# Spinors and Descartes’ Theorem

Daniel V. Mathews<EMAIL_ADDRESS>School of Mathematics, Monash
University, VIC 3800, Australia  Orion Zymaris<EMAIL_ADDRESS>School
of Mathematics, Monash University, VIC 3800, Australia

###### Abstract

Descartes’ circle theorem relates the curvatures of four mutually externally
tangent circles, three “petal” circles around the exterior of a central
circle, forming a “$3$-flower” configuration. We generalise this theorem to
the case of an “$n$-flower”, consisting of $n$ tangent circles around the
exterior of a central circle, and give an explicit equation satisfied by their
curvatures. The proof uses a spinorial description of horospheres in
hyperbolic geometry.

##  1 Introduction

_Descartes’ theorem_ is a classical result in 2-dimensional Euclidean
geometry, relating the curvatures of four mutually tangent circles (Figure 1)
which form a _3-flower_ in the following sense.

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Figure 1: Left, a $3$-flower. Right, a $5$-flower.

######  Definition 1.1.

Let $n\geq 3$. An _$n$ -flower_ consists of a central circle $C_{\infty}$, and
$n$ _petal_ circles $C_{j}$, over integers $j$ mod $n$, so that the $C_{j}$
are externally tangent to $C_{\infty}$ in order around $C_{\infty}$, and each
$C_{j}$ is externally tangent to $C_{j-1}$ and $C_{j+1}$.

Throughout, we denote the curvature of a circle $C_{\bullet}$ by
$\kappa_{\bullet}$. Descartes’ theorem gives an equation satisfied by the
curvatures in a 3-flower:

$(\kappa_{\infty}+\kappa_{1}+\kappa_{2}+\kappa_{3})^{2}=2(\kappa_{\infty}^{2}+\kappa_{1}^{2}+\kappa_{2}^{2}+\kappa_{3}^{2}).$
(1.2)

In this paper we present an explicit equation satisfied by the curvatures of
an $n$-flower.

###### Generalised Descartes Theorem.

Let the circles $C_{\infty}$ and $C_{j}$ ($j\in\mathbb{Z}/n\mathbb{Z}$) in an
$n$-flower have curvatures $\kappa_{\infty},\kappa_{j}$ respectively. Define
$m_{0}$ and $m_{j}$ for $1\leq j\leq n-1$ as

$m_{0}=\sqrt{\frac{\kappa_{0}}{\kappa_{\infty}}+1},\quad
m_{j}=\sqrt{\left(\frac{\kappa_{j}}{\kappa_{\infty}}+1\right)\left(\frac{\kappa_{j-1}}{\kappa_{\infty}}+1\right)-1}.$
(1.3)

Then for odd $n$, the following holds:

$\frac{m_{0}^{2}\,i}{2}\left(\prod_{j=1}^{n-1}(m_{j}-i)-\prod_{j=1}^{n-1}(m_{j}+i)\right)-\prod_{j=1}^{\frac{n-1}{2}}\left(m_{2j-1}^{2}+1\right)=0.$
(1.4)

For even $n$, the following holds:

$\frac{i}{2}\left(\prod_{j=1}^{n-1}(m_{j}-i)-\prod_{j=1}^{n-1}(m_{j}+i)\right)-\prod_{j=1}^{\frac{n-2}{2}}\left(m_{2j}^{2}+1\right)=0.$
(1.5)

In other words, from the curvatures $\kappa_{\bullet}$ we define auxiliary
variables $m_{\bullet}$ via (1.3), and the $m_{\bullet}$ satisfy polynomial
equations. Since all $\kappa_{\bullet}>0$, each $m_{\bullet}$ is the square
root of a manifestly positive number and we take $m_{\bullet}$ to be the
positive square root.

In (1.4) and (1.5), $i$ is the usual square root of $-1$. Although complex
numbers are crucial to the proof, in writing these equations they are merely a
convenience. After expanding and cancelling terms, the resulting polynomials
have integer coefficients. Indeed, writing $[n]$ for $\\{1,2,\ldots,n\\}$,
then for any subset $K\subseteq[n-1]$, the term in the product
$\prod_{j=1}^{n-1}(m_{j}\pm i)$ involving precisely the $m_{k}$ with $k\in K$
is given by $(\pm i)^{n-1-|K|}\prod_{k\in K}m_{k}$. Such terms come in pairs,
one from $\prod_{j=1}^{n-1}(m_{j}+i)$ and one from
$\prod_{j=1}^{n-1}(m_{j}-i)$. When $n-1-|K|$ is even, the terms in each pair
are real, equal and cancel; when $n-1-|K|$ is odd, say equal to $2l+1$, then
the terms in each pair are imaginary and conjugate, so since
$\frac{i}{2}\left((-i)^{n-1-|K|}-i^{n-1-|K|}\right)=(-1)^{l}$, (1.4) and (1.5)
can be written as

$m_{0}^{2}\sum_{\stackrel{{\scriptstyle
K\subseteq[n-1]}}{{|K|=n-2l-2}}}(-1)^{l}\prod_{k\in
K}m_{k}=\prod_{j=1}^{\frac{n-1}{2}}\left(m_{2j-1}^{2}+1\right)$ (1.6)

and

$\sum_{\stackrel{{\scriptstyle
K\subseteq[n-1]}}{{|K|=n-2l-2}}}(-1)^{l}\prod_{k\in
K}m_{k}=\prod_{j=1}^{\frac{n-2}{2}}\left(m_{2j}^{2}+1\right)$ (1.7)

respectively.

In any case, making the substitutions of (1.3) in (1.4) or (1.5) (or (1.6) or
(1.7)) yields an equation satisfied by the $\kappa_{\bullet}$, providing a
generalisation of Descartes’ equation (1.2). This equation involves square
roots, but upon multiplying by various conjugates (replacing various
$m_{\bullet}$ with $-m_{\bullet}$) one may obtain a polynomial relation among
the $m_{\bullet}^{2}$; after substituting and clearing denominators one
obtains a polynomial relation among the $\kappa_{\bullet}$.

In this way, from equation (1.4) with $n=3$ one can recover Descartes’
theorem. Similarly, from (1.5) with $n=4$ we obtain the following equation
relating the curvatures in a 4-flower:

$\displaystyle 16\kappa_{\infty}^{4}$
$\displaystyle-8\kappa_{\infty}^{2}(\kappa_{1}\kappa_{2}+\kappa_{2}\kappa_{3}+\kappa_{3}\kappa_{4}+\kappa_{4}\kappa_{1}+2\kappa_{1}\kappa_{3}+2\kappa_{2}\kappa_{4})$
$\displaystyle+(\kappa_{1}^{2}+\kappa_{3}^{2})(\kappa_{2}^{2}+\kappa_{4}^{2})-16\kappa_{\infty}(\kappa_{1}\kappa_{2}\kappa_{3}+\kappa_{2}\kappa_{3}\kappa_{4}+\kappa_{3}\kappa_{4}\kappa_{1}+\kappa_{4}\kappa_{1}\kappa_{2})$
$\displaystyle-12\kappa_{1}\kappa_{2}\kappa_{3}\kappa_{4}-2(\kappa_{1}\kappa_{2}+\kappa_{3}\kappa_{4})(\kappa_{2}\kappa_{3}+\kappa_{4}\kappa_{1})=0.$

For larger $n$ the polynomials grow rapidly in size and degree.

Flowers are a building block of circle packing theory [25]. It is known from
the general theory of circle packing that once the curvatures of the petals of
an $n$-flower are known, the curvature of the central circle is determined.
Theorem Theorem gives an explicit equation for that central curvature.

