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, 2015

Regular hexagonal circle packing Regular hexagonal triangulation Regular 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 others

Koebe (1936) -Andreev ( 1971 ) -Thurston (1978) theorem Any topological triangulation of a disk can be realized by a circle packing of the unit disk.

Thurston’s discrete Riemann mapping conjecture, Rodin-Sullivan’s theorem Koebe-Andreev-Thurston theorem Proof: 1. f n converges 2. limit is conformal rigidity of the hexagonal circle packing ƎK, all f n are K-quasiconformal fnfn f n →Riemann mapping

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.

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 d 1 =d, d 2, …, d k =d’ and T 1, …, T k on (S,V) s.t., (a)T i is Delaunay in d i, (b)If T i ≠ T i+1, then (S, d i ) ≅ (S,d i+1 ) by an isometry homotopic to id on (S,V), (c)If T i =T i+1, ∃ u:V -> R s.t., l d i+1 (vv’)= u(v) u(v’) l d i (vv’), vv’ edges in T i.

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., (a)d* is discrete conformal to d, (b)The discrete curvature of d* is K *. Furthermore, d* can be found by minimizing a convex function on R V.

Conformal and discrete conformal g = Riemannian metric on a compact S u: S →R >0 u 4 g conformal metric then |d u 4 g (p,q) –u(p)u(q)d g (p,q)| ≤Cd g (p,q) 3. Discrete conformality: L (pq) =u(p)u(q) l (pq).

. discrete uniformization thm Riemann mapping sending  to the triangle. fnfn Conj. f n → Riemann mapping

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., 1.it is Delaunay (a+b ≤ π at all edges), 2.∃ 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., (a)  n triangulated by equilateral triangles, (b) associated discrete uniformization maps f n → Riemann mapping for (  ;A,B,C).

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 =e πi/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’)=e u(v) +e u(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.

Max Principle (Thurston): If r 0 ≥ R 0 and r i ≤ R i, i=1,…,6, and K r (v 0 )=K R (v 0 ), then r i =R i for all i. Proof (Thurston) Fix r 2, r 3 and let r 1 ↗, 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.

A new proof of Rodin-Sullivan’s thm, cont. Let V= Z+Z(η), η =e πi/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’)=e u(v) +e u(v’), then u=const. Suppose u: V→ R is not a const. Then Ǝ δ  { 1, e πi/3 }, s.t., λ=sup{ u(v+δ)-u(v) } ≠ 0 and <∞. Take v n  V, s.t., u(v n +δ)-u(v n ) > λ-1/n u(v+δ)-u(v) ≤ λ, for all v  V |u(v)-u(v’)| ≤ C, v ~ v ’, ratio lemma Define, u n (v)= u(v+v n )-u(v n ): u n (0)=0, u n (δ)-u n (0) > λ -1/n, u n (v+δ)-u n (v)≤ λ, |u n (v)| ≤ C d(v,0). Combinatorial distance from v to 0. 0 vnvn

Taking a subsequence, lim n u n =u #, u #  R V, s.t., (1) the CP metric e u # 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(v n )  R V : u n (0)=0, u n (δ)-u n (0) >λ-1/n, u n (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.

Lemma (Doyle) If f: V → R non-constant linear, the CP metric e f is flat and the developing map sends to two disjoint circles to two circles in C with overlapping interiors. Doyle spiral circle packing (raduii=e u, u linear, implies flat) CP metric e u ## does not have injective developing map. CP metrics e u # and hence e u do not have injective developing maps, a contradiction. Developing map Need: 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.

A generalized triangle has 3 edge positive edge lengths l 1,l 2,l 3 and 3 angles a 1,a 2,a 3 A 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., l i +l j ≥l k

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 combinatorial cross ratio. w: V →R, w * L 0 (vv’)=e w(v)+w(v’), vv’  E

Maximum principle. Let (B 1 (v 0 ), L) and (B 1 (v 0 ), L ) be two Delaunay GPL, flat at v 0, s.t., L= u * L and u(v 0 )= max{ u(v 1 ), …., 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 B 2 (v), and has injective developing map on B 2 (v), then there is a triangle t in B 2 (v) s.t. all angles in t ≥

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 triangulations

A sketch of proof of the maximum principle Lemma (L, 2004 ) Fix a triangle  of lengths l 1, l 2, l 3, varying u 1,u 2,u 3, angles a i =a i (u 1,u 2,u 3 ). 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 then

Conjecture. Suppose (C, V, T, l ) and (C, V, T’, l’ ) are two geometric triangulations of the plane s.t., 1.both are Delaunay, 2.T, T’ combinatorially isomorphic, 3. w * l = l’. Then T and T’ differ by a scaling.

Thank you.