere $\hat\mathbb{C}$ with simple poles at
$q_{l,k}$ with residue $-c_l$ for each $1 \le k \le n_l$, at $q_{l-1,k}$ with
residue $c_{l-1}$ for each $1 \le k \le n_{l-1}$, at $0$ with residue
$\dot\theta_{l,0}$, and at $\infty$ with residue $\dot\theta_{l,\infty}$.  More
explicitly,
\[
	\psi_l =  \Big(\sum_{k = 1}^{n_l} \frac{-c_l}{z-q_{l,k}}
		+ \sum_{k = 1}^{n_{l-1}} \frac{c_{l-1}}{z-q_{l-1,k}}
		+ \frac{\dot\theta_{l,0}}{z}
	\Big) dz.
\]
By the Residue Theorem, it is necessary that
\begin{equation}\label{eq:resc}
	- n_l c_l + n_{l-1} c_{l-1} + \dot\theta_{l,0} + \dot\theta_{l,\infty} = 0,
	\quad 1 \le l \le L+1.
\end{equation}
A summation over $l$ yields
\begin{equation}\label{eq:Theta1}
	\Theta_1 = \sum_{h \in \graph{H}} \dot\theta_h = 0
\end{equation}
If~\eqref{eq:Theta1} is satisfied, the real numbers $c_l$ are determined
by~\eqref{eq:resc} as functions of $\dot\theta$.

\begin{remark}
  We may replace $\dot\theta$ by the parameters $(c_l, \dot\theta_{l+1,0} -
  \dot\theta_{l,0})_{1 \le l \le L}$.  Then $\theta_{l,0}$'s are defined up to
  an additive constant (corresponding to a rotation), $\theta_{l,\infty}$'s are
  determined by~\eqref{eq:resc}, and \eqref{eq:Theta1} is automatically
  satisfied.  To quotient out reparameterizations of the family, we may assume
  that $c_l = 1$ for some $1 \le l \le L$.
\end{remark}

We define the force $F_{l,k}$ by
\begin{equation}\label{eq:force0}
	F_{l,k}

	= \Res{\frac{\psi_l^2 + \psi_{l+1}^2}{2}\frac{z}{d z}}{q_{l,k}}.
\end{equation}
Or, more explicitly,
\begin{multline}\label{eq:force}
	F_{l,k} =
	\sum_{1 \le k \ne j \le n_l} \frac{2 c_l^2 q_{l,k}}{q_{l,k}-q_{l,j}}
	- \sum_{1 \le j \le n_{l+1}} \frac{c_l c_{l+1} q_{l,k}}{q_{l,k} - q_{l+1,j}}
	- \sum_{1 \le j \le n_{l-1}} \frac{c_l c_{l-1} q_{l,k}}{q_{l,k} - q_{l-1,j}} \\
	+ c_l^2 + c_l (\dot\theta_{l+1,0} - \dot\theta_{l,0}).
\end{multline}

\begin{remark}
	In~\cite{li2012}, the force had different expressions depending on the parity
	of $l$.  For $l$ odd, the force was given by (using our notation)
	\begin{multline}\label{eq:kevinodd}
		F_{l,k} =
		\sum_{1 \le k \ne j \le n_l} c_l^2 \frac{q_{l,k}+q_{l,j}}{q_{l,k}-q_{l,j}}
		-\sum_{1 \le j \le n_{l+1}} \frac{c_l c_{l+1} q_{l,k}}{q_{l,k} - q_{l+1.j}}
		-\sum_{1 \le j \le n_{l-1}} \frac{c_l c_{l-1} q_{l-1,j}}{q_{l,k} - q_{l-1.j}}\\
		+c_l(\dot\theta_{l,\infty}+\dot\theta_{l+1,0}).
	\end{multline}
	Then $\eqref{eq:kevinodd} - \eqref{eq:force}$ yields
	\[
		-n_l c_l^2 + n_{l-1}c_lc_{l-1}+c_l(\dot\theta_{l,0}+\dot\theta_{l,\infty}) = 0.
	\]
	For $l$ even, the force was given by
	\begin{multline}\label{eq:kevineven}
		F_{l,k} =
		\sum_{1 \le k \ne j \le n_l} c_l^2 \frac{q_{l,k}+q_{l,j}}{q_{l,k}-q_{l,j}}
		-\sum_{1 \le j \le n_{l+1}} \frac{c_l c_{l+1} q_{l+1,j}}{q_{l,k} - q_{l+1.j}}
		-\sum_{1 \le j \le n_{l-1}} \frac{c_l c_{l-1} q_{l,k}}{q_{l,k} - q_{l-1.j}}\\
		-c_l(\dot\theta_{l,0}+\dot\theta_{l+1,\infty}).
	\end{multline}
	Then $\eqref{eq:kevineven} - \eqref{eq:force}$ yields
	\[
		-n_l c_l^2 + n_{l+1}c_lc_{l+1}-c_l(\dot\theta_{l+1,0}+\dot\theta_{l+1,\infty}) = 0.
	\]
	So, independent of the parity of $l$, both expressions are equivalent to ours.
\end{remark}

\begin{remark}[Electrostatic interpretation]
	Adding~\eqref{eq:kevinodd} and~\eqref{eq:kevineven} yields
	\begin{multline*}
		2F_{l,k} =
		\sum_{1 \le k \ne j \le n_l} 2 c_l^2 \frac{q_{l,k}+q_{l,j}}{q_{l,k}-q_{l,j}}\\
		-\sum_{1 \le j \le n_{l+1}} c_l c_{l+1} \frac{q_{l,k} + q_{l+1,j}}{q_{l,k} - q_{l+1.j}}
		-\sum_{1 \le j \le n_{l-1}} c_l c_{l-1} \frac{q_{l,k} + q_{l-1,j}}{q_{l,k} - q_{l-1.j}}\\
		+c_l(\dot\theta_{l,\infty} - \dot\theta_{l,0} - \dot\theta_{l+1,\infty} + \dot\theta_{l+1,0}).
	\end{multline*}
	Note that
	\[
		\frac{a+b}{a-b} = \coth\frac{\ln a - \ln b}{2}
		= \sum_{m \in \mathbb{Z}} \frac{2}{\ln a - \ln b - 2m\pi{\rm i}}.
	\]
	So
	\begin{multline*}
		F_{l,k} =
		\sum_{0 \ne m \in \mathbb{Z}} \frac{2 c_l^2}{2m\pi{\rm i}}
		+\sum_{\substack{m \in \mathbb{Z}\\1 \le k \ne j \le n_l}} \frac{2 c_l^2}{\ln q_{l,k} - \ln q_{l,j} - 2m\pi{\rm i}}\\
		-\sum_{\substack{m \in \mathbb{Z}\\1 \le j \le n_{l+1}}} \frac{c_l c_{l+1}}{\ln q_{l,k} - \ln q_{l+1.j} - 2m\pi{\rm i}}
		-\sum_{\substack{m \in \mathbb{Z}\\1 \le j \le n_{l-1}}} \frac{c_l c_{l-1}}{\ln q_{l,k} - \ln q_{l-1.j} - 2m\pi{\rm i}}\\
		+c_l(\dot\theta_{l,\infty} - \dot\theta_{l,0} - \dot\theta_{l+1,\infty} + \dot\theta_{l+1,0}).
	\end{multline*}
	Recall that $\ln q_{l,k} + 2m\pi{\rm i}$ are the real positions of the necks.  So
	this expression has an electrostatic interpretation similar to those
	in~\cite{traizet2002} and~\cite{traizet2008}.  Here, each neck interacts not
	only with all other necks in the same or adjacent layers, but also with
	background constant fields given by $\dot\theta$.
\end{remark}

\begin{remark}[Another electrostatic interpretation]\label{rem:electrob}
	In fact, $\eqref{eq:force}/q_{l,k}$ has a similar electrostatic
	interpretation.  But this time, the necks are seen as placed at $q_{l,k}$.
	Each neck interacts with all other necks in the same and adjacent layers, as
	well as a virtual neck at $0$ with ``charge'' $c_l + \dot\theta_{l+1,0} -
	\dot\theta_{l,0}$.  This is no surprise, as electrostatic laws are known to be
	preserved under conformal mappings (such as $\ln(z)$).
\end{remark}

\subsection{Main result}

In the following, we write $F=(F_{l,k})_{1 \le l \le L, 1 \le k \le n_l}$.

\begin{definition}
	The configuration is \emph{balanced} if $F = 0$ and $\Theta_1=0$.
\end{definition}
Summing up all forces yields a necessary condition for the configuration to be
balanced, namely
\begin{multline*}
	\Theta_2 =\sum_{\substack{1 \le l \le L \\ 1 \le k \le n_l}} F_{l,k}
	= \sum_{1 \le l \le L+1} -\frac{1}{2} \Bigg(
		\Res{\frac{z\psi_l^2}{dz}}{0} + \Res{\frac{z\psi_l^2}{dz}}{\infty}
	\Bigg)\\
	= \sum_{1 \le l \le L+1} \frac{{\dot\theta_{l,\infty}^2 - \dot\theta_{l,0}^2}}{2} = 0.
\end{multline*}

\begin{lemma}\label{lem:rigid}
	The Jacobian matrix $\partial (\Theta_1, \Theta_2)/\partial \dot\theta$ has real
	rank $2$ as long as $c_l \ne 0$ for some $1 \le l \le L$.
\end{lemma}

The assumption of the lemma simply says that the surface does not remain a
degenerate plane to the first order.

\begin{proof}
	The proposition says that the matrix has an invertible minor of size $2
	\times 2$.  Explicitly, we have
	\[
		\frac{\partial (\Theta_1, \Theta_2)}{\partial(\dot\theta_{l,0}, \dot\theta_{l,\infty})}
		=\begin{pmatrix} 1 & 1\\ -\dot\theta_{l,0} & \dot\theta_{l,\infty} \end{pmatrix}.
	\]
	This minor is invertible if and only if the ends $\dot\theta_{l,0} +
	\dot\theta_{l,\infty} \ne 0$.  This must be the case for at least one $1 \le l
	\le L$ because, otherwise, we have $c_l = 0$ for all $1 \le l \le L$.
\end{proof}

\begin{definition}
	The configuration is \emph{rigid} if the complex rank of $\partial F
	/ \partial q$ is $N-1$.
\end{definition}

\begin{remark}
	In fact, the complex rank of $\partial F / \partial q$ is at most $N-1$.  We
	have seen that a complex scaling of $q$ corresponds to a translation of $\ln
	q_{l,k} + 2m\pi{\rm i}$, $m \in \mathbb{Z}$, which does not change the force.  It then
	makes sense to normalize $q$ by fixing $q_{1,1}=1$.
\end{remark}

\begin{theorem}
	Let $(q,\dot\theta)$ be a balanced and rigid configuration.  Then for $\tau >
	0$ sufficiently small, there exists a smooth family $M_\tau$ of complete
	singly periodic minimal surface of genus $g=N-L$, period $(0,0,2\pi)$, and
	$2(L+1)$ Scherk ends such that, as $\tau \to 0$,
	\begin{itemize}
		\item $M_\tau$ converges to an $(L+1)$-sheeted $xz$-plane with singular
			points at $\ln q_{l,k} + 2m\pi{\rm i}$, $m \in \mathbb{Z}$.  Here, the $xz$-plane is
			identified to the complex plane $\mathbb{C}$, with $x$-axis (resp.\ $z$-axis)
			identified to the real (resp.\ imaginary) axis.

		\item After suitable scaling and translation, each singular point opens up
			into a neck that converges to a catenoid.

		\item The unit vector in the direction of each Scherk end $h$ has the
			$y$-component $\tau\dot\theta_h + \mathcal{O}(\tau^2)$.
	\end{itemize}
			Moreover, $M_\tau$ is embedded if
	\begin{equation}\label{eq:embedcond}
		\dot\theta_{1,0} > \cdots > \dot\theta_{L+1,0}
		\quad\text{and}\quad
		\dot\theta_{1,\infty} > \cdots > \dot\theta_{L+1,\infty}.
	\end{equation}
\end{theorem}

\begin{remark}
	$M_\tau$ also depends smoothly on $\dot\theta$ belonging to the local smooth
	manifold defined by $\Theta_1=0$ and $\Theta_2 = 0$.  Up to
	reparameterizations of the family and horizontal rotations, we obtain
	families parameterized by $2L-1$ parameters.  Since we have $2(L+1)$ Scherk
	ends, this parameter count is compatible with the fact that Karcher--Scherk
	saddle towers with $2k$ ends form a family parameterized by $2k-3$
	parameters.
\end{remark}

\section{Examples}

\subsection{Surfaces of genus 0}\label{ssec:g0}

When the genus $g = N-L = 0$, we have $n_l = 1$ for all $1 \le l \le L$, i.e.\
there is only one neck on every layer.  It then makes sense to drop the
subscript $k$.  For instance, the position and the force for the neck on layer
$l$ are simply denoted by $q_l$ and $F_l$.

In this case, if $\Theta_1 = 0$, Equation~\eqref{eq:resc} can be explicitly
solved by
\[
	c_l = \sum_{i=1}^{l} (\dot\theta_{i,0} + \dot\theta_{i,\infty}),\quad 1 \le l \le L,
\]
and the force can be written in the form
\[
	F_{l} =
	- \widetilde{Q}_{l} + \widetilde{Q}_{l-1}
	+ c_l (\dot\theta_{l,\infty} + \dot\theta_{l+1,0}),\quad 1 \le l \le L,
\]
where we changed to the parameters
\[
	\widetilde{Q}_l = \frac{c_{l+1} c_l}{1-q_{l+1}/q_l}, \quad 1 \le l < L,
\]
with the convention that $\widetilde{Q}_0 = \widetilde{Q}_{L} = 0$.  Then the
forces are linear in $\widetilde Q$ and, if $\Theta_2 = 0$, the balance
condition $F=0$ is uniquely solved by
\[
	\widetilde{Q}_l
	= \sum_{i=1}^{l} c_i (\dot\theta_{i+1,0} + \dot\theta_{i,\infty})
	= - \sum_{i=l+1}^{L} c_i (\dot\theta_{i+1,0} + \dot\theta_{i,\infty}),
	\quad 1 \le l < L.
\]
Therefore, if we fix $q_1 = 1$, all other $q_l$, $1 < l \le L$ are uniquely
determined.

