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Two Connections betweenCombinatorial and Differential Geometry
John M. Sullivan
Institut fur Mathematik, Technische Universitat BerlinBerlin Mathematical School
DFG Research Group Polyhedral SurfacesDFG Research Center MATHEON
Discrete Differential Geometry2007 July 16–20 Berlin
The 5,7–triangulation problem
Triangulations of the torus T2
Average vertex degree 6
Exceptional vertices have d 6= 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 are5,7–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 preservedOld 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)
No!We prove this combinatorial statement geometrically
using curvature and holonomy
or complex function theory
Joint work with
Ivan Izmestiev (TU Berlin)
Rob Kusner (UMass/Amherst)
Gunter 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
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
or complex function theory
Joint work with
Ivan Izmestiev (TU Berlin)
Rob Kusner (UMass/Amherst)
Gunter 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 surfaceDouble-counting edges gives:
dV = 2E = 3F
χ
dV=
χ
2E=
χ
3F=
1
d− 1
2+
13
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 correspondingtriangulation of S2, after perhaps modifying v6.
Examples
512–triangulation exists for v6 6= 1
34–triangulation exists for v6 6= 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
DefinitionTriangulation on M induces equilateral metric:each face an equilateral euclidean triangle.
Exceptional vertices are cone points
DefinitionEuclidean cone metric on M is locally euclideanaway 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 theregular triangulation T0 of the plane, or equivalently a finite cover ofthe 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.
CorollaryAny 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
DefinitionMo := M r cone points
h : π1(Mo)→ SO2
H := h(π1)
LemmaFor a triangulation, H is a subgroup of C6 := 〈2π/6〉.
Proof.As we parallel transport a vector, look at the angle it makes with eachedge 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
TheoremA torus with two cone points p± of curvature κ = ±2π/nhas holonomy strictly bigger than Cn.
CorollaryThere 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 excessangle 2π/n there, consider holonomy of a pertubation. Otherwise, γ
avoids p− or makes one angle π there, so slide it to foliate a euclideancylinder. Complementary digon has two positive angles, so geodesicfrom p− to p− within the cylinder does split the excess 2π/n.
π πp−
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 aremultiples of 2π/n, when is its holonomy H contained in Cn?
Curvature as divisorCone 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 iswell-defined on M iff H < Cn. If so, its divisor is D. Conversely, if D isprincipal, corresponding meromorphic function is this nth power.
NoteThe case n = 2 is the classical correspondance between meromorphicquadratic 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 4761three-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
DefinitionA 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 endasymptotic 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 diskwith 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 diskwith 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 coplanark-unduloids to space Dk of spherical k-point metrics, which is aconnected (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 r {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 trianglewith three chains of slit spheres
D3∼= T3
∼= B3 ∼= R3
C3 := D3/Mob∼= {∗}
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 circleswith ≥ 3 boundary points
Mobius-invariant notion
John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 25 / 29
k-point metrics Associahedron
Metric trees
TheoremThe 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 Ra, b≥ 0, at most one positive
two rays join to form R 3 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
bc
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 pqending 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 Mobius-invariant
John M. Sullivan (TU Berlin) Combinatorial and Differential Geometry 2007 July 16 29 / 29