Transcript Polyhedral Geometry

An Introduction to Polyhedral
Geometry
Feng Luo
Rutgers undergraduate math club
Thursday, Sept 18, 2014
New Brunswick, NJ
Polygons and polyhedra
3-D Scanned pictures
The 2 most important theorems in Euclidean geometry
Homework
Curvatures
Gauss-Bonnet theorem
Theorem. a+b+c = π.
Area =(a+b)2 =a2+b2+2ab
Area = c2+4 (ab/2)=c2+2ab
Pythagorean Theorem
distances, inner product, Hilbert spaces,….
The 3rd theorem is Ptolemy
For a quadrilateral inscribed to a circle:
It holds in spherical geometry, hyperbolic geometry, Minkowski plane and di-Sitter space, …
Homework: prove the Euclidean space version using trigonometry.
It has applications to algebra (cluster algebra), geometry (Teichmuller theory),
computational geometry (Delaunay), ….
Q. Any unsolved problems for polygons?
Triangular Billiards Conjecture. Any triangular billiards board
admits a closed trajectory.
True: for any acute angled triangle.
Best known result (R. Schwartz at Brown): true for all triangles of angles < 100 degree!
Check: http://www.math.brown.edu/~res/
Polyhedral surfaces
Metric gluing of Eucildean triangles by isometries
along edges.
Metric d: = edge lengths
Curvature K at vertex v: (angles) =
metric-curvature:
determined by the cosine law
the Euler Characteristic V-E-F
4 faces
genus = 0
E = 12
F=6
V=8
V-E+F = 2
genus = 0
E = 15
F=7
V = 10
V-E+F = 2
genus = 1
E = 24
F = 12
V = 12
V-E+F = 0
A link between geometry and topology:
Gauss Bonnet Theorem
For a polyhedral surface S,
∑v Kv = 2π (V-E+F).
The Euler characteristic of S.
Q: How to determine a convex polyhedron?
Cauchy’s rigidity thm (1813)
If two compact convex polytopes
have isometric boundaries,
then they differ by a rigid motion of E3.
Assume the same combinatorics and triangular faces, same edge lengths
Thm
Dihedral angles the same.
Then the same in 3-D.
Thm(Rivin) Any polyhedral surface is determined, up to scaling, by the
quantity F sending each edge e to the sum of the two angles facing e.
F(e) = a+b
So far, there is no elementary proof of it.
h =0: a+b;
h=1: cos(a)+cos(b);
h=-2: cot(a)+cot(b);
h=-1: cot(a/2)cot(b/2);
Thm(L). For any h, any polyhedral surface is determined, up to scaling,
by the quantity Fh sending each edge e to :
Fh(e) =
𝒂
𝐡
𝐬𝐢𝐧
𝝅/𝟐
𝒕 𝒅𝒕 +
𝒃
𝐡
𝐬𝐢𝐧
𝝅/𝟐
𝒕 𝒅𝒕.
Basic lemma. If f: U R is smooth strictly convex and
U is an open convex set U in Rn,
then ▽f: U  Rn is injective.
Proof.
Eg 1. For a E2 triangle of lengths x and angles y,
the differential 1-form w is closed due to prop. 1,
w= Σi ln(tan(yi /2)) d xi.
Thus, we can integrate w and obtain a function of x,
x
F(x) = ∫ w
This function can be shown to be convex in x.
This function F, by the construction, satisfies:
∂F(x)/ ∂xi = ln(tan(yi /2)).
Thank you