# An explicit construction of Kaleidocycles

Shizuo  Kenji  Shota

(XXX; XXX)

###### Abstract

We model a family of closed kinematic chains, known as Kaleidocycles, with the
theory of discrete spatial curves. By leveraging the connection between the
deformation of discrete curves and the semi-discrete integrable systems, we
describe the motion of a Kaleidocycle by elliptic theta functions. This study
showcases an interesting example in which an integrable system generates an
orbit in the space of the real solutions of polynomial equations defined by
geometric constraints.

KajiFukuoka KajiwaraFukuoka ShigetomiFukuoka Institute of Mathematics for
Industry, Kyushu University, 744 Motooka, Nishi-ku, Fukuoka 819-0395,
<EMAIL_ADDRESS>Institute of Mathematics for Industry, Kyushu
University, 744 Motooka, Nishi-ku, Fukuoka 819-0395,
<EMAIL_ADDRESS>Institute of Mathematics for Industry, Kyushu
University, 744 Motooka, Nishi-ku, Fukuoka 819- 0395, Japans-
<EMAIL_ADDRESS>
##  1 Introduction

The primary motivation of this work comes from the analysis of the motion of a
particular class of linkage mechanisms called _Kaleidocycles_ (Fig. 1 (Left)).
It consists of (typically six) identical copies of equifacial tetrahedra
connected by their opposing edges as hinge joints to form a ring. A
Kaleidocycle can be folded from a sheet of paper and makes an example of rigid
origami; it turns like a bubble ring while keeping all faces rigid. The origin
of Kaleidocycle is obscure but one of the first appearances in the literature
is in Bricard’s paper [3], where he considers an essentially equivalent closed
kinematic chain. This kinematic chain, now known as the Bricard 6R, is an
_overconstrained_ linkage, which violates Maxwell’s degree-of-freedom counting
law [8]. For nearly a hundred years, the geometry and kinematics of this
object have attracted the attention of researchers (see [25] and the
references therein).

For a fixed number $k$ of tetrahedra and their shape, the state of a
Kaleidocycle is identified by the sequence of the turning angles at the
hinges, which we define in the following manner. Notice that the line segment,
which we call the axis, connecting the centres of hinge edges for each
equifacial tetrahedron is orthogonal to both hinge edges. By connecting the
axes, we obtain a spatial closed polygonal curve. The angles at the vertices
of this polygon determine the state of a Kaleidocycle. The angles between the
hinge edges are reminiscent of the torsion angle of a polygonal curve, and the
turning angles are reminiscent of the curvature angle. We will make this
connection more rigorous and formulate the state of a Kaleidocycle in the
language of a discrete spatial curve (§2.1).

In the geometric formulation as discrete spatial curves, the motion of a
Kaleidocycle is modelled by a segment-length and torsion-angle preserving
deformation of the corresponding closed polygonal curve of a constant
torsional angle. Here, we ignore the self-collision of the tetrahedra. There
is another algebraic perspective of Kaleidocycles. The state of a Kaleidocycle
is determined by the coordinates of the vertices of the tetrahedra satisfying
the distance condition specified by the edge lengths. Thus, the state space,
or the configuration space, is identified with the space of real solutions to
a system of quadratic equations. The motion corresponds to a one-dimensional
orbit in the configuration space. The analysis of the real algebraic set
obtained as the configuration space of a Kaleidocycle is mathematically
challenging as well as of practical importance. For example, a simple counting
of the number of variables and the constraints tells that the state space of a
Kaleidocycle with $k$ tetrahedra is generally $(k-6)$-dimensional (up to rigid
motion). The first-named author and collaborators [19] have numerically
discovered that the state space degenerates to a circle when the hinge angle
is set to be maximal or minimal in some sense.111The results of the patent are
also published in [26]. This class of Kaleidocycles, named the _Möbius
Kaleidocycles_ , provides a rare example of _underconstrained_ linkage
mechanisms whose state space has less dimension than the expected one.
Numerical experiments suggest that the Kaleidocycles constructed in this paper
are indeed Möbius Kaleidocycles (Conjecture 6.1). The first-named and second-
named authors have investigated the motion of Möbius Kaleidocycles and
identified a particular motion of Kaleidocycles (not restricted to Möbius
Kaleidocycles) that is governed by the semi-discrete modified KdV (mKdV) and
the sine-Gordon equations [18](see §3.2). Given an initial state, a one-
dimensional orbit in the generally $(k-6)$-dimensional state space is obtained
by the time evolution defined by a system of ordinary differential equations.
The main result of this paper can be seen as giving an explicit solution to
the system by using the explicit construction in terms of the $\tau$ functions
of the two-component KP hierarchy [10]. However, our approach is different
from that in [18]; we start with the functional identities of elliptic theta
functions in connection to the compatibility conditions of the framing of a
discrete spatial curve. It is thus not trivial that these two approaches yield
the same motions. We show that our construction of the motion of the
Kaleidocycle (§5) indeed satisfies the system of the ordinary differential
equations described in [18]. The sweeping surface of the motion corresponds to
a semi-discrete analogue of the discrete constant negative curvature surface
(_K-surface_) studied in [2, 15].

Our work connects seemingly distant fields of integrable systems and
kinematics of linkage via an interesting object, Kaleidocycle. It provides a
remarkable example, where a flow generated by an integrable system on a real
algebraic set is explicitly integrated using elliptic theta functions.

The structure of the paper is as follows. §2 introduces Kaleidocycles and
their representation as discrete space curves. As essential numerical
invariants to describe the state of a Kaleidocycle, the signed curvature angle
and torsion angle are defined using the Frenet frame. A concise
parametrisation of a Kaleidocycle as a solution to a system of quadratic
equations corresponding to the geometric constraints on the binormal vectors
is given. §3 reviews the connection between the theory of curves and
integrable systems. The construction of discrete space curves and their Frenet
frames in terms of the $\tau$ functions is presented. A previous result that
describes the motion of a Kaleidocycle using semi-discrete integrable
equations is recalled. §4 constructs a set of $\tau$ functions with three
parameters $v,r,y$ using elliptic theta functions. The resulting curve is
shown to have a constant torsion angle. A condition for the resulting curve to
be closed is also discussed; for a given $k$, the appropriate choice of
parameters $m,r,v,y$ yields a Kaleidocycle. By introducing another parameter,
a family of time-parametrised Kaleidocycles is constructed in §5. Then, we
investigate the time evolution equations satisfied by the vertex positions and
curvature angle of the Kaleidocycles and show connections to previous
research. §6 provides examples of Kaleidocycles and corresponding semi-
discrete surfaces with figures. We present two conjectures that are suggested
by numerical experiments. Most of the proofs of our theorems are deferred to
Appendix since they require technically-involved computations regarding
elliptic theta functions.

The second and third-named authors were partially supported by KAKENHI Grant-
in-Aid for Scientific Research (C) 21K3229. All authors are partially
supported by JST CREST Grant No. JPMJCR1911.

##  2 Geometric and algebraic models of Kaleidocycle

###  2.1 Definition of Kaleidocycle

We first define a generalisation of Kaleidocycle following [17]. We consider
$k$-copies of an equifacial tetrahedron in $\mathbb{R}^{3}$ connected by
opposing edges to form a ring. Denote the vertex set by
$\\{0,1,\ldots,2k-1\\}$ such that the vertices $2n$ and $2n+1$ form a hinge
edge and $2n,2n+1,2n+2$ form one of the triangle faces. Let
$l_{01},l_{12},l_{20}$ be the lengths of the sides of the triangle face. A
state of a $k$-Kaleidocycles of type $(l_{01},l_{12},l_{20})$ is identified by
the real solution $(x_{0},x_{1},\ldots,x_{2k-1})\in(\mathbb{R}^{3})^{2k}$ of
the quadratic equations

(1) $\displaystyle|x_{2n}-x_{2n+1}|^{2}$ $\displaystyle=l^{2}_{01},$
$\displaystyle|x_{2n}-x_{2n+2}|^{2}$
$\displaystyle=|x_{2n+1}-x_{2n+3}|^{2}=l^{2}_{02},$
$\displaystyle|x_{2n+1}-x_{2n+2}|^{2}$
$\displaystyle=|x_{2n}-x_{2n+3}|^{2}=l^{2}_{12}.$

We note that we ignore collision between tetrahedra. The solution space of (1)
admits an action of the Euclidean transformation group $E(3)$ of
$\mathbb{R}^{3}$, and the expected dimension of the quotient is $3\cdot
2k-5k-\dim(E(3))=k-6$.

The state of a Kaleidocycle can be equivalently described by a spatial polygon
or a _closed discrete spatial curve_. Let $\gamma_{n}=(x_{2n}+x_{2n+1})/2\
(0\leq n<k-1)$ be the mid-point of the hinge edge. We set
$\gamma_{k}=\gamma_{0}$. Since the tetrahedron is equifacial,
$\gamma_{n+1}-\gamma_{n}$ is orthogonal to both $\widetilde{B}_{n}$ and
$\widetilde{B}_{n+1}$, where
$\widetilde{B}_{n}=\dfrac{x_{2n+1}-x_{2n}}{|x_{2n+1}-x_{2n}|}$ is the hinge
direction. By reordering the vertices, we can assume the inner product
$\langle\widetilde{B}_{n-1},\widetilde{B}_{n}\rangle$ is constant for $1\leq
n<k$. Denote this constant by $\cos\lambda$ with $0<\lambda<\pi$, which we
call the _torsion angle_. We have two cases
$\langle\widetilde{B}_{k-1},\widetilde{B}_{0}\rangle=\pm\cos\lambda$. We say
the Kaleidocycle is oriented (resp. non-oriented) when the value is
$+\cos\lambda$ (resp. $-\cos\lambda$). We set
$\widetilde{B}_{k}=\pm\widetilde{B}_{0}$ so that
$\langle\widetilde{B}_{k-1},\widetilde{B}_{k}\rangle=\cos\lambda$.
Furthermore, since $\widetilde{B}_{n+1}\times\widetilde{B}_{n}$ is parallel to
$\gamma_{n+1}-\gamma_{n}$, we have
$\sum_{n=0}^{k-1}(\widetilde{B}_{n+1}\times\widetilde{B}_{n})=\sum_{n=0}^{k-1}\frac{\gamma_{n+1}-\gamma_{n}}{|\gamma_{n+1}-\gamma_{n}|}=0$.
Conversely, the vertex coordinates $x_{n}$ can be reconstructed from
$\\{\gamma_{n}\\}$, $\\{\widetilde{B}_{n}\\}$, and $(l_{01},l_{12},l_{20})$ up
to translation. Therefore, $\\{\widetilde{B}_{n}\in S^{2}\mid 0\leq n\leq
k\\}$ satisfying the conditions

(2) $\displaystyle\langle\widetilde{B}_{n-1},\widetilde{B}_{n}\rangle$
$\displaystyle=\cos\lambda,$
$\displaystyle\sum_{n=0}^{k-1}(\widetilde{B}_{n+1}\times\widetilde{B}_{n})$
$\displaystyle=0,$ $\displaystyle\widetilde{B}_{k}$
$\displaystyle=\pm\widetilde{B}_{0}$

define a Kaleidocycle. We call the space of solutions to (2) the state space
of the $k$-Kaleidocycles with the torsion angle $\lambda$.

From this perspective, a motion of a Kaleidocycle is identified with a one-
dimensional (time-parametrised) orbit of the space of the quadratic system
defined by (1) or (2).

Figure 1: (Left) The classical 6-Kaleidocycle. (Right) Kaleidocycle as a
discrete curve with its framing.

###  2.2 Discrete spatial curve of a constant torsion angle

A discrete spatial curve with non-vanishing torsion angles can be described in
terms of binormals. Let $\widetilde{B}:\mathbb{Z}\rightarrow S^{2}$ satisfy
$\langle\widetilde{B}_{n},\widetilde{B}_{n-1}\rangle\neq 0$ for all
$n\in\mathbb{Z}$. A unit segment-length discrete spatial curve having
$\widetilde{B}$ as the binormals can be constructed as follows. Define the
tangent $T:\mathbb{Z}\rightarrow S^{2}$ and the normal
$\widetilde{N}:\mathbb{Z}\rightarrow S^{2}$ by

(3)
$T_{n}=\frac{\widetilde{B}_{n+1}\times\widetilde{B}_{n}}{\left|\widetilde{B}_{n+1}\times\widetilde{B}_{n}\right|},\quad\widetilde{N}_{n}=\widetilde{B}_{n}\times
T_{n}.$

Furthermore, given $\gamma_{0}\in\mathbb{R}^{3}$, define
$\gamma:\mathbb{Z}\to\mathbb{R}^{3}$ by

(4)
$\gamma_{n+1}=\gamma_{n}+T_{n}=\gamma_{n}+\frac{\widetilde{B}_{n+1}\times\widetilde{B}_{n}}{\left|\widetilde{B}_{n+1}\times\widetilde{B}_{n}\right|}.$

The _signed curvature angle_ $\kappa$ and the _torsion angle_ are defined by

(5) $\begin{split}\langle
T_{n},T_{n-1}\rangle&=\cos\kappa_{n},\quad-\pi<\kappa_{n}<\pi,\\\
\langle\widetilde{B}_{n},\widetilde{B}_{n-1}\rangle&=\cos\lambda_{n},\quad
0<\lambda_{n}<\pi.\end{split}$

The _discrete Frenet frame_
$\Phi_{n}=[T_{n},\widetilde{N}_{n},\widetilde{B}_{n}]\in\mathrm{SO}(3)$
satisfies the _discrete Frenet-Serret formula_

(6) $\Phi_{n+1}=\Phi_{n}L_{n},\quad
L_{n}=M_{1}(-\lambda_{n+1})M_{3}(\kappa_{n+1}),$

where

(7) $M_{1}(\theta)=\left[\begin{array}[]{ccc}1&0&0\\\
0&\cos\theta&-\sin\theta\\\ 0&\sin\theta&\cos\theta\\\
\end{array}\right],\quad
M_{3}(\theta)=\left[\begin{array}[]{ccc}\cos\theta&-\sin\theta&0\\\
\sin\theta&\cos\theta&0\\\ 0&0&1\end{array}\right].$

######  Remark 2.1.

Note that our Frenet frame coincides with the usual one only up to sign. We
adopt this convention to avoid discontinuous change in the framing during the
motion of the Kaleidocycle; its motion passes the state when the curvature
vanishes.

As in the case of a smooth curve, a discrete spatial curve of a unit segment
length is determined by its torsion and curvature angles. Therefore, the
motion of a discrete curve of a constant torsion angle is determined by the
set of functions $\kappa_{n}(t)$, which can be freely chosen. The fact that a
Kaleidocycle is _closed_ puts interesting topological constraints. For
example, the self-linking number must be a half-integer, and hence, stay
constant during the motion. By Călugăreanu-Fuller-White’s theorem [1], the
writhing number of the curve should stay constant as well. The self-linking
number also tells that the state space of a Kaleidocycle of a fixed type is
not connected in general.

##  3 Curves and integrable systems

There are deep connections between differential geometry and the theory of
integrable systems, where the integrable differential or difference equations
arise as compatibility conditions of the geometric objects (see, for example,
[24]). In the context of the deformation of spatial and plane curves, the
Frenet frame and its deformation are described by the system of linear partial
differential equations. The compatibility condition yields so-called the
Ablowitz-Kaup-Newell-Segur (AKNS) hierarchy for the curvature and the torsion,
including the nonlinear Schrödinger (NLS) equation as a typical example. In
particular, if the torsion is a constant, the integrable deformation is
governed by the mKdV equation and its hierarchy [9, 21, 24]. The connections
are also investigated in a discrete setting, where curves are discretised as
piece-wise linear (polygonal) curves. For example, continuous deformations of
discrete plane and spatial curves have been considered in [5], where the
deformations are governed by the semi-discrete analogue of the mKdV and the
NLS equations. Discrete deformations of the discrete plane and spatial curves
have been studied in [15]. In principle, for a given solution to the nonlinear
differential/difference equation for the curvature and torsion, one can
reconstruct the curve by solving the associated system of linear
differential/difference equations for the Frenet frame and integrating the
tangent vector.

The theta functions naturally appear in so-called the quasi-periodic solutions
to the nonlinear integrable equations, and the corresponding solutions to the
associated linear problem are nothing but the Baker-Akhiezer functions.
Reconstruction of the deformation of the spatial curve with constant torsion
described by the theta function solutions to the mKdV equation has been
studied in this framework in [4, 22]. Also, an explicit formula of the spatial
curves in terms of the $\tau$ functions of the two-component KP hierarchy
based on Hirota’s bilinear formalism has been presented in [10].

Since the dynamics of the Kaleidocycles are modelled by the integrable
deformation of the closed discrete spatial curves with a constant torsion
angle which preserves the segment-length and torsion angle [18], we study the
theta function solutions to the corresponding integrable semi-discrete
equations, where the theta functions are nothing but the $\tau$ functions.
Therefore we construct the explicit formula of the Kaleidocycles based on the
method of direct reconstruction of the discrete spatial curves in terms of the
$\tau$ functions.

Hereafter, a (discrete) curve always indicates a polygonal spatial curve.

###  3.1 Expression of a curve by the $\tau$ functions

To describe Kaleidocycles, we introduce a special class of curves represented
by the $\tau$ functions, which is a slight modification of the one presented
in [10]. We construct a family of discrete Frenet frames
$\Phi_{n}=[T_{n},\widetilde{N}_{n},\widetilde{B}_{n}]$ indexed by
$n\in\mathbb{Z}$ and the corresponding curve $\gamma_{n}$ satisfying (6). Let
$F_{n},\widetilde{G}_{n}:\mathbb{C}\to\mathbb{R}$ and
$f_{n},g_{n},G_{n},H_{n}:\mathbb{C}\to\mathbb{C}$ be functions indexed by
$n\in\mathbb{Z}$ that satisfy the following equations

(8) $\displaystyle f_{n}f_{n}^{*}+g_{n}g_{n}^{*}=F_{n+1}F_{n}$ (9)
$\displaystyle\beta F_{n}f_{n}+G_{n}g_{n}^{*}=F_{n+1}f_{n-1}$ (10)
$\displaystyle\beta F_{n}g_{n}-G_{n}f_{n}^{*}=F_{n+1}g_{n-1}$ (11)
$\displaystyle H_{n+1}F_{n}-H_{n}F_{n+1}=f_{n}^{*}g_{n}$ (12) $\displaystyle
D_{z}F_{n+1}\cdot F_{n}=g_{n}g_{n}^{*}$ (13)
$\displaystyle\left|G_{n}\right|^{2}=\widetilde{G}_{n}^{2},$

where $\beta\in\mathbb{R}$ is some constant and $D_{z}$ is Hirota’s bilinear
differential operator [11]. These functions are called the $\tau$ functions.
Let $\varrho_{n}=\dfrac{1}{2}\dfrac{G_{n}}{\widetilde{G}_{n}}$ and

(17) $\displaystyle T_{n}$
$\displaystyle=\left.\frac{1}{f_{n}f_{n}^{*}+g_{n}g_{n}^{*}}\left(\begin{array}[]{c}\vspace{2mm}f_{n}^{*}g_{n}+f_{n}g_{n}^{*}\\\
\vspace{2mm}\frac{1}{i}\left(f_{n}^{*}g_{n}-f_{n}g_{n}^{*}\right)\\\
f_{n}f_{n}^{*}-g_{n}g_{n}^{*}\end{array}\right)\right|_{z=0}$ (21)
$\displaystyle\widetilde{N}_{n}$
$\displaystyle=\left.\frac{1}{f_{n}f_{n}^{*}+g_{n}g_{n}^{*}}\left(\begin{array}[]{c}\vspace{2mm}\left\\{(f_{n}^{*})^{2}-(g_{n}^{*})^{2}\right\\}\varrho_{n}+\left\\{(f_{n})^{2}-(g_{n})^{2}\right\\}\varrho_{n}^{*}\\\
\vspace{2mm}\frac{1}{i}\left[\left\\{(f_{n}^{*})^{2}+(g_{n}^{*})^{2}\right\\}\varrho_{n}-\left\\{(f_{n})^{2}+(g_{n})^{2}\right\\}\varrho_{n}^{*}\right]\\\
-2\left(f_{n}^{*}g_{n}^{*}\varrho_{n}+f_{n}g_{n}\varrho_{n}^{*}\right)\end{array}\right)\right|_{z=0}$
(25) $\displaystyle\widetilde{B}_{n}$
$\displaystyle=\left.\frac{1}{f_{n}f_{n}^{*}+g_{n}g_{n}^{*}}\left(\begin{array}[]{c}\vspace{2mm}\frac{1}{i}\left[-\left\\{(f_{n}^{*})^{2}-(g_{n}^{*})^{2}\right\\}\varrho_{n}+\left\\{(f_{n})^{2}-(g_{n})^{2}\right\\}\varrho_{n}^{*}\right]\\\
\vspace{2mm}\left\\{(f_{n}^{*})^{2}+(g_{n}^{*})^{2}\right\\}\varrho_{n}+\left\\{(f_{n})^{2}+(g_{n})^{2}\right\\}\varrho_{n}^{*}\\\
\frac{2}{i}\left(f_{n}^{*}g_{n}^{*}\varrho_{n}-f_{n}g_{n}\varrho_{n}^{*}\right)\end{array}\right)\right|_{z=0}$
(29) $\displaystyle\gamma_{n}$
$\displaystyle=\left.\left(\begin{array}[]{c}\vspace{2mm}\dfrac{H_{n}+H_{n}^{*}}{F_{n}}\\\
\vspace{2mm}\dfrac{1}{i}\dfrac{H_{n}-H_{n}^{*}}{F_{n}}\\\
n-2\dfrac{\partial\log F_{n}}{\partial z}\end{array}\right)\right|_{z=0},$

where we denote by $z$ the variable of the $\tau$ functions so that, for
example, $F_{n}=F_{n}(z)$.