Overview of proof. The proof proceeds in two steps. First, we convert the
problem to one in hyperbolic geometry. Inverting each petal $C_{j}$ of an
$n$-flower in the central circle $C_{\infty}$ yields a collection of circles
$\mathring{C}_{j}$ internally tangent to $C_{\infty}$: an “inverted flower”.
Viewing $C_{\infty}$ as the boundary of the disc model of the hyperbolic
plane, each $\mathring{C}_{j}$ appears as a horosphere. It is useful for our
purposes to convert to the upper half plane model, where $C_{\infty}$ becomes
$\mathbb{R}\cup\\{\infty\\}$, and each $\mathring{C}_{j}$ becomes a horosphere
$\overline{C}_{j}$ appearing as a circle tangent to
$\mathbb{R}\cup\\{\infty\\}$. It is not difficult to relate the Euclidean
curvatures $\kappa_{j},\mathring{\kappa}_{j},\overline{\kappa}_{j}$ of the
circles $C_{j},\mathring{C}_{j},\overline{C}_{j}$ to each other.

Second, we use the spinor-horosphere correspondence recently proved by the
first author in [15]. Building on work of Penrose–Rindler [20] and Penner
[19], this correspondence provides an explicit, smooth, bijective,
$SL(2,\mathbb{C})$-equivariant correspondence between nonzero 2-component
spinors, and horospheres in hyperbolic 3-space $\mathbb{H}^{3}$, with a
certain spinorial decoration. For us, spinors can be regarded simply as pairs
of complex numbers $(\xi,\eta)$. When $\xi,\eta$ are _real_ , the
correspondence essentially reduces to 2 dimensions, yielding a correspondence
between nonzero $(\xi,\eta)\in\mathbb{R}^{2}$, and horocycles in the
hyperbolic plane $\mathbb{H}^{2}$, which is particularly simple in the upper
half plane model. Under the correspondence, there is a simple way to calculate
the distance between horospheres, and hence to know when they are tangent,
using the bilinear form on spinors given by the $2\times 2$ determinant.
Moreover, the curvatures $\mathring{\kappa}_{j}$ and $\kappa_{j}$ are both
straightforward quadratic functions of $\xi,\eta$: they are each given, up to
a constant, by the Euclidean square length $\xi^{2}+\eta^{2}$ of $(\xi,\eta)$.

Thus, the problem reduces to finding a relation among the Euclidean lengths of
real 2-dimensional vectors $(\xi,\eta)$, given that they satisfy certain
bilinear conditions equivalent to the tangency relations of a flower. The
bilinear conditions are essentially that the vectors successively span
parallelograms of area $1$. The desired relation is found by introducing
complex variables $z_{j}=\xi_{j}+i\eta_{j}$.

The expressions for the $m_{j}$ in (1.3) arise naturally in calculations with
the spinors $(\xi_{j},\eta_{j})$, as do the various products arising in the
equations (1.4) and (1.5). We thus include some further calculations in
Section 5 which help to motivate and explain these expressions, and explain
how (1.4) and (1.5) were found.

Related work. Descartes’ equation dates back to 1643 and we do not attempt to
provide a summary of the history or developments over subsequent centuries; we
simply mention some notable and recent works, and those most closely related
to our approach.

Recent historical accounts of the the correspondence between Descartes and
Elisabeth of the Palatinate, in which equation (1.2) first appeared, include
[4], [18, pp. 31–33], and [21, pp. 37–38, 73–81]. Equation (1.2) was stated by
Yamaji Nushizumi in 1751 [17], and rediscovered several times, including by
Steiner (1826) [24], Beecroft (1842) [3], and Soddy (1936) [22]; the latter
restating (1.2) in poetry.

Numerous generalisations of Descartes’ theorem are known; we mention a few.
For instance, in 1936 Gosset generalised the result to $n+2$ mutually tangent
spheres in $n$ dimensions, appending a verse to Soddy’s poem [1]. In 1962
Mauldon generalised these results to spherical and hyperbolic space [16]. In
2002 Lagarias–Mallows–Wilks [14] further extended these results, relating not
only the curvatures but also the centres of the spheres involved. In 2007,
Kocik [9] extended the result for $n+2$ Euclidean spheres in $n$ dimensions to
the case where circles need not be tangent.

Apollonian circle packings, consisting of nested 3-flowers, are a research
field in their own right; see e.g. Graham–Lagarias–Mallows–Wilks–Yan [6] or
Aharonov–Stephenson [2] for general background. These packings have important
number-theoretic properties and have seen recent breakthroughs, e.g. [7]. A
special case closely related to our work, and mentioned by the first author in
[15], is that of Ford circles [5]: these arise from _integer_ spinors.

So far as we know, however, none of these works provide a generalisation of
Descartes’ theorem to $n$-flowers for $n>3$. There are certainly results
involving flower-like configurations, such as Soddy’s hexlet [23], but they do
not provide an equation relating curvatures.

Flowers being simple examples of circle packings, the general theory of
packing applies to them: see generally Stephenson [25]. In general, from a
simplicial 2-complex $K$ triangulating an oriented surface, circle packing
theory studies the existence and uniqueness of circle packings realising $K$,
in the sense that vertices of $K$ correspond to circles, edges of $K$
correspond to tangencies, and triangles of $K$ correspond to oriented tangent
circle triples. Flowers arise in the simple case when $K$ is a disc built from
$n$ triangles around a central vertex. Perhaps most relevantly for our
purposes, the “boundary value theorem” of [25, thm. 11.6] provides that when
$K$ is topologically a disc, the curvatures of circles at boundary vertices,
and branching structure, may be specified arbitrarily, and then a unique
packing (in Eudlicean or hyperbolic geometry) exists. Our result gives the
Euclidean curvature of the interior circle in the flower case.

To the best of our knowledge, perhaps the existing works most closely related
to our approach are those of Kocik, who in several papers uses spinors to
describe Descartes circle configurations and Apollonian packings [13, 12, 11,
10, 9, 8]. However in those works, spinors are complex numbers (defined up to
sign) describing tangencies between pairs of circles. This is significantly
different from our approach. We also mention that the linear algebra and
bilinear forms of Lagarias–Mallows–Wilks [14] and Aharonov–Stephenson [2] are
of a somewhat similar flavour.

Structure of this paper. Section 2 provides relevant background forming the
basis of this paper. Section 3 demonstrates how the $n$-flower can be linked
to horocycles in hyperbolic geometry. Section 4 then equips these horocycles
with spinor coordinates in $\mathbb{C}^{2}$ by applying a correspondence
between spinors and horospheres. Section 5 performs calculations on spinors,
which are not necessary for but motivate the main result. Finally, Section 6
proves the generalised Descartes theorem.

Acknowledgments. The authors are supported by Australian Research Council
grant DP210103136.

##  2 Background

###  2.1 Spinors and Horospheres

We state here results of the first author in [15] required in the sequel.

In [20], Penrose and Rindler consider 2-component spinors
$(\xi,\eta)\in\mathbb{C}^{2}$ and provide them with interpretations in
Minkowski space, and more generally in relativity theory. They associate to
such a spinor a _null flag_ , consisting of a ray on the future light cone,
together with a spinorial tangent plane to the light cone. There is a natural
bilinear form $\\{\cdot,\cdot\\}$ on such spinors, given by the standard
complex symplectic form on $\mathbb{C}^{2}$: given two spinors
$\alpha=(\xi,\eta)$ and $\alpha^{\prime}=(\xi^{\prime},\eta^{\prime})$,

$\\{\alpha,\alpha^{\prime}\\}=\det\begin{pmatrix}\xi&\xi^{\prime}\\\
\eta&\eta^{\prime}\end{pmatrix}=\xi\eta^{\prime}-\eta\xi^{\prime}.$

In [19], Penner associates to each point on the future light cone a horocycle
in the hyperboloid model of hyperbolic space. In [15], the first author
combined these associations to prove the following.

######  Theorem 2.1.

There is an explicit, smooth, bijective, $SL(2,\mathbb{C})$-equivariant
correspondence between nonzero spinors $(\xi,\eta)\in\mathbb{C}^{2}$, and
horospheres $H$ in hyperbolic 3-space decorated with spin-directions.

We briefly explain the notions in this theorem; see [15, sec. 4] for further
details.

