apollonian circle packings: dynamics and number theorygauss.math.yale.edu/~ho2/doc/oh_icwm.pdf138...
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Apollonian circle packings: Dynamics andNumber theory
Hee Oh
Yale University
ICWM, 2014
Apollonius of Perga
I Lived from about 262 BC to about 190 BC.I Known as “The Great Geometer”.I His famous book on Conics introduced the terms parabola,
ellipse and hyperbola.
Apollonius’ theorem
Theorem (Apollonius of Perga)
Given 3 mutually tangent circles, there exist exactly two circlestangent to all three.
Proof of Apollonius’ theorem
We give a modern proof of this ancient theorem using Mobiustransformations: For a,b, c,d ∈ C, ad − bc = 1,(
a bc d
)(z) =
az + bcz + d
, z ∈ C ∪ {∞}.
A Mobius transformation maps circles (including lines) tocircles, preserving angles between them.
In particular, it maps tangent circles to tangent circles.
Proof of Apollonius’ theorem
We give a modern proof of this ancient theorem using Mobiustransformations: For a,b, c,d ∈ C, ad − bc = 1,(
a bc d
)(z) =
az + bcz + d
, z ∈ C ∪ {∞}.
A Mobius transformation maps circles (including lines) tocircles, preserving angles between them.
In particular, it maps tangent circles to tangent circles.
Proof of Apollonius’ theorem
We give a modern proof of this ancient theorem using Mobiustransformations: For a,b, c,d ∈ C, ad − bc = 1,(
a bc d
)(z) =
az + bcz + d
, z ∈ C ∪ {∞}.
A Mobius transformation maps circles (including lines) tocircles, preserving angles between them.
In particular, it maps tangent circles to tangent circles.
Proof of Apollonius’ theorem
4 mutually tangent circles
Four possible configurations
Construction of Apollonian circle packings
Beginning with 4 mutually tangent circles, we can keep addingnewer circles tangent to three of the previous circles, providedby the Apollonius theorem. Continuing this process indefinitely,we arrive at an infinite circle packing called an
Apollonian circle packing .
We’ll show the first few generations of this process:
Construction of Apollonian circle packings
Beginning with 4 mutually tangent circles, we can keep addingnewer circles tangent to three of the previous circles, providedby the Apollonius theorem. Continuing this process indefinitely,we arrive at an infinite circle packing called an
Apollonian circle packing .
We’ll show the first few generations of this process:
Initial stage
2
23
Here each circle C is labelled with its curvature:
curv(C) =1
radius(C).
The curvature of the outermost circle is −1 (oriented to havedisjoint interiors).
First generation
2
23
Second generation
2
2315
6
6
3
Third generation
2
2315
38
38
35
623 14
11
623 14
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3 15
6
6
Example of bounded Apollonian circle packing
The outermost circle has curvature −1.
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Example of bounded Apollonian circle packing
The outermost circle has curvature −10.
Example of unbounded Apollonian circle packing
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41There are also other unbounded Apollonian packingscontaining either only one line or no line at all. Since circles willget enormously large, it is hard to draw them.
Example of unbounded Apollonian circle packing
1 141224406084
201201
129316
268385129
316
268385
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432
249465
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345
561
217580
616
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465
249
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616
433
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609
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169276
504537
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697
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156345
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412505
97280808
745
553
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681 489
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705
609
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9 1652108
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148396 417
297
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433
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313
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201412592
505
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705
169504276
537
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697
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681 489
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745
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577481
201412592
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705
609
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697
169504
276
537
97256768
489 681
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553
745
193576
600
321
92872136
220
409
433
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585
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417
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217
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144 81
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129
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100 121 2584
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153
201
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14481
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7373
3388
649733
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9 1652
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41There are also other unbounded Apollonian packingscontaining either only one line or no line at all. Since circles willget enormously large, it is hard to draw them.
Circle-counting question
For a bounded Apollonian packing P, there are only finitelymany circles of radius bigger than a given number.
For each T > 0, we set
NP(T ) := #{C ∈ P : curv(C) < T} <∞.
Clearly, NP(T )→∞ as T →∞.
Question
I Is there an asymptotic of NP(T ) as T →∞?I If so, can we compute?