Similarly to [10, §3], we obtain:

######  Proposition 3.1.

$\Phi_{n}=[T_{n},\widetilde{N}_{n},\widetilde{B}_{n}]$ defines the discrete
Frenet frame for the curve $\gamma$. Its curvature and torsion angles are
given by

(30) $\displaystyle\lambda_{n}$
$\displaystyle=\left.\frac{1}{i}\log\left(\frac{G_{n}^{*}G_{n-1}}{\widetilde{G}_{n}\widetilde{G}_{n-1}}\right)\right|_{z=0}=\left.\frac{1}{2i}\log\left(\frac{G_{n}^{*}G_{n-1}}{G_{n}G_{n-1}^{*}}\right)\right|_{z=0},$
$\displaystyle\kappa_{n}$
$\displaystyle=\left.2\arctan\left(\frac{\widetilde{G}_{n}}{\beta
F_{n}}\right)\right|_{z=0}.$

###  3.2 Motion of Kaleidocycle on the osculating plane

Let $\gamma$ be a Kaleidocycle, that is, a closed curve with a constant
torsion angle and a unit segment length. A motion of a Kaleidocycle is
identified with a time-parametrised family
$\gamma:\mathbb{Z}\times\mathbb{R}_{\geq 0}\to\mathbb{R}^{3}$ such that
$\gamma(t)$ is a Kaleidocycle for all $t\geq 0$. In [18], a particular motion
of a Kaleidocycle satisfying the following conditions is considered:

(31)
$\left\langle\dot{\gamma}_{n},\widetilde{B}_{n}\right\rangle=0,\quad\left\langle\dot{\gamma}_{n},\dot{\gamma}_{n}\right\rangle=\tilde{\rho}^{2},\quad|{\gamma}_{n+1}-{\gamma}_{n}|=1\quad(0\leq\forall
n\leq k),$

where $\tilde{\rho}>0$ is a constant and
$\dot{\gamma}_{n}=\dfrac{d\gamma_{n}}{dt}$. These conditions mean that the
velocity of each vertex is restricted within the osculating plane and has a
constant magnitude. It is shown in [18] that the Frenet-Serret formula (6)
together with (31) demand that there exists a function $\Theta_{n}$, called
the _potential function_ , with
$\kappa_{n}=\dfrac{\Theta_{n+1}-\Theta_{n-1}}{2}$ that satisfies either the
_semi-discrete potential mKdV equation_ :

(32)
$\frac{d}{dt}\left(\frac{\Theta_{n+1}+\Theta_{n}}{2}\right)=\left(1+\cos\lambda\right)\tilde{\rho}\sin\left(\frac{\Theta_{n+1}-\Theta_{n}}{2}\right),$

or the _semi-discrete sine-Gordon equation_

(33)
$\frac{d}{dt}\left(\frac{\Theta_{n+1}-\Theta_{n}}{2}\right)=\left(1-\cos\lambda\right)\tilde{\rho}\sin\left(\frac{\Theta_{n+1}+\Theta_{n}}{2}\right).$

The motions corresponding to these equations differ in general. However, the
explicit construction discussed in §5 satisfies both equations simultaneously.

As noted in Introduction, for a fixed $k$, the dimension of the configurations
of a Kaleidocycle defined by (2) is generally $k-6$ and (31) may look too
restrictive. However, for the non-oriented Kaleidocycle with the minimum
torsional angle (the Möbius Kaleidocycle), the dimension of the configurations
is conjectured to be one regardless of $k$. An extensive numerical simulation
suggests that the motion of the Möbius Kaleidocycle satisfies (2). Therefore,
(2) specifies a special yet essential motion of Kaleidocycles.

##  4 An explicit formula of closed discrete curve with a constant torsion
angle in terms of elliptic theta functions

Based on the representation of curves discussed in §3.1, we define certain
$\tau$ functions satisfying (8)-(13) using elliptic theta functions to
construct a family of discrete curves with a constant torsion angle. Then, we
give sufficient conditions for the curve to be closed and derive an explicit
expression for the motion of Kaleidocycles.

###  4.1 A family of open curves with a constant torsion angle

Throughout this section, $j$ runs through $\\{1,2,3,4\\}$. Fix
$w\in\mathbb{C}$ with $\mathrm{Im}(w)>0$. The elliptic theta functions
$\vartheta_{j}:\mathbb{C}\to\mathbb{C}$, are defined as follows [23, 20]:

(34) $\displaystyle\vartheta_{1}(v;w)$
$\displaystyle=-\sum_{n=-\infty}^{\infty}e^{\pi
iw\left(n+\frac{1}{2}\right)^{2}+2\pi
i\left(n+\frac{1}{2}\right)\left(v+\frac{1}{2}\right)},$
$\displaystyle\vartheta_{2}(v;w)$
$\displaystyle=\sum_{n=-\infty}^{\infty}e^{\pi
iw\left(n+\frac{1}{2}\right)^{2}+2\pi i\left(n+\frac{1}{2}\right)v},$
$\displaystyle\vartheta_{3}(v;w)$
$\displaystyle=\sum_{n=-\infty}^{\infty}e^{\pi iwn^{2}+2\pi inv},$
$\displaystyle\vartheta_{4}(v;w)$
$\displaystyle=\sum_{n=-\infty}^{\infty}e^{\pi iwn^{2}+2\pi
in\left(v+\frac{1}{2}\right)}.$

We often denote $\vartheta_{j}(v;w)$ by $\vartheta_{j}(v)$ when $w$ is fixed.

We construct the $\tau$ functions satisfying (8)-(13) using these elliptic
theta functions so that the torsion angle given by Proposition 3.1 is a
constant. For $x\in\mathbb{C}$ and $y>0$, we define
$d_{j}(x;y)=\frac{\vartheta^{{}^{\prime}}_{j}(x;iy)}{\vartheta_{j}(x;iy)}$,
where ${}^{{}^{\prime}}$ denotes the derivative with respect to the first
variable of elliptic theta functions. For
$r,v\in\mathbb{R}\setminus{(1/2\mathbb{Z})}$ and $y>0$, we define

(35)
$\displaystyle\alpha_{2}=\sqrt{\vartheta_{3}(2r;iy)\vartheta_{3}(0;iy)},\quad\beta=\frac{\vartheta_{3}(iv;iy)}{\vartheta_{3}(0;iy)},\quad
s_{2}=\vartheta_{3}(2r;iy)\vartheta_{1}(iv;iy),$ $\displaystyle
u_{2}=\vartheta_{1}\left(-\frac{1}{2}iv+r;iy\right)\vartheta_{3}\left(-\frac{1}{2}iv+r;iy\right)\vartheta_{1}\left(\frac{1}{2}iv+r;iy\right)\vartheta_{3}\left(\frac{1}{2}iv+r;iy\right),$
$\displaystyle\Delta_{j}=d_{j}\left(-\frac{1}{2}iv+r;y\right)-d_{j}\left(\frac{1}{2}iv+r;y\right),$
$\displaystyle
R_{j}=\frac{\vartheta_{j}\left(-\frac{1}{2}iv+r;iy\right)}{\vartheta_{j}\left(\frac{1}{2}iv+r;iy\right)}.$

Noting $\vartheta_{j}(z;iy)^{*}=\vartheta_{j}(z^{*};iy)$ when $y>0$, we have

(36) $\begin{split}&\alpha_{2}^{*}=\alpha_{2},\quad\beta^{*}=\beta,\quad
s_{2}^{*}=-s_{2},\quad u_{2}^{*}=u_{2},\\\ &\Delta_{j}^{*}=-\Delta_{j},\quad
R_{j}R_{j}^{*}=1.\end{split}$

######  Theorem 4.1.

Put $\mu_{n}=ivn+\frac{z}{\Delta_{3}-\Delta_{1}}$ and define

(37) $\displaystyle
F_{n}(z)=\alpha_{2}\exp\left(\frac{n\Delta_{3}}{\Delta_{3}-\Delta_{1}}z\right)\vartheta_{2}\left(\mu_{n}-\frac{1}{2}iv\right),$
$\displaystyle
f_{n}(z)=\vartheta_{3}\left(-\frac{1}{2}iv+r\right)R_{3}^{n}\exp\left(\frac{\left(n+\frac{1}{2}\right)\Delta_{3}}{\Delta_{3}-\Delta_{1}}z\right)\vartheta_{2}\left(\mu_{n}+r\right),$
$\displaystyle
g_{n}(z)=\vartheta_{1}\left(\frac{1}{2}iv+r\right)R_{1}^{-n}\exp\left(\frac{\left(n+\frac{1}{2}\right)\Delta_{3}}{\Delta_{3}-\Delta_{1}}z\right)\vartheta_{4}\left(\mu_{n}-r\right),$
$\displaystyle
G_{n}(z)=\alpha_{2}\frac{\vartheta_{1}(iv)}{\vartheta_{3}(0)}R_{1}^{-n}R_{3}^{n}\exp\left(\frac{n\Delta_{3}}{\Delta_{3}-\Delta_{1}}z\right)\vartheta_{4}\left(\mu_{n}-\frac{1}{2}iv\right),$
$\displaystyle
H_{n}(z)=\frac{u_{2}}{\alpha_{2}s_{2}}R_{1}^{-n}R_{3}^{-n}\exp\left(\frac{n\Delta_{3}}{\Delta_{3}-\Delta_{1}}z\right)\vartheta_{4}\left(\mu_{n}-\frac{1}{2}iv-2r\right),$
$\displaystyle\widetilde{G}_{n}(z)=-i\alpha_{2}\frac{\vartheta_{1}(iv)}{\vartheta_{3}(0)}\exp\left(\frac{n\Delta_{3}}{\Delta_{3}-\Delta_{1}}z\right)\vartheta_{4}\left(\mu_{n}-\frac{1}{2}iv\right).$

Then, they satisfy (8)-(13) to form $\tau$ functions. The corresponding
discrete curve $\gamma$ defined by (17) has a unit segment length
$\left|\gamma_{n+1}-\gamma_{n}\right|=1$. The torsion angle is given by

(38)
$\lambda_{n}=\frac{1}{i}\log\left(\frac{G_{n}^{*}(0)G_{n-1}(0)}{\widetilde{G}_{n}(0)\widetilde{G}_{n-1}(0)}\right)=\frac{1}{i}\log\left(R_{1}R_{3}^{-1}\right).$

In particular, this does not depend on $n$, and $\gamma$ has a constant
torsion angle $\lambda_{n}=\lambda_{0}$. Moreover, the signed curvature angle
is given by

(39) $\begin{split}&\kappa_{n}=2\arctan\left(\frac{\widetilde{G}_{n}(0)}{\beta
F_{n}(0)}\right)\\\
&=2\arctan\left(-i\frac{\vartheta_{1}(iv)\vartheta_{4}\left(ivn-\frac{1}{2}iv\right)}{\vartheta_{3}(iv)\vartheta_{2}\left(ivn-\frac{1}{2}iv\right)}\right),\end{split}$

where we take the primary value for $\arctan$ in $(-\pi/2,\pi/2)$.

###  4.2 Sufficient conditions for closedness

The constant torsion-angle and unit segment-length curves given in Theorem 4.1
are not necessarily closed. We derive a sufficient condition on the parameters
$v,r,y$ for the curve $\gamma$ to be closed to define a Kaleidocycle. We say a
curve $\gamma$ is closed with a period $k\in\mathbb{N}$ when
$\gamma_{n+k}=\gamma_{n}$ for all $n$. Rewriting this condition using the
$\tau$ functions, we obtain

(40) $\displaystyle\frac{H_{n+k}}{F_{n+k}}=\frac{H_{n}}{F_{n}},$
$\displaystyle n+k-2\displaystyle\frac{\partial\log F_{n+k}}{\partial
z}=n-2\displaystyle\frac{\partial\log F_{n}}{\partial z}.$

To give general solutions to (40) is difficult, but we can utilise the pseudo-
period of elliptic theta functions (98) to obtain particular solutions. For a
fixed $k$, choose $m\in\mathbb{Z}\setminus{\\{0\\}}$ and set $v=my/k$. Then,
from (98), we see

(41)
$\displaystyle\frac{H_{n+k}}{F_{n+k}}=\frac{H_{n}}{F_{n}}\left(R_{1}R_{3}\right)^{-k}\exp\left(4mr\pi
i\right)(-1)^{m},$ $\displaystyle
F_{n+k}=F_{n}\exp\left(k\frac{\Delta_{3}}{\Delta_{3}-\Delta_{1}}z\right)\exp\left[-\pi
i\left\\{2m\left(\mu_{n}-\frac{1}{2}iv\right)+m^{2}w\right\\}\right],$

where $\mu_{n}=ivn+\frac{z}{\Delta_{3}-\Delta_{1}}$. Using (41), (40) can be
rewritten as

(42) $\displaystyle\Lambda:=\exp\left\\{\pi
im\left(4r+1\right)\right\\}\left(R_{1}R_{3}\right)^{-k}-1=0,$
$\displaystyle\Xi:=i\left(\Delta_{3}+\Delta_{1}\right)+4\pi\frac{v}{y}=0.$

######  Theorem 4.2.

For a fixed $k$, let $m\in\mathbb{Z}\setminus{\\{0\\}}$,
$r,v\in\mathbb{R}\setminus{(1/2\mathbb{Z})}$, and $y>0$ satisfy $kv=my$ and
(42). Then, the curve defined by Theorem 4.1 is closed and forms a
$k$-Kaleidocycle with a torsion angle given by (38).

Hereafter, we assume that the parameters $v,r,y$ are taken to satisfy the
closure condition.

######  Example 4.3.

The classical 6-Kaleidocycle (Figure 1 Left) with $\lambda=\pi/2$ (e.g., see
[8]) is obtained by $(v,r,y,m)\approx(0.640,0.250,1.279,3)$.

######  Remark 4.4.

By using (42), we see that $\widetilde{B}_{0}=(-1)^{m}\widetilde{B}_{k}$. From
this, we also see that $m$ and $Lk$ have the same parity. Furthermore, under
the transformation $\widetilde{B}_{n}\rightarrow(-1)^{n+1}\widetilde{B}_{n}$,
$\widetilde{N}_{n}\rightarrow(-1)^{n+1}\widetilde{N}_{n}$, the curvature
becomes $(-1)^{n+1}\kappa_{n}$ and the torsion angle becomes $\pi+\lambda$.
This indicates that this transformation corresponds to the one described in
Proposition 5.16.

##  5 Construction of the motion of a Kaleidocycle

By introducing the deformation parameter to the curve constructed in §4.1, we
build an explicit formula for the motion of a Kaleidocycle in Theorem 5.1. We
then derive the evolution equations for the vertex positions of the curve and
the (potential of the) curvature. Connections to the previously-known
isoperimetric deformation of a curve by the semi-discrete mKdV equation [15]
and by the semi-discrete potential mKdV equation and the semi-discrete sine-
Gordon equation [18] are discussed and summarised in Table 2.

###  5.1 Explicit formula of the discrete curve with deformation parameter

The following theorem describes a Kaleidocycle together with its motion
explicitly in terms of elliptic theta functions.

######  Theorem 5.1.

Denote $\mu_{n}=ivn+\frac{z}{\Delta_{3}-\Delta_{1}}$. Consider the following
$\tau$ functions parameterised by $t\in\mathbb{R}$.

(43) $\displaystyle
F_{n}(t,z)=\alpha_{2}\exp\left(\frac{n\Delta_{3}}{\Delta_{3}-\Delta_{1}}z+\frac{C}{2}tz\right)\vartheta_{2}\left(\mu_{n}-\frac{1}{2}iv+it\right),$
$\displaystyle
f_{n}(t,z)=\vartheta_{3}\left(-\frac{1}{2}iv+r\right)R_{3}^{n}\exp\left(\frac{\left(n+\frac{1}{2}\right)\Delta_{3}}{\Delta_{3}-\Delta_{1}}z+\frac{C}{2}tz-\frac{\Gamma}{2}it\right)\vartheta_{2}\left(\mu_{n}+r+it\right),$
$\displaystyle
g_{n}(t,z)=\vartheta_{1}\left(\frac{1}{2}iv+r\right)R_{1}^{-n}\exp\left(\frac{\left(n+\frac{1}{2}\right)\Delta_{3}}{\Delta_{3}-\Delta_{1}}z+\frac{C}{2}tz+\frac{\Gamma}{2}it\right)\vartheta_{4}\left(\mu_{n}-r+it\right),$
$\displaystyle
G_{n}(t,z)=\alpha_{2}\frac{\vartheta_{1}(iv)}{\vartheta_{3}(0)}R_{1}^{-n}R_{3}^{n}\exp\left(\frac{n\Delta_{3}}{\Delta_{3}-\Delta_{1}}z+\frac{C}{2}tz\right)\vartheta_{4}\left(\mu_{n}-\frac{1}{2}iv+it\right),$
$\displaystyle
H_{n}(t,z)=\frac{u_{2}}{\alpha_{2}s_{2}}R_{1}^{-n}R_{3}^{-n}\exp\left(\frac{n\Delta_{3}}{\Delta_{3}-\Delta_{1}}z+\frac{C}{2}tz+\Gamma
it\right)\vartheta_{4}\left(\mu_{n}-\frac{1}{2}iv-2r+it\right),$
$\displaystyle\widetilde{G}_{n}(t,z)=-i\alpha_{2}\frac{\vartheta_{1}(iv)}{\vartheta_{3}(0)}\exp\left(\frac{n\Delta_{3}}{\Delta_{3}-\Delta_{1}}z+\frac{C}{2}tz\right)\vartheta_{4}\left(\mu_{n}-\frac{1}{2}iv+it\right),$

where $C,\Gamma\in\mathbb{R}$ are constants. Then, they satisfy (8)-(13) to
form $\tau$ functions. For each $t$, the corresponding discrete curve $\gamma$
(17) has a unit segment length $\left|\gamma_{n+1}-\gamma_{n}\right|=1$. The
torsion angle is given by

(44)
$\lambda_{n}=\frac{1}{i}\log\left(\frac{G_{n}^{*}(t,0)G_{n-1}(t,0)}{\widetilde{G}_{n}(t,0)\widetilde{G}_{n-1}(t,0)}\right)=\frac{1}{i}\log\left(R_{1}R_{3}^{-1}\right),$

In particular, this depends neither on $n$ nor on $t$, and $\gamma$ has a
constant torsion angle $\lambda_{n}(t)=\lambda_{0}(0)$. Moreover, the signed
curvature angle is given by

(45)
$\begin{split}&\kappa_{n}\left(t\right)=2\arctan\left(\frac{\widetilde{G}_{n}(t,0)}{\beta
F_{n}(t,0)}\right)\\\
&=2\arctan\left(-i\frac{\vartheta_{1}(iv)\vartheta_{4}\left(ivn-\frac{1}{2}iv+it\right)}{\vartheta_{3}(iv)\vartheta_{2}\left(ivn-\frac{1}{2}iv+it\right)}\right),\end{split}$

where we take the primary value for $\arctan$ in $(-\pi/2,\pi/2)$. If $r,v,y$
are chosen as in Theorem 4.2, $\gamma$ is closed for every $t$.

######  Proposition 5.2.

The parameter $C$ contributes only to the parallel translation of the curve
along the $Z$-axis, whereas $\Gamma$ contributes only to the rotation of the
curve around the $Z$-axis.

