The Parameter Space of
Apollonian Circle Packings

Summer Haag

University of Colorado Boulder

Aix Marseille Univerité Geometry Seminar

Introduction

We want to study Apollonian circle packings

  • what does it mean for two circle packings to be close to each other?
  • what type of packings can we generate?
  • there is a map from \(\PP^1(\C)\) to packings
  • we will find an alternative map that is equivalent

Apollonian Circle Packings

6-11 packing iteration 1
6-11 packing dotted

Apollonian Circle Packings

6-11 iteration 2

Apollonian Circle Packings

6-11 iteration 3

Apollonian Circle Packings

6-11 packing 4

Apollonian Circle Packings

6-11 curvatures

Apollonian Circle Packings

2-3 curvatures
\([-1,\,2,\,2,\,3]\)

Types of Packings

There are bounded packings, the strip packing, half-plane packings, and full-plane packings

types of packings
From Jan Holly

Where do the packings live?

strip
  • Looking at packings over \(\hat{\C}=\C\cup\{\infty\}\)
  • This means we have circles and straight lines
  • Two parallel straight lines are tangent at infinity and have curvature \(0\)
  • We consider packings equivalent under rotations, shifts, reflections, and scaling

Equivalence of Packings

6-11 packing 4 6-11 packing 4 6-11 packing 4 2-3 curvatures

How do we tell these packings apart?

  • Starting with 3 circles fixes a packing
  • choose four starting circles and call that a quadruple
  • from here we can generate a packing

Theorem of Elizabeth and Descartes

quad
quad swap

The two red circles are called Soddy Circles

Theorem of Elizabeth and Descartes

Curvatures \(a\), \(b\), \(c\), \(d\) of four mutually tangent circles (a Descartes Quadruple) satisfy

\[(a + b + c + d)^2 = 2(a^2 + b^2 + c^2 + d^2).\]

Theorem of Elizabeth and Descartes

Given \(a,b,c\), there exist \(d,d'\) such that

\[d+d'=2(a+b+c)\]

So if \(a,b,c,d\in\Z\) then curvatures \(\in\Z\)

Descartes Quadruples

Two different Desacrtes Quadruples can generate the same packing

for example \((3,-2,6,7)\) and \( (3,34,6,7) \)

So we want to know which quadruples generate the same packing and how to move around the packing

Apollonian Group

Acting on a quadruple of circles

circle swap
\(S_4\) (4 swaps)

\(\mathcal{A}=\la S_1, S_2, S_3, S_4 : S_i^2=1\ra < O_Q(\Z)\)   for example

\[S_1=\begin{bmatrix} -1 & 2 & 2 & 2 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\]

Apollonian Group example

Let's see the group act on \((-2,3,6,7)\)

\((-2,3,6,7)\)
\(S_1(-2,3,6,7)=(34,3,6,7)\)
\(S_2(-2,3,6,7)=(-2,19,6,7)\)
\(S_3(-2,3,6,7)=(-2,3,10,7)\)
\(S_4(-2,3,6,7)=(-2,3,6,7)\)

Apollonian Group

Acting on a quadruple of circles

circle swap
\(S_4\) (4 swaps)

\(\mathcal{A}=\la S_1, S_2, S_3, S_4 : S_i^2=1\ra < O_Q(\Z)\)

\(\mathcal{A}\) is a thin group so

  1. it has infinite index in \(O_Q(\Z)\)
  2. Zariski dense in \(O_Q(\Z)\)

Space of ACPs

  • We want to build a space who's points correspond to ACPs with a choice of circle
  • Let's start by fixing a circle in a fixed packing and looking at those quadruples
mother circle BQF

Why fix a circle?

  • For integral packings, if we fix an integer what packings contain it?
  • We want to know which Descartes quadruples can have this integer
  • So if we are in a packing and we fix a circle let's consider the curvatures tangent to the mother circle
mother circle BQF

\(n\)-quadruple

  • quadruples are ordered so we consider the group generated by \(\mathcal{A}\) and permutations
  • an \(n\)-quadruple is a Descartes quadruple \((n,a,b,c)\)
  • two \(n\)-quadruples are equivalent if they are in the same orbit of \(\la (23),(24),S_4\ra\)
  • Equivalent \(n\)-quadruples: \( (3,6,7,-2)\), \( (3,10,42,7) \), \( (3,10,7,42)\), \( (3,42,10,7)\), \( (3,7,42,10)\), \( (3,6,7,34)\), \( (3,15,10,-2)\)
[3,6,7,-2] [3,10,42,7] [3,42,7,6]
[3,42,10,7] [3,42,7,10] [3,6,7,34] [3,15,10,-2]