A _decoration_ on a horosphere is a parallel oriented line field on it, i.e.,
roughly speaking, a direction along it. Such fields or directions are well
defined because a horosphere is isometric to the Euclidean plane. A decoration
on a horosphere then lifts to two different _spin decorations_ as follows. A
parallel oriented line field on a horosphere $H$ can be equivalently described
by a parallel unit tangent vector field $v$ on $H$. From $v$ we can form
_inward_ and _outward_ right-handed orthonormal frame fields on $H$ given by
$f^{in}=(N^{in},v,N^{in}\times v)$ and $f^{out}=(N^{out},v,N^{out}\times v)$,
where $N^{in},N^{out}$ are respectively inward and outward normal vector
fields on $H$. Each of these two frame fields has two continuous lifts to the
spin double cover of the frame bundle, which we call _inward_ and _outward
spin decorations_. We define an inward spin decoration $W^{in}$ and an outward
spin decoration $W^{out}$ to be _associated_ if they are related by a specific
rotation around $v$. A _spin decoration_ on $H$ is then defined as a pair
$(W^{in},W^{out})$ of associated inward and outward spin decorations.

The action of $SL(2,\mathbb{C})$ on $\mathbb{C}^{2}$ is by matrix-vector
multiplication. The action of $SL(2,\mathbb{C})$ on horospheres is a lift of
the standard action of $PSL(2,\mathbb{C})$ on hyperbolic 3-space by isometries
(indeed, $PSL(2,\mathbb{C})\cong\operatorname{Isom}^{+}(\mathbb{H}^{3})$, the
orientation-preserving isometry group of hyperbolic 3-space). Each isometry in
$PSL(2,\mathbb{C})$ has two lifts to $SL(2,\mathbb{C})$, which are negatives
of each other. Just as with any universal cover, a lift of an isometry $\phi$
from $PSL(2,\mathbb{C})$ to $SL(2,\mathbb{C})$ may be specified by a path in
$PSL(2,\mathbb{C})$ from the identity to $\phi$; this path of isometries,
applied to a spin-decorated horosphere, yields a path of spin-decorated
horospheres; this determines the action of $SL(2,\mathbb{C})$ on spin-
decorated horospheres.

In the upper half space model, given in standard fashion as

$\mathbb{U}^{3}=\\{(x,y,z)\;\mid\;z>0\\}=\\{(x+yi,z)\;\mid\;z>0\\}=\mathbb{C}\times\mathbb{R}_{+}$

with Riemannian metric $ds^{2}=(dx^{2}+dy^{2}+dz^{2})/z^{2}$, the explicit
correspondence of Theorem 2.1 is particularly simple [15, prop. 3.9]. The
horosphere $H$ corresponding to the spinor $(\xi,\eta)$ has centre $\xi/\eta$,
the centre of a horosphere being its point at infinity. If
$\xi/\eta\in\mathbb{C}$, then $H$ appears in the model as a Euclidean sphere
of diameter $1/|\eta|^{2}$. A decoration on a horosphere can be given by a
tangent direction to $H$ at a point; at the point of $H$ with maximum
$z$-coordinate, tangent directions are parallel to $\mathbb{C}$ and are
conveniently specified by complex numbers (the “north pole specification” of a
direction). The decoration on the horosphere $H$ corresponding to $(\xi,\eta)$
is north-pole specified by $i/\eta^{2}$. (The spinor $(-\xi,-\eta)$ yields the
same decoration on the same horosphere, but a different spin decoration.) If
$\xi/\eta=\infty$ then $H$ appears in the model as a horizontal plane at
Euclidean height $|\xi|^{2}$. Since such a horosphere appears parallel to
$\mathbb{C}$, a direction can again be specified by an element of
$\mathbb{C}$. The decoration on this $H$, corresponding to $(\xi,\eta)$, is
then specified by $i\xi^{2}$.

We will be considering reducing this situation to 2 dimensions, as discussed
in [15, sec. 5]. The points $(x,y,z)$ of the upper half space model
$\mathbb{U}^{3}$ with $y=0$ form the upper half plane model of the hyperbolic
plane $\mathbb{U}^{2}$. The boundary at infinity of $\mathbb{U}^{2}$ is then
given by
$\partial\mathbb{U}^{2}=\mathbb{R}\cup\\{\infty\\}\subset\mathbb{C}\cup\\{\infty\\}=\partial\mathbb{U}^{3}$.
Horospheres in $\mathbb{U}^{3}$ centred at points of
$\mathbb{R}\cup\\{\infty\\}$ correspond bijectively to horocycles in
$\mathbb{U}^{2}$, with the bijection given by intersecting with
$\mathbb{U}^{2}$. In this way we may regard horocycles $H$ of $\mathbb{U}^{2}$
as arising from horospheres $\widetilde{H}$ of $\mathbb{U}^{3}$ centred at
$\mathbb{R}\cup\\{\infty\\}$.

Given a horocycle $H$ in $\mathbb{U}^{2}\subset\mathbb{U}^{3}$, a _planar spin
decoration_ on $H$ is a spin decoration on $\widetilde{H}$ whose decoration is
(north-pole) specified by the direction $i$. A spinor $(\xi,\eta)$ describes a
planar spin decoration on a horocycle if and only if $\xi$ and $\eta$ are both
_real_ [15, lem. 5.6]. Each horocycle $H$ of $\mathbb{U}^{2}$ has precisely
two planar spin decorations. These two spin decorations are related to each
other by a $2\pi$ rotation. The corresponding spinors are negatives of each
other. Thus, roughly speaking, reducing to 2 dimensions amounts to considering
_real_ spinors.

###  2.2 Lambda Lengths

Penner [19] introduced the notion of $\lambda$-length between horocycles in
the hyperbolic plane. Take the geodesic $\gamma$ connecting the centres of two
horocycles and consider the segment external to both; if they overlap, then
consider the segment within the overlap and call the distance negative. If the
hyperbolic length of this segment is $\rho$, the $\lambda$-length is defined
as $e^{\rho/2}$.

In [15] the first author generalised $\lambda$-lengths to 3 dimensions. Given
two spin-decorated horospheres $H_{1},H_{2}$ with distinct centres
$z_{1},z_{2}$, we may measure a _complex_ distance from $H_{1}$ to $H_{2}$ as
follows. Let $\gamma$ be the oriented geodesic from $z_{1}$ to $z_{2}$. Let
$\rho$ be the signed distance along $\gamma$ (using the orientation of
$\gamma$) from $p_{1}=\gamma\cap H_{1}$ to $p_{2}=\gamma\cap H_{2}$. The spin
decoration on $H_{1}$ contains an inward spin decoration $W^{in}_{1}$, and the
spin decoration $H_{2}$ contains an outward spin decoration $W^{out}_{2}$.
These spin decorations $W^{in}_{1},W^{out}_{2}$ project to frames
$f^{in}_{1},f^{out}_{2}$ at $p_{1},p_{2}$ respectively, whose first vectors
are positively tangent to $\gamma$. The two frames are then related by a
signed translation of length $\rho$ along $\gamma$, followed by a rotation of
some angle $\theta$ about $\gamma$. Lifting to spin decorations, this $\theta$
is well defined modulo $4\pi$. The complex distance from $H_{1}$ to $H_{2}$ is
then $\rho+i\theta$, and the _complex $\lambda$-length_ from $H_{1}$ to
$H_{2}$ is

$\lambda_{12}=\exp\left(\frac{\rho+i\theta}{2}\right)$

It turns out that the complex $\lambda$-length is _antisymmetric_ :
$\lambda_{12}=-\lambda_{21}$. See [15, sec. 4] for further details.

If we have two horocycles $H_{1},H_{2}$ in $\mathbb{U}^{2}$ with planar spin
decorations, then the frames $f^{in}_{1},f^{out}_{2}$ have their first vectors
pointing along $\gamma$ and their second vectors pointing the direction of the
decoration, i.e. normal to $\mathbb{U}^{2}$ in the direction specified by $i$,
or equivalently using the standard coordinates $(x,y,z)$, in the positive
$y$-direction. It follows that $\theta=0$ modulo $2\pi$. Then we observe that
the complex $\lambda$-length from $H_{1}$ to $H_{2}$ is real, taking a
positive value if $\theta=0$ mod $4\pi$ and a negative value if $\theta=2\pi$
mod $4\pi$. Penner’s 2-dimensional $\lambda$-length is the positive value.