###### Proof.

For $A\in\mathbb{C}$ and $B:\mathbb{C}\to\mathbb{C}$, we have

(46) $\begin{split}&\left(\begin{array}[]{ccc}\cos\Gamma t&-\sin\Gamma t\\\
\sin\Gamma t&\cos\Gamma t\\\
\end{array}\right)\left(\begin{array}[]{ccc}A+A^{*}\\\
\frac{1}{i}\left(A-A^{*}\right)\end{array}\right)\\\
&=\frac{1}{2}\left(\begin{array}[]{ccc}e^{i\Gamma t}+e^{-i\Gamma
t}&-\frac{1}{i}\left(e^{i\Gamma t}-e^{-i\Gamma t}\right)\\\
\frac{1}{i}\left(e^{i\Gamma t}-e^{-i\Gamma t}\right)&e^{i\Gamma t}+e^{-i\Gamma
t}\\\ \end{array}\right)\left(\begin{array}[]{ccc}A+A^{*}\\\
\frac{1}{i}\left(A-A^{*}\right)\end{array}\right)\\\
&=\left(\begin{array}[]{ccc}e^{i\Gamma t}A+e^{-i\Gamma t}A^{*}\\\
\frac{1}{i}\left(e^{i\Gamma t}A-e^{-i\Gamma
t}A^{*}\right)\end{array}\right)\end{split}$

and

(47)
$\begin{split}2\frac{\partial\log\left(e^{\frac{C}{2}tz}B\right)}{\partial
z}=Ct+2\frac{\partial\log B}{\partial z}.\end{split}$

The assertion follows from the definition (17). ∎

We will make convenient choices of these parameters to derive deformation
equations.

Table 1 summarises the relationship between various variables. The first
column lists the parameters that determine the curve $\gamma$ through the
variables listed in the first row. A circle in the table indicates that the
variable shown in the first row is dependent on the parameter shown in the
first column.

Table 1: Dependency of variables on the parameters $v,r,y,k,m,t,C,\Gamma$.  | $\gamma$ | $\kappa$ | $\lambda$ | $\Lambda$ | $\Xi$ | $u_{2}$ | $\beta$ | $s_{2}$ | $R_{j}$ | $\Delta_{j}$ | $\alpha_{2}$ | $\beta_{1,2}$ | $\epsilon_{1,2}$ | $\omega$ | $\delta_{1,2,3}$
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
$v$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$
$r$ | $\circ$ |  | $\circ$ | $\circ$ | $\circ$ | $\circ$ |  | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$
$y$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$
$k$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$
$m$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$
$t$ | $\circ$ | $\circ$ |  |  |  |  |  |  |  |  |  |  |  |  |
$\Gamma$ | $\circ$ |  |  |  |  |  |  |  |  |  |  | $\circ$ |  |  |
$C$ | $\circ$ |  |  |  |  |  |  |  |  |  |  |  |  |  |
| (17) | (45) | (44) | (42) | (42) | (35) | (35) | (35) | (35) | (35) | (35) | (48) | (48) | (48) | (48)

The constructed motion has a periodic curvature. The next proposition follows
from (98).

######  Proposition 5.3.

We have $\kappa_{n}\left(t+y\right)=-\kappa_{n}\left(t\right)$. In other
words, $2y$ is a period of the motion.

######  Remark 5.4.

There are certain relations between the curves constructed in Theorem 5.1 with
different sets of parameters. The transformation of parameters
$(v,m)\rightarrow(-v,-m)$ preserves the closure condition. We see that
$\kappa_{n}\left(t;-v\right)=-\kappa_{n}\left(-t;v\right),\lambda(-v,r)=-\lambda(v,r)$.
In what follows, We always choose $(v,m)$ or $(-v,-m)$ so that $\lambda\geq
0$.

###  5.2 Motion of vertex positions

The construction in Theorem 5.1 involves parameters $C,\Gamma$ that contribute
only to the global rigid transformation of the curve. In general, the velocity
of the vertex position $\dot{\gamma_{n}}$ has a binormal component, and its
magnitude is not constant as described in Theorem 5.6. However, with the right
choice of $C,\Gamma$, curves that satisfy (31) are obtained as described in
Theorem 5.9.

We derive bilinear equations satisfied by the $\tau$ functions given by
Theorem 5.1 to calculate $\dot{\gamma_{n}}$.

For $\Gamma\in\mathbb{R},r,v\in\mathbb{R}\setminus{(1/2\mathbb{Z})}$ and
$y>0$, we define

(48)
$\displaystyle\omega=\frac{\vartheta_{1}\left(2r;iy\right)\vartheta_{3}\left(0;iy\right)}{\vartheta_{3}\left(2r;iy\right)\vartheta_{1}\left(iv;iy\right)},$
$\displaystyle\delta_{1}=\frac{\vartheta_{3}\left(\frac{1}{2}iv+r;iy\right)^{2}\vartheta_{3}\left(-\frac{1}{2}iv+r;iy\right)^{2}}{\vartheta_{3}\left(2r;iy\right)\vartheta_{3}\left(0;iy\right)^{3}},$
$\displaystyle\delta_{2}=\frac{\vartheta_{1}\left(\frac{1}{2}iv+r;iy\right)^{2}\vartheta_{3}\left(-\frac{1}{2}iv+r;iy\right)^{2}}{\vartheta_{1}\left(iv;iy\right)^{2}\vartheta_{3}\left(2r;iy\right)\vartheta_{3}\left(0;iy\right)},$
$\displaystyle\delta_{3}=\frac{-iu_{2}}{s_{2}\vartheta_{3}(0;iy)\vartheta_{1}(iv;iy)}\left[\frac{\vartheta_{1}^{{}^{\prime}}\left(\frac{1}{2}iv+r;iy\right)\vartheta_{3}\left(-\frac{1}{2}iv+r;iy\right)}{\vartheta_{1}\left(-\frac{1}{2}iv+r;iy\right)\vartheta_{3}\left(\frac{1}{2}iv+r;iy\right)}\right.$
$\displaystyle\left.+\frac{\vartheta_{1}\left(\frac{1}{2}iv+r;iy\right)\vartheta_{3}^{{}^{\prime}}\left(-\frac{1}{2}iv+r;iy\right)}{\vartheta_{1}\left(-\frac{1}{2}iv+r;iy\right)\vartheta_{3}\left(\frac{1}{2}iv+r;iy\right)}\right],$
$\displaystyle\epsilon_{1}=\frac{2i}{\Delta_{3}-\Delta_{1}}\frac{\vartheta_{3}^{{}^{\prime\prime}}(0;iy)}{\vartheta_{3}(0;iy)},$
$\displaystyle\epsilon_{2}=\frac{2i}{\Delta_{3}-\Delta_{1}}\frac{\vartheta_{1}^{{}^{\prime}}(0;iy)^{2}}{\vartheta_{1}(iv;iy)^{2}},$
$\displaystyle\beta_{1}=\frac{1}{2}\left(\delta_{3}+\delta_{3}^{*}\right)+\frac{1}{2}i\Gamma\left(\delta_{2}-\delta_{2}^{*}\right),$
$\displaystyle\beta_{2}=\frac{1}{2}i\left(\delta_{3}-\delta_{3}^{*}\right)-\frac{1}{2}\Gamma\left(\delta_{2}+\delta_{2}^{*}\right).$

Then, it follows that

(49)
$\displaystyle\omega^{*}=-\omega,\quad\delta_{1}^{*}=\delta_{1},\quad\epsilon_{1}^{*}=\epsilon_{1},\quad\epsilon_{2}^{*}=\epsilon_{2},\quad\beta_{1}^{*}=\beta_{1},\quad\beta_{2}^{*}=\beta_{2}.$

A proof of the following proposition is given in Appendix A.

######  Proposition 5.5.

The $\tau$ functions (43) satisfy the following bilinear equations.

(50) $\displaystyle f_{n}f_{n-1}^{*}+g_{n}^{*}g_{n-1}=\beta F_{n}F_{n}+\omega
G_{n}G_{n}^{*},$ (51) $\displaystyle
g_{n}^{*}g_{n-1}-g_{n}g_{n-1}^{*}=f_{n}f_{n-1}^{*}-f_{n}^{*}f_{n-1}=\omega
G_{n}G_{n}^{*},$ (52)
$\displaystyle\delta_{2}f_{n}^{*}g_{n-1}-\delta_{2}^{*}f_{n-1}^{*}g_{n}=H_{n}F_{n},$
(53)
$\displaystyle\delta_{1}F_{n}F_{n}+\delta_{2}G_{n}G_{n}^{*}=f_{n}f_{n-1}^{*},$
(54)
$\displaystyle\left(\beta-\delta_{1}\right)F_{n}F_{n}+\left(\omega-\delta_{2}\right)G_{n}G_{n}^{*}=g_{n}^{*}g_{n-1},$
(55) $\displaystyle D_{t}H_{n}\cdot
F_{n}=\left(\delta_{3}+i\Gamma\delta_{2}\right)f_{n}^{*}g_{n-1}+\left(\delta_{3}^{*}-i\Gamma\delta_{2}^{*}\right)f_{n-1}^{*}g_{n},$
(56) $\displaystyle D_{t}D_{z}F_{n}\cdot
F_{n}=\left(\epsilon_{1}+C\right)F_{n}F_{n}+\epsilon_{2}G_{n}G_{n}^{*}.$

By using these equations, we obtain the evolution equation of the vertex
positions of the curve:

######  Theorem 5.6.

The curve $\gamma$ constructed in Theorem 5.1 satisfies the following
deformation equation.

(57)
$\begin{split}\dot{\gamma}_{n}(t)=&2\beta\left[\beta_{1}T_{n}-\beta_{1}\tan\frac{\kappa_{n}}{2}\widetilde{N}_{n}+\beta_{2}\tan\frac{\kappa_{n}}{2}\widetilde{B}_{n}\right]\\\
&+\left(\begin{array}[]{c}0\\\ 0\\\
2\beta_{1}\left(\beta-2\delta_{1}\right)-\epsilon_{1}-C\end{array}\right).\end{split}$

This is a deformation equation similar to the one presented in [15]. Starting
from (57), the compatibility condition of the Frenet frame yields the _semi-
discrete mKdV equation_.

######  Proposition 5.7.

When the curve with a unit segment length and a constant torsion angle deforms
according to the deformation equation (57), the following holds.

  1. 1.

The segment length is preserved.

  2. 2.

$\kappa$ satisfies the following semi-discrete mKdV equation:

(58)
$\dot{\kappa}_{n}=\left(2\beta\beta_{1}\cos\lambda+2\beta\beta_{2}\sin\lambda\right)\left(\tan\frac{\kappa_{n+1}}{2}-\tan\frac{\kappa_{n-1}}{2}\right).$

Furthermore, if the left-hand-side of the equation (69) does not depend on
$n$, the following holds:

  1. 3.

The torsion angle is preserved.

######  Proof.

The proofs of (i) and (ii) are given by the same calculation as in Section 5
of [15]. We prove (iii). From the deformation equation (57) and the discrete
Frenet-Serret formula (6), we see that

(59)
$\begin{split}&-\dot{\lambda}\sin\lambda=\left\langle\dot{\widetilde{B}}_{n},\widetilde{B}_{n+1}\right\rangle+\left\langle\widetilde{B}_{n},\dot{\widetilde{B}}_{n+1}\right\rangle\\\
&=-\sin\lambda\left(2\beta\beta_{1}\sin\lambda-2\beta\beta_{2}\cos\lambda\right)\times\left(\frac{1}{\sin\kappa_{n+1}}\tan\frac{\kappa_{n+2}}{2}\right.\\\
&\left.+\frac{1}{\tan\kappa_{n+1}}\tan\frac{\kappa_{n}}{2}-\frac{1}{\tan\kappa_{n}}\tan\frac{\kappa_{n+1}}{2}-\frac{1}{\sin\kappa_{n}}\tan\frac{\kappa_{n-1}}{2}\right),\end{split}$

holds, which vanishes because of (69). Therefore $\dot{\lambda}=0$. ∎

######  Remark 5.8.

Proposition 5.7 (iii) does not hold in general without the assumption on (69).
We can construct a motion of a Kaleidocycle satisfying (57) such that the
left-hand side of (69) depends both on $n$ and $t$. This suggests that the
conservation of the left-hand side of the recurrence equation (69) plays a
crucial role in the preservation of the torsion angle.

By adjusting the global motion of $\gamma$ by $C,\Gamma$, the deformation
equation of $\gamma$ can be made to satisfy the equation considered in the
previous studies presented in §3.2.

######  Theorem 5.9.

Set $C,\Gamma$ to be

(60)
$\begin{split}\Gamma_{+}=&\frac{-\left(\delta_{3}+\delta_{3}^{*}\right)+\left(\delta_{3}e^{i\lambda}+\delta_{3}^{*}e^{-i\lambda}\right)}{i\omega},\\\
C_{+}=&-\epsilon_{1}-i\Gamma_{+}\left(\omega\left(2\delta_{1}-\beta\right)-\frac{\delta_{2}+\delta_{2}^{*}}{\omega}\right)\\\
&-\left(\delta_{3}+\delta_{3}^{*}\right)\left(2\delta_{1}-\beta\right)+\frac{\delta_{3}-\delta_{3}^{*}}{\omega},\\\
\rm or\\\
\Gamma_{-}=&\frac{-\left(\delta_{3}+\delta_{3}^{*}\right)-\left(\delta_{3}e^{i\lambda}+\delta_{3}^{*}e^{-i\lambda}\right)}{i\omega},\\\
C_{-}=&-\epsilon_{1}-i\Gamma_{-}\left(\omega\left(2\delta_{1}-\beta\right)-\frac{\delta_{2}+\delta_{2}^{*}}{\omega}\right)\\\
&-\left(\delta_{3}+\delta_{3}^{*}\right)\left(2\delta_{1}-\beta\right)+\frac{\delta_{3}-\delta_{3}^{*}}{\omega}.\end{split}$

Then, the curve constructed in Theorem 5.1 satisfies the following deformation
equations.

(61)
$\begin{split}&\left\langle\dot{\gamma}_{n},\widetilde{B}_{n}\right\rangle=0,\quad\langle\dot{\gamma}_{n},\dot{\gamma}_{n}\rangle=\rho^{2},\end{split}$

where $\rho^{2}>0$ is given by

(62)
$\rho^{2}=\rho_{+}^{2}:=\left(i\frac{\vartheta_{1}^{{}^{\prime}}(0)}{\vartheta_{1}(iv)}\frac{u_{2}}{\vartheta_{3}(r)^{2}\vartheta_{1}(r)^{2}}\right)^{2}$

when $(C,\Gamma)=(C_{+},\Gamma_{+})$ and

(63)
$\rho^{2}=\rho_{-}^{2}:=\left(-i\frac{\vartheta_{1}^{{}^{\prime}}(0)}{\vartheta_{1}(iv)}\frac{u_{2}}{\vartheta_{4}(r)^{2}\vartheta_{2}(r)^{2}}\right)^{2}$

when $(C,\Gamma)=(C_{-},\Gamma_{-})$.

######  Remark 5.10.

A discrete analogue of the constant negative curvature surface, called the
discrete K-surface, is introduced in [2]. It is discussed that certain
discrete-time deformation of discrete curves with a constant torsion angle
generates discrete K-surfaces in [15]. Deformation equation (31) can be seen
as a continuous-time analogue of the discrete-time deformation. Theorem 5.9
shows that specialising $(C,\Gamma)$ to $(C_{-},\Gamma_{-})$ or
$(C_{+},\Gamma_{+})$, the curve constructed in Theorem 5.1 deforms according
to (31). Therefore, its sweeping surface forms a semi-discrete analogue of the
discrete K-surface (see figures in §6).

Since the curvature and torsion angles uniquely determine a curve up to a
rigid transformation, Proposition 5.3 shows that the motion constructed in
Theorem 5.1 is periodic up to a global rigid transform. Here, we control the
global orientation by choosing the parameters $C,\Gamma$ appropriately so that
the motion becomes periodic in terms of vertex positions.

######  Proposition 5.11.

Let $l\in\mathbb{Z}$. If we set

(64) $\displaystyle C$ $\displaystyle=\frac{4\pi
i}{y}\frac{1}{\Delta_{3}-\Delta_{1}},$ (65) $\displaystyle\Gamma$
$\displaystyle=-\frac{\pi}{y}\left(4r+2l+1\right),$

we have

(66) ${\gamma_{n}}(t+y)={\gamma_{n}}(t),$

for any $n\in\mathbb{Z}$ and $t\in\mathbb{R}$.

######  Proof.

First, we prove the periodicity for the $Z$-component. Denote the
$Z$-component of $\gamma$ by ${\gamma_{n}}_{Z}$. From the pseudo-period of
elliptic theta functions (98), we see

(67) $\begin{split}&{\gamma_{n}}_{Z}(t+y)-{\gamma_{n}}_{Z}(t)\\\
&=-2\frac{\partial}{\partial
z}\log\left\\{\exp\left(\frac{C}{2}yz\right)\frac{\vartheta_{2}\left(iv\left(n-\frac{1}{2}\right)+\frac{z}{\Delta_{3}-\Delta_{1}}+it+iy\right)}{\vartheta_{2}\left(iv\left(n-\frac{1}{2}\right)+\frac{z}{\Delta_{3}-\Delta_{1}}+it\right)}\right\\}\\\
&=-2\frac{\partial}{\partial z}\log\exp\left\\{\frac{C}{2}yz-\pi
i\left(2iv\left(n-\frac{1}{2}\right)+\frac{2z}{\Delta_{3}-\Delta_{1}}+2it\right)+\pi
y\right\\}\\\ &=-2\left(\frac{C}{2}y-\frac{2\pi
i}{\Delta_{3}-\Delta_{1}}\right)=0.\end{split}$

For the $X$ and $Y$ components, from (17) we have to show
$\frac{H_{n}(t+y)}{F_{n}(t+y)}\frac{F_{n}(t)}{H_{n}(t)}=1$.

If we write
$P_{n}=iv\left(n-\frac{1}{2}\right)+\frac{z}{\Delta_{3}-\Delta_{1}}+it$, the
left-hand side can be computed using (98) as

(68) $\begin{split}&\exp\left(i\Gamma
y\right)\frac{\vartheta_{4}\left(P_{n}-2r+iy\right)\vartheta_{2}\left(P_{n}\right)}{\vartheta_{2}\left(P_{n}+iy\right)\vartheta_{4}\left(P_{n}-2r\right)}\\\
&=-\exp\left(i\Gamma y\right)\frac{\exp\left\\{-\pi
i\left(2X_{n}-4r+iy\right)\right\\}}{\exp\left\\{-\pi
i\left(2X_{n}+iy\right)\right\\}}\\\ &=-\exp\left\\{i\left(\Gamma y+4\pi
r\right)\right\\}=1.\end{split}$

∎

###  5.3 Evolution of curvature

The curve constructed in Theorem 5.1 reveals an interesting recurrence
equation for $\kappa$.

######  Proposition 5.12.

The signed curvature angle of the curve obtained in Theorem 5.1 satisfies the
following recurrence equation for any $t$ and $n$:

(69)
$\tan\frac{\kappa_{n+1}}{2}\tan\frac{\kappa_{n-1}}{2}-\tan^{2}\frac{\kappa_{n}}{2}+\tan\frac{\kappa_{n+1}}{2}\tan^{2}\frac{\kappa_{n}}{2}\tan\frac{\kappa_{n-1}}{2}=-\frac{\vartheta_{1}\left(iv\right)^{4}}{\vartheta_{3}\left(iv\right)^{4}}.$

The right-hand side of this equation depends neither on $n$ nor on $t$.
Therefore, the quantity on the left-hand side can be considered a conserved
quantity of the signed curvature angle.

The evolution of the signed curvature angle satisfies the semi-discrete mKdV
equation in Proposition 5.7.

######  Theorem 5.13.

The signed curvature angle (45) satisfies

(70)
$\dot{\kappa}_{n}=2\beta^{2}\eta_{3}\left(\tan\frac{\kappa_{n+1}}{2}-\tan\frac{\kappa_{n-1}}{2}\right),$

where

(71)
$\eta_{3}=i\frac{\vartheta_{1}\left(iv\right)\vartheta_{3}\left(iv\right)}{\vartheta_{1}\left(2iv\right)\vartheta_{3}\left(0\right)}\left(d_{1}(iv)-d_{3}(iv)\right).$

Combined with Proposition 5.7, (48), and the fact that
$e^{i\lambda}\delta_{2}\in\mathbb{R}$, we see

(72)
$2\beta^{2}\eta_{3}=\left(2\beta\beta_{1}\cos\lambda+2\beta\beta_{2}\sin\lambda\right).$

###  5.4 Potential function of curvature

A family of functions $\Theta_{n}$ parameterised by $n\in\mathbb{Z}$ is said
to be a potential of the curvature $\kappa$ if the equation
$2\kappa_{n}=\Theta_{n+1}-\Theta_{n-1}$ holds. When $\gamma$ deforms by (61),
the potential function $\Theta$ is known to satisfy the semi-discrete
potential mKdV equation or the semi-discrete sine-Gordon equation [18]. Here,
we derive the evolution equations satisfied by the potential function for the
curve constructed by Theorem 5.1.

