A discrete uniformization theorem for polyhedral surfaces

A discrete uniformization theorem for polyhedral surfaces

Discrete conformal geometry of polyhedral surfaces Feng Luo Rutgers University, USA Discretization in Geometry and Dynamics Dllnsee-Schorfheide, Templin, Germany Joint work with D. Gu (Stony Brook), J. Sun (Tsinghua Univ.), and T. Wu (Courant) Oct. 8, 2018 A simple problem topologically isomorphic T1 and T2 are two hexagonal Delaunay triangulations of the plane. T1

infinite triangulations T2 Problem. If (C, T1, l1) and (C, T2 , l2 ) are related by a vertex scaling, then T1 and T2 differ by a complex linear map. This is a Cauchy rigidity type problem. : {vertices} vertices} R>0 s.t., l1(uv) = (u) (v) l2(uv) i.e., the same length cross ratio at corresponding pairs of edges. Recall Riemann surfaces and differential geometry Uniformization Thm (Poincare-Koebe 1907) Riemannian metric g on a surface S, : S R>0 s.t., (S, g) is a complete metric of curvature -1, 0, 1.

Key step: if S is simply connected, the (S, g) is conformal to S2, C, or D. Koebe (1908): If the genus of S is zero, then (S, g) is conformal to a domain (connected open set) in the S2. Koebe conjecture (1908): Every domain in C is conformal to circle domain C-X s.t. connected components of X are round disks and points (i.e., circle type closed set). Riemann mapping circle type closed set X circle domain True: if it has finitely many (Koebe) or countably many (He-Schramm) boundary components. Weyl Problem (1915): A metric g of positive Gaussian curvature on S2 is isometric the boundary of a compact convex set in R3. Levy, Alexandrov, Pogorelov and Nirenberg Question: What about the Weyl problem in H3 for non-compact convex surfaces?

CH(Y) convex hull in H3 Known: 1. convex surfaces CH(Y) in H3 have curvature at least -1 and genus zero. 2. Thurston: for any closed set Y in H3, CH(Y) is complete and curvature -1 (hyperbolic). Y Conj. (L-Sun-Wu): complete hyperbolic metric on a (non-cpt) genus zero surface is isometric to CH(X) where X is a circle type closed set in H3. this conjecture, discrete uniformization, and Koebe conjecture, progresses we made. Hyperbolic 3-space H3 and convex hull H3=C X R>0 Geodesics: Planes and half spaces:

X Convex hull in H3 of a subset X in C: removing maximal half-spaces missing X Isometries of C are naturally isometries of H3. Ideal triangles: convex hull in H3 of {vertices} a,b,c} in . Polyhedral surface PL metric d on (S,V) is a flat cone metric, cone points in V. Isometric gluing of E2 triangles along edges: (S, T, l ).

triangulation K(v)<0 Curvature K=Kd: V R, K(v)= 2-sum of angles at v = 2- cone angle at v A triangulated PL metric (S, T, l) is Delaunay: a+b at each edge e. at each edge e. a+b at each edge e. Delaunay triangulation always exists on (S, V, d). K(v)>0 discrete conformality: PL surface (S,V, d) and hyperbolic metric d* on S-V Given a PL metric d on (S,V), produce a Delaunay triangulation T of (S,V,d) t T is associated an ideal hyperbolic triangle t*

t s t If t, s T glued by isometry f along e, then t* and s* are glued by the same f* along e*. t* s* a hyperbolic metric d* on S-V. Def. (G-L-S-W). Two PL metrics d1, d2 on (S,V) are discrete conformal (d.c.) iff d1* are d2* are isometric by an isometry homotopic to id on S-V. Motivated by the important work of Bobenko-Pinkall-Springborn. Equivalent to the previous def. using vertex scaling and Delaunay condition.

An important example: (C, V, dst) (V discrete infinite set, infinite triangulations) dst* = CH(V) H3 Delaunay triangulations dst =standard flat metric on (C, V) Boundary of hyperbolic convex hull of V circumcircles are maximum disks missing V a+b at each edge e.