In [15], the first author showed that complex $\lambda$-lengths are given by
spinors using Penrose and Rindler’s bilinear form.

######  Theorem 2.2.

Consider two spinors $\kappa_{1},\ \kappa_{2}$, corresponding to spin-
decorated horospheres $H_{1},H_{2}$. Then the complex $\lambda$-length
$\lambda_{12}$ from $H_{1}$ to $H_{2}$ is given by

$\lambda_{12}=\\{\kappa_{1},\kappa_{2}\\}.$

Indeed, in the case where $\kappa_{1}$, $\kappa_{2}$ are real spinors,
describing horocycles with planar spin decorations, we observe that
$\\{\kappa_{1},\kappa_{2}\\}$ is real, as is $\lambda_{12}$. The complex
$\lambda$-length $\lambda_{12}$ is positive or negative. Replacing one of the
$\kappa_{j}$ with $-\kappa_{j}$ has the effect of changing the spin decoration
on $H_{j}$, adding $2\pi$ to $\theta$ (mod $4\pi$) and changing the sign of
$\lambda_{12}$ (by introducing a factor of $e^{i\pi}=-1$). Similarly,
replacing $(\kappa_{1},\kappa_{2})$ with $(\kappa_{2},\kappa_{1})$ changes the
sign of the complex $\lambda$-length.

Complex $\lambda$-lengths have several nice properties. For instance, given
four horocycles (whose $\lambda$-lengths measure line segments forming an
inscribed quadrilateral inside the disk), their $\lambda$-lengths satisfy
Ptolemy’s relation

$\lambda_{12}\lambda_{34}+\lambda_{23}\lambda_{14}=\lambda_{13}\lambda_{24}.$

Importantly for our purposes, the $\lambda$-length between two horocycles
(with planar spin decorations) is $\pm 1$ if and only if the horocycles are
tangent. This gives a simple condition for checking a purported $n$-flower.

##  3 From Flowers to Horocycles

Consider an $n$-flower ($n\geq 3$) in the Euclidean plane, as in Definition
1.1. Let each circle $C_{\bullet}$ have radius $r_{\bullet}$ and curvature
$\kappa_{\bullet}=1/r_{\bullet}$.

Dilating the entire configuration by a factor of $\kappa_{\infty}$ sends the
curvature $\kappa_{\infty}\mapsto 1$ and, for each petal circle,
$\kappa_{j}\mapsto\kappa_{j}/\kappa_{\infty}$. We may thus assume the central
circle has unit radius, so that $\kappa_{\infty}=1$, and for each petal circle
$C_{j}$ write $\kappa_{j}$ for the resulting curvature.

Invert the configuration in the unit circle $C_{\infty}$. Each $C_{j}$ is
mapped to a circle $\mathring{C}_{j}$ internally tangent to $C_{\infty}$. For
each $j$ mod $n$ then $\mathring{C}_{j}$ is externally tangent to
$\mathring{C}_{j-1}$ and $\mathring{C}_{j+1}$. See Figure 2. We denote by
$\mathring{r}_{j}$ and $\mathring{\kappa}_{j}$ the (Euclidean) radius and
curvature of $\mathring{C}_{j}$ respectively. (The superscript ring notation
is intended to indicate “inside the unit disc”.)

Figure 2: An inverted $3$-flower.

######  Lemma 3.1.

For each $j$,

$\mathring{\kappa}_{j}=\kappa_{j}+2.$

###### Proof.

Let $O$ denote the centre of $C_{\infty}$. The furthest point $A$ from $O$ on
$C_{j}$ has distance $1+2r_{j}$ from $O$. The closest point $A^{\prime}$ to
$O$ on $\mathring{C}_{j}$ has distance $1-2\mathring{r}_{j}$ from $O$. See
Figure 3.

$O$$A$$A^{\prime}$$\mathring{r}$$r$ Figure 3: Inversion of a petal circle.

Inversion in $C_{\infty}$ sends $A\mapsto A^{\prime}$, so

$(1+2r_{j})(1-2\mathring{r}_{j})=1,$

which upon substituting $r_{j}=1/\kappa_{j}$ and
$\mathring{r}_{j}=1/\mathring{\kappa}_{j}$ yields the desired equation.
$\hfill\square$

We now consider the unit circle $C_{\infty}$ as the boundary of the disc model
$\mathbb{D}^{2}$ of the hyperbolic plane. Each $\mathring{C}_{j}$ then appears
as a horocycle.

Next, we convert from the disc model $\mathbb{D}^{2}$ to the upper half plane
model $\mathbb{U}^{2}$ of the hyperbolic plane. Regarding $\mathbb{D}^{2}$ and
$\mathbb{U}^{2}$ as subsets of the complex plane in the standard way, the two
models are related by the Cayley transform
$\mathfrak{S}\colon\mathbb{U}^{2}\longrightarrow\mathbb{D}^{2}$ and its
inverse:

$\mathfrak{S}(z)=\frac{z-i}{z+i},\quad\mathfrak{S}^{-1}(z)=\frac{(z+1)i}{1-z}.$
(3.2)

Observe that $\mathfrak{S}^{-1}$ sends $\partial\mathbb{D}^{2}$ to
$\partial\mathbb{U}^{2}=\mathbb{R}\cup\\{\infty\\}$, and sends horocycles in
$\mathbb{D}^{2}$ to horocycles in $\mathbb{U}^{2}$. Moreover, the point $1$ on
$\partial\mathbb{D}^{2}$ corresponds to the point $\infty$ on
$\partial\mathbb{U}^{2}=\mathbb{R}\cup\\{\infty\\}$. And as the unit circle
$\partial\mathbb{D}^{2}$ is traversed anticlockwise, its image
$\mathbb{R}\cup\\{\infty\\}$ under $\mathfrak{S}^{-1}$ is traversed in the
increasing direction.

Before applying $\mathfrak{S}^{-1}$, if necessary we reflect the configuration
in an arbitrary hyperbolic line through the origin, so that the centres of
$\mathring{C}_{0},\mathring{C}_{1},\ldots,\mathring{C}_{n-1}$ are in
anticlockwise order around $\partial\mathbb{D}^{2}$. This is a reflection in
both hyperbolic and Euclidean geometry, so preserves all $\mathring{r}_{j}$
and $\mathring{\kappa}_{j}$. Then, if necessary, we rotate $\mathbb{D}^{2}$
about $0$ so that, proceeding anticlockwise from $1$ around
$\partial\mathbb{D}^{2}$, we encounter the centre of $\mathring{C}_{0}$ first,
then $\mathring{C}_{1},\ldots,\mathring{C}_{n-1}$, in order. Again, this is a
rotation in both hyperbolic and Euclidean geometry, so preserves all
$\mathring{r}_{j}$ and $\mathring{\kappa}_{j}$.

Now apply $\mathfrak{S}^{-1}$ to the entire configuration. Since no
$\mathring{C}_{j}$ is tangent to $\partial\mathbb{D}^{2}$ at $1$, then no
$\overline{C}_{j}$ is tangent to
$\partial\mathbb{U}^{2}=\mathbb{R}\cup\\{\infty\\}$ at $\infty$. Thus each
horocycle $\overline{C}_{j}$ has centre in
$\mathbb{R}\subset\partial\mathbb{U}^{2}$, and appears as a circle in
$\mathbb{U}^{2}$, with a finite radius $\overline{r}_{j}$ and curvature
$\overline{\kappa}_{j}$. Moreover, since the centres of $\mathring{C}_{j}$ are
in order around $\partial\mathbb{D}^{2}$ proceeding anticlockwise from $1$,
the centres of the $\overline{C}_{j}$ are in increasing order along
$\mathbb{R}\subset\partial\mathbb{U}^{2}$.

