Inflation and realization of polyhedral surfaces Igor Pak, MIT - - PDF document

Inflation and realization of polyhedral surfaces

Igor Pak, MIT

Berlin July 17, 2007

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Figure 1. Real cushion pillow and the ideal pillow shape.

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Take a pillow and start inflating it until it is no longer possible. The ideal pillow P is the resulting non-inflatable surface. Here are some basic questions we have about the ideal pillow surface:

• Does P have creases along the boundary?
• Is the boundary smooth (except at the vertices)?
• Are there any crimps and crumples?
• What is the area(P)?
• What is the vol(P)?

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Figure 2. Gammel’s computer simulation of the ideal pillow shape: an intermediate step and the final shape (view from the side).

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Figure 3. Computer simulation of the inflation of a cube.

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Let S ⊂ R3 be a 2-dimensional polyhedral surface. Example: S = ∂P, where P is a convex polytope. Two surfaces S1 ≃ S2 are isometric if there exists a piecewise-linear homeomorphism ϕ : S1 → S2 which preserves geodesic distances. Definition: Bending is a continuous piecewise-linear isometric deformation of S. Definition: Inflation is a volume-increasing bending.

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Figure 4. Bending of the surface of a cube.

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Figure 5. Simulation of an elastic cube inflation.

Warning: This is not a bending!

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Figure 6. Buckling of a cube and its unfolding (scaled down). Note: Volume decreases!

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Question: What is known about bendable polyhedral surfaces? Theorem: (Cauchy-Alexandrov) Every two isometric convex surfaces are equal up to a rigid motion. Corollary: Every nontrivial bendings is non-convex. Theorem: (Alexandrov) Every intrinsically convex 2-dimensional abstract polyhedral surface homeomorphic to a sphere can be realized in R3 as a convex surface.

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Theorem: (Burago-Zalgaller, P.) The surface ∂P of a convex poly- tope P ⊂ R3 can be embedded into the interior of a ball Bε ⊂ R3 or radius ε > 0. Moreover, there exists a bending that achieves this. Proof: Very technical. The first step is the following lemma (proved by BZ in 1960):

• Every abstract 2-dimensional polyhedral surface can be

subdivided into acute triangles. Note: The volume decreases once again...

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Theorem: (Le˘ ıbin) Let F be a face in a convex polytope P ⊂ R3. Then the polyhedral surface S = ∂P F is convexly bendable. Theorem: (Alexandrov) Let F be a face in a convex polytope P ⊂ R3, and let Q ⊂ F be a polygon strictly inside of F. Then the polyhedral surface S = ∂P Q is not convexly bendable. By convexly bendable we mean that the deformations of S are regions

• n convex polyhedra.

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Figure 7. Three sides of a tetrahedron form a convexly bendable surface.

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K e f i r

Y

• g

u r t

M i l k

M

• l
• k
• The first carton one is convexly bendable, while the second is not.

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So, is it ever possible to inflate a convex polyhedral surface?

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Figure 8. Inflation of a doubly covered square.

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Bleecker’s inflation of a tetrahedron can increase the volume by about 37%. Watch a movie...

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Inflation Theorem (P.) Every 2-dimensional polyhedral surface S ⊂ R3 admits a volume-increasing bending. Note: Surface S does not have to be convex! Before: Bleecker’s Conjecture (1996). Known only for simplicial convex polyhedra in R3. Proof idea: This requires a really good understanding of the ideal pillow shapes...

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Inflation of a cube:

ε ε √ 2ε 2ε 2ε

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Figure 9. Party balloon. Observation: the balloon surface is shrinking!

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Figure 10. Mylar party balloons.

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Definition: We say that surface S1 is submetric to S2 if there exists a piecewise-linear homeomorphism ϕ : S2 → S1 which non-increases the distances. (Such maps ϕ are also called short or Lipschitz 1). Definition: Shrinking is a continuous piecewise-linear submetric de- formation. (Bending is a special case.) Main Lemma: Every 2-dimensional polyhedral surface S ⊂ R3 admits a volume-increasing shrinking. Theorem (Burago-Zalgaller) Every submetric embedding S0 ⊂ R3 of a 2-dimensional polyhedral surface S can be approximated by isometric embeddings of S. This is a PL-analogue of the classical Nash and Kuiper results.