Moreover, if the embeddedness condition~\eqref{eq:embedcond} is satisfied, the
sequence $\dot\theta_{l,0} + \dot\theta_{l,\infty}$ is monotonically
decreasing, and changes sign at most once.  Then the sequence $c$ is unimodal,
i.e.\ there exists $1 \le l' \le L$ such that
\[
	0 = c_0 \le c_1 \le \cdots \le c_{l'} \ge \cdots \ge c_L \ge c_{L+1} = 0.
\]
Hence $c$ is positive.  The summands of $\widetilde Q$ also changes sign at most
once, so the sequence $\widetilde Q$ is also unimodal hence positive.  Moreover,
\begin{align*}
	\widetilde{Q}_l
	< \sum_{i=1}^{l} c_i (\dot\theta_{i,0} + \dot\theta_{i,\infty})
	&= \sum_{i=1}^{l} (c_i^2 - c_{i-1}c_i) \le c_l^2 \le c_{l+1}c_l
	& \text{if } l < l',\\
	\widetilde{Q}_l
	< - \sum_{i=l+1}^{L} c_i (\dot\theta_{i+1,0} + \dot\theta_{i+1,\infty})
	&= \sum_{i=l+1}^{L} (c_i^2 - c_{i+1}c_i) \le c_{l+1}^2 \le c_{l+1}c_l
	& \text{if } l \ge l'.
\end{align*}
So $q$ consists of real numbers and $q_{l+1}/q_{l} < 0$ for all $1 \le l <
L$.

We have proved that
\begin{proposition}\label{prop:rigidhandel}
	If the genus $g = N-L = 0$, and $\dot\theta$ satisfies the balancing
	condition $\Theta_1 = \Theta_2 = 0$ as well as the embeddedness
	condition~\eqref{eq:embedcond}, then up to complex scalings, there exist
	unique values for the parameters $q$, depending analytically on $\dot\theta$,
	such that the configuration $(q,\dot\theta)$ is balanced.  Moreover, all such
	configurations are rigid.  If we fix $q_1 = 1$, then $q$ consist of real
	numbers, and we have $q_l > 0$ (resp.\ $<0$) if $l$ is odd (resp.\ even).
\end{proposition}

\subsection{Surfaces with four ends}

When $L=1$, $\Theta_1 = \Theta_2 = 0$ implies that $\theta_{1,0} =
-\theta_{2,\infty}$ and $\theta_{2,0} = -\theta_{1, \infty}$.  Up to
reparameterizations of the family, we may assume that $c_1 = 1$.  It makes
sense to drop the subscript $l$ in other notations, and write $F_k$ for
$F_{1,k}$, $q_k$ for $q_{1,k}$, and $n$ for $n_1$.  The goal of this part is to
prove the following classification result.

\begin{proposition}
	Up to a complex scaling, a configuration with $L=1$ and $n$ nodes must be
	given by $q_k = \exp(2\pi{\rm i} k/n)$, and such a configuration is
	rigid.
\end{proposition}

\begin{proof}
	To find the positions $q_k$ such that
	\begin{equation}\label{eq:balance4ends}
		F_k=\sum_{1 \le k \ne j \le n} \frac{2q_k}{q_k-q_j} - (n-1) = 0,
		\qquad 1 \le k \le n,
	\end{equation}
	we use the polynomial method:  Consider the polynomial
	\[
		P(z) = \prod_{k = 1}^n (z - q_k).
	\]
	Then we have
	\begin{align*}
		P' &= P \sum_{k = 1}^n \frac{1}{z-q_k}\\
		P'' &= P \sum_{k = 1}^n \sum_{1 \le k \ne j \le n} \frac{1}{z-q_j}\frac{1}{z-q_k}
		= 2P \sum_{k = 1}^n \frac{1}{z-q_k} \sum_{1 \le k \ne j \le n} \frac{1}{q_k-q_j}\\
		&= P \sum_{k = 1}^n \frac{n-1}{q_k(z-q_k)}
		= (n-1) P \sum_{k=1}^n \frac{1}{z}\Big(\frac{1}{q_k} + \frac{1}{z-q_k}\Big)
		\qquad \text{by~\eqref{eq:balance4ends}}\\
		&= \frac{n-1}{z} \Big(P' - \frac{P'(0)}{P(0)} P \Big)
	\end{align*}
	For the last equation to have a polynomial solution, we must have $P'(0) =
	0$.  Otherwise, the left-hand side would be a polynomial of degree $n-2$, but
	the right-hand side would be a polynomial of degree $n-1$.

	Consequently, $F_k = 0$ if and only if
	\[
		zP''(z) -(n-1) P'(z) = 0,
	\]
	which, up to a complex scaling, is uniquely solved by
	\[
		P(z) = z^n - 1.
	\]
	So a balanced 4-end configuration must be given by the roots of unity $q_k =
	\exp(2\pi{\rm i} k/n)$, $0 \le k \le n-1$.

	We now verify that the configuration is rigid.  For this purpose, we
	compute
	\[
		\frac{\partial F_k}{\partial q_j} = \begin{dcases}
			2 \frac{q_k}{(q_k-q_j)^2}, & j \ne k;\\
			2 \sum_{1 \le k \ne i \le n} \frac{-q_i}{(q_k-q_i)^2}, & j = k.
		\end{dcases}
	\]
	Note that $\sum_{j=1}^n q_j\partial F_k/\partial q_j = 0$ while
	\[
		q_j\frac{\partial F_k}{\partial q_j} = 2\frac{q_j q_k}{(q_k - q_j)^2}
		= 2\frac{e^{2\pi{\rm i} \frac{j+k}{n}}}{(e^{2\pi{\rm i}\frac{j}{n}} - e^{2\pi{\rm i}\frac{k}{n}})^2}
		\in \mathbb{R}_{<0}
	\]
	when $j \ne k$, so the matrix
	\[
		\frac{\partial F}{\partial q} \operatorname{diag}(q_1, \cdots, q_n)
	\]
	has real entries, has a kernel of complex dimension $1$ (spanned by the
	all-one vector), and any of its principal submatrix is diagonally dominant.
	We then conclude that the matrix, as well as the Jacobian $\partial F /
	\partial q$, has a complex rank $n-1$.  This finishes the proof of
	rigidity.
\end{proof}

\begin{remark}
	The perturbation argument as in the proof of \cite[Proposition
	1]{traizet2002} also applies here, word by word, to prove the rigidity.
\end{remark}

\subsection{Gluing two saddle towers of different periods}

We want to construct a smooth family of configurations depending on a positive
real number $\lambda$ such that, for small $\lambda$, the configuration looks
like two columns of nodes far away from each other, one with period $2\pi/n_1$,
and the other with period $2\pi/n_2$.  If balanced and rigid, these
configurations would give rise to minimal surfaces that look like two Karcher
saddle towers with different periods that are glued along a pair of ends.

\begin{remark}
	The first named author was shown a video suggesting that, when two saddle
	towers are glued into a minimal surface, one can slide one saddle tower with
	respect to the other while the surface remains minimal.  The result of this
	section suggests that this is not possible.
\end{remark}

\begin{proposition}
	For a real number $\lambda > 0$ sufficiently small, there are balanced and
	rigid configurations $(q(\lambda), \dot\theta(\lambda))$ with $L=1$
	depending smoothly on $\lambda$ such that, as $\lambda \to 0$,
	\begin{align*}
		q_{1,k}(0) &= \exp\Big(2\pi{\rm i}\frac{k}{n_1}\Big),
		& 1 \le k \le n_1,\\
		\frac{q_{2,k}}{q_{2,1}}(0) &= \exp\Big(2\pi{\rm i}\frac{k}{n_2}\Big),
		& 1 \le k \le n_2,\\
		\frac{q_{2,1}(\lambda)}{\lambda} &\to \exp({\rm i}\phi)
	\end{align*}
	if and only if $\phi\operatorname{lcm}(n_1, n_2)$ is a multiple of $\pi$.
\end{proposition}

In other words, the construction only works if the configuration admits a
reflection symmetry.

\begin{proof}
	At $\lambda = 0$, the configurations $(q(\lambda),\dot\theta(\lambda))$ and
	$(\lambda q(\lambda), \dot\theta(\lambda))$ converge to balanced
	configurations with 4 ends.  We have seen that these configurations are
	rigid.  Moreover, the forces at $\lambda=0$ do not involve
	interactions across layers.  Hence for $\lambda$ close to $0$, there exist
	unique values for $(q_{l,k})_{l = 1, 2; 2 \le k \le n_l}$, depending smoothly
	on $\dot\theta$, $\phi$, and $\lambda$, such that $(F_{l,k})_{l = 1, 2; 2 \le
	k \le n_l} = 0$.

	Write $G_l = \sum_{k = 1}^{n_l} F_{l,k}$, $l = 1, 2$.  Note that $\operatorname{Re} G_l$ is
	linear in $\dot\theta$.  So, up to reparameterization of the family and
	horizontal rotations, there exist unique values for $\dot\theta$, depending
	smoothly on $\phi$ and $\lambda$, such that $\operatorname{Re} G_l = 0$ and $\Theta_1 =
	\Theta_2 = 0$.

	It remains to find $\phi$ that solves $\operatorname{Im} G_2 = 0$.  Once this is done, we
	would have $\operatorname{Im} G_1=0$ by the Residue Theorem, and the balancedness is proved.
	Note that, as $\lambda \to 0$, we have
	\[
		G_2 = c_1 c_2 \sum_{j=1}^{n_1} \sum_{k=1}^{n_2} \frac{x_{j,k}}{1-x_{j,k}}
		= c_1 c_2 \sum_{m=1}^\infty \sum_{j=1}^{n_1} \sum_{k=1}^{n_2} x_{j,k}^m
	\]
	where
	\[
		x_{j,k} = q_{2,1}\exp\Big(2\pi{\rm i}\frac{k}{n_2} - 2\pi{\rm i}\frac{j}{n_1}\Big).
	\]
	Seen as a power series of $\lambda$, the first non-zero term in $G_2$ is
	\[
		c_1 c_2 n_1 n_2 q_{2,1}^\mu = c_1 c_2 n_1 n_2 \lambda^\mu \exp({\rm i}\mu\phi),
	\]
	where $\mu = \operatorname{lcm}(n_1, n_2)$. So $\operatorname{Im} G_2/\lambda^\mu$ vanishes
	only if $\mu\phi$ is a multiple of $\pi$.

	For the rigidity, we need to prove that the matrix
	\[
		\begin{pmatrix}
			\big(\frac{\partial F_{1,k}}{\partial q_{1,j}}\big)_{2 \le j,k\le n_1} & & \\
			& \big(\frac{\partial F_{2,k}}{\partial q_{2,j}}\big)_{2 \le j,k\le n_2} & \\
			& & \frac{\partial G_2}{\partial q_{2,1}}
		\end{pmatrix}
	\]
	is invertible.  We know that the first two blocks are invertible at
	$\lambda=0$.  By continuity, they remain invertible for $\lambda$
	sufficiently small.  The last block vanishes at $\lambda = 0$, but is
	non-zero (and real) once $\lambda$ increases.  Hence the configuration is
	rigid for sufficiently small $\lambda > 0$.
\end{proof}

\subsection{Surfaces with six ends of type (n,1)} \label{ssec:n1}

In this section, we investigate examples with $L=2$ (hence six ends), $n_1 =
n$, $n_2 = 1$.  Up to a reparameterization of the family, we may assume that
$c_1 = 1$.  Up to a complex scaling, we may assume that $q_{2,1} = 1$.

We will prove that $q_{1,k}$'s are given by the roots of hypergeometric
polynomials.  Let us first recall their definitions.  A \emph{hypergeometric
function} is defined by
\[
	{_2F_1}(a,b;c;z)=\sum_{k=0}^{\infty}\frac{(a)_k(b)_k}{(c)_k}\frac{z^k}{k!}
\]
with $a,b,c\in\mathbb{C}$, $c$ is not a non-positive integer,
\[
	(a)_k=a(a+1)\cdots(a+k-1)=\frac{\Gamma(a+k)}{\Gamma(a)},
\]
and $(a)_0 = 1$.  The hypergeometric function $w={_2F_1}(a,b;c;z)$ solves the
\emph{hypergeometric differential equation}
\begin{equation}\label{eq:ode2f1}
	z(1-z)w''+[c-(a+b+1)z]w'-abw=0.
\end{equation}

If $a=-n$ is a negative integer,
\[
	{_2F_1}(-n, b; c; z) := \sum_{k = 0}^n (-1)^k \binom{n}{k} \frac{(b)_k}{(c)_k} z^k
\]
is a polynomial of degree $n$, and is referred to as a \emph{hypergeometric
polynomial}.

\begin{proposition} \label{thm:n1}
	Let $(q, \dot\theta)$ be a balanced configuration with $L=3$, $c_1 = 1$, $n_1
	= n$, $n_2 = 1$.  Then, up to a complex scaling, we have $q_{2,1} = 1$ and
	$(q_{1,k})_{1 \le k \le n}$ are the roots of the hypergeometric polynomial
	${_2F_1}(-n, b; c; z)$ with
	\[
		b := n - c_2 + \dot\theta_{2,0} - \dot\theta_{1,0},\qquad
		c := 1 + \dot\theta_{2,0} - \dot\theta_{1,0}.
	\]
	Moreover, as long as $b$ and $c$ are not non-positive integers, and $c-b$ is
	not a non-positive integer bigger than $-n$, the configuration is
	rigid.
\end{proposition}

\begin{proof}
	The force equations are
	\[
		\begin{split}
			F_{1,k}&=\sum_{1\le k \ne j \le n}\frac{2q_{1,k}}{q_{1,k}-q_{1,j}}-\frac{q_{1,k}c_2}{q_{1,k}-1}+c, 1\leq k\leq n,\\
			F_{2,1}&=-\sum_{j=1}^n\frac{c_2}{1-q_{1,j}}+c_2^2+c_2(\dot\theta_{3,0}-\dot\theta_{2,0}),
		\end{split}
	\]
	where $c := 1+\dot\theta_{2,0}-\dot\theta_{1,0}$.  To solve $F_{1,k}=0$ for
	$k=1,2,\ldots,n$, we use again the polynomial method: Let $P(z) =
	\prod_{i=1}^n (z-q_k)$.  Then we have
	\begin{align*}
		P' &= P \sum_{k = 1}^n \frac{1}{z-q_{1,k}};\\
		\label{eq:ode6ends}
		P'' &= 2P \sum_{k = 1}^n \frac{1}{z-q_{1,k}} \sum_{1 \le k \ne j \le n} \frac{1}{q_{1,k}-q_{1,j}}\\
		&= P \sum_{k = 1}^n \frac{1}{(z-q_{1,k})}\Big(\frac{c_2}{q_{1,k}-1} - \frac{c}{q_{1,k}}\Big) \qquad \text{by $F_{1,k} = 0$}\\
		&= P \sum_{k=1}^n \Big( \frac{c_2}{(z-1)(z-q_{1,k})} + \frac{c_2}{(z-1)(q_{1,k}-1)}
			- \frac{c}{z (z-q_{1,k})}
		- \frac{c}{z q_{1,k}} \Big).
	\end{align*}
	So the configuration is balanced if and only if
	\begin{equation}\label{eq:ode6ends}
		P''
		+ \Big(\frac{-c_2}{z-1} + \frac{c}{z}\Big)P'
		+ \Big(\frac{c_2}{z-1}\frac{P'(1)}{P(1)}-\frac{c}{z}\frac{P'(0)}{P(0)}\Big)P=0.
	\end{equation}
	Define
	\[
		b := n - 1 - c_2 + c.
	\]
	For~\eqref{eq:ode6ends} to have a polynomial solution of degree $n$, we must
	have
	\[
		c_2\frac{P'(1)}{P(1)} = c \frac{P'(0)}{P(0)} = -nb,
	\]
	so that the leading coefficients cancel.  Then~\eqref{eq:ode6ends} becomes
	the hypergeometric differential equation
	\[
		z(1-z)P''+[c-(-n+b+1)z]P'+nbP=0
	\]
	to which the only polynomial solution (up to a multiplicative
	constant) is given by the hypergeometric polynomial $P(z) = {_2F_1}(-n, b; c;
	z)$ of degree $n$.