We thus obtain a “flat $n$-flower” where the original central circle has
become $\mathbb{R}\cup\\{\infty\\}$, and each petal circle has become a
horocycle in $\mathbb{U}^{2}$, with each $\overline{C}_{j}$ externally tangent
to $\overline{C}_{j-1}$ and $\overline{C}_{j+1}$ (with indices taken mod $n$),
and the horocycles increasing in index from left to right. See Figure 4. (The
overline notation is intended to indicate “in a flat flower along the real
line”.)

$\frac{\xi_{0}}{\eta_{0}}$$\frac{\xi_{1}}{\eta_{1}}$$\frac{\xi_{2}}{\eta_{2}}$
Figure 4: A flat $3$-flower

Finally, observe that the circles $\overline{C}_{j}$ can be translated
horizontally without affecting the tangencies of circles or their or
curvatures. We apply such a translation so that $\overline{C}_{0}$ is tangent
to the real line at $0$. Since the horocycles $\overline{C}_{j}$ are tangent
to the real line in increasing order, for $j\geq 1$ each $\overline{C}_{j}$ is
tangent to the real line at a positive number.

##  4 From Horospheres to Spinors

We now introduce spinors for the horocycles $\overline{C}_{j}$.

As discussed in Section 2.1, each horocycle in the hyperbolic plane has two
planar spin decorations, corresponding to two real spinors which are negatives
of each other.

Thus each horocycle $\overline{C}_{j}$ corresponds to a real spinor
$\alpha_{j}=(\xi_{j},\eta_{j})$, well defined up to sign. The centre of
$\overline{C}_{j}$ is at $\xi_{j}/\eta_{j}$, which lies in $\mathbb{R}$, so
each $\eta_{j}$ is nonzero. The Euclidean radius $\overline{r}_{j}$ and
curvature $\overline{\kappa}_{j}$ of $\overline{C}_{j}$ are given by

$\overline{r}_{j}=\frac{1}{2\eta_{j}^{2}}\quad\text{and}\quad\overline{\kappa}_{j}=2\eta_{j}^{2}.$
(4.1)

(In general the Euclidean diameter of the horosphere corresponding to
$(\xi,\eta)$ is $\frac{1}{|\eta|^{2}}$. Here $\eta$ is real.)

We now choose each $\alpha_{j}$ so that $\eta_{j}$ is positive.

Recall $\overline{C}_{0}$ is constructed to be tangent to the real line at
$0$, and the centres of the $\overline{C}_{j}$ are in increasing order along
the real line. Thus we have

$0=\frac{\xi_{0}}{\eta_{0}}<\frac{\xi_{1}}{\eta_{1}}<\cdots<\frac{\xi_{n-1}}{\eta_{n-1}}.$

Thus $\xi_{0}=0$ and, since all $\eta_{j}$ are chosen positive, for $j\geq 1$
we have $\xi_{j}>0$.

Calculating bilinear forms, for all $0\leq j<k\leq n-1$ we have

$\\{\alpha_{j},\alpha_{k}\\}=\xi_{j}\eta_{k}-\xi_{k}\eta_{j}=\eta_{j}\eta_{k}\left(\frac{\xi_{j}}{\eta_{j}}-\frac{\xi_{k}}{\eta_{k}}\right)<0.$

(This is the opposite of the _totally positive_ notion of [15, sec. 6]; the
$\alpha_{j}$ here are “totally negative”.)

Now for each $j$ mod $n$ we know that $\overline{C}_{j}$ and
$\overline{C}_{j+1}$ are tangent. Thus the complex lambda length between
planar spin decorations of $\overline{C}_{j}$ and $\overline{C}_{j+1}$ is $\pm
1$. Combining these observations we have the following.

######  Lemma 4.2.

Choosing the spinors $\alpha_{j}=(\xi_{j},\eta_{j})$ for planar spin
decorations on each $\overline{C}_{j}$ so that all $\eta_{j}>0$, then for all
$0\leq j\leq n-2$ we have

$\\{\alpha_{j},\alpha_{j+1}\\}=\xi_{j}\eta_{j+1}-\xi_{j+1}\eta_{j}=-1,$

and $\\{\alpha_{0},\alpha_{n-1}\\}=-1$. $\hfill\square$

The following proposition gives a useful relationship between each spinor
$(\xi_{j},\eta_{j})$, and the Euclidean curvature $\mathring{\kappa}_{j}$ of
the corresponding horocycle in the disc model.

######  Proposition 4.3.

Let $(\xi,\eta)$ be a real spinor and $H$ the corresponding horocycle in
$\mathbb{D}^{2}$ with a planar spin decoration. Let the Euclidean curvature of
$H$ as it appears in the disc model $\mathbb{D}^{2}$ be $\mathring{\kappa}$.
Then

$\mathring{\kappa}=\xi^{2}+\eta^{2}+1.$ (4.4)

Combining Proposition 4.3 with Lemma 3.1, we observe that
$\kappa=\mathring{\kappa}-2=\xi^{2}+\eta^{2}-1$, so the curvatures of the
original circles are also usefully expressed in terms of spinor coordinates.

In order to prove Proposition 4.3 we use some lemmas.

######  Lemma 4.5.

Let $H$ be a horocycle in $\mathbb{U}^{2}$ with a planar spin decoration,
corresponding to the spinor $(\xi,\eta)$. Then the planar spin-decorated
horocycle $H^{\prime}$ obtained by rotating $H$ by $\pi$ about
$i\in\mathbb{U}^{2}$ corresponds to the spinor $(-\eta,\xi)$.

Let us briefly explain what is meant by rotating $H$ in the above proposition.
As $H$ has a planar spin decoration, it corresponds to a spin-decorated
horosphere $\widetilde{H}$ in $\mathbb{U}^{3}$. This $\widetilde{H}$ has a
spin decoration $W$ consisting of associated inward and outward spin
decorations, which are lifts of frame fields $f$ along $\widetilde{H}$. An
orientation-preserving isometry
$\phi\in\operatorname{Isom}^{+}(\mathbb{U}^{3})\cong PSL(2,\mathbb{C})$ may be
applied to the horosphere $\widetilde{H}$ and (via its derivative) the frame
fields $f$, yielding a decoration on a horosphere $\widetilde{H^{\prime}}$. If
we specify a lift of $\phi$ to the spin double cover $SL(2,\mathbb{C})$, then
there is a well-defined map of the spin decorations $W$ from $\widetilde{H}$
to $\widetilde{H^{\prime}}$. A lift of $\phi$ to $SL(2,\mathbb{C})$ can be
specified by a path from the identity to $\phi$ in $PSL(2,\mathbb{C})$. Here,
we have a rotation of $\pi$ about $i\in\mathbb{U}^{2}$, which naturally
extends to a rotation in the 3-dimensional model $\mathbb{U}^{3}$ about the
geodesic normal to $\mathbb{U}^{2}$ through $i$. This rotation is naturally
lifted to the spin double cover of $\operatorname{Isom}^{+}(\mathbb{U}^{3})$
by taking the path of isometries $\phi_{t}$ over $t\in[0,1]$, where $\phi_{t}$
is a rotation of angle $\pi t$ about the same axis as $\phi$. The isometries
$\phi_{t}$ then take the spin decoration on $\widetilde{H}$ to a spin
decoration on $\widetilde{H^{\prime}}$, corresponding to the planar spin-
decorated $H^{\prime}$ of the lemma.

