Rigidity of the hexagonal - PowerPoint Presentation

Slide1

Rigidity of the hexagonal triangulation of the plane and its applications

Feng Luo, Rutgers University

Joint with Jian Sun (Tsinghua Univ. China) and Tianqi Wu (Courant)

Workshop on Geometric Structures on

3-manifolds

IAS, Oct 5-9, 2015Slide2

Regular hexagonal circle packing

Regular hexagonal triangulation

R

egular hexagonal square tiling

circle packing of

C

:

Conjecture

: All hexagonal square

tilings

of C are regular.

Regular patterns in the plane C

Thurston’s conjecture (1985):All hexagonal circle packings of C are regular.

Theorem (Rodin-Sullivan 1987): Thurston’s conjecture holds.

Hexagonal = every square (circle) intersects exactly 6 othersSlide3

Koebe(1936)-Andreev(

1971)-Thurston

(1978) theorem

Any topological triangulation of

a

disk can be

realized

by a circle packing of the unit disk.Slide4

Thurston’s discrete Riemann mapping conjecture, Rodin-Sullivan’s theorem

Koebe

-Andreev-Thurston theorem

Proof:

1.

f

n

converges

2. limit is conformal

r

igidity of the hexagonal circle packing

ƎK, all

f

n

are K-

quasiconformal

f

n

f

n

→Riemann

mappingSlide5

PL metrics on surface S

= flat cone metrics on S

Metric gluing of Euclidean triangles by isometries along edges.

Metric is determined by edge lengths

l

: E={all edges} →

R

Curvature

k: V={all vertices}→

R

,

k(v)= 2π-sum of inner angles

at v.

A PL metric is

Delaunay

:

a+b

≤

π

at each edge e.

Fact: Every flat cone metric on S has a Delaunay triangulation T with V(T)=cone points. Slide6

Discrete conformality of polyhedral metrics on (S,V)

Def. Polyhedral metrics d, d’ on closed marked surface (S,V) are

discrete conformal, if Ǝ sequences d1=d, d

2, …, dk=d’ and T1, …,

T

k

on (S,V)

s.t.

,

T

i

is Delaunay in di,If T

i ≠ Ti+1, then (S, di

) ≅ (S,di+1) by an isometry homotopic to id on (S,V),If T

i=Ti+1, ∃ u:V -> R

s.t., l

di+1 (vv’)= u(v)

u(v’) ldi(

vv

’) ,

vv

’ edges in

T

i

.Slide7

Thm (Gu

-L-Sun-Wu) ∀ PL metric d on (S,V) and ∀

K*:V -> (-∞, 2π), s.t., ∑ k*(v) =2πχ(S),

Ǝ a PL metric d*, unique up to scaling on (S,V), s.t.,

d* is

discrete conformal

to d,

The

discrete curvature

of d* is

K

*.

Furthermore, d* can be found by minimizing a convex function on RV

.Slide8

Conformal and discrete conformal

g =

Riemannian metric on a compact S u: S →R>0

u4

g

conformal metric

t

hen

|d

u

4g(

p,q) –u(p)

u(q)dg(p,q)| ≤Cd

g(p,q)3. Discrete

conformality:

L(pq) =u(p)u(q) l

(pq). Slide9

.

discrete

uniformization

thm

Riemann mapping sending



to the triangle.

f

n

Conj.

f

n

→ Riemann mappingSlide10

Thm (Rodin-Sullivan). If T is a geometric hexagonal

triangulation of a simply connected domain in C s.t.,

∃ r: V -> R>0

satisfying length(

vv

’)=r(v)+r(v’)

for all edges

vv

’,

then r=constant.

Thm(L-Sun-Wu). If T is a geometric hexagonal triangulation of a simply connected domain in C

s.t.,it is Delaunay (a+b ≤ π at all edges),

∃ g: V -> R>0 satisfying

length(vv’)=g(v)g(v’) for all edges vv

’, then g = constant.Thm (L-Sun-Wu).

Given Jordan domain



and A,B,C

, Ǝ domains 

n

approximating it,

s.t.

,



n

triangulated by equilateral triangles,

(b) associated discrete

uniformization maps fn

→ Riemann mapping for (;A,B,C).Slide11

A new proof of Rodin-Sullivan’s thm

Liouville

type thm: A bounded discrete harmonic function u on

V is a constant.Proof: for δ



V

, show g(x)=u(x+

δ

)-u(x) is constant.

Let V =

Z

+Z

(η), η =eπi/3

:

Thm(Rodin-Sullivan) If T is a geometric hexagonal triangulation of a

simply connected domain in C s.t., Ǝ u:V→ R satisfying length(vv

’)=eu(v)+eu(v’), then u=const.

Ratio Lemma (R-S)

. Ǝ C

>0

s.t.

,

for all pairs of adjacent radii

r(v)/r(v’)

≤

e

C, i.e., |u(v)-u(v’)| ≤ C

.

Corollary. If d(v,v’) ≤ N in V, then, |u(v’)| ≤ |u(v)|+ NC.Slide12

Max Principle (Thurston): If r0 ≥ R

0 and r

i ≤ Ri

, i=1,…,6, and

K

r

(v

0

)=K

R

(v

0

),then

ri=R

i for all i.

Proof (Thurston)

Fix r

2, r3 and let r1 ↗,

then a

1

↘ and a

2

↗, a

3

↗.

smaller

larger

Corollary.

The ratio function r/R of two flat CP metrics has no max point unless r/R=constant

.Slide13

A new proof of Rodin-Sullivan’s thm, cont.