We construct a potential function which simultaneously satisfy (32) and (33)
in terms of elliptic theta functions.

######  Theorem 5.14.

For fixed $v,y$, define the function $\Theta$ as follows.

(73)
$\Theta_{n}\left(t\right)=\Theta_{n}\left(t;v,y\right)=4\arctan\frac{\vartheta_{1}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y};\frac{2i}{y}\right)}{\vartheta_{4}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y};\frac{2i}{y}\right)}.$

Then we see that $\kappa$ defined in (45) satisfies

(74) $\kappa_{n}=\frac{\Theta_{n+1}-\Theta_{n-1}}{2}.$

The potential function simultaneously solves the semi-discrete potential mKdV
equation and the semi-discrete sine-Gordon equation.

######  Theorem 5.15.

Let $v\in\mathbb{R}\setminus{(1/2\mathbb{Z})}$ and $y>0$. The potential
function $\Theta$ defined by (73) simultaneously satisfies the semi-discrete
potential mKdV and sine-Gordon equations

(75) $\displaystyle\frac{d}{dt}\left(\frac{\Theta_{n+1}+\Theta_{n}}{2}\right)$
$\displaystyle=\frac{L_{1}(v,y)}{y}\sin\left(\frac{\Theta_{n+1}-\Theta_{n}}{2}\right),$
(76) $\displaystyle\frac{d}{dt}\left(\frac{\Theta_{n+1}-\Theta_{n}}{2}\right)$
$\displaystyle=\frac{L_{2}(v,y)}{y}\sin\left(\frac{\Theta_{n+1}+\Theta_{n}}{2}\right),$

where $Q_{1},Q_{2},L_{1},L_{2}\in\mathbb{R}$ are defined by

(77) $\displaystyle
Q_{1}(v,y)=\frac{\vartheta^{{}^{\prime}}_{1}\left(0;\frac{2i}{y}\right)\vartheta_{1}\left(\frac{v}{y};\frac{2i}{y}\right)}{\vartheta_{4}\left(0;\frac{2i}{y}\right)\vartheta_{4}\left(\frac{v}{y};\frac{2i}{y}\right)},\quad
Q_{2}(v,y)=\frac{\vartheta^{{}^{\prime}}_{1}\left(0;\frac{2i}{y}\right)\vartheta_{4}\left(\frac{v}{y};\frac{2i}{y}\right)}{\vartheta_{1}\left(\frac{v}{y};\frac{2i}{y}\right)\vartheta_{4}\left(0;\frac{2i}{y}\right)},$
(78) $\displaystyle
L_{1}(v,y)=d_{1}\left(\frac{v}{y};\frac{2}{y}\right)-d_{4}\left(\frac{v}{y};\frac{2}{y}\right)+Q_{1}(v,y)+Q_{2}(v,y),$
(79) $\displaystyle
L_{2}(v,y)=d_{1}\left(\frac{v}{y};\frac{2}{y}\right)-d_{4}\left(\frac{v}{y};\frac{2}{y}\right)-Q_{1}(v,y)-Q_{2}(v,y).$

The following shows two equations are related by a parameter shift.

######  Proposition 5.16.

By the transformation of parameters $v\rightarrow v+y$ and
$r\rightarrow\frac{1}{2}-r$, the signed curvature angle $\kappa$, the torsion
angles $\lambda$, the potential functions $\Theta$, the coefficients of the
semi-discrete mKdV equation $L_{1}/y$ and of the semi-discrete sine-Gordon
equation $L_{2}/y$, the normal $\widetilde{N}$ and the binormal
$\widetilde{B}$ are transformed as

(80) $\displaystyle\kappa_{n}\left(t\right)$
$\displaystyle\rightarrow(-1)^{n+1}\kappa_{n}\left(t-\frac{y}{2}\right),\quad\lambda\rightarrow\pi+\lambda,$
(81) $\displaystyle\Theta_{n}\left(t\right)$
$\displaystyle\rightarrow(-1)^{n}\Theta_{n}\left(t-\frac{y}{2}\right),$ (82)
$\displaystyle\frac{L_{1}}{y}$
$\displaystyle\rightarrow\frac{L_{2}}{y},\quad\frac{L_{2}}{y}\rightarrow\frac{L_{1}}{y},$
(83) $\displaystyle\widetilde{N}_{n}(t)$
$\displaystyle\rightarrow(-1)^{n+1}\widetilde{N}_{n}\left(t-\frac{y}{2}\right),\quad\widetilde{B}_{n}(t)\rightarrow(-1)^{n+1}\widetilde{B}_{n}\left(t-\frac{y}{2}\right).$

The transformation
$\Theta_{n}\rightarrow\widetilde{\Theta}_{n}:=(-1)^{n}\Theta_{n}$ exchanges a
solution of the semi-discrete potential mKdV equation (75) with one of the
semi-discrete sine-Gordon equation (76). Proposition 5.16 shows that this
equation-level transformation is also achieved at the solution level by
transforming the parameters.

######  Remark 5.17.

Rewriting (70) in terms of the potential function $\Theta$, we obtain

(84)
$\frac{d}{dt}\Theta_{n}=4\beta^{2}\eta_{3}\tan\left(\frac{\Theta_{n+1}-\Theta_{n-1}}{4}\right),$

which is also called the _semi-discrete potential mKdV equation_ [13, 12, 6].
In other words, the potential function $\Theta$ defined by (73) is a
simultaneous solution of three differential-difference equations, (75), (76)
and (84). It does not usually happen that these equations are simultaneously
solved. For example, it can be checked that the 2 and 3-soliton solutions of
the semi-discrete sine-Gordon equation constructed in [6, 7] are not solutions
of the semi-discrete potential mKdV equation.

The following describes an explicit relation between our construction and the
previous work [18] reviewed in §3.2.

######  Proposition 5.18.

When $C_{+}$ and $\Gamma_{+}$ are chosen as in (60), we have

(85)
$\frac{L_{1}(v,y)}{y}\frac{1}{1+\cos\lambda}=i\frac{\vartheta_{1}^{{}^{\prime}}(0)}{\vartheta_{1}(iv)}\frac{u_{2}}{\vartheta_{3}(r)^{2}\vartheta_{1}(r)^{2}}=\rho_{+}.$

Similarly, when $C_{-}$ and $\Gamma_{-}$ are chosen as in (60), we have

(86)
$\frac{L_{2}(v,y)}{y}\frac{1}{1-\cos\lambda}=-i\frac{\vartheta_{1}^{{}^{\prime}}(0)}{\vartheta_{1}(iv)}\frac{u_{2}}{\vartheta_{4}(r)^{2}\vartheta_{2}(r)^{2}}=\rho_{-}.$

Note the difference in approaches of [18] and this work; [18] starts with a
geometric property observed in the motion of Kaleidocycles, and derive
evolution equations, whereas we aim to construct a particular motion of
Kaleidocycles and investigate the equations satisfied by them. They turn out
to arrive at the same place; the latter gives an explicit solution to the
equations derived in the former.

Table 2 summarises the relationships among the three deformation equations.

Table 2: Relation between the three deformation equations | Values of
---
$C,\Gamma$
Deformation equation of the curve | Equation for potential function
$C,\Gamma\in\mathbb{R}$ |  | $\dot{\gamma}_{n}=\footnotesize\left(\begin{array}[]{c}0\\\ 0\\\ 2\beta_{1}\left(\beta-2\delta_{1}\right)-\epsilon_{1}-C\end{array}\right)\normalsize$
---
$+2\beta\left(\beta_{1}T_{n}-\beta_{1}\tan\frac{\kappa_{n}}{2}\widetilde{N}_{n}+\beta_{2}\tan\frac{\kappa_{n}}{2}\widetilde{B}_{n}\right)$
|
$\displaystyle\dot{\Theta}_{n}=\left(4\beta\beta_{1}\cos\lambda+4\beta\beta_{2}\sin\lambda\right)$
---
$\displaystyle\times\tan\left(\frac{\Theta_{n+1}-\Theta_{n-1}}{4}\right)$
$\downarrow$ Specialise the deformation equation by choosing rigid
transformation parameters $C,\Gamma$.
$C_{+},\Gamma_{+}$ | $\displaystyle\left\langle\dot{\gamma}_{n},\widetilde{B}_{n}\right\rangle=0,\quad\langle\dot{\gamma}_{n},\dot{\gamma}_{n}\rangle=\rho^{2}_{+}$ |  | $\displaystyle\frac{\dot{\Theta}_{n+1}+\dot{\Theta}_{n}}{2}=\left(1+\cos\lambda\right)\rho_{+}$
---
$\displaystyle\times\sin\left(\frac{\Theta_{n+1}-\Theta_{n}}{2}\right)$
|  $\updownarrow$ By replacing the parameters of the solution with
$v\rightarrow v+y,r\rightarrow 1/2-r$,
---
these two deformation equations transform to each other.
$C_{-},\Gamma_{-}$ | $\displaystyle\left\langle\dot{\gamma}_{n},\widetilde{B}_{n}\right\rangle=0,\quad\langle\dot{\gamma}_{n},\dot{\gamma}_{n}\rangle=\rho^{2}_{-}$ |  | $\displaystyle\frac{\dot{\Theta}_{n+1}-\dot{\Theta}_{n}}{2}=\left(1-\cos\lambda\right)\rho_{-}$
---
$\displaystyle\times\sin\left(\frac{\Theta_{n+1}+\Theta_{n}}{2}\right)$

##  6 Numerical Experiments

In this section, we present examples of Kaleidocycles constructed by Theorem
4.2 and the semi-discrete K-surfaces obtained as their motion trajectory as
explained in Remark 5.10.

Numerical experiments using our computer codes [16] suggest that the $8$-,
$9$\- and $15$-Kaleidocycles in Figs. 2, 3, and 4 have the minimum torsion
angles among non-orientable $8$-, $9$-, and $15$-Kaleidocycles, respectively
[17]. In other words, they are the Möbius Kaleidocycles [19].

For Fig. 5, the torsion angle is larger than the above minimum one. However,
it has a self-linking number $7.5$. As the self-linking number of a
Kaleidocycle is a half-integer, these $15$-Kaleidocycles depicted in Figs. 4
and 5 belong to different connected components of the configuration space of
$15$-Kaleidocycles. We numerically find that the latter attains the local
minima of the torsion angle. We present the following two conjectures based on
our extensive numerical experiments:

######  Conjecture 6.1.

Any Kaleidocycle obtained by Theorem 4.2 has a minimal or maximal torsion
angle. In other words, the projection from the space defined by (2) onto the
$\lambda$-coordinate attains a critical value.

######  Conjecture 6.2.

The self-linking number of the closed curve $\gamma$ constructed in Theorem
4.2 is $m/2$ when the curve is unknotted.

As briefly discussed in §1, these Kaleidocycles with critical torsion angles
numerically exhibit special properties, including the single degree of freedom
[17, 19, 26]. Our construction produces this interesting family of
Kaleidocycles and may shed light on the rigorous proof for some of their
unique properties.

Figure 2: (Left) An example of 8-Kaleidocycle with
$(v,r,y,m)\approx(0.400,0.717,1.067,3)$ and $\cos\lambda\approx 0.4700$.
(Middle) 8-Kaleidocycle as a discrete curve. (Right) the semi-discrete
K-surface obtained as its trajectory.

Figure 3: (Left) An example of 9-Kaleidocycle with
$(v,r,y,m)\approx(0.342,0.289,1.027,3)$ and $\cos\lambda\approx 0.5852$.
(Middle) 9-Kaleidocycle as a discrete curve. (Right) the semi-discrete
K-surface obtained as its trajectory.

Figure 4: (Left) An example of 15-Kaleidocycle with
$(v,r,y,m)\approx(0.189,0.300,0.947,3)$ and $\cos\lambda\approx 0.8533$.
(Middle) 15-Kaleidocycle as a discrete curve. (Right) the semi-discrete
K-surface obtained as its trajectory.

Figure 5: (Left) An example of 15-Kaleidocycle with
$(v,r,y,m)\approx(0.401,0.689,1.204,5)$ and $\cos\lambda\approx 0.6497$.
(Middle) 15-Kaleidocycle as a discrete curve. (Right) the semi-discrete
K-surface obtained as its trajectory.

##  Appendix A Appendix

In the Appendix, we prove the theorems and propositions in the paper by using
the properties of elliptic theta functions listed below.

###  A.1 Functional identities of elliptic theta functions

We collect some properties of elliptic theta functions which we rely heavily
on our calculations.

####  A.1.1 Four term identities

Let

(87) $\begin{split}&X_{1}=\frac{1}{2}(X+Y+U+V),\quad
Y_{1}=\frac{1}{2}(X+Y-U-V),\\\ &U_{1}=\frac{1}{2}(X-Y+U-V),\quad
V_{1}=\frac{1}{2}(X-Y-U+V),\end{split}$

then for any $X,Y,U,V\in\mathbb{C}$, elliptic theta functions satisfy the
following identities [23, 20].

(88)
$\displaystyle\vartheta_{2}(X_{1})\vartheta_{2}(Y_{1})\vartheta_{3}(U_{1})\vartheta_{3}(V_{1})-\vartheta_{4}(X_{1})\vartheta_{4}(Y_{1})\vartheta_{1}(U_{1})\vartheta_{1}(V_{1})$
$\displaystyle=\vartheta_{2}(X)\vartheta_{2}(Y)\vartheta_{3}(U)\vartheta_{3}(V)-\vartheta_{4}(X)\vartheta_{4}(Y)\vartheta_{1}(U)\vartheta_{1}(V),$
(89)
$\displaystyle\vartheta_{4}(X_{1})\vartheta_{3}(Y_{1})\vartheta_{1}(U_{1})\vartheta_{2}(V_{1})-\vartheta_{2}(X_{1})\vartheta_{1}(Y_{1})\vartheta_{3}(U_{1})\vartheta_{4}(V_{1})$
$\displaystyle=\vartheta_{4}(X)\vartheta_{3}(Y)\vartheta_{1}(U)\vartheta_{2}(V)-\vartheta_{2}(X)\vartheta_{1}(Y)\vartheta_{3}(U)\vartheta_{4}(V),$
(90)
$\displaystyle\vartheta_{2}(X_{1})\vartheta_{2}(Y_{1})\vartheta_{4}(U_{1})\vartheta_{4}(V_{1})+\vartheta_{1}(X_{1})\vartheta_{1}(Y_{1})\vartheta_{3}(U_{1})\vartheta_{3}(V_{1})$
$\displaystyle=\vartheta_{4}(X)\vartheta_{4}(Y)\vartheta_{2}(U)\vartheta_{2}(V)-\vartheta_{3}(X)\vartheta_{3}(Y)\vartheta_{1}(U)\vartheta_{1}(V),$
(91)
$\displaystyle\vartheta_{3}(X_{1})\vartheta_{3}(Y_{1})\vartheta_{2}(U_{1})\vartheta_{2}(V_{1})+\vartheta_{1}(X_{1})\vartheta_{1}(Y_{1})\vartheta_{4}(U_{1})\vartheta_{4}(V_{1})$
$\displaystyle=\vartheta_{3}(X)\vartheta_{3}(Y)\vartheta_{2}(U)\vartheta_{2}(V)+\vartheta_{1}(X)\vartheta_{1}(Y)\vartheta_{4}(U)\vartheta_{4}(V),$
(92)
$\displaystyle\vartheta_{4}(X_{1})\vartheta_{4}(Y_{1})\vartheta_{4}(U_{1})\vartheta_{4}(V_{1})-\vartheta_{1}(X_{1})\vartheta_{1}(Y_{1})\vartheta_{1}(U_{1})\vartheta_{1}(V_{1})$
$\displaystyle=\vartheta_{4}(X)\vartheta_{4}(Y)\vartheta_{4}(U)\vartheta_{4}(V)-\vartheta_{1}(X)\vartheta_{1}(Y)\vartheta_{1}(U)\vartheta_{1}(V),$
(93)
$\displaystyle\vartheta_{4}(X_{1})\vartheta_{4}(Y_{1})\vartheta_{1}(U_{1})\vartheta_{1}(V_{1})-\vartheta_{1}(X_{1})\vartheta_{1}(Y_{1})\vartheta_{4}(U_{1})\vartheta_{4}(V_{1})$
$\displaystyle=\vartheta_{4}(X)\vartheta_{4}(Y)\vartheta_{1}(U)\vartheta_{1}(V)-\vartheta_{1}(X)\vartheta_{1}(Y)\vartheta_{4}(U)\vartheta_{4}(V),$
(94)
$\displaystyle\vartheta_{4}(X_{1})\vartheta_{4}(Y_{1})\vartheta_{1}(U_{1})\vartheta_{1}(V_{1})+\vartheta_{2}(X_{1})\vartheta_{2}(Y_{1})\vartheta_{3}(U_{1})\vartheta_{3}(V_{1})$
$\displaystyle=\vartheta_{3}(X)\vartheta_{3}(Y)\vartheta_{2}(U)\vartheta_{2}(V)-\vartheta_{1}(X)\vartheta_{1}(Y)\vartheta_{4}(U)\vartheta_{4}(V),$
(95)
$\displaystyle\vartheta_{3}(X_{1})\vartheta_{3}(Y_{1})\vartheta_{1}(U_{1})\vartheta_{1}(V_{1})-\vartheta_{1}(X_{1})\vartheta_{1}(Y_{1})\vartheta_{3}(U_{1})\vartheta_{3}(V_{1})$
$\displaystyle=\vartheta_{3}(X)\vartheta_{3}(Y)\vartheta_{1}(U)\vartheta_{1}(V)-\vartheta_{1}(X)\vartheta_{1}(Y)\vartheta_{3}(U)\vartheta_{3}(V),$
(96)
$\displaystyle\vartheta_{3}(X_{1})\vartheta_{3}(Y_{1})\vartheta_{4}(U_{1})\vartheta_{4}(V_{1})-\vartheta_{1}(X_{1})\vartheta_{1}(Y_{1})\vartheta_{2}(U_{1})\vartheta_{2}(V_{1})$
$\displaystyle=\vartheta_{3}(X)\vartheta_{3}(Y)\vartheta_{4}(U)\vartheta_{4}(V)-\vartheta_{1}(X)\vartheta_{1}(Y)\vartheta_{2}(U)\vartheta_{2}(V).$

####  A.1.2 Shifts by periods and half-periods

For any $X\in\mathbb{C}$, $n\in\mathbb{Z}$ and $y>0$, the following holds [23,
20].

(97)
$\vartheta_{1}\left(X+n\right)=(-1)^{n}\vartheta_{1}\left(X\right),\quad\vartheta_{4}\left(X+n\right)=\vartheta_{4}\left(X\right).$
(98) $\begin{split}&\vartheta_{1}\left(X+niy;iy\right)=(-1)^{n}\exp\\{-\pi
i\left(2nX+n^{2}iy\right)\\}\vartheta_{1}\left(X;iy\right),\\\
&\vartheta_{2}\left(X+niy;iy\right)=\exp\\{-\pi
i\left(2nX+n^{2}iy\right)\\}\vartheta_{2}\left(X;iy\right),\\\
&\vartheta_{3}\left(X+niy;iy\right)=\exp\\{-\pi
i\left(2nX+n^{2}iy\right)\\}\vartheta_{3}\left(X;iy\right),\\\
&\vartheta_{4}\left(X+niy;iy\right)=(-1)^{n}\exp\\{-\pi
i\left(2nX+n^{2}iy\right)\\}\vartheta_{4}\left(X;iy\right).\end{split}$ (99)
$\begin{split}&\vartheta_{1}\left(X+\frac{1}{2}\right)=\vartheta_{2}\left(X\right),\quad\vartheta_{2}\left(X+\frac{1}{2}\right)=-\vartheta_{1}\left(X\right),\\\
&\vartheta_{3}\left(X+\frac{1}{2}\right)=\vartheta_{4}\left(X\right),\quad\vartheta_{4}\left(X+\frac{1}{2}\right)=\vartheta_{3}\left(X\right).\end{split}$
(100)
$\begin{split}&\vartheta_{1}\left(X+\frac{iy}{2};iy\right)=i\exp\left\\{-\pi
i\left(X+\frac{iy}{4}\right)\right\\}\vartheta_{4}\left(X;iy\right),\\\
&\vartheta_{2}\left(X+\frac{iy}{2};iy\right)=\exp\left\\{-\pi
i\left(X+\frac{iy}{4}\right)\right\\}\vartheta_{3}\left(X;iy\right),\\\
&\vartheta_{3}\left(X+\frac{iy}{2};iy\right)=\exp\left\\{-\pi
i\left(X+\frac{iy}{4}\right)\right\\}\vartheta_{2}\left(X;iy\right),\\\
&\vartheta_{4}\left(X+\frac{iy}{2};iy\right)=i\exp\left\\{-\pi
i\left(X+\frac{iy}{4}\right)\right\\}\vartheta_{1}\left(X;iy\right).\end{split}$

####  A.1.3 Identity for the theta constants

For any $y>0$, elliptic theta functions satisfy the following identity [23,
20].

