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Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work Thin Groups and Fractals Daniel Lautzenheiser and Arthur Baragar Department of Mathematical Sciences University of Nevada Las Vegas December 14, 2016 1 / 23

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Page 1: Thin Groups and Fractals - WordPress.com · Introduction: Hyperbolic GeometryRulesTranslationsApollonian PackingK3’sFurther Work WCNT 2016 Thin Groups and Fractals Daniel Lautzenheiser

Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

WCNT 2016

Thin Groups and Fractals

Daniel Lautzenheiser and Arthur Baragar

Department of Mathematical SciencesUniversity of Nevada Las Vegas

December 14, 2016

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Page 2: Thin Groups and Fractals - WordPress.com · Introduction: Hyperbolic GeometryRulesTranslationsApollonian PackingK3’sFurther Work WCNT 2016 Thin Groups and Fractals Daniel Lautzenheiser

Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

Spherical Geometry

S = {~x : ||~x || = 1} = {~x : ~x · ~x = 1} = {~x : ~x t I~x = 1}Lines: S ∩ PAngles: ~x · ~y = ||~x ||||~y || cos θWeighted dot product: I 7→ A, ( A positive-definite, symmetric)

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Page 3: Thin Groups and Fractals - WordPress.com · Introduction: Hyperbolic GeometryRulesTranslationsApollonian PackingK3’sFurther Work WCNT 2016 Thin Groups and Fractals Daniel Lautzenheiser

Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

Lorentz Product

~x ◦ ~y = ~x tJ~y , J =

1 0 00 1 00 0 −1

” ◦ ” is bi-linear and symmetric, but not an inner product.

Light Cone:

L+ = {~x : ~x ◦ ~x = 0}

Outer Hyperboloid (1 sheet) :

~x ◦ ~x = 1

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Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

Pseudosphere

Pseudosphere model:

V = {~x : ~x ◦ ~x = −1}

V+ = {~x = (x , y , z) ∈ V : z > 0}

Lines: P ∩ V+

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Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

Disk Model

Stereographic Projection:

π : V+ −→ D

D = {(x , y , 0) : x2 + y2 < 1}

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Page 6: Thin Groups and Fractals - WordPress.com · Introduction: Hyperbolic GeometryRulesTranslationsApollonian PackingK3’sFurther Work WCNT 2016 Thin Groups and Fractals Daniel Lautzenheiser

Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

Disk Model Cont.

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Page 7: Thin Groups and Fractals - WordPress.com · Introduction: Hyperbolic GeometryRulesTranslationsApollonian PackingK3’sFurther Work WCNT 2016 Thin Groups and Fractals Daniel Lautzenheiser

Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

Disk Model: Lines

~n1 = (1, 0, 0)~n2 = (−1, 1, 0)~n3 = (1, 1, 1)Intersect

planes :

~x ◦ ~n1 = 0~x ◦ ~n2 = 0~x ◦ ~n3 = 0

with V+ to get lines. Apply π to theselines.

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Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

Disk Model: Moving Points

The vectors ~n1, ~n2, and ~n3 define 3planes. The mapping

Rj (~x) = ~x − 2Proj~nj~x = ~x − 2

~x ◦ ~nj

~nj ◦ ~nj~nj

is reflection through the plane definedby ~nj . Rj is an isometry since

Rj (~x) ◦ Rj (~y) = ~x ◦ ~y

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Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

Disk Model: Lattice Points

(8, 4, 9) 7→R2 (4, 8, 9) 7→R3 (−2, 2, 3) 7→R1 (2, 2, 3) 7→R3 (0, 0, 1)

I F is the fundamental domain forthe group Γ = {R1,R2,R3}

I Γ acts transitively on the latticepoints of V+

I F is a Coxeter polyhedronI Method of descent to F

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Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

What does a picture look like after inversion in a circle?