Circle-counting question
For a bounded Apollonian packing P, there are only finitelymany circles of radius bigger than a given number.
For each T > 0, we set
NP(T ) := #{C ∈ P : curv(C) < T} <∞.
Clearly, NP(T )→∞ as T →∞.
Question
I Is there an asymptotic of NP(T ) as T →∞?I If so, can we compute?
Apollonian circle packing
2
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The study of this question involves notions related to metricproperties of the underlying fractal set called residual set.
Residual set
Definition (Residual set of P)
Res(P) := ∪C∈PC.
Equivalently, the residual set of P is the fractal set which is leftin the plane after removing all the open disks enclosed bycircles in P.
2
2315
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110287762
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362
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374
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738
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662347
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338
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102263798
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338
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134
278
275
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302443
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294
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170447330
507
179471522
354
59111
338350
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162423
486
315
174455
506 347 1447122231
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36398167
290
299
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378
354
2687182
218
251
42143
62
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75231
150
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22790279
234
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194
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294
267
110215314
299
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242254
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54167
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131
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3095266
194251
74231
138
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179471354
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302443
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59111
350338
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506 347162
423
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315144798167
290
299
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330
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131255
354
378
2687218
182251
42143
62
131
75231
186
150
35111290
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194
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27
Residual dimension
The Hausdorff dimension of the residual set of P is called theResidual dimension of P, which we denote by α.
Usually the dimension is an integer and defined for a smoothmanifold. But the Hausdorff dimension can be defined for anyset; and it does not have to be an integer.
In fact, the definition of a fractal is a set with non-integralHausdorff dimension.
Residual dimension
The Hausdorff dimension of the residual set of P is called theResidual dimension of P, which we denote by α.
Usually the dimension is an integer and defined for a smoothmanifold. But the Hausdorff dimension can be defined for anyset; and it does not have to be an integer.
In fact, the definition of a fractal is a set with non-integralHausdorff dimension.
Hausdorff dim. (Hausdorff and Caratheodory (1914))
DefinitionLet s ≥ 0. F ⊂ Rn. The s-dim. Hausdorff meas. of F is def. by:
Hs(F ) := limε→0
(inf{∑
d(Bi)s : F ⊂ ∪iBi ,d(Bi) < ε}
)where d(Bi) is the diameter of Bi .
It can be shown that as s increases, the s-dim Haus measureof F will be∞ up to a certain value and then jumps to 0.
DefinitionThe Hausdorff dim of F is this critical value of s:
dimH(F ) = sup{s : Hs(F ) =∞} = inf{s : Hs(F ) = 0}.
Hausdorff dim. (Hausdorff and Caratheodory (1914))
DefinitionLet s ≥ 0. F ⊂ Rn. The s-dim. Hausdorff meas. of F is def. by:
Hs(F ) := limε→0
(inf{∑
d(Bi)s : F ⊂ ∪iBi ,d(Bi) < ε}
)where d(Bi) is the diameter of Bi .
It can be shown that as s increases, the s-dim Haus measureof F will be∞ up to a certain value and then jumps to 0.
DefinitionThe Hausdorff dim of F is this critical value of s:
dimH(F ) = sup{s : Hs(F ) =∞} = inf{s : Hs(F ) = 0}.
α = dimH(Res(P)): Residual dim
We observeI 1 ≤ α ≤ 2
I α is independent of P: any two Apollonian packings areequivalent to each other by a Mobius transformation.
I The precise value of α is unknown, but approximately,α = 1.30568(8) (McMullen 1998)
In particular, Res(P) is much bigger than a c’ble union ofcircles of P, but not too big in the sense that its Leb. area (=2dimensional Haus. measure) is zero.
α = dimH(Res(P)): Residual dim
We observeI 1 ≤ α ≤ 2
I α is independent of P: any two Apollonian packings areequivalent to each other by a Mobius transformation.
I The precise value of α is unknown, but approximately,α = 1.30568(8) (McMullen 1998)
In particular, Res(P) is much bigger than a c’ble union ofcircles of P, but not too big in the sense that its Leb. area (=2dimensional Haus. measure) is zero.