(101)
$\vartheta_{1}^{{}^{\prime}}\left(0;iy\right)=\pi\vartheta_{2}\left(0;iy\right)\vartheta_{3}\left(0;iy\right)\vartheta_{4}\left(0;iy\right).$

####  A.1.4 Modular transformations

For any $X\in\mathbb{C}$ and $y>0$, elliptic theta functions satisfy the
following identities [23, 20].

(102)
$\begin{split}&\vartheta_{1}\left(\frac{X}{y};i\frac{2}{y}\right)=-i\sqrt{\frac{y}{2}}\exp\left(-\pi\frac{X^{2}}{2y}\right)\vartheta_{1}\left(\frac{iX}{2};i\frac{y}{2}\right),\\\
&\vartheta_{2}\left(\frac{X}{y};i\frac{2}{y}\right)=\sqrt{\frac{y}{2}}\exp\left(-\pi\frac{X^{2}}{2y}\right)\vartheta_{4}\left(\frac{iX}{2};i\frac{y}{2}\right),\\\
&\vartheta_{3}\left(\frac{X}{y};i\frac{2}{y}\right)=\sqrt{\frac{y}{2}}\exp\left(-\pi\frac{X^{2}}{2y}\right)\vartheta_{3}\left(\frac{iX}{2};i\frac{y}{2}\right),\\\
&\vartheta_{4}\left(\frac{X}{y};i\frac{2}{y}\right)=\sqrt{\frac{y}{2}}\exp\left(-\pi\frac{X^{2}}{2y}\right)\vartheta_{2}\left(\frac{iX}{2};i\frac{y}{2}\right).\end{split}$
(103)
$\frac{\vartheta_{1}\left(X;iy\right)}{\vartheta_{2}\left(X;iy\right)}=-i\frac{\vartheta_{1}\left(\frac{iX}{y};\frac{i}{y}\right)}{\vartheta_{4}\left(\frac{iX}{y};\frac{i}{y}\right)}.$

####  A.1.5 Bilinear identities connecting theta functions with $y$ and $y/2$

For any $X,Y\in\mathbb{C}$, and $y>0$, elliptic theta functions satisfy the
following identities [23, 20].

$\displaystyle
2\vartheta_{1}\left(X+Y;iy\right)\vartheta_{4}\left(X-Y;iy\right)$ (104)
$\displaystyle=\vartheta_{1}\left(X;\frac{iy}{2}\right)\vartheta_{2}\left(Y;\frac{iy}{2}\right)+\vartheta_{2}\left(X;\frac{iy}{2}\right)\vartheta_{1}\left(Y;\frac{iy}{2}\right),$
$\displaystyle
2\vartheta_{2}\left(X+Y;iy\right)\vartheta_{3}\left(X-Y;iy\right)$ (105)
$\displaystyle=\vartheta_{2}\left(X;\frac{iy}{2}\right)\vartheta_{2}\left(Y;\frac{iy}{2}\right)-\vartheta_{1}\left(X;\frac{iy}{2}\right)\vartheta_{1}\left(Y;\frac{iy}{2}\right).$
(106)
$\begin{split}&\vartheta_{3}\left(\frac{X}{2};\frac{iy}{2}\right)\vartheta_{4}\left(\frac{X}{2};\frac{iy}{2}\right)=\vartheta_{4}\left(X;iy\right)\vartheta_{4}\left(0;iy\right),\\\
&\vartheta_{1}\left(\frac{X}{2};\frac{iy}{2}\right)\vartheta_{2}\left(\frac{X}{2};\frac{iy}{2}\right)=\vartheta_{1}\left(X;iy\right)\vartheta_{4}\left(0;iy\right),\\\
-&\vartheta_{1}\left(\frac{X}{2};\frac{iy}{2}\right)^{2}+\vartheta_{2}\left(\frac{X}{2};\frac{iy}{2}\right)^{2}=2\vartheta_{2}\left(X;iy\right)\vartheta_{3}\left(0;iy\right).\\\
\end{split}$

###  A.2 Proofs of Theorem 4.1 and Theorem 5.1

Here we prove that the $\tau$ functions given in Theorem 5.1 satisfy (8)-(13).
Theorem 4.1 follows from Theorem 5.1 by setting $t=0$. We assume $z=0$ without
loss of generality except for the proof of (12). Putting

(111) $\displaystyle\left\\{\begin{array}[]{l}\vspace{2mm}X=ivn+r+it,\\\
\vspace{2mm}Y=ivn-r+it,\\\ \vspace{2mm}U=\frac{1}{2}iv+r,\\\
V=\frac{1}{2}iv-r,\\\ \end{array}\right.$

in (88) yields (8). Putting

(116)
$\displaystyle\left\\{\begin{array}[]{l}\vspace{2mm}X=iv\left(n-\frac{1}{2}\right)+it,\\\
\vspace{2mm}Y=ivn+r+it,\\\ \vspace{2mm}U=iv,\\\ V=\frac{1}{2}iv-r,\\\
\end{array}\right.$

in (88) yields (9). Putting

(121)
$\displaystyle\left\\{\begin{array}[]{l}\vspace{2mm}X=iv\left(n-\frac{1}{2}\right)+it,\\\
\vspace{2mm}Y=\frac{1}{2}iv+r,\\\ \vspace{2mm}U=iv,\\\ V=ivn-r+it,\\\
\end{array}\right.$

in (89) yields (10). Putting

(126) $\displaystyle\left\\{\begin{array}[]{l}\vspace{2mm}X=ivn-r+it,\\\
\vspace{2mm}Y=-\frac{1}{2}iv+r,\\\ \vspace{2mm}U=ivn-r+it,\\\
V=-\frac{1}{2}iv+r,\\\ \end{array}\right.$

in (90) yields

(127)
$\begin{split}&\frac{\vartheta_{2}\left(0\right)\vartheta_{4}\left(0\right)}{\vartheta_{1}\left(-\frac{1}{2}iv+r\right)\vartheta_{3}\left(-\frac{1}{2}iv+r\right)}\vartheta_{2}\left(iv\left(n-\frac{1}{2}\right)+it\right)\vartheta_{4}\left(iv\left(n+\frac{1}{2}\right)-2r+it\right)\\\
&=\frac{\vartheta_{2}\left(-\frac{1}{2}iv+r\right)\vartheta_{4}\left(-\frac{1}{2}iv+r\right)}{\vartheta_{1}\left(-\frac{1}{2}iv+r\right)\vartheta_{3}\left(-\frac{1}{2}iv+r\right)}\\\
&\times\vartheta_{2}\left(ivn-r+it\right)\vartheta_{4}\left(ivn-r+it\right)-\vartheta_{3}\left(ivn-r+it\right)\vartheta_{1}\left(ivn-r+it\right).\end{split}$

Also, putting

(132) $\displaystyle\left\\{\begin{array}[]{l}\vspace{2mm}X=ivn-r+it,\\\
\vspace{2mm}Y=\frac{1}{2}iv+r,\\\ \vspace{2mm}U=ivn-r+it,\\\
V=\frac{1}{2}iv+r,\\\ \end{array}\right.$

in (90) yields

(133)
$\begin{split}&\frac{\vartheta_{2}\left(0\right)\vartheta_{4}\left(0\right)}{\vartheta_{1}\left(\frac{1}{2}iv+r\right)\vartheta_{3}\left(\frac{1}{2}iv+r\right)}\vartheta_{2}\left(iv\left(n+\frac{1}{2}\right)+it\right)\vartheta_{4}\left(iv\left(n-\frac{1}{2}\right)-2r+it\right)\\\
&=\frac{\vartheta_{2}\left(\frac{1}{2}iv+r\right)\vartheta_{4}\left(\frac{1}{2}iv+r\right)}{\vartheta_{1}\left(\frac{1}{2}iv+r\right)\vartheta_{3}\left(\frac{1}{2}iv+r\right)}\vartheta_{2}\left(ivn-r+it\right)\vartheta_{4}\left(ivn-r+it\right)\\\
&-\vartheta_{3}\left(ivn-r+it\right)\vartheta_{1}\left(ivn-r+it\right).\end{split}$

Subtracting (133) from (127) and substituting

(134)
$\begin{split}&\vartheta_{3}\left(2r\right)\vartheta_{4}\left(0\right)\vartheta_{2}\left(0\right)\vartheta_{1}\left(iv\right)\\\
&=\vartheta_{1}\left(\frac{1}{2}iv+r\right)\vartheta_{3}\left(\frac{1}{2}iv+r\right)\vartheta_{2}\left(-\frac{1}{2}iv+r\right)\vartheta_{4}\left(-\frac{1}{2}iv+r\right)\\\
&-\vartheta_{1}\left(-\frac{1}{2}iv+r\right)\vartheta_{3}\left(-\frac{1}{2}iv+r\right)\vartheta_{2}\left(\frac{1}{2}iv+r\right)\vartheta_{4}\left(\frac{1}{2}iv+r\right),\end{split}$

in the equation, we obtain

(135)
$\begin{split}&\frac{\vartheta_{1}\left(-\frac{1}{2}iv+r\right)\vartheta_{3}\left(-\frac{1}{2}iv+r\right)\vartheta_{1}\left(\frac{1}{2}iv+r\right)\vartheta_{3}\left(\frac{1}{2}iv+r\right)}{\vartheta_{3}(2r)\vartheta_{1}(iv)}\\\
&\times\left\\{\frac{\vartheta_{1}\left(\frac{1}{2}iv+r\right)\vartheta_{3}\left(\frac{1}{2}iv+r\right)}{\vartheta_{1}\left(-\frac{1}{2}iv+r\right)\vartheta_{3}\left(-\frac{1}{2}iv+r\right)}\vartheta_{2}\left(iv\left(n-\frac{1}{2}\right)+it\right)\vartheta_{4}\left(iv\left(n+\frac{1}{2}\right)-2r+it\right)\right.\\\
&\left.-\vartheta_{2}\left(iv\left(n+\frac{1}{2}\right)+it\right)\vartheta_{4}\left(iv\left(n-\frac{1}{2}\right)-2r+it\right)\right\\}\\\
&=\vartheta_{3}\left(\frac{1}{2}iv+r\right)\vartheta_{2}\left(ivn-r\right)+it\vartheta_{1}\left(\frac{1}{2}iv+r\right)\vartheta_{4}\left(ivn-r+it\right),\end{split}$

from which we see (11) holds. Putting

(140) $\displaystyle\left\\{\begin{array}[]{l}\vspace{2mm}X=ivn+z+r+it,\\\
\vspace{2mm}Y=ivn+z-r+it,\\\ \vspace{2mm}U=\frac{1}{2}iv+r+h,\\\
V=\frac{1}{2}iv-r+h,\\\ \end{array}\right.$

in (88) yields

(141)
$\begin{split}&\vartheta_{2}\left(iv\left(n+\frac{1}{2}\right)+z+it+h\right)\vartheta_{2}\left(iv\left(n-\frac{1}{2}\right)+z+it-h\right)\vartheta_{3}(2r)\vartheta_{3}(0)\\\
&=\vartheta_{2}\left(ivn+z+r+it\right)\vartheta_{2}\left(ivn+z-r+it\right)\vartheta_{3}\left(\frac{1}{2}iv+r+h\right)\vartheta_{3}\left(\frac{1}{2}iv-r+h\right)\\\
&-\vartheta_{4}\left(ivn+z+r+it\right)\vartheta_{4}\left(ivn+z-r+it\right)\vartheta_{1}\left(\frac{1}{2}iv+r+h\right)\vartheta_{1}\left(\frac{1}{2}iv-r+h\right).\end{split}$

Differentiating (141) with $h$ and then putting $h=0$, we obtain

(142)
$\begin{split}&\vartheta_{3}(2r)\vartheta_{3}(0)D_{z}\vartheta_{2}\left(iv\left(n+\frac{1}{2}\right)+z+it\right)\cdot\vartheta_{2}\left(iv\left(n-\frac{1}{2}\right)+z+it\right)\\\
&=\left(\frac{\vartheta^{{}^{\prime}}_{3}\left(\frac{1}{2}iv+r\right)}{\vartheta_{3}\left(\frac{1}{2}iv+r\right)}-\frac{\vartheta^{{}^{\prime}}_{3}\left(-\frac{1}{2}iv+r\right)}{\vartheta_{3}\left(-\frac{1}{2}iv+r\right)}\right)\\\
&\times\vartheta_{3}\left(-\frac{1}{2}iv+r\right)\vartheta_{3}\left(\frac{1}{2}iv+r\right)\vartheta_{2}\left(ivn+z+r+it\right)\vartheta_{2}\left(ivn+z-r+it\right)\\\
&+\left(\frac{\vartheta^{{}^{\prime}}_{1}\left(\frac{1}{2}iv+r\right)}{\vartheta_{1}\left(\frac{1}{2}iv+r\right)}-\frac{\vartheta^{{}^{\prime}}_{1}\left(-\frac{1}{2}iv+r\right)}{\vartheta_{1}\left(-\frac{1}{2}iv+r\right)}\right)\\\
&\times\vartheta_{1}\left(-\frac{1}{2}iv+r\right)\vartheta_{1}\left(\frac{1}{2}iv+r\right)\vartheta_{4}\left(ivn+z+r+it\right)\vartheta_{4}\left(ivn+z-r+it\right).\end{split}$

Using the equation in (141) with $h=0$ and substituting (142), we obtain

(143)
$\begin{split}&\vartheta_{3}(2r)\vartheta_{3}(0)\left\\{D_{z}\vartheta_{2}\left(iv\left(n+\frac{1}{2}\right)+z+it\right)\cdot\vartheta_{2}\left(iv\left(n-\frac{1}{2}\right)+z+it\right)\right.\\\
&\left.+\left(\frac{\vartheta^{{}^{\prime}}_{3}\left(-\frac{1}{2}iv+r\right)}{\vartheta_{3}\left(-\frac{1}{2}iv+r\right)}-\frac{\vartheta^{{}^{\prime}}_{3}\left(\frac{1}{2}iv+r\right)}{\vartheta_{3}\left(\frac{1}{2}iv+r\right)}\right)\right.\\\
&\left.\times\vartheta_{2}\left(iv\left(n+\frac{1}{2}\right)+z+it\right)\vartheta_{2}\left(iv\left(n-\frac{1}{2}\right)+z+it\right)\right\\}\\\
&=\left(\frac{\vartheta^{{}^{\prime}}_{1}\left(\frac{1}{2}iv+r\right)}{\vartheta_{1}\left(\frac{1}{2}iv+r\right)}-\frac{\vartheta^{{}^{\prime}}_{1}\left(-\frac{1}{2}iv+r\right)}{\vartheta_{1}\left(-\frac{1}{2}iv+r\right)}+\frac{\vartheta^{{}^{\prime}}_{3}\left(-\frac{1}{2}iv+r\right)}{\vartheta_{3}\left(-\frac{1}{2}iv+r\right)}-\frac{\vartheta^{{}^{\prime}}_{3}\left(\frac{1}{2}iv+r\right)}{\vartheta_{3}\left(\frac{1}{2}iv+r\right)}\right)\\\
&\times\vartheta_{1}\left(-\frac{1}{2}iv+r\right)\vartheta_{4}\left(ivn+z+r+it\right)\vartheta_{1}\left(\frac{1}{2}iv+r\right)\vartheta_{4}\left(ivn+z-r+it\right),\end{split}$

from which we see (12) holds. By the definition of $G$ and $\widetilde{G}$,
(13) follows.∎

###  A.3 Proof of Proposition 5.5

For the proof of Proposition 5.5, we use the following lemma:

######  Lemma A.1 ([10, §3]).

The following equation holds.

(144) $\displaystyle
f_{n}f_{n-1}^{*}+g_{n}g_{n-1}^{*}=f_{n}^{*}f_{n-1}+g_{n}^{*}g_{n-1}=\beta
F_{n}F_{n},$

####  A.3.1 Proof of Proposition 5.5

Here we prove that the $\tau$ functions given in Theorem 5.1 satisfy
(50)-(56). We assume $z=0$ without loss of generality except for the proof of
(56). Putting

(149)
$\displaystyle\left\\{\begin{array}[]{l}\vspace{2mm}X=-\frac{1}{2}iv+r,\\\
\vspace{2mm}Y=\frac{1}{2}iv-r,\\\ \vspace{2mm}U=ivn+r+it,\\\
V=iv\left(n-1\right)-r+it,\\\ \end{array}\right.$

in (94) yields (50). Also, (51) follows from (144) and (50). Putting

(154) $\displaystyle\left\\{\begin{array}[]{l}\vspace{2mm}X=ivn-r+it,\\\
\vspace{2mm}Y=-\frac{1}{2}iv-r+h,\\\ \vspace{2mm}U=\frac{1}{2}iv-r+h,\\\
V=iv\left(n-1\right)-r+it,\\\ \end{array}\right.$

in (89) yields

(155)
$\begin{split}&\vartheta_{4}\left(iv\left(n-\frac{1}{2}\right)-2r+it+h\right)\vartheta_{3}\left(0\right)\vartheta_{1}\left(iv\right)\vartheta_{2}\left(iv\left(n-\frac{1}{2}\right)+it-h\right)\\\
&=\vartheta_{4}\left(ivn-r+it\right)\vartheta_{3}\left(-\frac{1}{2}iv-r+h\right)\\\
&\times\vartheta_{1}\left(\frac{1}{2}iv-r+h\right)\vartheta_{2}\left(iv\left(n-1\right)-r+it\right)\\\
&-\vartheta_{2}\left(ivn-r+it\right)\vartheta_{1}\left(-\frac{1}{2}iv-r+h\right)\\\
&\times\vartheta_{3}\left(\frac{1}{2}iv-r+h\right)\vartheta_{4}\left(iv\left(n-1\right)-r+it\right),\end{split}$

After differentiating (155) by $h$ and putting $h=0$, we obtain (55) by
rearrangement. Also, rearranging the equation with $h=0$ in (155) yields (52).
Putting

(160)
$\displaystyle\left\\{\begin{array}[]{l}\vspace{2mm}X=\frac{1}{2}iv+r,\\\
\vspace{2mm}Y=-\frac{1}{2}iv-r,\\\
\vspace{2mm}U=iv\left(n-\frac{1}{2}\right)+it,\\\
V=-iv\left(n-\frac{1}{2}\right)-it,\\\ \end{array}\right.$

in (91) yields (53). Also, (54) follows from (50) and (53). Putting

(165)
$\displaystyle\left\\{\begin{array}[]{l}\vspace{2mm}X=iv\left(n-\frac{1}{2}\right)+it+\frac{z}{\Delta_{3}-\Delta_{1}},\\\
\vspace{2mm}Y=iv\left(n-\frac{1}{2}\right)+it+\frac{z}{\Delta_{3}-\Delta_{1}},\\\
\vspace{2mm}U=h,\\\ V=h,\\\ \end{array}\right.$

in (88) yields

(166)
$\begin{split}&\vartheta_{2}\left(iv\left(n-\frac{1}{2}\right)+it+\frac{z}{\Delta_{3}-\Delta_{1}}+h\right)\vartheta_{2}\left(iv\left(n-\frac{1}{2}\right)+it+\frac{z}{\Delta_{3}-\Delta_{1}}-h\right)\\\
&=-\vartheta_{4}\left(iv\left(n-\frac{1}{2}\right)+it+\frac{z}{\Delta_{3}-\Delta_{1}}\right)^{2}\frac{\vartheta_{1}\left(h\right)^{2}}{\vartheta_{3}\left(0\right)^{2}}\\\
&+\vartheta_{2}\left(iv\left(n-\frac{1}{2}\right)+it+\frac{z}{\Delta_{3}-\Delta_{1}}\right)^{2}\frac{\vartheta_{3}\left(h\right)^{2}}{\vartheta_{3}\left(0\right)^{2}}.\end{split}$

After differentiating (166) twice with $h$ and putting $h=0$, we obtain (56).
∎

###  A.4 Proofs of Theorem 5.6 and Theorem 5.9

For the proofs of Theorem 5.6 and Theorem 5.9, we need the following lemmas.