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Proof outline (of the Inflation Theorem)

1) Construct a volume-increasing shrinking. 2) Use the BZ-theorem to construct isometric embeddings of larger volume. 3) Check that the proof of the BZ-theorem is robust enough so that the resulting isometries form an bending (at least in the beginning).

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Convex Shrinking Theorem: Every d-dimensional convex polyhe- dral surface S ⊂ Rd+1 admits a volume-increasing convex shrinking. Corollary: For every convex polytope P ⊂ Rd there exists a convex polytope Q ⊂ Rd such that vol(Q) > vol(P) and ∂Q is submetric to ∂P. In other words, one can simultaneously shrink the surface and increase the volume, while keeping the polytope convex.

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Proof idea:

v v′ e F F ′ XF XF ′ Figure 11. Subdivision of the surface S = ∂P into regions around edge e = (v, v′).

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e e1 L Figure 12. Cutting and projecting in case of an acute angle β1.

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e e1 e2 e3 e4 e5 Figure 13. Iterated cutting and projecting.

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Back to the pillows:

Figure 14. Mylar party balloon and NASA’s Ultra Long Duration Balloon.

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Mylar Balloon:

r = 4 √ 2π Γ 1

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2 ≈ 0.7627 area(K) = π2r2 ≈ 5.7422 vol(K) = 2πr2 3 ≈ 1.2185 crimping ratio = area(K) 2π ≈ 0.9139 = balloon width

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Mylar Balloon: the exact shape

(after Paulsen and Mladenov-Oprea)

K =

• x(u, v), y(u, v), z(u, v)
• , u ∈ [−A, A], v ∈ [0, 2π]
• x(u, v) = r cn(u, a) cos(v),

y(u, v) = r cn(u, a) sin(v) z(u, v) = r a

• E
• sn(u, a), a
• − 1

2 F

• sn(u, a), a
• a = 1

√ 2 and A = F(1, a) The elliptic integrals of the first and second kind: F(z, k) = z dt √ 1 − t2 √ 1 − k2t2 and E(z, k) = z √ 1 − k2t2 √ 1 − t2 dt. The Jacobi sine function sn(u, k) is defined as the inverse to F(z, k), for all k ∈ R. The Jacobi cosine function cn(u, k) is defined by sn(u, k)2 + cn(u, k)2 = 1.

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Open Problems and Conjectures

The Volume Problem. For a convex polyhedral surface S in R3, compute: η(S) := sup

S′∼S

vol(S′) , where the supremum is over all immersed polyhedral surfaces S′ iso- metric to S. Conjecture 1. For every convex polyhedral surface S ⊂ R3 there exist a unique (up to rigid motions) piecewise-smooth surface K = K(S) which is embedded into R3, submetric to S and such that vol(K) = η(S).

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Conjecture 2. The surface K = K(S) is strictly non-convex and is smooth everywhere except at the vertices. Conjecture 3. The surface K = K(S) has smaller area: area(K) < area(S). Moreover, the submetry is strict almost everywhere: |x, y|K < |x′, y′|S a.s. for x, y ∈ S , where x′ = ϕ(x), y′ = ϕ(y), and ϕ is the submetry map ϕ : S → K(S).

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Conjecture 4. Let S0 be a convex polyhedral surface in R3 and let S1 be a convex polyhedral surface submetric to S0 of greater volume: vol(S1) > vol(S0). Then there exist a volume-increasing shrinking from S0 to S1. Similarly, if vol(S1) = vol(S0), there exist a volume-constant shrinking from S0 to S1.

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For a convex piecewise-smooth surface S, define the crimping ratio cr(S): cr(S) = area

• K(S)
• area(S)

. Conjecture 5. Among all convex piecewise-smooth surfaces S, the crimping ratio minimizes on a doubly covered disc. In other words, we claim that cr(S) ≤ 0.9139 computed exactly for the doubly covered disc case (mylar balloon).