	Moreover, in order for $F_{2,1}=0$, we must have
	\begin{equation} \label{eq:F21}
		\dot\theta_{3,0}-\dot\theta_{2,0}=\sum_{j=1}^n\frac{1}{1-q_{1,j}}-c_
		= \frac{P'(1)}{P(1)} - c_2 = -\frac{nb}{c_2} - c_2
	\end{equation}

	Note that $b$ and $c$ are real.  If $b$ is not a non-positive integer, and
	$c-b$ is not a non-positive integer bigger than $-n$, then all the $n$ roots
	of $P(z) = {_2F_1}(-n,b;c;z)$ are simple.  Indeed, under these assumptions,
	we have $P(0) = 1$ and $P(1) = (c-b)_n/(c)_n \ne 0$ by the Chu--Vandermonde
	identity.  Let $z_0$ be a root of $P(z)$, then $z_0 \ne 0, 1$.  In view of
	the hypergeometric differential equation, if $z_0$ is not simple, we have
	$P(z_0) = P'(z_0) = 0$ hence $P(z) \equiv 0$ by uniqueness theorem.

	The rigidity means that no perturbation of $q_{1,k}$ preserve the
	balance to the first order.  To prove this fact, we use a perturbation
	argument similar to that in the proof of \cite[Proposition 1]{traizet2002}.

	Let $(q_{1,k}(t))_{1 \le k \le n}$ be a deformation of the configuration such
	that $q_{1,k}(0) = q_{1,k}$ and $(\dot F_{1,k}(0))_{1 \le k \le n} = 0$,
	where dot denotes derivative with respect to $t$.  Define
	\[
		P_t(z) = \sum_{j = 0}^n a_j(t) z^j := \prod_{k = 1}^n (z-q_{1,k}(t)).
	\]
	Then we have
	\[
		z(1-z)P_t''+[c-(-n+b+1)z]P_t'+nbP_t = o(t),
	\]
	meaning that the coefficients from the left side are all $o(t)$.  So the
	coefficients of $P_t$ must satisfy
	\begin{equation}\label{eq:hypergeometriccoeff}

		(b+j)(n-j) a_j(t) + (j^2 + j + cj) a_{j+1}(t) = o(t), \qquad 0 \le j \le n.
	\end{equation}
	Note that $P_t(z)$ is monic by definition, meaning that $a_n(t) \equiv 1$.
	Since $b$ and $c$ are not non-positive integers, we conclude that $a_j(t) =
	o(t)$ for all $0 \le j \le n$.  The simple roots depend analytically on the
	coefficients, so $q_{1,k}(t) = q_{1,k} + o(t)$.
\end{proof}

The simple roots of ${_2F_1}(-n,b;c;z)$ are either real or form conjugate pairs.  As
a consequence, if rigid, the configurations in the proposition above
will give rise to minimal surfaces with horizontal symmetry planes.

\begin{example}
	For each integer $n \ge 2$, the real parameters $(b, c)$ for which
	${_2F_1}(-n, b; c; z)$ has only real simple roots has been enumerated
	in~\cite{dominici2013}.  The results are plotted in blue in
	Figure~\ref{fig:bcplot}.  The embeddedness conditions \eqref{eq:embedcond}
	are
	\begin{align*}
		\dot\theta_{1,0} > \dot\theta_{2,0} & \Rightarrow c < 1, \\
		\dot\theta_{1,\infty} > \dot\theta_{2,\infty} & \Rightarrow b > -n, \\
		\dot\theta_{2,0} > \dot\theta_{3,0} & \Rightarrow c_2^2 > -nb, \\
		\dot\theta_{2,\infty} > \dot\theta_{3,\infty} & \Rightarrow c_2^2 > n(c_2+b),
	\end{align*}
	where $c_2 = n-1-b+c$.  The region defined by these is plotted in red in
	Figure~\ref{fig:bcplot}.  Then non-integer parameters $(b,c)$ in the
	intersection of red and blue regions give rise to balanced and rigid
	configurations with real $q_{1,k}$. \qed
\end{example}

\begin{figure}
	\includegraphics[width=\textwidth]{bcPlot}
	\caption{
		The set of $b$ and $c$ for which ${_2F_1}(-n,b;c;z)$ has only real simple
		roots (blue), and for which the embeddedness conditions are satisfied
		(red).\label{fig:bcplot}
	}
\end{figure}

\begin{remark}
	As $c \to 0$, ${_2F_1}(-n,b;c;z)/\Gamma(c)$ converges to a polynomial with a
	root at $0$.  One may interprete that, as $c$ increases across $0$, a root
	moves from the interval $(-\infty, 0)$ to the inverval $(0,1)$ through $0$.

	When $b = 1-n$, ${_2F_1}(-n,b;c;z)$ becomes a polynomial of degree $n-1$.
	One may interprete that, as $b$ increases across $1-n$, a root moves from the
	interval $(-\infty,0)$ to the interval $(1,\infty)$ through the infinity.
\end{remark}

\begin{example}\label{ex:2}
	Assume that $b+c=1-n$ (hence $c_2 = -2b$). Then by the identity
	\[
		{_2F_1}(-n,b;c;z)=\frac{(b)_n}{(c)_n}(-z)^n{_2F_1}\left(-n,1-c-n;1-b-n;\frac{1}{z}\right),
	\]
	the simple roots must be symmetrically placed.  That is, if $z_0$ is a
	root, so is $1/z_0$.  This symmetry appears in the resulting minimal surfaces
	as a rotational symmetry.  If the simple roots are real, the rotation reduces
	to a vertical reflectional.  In view of Figure~\ref{fig:bcplot}, we obtain
	the following concrete examples.
	\begin{itemize}
		\item $n \ge 2$ and $0 < c < 1$. In this case ${_2F_1}(-n, b; c; z)$ has
			$n$ simple negative roots.  See Figure \ref{fig:21} for an example of
			this type	with $n=2$.

		\item $n \ge 3$ and $-1 < c < 0$, or $n=3$ and $-5/4<c<-1$, or $n
			= 2$ and $-1/2 < c < 0$. In these case, ${_2F_1}(-n, b; c; z)$ has $n-2$
			simple negative roots, one root $0 < z_0 < 1$, and another root $1/z_0 >
			1$. \qed
	\end{itemize}

	\begin{figure}[htbp]
		\centering
 		\includegraphics[height=1.7in]{singly21}
		\caption{Genus 1 example with $n=2$ and $0<c<1$.}
		\label{fig:21}
	\end{figure}

\end{example}

\begin{remark}
	Saddle towers with six ends are parameterized by three real parameters, here
	by $b$, $c$, and the family parameter $\tau$.  We see that the relation
	$b+c=1-n$ imposes a rotational symmetry.  It can be imagined that removing
	the relation would break this symmetry.
\end{remark}

\begin{remark}
	The polynomial method is often used to find balanced configurations of
	interacting points in the plane.  In minimal surface theory, it has been
	employed in many implementations of Traizet's node-opening technique
	\cite{traizet2002, traizet2002b, traizet2005, li2012, connor2012, connor2017,
	connor2017b, helicoids}.

\end{remark}

\subsection{Surfaces with eight ends of type (1,n,1)} \label{ssec:1n1}
Theorem~\ref{thm:n1} generalizes to the following lemma with similar proof

\begin{lemma} \label{lemma:1n1}
	If we fix $q_{l\pm 1,k}$'s and assume that $c_l = 1$.  Then $q_{l,k}$'s in a
	balanced configuration are given by the roots of a Stieltjes polynomial
	$P(z)$ of degree $n_l$ that solves the generalized Lam\'e equation (a.k.a.\
	second-order Fuchsian equation)~\cite{marden1966}
	\begin{align}
		P''
		&+ \Big(
			\frac{c}{z}
			+ \sum_{k = 1}^{n_{l-1}} \frac{-c_{l-1}}{z-q_{l-1,k}}
			+ \sum_{k = 1}^{n_{l+1}} \frac{-c_{l+1}}{z-q_{l+1,k}}
		\Big)P'\label{eq:lame}\\
		&+ \Big(
			\frac{\gamma_0}{z}
			+ \sum_{k=1}^{n_{l-1}} \frac{\gamma_{l-1,k}}{z-q_{l-1,k}}
			+ \sum_{k=1}^{n_{l+1}} \frac{\gamma_{l+1,k}}{z-q_{l+1,k}}
		\Big)P =0,\nonumber
	\end{align}
	where $c = 1 + \dot\theta_{l+1,0} - \dot\theta_{l,0}$,

	subject to conditions
	\begin{align*}
		\gamma_0
		+ \sum_{k=1}^{n_{l-1}} \gamma_{l-1,k}
		+ \sum_{k=1}^{n_{l+1}} \gamma_{l+1,k}
		&=0,\\
		\sum_{k=1}^{n_{l-1}} \gamma_{l-1,k} q_{l-1,k}
		+ \sum_{k=1}^{n_{l+1}} \gamma_{l+1,k} q_{l+1,k}.
 		&=: -n_l b,
	\end{align*}
	and
	\[
		c - n_{l-1}c_{l-1} - n_{l+1}c_{l+1}
		= 1 - n_l + b.
	\]
	Moreover, the matrix $(\partial F_{l,k} / \partial q_{l,j})_{1 \le j,k\le
	n_l}$ is nonsingular as long as $b$ is not a non-positive integer bigger than
	$n_l$.
\end{lemma}

Indeed, a root of $P(z)$ is simple if and only if it does not coincide with $0$
or any $q_{l\pm 1, k}$.  If the roots $(q_{l,k})$ of $P(z)$ are all simple,
then they solve the equations~\cite{marden1966}
\begin{multline*}
	\sum_{1 \le k \ne j \le n_l} \frac{2}{q_{l,k}-q_{l,j}}
	+\sum_{1 \le j \le n_{l+1}} \frac{-c_{l+1}}{q_{l,k} - q_{l+1,j}}\\
	+\sum_{1 \le j \le n_{l-1}} \frac{-c_{l-1}}{q_{l,k} - q_{l-1,j}}
	+\frac{c}{q_{l,k}}
	= \frac{F_{l,k}}{q_{l,k}} = 0,
\end{multline*}
which is exactly our balance condition; see Remark~\ref{rem:electrob}.
Moreover, an equation system generalizing~\eqref{eq:hypergeometriccoeff} has
been obtained in~\cite[\S 136]{heine1878}, from which we may conclude the
non-singularity of the Jacobian.  In fact, there are
\[
	\binom{n_{l-1} + n_l + n_{l+1} - 1}{n_{l-1} + n_{l+1} - 1}
\]
choices of $\gamma$ for which~\eqref{eq:lame} has a polynomial solution of
degree $n_l$~\cite[\S 135]{heine1878}.

This observation allows us to easily construct balanced and rigid
configurations of type $(1,n,1)$:  Up to reparametrizations and complex
scalings, we may assume that $c_2 = 1$ and $q_{1,1} = 1$.  Then $q_{3,1}$ must
be real, and $(q_{2,k})_{1 \le k \le n}$ are given by roots of a Heun
polynomial.  Such a configuration depends locally on four real parameters,
namely $q_{3,1}$, $c_1$, $c_3$ and $c$ (or $b$).  When these are given, we have
$n+1$ Heun polynomials, each of which gives balanced positions of $q_{2,k}$'s.
For each of the Heun polynomials $P$, we have
\begin{align*}
	\dot\theta_{2,0} - \dot\theta_{1,0} &= \frac{P'(1)}{P(1)} - c_1,\\
	\dot\theta_{3,0} - \dot\theta_{2,0} &= c - 1 = b + c_1 + c_3 - n,\\
	\dot\theta_{4,0} - \dot\theta_{3,0} &= \frac{P'(q_{3,1})}{P(q_{3,1})} - c_3.
\end{align*}
Together with the family parameter $\tau$, the surface depends locally on five
parameters, which is expected because there are eight ends.

\begin{example}[Symmetric examples]
	When $q_{3,1} = q_{1,1} = 1$, the Heun polynomial reduces to a hypergeometric
	polynomial ${_2F_1}(-n, b; c; z)$, where $c_1 + c_3 = n-1-b+c$.  Assume
	further that $b+c=1-n$, so $c_1+c_3=-2b$.  This imposes a symmetry in the
	configuration.  Because $(c_1+c_3)P'(1)/P(1) = -nb$, the embeddedness
	conditions simplify to
	\[
		c_1 > n/2,\quad c_3 > n/2,\quad 1-n/2 < c < 1,\quad -n < b < -n/2.
	\]
	As explained in Example \ref{ex:2}, the hypergeometric polynomial has real
	roots if $b$ and $c$ lie in the blue regions of Figure~\ref{fig:bcplot}.
	More specifically:
	\begin{itemize}
		\item When $n \ge 2$ and $0 < c < 1$, ${_2F_1}(-n, b; c; z)$ has $n$
			simple negative roots.  See Figure \ref{fig:151} for an example with $n=5$.