Confirming Wilker’s prediction based on computer experiments,Boyd showed: (NP(T ) := #{C ∈ P : curv(C) < T})
Theorem (Boyd 1982)
limT→∞
log NP(T )
log T= α.
Boyd asked whether NP(T ) ∼ cTα as T →∞, and wrote thathis numerical experiments suggest this may be false andperhaps
NP(T ) ∼ c · Tα(log T )β
might be more appropriate.
Confirming Wilker’s prediction based on computer experiments,Boyd showed: (NP(T ) := #{C ∈ P : curv(C) < T})
Theorem (Boyd 1982)
limT→∞
log NP(T )
log T= α.
Boyd asked whether NP(T ) ∼ cTα as T →∞, and wrote thathis numerical experiments suggest this may be false andperhaps
NP(T ) ∼ c · Tα(log T )β
might be more appropriate.
Theorem (Kontorovich-O. 2009)
For a bounded Apollonian packing P, there exists a constantcP > 0 such that
NP(T ) ∼ cP · Tα
where α = 1.30568(8) is the residual dimension of P.
Theorem (Lee-O. 2012)
There exists η > 0 such that for any bounded Apollonianpacking P,
NP(T ) = cP · Tα + O(Tα−η)
where α = 1.30568(8) is the residual dimension of P.
Theorem (Kontorovich-O. 2009)
For a bounded Apollonian packing P, there exists a constantcP > 0 such that
NP(T ) ∼ cP · Tα
where α = 1.30568(8) is the residual dimension of P.
Theorem (Lee-O. 2012)
There exists η > 0 such that for any bounded Apollonianpacking P,
NP(T ) = cP · Tα + O(Tα−η)
where α = 1.30568(8) is the residual dimension of P.
Counting inside Triangle
For unbounded Apollonian packing P, NP(T ) =∞.
Consider a curvilinear triangle R whose sides are given bythree mutually tangent circles in any Apollonian packing (eitherbounded or unbounded):
SetNR(T ) := #{C ∈ R : curv(C) < T} <∞.
Counting inside Triangle
For unbounded Apollonian packing P, NP(T ) =∞.
Consider a curvilinear triangle R whose sides are given bythree mutually tangent circles in any Apollonian packing (eitherbounded or unbounded):
SetNR(T ) := #{C ∈ R : curv(C) < T} <∞.
Theorem (O.-Shah 10)
For a curvilinear triangle R of any Apollonian packing P,
NR(T ) ∼ cR · Tα.
Distribution of circles in Apollonian packing
QuestionCan we describe the asymptotic distribution of circles in P ofcurvature at most T as T →∞?
For a bounded Borel subset E , set
NT (P,E) := #{C ∈ P : curv(C) < T ,C ∩ E 6= ∅}.
As we vary E ⊂ C, how does NT (P,E) depend on E?
Distribution of circles in Apollonian packing
QuestionCan we describe the asymptotic distribution of circles in P ofcurvature at most T as T →∞?
For a bounded Borel subset E , set
NT (P,E) := #{C ∈ P : curv(C) < T ,C ∩ E 6= ∅}.
As we vary E ⊂ C, how does NT (P,E) depend on E?
Distribution of circles
QuestionDoes there exist a measure ωP on C such that for any bddBorel E ⊂ C,
limT→∞
NT (P,E)
Tα= ωP(E)?
Note that all the circles in P lie on the residual set of P.
Hence any measure describing the asymptotic distribution ofcircles of P must be supported on the residual set of P.
What measure could that be?
Distribution of circles
QuestionDoes there exist a measure ωP on C such that for any bddBorel E ⊂ C,
limT→∞
NT (P,E)
Tα= ωP(E)?
Note that all the circles in P lie on the residual set of P.
Hence any measure describing the asymptotic distribution ofcircles of P must be supported on the residual set of P.
What measure could that be?
Distribution of circles
QuestionDoes there exist a measure ωP on C such that for any bddBorel E ⊂ C,
limT→∞
NT (P,E)
Tα= ωP(E)?
Note that all the circles in P lie on the residual set of P.
Hence any measure describing the asymptotic distribution ofcircles of P must be supported on the residual set of P.
What measure could that be?