###### Proof.

Consider the matrices $M(\theta)$ for $\theta\in\mathbb{R}$ given by

$M(\theta)=\begin{pmatrix}\cos(\theta)&-\sin(\theta)\\\
\sin(\theta)&\cos(\theta)\\\ \end{pmatrix}\in SL(2,\mathbb{R})\subset
SL(2,\mathbb{C}).$

Each $M(\theta)$ describes an orientation-preserving isometry of
$\mathbb{U}^{2}$ or $\mathbb{U}^{3}$ as a Möbius transformation: a rotation of
$2\theta$ about $i$ in $\mathbb{U}^{2}$, or a rotation of $2\theta$ about the
geodesic normal to $\mathbb{U}^{2}$ through $i$ in $\mathbb{U}^{3}$. Then
$M(\frac{t\pi}{2})$, over $t\in[0,1]$, describes a path of isometries from the
identity to the rotation described in the lemma. As in fact all $M(\theta)$
lie in $SL(2,\mathbb{C})$, the double cover of the isometry group
$PSL(2,\mathbb{C})$, then $M(\frac{\pi}{2})$ is the desired spin lift of the
isometry described in the lemma: it is a rotation of $\pi$ (not $-\pi$) about
$i$.

We now use the equivariance of Theorem 2.1. Since $(\xi,\eta)$ corresponds to
$H$, and $H^{\prime}$ is obtained from $H$ by applying $M(\pi/2)$, then
$H^{\prime}$ corresponds to the spinor

$M\left(\frac{\pi}{2}\right)\begin{pmatrix}\xi\\\
\eta\end{pmatrix}=\begin{pmatrix}0&-1\\\ 1&0\end{pmatrix}\begin{pmatrix}\xi\\\
\eta\end{pmatrix}=\begin{pmatrix}-\eta\\\ \xi\end{pmatrix}.$

$\hfill\square$

Translating Lemma 4.5 into the disc model $\mathbb{D}^{2}$ via the Cayley
transform (3.2), we immediately obtain the following.

######  Lemma 4.6.

Let $H$ be a horocycle in $\mathbb{D}^{2}$ with a planar spin decoration,
corresponding to the spinor $(\xi,\eta)$. Then the planar spin-decorated
horocycle $H^{\prime}$ obtained by rotating $H$ by $\pi$ about
$0\in\mathbb{D}^{2}$ corresponds to the spinor $(-\eta,\xi)$. $\hfill\square$

###### Proof of Proposition 4.3.

Let $H^{\prime}$ be the planar spin-decorated horocycle obtained by rotating
$H$ by $\pi$ about $0\in\mathbb{D}^{2}$. This is a Euclidean rotation as well
as a hyperbolic rotation, so $H^{\prime}$ has the same radius and curvature as
$H$. By Lemma 4.6, $H^{\prime}$ corresponds to the spinor $(-\eta,\xi)$.

Let $X$ be the hyperbolic distance from $0\in\mathbb{D}^{2}$ to $H$. By
symmetry, the hyperbolic distance from $0$ to $H^{\prime}$ is also $X$, so the
hyperbolic distance from $H$ to $H^{\prime}$ is $2X$. The metric in
$\mathbb{D}^{2}$ is given by

$ds^{2}=\frac{4\ dr^{2}}{(1-r^{2})^{2}},$

where $r$ is a Euclidean radial coordinate. Since $H$ and $H^{\prime}$ have
Euclidean radius $\mathring{r}$, the line from $0$ to $H$ is a Euclidean line
from $r=0$ to $r=1-2\mathring{r}$. Thus

$X=\int_{0}^{1-2\mathring{r}}\frac{2dr}{1-r^{2}}=\ln\left(\frac{1-\mathring{r}}{\mathring{r}}\right).$

$\mathring{r}$$X$$0$$H$$H^{\prime}$ Figure 5: Computing the radius of $H$

Let us now consider the complex lambda length from $H$ to $H^{\prime}$. We
thus consider the 3-dimensional spin-decorated horospheres
$\widetilde{H},\widetilde{H^{\prime}}$ corresponding to $H,H^{\prime}$. The
distance between $\widetilde{H}$ and $\widetilde{H^{\prime}}$ is $\rho=2X$.
Consider the angle $\theta$ between the inward spin frame $W_{in}$ of
$\widetilde{H}$ and the outward spin frame $W^{\prime}_{out}$ of
$\widetilde{H^{\prime}}$ along the common perpendicular from $H$ to
$H^{\prime}$. Since both spin decorations are planar, we have $\theta=0$ or
$2\pi$ mod $4\pi$. Following [15, defn. 4.3], the outward spin frame
associated to $W_{in}$ is obtained from $W_{in}$ by a rotation of angle $-\pi$
about the decoration direction normal to $\mathbb{D}^{2}$. The outward spin
decoration of $W^{\prime}_{out}$ is obtained from this associated frame by a
rotation of angle $\pi$ about the same normal direction to $\mathbb{D}^{2}$
(i.e. the rotation which takes $\widetilde{H}$ to $\widetilde{H^{\prime}}$).
We conclude that $\theta=0$. Thus the complex lambda length $\lambda$ from $H$
to $H^{\prime}$ is

$\lambda=\exp\left(\frac{\rho+i\theta}{2}\right)=e^{X}=\frac{1-\mathring{r}}{\mathring{r}}.$
(4.7)

On the other hand, applying Theorem 2.2 to the two spinors involved yields

$\lambda=\\{(\xi,\eta),(-\eta,\xi)\\}=\xi^{2}+\eta^{2}.$ (4.8)

Equating (4.7) and (4.8) and using $\mathring{\kappa}=1/\mathring{r}$ then
gives the desired result. $\hfill\square$

##  5 Spinor Calculations

We now present some calculations that are not strictly necessary for the main
result, but may be of interest, and may motivate some expressions arising in
the main result. The reader interested only in the proof of the main result
may skip this section.

###  5.1 Equation Relating $\eta_{j}$

It follows from Lemma 4.2 that, for $0\leq j\leq n-2$,

$\frac{\xi_{j+1}}{\eta_{j+1}}-\frac{\xi_{j}}{\eta_{j}}=\
\frac{1}{\eta_{j}\eta_{j+1}},$ (5.1)

which is the distance along the real line between the centres of the
horospheres $\overline{C}_{j}$ and $\overline{C}_{j+1}$. Summing these
distances over all $j$ from $0$ to $n-2$, we obtain

$\frac{\xi_{n-1}}{\eta_{n-1}}-\frac{\xi_{0}}{\eta_{0}}=\sum_{j=0}^{n-2}\frac{1}{\eta_{j}\eta_{j+1}}.$
(5.2)

Since $\\{\alpha_{0},\alpha_{n-1}\\}=\xi_{0}\eta_{n-1}-\xi_{n-1}\eta_{0}=-1$,
we also have

$\sum_{j=0}^{n-2}\frac{1}{\eta_{j}\eta_{j+1}}=\frac{1}{\eta_{0}\eta_{n-1}}.$
(5.3)

###  5.2 Equation Relating $\overline{\kappa}_{j}$

Substituting $\eta_{j}$ for $\overline{\kappa}_{j}$ in (5.3) using (4.1)
yields an equation relating the curvatures $\overline{\kappa}_{j}$:

$\sum_{j=0}^{n-2}\frac{1}{\sqrt{\overline{\kappa}_{j}\overline{\kappa_{j+1}}}}=\frac{1}{\sqrt{\overline{\kappa}_{0}\overline{\kappa}_{n-1}}}.$
(5.4)

We can obtain a polynomial relation between the $\overline{\kappa}_{j}$ by
clearing denominators and squaring out roots.

This is a relation between the curvatures in a flat $n$-flower, i.e. when
$\kappa_{\infty}=0$.

For example in the case $n=3$, the equation

$\frac{1}{\sqrt{\overline{\kappa}_{0}\overline{\kappa}_{1}}}+\frac{1}{\sqrt{\overline{\kappa}_{1}\overline{\kappa}_{2}}}=\frac{1}{\sqrt{\overline{\kappa}_{0}\overline{\kappa}_{2}}}$

leads to the polynomial

$\overline{\kappa}_{0}^{2}+\overline{\kappa}_{1}^{2}+\overline{\kappa}_{2}^{2}-2\overline{\kappa}_{0}\overline{\kappa}_{1}-2\overline{\kappa}_{1}\overline{\kappa}_{2}-2\overline{\kappa}_{2}\overline{\kappa}_{0}=0,$

which is equivalent to Descartes’ equation (1.2) with $\kappa_{\infty}=0$.