		\item When $n\ge 4$ and $-1<c<0$, or $n = 3$ and $-1/2 < c < 0$,
			${_2F_1}(-n, b; c; z)$ has $n-2$ simple negative roots, one root $0 < z_0
			< 1$, and another root $1/z_0 > 1$. \qed
	\end{itemize}

	\begin{figure}[htbp]
		\centering
 		\includegraphics[height=2in]{singly151}
		\caption{Genus 4 example with $n=5$ and $0<c<1$.}
		\label{fig:151}
	\end{figure}
\end{example}

\begin{example}[Offset handles]
	There are embedded examples in which the handles are not symmetrically
	placed.  For instance, one balanced configuration of type $(1,2,1)$ is given
	by
	\begin{align*}
		&q_{3,1} = 2/3,&& q_{2,1} = -1/3, && q_{2,2} = -23,\\
		&c_1 = 8/5, && c_3 = 4189/2890, && b = -9857/8670,
	\end{align*}
	so $c = 3956/4335$. \qed

\end{example}

\subsection{Concatenating surfaces of type (1,n,1)} \label{ssec:1n1n1}

Assume that we are in possession of $R$ configurations of type $(1,n^{(r)},1)$,
$n^{(r)} > 1$, $1 \le r \le R$.  In the following, we use superscript $(r)$ to
denote the parameters of the $r$-th configuration.  Up to reparameterizations
and complex scalings, we may assume that $c_2^{(r)} = 1$ and $q_{1,1}^{(r)}=1$.
Then we may concatenate these configurations into one of type
\[
	(1, n_2, 1, n_4, 1, \cdots, 1, n_{2R}, 1)
\]
such that $q_{1,1}=1$, $c_1 = 1$, and for $1 \le r \le R$, we have $n_{2r} =
n^{(r)}$,
\begin{align*}
	q_{2r,k} &= q_{2r-1,1} q_{2,k}^{(r)}, &
	q_{2r+1,1} &= q_{2r-1,1} q_{3,1}^{(r)},\\
	c_{2r} &= c_{2r-1}/c_1^{(r)}, &
	c_{2r+1} &= c_{2r-1} c_3^{(r)}/c_1^{(r)}, &
\end{align*}
and
\[
	\dot\theta_{2r+1, 0} - \dot\theta_{2r,0} = c_{2r} (c^{(r)} - 1).
\]
The balance of even layers then follows from the balance of each
sub-configuration.  The balance of odd layers leads to
\[
	\dot\theta_{2r, 0} - \dot\theta_{2r-1,0} =
	c_{2r} ( \dot\theta_{2,0}^{(r)} - \dot\theta_{1,0}^{(r)} + c_1^{(r)} )
	+ c_{2r-2} ( \dot\theta_{4,0}^{(r-1)} - \dot\theta_{3,0}^{(r-1)} + c_3^{(r-1)} )
	- c_{2r-1}

\]
for $1 \le r \le R+1$.  As expected, such a configuration depends locally on
$4R$ real parameters, namely $q_{3,1}^{(r)}$, $c_1^{(r)}$, $c_3^{(r)}$, and
$c^{(r)}$, $1 \le r \le R$.

We may impose symmetry by assuming that $q_{3,1}^{(r)} = 1$, so $q_{2r+1,1}=1$
for all $0 \le r \le R$, and that $b^{(r)}+c^{(r)} = 1 - n^{(r)}$, so
$c_1^{(r)} + c_3^{(r)} = n^{(r)} - 1 - b^{(r)} + c^{(r)} = -2b^{(r)}$.  Then
$q_{2r,k} = q_{2,k}^{(r)}$, $1 \le k \le n^{(r)}$, are given by the roots of
${_2F_1}(-n^{(r)}, b^{(r)}; c^{(r)}; z)$, $1 \le r \le R$.  Because
\[
 	\dot\theta_{2,0}^{(r)} - \dot\theta_{1,0}^{(r)} + c_1^{(r)} =
 	\dot\theta_{4,0}^{(r)} - \dot\theta_{3,0}^{(r)} + c_1^{(r)} = n^{(r)}/2,
\]
the embeddedness conditions simplify to
\[
	2 n_l c_l > n_{l-1} c_{l-1} + n_{l+1} c_{l+1}
\]
for $1 \le l \le L$ under the condition that $n_0 = n_{L+1} = 0$.  That is, the
sequence $(n_l c_l)_{1 \le l \le L}$ must be concave.  For even $l$, the
concavity implies that $b^{(r)} > -n$ hence $c^{(r)} < 1$ for all $1 \le r \le
R$.  We may choose, for instance, $n_l c_l = \ln(1+l)$ or $n_l c_l =
(\exp l - 1)/\exp(l-1)$ to obtain embedded minimal surfaces.

\begin{remark}
We can also append a configuration of type $(1,n^{(r)})$ to the sequence of
$(1, n^{(r)}, 1)$-configurations to obtain a configuration of type
\[
	(1,n_2,1,n_4,1,\ldots,1,n_{2R-2},1,n_{2R}),
\]
where the $q_{l,k}, c_l,\dot\theta_{l,0}$ terms are defined as above.
Therefore, an embedded example of any genus with any even number ($>2$) of ends
can be constructed.
\end{remark}

\subsection{Numerical examples}

The balance equations can be combined into one differential equation that is
much easier to solve.  A solution of this differential equation corresponds to
a lot of balance configurations that are equivalent by permuting the locations
of the nodes.

\begin{lemma}
	Let $L$ be a positive integer, $n_1,n_2,\ldots,n_L\in\mathbb{N}$, and suppose $\{q_{l,k}\}$ is a configuration such that the $q_{l,k}$ are distinct.  Let
	\[
		P_l(z)=\prod_{k=1}^{n_l}(z-q_{l,k}),\quad P(z)=\prod_{l=1}^LP_l(z), \quad P_0(z)=P_{L+1}(z)=1,\\
	\]
	and
	\[
		\begin{split}
			\mathcal{F} P(z)=\sum_{l=1}^L&\left(\frac{c_l^2zP_l''(z)P(z)}{P_l(z)}-\frac{c_lc_{l+1}zP_l'(z)P_{l+1}'(z)P(z)}{P_l(z)P_{l+1}(z)}\right.\\
			&\hspace{.2in}\left.+\left(c_l^2+c_l(\dot\theta_{l+1,0}-\dot\theta_{l,0})\right)\frac{P_l'(z)P(z)}{P_l(z)}\right).
		\end{split}
	\]
	Then the configuration $\{q_{l,k}\}$ is balanced if and only if $\mathcal{F}P(z)\equiv 0$.
\end{lemma}

\begin{proof}
	We have seen that,
	\[
		\frac{P_l''(z)}{P'_l(z)}=\sum_{1\leq k\neq j\leq n_l}\frac{2}{q_{l,k}-q_{l,j}},
		\qquad
		\frac{P_{l\pm1}'(q_{l,k})}{P_{l\pm1}(q_{l,k})}=\sum_{j=1}^{n_{l\pm1}}\frac{1}{q_{l,k}-q_{l\pm1,j}}.
	\]
	Define
	\[
		F_l(z) = \frac{c_l^2 z P_l''(z)}{P_l'(z)}-\frac{c_lc_{l+1} z P_{l+1}'(z)}{P_{l+1}(z)}-\frac{c_lc_{l-1} z P_{l-1}'(z)}{P_{l-1}(z)}+c_l^2+c_l(\dot\theta_{l+1,0}-\dot\theta_{l,0}).
	\]
	Then $F_{l,k} = F_l(q_{l,k})$.

	Set
	\[
		\begin{split}
			Q_l(z)&=\frac{P_l'(z)P(z)}{P_l(z)}F_l(z)\\
			&=\frac{c_l^2zP_l''(z)P(z)}{P_l(z)}-\frac{c_lc_{l+1}zP_l'(z)P_{l+1}'(z)P(z)}{P_l(z)P_{l+1}(z)}-\frac{c_lc_{l-1}zP_{l-1}'(z)P_l'(z)P(z)}{P_{l-1}(z)P_l(z)}\\
			&\hspace{.2in}+\left(c_l^2+c_l(\dot\theta_{l+1,0}-\dot\theta_{l,0})\right)\frac{P_l'(z)P(z)}{P_l(z)}.
		\end{split}
	\]
	Then $F_{l,k}=0$ if and only if $Q_l(q_{l,k})=0$.

	Now observe that $Q_l(z)$ and $Q(z) = \mathcal{F} P(z)$ are polynomials with
	degree strictly less than
	\[
		\deg P = N = \sum_{l=1}^Ln_l,
	\]
	and $Q(q_{l,k})=Q_l(q_{l,k})$ for $1 \le k \le n_l$ and $1 \le l \le L$.  If
	$Q\equiv 0$ then $Q_l(q_{l,k})=0$ and so $\{q_{l,k}\}$ is a balanced
	configuration.  If $\{q_{l,k}\}$ is a balanced configuration then
	$Q(q_{l,k})=Q_l(q_{l,k})=F_{l,k}=0$.  Hence, $Q$ has at least $N$ distinct
	roots.  Since the degree of $Q$ is strictly less than $N$, we must have
	$Q\equiv 0$.
\end{proof}

It is relatively easy to numerically solve $\mathcal{F} P(z)\equiv 0$ as long
as we don't have too many levels and necks.  So we use this lemma to find
balanced configurations.  Since all previous examples admit a horizontal
reflection symmetry, we are most interested examples without this symmetry, or
with no non-trivial symmetry at all.

Figure \ref{fig:132} shows an example with $L=3$,
\begin{gather*}
	n_1=1, n_2=3, n_3=2\\
	c_1=2, c_2=1, c_3=13/16\\
	\theta_{1,0}=0, \theta_{2,0}=-1/2, \theta_{3,0}=-27/16, \theta_{4,0}=-29/16.
\end{gather*}
This configurations correspond to embedded minimal surfaces with eight ends and
genus three in the quotient. It has no horizontal reflectional symmetry, but
does have a rotational symmetry.

\begin{figure}[h]
  \centering
  \includegraphics[width=0.5\textwidth]{132.pdf}
  \caption{
  	A $(1,3,2)$ balanced configuration with no horizontal reflectional
  	symmetry.  The circles, squares, and diamonds represent the necks at levels
  	one, two, and three respectively.
	}\label{fig:132}
\end{figure}

Figure \ref{fig:143} shows two examples with $L=3$,
\begin{gather*}
	n_1=1, n_2=4, n_3=3\\
	c_1=7/2, c_2=1, c_3=3/4\\
	\theta_{1,0}=0, \theta_{2,0}=-2, \theta_{3,0}=-13/5, \theta_{4,0}=-541/180.
\end{gather*}
These configurations correspond to embedded minimal surfaces with eight ends
and genus five in the quotient, with no nontrivial symmetry.

\begin{figure}[h]
  \centering
  \begin{subfigure}[b]{0.5\textwidth}
    \centering
    \includegraphics[width=\textwidth]{143c.pdf}
  \end{subfigure}
  ~

  \begin{subfigure}[b]{0.5\textwidth}
    \centering
    \includegraphics[width=\textwidth]{143g.pdf}
  \end{subfigure}
  ~

  \caption{
  	$(1,4,3)$ balanced configurations with no symmetries.  The circles, squares,
  	and diamonds represent the necks at levels one, two, and three respectively.
	}\label{fig:143}
\end{figure}

Figure \ref{fig:173} shows two examples with $L=3$,
\begin{gather*}
	n_1=1, n_2=7, n_3=3,\\
	c_1=17/7, c_2=1, c_3=3/2,\\
	\theta_{1,0}=0, \theta_{2,0}=-1/2, \theta_{3,0}=-3/2, \theta_{4,0}=-2468/441.
\end{gather*}
These configurations correspond to embedded minimal surfaces with eight ends
and genus eight in the quotient, with no nontrivial symmetry.

\begin{figure}[h]
  \centering
  \begin{subfigure}[b]{0.5\textwidth}
    \centering
    \includegraphics[width=\textwidth]{173a.pdf}
  \end{subfigure}
  ~

  \begin{subfigure}[b]{0.5\textwidth}
    \centering
    \includegraphics[width=\textwidth]{173b.pdf}
  \end{subfigure}
  ~

  \caption{
  	$(1,7,3)$ balanced configurations with no symmetries.  The circles, squares,
  	and diamonds represent the necks at levels one, two, and three respectively.
	}\label{fig:173}
\end{figure}

\section{Construction}

\subsection{Opening nodes}

To each vertical plane is associated a punctured complex plane $\mathbb{C}^\times_l
\simeq \mathbb{C} \setminus \{0\}$, $1 \le l \le L+1$.  They can be seen as Riemann
spheres $\hat\mathbb{C}_l \simeq \mathbb{C} \cup \{\infty\}$ with two fixed punctures at
$p_{l,0} = 0$ and $p_{l,\infty} = \infty$, corresponding to the two ends.

To each neck is associated a puncture $\cent p_{l,k} \in \mathbb{C}^\times_l$ and a
puncture $\cent{p'}_{l,k} \in \mathbb{C}^\times_{l+1}$.  Our initial surface at
$\tau=0$ is the noded Riemann surface $\Sigma_0$ obtained by identifying
$\cent p_{l,k}$ and $\cent{p'}_{l,k}$ for $1 \le l \le L$ and $1 \le k \le
n_l$.

As $\tau$ increases, we open the nodes into necks as follows:
Fix local coordinates $w_{l,0} = z$ in the neighborhood of $0 \in \hat\mathbb{C}_l$
and $w_{l,\infty} = 1/z$ in the neighborhood of $\infty \in \hat\mathbb{C}_l$.
For each neck, we consider parameters $(p_{l,k}, p'_{l,k})$ in the
neighborhoods of $(\cent p_{l,k}, \cent{p'}_{l,k})$ and local coordinates
\[

	w_{l,k} = \ln(z/p_{l,k})
	\quad\text{and}\quad

	w'_{l,k} = \ln(z/p'_{l,k})
\]
in a neighborhood of $p_{l,k}$ and $p'_{l,k}$, respectively.  In this paper,
the branch cut of $\ln(z)$ is along the negative real axis, and we use the
principal value of $\ln(z)$ with imaginary part in the interval $(-\pi, \pi]$.