Distribution of circles
QuestionDoes there exist a measure ωP on C such that for any bddBorel E ⊂ C,
limT→∞
NT (P,E)
Tα= ωP(E)?
Note that all the circles in P lie on the residual set of P.
Hence any measure describing the asymptotic distribution ofcircles of P must be supported on the residual set of P.
What measure could that be?
We show that the α-dim. Hausd. measure Hα on Res(P) doesthe job.
Theorem (O.-Shah, 10)
For any bdd. Borel E ⊂ C with smooth bdry,
NT (P,E) ∼ cA · Hα(E ∩ Res(P)) · Tα
where 0 < cA <∞ is an absolute constant independent of P.
We show that the α-dim. Hausd. measure Hα on Res(P) doesthe job.
Theorem (O.-Shah, 10)
For any bdd. Borel E ⊂ C with smooth bdry,
NT (P,E) ∼ cA · Hα(E ∩ Res(P)) · Tα
where 0 < cA <∞ is an absolute constant independent of P.
Distribution of circles
Thm says that circles in an Apollonian packing are uniformlydistributed w.r.t the α-dim. Hausdorff meas. on its residual set:
NT (P,E1)
NT (P,E2)∼ H
α(E1 ∩ Res(P))
Hα(E2 ∩ Res(P)).
Integral Apollonian circle packings
We call an Apollonian packing P integral if every circle in Phas integral curvature.
Does there exist any integral P?
The answer is positive thanks to the following beautiful thm ofDescartes:
Integral Apollonian circle packings
We call an Apollonian packing P integral if every circle in Phas integral curvature.
Does there exist any integral P?
The answer is positive thanks to the following beautiful thm ofDescartes:
Descartes circle theorem, 1643
Theorem (in a letter to Princess Elisabeth of Bohemia)
A quadruple (a,b, c,d) is the curvatures of four mutuallytangent circles if and only if it satisfies the quadratic equation:
2(a2 + b2 + c2 + d2) = (a + b + c + d)2.
2
2315
3871114
167
374
371
222455
554659
219447
546
654
110287762
546
867
218647
362
683
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531
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554659
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374
371
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546
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867
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323927642
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738
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662347
3563198
407
590
491
198407
590
491
99143326
326
102263498
798
699
302863
602
803
203599
338
638
102263798
498
699
302863
602
803
203599
638
338
623
5087
134
278
275
150383
302443
147375
294
438
62 119362
194
371
170447330
507
179471522
354
59111
338350
179
162423
486
315
174455
506 347 1447122231
330
36398167
290
299
131255
378
354
2687182
218
251
42143
62
131
75231
150
186
35111314290
22790279
234
171
66215
194
107
1139
98183
294
267
110215314
299
83143
242254
1863134
182155
54167
110
131
2795
3886
3095266
194251
74231
138
195
59191
170
98
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371
170447330
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179471354
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302443
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423
486
315144798167
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299
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330
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354
378
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42143
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210318
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114
219
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218
323
107279
318210
35 63198
198
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203
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203
623
5087
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14762119
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162174
14 47122
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66
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98110
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E.g: 2((−1)2 + 22 + 22 + 32) = 36 = (−1 + 2 + 2 + 3)2
E.g. 2(22 + 32 + 62 + 232) = 1156 = (2 + 3 + 6 + 23)2
Given three mutually tangent circles of curvatures a,b, c, if wedenote by d and d ′ for the curvatures of the two circles tangentto all three, then
2(a2 + b2 + c2 + d2) = (a + b + c + d)2
and2(a2 + b2 + c2 + (d ′)2) = (a + b + c + d ′)2.
By subtracting one from the other, we obtain
d + d ′ = 2(a + b + c).
So, if a,b, c,d are integers, so is d ′.
Given three mutually tangent circles of curvatures a,b, c, if wedenote by d and d ′ for the curvatures of the two circles tangentto all three, then
2(a2 + b2 + c2 + d2) = (a + b + c + d)2
and2(a2 + b2 + c2 + (d ′)2) = (a + b + c + d ′)2.
By subtracting one from the other, we obtain
d + d ′ = 2(a + b + c).
So, if a,b, c,d are integers, so is d ′.
Theorem (Soddy 1936)
If the initial 4 circles in an Apollonian packing P have integralcurvatures, P is integral.