######  Lemma A.2 ([10, §3]).

Using (9) and (10), we see that the following equation holds.

(167) $\displaystyle g_{n}f_{n-1}-f_{n}g_{n-1}=G_{n}F_{n}.$

######  Lemma A.3.

The following relation holds.

(168) $\omega=\delta_{2}-\delta_{2}^{*}.$

######  Proof.

Putting

(173)
$\displaystyle\left\\{\begin{array}[]{l}\vspace{2mm}X=-\frac{1}{2}iv+r,\\\
\vspace{2mm}Y=-\frac{1}{2}iv+r,\\\ \vspace{2mm}U=\frac{1}{2}iv+r,\\\
V=\frac{1}{2}iv+r,\\\ \end{array}\right.$

in (95) yields (168). ∎

######  Lemma A.4.

The following identity holds.

(174)
$2\beta_{1}\left(\delta_{2}+\delta_{2}^{*}\right)+2i\omega\beta_{2}+\epsilon_{2}=2\delta_{2}\delta_{3}^{*}+2\delta_{2}^{*}\delta_{3}+\epsilon_{2}=0.$

######  Proof.

Putting

(179)
$\displaystyle\left\\{\begin{array}[]{l}\vspace{2mm}X=\frac{1}{2}iv+r+h,\\\
\vspace{2mm}Y=-\frac{1}{2}iv+r-h,\\\ \vspace{2mm}U=\frac{1}{2}iv+r-h,\\\
V=-\frac{1}{2}iv+r+h,\\\ \end{array}\right.$

in (95) yields

(180)
$\begin{split}&\vartheta_{3}\left(\frac{1}{2}iv+r+h\right)\vartheta_{3}\left(-\frac{1}{2}iv+r-h\right)\vartheta_{1}\left(\frac{1}{2}iv+r-h\right)\vartheta_{1}\left(-\frac{1}{2}iv+r+h\right)\\\
&-\vartheta_{1}\left(\frac{1}{2}iv+r+h\right)\vartheta_{1}\left(-\frac{1}{2}iv+r-h\right)\vartheta_{3}\left(\frac{1}{2}iv+r-h\right)\vartheta_{3}\left(-\frac{1}{2}iv+r+h\right)\\\
&=\vartheta_{3}\left(2r\right)\vartheta_{3}\left(0\right)\vartheta_{1}\left(iv\right)\vartheta_{1}\left(2h\right),\end{split}$

After differentiating (180) by $h$ and setting $h=0$, we obtain (174). ∎

For simplicity, we define $p$ and $q$ as follows

(181) $p_{n}=g_{n}^{*}f_{n-1}-f_{n}g_{n-1}^{*},$ (182)
$q_{n}=g_{n}^{*}f_{n-1}+f_{n}g_{n-1}^{*}.$

Then we rewrite $D_{t}H_{n}\cdot F_{n}$ by using (48), (181) and (182) as
follows

(183) $D_{t}H_{n}\cdot F_{n}=\beta_{1}q_{n}^{*}+i\beta_{2}p_{n}^{*}.$

Also, we see the following hold.

(184) $\tan\frac{\kappa_{n}}{2}\widetilde{N}_{n}=T_{n}-\frac{1}{2\beta
F_{n}^{2}}\left(\begin{array}[]{c}\vspace{2mm}q_{n}^{*}+q_{n}\\\
\vspace{2mm}\frac{1}{i}\left(q_{n}^{*}-q_{n}\right)\\\
-2\left(\beta-2\delta_{1}\right)F_{n}^{2}+2\left(\delta_{2}+\delta_{2}^{*}\right)|G_{n}|^{2}\end{array}\right),$
(185) $\tan\frac{\kappa_{n}}{2}\widetilde{B}_{n}=\frac{1}{2\beta
F_{n}^{2}}\left(\begin{array}[]{c}\vspace{2mm}\frac{1}{i}\left(p_{n}-p_{n}^{*}\right)\\\
\vspace{2mm}p_{n}+p_{n}^{*}\\\
-\frac{2}{i}\omega|G_{n}|^{2}\end{array}\right).$

We have the following Lemma.

######  Lemma A.5.

The following relations hold.

(186) $p_{n}p_{n}^{*}=F_{n}^{2}|G_{n}|^{2}+\omega^{2}|G_{n}|^{4},$ (187)
$\begin{split}p_{n}q_{n}^{*}-p_{n}^{*}q_{n}=2\left(2\delta_{1}-\beta\right)\omega
F_{n}^{2}|G_{n}|^{2}+2\omega\left(\delta_{2}+\delta_{2}^{*}\right)|G_{n}|^{4},\end{split}$
(188)
$\begin{split}&q_{n}q_{n}^{*}=4\delta_{1}\left(\beta-\delta_{1}\right)F_{n}^{4}+\left\\{1+2\left(\beta-2\delta_{1}\right)\left(\delta_{2}+\delta_{2}^{*}\right)\right\\}F_{n}^{2}|G_{n}|^{2}\\\
&-\left(\delta_{2}+\delta_{2}^{*}\right)^{2}|G_{n}|^{4}.\end{split}$

######  Proof.

By definition of $p$, we see that

(189)
$\begin{split}p_{n}p_{n}^{*}&=\left(g_{n}^{*}f_{n-1}-f_{n}g_{n-1}^{*}\right)\left(g_{n}f_{n-1}^{*}-f_{n}^{*}g_{n-1}\right)\\\
&=\left(f_{n}^{*}g_{n-1}^{*}-g_{n}^{*}f_{n-1}^{*}\right)\left(f_{n}g_{n-1}-g_{n}f_{n-1}\right)\\\
&+\left(f_{n}f_{n-1}^{*}-f_{n}^{*}f_{n-1}\right)\left(g_{n}^{*}g_{n-1}-g_{n}g_{n-1}^{*}\right),\end{split}$

holds. Substituting (167) and (51) into (189) yields (186). By definition of
$p$ and $q$, we see that

(190) $\begin{split}&p_{n}q_{n}^{*}-p_{n}^{*}q_{n}\\\
&=\left(g_{n}^{*}f_{n-1}-f_{n}g_{n-1}^{*}\right)\left(g_{n}f_{n-1}^{*}+f_{n}^{*}g_{n-1}\right)\\\
&-\left(g_{n}f_{n-1}^{*}-f_{n}^{*}g_{n-1}\right)\left(g_{n}^{*}f_{n-1}+f_{n}g_{n-1}^{*}\right)\\\
&=2\left(g_{n}^{*}f_{n-1}f_{n}^{*}g_{n-1}-g_{n}f_{n-1}^{*}f_{n}g_{n-1}^{*}\right),\\\
\end{split}$

holds. On the other hand, from (144) and (51), the following holds.

(191) $\begin{split}&\beta\omega F_{n}^{2}|G_{n}|^{2}\\\
&=\left(f_{n}f_{n-1}^{*}+g_{n}g_{n-1}^{*}\right)\left(g_{n}^{*}g_{n-1}-g_{n}g_{n-1}^{*}\right)\\\
&=\left(f_{n}^{*}f_{n-1}+g_{n}^{*}g_{n-1}\right)\left(g_{n}^{*}g_{n-1}-g_{n}g_{n-1}^{*}\right)\\\
&=\left(f_{n}f_{n-1}^{*}+g_{n}g_{n-1}^{*}\right)\left(f_{n}f_{n-1}^{*}-f_{n}^{*}f_{n-1}\right)\\\
&=\left(f_{n}^{*}f_{n-1}+g_{n}^{*}g_{n-1}\right)\left(f_{n}f_{n-1}^{*}-f_{n}^{*}f_{n-1}\right).\\\
\end{split}$

Using (191), we obtain

(192) $\begin{split}&p_{n}q_{n}^{*}-p_{n}^{*}q_{n}\\\
&=f_{n}f_{n-1}^{*}f_{n}f_{n-1}^{*}-f_{n}^{*}f_{n-1}f_{n}^{*}f_{n-1}+g_{n}g_{n-1}^{*}g_{n}g_{n-1}^{*}-g_{n}^{*}g_{n-1}g_{n}^{*}g_{n-1}\\\
&=\left(f_{n}f_{n-1}^{*}\right)^{2}-\left(f_{n}^{*}f_{n-1}\right)^{2}+\left(g_{n}g_{n-1}^{*}\right)^{2}-\left(g_{n}^{*}g_{n-1}\right)^{2}\\\
&=\left(f_{n}f_{n-1}^{*}-f_{n}^{*}f_{n-1}\right)\left(f_{n}f_{n-1}^{*}+f_{n}^{*}f_{n-1}\right)\\\
&+\left(g_{n}g_{n-1}^{*}-g_{n}^{*}g_{n-1}\right)\left(g_{n}g_{n-1}^{*}+g_{n}^{*}g_{n-1}\right).\end{split}$

Substituting (51), (53) and (54) into (192) gives

(193) $\begin{split}&p_{n}q_{n}^{*}-p_{n}^{*}q_{n}\\\
&=\omega|G_{n}|^{2}\left(f_{n}f_{n-1}^{*}+f_{n}^{*}f_{n-1}-g_{n}g_{n-1}^{*}-g_{n}^{*}g_{n-1}\right)\\\
&=2\left(2\delta_{1}-\beta\right)\omega
F_{n}^{2}|G_{n}|^{2}+2\omega\left(\delta_{2}+\delta_{2}^{*}\right)|G_{n}|^{4},\end{split}$

thus (187). Also from (167), (53) and (54), we see that

(194) $\begin{split}&q_{n}q_{n}^{*}-F_{n}^{2}|G_{n}|^{2}\\\
&=\left(g_{n}^{*}f_{n-1}+f_{n}g_{n-1}^{*}\right)\left(g_{n}f_{n-1}^{*}+f_{n}^{*}g_{n-1}\right)\\\
&-\left(g_{n}f_{n-1}-f_{n}g_{n-1}\right)\left(g_{n}^{*}f_{n-1}^{*}-f_{n}^{*}g_{n-1}^{*}\right)\\\
&=\left(f_{n}f_{n-1}^{*}+f_{n}^{*}f_{n-1}\right)\left(g_{n}g_{n-1}^{*}+g_{n}^{*}g_{n-1}\right)\\\
&=\left(2\delta_{1}F_{n}^{2}+\left(\delta_{2}+\delta_{2}^{*}\right)|G_{n}|^{2}\right)\left(2\left(\beta-\delta_{1}\right)F_{n}^{2}-\left(\delta_{2}+\delta_{2}^{*}\right)|G_{n}|^{2}\right)\\\
&=4\delta_{1}\left(\beta-\delta_{1}\right)F_{n}^{4}+2\left(\beta-2\delta_{1}\right)\left(\delta_{2}+\delta_{2}^{*}\right)F_{n}^{2}|G_{n}|^{2}-\left(\delta_{2}+\delta_{2}^{*}\right)^{2}|G_{n}|^{4}.\end{split}$

holds. From (194), we get (188). ∎

Using Proposition 5.5, Lemma A.3, Lemma A.4 and Lemma A.5, Theorem 5.6 and
Theorem 5.9 can be proved as follows.

####  A.4.1 Proof of Theorem 5.6

By differentiating $\gamma$ with respect to $t$, we get

(195)
$\begin{split}&\dot{\gamma}_{n}(t)=\frac{1}{F_{n}^{2}}\left(\begin{array}[]{c}\vspace{2mm}\displaystyle
D_{t}H_{n}\cdot F_{n}+D_{t}H_{n}^{*}\cdot F_{n}\\\
\vspace{2mm}\displaystyle\frac{1}{i}\left(D_{t}H_{n}\cdot
F_{n}-D_{t}H_{n}^{*}\cdot F_{n}\right)\\\ -D_{t}D_{z}F_{n}\cdot
F_{n}\end{array}\right)\\\
&=\frac{1}{F_{n}^{2}}\left(\begin{array}[]{c}\vspace{2mm}\displaystyle\beta_{1}q_{n}^{*}+i\beta_{2}p_{n}^{*}+\beta_{1}q_{n}-i\beta_{2}p_{n}\\\
\vspace{2mm}\displaystyle\frac{1}{i}\left(\beta_{1}q_{n}^{*}+i\beta_{2}p_{n}^{*}-\beta_{1}q_{n}+i\beta_{2}p_{n}\right)\\\
-\left(\epsilon_{1}+C\right)F_{n}^{2}-\epsilon_{2}|G_{n}|^{2}\end{array}\right).\end{split}$

On the other hand since

(196)
$\begin{split}&\beta_{1}\left(T_{n}-\tan\frac{\kappa_{n}}{2}\widetilde{N}_{n}\right)+\beta_{2}\tan\frac{\kappa_{n}}{2}\widetilde{B}_{n}\\\
&=\frac{1}{2\beta
F_{n}^{2}}\left(\begin{array}[]{c}\vspace{2mm}\displaystyle\beta_{1}q_{n}^{*}+i\beta_{2}p_{n}^{*}+\beta_{1}q_{n}-i\beta_{2}p_{n}\\\
\vspace{2mm}\displaystyle\frac{1}{i}\left(\beta_{1}q_{n}^{*}+i\beta_{2}p_{n}^{*}-\beta_{1}q_{n}+i\beta_{2}p_{n}\right)\\\
-2\beta_{1}\left(\beta-2\delta_{1}\right)F_{n}^{2}+2\beta_{1}\left(\delta_{2}+\delta_{2}^{*}\right)|G_{n}|^{2}+2i\beta_{2}\omega|G_{n}|^{2}\end{array}\right).\end{split}$

holds. From (174), this equation can be rewritten as follows

(197)
$\begin{split}&\beta_{1}\left(T_{n}-\tan\frac{\kappa_{n}}{2}\widetilde{N}_{n}\right)+\beta_{2}\tan\frac{\kappa_{n}}{2}\widetilde{B}_{n}\\\
&=\frac{1}{2\beta
F_{n}^{2}}\left(\begin{array}[]{c}\vspace{2mm}\displaystyle\beta_{1}q_{n}^{*}+i\beta_{2}p_{n}^{*}+\beta_{1}q_{n}-i\beta_{2}p_{n}\\\
\vspace{2mm}\displaystyle\frac{1}{i}\left(\beta_{1}q_{n}^{*}+i\beta_{2}p_{n}^{*}-\beta_{1}q_{n}+i\beta_{2}p_{n}\right)\\\
-2\beta_{1}\left(\beta-2\delta_{1}\right)F_{n}^{2}-\epsilon_{2}|G_{n}|^{2}\end{array}\right).\end{split}$

Thus (57) is obtained. ∎

####  A.4.2 Proof of Theorem 5.9

We compute the condition satisfied by $C,\Gamma$ when

(198) $\left\langle\dot{\gamma}_{n},\widetilde{B}_{n}\right\rangle=0,$ (199)
$\left\langle\dot{\gamma}_{n},\dot{\gamma}_{n}\right\rangle=\rho^{2},$

where $\rho\in\mathbb{R}$ is a constant. From (185) and (195) we see that

(200)
$\begin{split}2iF_{n}^{3}\widetilde{G}_{n}\left\langle\dot{\gamma}_{n},\widetilde{B}_{n}\right\rangle=&\left(p_{n}-p_{n}^{*}\right)\left(\beta_{1}q_{n}^{*}+i\beta_{2}p_{n}^{*}+\beta_{1}q_{n}-i\beta_{2}p_{n}\right)\\\
&+\left(p_{n}+p_{n}^{*}\right)\left(\beta_{1}q_{n}^{*}+i\beta_{2}p_{n}^{*}-\beta_{1}q_{n}+i\beta_{2}p_{n}\right)\\\
&+2\omega|G_{n}|^{2}\left\\{\left(\epsilon_{1}+C\right)F_{n}^{2}+\epsilon_{2}|G_{n}|^{2}\right\\}\\\
=&2F_{n}^{2}|G_{n}|^{2}\left\\{2\beta_{1}\left(2\delta_{1}-\beta\right)\omega+2i\beta_{2}+\omega\left(\epsilon_{1}+C\right)\right\\}\\\
&+2\omega|G_{n}|^{4}\left\\{2\beta_{1}\left(\delta_{2}+\delta_{2}^{*}\right)+2i\omega\beta_{2}+\epsilon_{2}\right\\},\end{split}$

holds. For (198) to be satisfied, it is sufficient that the following
equations hold.

(201)
$2\beta_{1}\left(2\delta_{1}-\beta\right)\omega+2i\beta_{2}+\omega\left(\epsilon_{1}+C\right)=0,\\\
$ (202)
$2\beta_{1}\left(\delta_{2}+\delta_{2}^{*}\right)+2i\omega\beta_{2}+\epsilon_{2}=0.$

By comparing (202) and (174) we see that (201) is a necessary and sufficient
condition for the realization of (198). By substituting (48) into (201), it
can be rewritten in terms of $C,\Gamma$ as follows.

(203)
$i\Gamma\left(\omega\left(2\delta_{1}-\beta\right)-\frac{\delta_{2}+\delta_{2}^{*}}{\omega}\right)+\epsilon_{1}+C+\left(\delta_{3}+\delta_{3}^{*}\right)\left(2\delta_{1}-\beta\right)-\frac{\delta_{3}-\delta_{3}^{*}}{\omega}=0.$

Also, since

(204)
$\begin{split}&F_{n}^{4}\langle\dot{\gamma}_{n},\dot{\gamma}_{n}\rangle\\\
=&\left(\beta_{1}q_{n}^{*}+i\beta_{2}p_{n}^{*}+\beta_{1}q_{n}-i\beta_{2}p_{n}\right)^{2}\\\
&-\left(\beta_{1}q_{n}^{*}+i\beta_{2}p_{n}^{*}-\beta_{1}q_{n}+i\beta_{2}p_{n}\right)^{2}\\\
&+\left\\{\left(\epsilon_{1}+C\right)F_{n}^{2}+\epsilon_{2}|G_{n}|^{2}\right\\}^{2}\\\
=&F_{n}^{4}\left\\{16\beta_{1}^{2}\delta_{1}\left(\beta-\delta_{1}\right)+\left(\epsilon_{1}+C\right)^{2}\right\\}\\\
&+F_{n}^{2}|G_{n}|^{2}\left\\{4\beta_{1}^{2}\left(1+2\left(\beta-2\delta_{1}\right)\left(\delta_{2}+\delta_{2}^{*}\right)\right)\right.\\\
&\left.-8i\beta_{1}\beta_{2}\omega\left(2\delta_{1}-\beta\right)+4\beta_{2}^{2}+2\epsilon_{2}\left(\epsilon_{1}+C\right)\right\\}\\\
&+|G_{n}|^{4}\left(-4\beta_{1}^{2}\left(\delta_{2}+\delta_{2}^{*}\right)^{2}-8i\beta_{1}\beta_{2}\omega\left(\delta_{2}+\delta_{2}^{*}\right)+4\beta_{2}^{2}\omega^{2}+\epsilon_{2}^{2}\right),\end{split}$

holds, we see that for (199) holds, it is sufficient that

(205)
$4\beta_{1}^{2}\left(1+2\left(\beta-2\delta_{1}\right)\left(\delta_{2}+\delta_{2}^{*}\right)\right)-8i\beta_{1}\beta_{2}\omega\left(2\delta_{1}-\beta\right)+4\beta_{2}^{2}+2\epsilon_{2}\left(\epsilon_{1}+C\right)=0,$
(206)
$-4\beta_{1}^{2}\left(\delta_{2}+\delta_{2}^{*}\right)^{2}-8i\beta_{1}\beta_{2}\omega\left(\delta_{2}+\delta_{2}^{*}\right)+4\beta_{2}^{2}\omega^{2}+\epsilon_{2}^{2}=0,$

hold. Since

(207)
$\begin{split}&4\beta_{1}^{2}\left(\delta_{2}+\delta_{2}^{*}\right)^{2}+8i\beta_{1}\beta_{2}\omega\left(\delta_{2}+\delta_{2}^{*}\right)-4\beta_{2}^{2}\omega^{2}-\epsilon_{2}^{2}\\\
=&\left\\{2\beta_{1}\left(\delta_{2}+\delta_{2}^{*}\right)+2i\omega\beta_{2}+\epsilon_{2}\right\\}\left\\{2\beta_{1}\left(\delta_{2}+\delta_{2}^{*}\right)+2i\omega\beta_{2}-\epsilon_{2}\right\\},\end{split}$

holds, (206) is identically satisfied. Therefore, (205) is a necessary and
sufficient condition for the realization of (199). By using (174), (48) and
(203), (205) can be written in terms of $\Gamma$ as follows.