###  5.3 Recursive Computation of Spinor Coordinates

We show that, starting from $(\xi_{0},\eta_{0})$, the remaining
$(\xi_{j},\eta_{j})$ can be calculated recursively using previous
$(\xi_{j},\eta_{j})$ and the Euclidean curvatures $\mathring{\kappa}_{j}$ in
the disc model. (Using Lemma 3.1, one could use the $\kappa_{j}$ instead of
the $\mathring{\kappa}_{j}$.)

######  Lemma 5.5.

The spinors $(\xi_{j},\eta_{j})$ satisfy the following:

$\xi_{0}=0,\quad\eta_{0}=\sqrt{\mathring{\kappa}_{0}-1},$

and for $0\leq j\leq n-2$,

$\displaystyle\xi_{j+1}$
$\displaystyle=\left(\frac{-\xi_{j}+\eta_{j}\sqrt{(\mathring{\kappa}_{j+1}-1)(\mathring{\kappa}_{j}-1)-1}}{\eta_{j}(\mathring{\kappa}_{j}-1)}\right)\xi_{j}+\frac{1}{\eta_{j}},$
(5.6) $\displaystyle\eta_{j+1}$
$\displaystyle=\frac{-\xi_{j}+\eta_{j}\sqrt{(\mathring{\kappa}_{j+1}-1)(\mathring{\kappa}_{j}-1)-1}}{\mathring{\kappa}_{j}-1}.$
(5.7)

By Lemma 3.1 each $\mathring{\kappa}_{j}>2$, so each quantity under a square
root sign is positive, and we take the positive square root in each case.

###### Proof.

Since $\overline{C}_{0}$ is tangent to the real line at $\xi_{0}/\eta_{0}=0$,
we have $\xi_{0}=0$. From Proposition 4.3 then
$\mathring{\kappa}_{0}=\eta_{0}^{2}+1$, so
$\eta_{0}^{2}=\mathring{\kappa}_{0}-1$. Since all $\eta_{j}$ are taken to be
positive then $\eta_{0}$ is as claimed.

From (5.1) we have

$\xi_{j+1}=\frac{\eta_{j+1}}{\eta_{j}}\xi_{j}+\frac{1}{\eta_{j}},$ (5.8)

which shows that (5.6) follows from (5.7).

Squaring (5.8) and substituting the resulting expression for $\xi_{j+1}^{2}$
into (4.4) for $\mathring{\kappa}_{j+1}$ yields

$\mathring{\kappa}_{j+1}=\eta_{j+1}^{2}+\frac{\eta_{j+1}^{2}}{\eta_{j}^{2}}\xi_{j}^{2}+2\frac{\eta_{j+1}}{\eta_{j}^{2}}\xi_{j}+\frac{1}{\eta_{j}^{2}}+1.$

This is a quadratic equation for $\eta_{j+1}$; solving in the standard way and
using (4.4) for $\mathring{\kappa}_{j}$, we obtain

$\eta_{j+1}=\frac{-\xi_{j}\pm\eta_{j}\sqrt{(\mathring{\kappa}_{j+1}-1)(\mathring{\kappa}_{j}-1)-1}}{\mathring{\kappa}_{j}-1}.$

As all $\xi_{j},\eta_{j}$, and $\mathring{\kappa}_{j}-1$ are non-negative, we
must take the $+$ in the $\pm$ in order for $\eta_{j+1}$ to be positive,
yielding (5.7) as desired. $\hfill\square$

###  5.4 Variables $m_{j}$

The square root expressions in the recursive equations (5.6) and (5.7) for
$\xi$ and $\eta$ involve square root expressions which arise throughout; they
are in fact the $m_{j}$ of our main theorem. By Lemma 3.1 we have
$\mathring{\kappa}_{j}-1=\kappa_{j}+1$, so the two expressions given in the
definition below are equal.

######  Definition 5.9.

We define

$m_{0}=\sqrt{\mathring{\kappa}_{0}-1}=\sqrt{\kappa_{0}+1}$

and for $1\leq j\leq n-1$,

$m_{j}=\sqrt{\left(\mathring{\kappa}_{j}-1\right)\left(\mathring{\kappa}_{j-1}-1\right)-1}=\sqrt{\left(\kappa_{j}+1\right)\left(\kappa_{j-1}+1\right)-1}.$

From this definition, a straightforward induction allows us to express each
$\mathring{\kappa}_{j}$ or $\kappa_{j}$ in terms of the $m_{j}$; we state this
in the following lemma and omit the proof. We observe that products arising
here also appear in the main theorem. As usual, the empty product is taken to
be $1$.

######  Lemma 5.10.

For $j>0$ we have

$\kappa_{j}+1=\mathring{\kappa}_{j}-1=\begin{cases}\frac{m_{0}^{2}\prod_{k=1}^{\frac{j}{2}}(m_{2k}^{2}+1)}{\prod_{k=1}^{\frac{j}{2}}(m_{2k-1}^{2}+1)}\text{,
if j is even and }j>0\\\
\frac{\prod_{k=0}^{\frac{j-1}{2}}(m_{2k+1}^{2}+1)}{m_{0}^{2}\prod_{k=1}^{\frac{j-1}{2}}(m_{2k}^{2}+1)}\text{,
if j is odd and }j>0.\end{cases}$ (5.11)

$\hfill\square$

For convenience, we express the above expansion of $\kappa_{j}+1$ in terms of
the $m_{k}$ as $g_{j}$.

###  5.5 Closed Form for Spinor Coordinates

Between them, (5.6) and (5.7) allow us to iteratively find
$(\xi_{j},\eta_{j})$ in terms of the $\mathring{\kappa}_{j}$, or, using
Definition 5.9, in terms of the $m_{j}$. In this section we describe closed
forms for $\xi_{j}$ and $\eta_{j}$ in terms of the $m_{j}$.

Since $\overline{C}_{0}$ is tangent to the real line at $0$, so that
$\xi_{0}=0$, then forming a telescoping sum from (5.1), as in (5.2), we have

$\xi_{j}=\frac{1}{\eta_{j-1}}+\eta_{j}\sum_{k=1}^{j-1}\frac{1}{\eta_{k-1}\eta_{k}},$
(5.12)

so it suffices to find a closed form for $\eta$.

######  Lemma 5.13 (A Closed Form for $\eta_{j}$).

$\eta_{j}=\begin{cases}\frac{m_{0}\left(\prod_{n=1}^{j}(m_{n}-i)+\prod_{n=1}^{j}(m_{n}+i)\right)}{2\prod_{n=1}^{\frac{j}{2}}\left(m_{2n-1}^{2}+1\right)},&\text{
if j is even}\\\
\frac{\left(\prod_{n=1}^{j}(m_{n}-i)+\prod_{n=1}^{j}(m_{n}+i)\right)}{2m_{0}\prod_{n=1}^{\frac{j-1}{2}}\left(m_{2n}^{2}+1\right)},&\text{
if j is odd.}\end{cases}$

###### Proof.

Substituting $m_{j}$ into (5.7) yields

$\eta_{j+1}=\frac{-\xi_{j}+\eta_{j}m_{j+1}}{\mathring{\kappa}_{j}-1}.$

Replacing $\xi_{j}$ using (5.12), and using the definition of $g_{j}$ after
Lemma 5.10 gives

$\eta_{j}=\frac{-1+\eta_{j-1}\eta_{j-2}\left(m_{j}-\sum_{n=1}^{j-2}\frac{1}{\eta_{n-1}\eta_{n}}\right)}{\eta_{j-2}g_{j-1}}.$
(5.14)

Rearrange to solve for the sum:

$\sum_{n=1}^{j-2}\frac{1}{\eta_{n-1}\eta_{n}}=m_{j}-\frac{\eta_{j}\eta_{j-2}g_{j-1}+1}{\eta_{j-1}\eta_{j-2}}.$