As we only open finitely many necks, we may choose $\delta > 0$ independent of
$k$ and $l$ such that the disks $|w_h| < 2\delta$, $|w_{l,k}| < 2\delta$, and
$|w'_{l,k}| < 2\delta$ are all disjoint.  For parameters $t = (t_{l,k})_{1 \le
l \le L, 1 \le k \le n_l}$ in a neighborhood of $0$ with $|t_{l,k}| <
\delta^2$, we remove the disks
\[
	|w_{l,k}|<|t_{l,k}|/\delta
	\quad\text{and}\quad
	|w'_{l,k}| < |t_{l,k}|/\delta
\]
and identify the annuli
\[
	|t_{l,k}|/\delta \le |w_{l,k}| \le \delta
	\quad\text{and}\quad
	|t_{l,k}|/\delta \le |w'_{l,k}| \le \delta
\]
by
\[
	w_{l,k} w'_{l,k} = t_{l,k}.
\]
If $t_{l,k} \ne 0$ for all $1 \le l \le L$ and $1 \le k \le n_l$, we obtain a
Riemann surface denoted by $\Sigma_t$.

\subsection{Weierstrass data}

We construct a conformal minimal immersion using the Weierstrass
parameterization in the form
\[
	z\mapsto	\operatorname{Re} \int^z (\Phi_1, \Phi_2, \Phi_3) ,
\]
where $\Phi_i$ are meromorphic 1-forms on $\Sigma_t$ satisfying the
conformality equation
\begin{equation}\label{eq:conformal}
	Q := \Phi_1^2 + \Phi_2^2 + \Phi_3^2 = 0.
\end{equation}

\subsubsection{A-periods}

We consider the following fixed domains in all $\Sigma_t$:
\begin{align*}
	U_{l,\delta}=\{ z \in \hat\mathbb{C}_v \colon
		&|\cent w_{l,k}(z)| > \delta/2 \quad \forall 1 \le k \le n_l \quad \text{if}\; 1 \le l \le L \\
	\text{and}\quad &|\cent{w'}_{l,k}(z)| > \delta/2 \quad \forall 1 \le k \le n_{l-1} \quad \text{if}\; 2 \le l \le L+1\}
\end{align*}
and $U_{\delta}=\bigsqcup_{1 \le l \le L} U_{l,\delta}$.

Let $A_{l,k}$ denote a small counterclockwise circle in $U_{l,\delta}$ around
$p_{l,k}$; it is then homologous in $\Sigma_t$ to a clockwise circle in
$U_{l+1,\delta}$ around $p'_{l,k}$.  Moreover, let $A_{l,0}$ (resp.\
$A_{l,\infty}$) denote a small counterclockwise circle in $U_{l,\delta}$ around
$0$ (resp.\ $\infty$).

Recall that the vertical period vector is assumed to be $(0,0,2\pi)$, so we
need to solve the A-period problems
\[
	\operatorname{Re} \int_{A_h} (\Phi_1, \Phi_2, \Phi_3) = (0,0, 2\pi\sigma_h)
	\quad\text{and}\quad
	\operatorname{Re} \int_{A_{l,k}} (\Phi_1, \Phi_2, \Phi_3) = (0,0,0)
\]
for $h \in \graph{H}$, $1 \le l \le L$, and $1 \le k \le n_l$.  Here, the orientation
$\sigma_h = \pm 1$ satisfies
\[
	\sigma_h = - \sigma_{\varsigma(h)},
\]
where the ``counterclockwise rotation'' $\varsigma$ on $\graph{H}$ is defined by
\begin{equation}\label{eq:degrotate}
	\begin{cases}
		\varsigma(0_l) = 0_{l-1}, & 2 \le l \le L+1,\\
		\varsigma(0_1)= \infty_1,\\
		\varsigma(\infty_l) = \infty_{l+1}, & 1 \le l \le L,\\
		\varsigma(\infty_{L+1}) = 0_{L+1}.
	\end{cases}
\end{equation}
In particular, we have $\sigma_{l,0} = - \sigma_{l,\infty}$ for all $1 \le l
\le L+1$.

Recall that the surface tends to an $(L+1)$-sheeted $xz$-plane in the limit
$\tau \to 0$.  So we define the meromorphic functions $\Phi_1$, $\Phi_2$ and
$\Phi_3$ as the unique regular 1-forms on $\Sigma_t$ with simple poles at
$p_h$, $h \in \graph{H}$, and the A-periods
\begin{align*}
	\int_{A_h}{(\Phi_1, \widetilde\Phi_2, \Phi_3)} &=
	2 \pi {\rm i} ( \alpha_h, \beta_h, \gamma_h - {\rm i} \sigma_h), && h \in \graph{H}, \\
	\int_{A_{l,k}} (\Phi_1, \widetilde\Phi_2, \Phi_3) &=
	2 \pi {\rm i} (\alpha_{l,k}, \beta_{l,k}, \gamma_{l,k}), && 1 \le l \le L, 1 \le k \le n_l,
\end{align*}
where $\Phi_2 = \tau\widetilde\Phi_2$.  Then the A-period problems are
solved by definition.

\subsubsection{Balance of ends}

By Residue Theorem, it is necessary that
\begin{align}
	\alpha_{l,0} + \alpha_{l,\infty} + \sum_{1 \le k \le n_l} \alpha_{l,k} -
	\sum_{1 \le k \le n_{l-1}} \alpha_{l-1,k}&= 0\label{eq:resalpha},\\
	\beta_{l,0} + \beta_{l,\infty} + \sum_{1 \le k \le n_l} \beta_{l,k} -
	\sum_{1 \le k \le n_{l-1}} \beta_{l-1,k}&= 0\label{eq:resbeta},\\
	\gamma_{l,0} + \gamma_{l,\infty} + \sum_{1 \le k \le n_l} \gamma_{l,k} -
	\sum_{1 \le k \le n_{l-1}} \gamma_{l-1,k}&= 0\label{eq:resgamma},
\end{align}
for $1 \le l \le L+1$.  Summing up~\eqref{eq:resbeta} over $l$ gives
\begin{equation}\label{eq:endbalanceb}
	\sum_{h \in \graph{H}} \beta_h = 0,
\end{equation}
which we use to replace Equation \eqref{eq:resbeta} with $l = L+1$.

In this paper, the punctures $p_{l,0}$ and $p_{l,\infty}$ correspond to ends
asymptotic to vertical planes.  Hence we fix
\[
	\alpha^2_h + \tau^2 \beta_h^2 \equiv 1
	\quad\text{and}\quad
	\gamma_h \equiv 0
\]
for all $h \in \graph{H}$, so that (the stereographic projection of) the Gauss map $G
= -(\Phi_1 + {\rm i} \Phi_2)/\Phi_3$ extends holomorphically to the punctures $p_h$
with unitary values.  Then Equations \eqref{eq:resgamma} are not independent:
if it is solved for $1 \le l \le L$, it is automatically solved for $l = L+1$.

In particular, at $\tau=0$, we have $\alpha_h^2=1$.  In view of the orientation
of the ends, we choose $\alpha_{l,0} = 1$ and $\alpha_{l,\infty} = -1$
so that $G(p_{l,\infty}) = G(p_{l,0}) = {\rm i}\sigma_{l,0}$.  Then, at $\tau = 0$,
we have
\begin{equation}\label{eq:dalphagammah}
	\frac{\partial \alpha_{l,0}}{\partial(\tau^2)} = -\frac{1}{2} \beta_{l,0}^2\,,
	\quad
	\frac{\partial \alpha_{l,\infty}}{\partial(\tau^2)} = \frac{1}{2} \beta_{l,\infty}^2\,,
\end{equation}
Then summing up~\eqref{eq:resalpha} over $l$ and taking its derivative over
$\tau^2$ gives
\begin{equation}\label{eq:endbalancea}
	\sum_{1 \le l \le L+1} \frac{{\beta_{l,\infty}^2 - \beta_{l,0}^2}}{2} = 0,
\end{equation}
which we use to replace Equation~\eqref{eq:resalpha} with $l = L+1$.

\begin{remark}
	The conditions~\eqref{eq:endbalanceb} and~\eqref{eq:endbalancea} are
	disguises of the balance condition of saddle towers, namely that the unit
	vectors in the directions of the ends sum up to $0$.
\end{remark}

\subsubsection{B-periods}

For $1 \le l \le L+1$, we fix a point $O_l \in U_{l,\delta}$.  For every $1 \le
l \le L$ and $1 \le k \le n_l$ and $t_{l,k}\neq 0$, let $B_{l,k}$ be the
concatenation of
\begin{enumerate}
	\item a path in $U_{l,\delta}$ from $O_l$ to $w_{l,k} = \delta$,
	\item the path parameterized by $w_{l,k} = \delta^{1-2s}\,t_{l,k}^s$ for $s \in
		[0,1]$, from $w_{l,k} = \delta$ to $w_{l,k} = t_h/\delta$, which is identified with
		$w'_{l,k}=\delta$, and
	\item a path in $U_{l+1,\delta}$ from $w'_{l,k}=\delta$ to $O_{l+1}$.
\end{enumerate}

We need to solve the B-period problem, namely that
\begin{equation}\label{eq:Bperiod}
	\operatorname{Re} \int_{B_{l,k}} (\Phi_1, \Phi_2, \Phi_3) =
	\operatorname{Re} \int_{B_{l,1}} (\Phi_1, \Phi_2, \Phi_3).
\end{equation}

\subsubsection{Conformality}

\begin{lemma}
	For $t$ sufficiently close to $0$, the conformality condition
	\eqref{eq:conformal} is equivalent to
	\begin{align}
		\mathfrak{E}_{l,k} := \int_{A_{l,k}} \frac{w_{l,k} Q}{dw_{l,k}} &= 0, \quad 1 \le l \le L, \quad 1 \le k \le n_l, \label{eq:conform1}\\
		\mathfrak{F}_{l,k} := \int_{A_{l,k}} \frac{Q}{dw_{l,k}} &= 0, \quad 1 \le l \le L,\quad 2 \le k \le n_l, \label{eq:conform2}\\
		\mathfrak{F}'_{l,k} := \int_{A'_{l,k}} \frac{Q}{dw'_{l,k}} &= 0, \quad 1 \le l \le L,\quad 1 + \delta_{l,L} \le k \le n_l, \label{eq:conform3}
	\end{align}
	where $A'_{l,k}$ in~\eqref{eq:conform3} denotes a small counterclockwise
	circle in $U_{l+1,\delta}$ around $p'_{l,k}$, and $\delta_{l,L}=1$ if $l = L$
	and $0$ otherwise.
\end{lemma}

\begin{proof}
	By our choice of $\alpha_h$ and $\gamma_h$, the quadratic differential $Q$
	has at most simple poles at the $2L+2$ punctures $p_h$, $h \in \graph{H}$.  The
	space of such quadratic differentials is of complex dimension
	$3(N-L)-3+(2L+2) = 3N-L-1$.  We will prove that
	\[
		Q \mapsto (\mathfrak{E}, \mathfrak{F}, \mathfrak{F}')
	\]
	is an isomorphism.  We prove the claim at $t=0$; then the claim follows by
	continuity.

	Consider $Q$ in the kernel.  Recall from~\cite{traizet2008} that a regular
	quadratic differential on $\Sigma_0$ has at most double poles at the nodes
	$p_{l,k}$ and $p'_{l,k}$.  Then~\eqref{eq:conform1} guarantees that $Q$ has
	at most simple poles at the nodes.  By~\eqref{eq:conform2}
	and~\eqref{eq:conform3}, $Q$ may only have simple poles at $p_{l,1}$, $1 \le
	l \le L$, and $p'_{L+1,1}$.  So, on each Riemann sphere $\hat\mathbb{C}_l$, $Q$ is a
	quadratic differential with at most simple poles at three punctures; the
	other two being $0$, $\infty$.  But such a quadratic differential must be
	$0$.
\end{proof}

\subsection{Using the Implicit Function Theorem}

All parameters varies in a neighborhood of their central values, denoted by a
superscript $\circ$.  We will see that
\[
	\cent\alpha_{l,k}=\cent\gamma_{l,k} = 0,\;
	\cent\beta_{l,k} = -c_l,\;
	\cent{p'}_{l,k} = \overline{p_{l,k}}.
\]

\subsubsection{Solving conformality problems}

\begin{proposition}\label{prop:talphagamma}
	For $\tau$ sufficiently small and $\beta_{l,k}$, $p_{l,k}$, and
	$p'_{l,k}$ in a neighborhood of their central values, there exist unique
	values of $t_{l,k}$, $\alpha_{l,k}$, and $\gamma_{l,k}$, depending
	real-analytically on $(\tau, \beta, p, p')$, such that the balance
	equations~\eqref{eq:resalpha} and~\eqref{eq:resgamma} with $1 \le l \le L$
	and the conformality Equations \eqref{eq:conform1} and~\eqref{eq:conform2}
	are solved.  Moreover, at $\tau=0$, we have $\alpha_{l,k} =
	\gamma_{l,k} = 0$,
	\[
		\frac{\partial t_{l,k}}{\partial(\tau^2)} = \frac{1}{4}\beta_{l,k}^2,
	\]
	and, for $2 \le k \le n_l$,
	\begin{equation}\label{eq:dalphagamma}
		\frac{\partial}{\partial(\tau^2)} (\alpha_{l,k} - {\rm i}\sigma_{l,0} \gamma_{l,k})
		= -\frac{1}{2} \Res{\frac{\widetilde\Phi_2^2}{dw_{l,k}}}{p_{l,k}}
		= -\frac{1}{2} \Res{\frac{z\widetilde\Phi_2^2}{dz}}{p_{l,k}}.
	\end{equation}
\end{proposition}

\begin{proof}
	At $\tau=0$, for $2 \le k \le n_l$ we have
	\[
		\mathfrak{G}_{l,k} = \int_{A_{l,k}} \frac{w_{l,k}Q}{dw_{l,k}} =
		2\pi{\rm i}(\alpha_{l,k}^2 + \gamma_{l,k}^2) = 0
	\]
	which vanishes when
	\[
		\alpha_{l,k} = \gamma_{l,k} = 0.
	\]
	Recall that $\alpha_h = \pm 1$ at $\tau=0$ and that $\gamma_h \equiv
	0$.  Then by Residue Theorem, we have
	\[
		\alpha_{l,1} = \gamma_{l,1} = 0.
	\]
	As a consequence, we have at $\tau=0$
	\[
		\cent\Phi_1 = dz/z,\quad
		\cent\Phi_2 = 0,\quad\text{and}\quad
		\cent\Phi_3 = - {\rm i} \sigma_{l,0} dz/z,
	\]
	so $Q = 0$ as we expect.

