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

Inflation and realization of polyhedral surfaces Igor Pak, MIT Berlin July 17, 2007 1

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

3 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 )?

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

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

6 Let S ⊂ R 3 be a 2-dimensional polyhedral surface. Example: S = ∂P , where P is a convex polytope. Two surfaces S 1 ≃ S 2 are isometric if there exists a piecewise-linear homeomorphism ϕ : S 1 → S 2 which preserves geodesic distances. is a continuous piecewise-linear Definition: Bending isometric deformation of S . Definition: Inflation is a volume-increasing bending.

7 Figure 4. Bending of the surface of a cube.

8 Figure 5. Simulation of an elastic cube inflation. Warning: This is not a bending!

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

10 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 R 3 as a convex surface.

11 Theorem: (Burago-Zalgaller, P.) The surface ∂P of a convex poly- tope P ⊂ R 3 can be embedded into the interior of a ball B ε ⊂ R 3 or radius ε > 0 . Moreover, there exists a bending that achieves this. Very technical. The first step is the following lemma (proved Proof: by BZ in 1960): • Every abstract 2-dimensional polyhedral surface can be subdivided into acute triangles. Note: The volume decreases once again...

12 ıbin) Let F be a face in a convex polytope P ⊂ R 3 . Then Theorem: (Le˘ the polyhedral surface S = ∂P � F is convexly bendable. Theorem: (Alexandrov) Let F be a face in a convex polytope P ⊂ R 3 , 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 on convex polyhedra.

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

14 o r k i o l f o t M i l k r u g e M o Y K The first carton one is convexly bendable, while the second is not.

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

16 Figure 8. Inflation of a doubly covered square.

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

18 Inflation Theorem (P.) Every 2 -dimensional polyhedral surface S ⊂ R 3 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 R 3 . This requires a really good understanding of the ideal Proof idea: pillow shapes...

19 Inflation of a cube: 2 ε ε √ ε 2 ε 2 ε

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

21 Figure 10. Mylar party balloons.

22 Definition: We say that surface S 1 is submetric to S 2 if there exists a piecewise-linear homeomorphism ϕ : S 2 → S 1 which non-increases the distances. (Such maps ϕ are also called short or Lipschitz 1). Shrinking is a continuous piecewise-linear submetric de- Definition: formation. (Bending is a special case.) Main Lemma: Every 2 -dimensional polyhedral surface S ⊂ R 3 admits a volume-increasing shrinking. Every submetric embedding S 0 ⊂ R 3 of Theorem (Burago-Zalgaller) 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.

23 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).

24 Convex Shrinking Theorem: Every d -dimensional convex polyhe- dral surface S ⊂ R d +1 admits a volume-increasing convex shrinking. Corollary: For every convex polytope P ⊂ R d there exists a convex polytope Q ⊂ R d 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.

25 Proof idea: v X F ′ e X F F ′ F v ′ Figure 11. Subdivision of the surface S = ∂P into regions around edge e = ( v, v ′ ).

26 e e 1 L Figure 12. Cutting and projecting in case of an acute angle β 1 .

27 e e 1 e 2 e 3 e 4 e 5 Figure 13. Iterated cutting and projecting.

28 Back to the pillows: Figure 14. Mylar party balloon and NASA’s Ultra Long Duration Balloon.

29 Mylar Balloon: √ r = 4 2 π � 2 ≈ 0 . 7627 � 1 Γ 4 area( K ) = π 2 r 2 ≈ 5 . 7422 vol( K ) = 2 πr 2 ≈ 1 . 2185 3 crimping ratio = area( K ) ≈ 0 . 9139 2 π = balloon width

30 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 , 2 π ] x ( u, v ) = r cn( u, a ) cos( v ) , y ( u, v ) = r cn( u, a ) sin( v ) � − 1 �� z ( u, v ) = r � � � sn( u, a ) , a sn( u, a ) , a E 2 F a a = 1 √ and A = F (1 , a ) 2 The elliptic integrals of the first and second kind: √ � z � z dt 1 − k 2 t 2 F ( z, k ) = √ 1 − t 2 √ and E ( z, k ) = √ 1 − t 2 dt. 1 − k 2 t 2 0 0 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 .

31 Open Problems and Conjectures The Volume Problem. For a convex polyhedral surface S in R 3 , compute: vol( S ′ ) , η ( S ) := sup S ′ ∼ S where the supremum is over all immersed polyhedral surfaces S ′ iso- metric to S . Conjecture 1. For every convex polyhedral surface S ⊂ R 3 there exist a unique (up to rigid motions) piecewise-smooth surface K = K ( S ) which is embedded into R 3 , submetric to S and such that vol( K ) = η ( S ) .

32 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 ).

33 Conjecture 4. Let S 0 be a convex polyhedral surface in R 3 and let S 1 be a convex polyhedral surface submetric to S 0 of greater volume: vol( S 1 ) > vol( S 0 ) . Then there exist a volume-increasing shrinking from S 0 to S 1 . Similarly, if vol( S 1 ) = vol( S 0 ), there exist a volume-constant shrinking from S 0 to S 1 .

34 For a convex piecewise-smooth surface S , define the crimping ratio cr( S ): � � cr( S ) = area K ( S ) . area( S ) Among all convex piecewise-smooth surfaces S , the Conjecture 5. 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).