Two Connections between Combinatorial and Differential Geometry - - PowerPoint PPT Presentation

Two Connections between Combinatorial and Differential Geometry

John M. Sullivan

Institut f¨ ur Mathematik, Technische Universit¨ at Berlin Berlin Mathematical School DFG Research Group Polyhedral Surfaces DFG Research Center MATHEON

Discrete Differential Geometry 2007 July 16–20 Berlin

The 5,7–triangulation problem

Triangulations of the torus T2

Average vertex degree 6 Exceptional vertices have d = 6 Regular triangulations have d ≡ 6

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 2 / 29

The 5,7–triangulation problem

Edge flips give new triangulations

Flip changes four vertex degrees Can produce 5272–triangulations (four exceptional vertices) Quotients of some such tori are 5,7–triangulations of Klein bottle

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 3 / 29

The 5,7–triangulation problem

Two-vertex torus triangulations

regular 4,8 3,9 2,10 1,11

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 4 / 29

The 5,7–triangulation problem

Refinement or subdivision schemes

√ 3–fold 2–fold √ 7–fold 3–fold

Exceptional vertices preserved

Old vertex degrees fixed New vertices regular Lots more 4,8–, 3,9–, 2,10– and 1,11–triangulations

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 5 / 29

The 5,7–triangulation problem

Is there a 5,7–triangulation of the torus?

(any number of regular vertices allowed)

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 6 / 29

The 5,7–triangulation problem

Is there a 5,7–triangulation of the torus?

(any number of regular vertices allowed)

No!

We prove this combinatorial statement geometrically using curvature and holonomy

• r complex function theory

Joint work with Ivan Izmestiev (TU Berlin) Rob Kusner (UMass/Amherst) G¨ unter Rote (FU Berlin) Boris Springborn (TU Berlin)

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 6 / 29

The 5,7–triangulation problem

Combinatorics and topology

Triangulation of any surface

Double-counting edges gives:

˜ dV = 2E = 3F χ ˜ dV = χ 2E = χ 3F = 1 ˜ d − 1 2 + 1 3 6χ =

• d

(6 − d)vd Notation

˜ d := average vertex degree vd := number of vertices of degree d

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 7 / 29

The 5,7–triangulation problem

Eberhard’s theorem

Triangulation of S2

12 =

• d

(6 − d)vd

Theorem (Eberhard, 1891)

Given any (vd) satisfying this condition, there is a corresponding triangulation of S2, after perhaps modifying v6.

Examples

512–triangulation exists for v6 = 1 34–triangulation exists for v6 = 2 and even

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 8 / 29

The 5,7–triangulation problem

Torus triangulations

The condition 0 = (6 − d)vd is simply ˜ d = 6. Analog of Eberhard’s Theorem would say ∃ 5,7–triangulation for some v6 Instead, we show there are none

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 9 / 29

The 5,7–triangulation problem Euclidean cone metrics

Euclidean cone metrics

Definition

Triangulation on M induces equilateral metric: each face an equilateral euclidean triangle. Exceptional vertices are cone points

Definition

Euclidean cone metric on M is locally euclidean away from discrete set of cone points. Cone of angle ω > 0 has curvature κ := 2π − ω. Vertex of degree d has curvature (6 − d)π/3

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 10 / 29

The 5,7–triangulation problem Euclidean cone metrics

Regular triangulations on the torus

Theorem (cf. Alt73, Neg83, Tho91, DU05, BK06)

A triangulation of T2 with no exceptional vertices is a quotient of the regular triangulation T0 of the plane, or equivalently a finite cover of the 1-vertex triangulation.

Proof.

Equilateral metric is flat torus R2/Λ. The triangulation lifts to the cover, giving T0. Thus Λ ⊂ Λ0, the triangular lattice.

Corollary

Any degree-regular triangulation has vertex-transitive symmetry.

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 11 / 29

The 5,7–triangulation problem Euclidean cone metrics

Holonomy of a cone metric

Definition

Mo := M cone points h : π1(Mo) → SO2 H := h(π1)

Lemma

For a triangulation, H is a subgroup of C6 := 2π/6.

Proof.

As we parallel transport a vector, look at the angle it makes with each edge of the triangulation.

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 12 / 29

The 5,7–triangulation problem Holonomy theorem

Holonomy theorem

Theorem

A torus with two cone points p± of curvature κ = ±2π/n has holonomy strictly bigger than Cn.

Corollary

There is no 5,7–triangulation of the torus.

Proof.

Lemma says H contained in C6; theorem says H strictly bigger.

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 13 / 29

The 5,7–triangulation problem Holonomy theorem

Proof of Holonomy theorem.