(208)
$i\delta_{2}\delta_{2}^{*}\left(\delta_{2}-\delta_{2}^{*}\right)\Gamma^{2}+2\delta_{2}\delta_{2}^{*}\left(\delta_{3}+\delta_{3}^{*}\right)\Gamma+i\left(\delta_{2}{\delta_{3}^{*}}^{2}-\delta_{2}^{*}\delta_{3}^{2}\right)=0.$

(208) has the following solutions.

(209)
$\Gamma=\frac{-\delta_{2}\delta_{2}^{*}\left(\delta_{3}+\delta_{3}^{*}\right)\pm\sqrt{\delta_{2}\delta_{2}^{*}\left(\delta_{2}\delta_{3}^{*}+\delta_{2}^{*}\delta_{3}\right)^{2}}}{i\delta_{2}\delta_{2}^{*}\left(\delta_{2}-\delta_{2}^{*}\right)}.$

Since

(210)
$\begin{split}\delta_{2}\delta_{2}^{*}\left(\delta_{2}\delta_{3}^{*}+\delta_{2}^{*}\delta_{3}\right)^{2}=\left(\delta_{2}\delta_{2}^{*}\right)^{2}\left(\delta_{3}e^{i\lambda}+\delta_{3}^{*}e^{-i\lambda}\right)^{2}=\left(\delta_{2}\delta_{2}^{*}\right)^{2}\left(i\frac{\vartheta_{1}^{{}^{\prime}}(0)}{\vartheta_{1}(iv)}\right)^{2},\end{split}$

holds, (209) can be rewritten as

(211)
$\Gamma=\frac{-\left(\delta_{3}+\delta_{3}^{*}\right)\pm\left(\delta_{3}e^{i\lambda}+\delta_{3}^{*}e^{-i\lambda}\right)}{i\omega}.$

Now, we prove equations (62) and (63). From (35) and (48), we see that when
$C_{+},\Gamma_{+}$ are chosen as in (60), the following relation holds

(212)
$16\beta_{1}^{2}\left(\beta-\delta_{1}\right)\delta_{1}=4\left(i\frac{\vartheta_{1}^{{}^{\prime}}(0)}{\vartheta_{1}(iv)}\right)^{2}\frac{u_{2}^{2}}{\vartheta_{3}(2r)^{2}\vartheta_{3}(0)^{6}}.$

Also, from (35), (48), (201) and (202), we see that when $C_{+},\Gamma_{+}$
are chosen as in (60), the following relation holds

(213)
$\left(\epsilon_{1}+C\right)^{2}=\left(i\frac{\vartheta_{1}^{{}^{\prime}}(0)}{\vartheta_{1}(iv)}\right)^{2}\frac{u_{2}^{2}}{\vartheta_{3}(2r)^{2}\vartheta_{3}(0)^{6}}\frac{\left(\vartheta_{3}(r)^{4}-\vartheta_{1}(r)^{4}\right)^{2}}{\vartheta_{3}(r)^{4}\vartheta_{1}(r)^{4}}.$

From (204), (212) and (213), we see that

(214)
$\rho^{2}=16\beta_{1}^{2}\left(\beta-\delta_{1}\right)\delta_{1}+\left(\epsilon_{1}+C\right)^{2}=\left(i\frac{\vartheta_{1}^{{}^{\prime}}(0)}{\vartheta_{1}(iv)}\right)^{2}\frac{u_{2}^{2}}{\vartheta_{3}(r)^{4}\vartheta_{1}(r)^{4}}=\rho_{+}^{2},$

holds. (214) is equivalent to (62). From (35) and (48), we see that when
$C_{-},\Gamma_{-}$ are chosen as in (60), the following relation holds

(215)
$16\beta_{1}^{2}\left(\beta-\delta_{1}\right)\delta_{1}=4\left(i\frac{\vartheta_{1}^{{}^{\prime}}(0)}{\vartheta_{1}(iv)}\right)^{2}\frac{u_{2}^{2}}{\vartheta_{3}(2r)^{2}\vartheta_{3}(0)^{6}}.$

Also, from (35), (48), (201) and (202), we see that when $C_{-},\Gamma_{-}$
are chosen as in (60), the following relation holds

(216)
$\left(\epsilon_{1}+C\right)^{2}=\left(i\frac{\vartheta_{1}^{{}^{\prime}}(0)}{\vartheta_{1}(iv)}\right)^{2}\frac{u_{2}^{2}}{\vartheta_{3}(2r)^{2}\vartheta_{3}(0)^{6}}\frac{\left(\vartheta_{4}(r)^{4}-\vartheta_{2}(r)^{4}\right)^{2}}{\vartheta_{4}(r)^{4}\vartheta_{2}(r)^{4}}.$

From (204), (215) and (216), we see that

(217)
$\rho^{2}=16\beta_{1}^{2}\left(\beta-\delta_{1}\right)\delta_{1}+\left(\epsilon_{1}+C\right)^{2}=\left(i\frac{\vartheta_{1}^{{}^{\prime}}(0)}{\vartheta_{1}(iv)}\right)^{2}\frac{u_{2}^{2}}{\vartheta_{4}(r)^{4}\vartheta_{2}(r)^{4}}=\rho_{-}^{2},$

holds. (217) is equivalent to (63).∎

###  A.5 Proof of Proposition 5.12

Here we prove that the signed curvature angle $\kappa$ satisfies (69). First,
we state Lemma A.6.

######  Lemma A.6.

The $\tau$ functions (43) satisfy the following bilinear equations.

(218) $\displaystyle\widetilde{G}_{n}^{2}+\beta^{2}F_{n}^{2}=F_{n+1}F_{n-1},$
(219)
$\displaystyle\beta^{2}\widetilde{G}_{n}^{2}-\frac{\vartheta_{1}\left(iv\right)^{4}}{\vartheta_{3}\left(iv\right)^{4}}\beta^{4}F_{n}^{2}=\widetilde{G}_{n+1}\widetilde{G}_{n-1}.$

######  Proof.

Using (8)-(10), we see that (218) hold. Putting

(224) $\displaystyle\left\\{\begin{array}[]{l}\vspace{2mm}X=0,\\\
\vspace{2mm}Y=0,\\\
\vspace{2mm}U=iv\left(n+\frac{1}{2}\right)+it+\frac{z}{\Delta_{3}-\Delta_{1}},\\\
V=-iv\left(n-\frac{3}{2}\right)-it-\frac{z}{\Delta_{3}-\Delta_{1}},\\\
\end{array}\right.$

in (96), we obtain (219). ∎

####  A.5.1 Proof of Proposition 5.12

Using (218) and (219), we see that (69) holds.∎

###  A.6 Proof of Theorem 5.13

Here, we show that the signed curvature angle satisfies the semi-discrete mKdV
equation (70). For the proof of Theorem 5.13, we first state Lemma A.7.

######  Lemma A.7.

The $\tau$ functions (43) satisfy the following bilinear equation.

(225) $D_{t}\widetilde{G}_{n}\cdot
F_{n}=\eta_{3}\left(\widetilde{G}_{n+1}F_{n-1}-\widetilde{G}_{n-1}F_{n+1}\right).$

######  Proof.

We assume $z=0$ without loss of generality. After differentiating the equation
in (89) with

(230)
$\displaystyle\left\\{\begin{array}[]{l}\vspace{2mm}X=iv\left(n+\frac{1}{2}\right)+it,\\\
\vspace{2mm}Y=iv+h,\\\ \vspace{2mm}U=-iv+h,\\\
V=iv\left(n-\frac{3}{2}\right)+it,\\\ \end{array}\right.$

by $h$ and putting $h=0$, we obtain

(231)
$\begin{split}&D_{t}\left\\{-i\vartheta_{1}(iv)\vartheta_{4}\left(iv\left(n-\frac{1}{2}\right)+it\right)\right\\}\cdot\left\\{\vartheta_{3}(iv)\vartheta_{2}\left(iv\left(n-\frac{1}{2}\right)+it\right)\right\\}\\\
&=\frac{\vartheta_{1}(iv)\vartheta_{3}(iv)}{\vartheta_{1}(2iv)\vartheta_{3}(0)}\frac{\vartheta^{{}^{\prime}}_{1}\left(iv\right)\vartheta_{3}\left(iv\right)-\vartheta_{1}\left(iv\right)\vartheta^{{}^{\prime}}_{3}\left(iv\right)}{i\vartheta_{1}\left(iv\right)\vartheta_{3}\left(iv\right)}\\\
&\times\left\\{-i\vartheta_{1}(iv)\vartheta_{4}\left(iv\left(n-\frac{3}{2}\right)+it\right)\right\\}\left\\{\vartheta_{3}(iv)\vartheta_{2}\left(iv\left(n+\frac{1}{2}\right)+it\right)\right\\}\\\
&-\frac{\vartheta_{1}(iv)\vartheta_{3}(iv)}{\vartheta_{1}(2iv)\vartheta_{3}(0)}\frac{\vartheta^{{}^{\prime}}_{1}\left(iv\right)\vartheta_{3}\left(iv\right)-\vartheta_{1}\left(iv\right)\vartheta^{{}^{\prime}}_{3}\left(iv\right)}{i\vartheta_{1}\left(iv\right)\vartheta_{3}\left(iv\right)}\\\
&\times\left\\{-i\vartheta_{1}(iv)\vartheta_{4}\left(iv\left(n+\frac{1}{2}\right)+it\right)\right\\}\left\\{\vartheta_{3}(iv)\vartheta_{2}\left(iv\left(n-\frac{3}{2}\right)+it\right)\right\\}.\end{split}$

After rearranging (231), we obtain (225). ∎

####  A.6.1 Proof of Theorem 5.13

Using (218) and (225), we see that the signed curvature angle satisfies the
semi-discrete mKdV equation (70).∎

###  A.7 Proof of Theorem 5.14

From (103), (104) and (105), we see that

(232)
$\begin{split}&\kappa_{n}=2\arctan\left(-i\frac{\vartheta_{1}(iv;iy)\vartheta_{4}\left(iv\left(n-\frac{1}{2}\right)+it;iy\right)}{\vartheta_{3}(iv;iy)\vartheta_{2}\left(iv\left(n-\frac{1}{2}\right)+it;iy\right)}\right)\\\
&=2\arctan\left(-i\frac{\frac{\vartheta_{1}\left(i\frac{v}{2}\left(n+\frac{1}{2}\right)+\frac{it}{2};\frac{iy}{2}\right)}{\vartheta_{2}\left(i\frac{v}{2}\left(n+\frac{1}{2}\right)+\frac{it}{2};\frac{iy}{2}\right)}-\frac{\vartheta_{1}\left(i\frac{v}{2}\left(n-\frac{3}{2}\right)+\frac{it}{2};\frac{iy}{2}\right)}{\vartheta_{2}\left(i\frac{v}{2}\left(n-\frac{3}{2}\right)+\frac{it}{2};\frac{iy}{2}\right)}}{1-\frac{\vartheta_{1}\left(i\frac{v}{2}\left(n+\frac{1}{2}\right)+\frac{it}{2};\frac{iy}{2}\right)\vartheta_{1}\left(i\frac{v}{2}\left(n-\frac{3}{2}\right)+\frac{it}{2};\frac{iy}{2}\right)}{\vartheta_{2}\left(i\frac{v}{2}\left(n+\frac{1}{2}\right)+\frac{it}{2};\frac{iy}{2}\right)\vartheta_{2}\left(i\frac{v}{2}\left(n-\frac{3}{2}\right)+\frac{it}{2};\frac{iy}{2}\right)}}\right)\\\
&=2\arctan\left(\frac{\frac{\vartheta_{1}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y};\frac{2i}{y}\right)}{\vartheta_{4}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y};\frac{2i}{y}\right)}-\frac{\vartheta_{1}\left(\frac{v}{y}\left(n-\frac{3}{2}\right)+\frac{t}{y};\frac{2i}{y}\right)}{\vartheta_{4}\left(\frac{v}{y}\left(n-\frac{3}{2}\right)+\frac{t}{y};\frac{2i}{y}\right)}}{1+\frac{\vartheta_{1}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y};\frac{2i}{y}\right)\vartheta_{1}\left(\frac{v}{y}\left(n-\frac{3}{2}\right)+\frac{t}{y};\frac{2i}{y}\right)}{\vartheta_{4}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y};\frac{2i}{y}\right)\vartheta_{4}\left(\frac{v}{y}\left(n-\frac{3}{2}\right)+\frac{t}{y};\frac{2i}{y}\right)}}\right)\\\
&=2\arctan\left(\frac{\vartheta_{1}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y};\frac{2i}{y}\right)}{\vartheta_{4}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y};\frac{2i}{y}\right)}\right)\\\
&-2\arctan\left(\frac{\vartheta_{1}\left(\frac{v}{y}\left(n-\frac{3}{2}\right)+\frac{t}{y};\frac{2i}{y}\right)}{\vartheta_{4}\left(\frac{v}{y}\left(n-\frac{3}{2}\right)+\frac{t}{y};\frac{2i}{y}\right)}\right)\\\
&=\frac{\Theta_{n+1}-\Theta_{n-1}}{2},\end{split}$

holds. ∎

###  A.8 Proof of Theorem 5.15

We first construct the solution to the semi-discrete potential mKdV equation
(75) in two stages by proving Lemma A.8 and Lemma A.9. Then we construct the
solution to the semi-discrete sine-Gordon equation (76) in two stages: Lemma
A.10 and Lemma A.11. This is based on the construction of the solution to the
mKdV equation using the theory of $\tau$ functions [14].

For fixed $v,y$, define the following real-valued functions

(233)
$\begin{split}&A_{n}\left(t\right)=\vartheta_{4}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y};\frac{2i}{y}\right),\\\
&B_{n}\left(t\right)=\vartheta_{1}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y};\frac{2i}{y}\right).\end{split}$

Then the potential function can be written as follows

(234)
$\Theta_{n}\left(t\right)=4\arctan\frac{B_{n}}{A_{n}}=\frac{2}{i}\log\frac{A_{n}+iB_{n}}{A_{n}-iB_{n}}=\frac{2}{i}\log\frac{\widetilde{\tau}_{n}}{\widetilde{\tau}^{*}_{n}},$

where $\widetilde{\tau}_{n}=A_{n}+iB_{n}$. $\widetilde{\tau}$ corresponds to
the $\tau$ functions of the one-component KP hierarchy, while $A$ and $B$
correspond to the $\tau$ functions of the two-component KP hierarchy.

######  Lemma A.8.

Assuming that the solution of (75) is expressed as (234), we obtain the
following bilinear equations as a sufficient condition for satisfying (75).

(235) $D_{t}A_{n+1}\cdot A_{n}+D_{t}B_{n+1}\cdot
B_{n}=\frac{L_{1}}{2y}\left(\eta_{1}-1\right)\left(A_{n+1}A_{n}+B_{n+1}B_{n}\right),$
(236) $D_{t}A_{n+1}\cdot B_{n}-D_{t}B_{n+1}\cdot
A_{n}=\frac{L_{1}}{2y}\left(\eta_{1}+1\right)\left(A_{n+1}B_{n}-B_{n+1}A_{n}\right),$

where $\eta_{1}$ is a real constant.

######  Proof.

Since

(237)
$\begin{split}&\frac{d}{dt}\left(\frac{\Theta_{n+1}+\Theta_{n}}{2}\right)=\frac{1}{i}\frac{d}{dt}\left(\log\frac{\widetilde{\tau}_{n+1}}{\widetilde{\tau}^{*}_{n+1}}+\log\frac{\widetilde{\tau}_{n}}{\widetilde{\tau}^{*}_{n}}\right)\\\
&=\frac{1}{i}\frac{d}{dt}\left(\log\frac{\widetilde{\tau}_{n+1}}{\widetilde{\tau}^{*}_{n}}-\log\frac{\widetilde{\tau}^{*}_{n+1}}{\widetilde{\tau}_{n}}\right)\\\
&=\frac{1}{i}\left(\frac{D_{t}\widetilde{\tau}_{n+1}\cdot\widetilde{\tau}^{*}_{n}}{\widetilde{\tau}_{n+1}\widetilde{\tau}^{*}_{n}}-\frac{D_{t}\widetilde{\tau}^{*}_{n+1}\cdot\widetilde{\tau}_{n}}{\widetilde{\tau}^{*}_{n+1}\widetilde{\tau}_{n}}\right).\end{split}$
(238)
$\begin{split}&\sin\left(\frac{\Theta_{n+1}-\Theta_{n}}{2}\right)=\frac{1}{2i}\left(\frac{\widetilde{\tau}_{n+1}\widetilde{\tau}^{*}_{n}}{\widetilde{\tau}^{*}_{n+1}\widetilde{\tau}_{n}}-\frac{\widetilde{\tau}^{*}_{n+1}\widetilde{\tau}_{n}}{\widetilde{\tau}_{n+1}\widetilde{\tau}^{*}_{n}}\right),\end{split}$

hold, it can be seen that a sufficient condition for (234) to be a solution of
(75) is that

(239)
$\begin{split}&D_{t}\widetilde{\tau}_{n+1}\cdot\widetilde{\tau}^{*}_{n}=-\frac{L_{1}}{2y}\widetilde{\tau}^{*}_{n+1}\widetilde{\tau}_{n}+\frac{L_{1}}{2y}\eta_{1}\widetilde{\tau}_{n+1}\widetilde{\tau}_{n}^{*},\end{split}$

($\eta_{1}$ is a real constant) is satisfied. Substituting
$\widetilde{\tau}_{n}=A_{n}+iB_{n}$ for this, we get

(240) $\begin{split}&D_{t}\left(A_{n+1}\cdot A_{n}+B_{n+1}\cdot
B_{n}\right)-iD_{t}\left(A_{n+1}\cdot B_{n}-B_{n+1}\cdot A_{n}\right)\\\
&=\frac{L_{1}}{2y}\left(\eta_{1}-1\right)\left(A_{n+1}A_{n}+B_{n+1}B_{n}\right)-i\frac{L_{1}}{2y}\left(\eta_{1}+1\right)\left(A_{n+1}B_{n}-B_{n+1}A_{n}\right).\end{split}$

The real part of this equation yields (235), and the imaginary part yields
(236). ∎

In Lemma A.9, we construct solutions of (235) and (236) in terms of elliptic
theta functions.