	We then compute the partial derivatives
	\[
		\frac{\partial}{\partial \alpha_{l,k}} \mathfrak{F}_{l,k} = 4\pi{\rm i}\,,\quad
		\frac{\partial}{\partial \gamma_{l,k}} \mathfrak{F}_{l,k} = 4\pi\sigma_{l,0}\,,\quad
		\text{and}\quad
		\frac{\partial}{\partial t_{l,k}} \mathfrak{G}_{l,k} = -8\pi{\rm i}\;.
	\]
	All other partial derivatives vanish.  Therefore, by the Implicit Function
	Theorem, there exist unique values of $\alpha_{l,k}$, $\gamma_{l,k}$ (with $2
	\le k \le n_l$), and $t_{l,k}$ (with $1 \le k \le n_l$) that solve the
	conformality Equations \eqref{eq:conform1} and~\eqref{eq:conform2}.  Then
	$\alpha_{l,1}$ and $\gamma_{l,1}$ are uniquely determined by the linear
	balance equations~\eqref{eq:resalpha} and~\eqref{eq:resgamma}.

	Moreover,
	\[
		\frac{\partial}{\partial(\tau^2)} \mathfrak{F}_{l,k} = \int_{A_{l,k}} \frac{\widetilde\Phi_2^2}{dw_{l,k}},\quad
		\frac{\partial}{\partial(\tau^2)} \mathfrak{G}_{l,k} = 2\pi{\rm i} \beta_{l,k}^2.
	\]
	Hence the total derivatives
	\[
		\frac{d}{d(\tau^2)} \mathfrak{F}_{l,k} =
		4\pi{\rm i} \frac{\partial\alpha_{l,k}}{\partial(\tau^2)} +
		4\pi\sigma_{l,0} \frac{\partial\gamma_{l,k}}{\partial(\tau^2)} +
		2\pi{\rm i} \Res{\frac{\widetilde\Phi_2^2}{dw_{l,k}}}{p_{l,k}} = 0
	\]
	and
	\[
		\frac{d}{d(\tau^2)} \int_{A_{l,k}} \mathfrak{G}_{l,k} =
 		-8\pi{\rm i}\frac{\partial t_{l,k}}{\partial(\tau^2)} + 2\pi{\rm i} \beta_{l,k}^2 = 0.
 	\]
 	This proves the claimed partial derivatives with respect to $\tau^2$.
\end{proof}

From now on, we assume that the parameters $t_{l,k}$, $\alpha_{l,k}$, and
$\gamma_{l,k}$ are given by Proposition~\ref{prop:talphagamma}.

\subsubsection{Solving B-period problems}

\begin{proposition}\label{prop:beta}
	For $\tau$ sufficiently small and $p_{l,k}$ and $p'_{l,k}$ in a
	neighborhood of their central values, there exist unique values of
	$\beta_{l,k}$, depending real-analytically on $(\tau,p,p')$ and
	$(\beta_h)_{h \in \graph{H}}$, such that the balance equation~\eqref{eq:resbeta}
	with $1 \le l \le L$ and the $y$-component of the B-period
	problem~\eqref{eq:Bperiod} are solved.  Moreover, at $\tau=0$, we have
	$\beta_{l,k} = \beta_{l,1} = -c_l$ where $c_l$ is given by~\eqref{eq:resc}.
\end{proposition}

\begin{proof}
	By Lemma 8.3 of~\cite{saddle1},
	\[
		\Big(\int_{B_{l,k}} \widetilde\Phi_2\Big)- \beta_{l,k} \ln t_{l,k}
	\]
	extends holomorphically to $t=0$ as bounded analytic functions of other
	parameters.  We have seen that $t_{l,k} \sim \tau^2\beta^2_{l,k}/4$.  So
	\[
		\mathfrak{H} := \frac{1}{2\tau\ln\tau} \operatorname{Re} \Big( \int_{B_{l,k}}
		\Phi_2 - \int_{B_{l,1}} \Phi_2 \Big) = \beta_{l,k} - \beta_{l,1}
	\]
	at $\tau=0$.  Therefore, $\mathfrak{H}=0$ is solved at $\tau = 0$ by
	$\beta_{l,k} = \beta_{l,1}$ for all $2 \le k \le n_l$, and $\beta_{l,1}=
	-c_l$ follows as \eqref{eq:resc} is just a reformulation
	of~\eqref{eq:resbeta}.  The proposition then follows by the Implicit Function
	Theorem.
\end{proof}

From now on, we assume that the parameters $\beta_{l,k}$ are given by
Proposition~\ref{prop:beta}.

\begin{proposition}\label{prop:pprime}
	For $\tau$ sufficiently small and $p_{l,k}$ in a neighborhood of their
	central values, there exist unique values of $p'_{l,k}$, depending
	real-analytically on $\tau$, $p$, and $(\beta_h)_{h \in \graph{H}}$, such that the
	$x$- and $z$-components of the B-period problem~\eqref{eq:Bperiod} are
	solved.  Moreover, up to complex scalings on $\mathbb{C}^\times_{l+1}$, $1 \le l \le
	L$, we have $p'_{l,k} = \overline{p_{l,k}}$ at $\tau=0$ for any $1 < k \le
	n_l$.
\end{proposition}

\begin{proof}
	At $\tau=0$, recall that $\Phi_1 = dz/z$ and $\Phi_3 =
	-{\rm i}\sigma_{l,0} dz/z$.  So
	\begin{align*}
		\operatorname{Re} \int_{B_{l,k}} \Phi_1 - \operatorname{Re}\int_{B_{l,1}} \Phi_1
		&= \operatorname{Re}\ln(p_{l,k}/p_{l,1}) - \operatorname{Re}\ln(p'_{l,k}/p'_{l,1}); \\
		\operatorname{Re} \int_{B_{l,k}} \Phi_3 - \operatorname{Re} \int_{B_{l,1}} \Phi_3
		&= \sigma_{l,0} (\operatorname{Im}\ln(p_{l,k}/p_{l,1}) + \operatorname{Im}\ln(p'_{l,k}/p'_{l,1})).
	\end{align*}
	They vanish if and only if $\ln(p_{l,k}/p_{l,1}) =
	\overline{\ln(p'_{l,k}/p'_{l,1})}$.  We normalize the complex scaling on
	$\mathbb{C}^\times_{l+1}$, $1 \le l \le L$, by fixing $p'_{l,1} =
	\overline{p_{l,1}}$. Then the B-period problem is solved at $\tau=0$ with
	$p'_{l,k} = \overline{p_{l,k}}$.  By the same argument as
	in~\cite{traizet2008}, the integrals are smooth functions of $\tau$ and other
	parameters, so the proposition follows by the Implicit Function Theorem.
\end{proof}

From now on, we assume that the parameters $p'$ are given by
Proposition~\ref{prop:beta}.

\subsubsection{Balancing conditions}

Define
\[
	\mathfrak{R}_{l,k} = \Res{\frac{z\widetilde\Phi_2^2}{dz}}{p_{l,k}}
	\quad\text{and}\quad
	\mathfrak{R}'_{l,k} = \Res{\frac{z\widetilde\Phi_2^2}{dz}}{p'_{l,k}}.
\]
Let the central values $\cent p_{l,k} = \operatorname{conj}^l q_{l,k}$, where $q$ is from a
balanced configuration.  So the central values $\cent{p'}_{l,k} = \operatorname{conj}^{l+1}
q_{l,k}$ and
\[
	\cent{\widetilde\Phi}_2 = \begin{cases}
		\overline{\operatorname{conj}^* \psi_l} & \text{on $\mathbb{C}^\times_l$ for $l$ odd,}\\
		\psi_l & \text{on $\mathbb{C}^\times_l$ for $l$ even.}
	\end{cases}.
\]
Then we have
\[
	\overline{\mathfrak{R}_{l,k}} + \mathfrak{R}'_{l,k} = 2 \operatorname{conj}^{l+1} F_{l,k}
\]
at the central values.  Moreover, by Residue Theorem on $\mathbb{C}^\times_l$,
\begin{equation}\label{eq:residuephi2}
	\sum_{k=1}^{n_{l-1}}\mathfrak{R}'_{l-1,k} + \sum_{k=1}^{n_l}\mathfrak{R}_{l,k} +
	\beta_{l,0}^2 - \beta_{l,\infty}^2 = 0.
\end{equation}

\begin{proposition}\label{prop:p}
	Assume that the central values $q_{l,k} = \operatorname{conj}^l\cent p_{l,k}$ and
	$\dot\theta_h = \cent\beta_h$ form a balanced and rigid configuration $(q,
	\dot\theta)$.  Then for $(\tau,\beta)$ in a neighborhood of $(0,\dot\theta)$,
	there exists values for $p_{l,k}$, unique up to a complex scaling, depending
	smoothly on $\tau$ and $(\beta_h)_{h \in \graph{H}}$, such that
	$p_{l,k}(0,\dot\theta) = \cent p_{l,k}$ and the conformality
	condition~\eqref{eq:conform3} is solved.
\end{proposition}

\begin{proof}
	We compute at $\tau = 0$
	\[
		\frac{\partial}{\partial \alpha} \mathfrak{F}'_{l,k} = -4\pi{\rm i}
		\quad\text{and}\quad
		\frac{\partial}{\partial \gamma} \mathfrak{F}'_{l,k} = 4\pi\sigma_{l,0}.
	\]
	Then
	\[
		\frac{d}{d(\tau^2)} \mathfrak{F}'_{l,k}
		= -4\pi{\rm i} \frac{\partial\alpha_{l,k}}{\partial(\tau^2)} +
		4\pi\sigma_{l,0} \frac{\partial\gamma_{l,k}}{\partial(\tau^2)} +
		2\pi{\rm i} \mathfrak{R}'_{l,k}.
	\]
	For $2 \le k \le n_l$, by~\eqref{eq:dalphagamma}, $\widetilde\mathfrak{F}'_{l,k} :=
	\tau^{-2} \mathfrak{F}'_{l,k}$ extends to $\tau=0$ with the value
	\[
		\frac{d}{d(\tau^2)} \mathfrak{F}'_{l,k}
	 	= 2\pi{\rm i} (\overline{\mathfrak{R}_{l,k}} + \mathfrak{R}'_{l,k})
		= 4\pi{\rm i} \operatorname{conj}^{l+1} F_{l,k}.
	\]
	As for $k = 1$, we compute
	\begin{align*}
		&\sum_{k=1}^{n_l} \frac{d\mathfrak{F}'_{l,k}}{d(\tau^2)}
		+ \sum_{k=1}^{n_{l-1}} \operatorname{conj}\Big(\frac{d\mathfrak{F}'_{l-1,k}}{d(\tau^2)}\Big)\\
		=& 2\pi{\rm i}\Big(
			- \beta_{l,0}^2 + \beta_{l,\infty}^2
			+ \sum_{k=1}^{n_l} \mathfrak{R}'_{l,k}
			- \sum_{k=1}^{n_{l-1}} \overline{\mathfrak{R}'_{l-1,k}}
		\Big)
		&& \text{(by~\eqref{eq:resalpha} and~\eqref{eq:dalphagammah})}\\
		=& 2\pi{\rm i}\Big(
			- \beta_{l,0}^2 + \beta_{l,\infty}^2
			+ \sum_{k=1}^{n_l} \mathfrak{R}'_{l,k}
			+ \sum_{k=1}^{n_l} (\overline{\mathfrak{R}_{l,k}} + \beta_{l,0}^2 - \beta_{l,\infty}^2)
		\Big)
		&& \text{(by~\eqref{eq:residuephi2})}\\
		=& 2\pi{\rm i} \sum_{k=1}^{n_l}(\overline{\mathfrak{R}_{l,k}} + \mathfrak{R}'_{l,k})
		= 4\pi{\rm i} \sum_{k=1}^{n_l} \operatorname{conj}^{l+1} F_{l,k}.
	\end{align*}
	So $\widetilde\mathfrak{F}'_{l,1} := \tau^{-2} \mathfrak{F}'_{l,1}$ extends to
	$\tau=0$ with the value
	\[
		\frac{d\mathfrak{F}'_{l,1}}{d(\tau^2)} =
		4\pi{\rm i} \operatorname{conj}^{l+1} \Big(
			F_{l,1} + \sum_{j=1}^{l-1}\sum_{k=1}^{n_j} (-1)^{l-j} F_{l,k}
		\Big).
	\]
	Hence, if $(q,\dot\theta)$ is balanced, $\widetilde\mathfrak{F}'=0$ is solved at
	$\tau=0$.

	Recall that we normalize the complex scaling on $\mathbb{C}^\times_1$ by fixing
	$p_{1,1}$.  If $(q,\dot\theta)$ is rigid, because $\Theta_2 = \sum F_{l,k} =
	0$ independent of $p$, the partial derivative of $(\widetilde\mathfrak{F}')_{(l,k) \ne
	(L,1)}$ with respect to $(p_{l,k})_{(l,k) \ne (1,1)}$ is an isomorphism from
	$\mathbb{C}^{N-1}$ to $\mathbb{C}^{N-1}$.  The proposition then follows by the Implicit
	Function Theorem.
\end{proof}

From now on, we assume that the parameters $p$ are given by
Proposition~\ref{prop:p}.

\begin{proposition}\label{prop:theta}
	Assume that the central values $q_{l,k} = \operatorname{conj}^l\cent p_{l,k}$ and
	$\dot\theta_h = \cent\beta_h$ form a balanced and rigid configuration $(q,
	\dot\theta)$.  Then, in the neighborhood of $(0,\dot\theta)$, the solutions
	$(\tau,\beta)$ to $(\Theta_1,\Theta_2)=0$ form a smooth manifold of dimension
	$2L+1$.
\end{proposition}

This follows directly from Lemma~\ref{lem:rigid} and the Implicit Function
Theorem.