Let V=

Z+Z(η

), η =eπ

i

/3

.

Thm

(Rodin-Sullivan)

If T is a geometric hexagonal triangulation of

a simply connected domain in

C

s.t.

, Ǝ u:V -> R satisfying

length(vv’)=eu(v)

+eu(v’), then u=const.

Suppose u: V→ R is not a const. Then Ǝ

δ { 1, eπi/3}, s.t.,

λ

=sup{ u(v+

δ

)-u(v) } ≠ 0 and <∞.

Take

v

n

 V,

s.t.

,

u(vn+δ

)-u(vn) > λ-1/n

u(v+δ)-u(v) ≤ λ, for all v  V |u(v)-u(v’)| ≤

C, v~v’,

ratio lemmaDefine, u

n(v)= u(v+vn)-u(

vn): un

(0)=0, un(δ)-u

n(0) > λ -1/n, un(v+

δ)-un

(v)≤ λ, |un(v)| ≤

C

d(v,0).

Combinatorial distance from v to 0.

0

v

nSlide14

Taking a subsequence, limn un =u

# , u#

 RV , s.t.,

(1) the CP metric eu

#

is still flat

(may not be complete).

(2) 

u

#

(v) =u#(v+δ)-u

#(v) achieves maximum point at v=0. By the

max principle, u#(v+δ)-

u#(v) ≡ λ.

Repeat it for u# (instead of u), taking limit to get u##: (

δ, δ’ generate V) u

##

(v+

δ

’)-u

##

(v)= constant

u

##

(

v+

δ)-

u

##(v) ≡

λ. So u## is linear on V, not a constant.

Recall u

n(v)= u(v+v

n)-u(vn

)  RV:

un(0

)=0, un(

δ)-un(0)

>λ-1/n

, un(v+

δ

)-u

n

(v

)≤

λ

, |

u

n

(v

)| ≤ C

d(v,0).

F: V → R is

linear

if it is a restriction of a linear map on

R

2

.Slide15

Lemma (Doyle) If f: V → R non-constant linear, the CP metric ef

is flat and the developing map sends to two disjoint circles to two circles in C

with overlapping interiors.Doyle spiral circle packing (

raduii=eu

, u linear, implies flat

)

CP metric

e

u

##

does not have injective developing map.

CP metrics

e

u# and hence eu do not have injective developing maps, a contradiction.

Developing mapNeed: a ratio lemma (for taking limit),

a maximum principle,

a spiral situation (log(radius) linear) producing self intersections.

All of them hold in the new setting.Slide16

A generalized triangle has 3 edge positive edge lengths l1,l2

,l3 and 3 angles a1,a2

,a3A

generealized PL (GPL) metric on (S, T):

l

: E(T) →R

>0

,

s.t.

each triangle is generalized.

The discrete curvature

K of l is defined as before: K(v)=2π-sum of angles at v.

It is flat at v if K(v)=0.

A GPL metric is Delaunay: a+b ≤ π at each edge e.

If l: E → R>0, and w: V(T)→ R, then w

*l: E → R>0 is

w*l(vv

’)=

e

w

(v)+w(v’)

l

(

vv

’)

Vertex scaling

Lemma.

If

π

1

(S)=1 and (S,T,

l) is flat, then there is a developing map φ: S →C.

s.t., li+lj

≥lkSlide17

T = standard topological hexagonal triangulation.

L

0

: E=E(T) →

R

sending e to 1.

Thm

(L-Sun-Wu).

If (

C

, T, w

*

L

0

) is a flat, Delaunay, PL metric with injective developing map, then w=constant.

Ratio Lemma.

If w

*

L

0

is a generalized PL metric, flat at v, on B

1

(v), then x/y ≤6.

Proof.

Triangle inequality and

c

ombinatorial cross ratio.

w: V →

R

,

w

*

L0(vv

’)=ew(v)+w(v’), vv

’  ESlide18

Maximum principle. Let (B1(v0), L) and (B

1(v0), L

) be two Delaunay GPL, flat at v0 , s.t., L= u*

L and u(v0)= max{ u(v1), …., u(v

6

)}. Then u=constant.

Angle lemma.

Fix v

 V and B

2

(v). If w

*

L

0 is a Delaunay GPL, flat at interior vertices in B2(v),

and has injective developing map on B2(v), then there is a triangle t in B2(v) s.t.

all angles in t ≥ 0.0001.Slide19

Spiral Lemma (Gu-Sun-Wu (2011)

). Suppose w: V → R

is non-constant linear s.t. w*L

0 is GPL. Then (1) w

*

L

0

is flat.

(2) If Ǝ a non-degenerate triangle in w

*

L

0

, then Ǝ two non-degenerated triangles in T whose images under the develop map intersect in their interiors.

Spiral triangulationsSlide20

A sketch of proof of the maximum principle

Lemma (L, 2004)

Fix a triangle  of lengths l1, l

2, l3,

varying u

1

,u

2

,u

3

,

angles

ai

=ai (u

1,u2,u3

).

,

.

Rate of change of cone angle at v

0

= -[cot(

a

i

)+cot(b

i

)]+negative <0

a

i

+b

i

≤

π

due to Delaunay

Non-positive

thenSlide21

Conjecture. Suppose (C, V, T,

l) and (C, V, T’,

l’) are two geometric triangulations of the plane s.t.,b

oth are Delaunay,T, T’ combinatorially

isomorphic,

w

*

l = l’

.

Then T and T’ differ by a scaling.Slide22

Thank you.