Therefore, for any integral solution of2(a2 + b2 + c2 + d2) = (a + b + c + d)2, ∃ an integralApollonian packing!
Theorem (Soddy 1936)
If the initial 4 circles in an Apollonian packing P have integralcurvatures, P is integral.
Therefore, for any integral solution of2(a2 + b2 + c2 + d2) = (a + b + c + d)2, ∃ an integralApollonian packing!
Integral Apollonian circle packings
Any integral Apollonian packing is either bounded or liesbetween two parallel lines:
2
2315
3871114
167
374
371
222455
554659
219447
546
654
110287762
546
867
218647
362
683
323927
858
642
107279738
846
531
210623
662347318911
842635
3871222455
554659
114167
374
371
219447 654
546
110287546
762
867
218647
362
683
323927642
858
107279846
738
531
318911
842635
210623
662347
3563198
407
590
491
198407
590
491
99143326
326
102263498
798
699
302863
602
803
203599
338
638
102263798
498
699
302863
602
803
203599
638
338
623
5087
134
278
275
150383
302443
147375
294
438
62 119362
194
371
170447330
507
179471522
354
59111
338350
179
162423
486
315
174455
506 347 1447122231
330
36398167
290
299
131255
378
354
2687182
218
251
42143
62
131
75231
150
186
35111314290
22790279
234
171
66215
194
107
1139
98183
294
267
110215314
299
83143
242254
1863134
182155
54167
110
131
2795
3886
3095266
194251
74231
138
195
59191
170
98
623 62 119
194362
371
170447330
507
179471354
522
5087 278
134
275
150383
302443
147375 438
294
59111
350338
179
174455
506 347162
423
486
315144798167
290
299
122231
330
363
131255
354
378
2687218
182251
42143
62
131
75231
186
150
35111290
314
227
66215
194
107
90279
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171
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98183
294
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218
323
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318210
35 63198
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Diophantine questions
For a given integral Apollonian packing P, it is natural to inquireabout its the Diophantine properties such as
Question
I Are there infinitely many circles with prime curvatures?I Which integers appear as curvatures?
Assume that P is primitive, i.e.,g. c.dC∈P(curv(C)) = 1.
Definition
1. A circle is prime if its curvature is a prime number.2. A pair of tangent prime circles is a tangent prime.
Diophantine questions
For a given integral Apollonian packing P, it is natural to inquireabout its the Diophantine properties such as
Question
I Are there infinitely many circles with prime curvatures?I Which integers appear as curvatures?
Assume that P is primitive, i.e.,g. c.dC∈P(curv(C)) = 1.
Definition
1. A circle is prime if its curvature is a prime number.2. A pair of tangent prime circles is a tangent prime.
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prime circles: 2,3, 11, 23,... Tangent prime circles: (2,3), (2,11),(3, 23), ...
Infinitude of prime circles
Theorem (Sarnak 07)
In any primitive integral P, there are infinitely many primecircles as well as tangent prime circles.
SetΠT (P) := #{prime C ∈ P : curv(C) < T}
and
Π(2)T (P) := #{tangent primes C1,C2 ∈ P : curv(Ci) < T}.
Infinitude of prime circles
Theorem (Sarnak 07)
In any primitive integral P, there are infinitely many primecircles as well as tangent prime circles.
SetΠT (P) := #{prime C ∈ P : curv(C) < T}
and
Π(2)T (P) := #{tangent primes C1,C2 ∈ P : curv(Ci) < T}.
Analogue of Prime number theorem?
Using the sieve method based on heuristics on the randomnessof the Mobius function, Fuchs and Sanden formulated aconjecture analogous to the prime number theorem:
Conjecture (Fuchs-Sanden)
ΠT (P) ∼ c1NT (P)
log T; Π
(2)T (P) ∼ c2
NT (P)
(log T )2
where c1 and c2 can be given explicitly.
Using the breakthrough work of Bourgain, Gamburd, Sarnak onexpanders together with Selberg’s upper bound sieve, weobtain upper bounds of true order of magnitude:
Theorem (Kontorovich-O. 09)
1. ΠT (P)� NT (P)log T
2. Π(2)T (P)� NT (P)
(log T )2 .