\subsection{Embeddedness}

It remains to prove that
\begin{proposition}
	The minimal immersion given by the Weierstrass parameterization is regular
	and embedded.
\end{proposition}

\begin{proof}
	The immersion is regular if $|\Phi_1|^2 + |\Phi_2|^2 + |\Phi_3|^2 > 0$.  This
	is easily verified on $U_\delta$.  On the necks and the ends, the
	regularity follows if we prove that $\widetilde\Phi_2$ has no zeros outside
	$U_\delta$.  At $\tau=0$, $\widetilde\Phi_2$ has $n_l + n_{l-1}+2$ poles on
	$\hat\mathbb{C}_l$, hence $n_l + n_{l-1}$ zeros.  By taking $\delta$ sufficiently
	small, we may assume that all these zeros lie in $U_{l,\delta}$.  By
	continuity, $\widetilde\Phi_2$ has $n_l + n_{l-2}$ zeros in $U_{l,\delta}$
	also for $\tau$ sufficiently small.  But for $\tau \ne 0$, $\widetilde\Phi_2$
	is meromorphic on a Riemann surface $\Sigma_\tau$ of genus $g = N-L$ and has
	$2L+2$ simple poles, hence has $2(N-L) - 2 + 2L + 2 = 2N$ zeros.  So
	$\widetilde\Phi_2$ has no further zeros, in $\Sigma_t$, in particular not
	outside $U_\delta$.

	We now prove that the immersion
	\[
		z\mapsto	\operatorname{Re} \int^z (\Phi_1, \widetilde\Phi_2, \Phi_3)
	\]
	is an embedding, and the limit positions of the necks are as prescribed.

	On $U_{l,\delta}$, the Gauss map $G = -(\Phi_1 + {\rm i} \Phi_2)/\Phi_3$ converges
	to ${\rm i}\sigma_{l,0}$, so the immersion is locally a graph over the $xz$-plane.
	Fix an orientation $\sigma_{1,0} = -1$, then up to translations, we have
	\[
		\lim_{\tau \to 0} \Big(\operatorname{Re} \int^z \Phi_1 + {\rm i} \operatorname{Re} \int^z \Phi_3\Big)
		= \operatorname{conj}^l(\ln z) + 2m\pi{\rm i}
	\]
	where $m$ depends on the integral path, and
	\[
		\lim_{\tau \to 0} \operatorname{Re} \int^z \widetilde\Phi_2 = \operatorname{Re} \int^z (\operatorname{conj}^*)^l \psi_l
		=: \Psi_l(\operatorname{conj}^l z)
	\]
	which is well defined for $z \in U_{l,\delta}$ because the residues of $\psi_l$
	are all real.

	With a change of variable $z \mapsto \ln z$, we see that the immersion
	restricted to $U_{l,\delta}$ converges to a periodic graph over the
	$xz$-planes, defined within bounded $x$-coordinate and away from the points
	$\ln q_{l,k} + 2m\pi{\rm i}$, and the period is $2\pi{\rm i}$.  Here, again, we
	identified the $xz$-plane with the complex plane.

	This graph must be included in a slab parallel to the $xz$-plane with bounded
	thickness.  We have seen from the integration along $B_k$ that the distance
	between adjacent slabs is of the order $\mathcal{O}(\ln \tau)$.  So the slabs
	are disjoint for $\tau$ sufficiently small.

	As for the necks and ends, note that there exists $Y > 0$ such that
	$\Psi_l^{-1}([-Y,Y])$ is bounded by $n_l+n_{l-1}+2$ convex curves.  After the
	change of variable $z \mapsto \ln z$, all but two of these curves remain
	convex; those around $0$ and $\infty$ become periodic infinite curves.  If
	$Y$ is chosen sufficiently large, there exists $X > 0$ independent of $l$
	such that the curves $|z| = \exp(\pm X)$ are included in
	$\Psi_l^{-1}([-Y,Y])$ for every $1 \le l \le L+1$.  After the change of
	variable $z \mapsto \ln z$, these curves become curves with $\operatorname{Re} z = \pm X$.

	Hence for $\tau$ sufficiently small, we may find $Y^+_l$ and $Y^-_l$, with
	$Y^-_l < Y^+_l < Y^-_{l+1}$, and $X>0$, such that
	\begin{itemize}
		\item The immersion with $Y^-_l < y < Y^+_l$ and $-X < x < X$ is a graph
			bounded by $n_l + n_{l-1}$ planar convex curves parallel to the $xz$-plane,
			and two periodic planar infinite curves parallel to the $yz$-plane.

		\item The immersion with $Y^+_l < y < Y^-_{l+1}$ and $-X < x < X$
			consists of annuli, each bounded by two planar convex curves parallel to
			the $xz$-plane.  These annuli are disjoint and, by a Theorem of
			Schiffman~\cite{schiffman1956}, all embedded.

		\item The immersion with $|x| > X$ are ends, i.e.\ graphs over
			vertical half-planes, extending in the direction $(-1, -\dot\theta_{l,0})$ and
			$(+1, -\dot\theta_{l,\infty})$, $1 \le l \le L+1$.  If the
			inequality~\eqref{eq:embedcond} is satisfied, these graphs are disjoint.
	\end{itemize}
	This finishes the proof of embeddedness.
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
Ministerial Human Rights Meeting: Canada's Opportunity to Get Human Rights Implementation Right
From Promise to Reality - EN FINAL.PDF
Governments across Canada have an unprecedented opportunity to begin addressing long-standing deficiencies in the implementation of the country's international human rights obligations by laying the groundwork for a coordinated approach to upholding rights across all levels of government, says Amnesty International Canada in a briefing paper submitted to federal, provincial and territorial Ministers. For the first time in 29 years, Ministers will meet to discuss cross-jurisdictional weaknesses and challenges in implementing Canada's commitments under an array of binding international human rights instruments.
"Amnesty International welcomes this urgently-needed and long-overdue meeting. It is disconcerting that it has taken nearly three decades to bring Ministers back together to discuss human rights. People across Canada suffer when human rights are not taken seriously and all will stand to benefit from these vital reforms," says Alex Neve, Secretary General of Amnesty International Canada's English Branch. "For decades, measures needed to achieve full implementation of critical human rights agreements have been neglected, fallen through inter-jurisdictional cracks, or suffered from lack of political will within governments at all levels. This has deprived people of their rights, often with grave consequences, and hampered Canada's ability to be a global human rights champion and press other countries to follow our lead."
Amnesty International Canada has called for a Ministerial Human Rights Meeting for more than 15 years. In its briefing paper, From Promise to Reality, Amnesty has presented federal, provincial and territorial ministers with ten recommended principles to guide their efforts to close Canada's long-standing human rights implementation gap. The briefing notes that, for decades, Canada has championed and acceded to crucial human rights agreements, but has fallen notably short when it comes to domestic implementation. Poor coordination between different departments and levels of government, inadequate partnerships with First Nations, Inuit and Metis peoples, and lack of clear accountability structures within and across government have all contributed to Canada's shortcomings. As a result, successive reviews by United Nations and Inter-American human rights bodies have repeatedly expressed concern with Canada's lack of progress on often-repeated recommendations spanning many years.
Among the consistent deficiencies noted in these reviews are failure to uphold Indigenous rights, the rights of women and girls, people living with disabilities, migrants and refugees, racialized groups, economically and socially marginalized groups, and communities abroad affected by destructive Canadian mining practices. Some of the specific shortcomings include the failure to recognise Indigenous peoples' right to free prior and informed consent, address over-representation of Indigenous peoples in prison, ensure the safety of Indigenous women and girls, restrict the use of solitary confinement in Canadian prisons, provide effective remedies for violations of economic, social and cultural rights, and prevent Canadian complicity in torture.
"It is critical that Ministers walk away from next weeks' Meeting with a roadmap and clear commitment to address these long-standing and unacceptable deficiencies," says Béatrice Vaugrante, Directrice générale of Amnistie internationale Canada francophone. "This meeting must be the start of a continuing process to right these wrongs, including by establishing clear monitoring and accountability mechanisms to ensure continued progress toward upholding the rights of all. Amnesty International's ten recommendations which we have submitted to the 16 Ministers attending this historic meeting are a strong starting point in defining that path."
Amnesty International's recommendations to the 2017 Federal, Provincial, Territorial Ministerial Human Rights Meeting include:
Recognize and respect the rights of Indigenous peoples: Indigenous peoples should be full partners in the development of priorities and plans for implementation of Canada's international obligations.
Recognize universality and interdependence of rights: The implementation of Canada's international human rights obligations cannot be selective. It must apply fully and equally to the entirety of the country's international obligations, including economic, social and cultural rights and the rights enshrined in the UN Declaration on the Rights of Indigenous Peoples; and must be grounded in a commitment to substantive equality for all disadvantaged groups.
Adopt a feminist approach to upholding human rights: Canada must fully integrate a gender perspective into all aspects of Canada's approach to implementing international human rights obligations.
Strengthen and streamline coordination: The Continuing Committee of Officials on Human Rights, established in 1975, remains the only body that brings federal, provincial and territorial governments together to discuss international human rights matters. It must be replaced by a more effective and accountable mechanism to ensure meaningful accountability and facilitation of human rights coordination among federal, provincial and territorial governments.
Engage Parliament and legislatures: Parliament and legislatures across Canada should be centrally engaged in ensuring compliance with the country's international human rights obligations. Parliament and legislatures must broaden and deepen their engagement in overseeing implementation of Canada's international human rights obligations.
Strengthened role for federal, provincial and territorial human rights institutions: All human rights commissions and tribunals in Canada should have resources, a clear mandate and responsibility to monitor and promote implementation of Canada's international human rights obligations.
Engage with civil society: Federal, provincial and territorial governments should commit to meaningful and timely engagement with civil society with respect to the implementation of the country's international human rights obligations, including by ensuring an enabling environment with adequate resources to support human rights work.
Commit to transparency, monitoring and reporting: There must be regular public reporting about the status and progress of implementation, backed up by clear goals and timelines in rights-based strategies and action plans, with independent monitoring and complaints procedures.
Provide access to justice and effective remedies: Commit to closing Canada's remedies gap by ensuring access to justice for all human rights violations, including violations of economic, social and cultural rights. Effective remedies must also be available to address the human rights impact of Canadian companies operating abroad.
Ensure political accountability and regular ministerial meetings: To demonstrate ongoing accountability for human rights implementation in the country, Amnesty International urges Ministers to meet again in December 2018 on the occasion of the 70th anniversary of the Universal Declaration of Human Rights and at intervals of at least once every two years thereafter.
For further information, please contact: Elizabeth Berton-Hunter, Media Relations, 416-363-9933 ext. 332 bberton-hunter@amnesty.ca

redpajama_set_name: RedPajamaCommonCrawl


A yahoo is a powerful platform. After it came to existence it has continued to serve as a best platform for mailing needs. Yahoo was already a self made & popular identity for all its millions of users worldwide but with mergers it has always to deliver best possible services with its extra friendly environment. For yahoo its users are always important to serve them with best facilities and features it accepts changes and work for innovation in its platform. With yahoo help is also possible and to obtain easy help you can make a call to Yahoo support number.
Yahoo and its services have always been a popular source for communication and mail transfer but recently yahoo has ahead to join hands with very famous Verizon. Verizon which is already a very famous name in internet world has agree to acquire Yahoo's core internet business. By pursuing this from yahoo Verizon and yahoo are looking forward to deliver highest possible efficiency with its services. This purchasing has been planned at $4.8 billion. Hopefully this will be done in July after which Verizon will be delivering the core internet services of yahoo. The new name to this will be Altaba.
The core internet services of yahoo purchased by yahoo includes its famous services like it popular mail services, sports verticals and various other apps. The rest of the technology enabled yahoo ownership will be looked after by very famous Chinese internet giant Alibaba. It is said by top leaders of yahoo that once the deal will be done and implementation will be passed all the services will be known with the name as Altaba.
The meaning of name is although not known completely but it is said that it is the combination of Alternative and Alibaba that will deliver alternative services of yahoo from now. As per recent news yahoo is currently the owner of about 15% share of alibaba that is worth around $ 35 billion. With this now the name will be tracked as alternative of Alibaba and this is because of the fact that yahoo owns small chunk of Chinese company.
With this there has been a shift in various top post of yahoo as well. With this soon many services of yahoo will be acquired by other leading companies and we hope that it will continue to serve as a leading platform for useful services. For knowing more about this, stay in touch with Yahoo contact number.
The Yahoo Came up with its New Name: Altaba that is definitely discussed here is really leaving.
Extremely helpful information specially the last phase 🙂 I deal with such info much.
Anyway keep up the excellent quality writing, it is rare to seee a nice blog like this one nowadays.


punwadi	Leave a Comment on Malls – The Next Undestined Destination!
Festivals, ordinary days or any special event! Malls are the prior choice of common people to hang out…..
From Berwick Street of London, Chinatown Shopping Area of Bangkok, Shimla's Mall Road, Connaught Place of Delhi, from Mumbai to Manali, Singapore to Switzerland; shopping in large malls is a year-round priority.
No doubt shopping centers are good. They are the direct statement of dizzying economic growth representing comfort, choice, luxury and better times teeming with outlets of Marks & Spencer, Gap, Starbucks, Bulgari, Cartier, Mont Blanc and many others. Moreover, they have become important landmarks or tourist destinations. Besides home, school, temples; malls have become the most natural places to be. Shopping, eateries, movies, dating whatever! All at one place.
My dissension is not about just malls! My main gripe is that – Are they really making us happy and contented?
Let's keep shopping as the only reason to visit a mall aside, the large advertisements flashing on the walls of stores, 25% flat off, 40% off, or simply air conditioned aroma 24/7 during these days of scorching heat; is more than enough to attract people towards these large concrete blocks. And for shopaholics; little shopping and we always want to shop more! Our appetites always demand for having more chicken kebabs or pizza …or donuts!!
Even though these building blocks try to look as welcoming as possible; they play on human feelings of inadequacy and competitiveness. We must earn this much, we must consume this much, we lack something, or that everyone else has something we do not yet own. How can this constant flow of competitiveness bring us true joy? We could think of spending our weekends visiting a nearby waterfall, reading a novel, writing poetry, making music, taking a long walk, playing a game, dance or it can be anything…. Every spare piece of land gives us idea of building a mall instead of building libraries, theaters, art galleries, sports halls, public swimming pools, parks, animal sanctuaries or simply leaving the land alone to flourish as it will.
Doesn't it show a poverty of imagination? Or simply this is the sole criterion to show one's wellness?
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text:
One of theatre group's longest-serving members turns director for 132nd production
Geraldine Scott
geraldine.scott@archant.co.uk
@Geri_E_L_Scott
Asher Girling and Lucy Finnie, who play Morris Townsend and Catherine Sloper in Saxlingham Players' production of the Heiress. Photo: Ross Jolliffe
Ross Jolliffe
One of the oldest amateur theatre groups in Norfolk is set to perform their 132nd production - The Heiress by Ruth and Augustus Goetz.
Saxlingham Players, founded in 1950, proudly holds a complete archive of programmes, press cuttings, and recordings of more than 50 live performances - and from November 29 will have another to add, as the curtain rises on their latest run.
Rachel Girling, has been a member of the Players since she was a child and for the first time is directing a production.
She said: "I am thrilled to be able to put on my all-time favourite play, The Heiress. The play is based on the Henry James novel, Washington Square, and is set in 1850s New York. Dr Sloper's adored wife died many years ago, and his disappointment with his charmless, plain daughter is obvious. When a handsome, charming young man appears and pursues her can he be honest in his feelings for her? It is a fascinating play dealing with love, loss, expectations and disappointment, all things that are so relevant today."
She added: "Amateur dramatic groups such as ours have wonderfully loyal followers and it is great to be able to share with them plays that they may not be familiar with, as well as the crowd-pleasing authors such as Alan Ayckbourn. We are blessed to be a thriving group who meet, have fun and are passionate about performing great plays, but so many groups are having to close due to lack of support - please do get behind your local drama group, because as the saying goes - use them or lose them. We produce great quality theatre at a fraction of professional ticket cost. Come along and judge for yourself."
There will be four performances at Saxlingham Nethergate Village Hall from Wednesday, November 29 to Saturday, December 2 at 7.30pm. For tickets, call on 01508 499631 between 9am and 9pm.