Shortest nontrivial geodesic γ avoids p+. If it hits p− and splits excess angle 2π/n there, consider holonomy of a pertubation. Otherwise, γ avoids p− or makes one angle π there, so slide it to foliate a euclidean

• cylinder. Complementary digon has two positive angles, so geodesic

from p− to p− within the cylinder does split the excess 2π/n. π π p− p+ γ γ′

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 14 / 29

The 5,7–triangulation problem Holonomy theorem

Quadrangulations and hexangulations

Theorem

The torus T2 has no 3,5–quadrangulation no bipartite 2,4–hexangulation 2,6–quad 3252–quad 2,4–hex 1,5–hex bip 1,5–hex

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 15 / 29

The 5,7–triangulation problem Riemann surfaces

Generalizing the holonomy theorem

Question

Given n > 0 and a euclidean cone metric on T2 whose curvatures are multiples of 2π/n, when is its holonomy H contained in Cn?

Curvature as divisor

Cone metric induces Riemann surface structure Cone point pi has curvature mi2π/n Divisor D = mipi has degree 0 Cone metric gives developing map from universal cover of Mo to C.

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 16 / 29

The 5,7–triangulation problem Riemann surfaces

Main theorem

Theorem

H < Cn ⇐ ⇒ D principal

Proof.

Consider the nth power of the derivative of the developing map. This is well-defined on M iff H < Cn. If so, its divisor is D. Conversely, if D is principal, corresponding meromorphic function is this nth power.

Note

The case n = 2 is the classical correspondance between meromorphic quadratic differentials and “singular flat structrues”.

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 17 / 29

The 5,7–triangulation problem Three dimensions

Combinatorics − → geometry in three dimensions

Triangulated 3-manifold: make each tetrahedron regular euclidean Edge valence ≤ 5 ⇐ ⇒ curvature bounded below by 0

Enumeration (with Frank Lutz, TU Berlin)

Enumerate simplicial 3-manifolds with edge valence ≤ 5 Exactly 4761 three-spheres plus 26 finite quotients [Matveev, Shevchishin]: Can smooth to get positive curvature

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 18 / 29

k-point metrics CMC Surfaces

CMC Surfaces

Definition

A coplanar k-unduloid is an Alexandrov-embedded CMC (H ≡ 1) surface M with k ends and genus 0, contained in a slab in R3. Note: each end asymptotic to unduloid

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 19 / 29

k-point metrics CMC Surfaces

Classifying map

M has mirror symmetry Upper half M+ is a topological disk with k boundary curves in mirror plane Conjugate cousin M+ is minimal in S3 with k boundary Hopf great circles Hopf projection gives spherical metric on open disk with k completion boundary points

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 20 / 29

k-point metrics CMC Surfaces

Classification

Theorem (with Karsten Große-Brauckmann and Rob Kusner)

Classifying map is homeomorphism from moduli space of coplanar k-unduloids to space Dk of spherical k-point metrics, which is a connected (2k − 3)–manifold.

New work (also with Nick Korevaar)

Dk ∼ = R2k−3

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 21 / 29

k-point metrics 2- and 3-point metrics

2-point metrics

Universal cover of S2 {p, q} Bi-infinite chain of slit spheres D2 ∼ = (0, π]

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 22 / 29

k-point metrics 2- and 3-point metrics

Triunduloids classified by spherical triples

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 23 / 29

k-point metrics 2- and 3-point metrics

3-point metrics

Spherical triangle with three chains of slit spheres D3 ∼ = T3 ∼ = B3 ∼ = R3 C3 := D3/M¨

• b ∼

= {∗}

General case

We show Ck ∼ = Ck−3, so Dk ∼ = R2k−3

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 24 / 29

k-point metrics Medial axis

Medial axis

Maximal balls within ¯ D Touch ≥ 2 boundary points Medial axis is tree (retract of D) k ends are infinite rays nodes are maximal circles with ≥ 3 boundary points M¨

• bius-invariant notion

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 25 / 29

k-point metrics Associahedron

Metric trees

Theorem

The space of planar trees with k ends (labeled in order) length ≥ 0 on each midsegment is cone over dual associahedron, homeomorphic to Rk−3.

k = 4 gives R

a, b ≥ 0, at most one positive two rays join to form R ∋ a − b a b

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 26 / 29

k-point metrics Associahedron

k = 5

a b a b

c d e

a b c d e (a, b) Combinatorial trees ← → triangulations of pentagon Five quadrants glue together to form plane

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 27 / 29

k-point metrics Associahedron

k = 6

14 generic trees 14 triangulations of hexagon 14 octants fit together to form R3

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 28 / 29

k-point metrics Associahedron

Complexification

Ck ∼ = Ck−3

Medial axis is a tree Midsegment is circles through pq ending with ones touching r, s Length is angle (in R≥0, not mod 2π) between end circles But also have a “twist” Together these are log of cross-ratio(p, q, r, s) All these notions are M¨

• bius-invariant

John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 29 / 29