The lower bounds are open and seem very challenging.
Using the breakthrough work of Bourgain, Gamburd, Sarnak onexpanders together with Selberg’s upper bound sieve, weobtain upper bounds of true order of magnitude:
Theorem (Kontorovich-O. 09)
1. ΠT (P)� NT (P)log T
2. Π(2)T (P)� NT (P)
(log T )2 .
The lower bounds are open and seem very challenging.
QuestionFor a primitive integral P, how many integers appear ascurvatures of circles in P?
I.e., how big is #{curv (C) ≤ T ,C ∈ P} compared to T ?
Our counting result for circles says
#{curv (C) ≤ T counted with multiplicity : C ∈ P} ∼ c·T 1.305...
So we may hope that a positive density (=proportion) ofintegers arise as curvatures, conjectured by Graham, Lagarias,Mallows, Wilkes, Yan (Positive density conjecture):
QuestionFor a primitive integral P, how many integers appear ascurvatures of circles in P?
I.e., how big is #{curv (C) ≤ T ,C ∈ P} compared to T ?
Our counting result for circles says
#{curv (C) ≤ T counted with multiplicity : C ∈ P} ∼ c·T 1.305...
So we may hope that a positive density (=proportion) ofintegers arise as curvatures, conjectured by Graham, Lagarias,Mallows, Wilkes, Yan (Positive density conjecture):
QuestionFor a primitive integral P, how many integers appear ascurvatures of circles in P?
I.e., how big is #{curv (C) ≤ T ,C ∈ P} compared to T ?
Our counting result for circles says
#{curv (C) ≤ T counted with multiplicity : C ∈ P} ∼ c·T 1.305...
So we may hope that a positive density (=proportion) ofintegers arise as curvatures, conjectured by Graham, Lagarias,Mallows, Wilkes, Yan (Positive density conjecture):
(Strong) Positive density conjecture
P: primitive integral Apollonian packing
Theorem (Bourgain-Fuchs 10)
#{curv (C) < T : C ∈ P} � T .
Theorem (Bourgain-Kontorovich 12)
#{curv (C) < T : C ∈ P} ∼ κ(P)
24· T
where κ(P) > 0 is the number of residue classes mod 24 ofcurvatures of P.
• There are congruence restriction: modulo 24, not all residueclasses appear.
(Strong) Positive density conjecture
P: primitive integral Apollonian packing
Theorem (Bourgain-Fuchs 10)
#{curv (C) < T : C ∈ P} � T .
Theorem (Bourgain-Kontorovich 12)
#{curv (C) < T : C ∈ P} ∼ κ(P)
24· T
where κ(P) > 0 is the number of residue classes mod 24 ofcurvatures of P.
• There are congruence restriction: modulo 24, not all residueclasses appear.
Improving Sarnak’s result on the infinitude of prime circles,Bourgain showed that a positive fraction of prime numbersappear as curvatures in P.
Theorem (Bourgain, 2011)
#{prime curv (C) ≤ T : C ∈ P} � Tlog T
.
Hidden symmetries
QuestionHow are we able to count circles in an Apollonian packing?
We exploit the fact that
an Apollonian packing has lots of hidden symmetries.
Explaining these hidden symmetries will lead us to explain therelevance of the packing with a Kleinian group, called theApollonian group.
Hidden symmetries
QuestionHow are we able to count circles in an Apollonian packing?
We exploit the fact that
an Apollonian packing has lots of hidden symmetries.
Explaining these hidden symmetries will lead us to explain therelevance of the packing with a Kleinian group, called theApollonian group.
Symmetry group of P
Fixing 4 mutually tangent (black) circles in P, we obtain fourdual (red) circles, each passing through 3 tangent points.
Inverting w.r.t a dual circle fixes the three circles that it meetsperpendicularly and interchanges the two circles which aretangent to the three circles; indeed, it preserves P.
Apollonian group
The Apollonian group A associated to P is generated by 4inversions w.r.t those dual circles:
A = 〈τ1, τ2, τ3, τ4〉 < Mob(C)
where Mob(C) = PSL2(C)± the gp of Mobius transf. of C.