text:
Your Microdosing Resource Page
6 Days to Go – Special Microdose Countdown to Your Best Year Ever – Do This Every Morning
December 26, 2022 /in Podcast /by David Madow
Please meet up with me over in my new private FACEBOOK group to continue the discussion: https://www.facebook.com/groups/853347769006296/
New to microdosing? Check out our resource page at https://davidmadow.com/your-microdosing-resource-page/
A MUST watch! "The Absolute Beginners Guide To Microdosing Magic Mushrooms" https://www.youtube.com/watch?v=3tqA12NfFlA
Nothing here is to be taken as medical advice. Dr. Dave is sharing his personal story with you. Please contact your healthcare professional to find out if this is right for you.
Tags: anxiety, depression, magic mushrooms, microdose, microdosing, mushrooms, psilocybin, psychedelic, psychedelics
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text:
First Two Same Sex Couples to Be Married in Washington, D.C..
Gay Couples in Nation's Capital to be Married as "Spouse" and "Spouse"
ByBy: MARY KATHRYN BURKE
Mar. 8, 2010— -- A simple update headlined the Superior Court of D.C. Marriage Bureau Web site this week: NOTE: Pursuant to the Religious Freedom and Civil Marriage Equality Amendment Act of 2009, A18-248, effective March 3, 2010, same sex couples may apply for marriage licenses in the District of Columbia.
After applying on forms that now read "spouse" and "spouse," some same-sex couples will begin to receive marriage licenses Tuesday morning with wedding ceremonies following the same day, even as some opponents vow to work against the measure.
First Couples
Angelisa Young, 47, and partner Sinjoyla Townsend, 41, will be the first D.C., same-sex couple to wed.
Young and Townsend met 12 years ago at the University of the District of Columbia as undergraduates in a constitutional law class. Both are lifetime residents of the metropolitan area and each of them is a D.C. government employee. They will be married at the headquarters for the Human Rights Campaign, a lesbian, gay, bisexual, transgender lobbying group.
The days of awaiting their marriage have been transformative, they said.
"This week has been a whirlwind," Young told the ABC News Law & Justice Unit. "We are in the process of getting dresses and flowers. We got in line for an application Wednesday at six in the morning and they said we were No. 1. It is just a dream come true."
Making it Legal
Most importantly, Townsend said, they share a sense of gratitude.
"I feel proud, elated," she said. "I am very thankful for the effort the City Council put in."
"I work with a lot of sadness," Young, who's employed by the attorney general's office, said, "and I just felt pride and I felt respected. I felt all of those emotions at once."
Although they had a commitment ceremony five years ago and are registered domestic partners, the couple said, marriage, to them, is different.
"We feel gratitude that we are being included," Young said. "We have the legal right to love each other and to provide. I think everyone can understand that."
Few Protestors
The Young-Townsend nuptials will not be the only one at the Human Rights Campaign headquarters Tuesday. The couple will be joined by Reggie Stanley, 50, and Rocky Galloway, 50. Stanley works in financial services and Galloway is a project manager at a tech firm.
Stanley and Galloway are new fathers of 15-months-old twin girls who they have parented since birth. They have been together for six years, have also had a commitment ceremony and are registered domestic partners.
The most striking part of the reaction since Wednesday's ruling has been the conspicuous lack of widespread opposition, Stanley said.
"It just feels like D.C. has this sense of support," Stanley says, "Gay, straight, black, white, Asian, young, old, it is the best of the best of D.C. First and foremost, this is about us as human beings and that we appreciate the diversity among us. We are in a country built on respect and love and responsibility.
"There were only a few protesters," Stanley added. "This will be a recommitment for us and will be for securing protection for our daughters and our family. It's a phenomenal inspiration for the community. I don't mean the gay community. I mean the human community."
Sixth in the Nation
The District of Columbia joined Connecticut, Iowa, Massachusetts, New Hampshire and Vermont in legalizing marriage between same-sex partners.
Same-sex couples filing for licenses in the District were congratulated by Mayor Adrian M. Fenty, who wished each couple "a long and fulfilling marriage." And D.C. Attorney General Peter Nickles said, legally, he believes "it was the right decision."
The decision is a victory for gay-rights advocacy groups who last year faced the possibility of ballot initiatives blocking same-sex marriages.
An elated Aisha Mills, president of the Campaign for All DC Families, said she is "proud of our officials here who continue the progressive legacy of the District of Columbia."
"We're all doing great on the legal front," Mills added. "Judges have voted to deny discrimination on the ballot. We are looking to neutralize any threats and keep Congress at bay."
But opponents such as the National Organization for Marriage issued statements saying that they will continue to fight.
"It is very disappointing that voters in the District of Columbia have been precluded from exercising their constitutional right to a referendum," executive director Brian Brown said in a statement, "However, this battle is far from over. While same-sex marriages will be permitted in the District for the time being, NOM will continue to fight to get an initiative on the ballot."
Most Vocal Opponent
One of the most vocal opponents of same-sex marriage has been Bishop Harry Jackson, the senior pastor of Hope Christian Church in Beltsville, Maryland. Bishop Jackson is a member of the coalition Stand4Marriage.
"I think it is a real tragedy that the people of Washington, D.C., never got a chance to weigh in on this," Jackson said, referring to the movement for a ballot initiative defining marriage.
"Our whole thing has always been, 'Let the people vote.' You haven't heard the end of this matter from the people who are fighting. This is a civil-rights issue. People have the right to vote."
Jackson, who is black, added that he rejects the idea of gay marriage as civil-rights issue and believes the true civil-rights violation has been the lack of a ballot initiative.
"My father knew civil rights," he said. "We knew civil rights when they said, 'Go to the back of the bus.' I have a strong opinion and I am against same-sex marriage. I am praying for everyone getting married on Tuesday."
Although much of the opposition come from religious groups, many in the religious community came together to support gay marriage, including more than 200 clergy known as D.C. Clergy United.
"We do not believe sexual orientation is chosen but is how you are created," Pastor Dennis W. Wiley of Covenant Baptist Church in Washington DC said. "We believe that it is possible to support same sex marriage and still be faithful to our religious beliefs and tenets."
Author of the Act
Councilman David Catania, who authored the act legalizing same-sex marriage, is perhaps Jackson's most staunch opponent
"This is a journey we have been involved in for decades," Catania said. "What we did was profoundly American. Our city has a rich history of embracing human rights. The population is extremely perceptive to how we expand our human rights."
Catania, who is gay, said the issue of marriage is so important to him that it prompted him to part with the Republican Party in 2004.
"I was the first openly gay member voted to this council," Catania said. "I left the Republican party. I was a gay Republican. An anomaly. But I broke with President Bush over this issue. To make same-sex marriage a violation of the Constitution? What an affront."
Catania vs. Jackson
Catania, who arrived at the city courthouse with cupcakes for gay couples Wednesday, said he saw gay marriage in D.C. not only a personal victory but as a symbolic measure for the country.
"We are shaping and reshaping families and communities and it is perfecting what 'equal rights' means," he said. "In D.C., we have stared down a monster and said, 'Not here.' I was cautious about the national implications but we are the nation's capital. And this is where people come to understand our democracy and to be a part of that understanding."
"D.C. is a tolerant place," Catania added. "We did not have our own indigenous tolerance. It was imported, especially by Bishop Jackson."
Catania said that Jackson relocated from Maryland for the specific purpose of fighting gay marriage, setting up a residence in the District. He said Jackson and his supporters leave many people in D.C. "shocked and chagrined."
"Washingtonians do not like being told how to live our lives by people who are not among us," Catania said. "They blew it. We are eagerly awaiting Jackson's D.C. tax filing, eagerly."
"Lock on the South"
While he waits, some academics looked to the other economic implications of same-sex marriages coming to fruition in the capital.
Lee Badgett of the Williams Institute at UCLA said she anticipates about 50 percent of D.C.'s same-sex couples seeking marriage, about 1,900 couples.
"We see a pretty clear pattern that about half of couples will get married," Badgett said. "We expect they will spend about $10,000 a piece, so right there is about $19 million. D.C. pretty much has a lock on the south. All the same sex couples below the Mason-Dixon line have nowhere else to go."
The Williams Institute also estimated that gay marriage in the city will create about 700 jobs through additional needed hospitality personnel and government employees.
Looking Forward to Tuesday
As for the couples who will spend that money Tuesday, they have been stepping away from politics for the moment and are worrying instead about guest lists, seating charts and the legacy they will leave.
"My children and grandchildren will look back and say, 'Our country saw the light." Young said, with an eye toward her Tuesday wedding. "Right now, reporters are calling us … but outside of this moment, we are really kind of boring people."
Not so, said Townsend, laughing like a long-married spouse: "I don't think I'm boring."
Georgia declares state of emergency after protests, prepares state National Guard
Authorities investigate man's death following arrest during traffic stop


I am thinking the next is in the wrong place or I need to clear the contents of the strNewText variable (but cannot see why if the next is in the correct place.
Can someone kindly show me where the above script ig going wrong?
This is closer to what you want.


With all of the Apple Television rumors, it was only a matter of time before someone would start designing some concept images of what an integrated Apple Television, now being dubbed the iTV, might look like.
Artist Guilherme M. Schasiepen has designed what he envisions the iTV to look like, and it actually looks pretty sleek in his concept images.
If the iTV is real, it will sport an iOS feel with multitouch capabilities, Siri support, and possibly 1TB of built in storage. In addition to these features, the Apple TV set is also said to have some sort of iCloud integration, FaceTime integration, and a possibility of glasses-less 3D. If this Apple iTV actually becomes a reality it will be available in 32″, 42″, and 50″ models.
Just knowing that almost all concepts of Apple products are way off, this will most likely not be real on any terms. Although, it does seem reasonable and the images look promising, you have to remember, this is just another rumor.
Do you think the iTV could look like this?


Myanmar Best Tour by Local Operator 2017/2018/2019. View details with 1950 Tripadvisor Reviews.
Arrival in Yangon where you will be welcomed and transferred to your hotel. In the afternoon, a walk through "downtown" Yangon, a diverse mix of faded colonial architecture from the British times, and high-rises built by more recent investors; Sule Pagoda, built in the early 3rd century is a landmark in the city centre.
In the afternoon, after a visit to lacquer ware workshops; continue to Manuha Temple built by exile King Manuha; Nanphaya, said to have been the residence of King Manuha; Ananda Temple, architectural masterpiece and Oak-Kyaung monastery decorated with multi colour mural paintings. Enjoy the breathtaking sunset from one of the temple of Bagan. Overnight in Bagan.
After lunch, return drive back to Bagan. On the way back, enroute stop at village cottage industry to learn the regional product as well as to watch the collection of juice from the palm trees, and its conversion to brown sugar. Overnigh in Bagan.
After refreshment, you will take a sightseeing tour of Mandalay which includes Mahanumi, a life like Buddha image. Mandalay is the last capital of Myanmar and that make her to be the best place to see all the traditional arts and crafts. Start visiting gold leaf making work-place and see how the gold leaf is being made by hand in a traditional way. Continue visit the Old Palace grounds; Kuthodaw Pagoda, known as the world's largest book for its 729 marble slabs inscribed with the Buddha's Doctrine; Golden Palace Monastery, noted for its exquisite wood carvings and enjoy the panoramic view of the city from Mandalay Hill. Overnight in Mandalay.
After lunch, drive to Ava, 30-minute to the south of Mandalay, was known in the west as the Kingdom of Ava. Stop by a tributary and cross it with a ferry. Travel by pony-cart around Ava to Nanmyint leaning tower which gives a birds' eye view of the mighty Ayeyarwaddy River; Bagaya wooden monastery; Aungmyaybonza Monastery made of brick and stucco. Though the Royal Palace has gone, the Pagodas, Temples and Monasteries still remain. Return drive to Mandalay and see the silversmith work place on our way back to hotel. Overnight in Mandalay.
After breakfast, depart the hotel for flight to Heho on the Shan plateau. Transfer by road to Inle Lake and transfer to our hotel by long tail boat across the lake. Upon arrival, check in at hotel. In the afternoon, we plan to spend the day visiting villages and sampling the spectacular array of traditional handicrafts. You will see the unique method locals use to row their boats, fishermen with their own style of fishing, floating farms and gardens and a handloom weaving cottage with its own showroom. You will visit Phaungdawoo Pagoda which houses the five most revered Buddha Images in the southern Shan State. Overnight in Inle Lake.
We are heading back down country today to return to the airport at Heho for our flight to Yangon. Our route takes us back to Heho and down the escarpment to Nyaung-Shwe for a busy local market. Check in at airport and fly back to Yangon. Upon arrival, check in at hotel. In the afternoon, we plan to visit by car to Karaweik, a royal barge for fantastic photo shot along the "pilgrims" causeway across the Royal Lake with stunning views of Schwedagon Pagoda and visit the Bogyoke Aung San Market (Scott Market) for souvenir shopping. Overnight in Yangon.