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Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

Process: Pictures to/from Lorentz Product

Rules:

1. Choose a basis: {e1, · · · , en}2. Construct J = [ei · ej ]...

Example:

I Think of ” · ” as ” ◦ ”

I Choose e1 · e1 = e2 · e2 = −2I Since e3 is the point at infinity,

e3 · e3 = 0I e1 · e2 = ±||e1||||e2|| cos θ or...

e1 · e2 = ±||e1||||e2|| coshψI e1 · e3 = curvature (1/radius)

J =

−2 d 1d −2 11 1 0

I J induces "◦"

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Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

Fundamental Domain

{l1 · e1 = 0⇐⇒~l1 ◦ ~e1 = 0l1 · e3 = 0⇐⇒~l1 ◦ ~e1 = 0{

n1 · e1 = 0n1 · e2 = 0

{l3 · n1 = 0l3 · e3 = 0

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Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

Fundamental Domain

~l1 = (1,−1, 2 + d), ~n1 = (1, 1, 2− d)

L1 =

1 2 00 −1 00 2d + 4 1

,N1 =

1 0 2d−2

0 1 2d−2

0 0 −1

, L3 =

0 1 01 0 00 0 1

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Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

Hyperbolic Translations

L1L3 ∼

1 1 00 1 10 0 1

N1L1 ∼

d+6+4

√d+2

d−2 0 0

0 d+6+4√

d+2d−2 0

0 0 1

I Since N1L1 is an isometry,det(N1L1) = 1 = product ofeigenvalues.

I If d = 1, 3, 4 then we getfundamental units in the ring ofintegers Z[

√D] ⊆ Q[

√D], some

D ∈ N.I Example: If d = 4, then

Spec(N1L1L3) = {7± 4√

3,−1},Spec(N1(L1L3)2) ={25± 4

√39,−1}.

(Pell’s eqn. for D=3,39)

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Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

Translations cont.Let Λ = {N1(L1L3)k}∞k=1. How big is Spec(Λ)?

Values of the Discriminant (Within Eigenvalue)

k d=1 d=3 d=41 6 30 32 33 105 393 78 230 214 141 5 35 222 70 576 321 905 3277 438 1230 1118 573 1605 5799 6 2030 18310 897 2505 903

Guess (based on first 10,000 cases): Spec(Λ) has units in infinitely many D,but misses some values e.g. D = 7.

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Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

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Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

Ji = [ei ·ej ] =

−2 2 2 42 −2 2 42 2 −2 04 4 0 0

, Γ = 〈R1,R2,R3,R4,R5〉 ,T = 〈R1,R2,R3,R4〉

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Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

Connection to K3 surfaces

Given a (class of) K3 surface, we may describe a hyperbolic cross section ofthe ample cone.

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Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

I Each previous J is even (even entries on main diagonal), symmetric,and has signature (1, n − 1).

I By [Mor84] and the Hodge Index Theorem, there exists a class of K3surfaces with intersection matrix J = [ei · ej ].

I So, these fractals are ample cones of K3 surfaces.I "Apollonian K3 surfaces"

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Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

Further Work

1. Can we modify or generalize the matrix J in the hyperbolic translationscase to hit more values of D?

2. Generalized method of descent in "non-Coxeter" cases

3. Using kJ−1 as a Lorentz product, some k ∈ N.

"Dual" belt packing:

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Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

References

[1] Stange, Katherine. The Apollonian Structure of Bianchi Groups.Department of Mathematics, University of Colorado. 2015[2] Dolgachev, Igor. Orbital Counting on Algebraic Surfaces and SpherePackings. Springer International Publishing Switzerland 2016.[3] Baragar, Arthur. Automorphisms on K3 Surfaces with Small PicardNumber. Department of mathematical sciences, UNLV. 2014[4] Morrison, D. R. "On K3 Surfaces with Large Picard Number." InventionesMathematicae Invent Math 75.1 (1984): 105-21. Web.[5] Boyd, David W. A New Class of Infinite Sphere Packings. Pacific Journalof Mathematics. 1974[6] Baragar, Arthur. The Neron-Tate Pairing and Elliptic K3 Surfaces.Preprint. 2016[7] Baragar, Arthur. Lattice Points on Hyperboloids of One Sheet. New YorkJournal of Mathematics. 2013.

Special thanks to Sarah Glaser for helping me with some of the graphics

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Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

Connections to K3 surfaces

I K3 surfaces are examples of elliptic fibrations There are smooth curves(divisor classes) passing through each elliptic curve.

I The isogeny P 7→ −P sends one smooth curve to another and inducesan automorphism of the K3 surface.

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Introduction: Hyperbolic Geometry Rules Translations Apollonian Packing K3’s Further Work

Ample Cone Symmetries

Ample cone symmetries←→arithmetic on elliptic curves

I O 7→ O + P (horizontaltranslation fixing∞)

I O 7→ O + Q (diagonal translation)I P 7→ −P (−1 map through O)

Jamp =

−2 2 2 42 −2 2 42 2 −2 44 4 4 0

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