Delaunay triangulation = projection of boundary triangulation of the hyperbolic convex hull = Thm 1. (Gu-L-Sun-Wu). PL metric d on a closed (S,V) is discrete conformal unique, up to scaling, a PL metric d* of constant curvature 2 ( ) . | | Question 1. Do the dis. unif. metrics d* converge to the smooth Poincare metric? U. Bcking, the torus case by Gu-L-Wu Question 2. Non-compact infinite triangulated polyhedral surfaces? Especially, simply connected ones? Why dis. conf. geom. of infinite triangulated PL surfaces?

Needed to prove convergence of discrete conformal maps to conformal maps. . PL approximations Convergence of fn Riemann mapping F? dis cre te u quasi-conformality implies fn G. nif fn or m iza tio

nt hm Is G conformal? Riemann mapping F Problem. A Delaunay hexagonal triangulation of C, dis.conf. to the equilateral hexagonal triangulation, is unique up to complex linear maps. ANS: Yes, L-Sun-Wu, Dai-Ge-Ma Discrete uniformization for non-cpt simply connected PL surfaces (S, V) Uniformization Thm. Riemannian metric d, (S,d) is conformal to Riemannian metric d, (S,d) is conformal to C or D. Discrete uniformization conjecture Every PL surface (S,V,d) is d.c. to a unique (C, V, dst) or (D, V, dst).

Associated hyperbolic metric Weyls problem (S-V, d*) complete hyperbolic w/ cusp ends at V isometric to a unique CH(V) in H3 CH(V (S2-D)) Geometry of convex hulls in H3 and conjectures Thurston. If Y closed in S2, then CH(Y) H3 is complete hyperbolic. Eg. simply connected domain in S2, Y=S2-, |Y|>1, Y|Y|>1, >1, CH(Y) isometric to H2. Thurstons isometry

convex hull & hyperbolic geom. Riemann mapping, conformal geom. Q. Not simply connected ? Koebe Conj. domain domain in S2 is conformal to a circle domain S2-X . Conj (L-S-W) 1. complete hyperbolic surf (, d) of genus 0 is isometric to CH(X) for a circle type closed set X. Conj.(L-S-W) 2. If X and Y are two circle type closed sets s.t. CH(X) isometric CH(Y) , then X, Y differ by a Moebius transf. Conj (L-S-W) 1. complete hyperbolic surf (, d) of genus 0 is isometric to CH(X) for a circle type closed set X. Conj.(L-S-W) 2. If X and Y are two circle type closed sets s.t. CH(X) isometric CH(Y) ,

then X, Y differ by a Moebius transformation. Known 1. Rivin: 2. Schlenker: 3. L-Tillmann: Thm (L-Wu) 2. Koebe conjecture implies Conjecture 1. Combining the work of He-Schramm on Koebe conjecture, Corollary (L-Wu) 3. domain simply connected PL surface (S,V,d) is d.c. to (C, V, dst) or (D, V, dst). Sketch of proof of Thm 2 KC implies Conj. 1. Goal: circle domain C-X with Poincare metric dX, find another circle domain C-Y

s.t., (C-X, dX) is isometric to CH(Y). Step 1. Take circle type closed sets Xn converging to X in Hausdorff metric s.t. each Xn has only finitely many connected components and components of Xn are circles. Step 2. By Schlenkers theorem, find a circle type closed set Y n s.t. isometry Fn: (C-Xn, dn) CH(Yn). Normalize Yn. Step 3. May assume that Yn converges in Hausdorff metric to a closed set Y. Easy to show (C-X, dX) is isometric to CH(Y) using Alexandrovs work. Claim: Each connected component of Y is the Hausdorff limit of connected components of Yns. Step 4 (Main) which implies the Claim. Prop. The sequence Fn is equicontinuous from (C-Xn, dE) to (CH(Yn), dE) in the Klein model where dE is Euclidean metric. If not, use Schramms transboundary extremal length and Schramms theorem for some curve families whose extremal lengths tending to zero on circle domains.

But Fn is conformal, so the corresponding curve families in CH(Yn) has extremal lengths tending to zero. However, this is not the case mainly due to: 1. The Euclidean area of any convex surface in the Poincare model of H3 is at most 16.. 2. The map g(x) = sending the Poincare model to the Klein model is 2-Lipschitz: g(x) g(y) at each edge e. 2 x-y. Thank you.

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