Descartes quadruples to BQF quadruples

Theorem: (Sarnack, Graham-Lagarias-Mallows-Wilks-Yan) We have the bijection \[ \begin{Bmatrix} \text{curvatures tangent to a fixed}\\ \text{circle with curvature }n \end{Bmatrix} \;\longrightarrow\; \begin{Bmatrix} Q_n(x,y)-n:\;\gcd(x,y)=1 \end{Bmatrix} \]

This is true regardless of integrality

Descartes quadruples to BQFs

Explicitly we have the map:

\[ \phi:\begin{Bmatrix} n\text{-quadruples up to}\\ \text{scaling by }\R^+ \end{Bmatrix} \longrightarrow \begin{Bmatrix} \text{BQF quadruples up to}\\ \text{scaling by }\R^+ \end{Bmatrix} \] \[\phi\bigl((n,a,b,c)\bigr)=[n,\;n+a,\;n+a+b-c,\;n+b]\]

\[\phi^{-1}([n,A,B,C])=[n,\;A-n,\;C-n,\;A+C-B-n]\]

Theorem: (Rickards) \(\phi\bigl(\la(23),(24),S_4\ra\cdot(n,a,b,c)\bigr) =\PGL_2(\Z)\cdot\phi\bigl((n,a,b,c)\bigr)\)

Essentially, we can decide if two \(n\)-quadruples are equivalent if their BQFs are

This is nice because in the integral case we can explicitly say which packings contain a fixed integer

Binary Quadratic forms

Def: A binary quadratic form (BQF) is \[\begin{align} Q(x,y) &= Ax^2 + Bxy + Cy^2 \\ &= \begin{pmatrix}x&y\end{pmatrix} \begin{pmatrix}A&B/2\\B/2&C\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix} \\ &= [A,B,C] \end{align}\] with \(x,y\in\Z\) coprime and discriminant \(\Delta=B^2-4AC\)
Def: \(\PGL_2(\Z)\) acts on BQFs by \[\begin{align} MQ(x,y) &= Q(ax+by,\,cx+dy) \\ &= \begin{pmatrix}x&y\end{pmatrix} M^T\begin{pmatrix}A&B/2\\B/2&C\end{pmatrix}M \begin{pmatrix}x\\y\end{pmatrix} \end{align}\] where \(M=\begin{pmatrix}a&b\\c&d\end{pmatrix}\)

BQF actions

Def: Two BQFs \(Q,Q'\) are equivalent if there is some \(M\in\PGL_2(\Z)\) such that \(MQ(x,y)=Q'(x,y)\)
Def: A BQF quadruple is of the form \([n,A,B,C]\) where \(\Delta=-4n^2\)

BQFs

We know more about equivalences of BQFs than of \(n\)-quadruples

So let's use things we know about BQFs to learn about ACPs and quadruples

BQF quadruples to \(\PP^1(\C)\)

We have a bijection! (Rickards)

\[ \rho:\begin{Bmatrix} \text{BQF quadruples up to}\\ \text{scaling by }\R^+ \end{Bmatrix} \longrightarrow \PP^1(\C) \] \[ \rho([n,A,B,C])=\begin{cases} \dfrac{B}{2A}+\dfrac{n}{A}\,i & \text{if }A\neq 0\\[6pt] \infty & \text{if }A=0 \end{cases} \]

\[ \rho^{-1}(x+iy)=\begin{cases} [0,0,0,1] & x+iy=\infty\\ [y,1,-2x,x^2+y^2] & \text{else} \end{cases} \]

This means we can send an \(n\)-quadruple to a point in \(\PP^1(\C)\) which is the map we want!

\(\PGL_2(\Z)\) with \(\rho\)

Def: \(M\in\PGL_2(\Z)\) acts on \(z\in\PP^1(\C)\) by

\[ M\cdot z:=\begin{cases} Mz & \text{if }\det(M)=1\\ M\bar z & \text{if }\det(M)=-1 \end{cases} \]

Theorem: (Rickards) Let \(Q\) be a BQF quadruple and \(M\in\PGL_2(\Z)\). Then \(\rho(MQ)=M^{-1}\rho(Q)\).