######  Lemma A.9.

The real functions $A$ and $B$ defined by (233) are the solutions of (235) and
(236) with

(241)
$\eta_{1}=\frac{d_{1}\left(\frac{v}{y};\frac{2}{y}\right)+d_{4}\left(\frac{v}{y};\frac{2}{y}\right)-Q_{1}(v,y)+Q_{4}(v,y)}{L_{1}(v,y)}.$

######  Proof.

In this proof, all elliptic theta functions have $w=2i/y$ and we denote
$\vartheta_{j}(v;2i/y)$ by $\vartheta_{j}(v)$. After differentiating the
equation in (92) with

(246)
$\displaystyle\left\\{\begin{array}[]{l}\vspace{2mm}X=\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y},\\\
\vspace{2mm}Y=\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y},\\\
\vspace{2mm}U=\frac{v}{y}+h,\\\ V=h,\\\ \end{array}\right.$

by $h$ and putting $h=0$, we obtain

(247)
$\begin{split}&D_{t}\vartheta_{4}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\cdot\vartheta_{4}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right)\\\
&=\frac{1}{y}\frac{\vartheta^{{}^{\prime}}_{4}\left(\frac{v}{y}\right)}{\vartheta_{4}\left(\frac{v}{y}\right)}\vartheta_{4}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\vartheta_{4}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right)\\\
&-\frac{1}{y}\frac{\vartheta_{1}\left(\frac{v}{y}\right)\vartheta^{{}^{\prime}}_{1}\left(0\right)}{\vartheta_{4}\left(\frac{v}{y}\right)\vartheta_{4}\left(0\right)}\vartheta_{1}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\vartheta_{1}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right)\\\
&=\frac{1}{y}d_{4}\left(\frac{v}{y};\frac{2}{y}\right)\vartheta_{4}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\vartheta_{4}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right)\\\
&-\frac{1}{y}Q_{1}(v,y)\vartheta_{1}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\vartheta_{1}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right).\end{split}$

Also, after differentiating the equation in (93) with

(252)
$\displaystyle\left\\{\begin{array}[]{l}\vspace{2mm}X=\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y},\\\
\vspace{2mm}Y=\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y},\\\
\vspace{2mm}U=\frac{v}{y}+h,\\\ V=h,\\\ \end{array}\right.$

by $h$ and putting $h=0$, we obtain

(253)
$\begin{split}&D_{t}\vartheta_{1}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\cdot\vartheta_{1}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right)\\\
&=\frac{1}{y}\frac{\vartheta^{{}^{\prime}}_{4}\left(\frac{v}{y}\right)}{\vartheta_{4}\left(\frac{v}{y}\right)}\vartheta_{1}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\vartheta_{1}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right)\\\
&-\frac{1}{y}\frac{\vartheta_{1}\left(\frac{v}{y}\right)\vartheta^{{}^{\prime}}_{1}\left(0\right)}{\vartheta_{4}\left(\frac{v}{y}\right)\vartheta_{4}\left(0\right)}\vartheta_{4}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\vartheta_{4}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right)\\\
&=\frac{1}{y}d_{4}\left(\frac{v}{y};\frac{2}{y}\right)\vartheta_{1}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\vartheta_{1}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right)\\\
&-\frac{1}{y}Q_{1}(v,y)\vartheta_{4}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\vartheta_{4}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right).\end{split}$

Also, after differentiating the equation in (93) with

(258)
$\displaystyle\left\\{\begin{array}[]{l}\vspace{2mm}X=\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y},\\\
\vspace{2mm}Y=\frac{v}{y}+h,\\\
\vspace{2mm}U=\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y},\\\ V=h,\\\
\end{array}\right.$

by $h$ and putting $h=0$, we obtain

(259)
$\begin{split}&D_{t}\vartheta_{1}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\cdot\vartheta_{4}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right)\\\
&=\frac{1}{y}\frac{\vartheta^{{}^{\prime}}_{1}\left(\frac{v}{y}\right)}{\vartheta_{1}\left(\frac{v}{y}\right)}\vartheta_{1}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\vartheta_{4}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right)\\\
&-\frac{1}{y}\frac{\vartheta_{4}\left(\frac{v}{y}\right)\vartheta^{{}^{\prime}}_{1}\left(0\right)}{\vartheta_{1}\left(\frac{v}{y}\right)\vartheta_{4}\left(0\right)}\vartheta_{4}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\vartheta_{1}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right)\\\
&=\frac{1}{y}d_{1}\left(\frac{v}{y};\frac{2}{y}\right)\vartheta_{1}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\vartheta_{4}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right)\\\
&-\frac{1}{y}Q_{2}(v,y)\vartheta_{4}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\vartheta_{1}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right).\end{split}$

Also, after differentiating the equation in (93) with

(264)
$\displaystyle\left\\{\begin{array}[]{l}\vspace{2mm}X=\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y},\\\
\vspace{2mm}Y=\frac{v}{y}+h,\\\
\vspace{2mm}U=\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y},\\\ V=h,\\\
\end{array}\right.$

by $h$ and putting $h=0$, we obtain

(265)
$\begin{split}&D_{t}\vartheta_{4}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\cdot\vartheta_{1}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right)\\\
&=\frac{1}{y}\frac{\vartheta^{{}^{\prime}}_{1}\left(\frac{v}{y}\right)}{\vartheta_{1}\left(\frac{v}{y}\right)}\vartheta_{4}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\vartheta_{1}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right)\\\
&-\frac{1}{y}\frac{\vartheta_{4}\left(\frac{v}{y}\right)\vartheta^{{}^{\prime}}_{1}\left(0\right)}{\vartheta_{1}\left(\frac{v}{y}\right)\vartheta_{4}\left(0\right)}\vartheta_{1}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\vartheta_{4}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right)\\\
&=\frac{1}{y}d_{1}\left(\frac{v}{y};\frac{2}{y}\right)\vartheta_{4}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\vartheta_{1}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right)\\\
&-\frac{1}{y}Q_{2}(v,y)\vartheta_{1}\left(\frac{v}{y}\left(n+\frac{1}{2}\right)+\frac{t}{y}\right)\vartheta_{4}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right).\end{split}$

Adding (247) and (253) yields (235), and subtracting (259) from (265) yields
(236). ∎

Next, we construct the solution to the semi-discrete sine-Gordon equation
(76).

######  Lemma A.10.

Assuming that the solution of (76) is expressed as (234), we obtain the
following bilinear equations as a sufficient condition for satisfying (76).

(266) $D_{t}A_{n+1}\cdot A_{n}-D_{t}B_{n+1}\cdot
B_{n}=\frac{L_{2}}{2y}\left(\eta_{2}-1\right)\left(A_{n+1}A_{n}-B_{n+1}B_{n}\right),$
(267) $D_{t}A_{n+1}\cdot B_{n}+D_{t}B_{n+1}\cdot
A_{n}=\frac{L_{2}}{2y}\left(\eta_{2}+1\right)\left(A_{n+1}B_{n}+B_{n+1}A_{n}\right),$

where $\eta_{2}$ is a real constant.

######  Proof.

Since

(268)
$\begin{split}&\frac{d}{dt}\left(\frac{\Theta_{n+1}-\Theta_{n}}{2}\right)=\frac{1}{i}\frac{d}{dt}\left(\log\frac{\widetilde{\tau}_{n+1}}{\widetilde{\tau}^{*}_{n+1}}-\log\frac{\widetilde{\tau}_{n}}{\widetilde{\tau}^{*}_{n}}\right)\\\
&=\frac{1}{i}\frac{d}{dt}\left(\log\frac{\widetilde{\tau}_{n+1}}{\widetilde{\tau}_{n}}-\log\frac{\widetilde{\tau}^{*}_{n+1}}{\widetilde{\tau}^{*}_{n}}\right)\\\
&=\frac{1}{i}\left(\frac{D_{t}\widetilde{\tau}_{n+1}\cdot\widetilde{\tau}_{n}}{\widetilde{\tau}_{n+1}\widetilde{\tau}_{n}}-\frac{D_{t}\widetilde{\tau}^{*}_{n+1}\cdot\widetilde{\tau}^{*}_{n}}{\widetilde{\tau}^{*}_{n+1}\widetilde{\tau}^{*}_{n}}\right).\end{split}$
(269)
$\begin{split}&\sin\left(\frac{\Theta_{n+1}+\Theta_{n}}{2}\right)=\frac{1}{2i}\left(\frac{\widetilde{\tau}_{n+1}\widetilde{\tau}_{n}}{\widetilde{\tau}^{*}_{n+1}\widetilde{\tau}^{*}_{n}}-\frac{\widetilde{\tau}^{*}_{n+1}\widetilde{\tau}^{*}_{n}}{\widetilde{\tau}_{n+1}\widetilde{\tau}_{n}}\right),\end{split}$

hold, it can be seen that a sufficient condition for (234) to be a solution of
(76) is that

(270)
$\begin{split}&D_{t}\widetilde{\tau}_{n+1}\cdot\widetilde{\tau}_{n}=-\frac{L_{2}}{2y}\widetilde{\tau}^{*}_{n+1}\widetilde{\tau}^{*}_{n}+\frac{L_{2}}{2y}\eta_{2}\widetilde{\tau}_{n+1}\widetilde{\tau}_{n},\end{split}$

($\eta_{2}$ is a real constant) is satisfied. Substituting
$\widetilde{\tau}_{n}=A_{n}+iB_{n}$ for this, we get

(271) $\begin{split}&D_{t}\left(A_{n+1}\cdot A_{n}-B_{n+1}\cdot
B_{n}\right)+iD_{t}\left(A_{n+1}\cdot B_{n}+B_{n+1}\cdot A_{n}\right)\\\
&=\frac{L_{2}}{2y}\left(\eta_{2}-1\right)\left(A_{n+1}A_{n}-B_{n+1}B_{n}\right)\\\
&+i\frac{L_{2}}{2y}\left(\eta_{2}+1\right)\left(A_{n+1}B_{n}+B_{n+1}A_{n}\right).\end{split}$

The real part of (271) yields (266), and the imaginary part yields (267). ∎

In Lemma A.11, we construct solutions of (266) and (267) in terms of elliptic
theta functions.

######  Lemma A.11.

The real functions $A$ and $B$ defined by (233) are the solutions of (266) and
(267) with

(272)
$\eta_{2}=\frac{d_{1}\left(\frac{v}{y};\frac{2}{y}\right)+d_{4}\left(\frac{v}{y};\frac{2}{y}\right)+Q_{1}(v,y)-Q_{4}(v,y)}{L_{2}(v,y)}.$

######  Proof.

In this proof, we set $w=2i/y$ similarly to Lemma A.9. Subtracting (253) from
(247) yields (266), and adding (265) and (259) yields (267). ∎

###  A.9 Proof of Proposition 5.16

We first show (82) and (81) by using the periodicity of elliptic theta
functions (97). Consider replacing the parameter with $\tilde{v}=v+y$. Then we
see

(273) $\begin{split}&Q_{1}(\tilde{v},y)=-Q_{1}(v,y),\quad
Q_{2}(\tilde{v},y)=-Q_{2}(v,y),\\\ &d_{1}(\tilde{v},y)=d_{1}(v,y),\quad
d_{4}(\tilde{v},y)=d_{4}(v,y),\\\ &L_{1}(\tilde{v},y)=L_{2}(v,y).\end{split}$

and

(274)
$\begin{split}&\Theta_{n}\left(t;\tilde{v},y\right)=4\arctan\frac{\vartheta_{1}\left(\frac{\tilde{v}}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right)}{\vartheta_{4}\left(\frac{\tilde{v}}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}\right)}\\\
&=4\arctan\frac{\vartheta_{1}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}+n-\frac{1}{2}\right)}{\vartheta_{4}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}+n-\frac{1}{2}\right)}\\\
&=(-1)^{n}4\arctan\frac{\vartheta_{1}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}-\frac{1}{2}\right)}{\vartheta_{4}\left(\frac{v}{y}\left(n-\frac{1}{2}\right)+\frac{t}{y}-\frac{1}{2}\right)}\\\
&=(-1)^{n}\Theta_{n}\left(t-\frac{y}{2};v,y\right).\end{split}$

From the relation between the signed curvature angle $\kappa$ and the
potential function $\Theta$ (74), we see

(275)
$\begin{split}&\kappa_{n}\left(t;\tilde{v}\right)=(-1)^{n+1}\kappa_{n}\left(t-\frac{y}{2};v\right).\end{split}$

Next, we check the transformation of $\lambda$ (44). From (99) and (100), we
have

(276) $\begin{split}&\lambda\left(\tilde{v},\frac{1}{2}-r\right)\\\
&=\frac{1}{i}\log\left(\frac{\vartheta_{1}\left(-\frac{1}{2}iv+\frac{1}{2}-r-\frac{1}{2}iy\right)}{\vartheta_{1}\left(\frac{1}{2}iv+\frac{1}{2}-r+\frac{1}{2}iy\right)}\frac{\vartheta_{3}\left(\frac{1}{2}iv+\frac{1}{2}-r+\frac{1}{2}iy\right)}{\vartheta_{3}\left(-\frac{1}{2}iv+\frac{1}{2}-r-\frac{1}{2}iy\right)}\right)\\\
&=\frac{1}{i}\log\left(\frac{\vartheta_{2}\left(-\frac{1}{2}iv-r-\frac{1}{2}iy\right)}{\vartheta_{2}\left(\frac{1}{2}iv-r+\frac{1}{2}iy\right)}\frac{\vartheta_{4}\left(\frac{1}{2}iv-r+\frac{1}{2}iy\right)}{\vartheta_{4}\left(-\frac{1}{2}iv-r-\frac{1}{2}iy\right)}\right)\\\
&=\frac{1}{i}\log\left(-\frac{\vartheta_{3}\left(-\frac{1}{2}iv-r\right)}{\vartheta_{3}\left(\frac{1}{2}iv-r\right)}\frac{\vartheta_{1}\left(\frac{1}{2}iv-r\right)}{\vartheta_{1}\left(-\frac{1}{2}iv-r\right)}\right)\\\
&=\pi+\frac{1}{i}\log\left(\frac{\vartheta_{1}\left(-\frac{1}{2}iv+r\right)}{\vartheta_{1}\left(\frac{1}{2}iv+r\right)}\frac{\vartheta_{3}\left(\frac{1}{2}iv+r\right)}{\vartheta_{3}\left(-\frac{1}{2}iv+r\right)}\right)\\\
&=\pi+\lambda\left(v,r\right).\\\ \end{split}$

Finally, we see that the transformation of the normal $\widetilde{N}$ and the
binormal $\widetilde{B}$ follows from the Frenet-Serret formula (6). ∎

###  A.10 Proof of Proposition 5.18

First, we calculate $L_{1}(v,y)/y$ and $L_{2}(v,y)/y$. From (78), (101), (102)
and (106), we see

(277)
$\begin{split}\frac{L_{1}(v,y)}{y}&=\frac{1}{y}\frac{\vartheta_{1}^{{}^{\prime}}\left(\frac{v}{y},i\frac{2}{y}\right)\vartheta_{4}\left(\frac{v}{y},i\frac{2}{y}\right)-\vartheta_{1}\left(\frac{v}{y},i\frac{2}{y}\right)\vartheta_{4}^{{}^{\prime}}\left(\frac{v}{y},i\frac{2}{y}\right)}{\vartheta_{1}\left(\frac{v}{y},i\frac{2}{y}\right)\vartheta_{4}\left(\frac{v}{y},i\frac{2}{y}\right)}\\\
&+\frac{1}{y}\frac{\vartheta_{1}^{{}^{\prime}}\left(0,i\frac{2}{y}\right)\vartheta_{1}\left(\frac{v}{y},i\frac{2}{y}\right)}{\vartheta_{4}\left(0,i\frac{2}{y}\right)\vartheta_{4}\left(\frac{v}{y},i\frac{2}{y}\right)}+\frac{1}{y}\frac{\vartheta_{1}^{{}^{\prime}}\left(0,i\frac{2}{y}\right)\vartheta_{4}\left(\frac{v}{y},i\frac{2}{y}\right)}{\vartheta_{4}\left(0,i\frac{2}{y}\right)\vartheta_{1}\left(\frac{v}{y},i\frac{2}{y}\right)}\\\
&=i\frac{\vartheta_{1}^{{}^{\prime}}\left(0,iy\right)}{\vartheta_{4}\left(0,iy\right)}\frac{\vartheta_{4}\left(iv,iy\right)}{\vartheta_{1}\left(iv,iy\right)}+i\frac{\vartheta_{1}^{{}^{\prime}}\left(0,iy\right)}{\vartheta_{2}\left(0,iy\right)}\frac{\vartheta_{2}\left(iv,iy\right)}{\vartheta_{1}\left(iv,iy\right)}\\\
&=i\frac{\vartheta_{1}^{{}^{\prime}}\left(0,iy\right)}{\vartheta_{1}\left(iv,iy\right)}\frac{\vartheta_{4}\left(iv,iy\right)\vartheta_{2}\left(0,iy\right)+\vartheta_{2}\left(iv,iy\right)\vartheta_{4}\left(0,iy\right)}{\vartheta_{4}\left(0,iy\right)\vartheta_{2}\left(0,iy\right)}.\end{split}$

Similarly, from (79), (101), (102) and (106), we see

(278)
$\begin{split}\frac{L_{2}(v,y)}{y}&=\frac{1}{y}\frac{\vartheta_{1}^{{}^{\prime}}\left(\frac{v}{y},i\frac{2}{y}\right)\vartheta_{4}\left(\frac{v}{y},i\frac{2}{y}\right)-\vartheta_{1}\left(\frac{v}{y},i\frac{2}{y}\right)\vartheta_{4}^{{}^{\prime}}\left(\frac{v}{y},i\frac{2}{y}\right)}{\vartheta_{1}\left(\frac{v}{y},i\frac{2}{y}\right)\vartheta_{4}\left(\frac{v}{y},i\frac{2}{y}\right)}\\\
&-\frac{1}{y}\frac{\vartheta_{1}^{{}^{\prime}}\left(0,i\frac{2}{y}\right)\vartheta_{1}\left(\frac{v}{y},i\frac{2}{y}\right)}{\vartheta_{4}\left(0,i\frac{2}{y}\right)\vartheta_{4}\left(\frac{v}{y},i\frac{2}{y}\right)}-\frac{1}{y}\frac{\vartheta_{1}^{{}^{\prime}}\left(0,i\frac{2}{y}\right)\vartheta_{4}\left(\frac{v}{y},i\frac{2}{y}\right)}{\vartheta_{4}\left(0,i\frac{2}{y}\right)\vartheta_{1}\left(\frac{v}{y},i\frac{2}{y}\right)}\\\
&=i\frac{\vartheta_{1}^{{}^{\prime}}\left(0,iy\right)}{\vartheta_{4}\left(0,iy\right)}\frac{\vartheta_{4}\left(iv,iy\right)}{\vartheta_{1}\left(iv,iy\right)}-i\frac{\vartheta_{1}^{{}^{\prime}}\left(0,iy\right)}{\vartheta_{2}\left(0,iy\right)}\frac{\vartheta_{2}\left(iv,iy\right)}{\vartheta_{1}\left(iv,iy\right)}\\\
&=i\frac{\vartheta_{1}^{{}^{\prime}}\left(0,iy\right)}{\vartheta_{1}\left(iv,iy\right)}\frac{\vartheta_{4}\left(iv,iy\right)\vartheta_{2}\left(0,iy\right)-\vartheta_{2}\left(iv,iy\right)\vartheta_{4}\left(0,iy\right)}{\vartheta_{4}\left(0,iy\right)\vartheta_{2}\left(0,iy\right)}.\end{split}$

On the other hand, from (35) and (44), we see

(279) $\begin{split}&1+\cos\lambda\\\
&=\frac{\left\\{\vartheta_{1}\left(\frac{iv}{2}+r,iy\right)\vartheta_{3}\left(-\frac{iv}{2}+r,iy\right)+\vartheta_{1}\left(-\frac{iv}{2}+r,iy\right)\vartheta_{3}\left(\frac{iv}{2}+r,iy\right)\right\\}^{2}}{2u_{2}},\end{split}$
(280) $\begin{split}&1-\cos\lambda\\\
&=-\frac{\left\\{\vartheta_{1}\left(\frac{iv}{2}+r,iy\right)\vartheta_{3}\left(-\frac{iv}{2}+r,iy\right)-\vartheta_{1}\left(-\frac{iv}{2}+r,iy\right)\vartheta_{3}\left(\frac{iv}{2}+r,iy\right)\right\\}^{2}}{2u_{2}}.\end{split}$

From (277) and (279), we see that (85) holds. Similarly, from (278) and (280),
we see that (86) holds. ∎

###  A.11 Special property of elliptic theta functions when $w$ is a pure
imaginary number

######  Lemma A.12.

The following holds for any $z\in\mathbb{C}$ when $w=iy$.

(281) $\vartheta_{j}(z)^{*}=\vartheta_{j}(z^{*}),\quad j=1,2,3,4.$

######  Proof.

Consider the following theta function with characteristics.

(282) $\vartheta_{a,b}(z):=\sum_{n\in\mathbb{Z}}\exp\left(2\pi i(n+a)(z+b)+\pi
i(n+a)^{2}\omega\right),\quad a,b\in\left\\{0,\frac{1}{2}\right\\},$

then,

(283)
$\vartheta_{1}(z)=-\vartheta_{\frac{1}{2},\frac{1}{2}}(z),\quad\vartheta_{2}(z)=\vartheta_{\frac{1}{2},0}(z),\quad\vartheta_{3}(z)=\vartheta_{0,0}(z),\quad\vartheta_{4}(z)=\vartheta_{0,\frac{1}{2}}(z),$

hold. Since $w^{*}=-w$, we have

(284)
$\begin{split}\vartheta_{a,b}(z)^{*}=&\sum_{n\in\mathbb{Z}}\exp\left(-2\pi
i(n+a)(z^{*}+b)-\pi i(n+a)^{2}w^{*}\right)\\\
=&\sum_{n\in\mathbb{Z}}\exp\left(2\pi i(-n-a)(z^{*}+b)+\pi
i(n+a)^{2}w\right)\\\ =&\sum_{n\in\mathbb{Z}}\exp\left(2\pi
i(n+a)(z^{*}+b)+\pi i(n+a)^{2}w\right)\\\
=&\vartheta_{a,b}(z^{*}).\end{split}$

∎

This lemma guarantees that some $\tau$ functions, such as $F_{n}$, are real-
valued functions.

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