The Apollonian group A is a Kleinian group (= discretesubgroup of PSL±2 (C)) and satisfies
P = ∪4i=1A(Ci),
i.e., inverting the initial four (black) circles in P w.r.t the (red)dual circles generate the whole packing P.
3 dim. hyp. geometry
The upper-half space model for hyp. 3 space H3:
H3 = {(x1, x2, y) : y > 0} with ds =
√dx2
1 + dx22 + dy2
y
and ∂∞(H3) = C.
Via the Poincare extension thm,
PSL2(C)± = Isom(H3).
Note that PSL2(C)± acts on C by linear fractionaltransformations and an inversion w.r.t a circle C in Ccorresponds to the inversion w.r.t the vertical hemisphere in H3
above C.
The Apollonian gp A (now considered as a discrete subgp ofIsom(H3)) has a fund. domain in H3, given by the exterior ofthe hemispheres above the dual circles to P:
In particular, A\H3 is an infinite vol. hyperbolic 3-mfld and hasfinitely many sided fund. domain.
Orthogonal translates of geodesic surface
Counting circles of curvature at most T is same as countingvertical hemispheres of height at least 1/T .
Noting that vertical hemispheres in H3 are totally geodesicsubspaces, we relate the circle-counting problem with theequidistribution of translates of a closed totally geodesicsurface in A\H3.
Orthogonal translates of geodesic surface
Counting circles of curvature at most T is same as countingvertical hemispheres of height at least 1/T .
Noting that vertical hemispheres in H3 are totally geodesicsubspaces, we relate the circle-counting problem with theequidistribution of translates of a closed totally geodesicsurface in A\H3.
For a tot. geo. surface S of T1(A\H3), what is the asymptoticdist. of its orthogonal translates gt (S) as t →∞?
Difficulties lie in the fact that the Apollonian mfld is of infinitevolume, as the dynamics of flows in inf. volume hyp. mflds arevery little understood. (if it were of fint vol, this type of questionis well-understood due to Margulis, Duke-Rudnick-Sarnak andEskin-McMullen...)
We show that this distribution in T1(A\H3) is described by asingular measure, called the Burger-Roblin measure, whoseconditional measures on horizontal planes turn out to be equalto the α-dim’l Haus. measures in this case, and this is why wehave the α-dim’l Haus. measure in our counting thm.
Difficulties lie in the fact that the Apollonian mfld is of infinitevolume, as the dynamics of flows in inf. volume hyp. mflds arevery little understood. (if it were of fint vol, this type of questionis well-understood due to Margulis, Duke-Rudnick-Sarnak andEskin-McMullen...)
We show that this distribution in T1(A\H3) is described by asingular measure, called the Burger-Roblin measure, whoseconditional measures on horizontal planes turn out to be equalto the α-dim’l Haus. measures in this case, and this is why wehave the α-dim’l Haus. measure in our counting thm.
Main ingredients of our proofs includeI the Lax-Phillips spectral theory for the Laplacian on A\H3;
I Ergodic properties of flows on T1(A\H3) based on thePatterson-Sullivan theory and the work of Burger-Roblin.
Main ingredients of our proofs includeI the Lax-Phillips spectral theory for the Laplacian on A\H3;
I Ergodic properties of flows on T1(A\H3) based on thePatterson-Sullivan theory and the work of Burger-Roblin.
More circle packings
This viewpoint via the study of Kleinian groups allows us to dealwith more general circle packings, provided they are invariantunder a finitely generated Kleinian group Γ.
Here are some other pictures of circle packings for which wecan count circles of bounded curvature:
Ex. of Sierpinski curve (McMullen)
Here the symmetry group isπ1(cpt. hyp. 3-mfd with tot. geod. bdry).
The next pictures are reproduced from the book ”Indra’s pearls”by Mumford, Series and Wright (Cambridge Univ. Press 2002).
For these circle packings, or more generally for any circlepackings which is invariant under a (geometrically finite)Kleinian group, we have the following:
Theorem (O.-Shah)
Let δ := dimH(Res(P)). For any bdd. Borel E ⊂ C with smoothboundary,
NT (P,E) ∼ c · Hδ(E ∩ Res(P)) · T δ
for some absolute constant c > 0.