What do we have now?

  1. For an \(n\)-quadruple, \((n,a,b,c) \), equivalence class we have a BQF quadruple, \([n,A,B,C]\), equivalence class
  2. the orbit of \(\la (23),(24),S_4\ra\) on \((n,a,b,c) \) is equivalent to the orbit of \(\PGL_2(\Z)\) on \([n,A,B,C]\)
  3. For a BQF quadruple, \([n,A,B,C]\), equivalence class we have a point \(z_0\) in \(\PP^1(\C)\)
  4. the orbit of \(\PGL_2(\Z)\) on \([n,A,B,C]\) is equivalent to the orbit of \(\PGL_2(\Z)\) on \(z_0\)
  5. So we have a way of going from \(n\)-quadruples to points in \(\PP^1(\C)\) by composing these maps

Descartes quadruples and \(\PP^1(\C)\)

Last bijection!

\[ \rho\circ\phi:\begin{Bmatrix} n\text{-quadruples up to}\\ \text{scaling by }\R^+ \end{Bmatrix} \longrightarrow \PP^1(\C) \]

\[ \rho\circ\phi\bigl((n,a,b,c)\bigr) =\frac{n+a+b-c}{2(n+a)}+\frac{n}{n+a}\,i \]

\[ \phi^{-1}\circ\rho^{-1}(x+iy) =\bigl(y,\;1-y,\;x^2+y^2-y,\;1+x+y^2+2x-y\bigr) \]

How can we separate this out?

  1. We now have \(\PGL_2(\Z)\)-equivalent points mapping to equivalent \(n\)-quadruples
  2. Let's try to draw the sets that seperate out the types of packings, i.e. bounded, strip, half-plane, and full-plane
parameter space depth coloring
James Rickards

Parameter Space

parameter space depth coloring
\((0,0,2,2)\) — James Rickards
Theorem: (Holly, Kocik, Rickards) The space of circle packings is a circle packing! Say \(\phi^{-1}\circ\rho^{-1}(z)=q\). Then \(q\) generates a
  1. bounded packing iff \(z\) is in the interior of a circle
  2. half-plane packing iff \(z\) is on the boundary but not a tangent point
  3. strip packing iff \(z\) is a tangency point of two circles
  4. full-plane packing iff \(z\) is not in or on any circle

What does this look like?

Let's make something that moves around the parameter space and gives you the packing

Parameter Space Shader

Move cursor over the left panel (parameter space) to see the corresponding packing on the right

New Theorem

Let's take a different perspective on this space

The map from the parameter space to a circle packing is explicit in a surprising way: it's an inversion!

Geometrically, fix a packing in \( \hat\C\) and invert in different points to get different packings and all packings arise from this!

Let's prove this...

Möbius transformations and Circles

We know

\[ \begin{pmatrix}\alpha&\beta\\\gamma&\delta\end{pmatrix}\hat{\R} \;\longrightarrow\; \begin{cases} \text{curvature} = 2\,\Im(\bar\gamma\delta) = i(\gamma\bar\delta - \bar\gamma\delta)\\[4pt] \text{curvature-center} = i(\alpha\bar\delta - \bar\gamma\beta) \end{cases} \]

Take the strip packing \((0,0,2,2)\) — we can represent these 4 circles by Möbius transformations

parameter space depth coloring
James Rickards

Parameter Space and circle inversion

  • Take a point in the parameter space \(x+iy=z_0\)
  • Invert our quadruple \((0,0,2,2)\) through this point using the transformation \(\begin{pmatrix}0&1\\1&-z_0\end{pmatrix}\)
  • Computing this new quadruple gives \[2\bigl(y,\;1-y,\;x^2+y^2-y,\;1+x+y^2+2x-y\bigr)\]

Parameter Space and Inversion Shader

Move cursor over the left panel — middle shows resulting packing, right shows inversion of the strip

New Map

This complicated bijection

\[ \PP^1(\C) \;\longrightarrow\; \begin{Bmatrix} \text{BQF quadruples up to}\\ \text{scaling by }\R^+ \end{Bmatrix} \;\longrightarrow\; \begin{Bmatrix} n\text{-quadruples up to}\\ \text{scaling by }\R^+ \end{Bmatrix} \]

is actually just inversion \(\displaystyle z\mapsto\frac{1}{z-z_0}\) applied to the strip packing!

strip

Thank you!

colorful strip packing