Then, taking one term out of the sum, we have

$\sum_{n=1}^{j-2}\frac{1}{\eta_{n-1}\eta_{n}}=\sum_{n=1}^{j-3}\frac{1}{\eta_{n-1}\eta_{n}}+\frac{1}{\eta_{j-3}\eta_{j-2}}=m_{j-1}-\frac{\eta_{j-1}g_{j-2}}{\eta_{j-2}}.$

Substituting this expression for the sum back into (5.14), we obtain a
recursive relation on $\eta_{j}$:

$\eta_{j}=\frac{-1+\eta_{j-1}^{2}g_{j-2}+\eta_{j-1}\eta_{j-2}(m_{j}-m_{j-1})}{\eta_{j-2}g_{j-1}}.$

The closed form is then obtained by induction, taking $\eta_{0}=m_{0}$,
$\eta_{1}=m_{1}/m_{0}$ for the base cases. $\hfill\square$

###  5.6 Polynomial Relations Between $m_{j}$

Multiplying (5.3) by $\eta_{0}\eta_{1}...\eta_{n-1}$ to clear denominators
yields

$\left(\sum_{j=0}^{n-2}\eta_{0}\eta_{1}\cdots\hat{\eta}_{j}\hat{\eta}_{j+1}\cdots\eta_{n-1}\right)-\eta_{1}\eta_{2}...\eta_{n-2}=0,$

where the hats indicate excluding those terms from the product. Substitution
using Lemma 5.13 yields polynomial relations among the $m_{j}$, and these
polynomials contain as factors the equations of our main result.

For example, when $n=3$ and $n=4$ we obtain, respectively,

$\displaystyle\frac{m_{1}\left(m_{1}m_{0}^{2}+m_{2}m_{0}^{2}-m_{1}^{2}-1\right)}{m_{0}\left(m_{1}^{2}+1\right)}=0,$
$\displaystyle\frac{m_{1}\left(m_{1}m_{2}-1\right)\left(-m_{2}^{2}+m_{1}m_{2}+m_{3}m_{2}+m_{1}m_{3}-2\right)}{\left(m_{1}^{2}+1\right)\left(m_{2}^{2}+1\right)}=0,$

and we expect that only the factor containing all the curvatures can be zero
in general. Indeed, in the $n=3$ case this factor is an instance of (1.4) (or
(1.6)) from the main theorem; and in the $n=4$ case an instance of (1.5) (or
(1.7)). This is how the equations of the main theorem were found.

##  6 Proof of the Generalised Descartes Theorem

After the constructions of Section 3 and Section 4, we have real spinors
$(\xi_{j},\eta_{j})$ for $0\leq j\leq n-1$, describing horocycles with planar
spin decorations arising from an $n$-flower.

For each $j$, we define a complex number

$z_{j}=\xi_{j}+i\eta_{j}$

(here $i^{2}=-1$ as usual).

From Lemma 4.2 we have, for $1\leq j\leq n-1$,

$\xi_{j-1}\eta_{j}-\xi_{j}\eta_{j-1}=-1,\quad\text{or
equivalently,}\quad\operatorname{Im}\left(z_{j-1}\,\overline{z_{j}}\right)=1.$

which, being the nonzero component of the cross product
$(\xi_{j-1},\eta_{j-1},0)\times(\xi_{j},\eta_{j},0)$, means that the $z_{j}$
occur in clockwise order around $0$ in the complex plane and each successive
pair $z_{j-1},z_{j}$ spans a parallelogram of area $1$. Since we chose
$\xi_{0}=0$ and all other $\xi_{j},\eta_{j}>0$ in Section 4, we have $z_{0}$
on the positive imaginary axis, and all the other $z_{j}$ lying in the top
right quadrant of the complex plane as in Figure 6. The sequence $\arg z_{j}$
for $0\leq j\leq n-1$ is thus strictly decreasing in $(0,\pi/2]$.

Moreover, Lemma 4.2 says that
$\xi_{0}\eta_{n-1}-\xi_{n-1}\eta_{0}=\operatorname{Im}\left(z_{0}\,\overline{z_{n-1}}\right)=-1$,
so the first and last of the $z_{j}$ also span a parallelogram of area 1.

$z_{0}$$z_{1}$$z_{2}$$z_{3}$$z_{4}$$\xi$$\eta$ Figure 6: Parallelograms formed
by the $z_{j}$.

From Proposition 4.3 (and the remark afterward applying Lemma 3.1) we have

$\kappa_{j}=\xi_{j}^{2}+\eta_{j}^{2}-1=|z_{j}|^{2}-1$

so the problem of finding a relation among the $\kappa_{j}$ (i.e. generalising
Descartes’ theorem) is reduced to a plane Euclidean geometry problem: given
vectors spanning parallelograms of area 1 in clockwise order around the origin
as described above, find a relation among the lengths of those vectors.

We now write the $m_{j}$ (Definition 5.9) for $1\leq j\leq n-1$ in terms of
the $z_{j}$:

$m_{j}=\sqrt{\left(\kappa_{j}+1\right)\left({\kappa}_{j-1}+1\right)-1}=\sqrt{|z_{j}|^{2}|z_{j-1}|^{2}-1}$
(6.1)

Since $\operatorname{Im}\left(z_{j-1}\overline{z_{j}}\right)=1$, it follows
that $\operatorname{Re}\left(z_{j-1}\overline{z_{j}}\right)=\pm m_{j}$. But we
saw above that the $\arg z_{j}$ form a decreasing sequence in $(0,\pi/2]$, so
$0<\arg(z_{j-1}\overline{z_{j}})<\pi/2$. Hence each $z_{j-1}\overline{z_{j}}$
has positive real part, and we obtain

$z_{j-1}\overline{z_{j}}=m_{j}+i.$

We now observe that the desired equations (1.4) and (1.5) contain the products

$\displaystyle\prod_{j=1}^{n-1}\left(m_{j}-i\right)$
$\displaystyle=\prod_{j=1}^{n-1}\overline{z_{j-1}}z_{j}=\overline{z_{0}}\,\left|z_{1}\cdots
z_{n-2}\right|^{2}\,z_{n-1},$
$\displaystyle\prod_{j=1}^{n-1}\left(m_{j}+i\right)$
$\displaystyle=\prod_{j=1}^{n-1}z_{j-1}\overline{z_{j}}=z_{0}\,\left|z_{1}\cdots
z_{n-2}\right|^{2}\,\overline{z_{n-1}}.$

In the case of even $n$ then we find

$\displaystyle\frac{i}{2}\left(\prod_{j=1}^{n-1}\left(m_{j}-i\right)-\prod_{j=1}^{n-1}\left(m_{j}+i\right)\right)$
$\displaystyle=\frac{i}{2}\left|z_{1}\cdots
z_{n-2}\right|^{2}\left(\overline{z_{0}}z_{n-1}-z_{0}\overline{z_{n-1}}\right)$
$\displaystyle=\left|z_{1}\cdots
z_{n-2}\right|^{2}\,\operatorname{Im}\left(\overline{z_{0}}z_{n-1}\right)$
$\displaystyle=\prod_{j=1}^{\frac{n-2}{2}}\left(m_{2j}^{2}+1\right),$

where in the second line we used the fact that
$(\alpha-\overline{\alpha})=2i\,\operatorname{Im}(\alpha)$ for any
$\alpha\in\mathbb{C}$, and in the third line we used
$\operatorname{Im}\left(z_{0}\overline{z_{n-1}}\right)=-1$ and
$m_{2j}^{2}+1=|z_{2j-1}|^{2}\,|z_{2j}|^{2}$ from (6.1).

The odd case is similar, with an extra factor of
$m_{0}^{2}=\kappa_{0}+1=|z_{0}|^{2}$ appearing. This proves the result with
$\kappa_{\infty}$ set to $1$. Upon reversing the original dilation, we replace
each $\kappa_{j}$ with $\frac{\kappa_{j}}{\kappa_{\infty}}$, completing the
proof.

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