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Rigidity and Polyhedral Combinatorics the American Institute of Mathematics Rigidity and polyhedral combinatorics The American Institute of Mathematics The following compilation of participant contributions is only intended as a lead-in to the AIM workshop \Rigidity and polyhedral combinatorics." This material is not for public distribution. Corrections and new material are welcomed and can be sent to [email protected] Version: Mon Nov 26 14:43:28 2007 1 2 Table of Contents A. Participant Contributions . 3 1. Alexandrov, Victor 2. Bezdek, Karoly 3. Bobenko, Alexander 4. Dolbilin, Nikolay 5. Fernandez, Silvia 6. Fillastre, Francois 7. Graver, Jack 8. Izmestiev, Ivan 9. Martin, Jeremy 10. Nevo, Eran 11. Panina, Gayane 12. Rote, Gunter 13. Sabitov, Idzhad 14. Schlenker, Jean-Marc 15. Schulze, Bernd 16. Servatius, Brigitte 17. Stachel, Hellmuth 18. Tabachnikov, Sergei 19. Tarasov, Alexey 20. Thurston, Dylan 21. Whiteley, Walter 22. Ye, Yinyu 3 Chapter A: Participant Contributions A.1 Alexandrov, Victor In the moment my research interests are focused at the Strong Bellows Conjecture which reads as follows: every two polyhedra in Euclidean 3-space, such that one of them is obtained from the other by a continuous flex, are scissor congruent, that is, one of them can be cut into \small" simplices which, being moved independently one from another, give a partition of the other polyhedron. A.2 Bezdek, Karoly Recall that the intersection of finitely many closed unit balls of E3 is called a ball- polyhedron (for more details see [blnp]). Also, it is natural to assume that removing any of the balls in question yields the intersection of the remaining balls to become a larger set. One can represent the boundary of a ball-polyhedron in E3 as the union of vertices, edges and faces defined in a rather natural way as follows. A boundary point is called a vertex if it belongs to at least three of the closed balls defining the ball-polyhedron. A face of the ball-polyhedron is the intersection of one of the generating closed balls with the boundary of the ball-polyhedron. Finally, if the intersection of two faces is non-empty, then it is the union of (possibly degenerate) circular arcs. The non-degenerate arcs are called edges of the ball-polyhedron. Obviously, if a ball-polyhedron in E3 is generated by at least three balls, then it possesses vertices, edges and faces. Finally, a ball-polyhedron is called a standard ball-polyhedron if its vertices, edges and faces (together with the empty set and the ball- polyhedron itself) form an algebraic lattice with respect to containment. We note that not every ball-polyhedron of E3 is a standard one a fact, that is somewhat surprising and is responsible for some of the difficulties arising at studying ball-polyhedra in general (for more details see [bn] as well as [blnp]). To each edge of a ball-polyhedron in E3 we can assign an inner dihedral angle. Namely, take any point p in the relative interior of the edge and take the two balls that contain the two faces of the ball-polyhedron meeting along that edge. Now, the inner dihedral angle along this edge is the angle of the two half-spaces supporting the two balls at p. The angle in question is obviously independent of the choice of p. Finally, at each vertex of a face of a ball-polyhedron there is a face angle formed by the two edges meeting at the given vertex (which is in fact, the angle between the two tangent half-lines of the two edges meeting at the given vertex). We say that the standard ball-polyhedron P in E3 is globally rigid with respect to its face angles (resp. its inner dihedral angles) if the following holds: If Q is another standard ball-polyhedron in E3 whose face-lattice is isomorphic to that of P and whose face angles (resp. inner dihedral angles) are equal to the corresponding face angles (resp. inner dihedral angles) of P , then Q is congruent to P . A ball-polyhedron of E3 is called triangulated if all its faces are bounded by three edges. It is easy to see that any triangulated ball-polyhedron is in fact, a standard one. The following statement has been proved in [bn]. Theorem. Let P be a triangulated ball-polyhedron in E3. Then P is globally rigid with respect to its face angles. The related question below is still open. 4 Problem. Let P be a triangulated ball-polyhedron in E3. Prove or disprove that P is globally rigid with respect to its dihedral angles. It would be natural and interesting to investigate the same problems in spherical as well as in hyperbolic 3−space. Bibliography [blnp] K. Bezdek, Zs. L´angi, M. Nasz´odi and P. Papez, Ball-polyhedra, Discrete Comput. Geom. 38 (2007), 201{230. [bn] K. Bezdek and M. Nasz´odi, Rigidity of ball-polyhedra in Euclidean 3-space, European J. Combin. 27 (2006), 255{268. A.3 Bobenko, Alexander Here I select several problems in rigidity I am interested in: 1. Which polyhedral surfaces composed of planar polygons (in particular of planar quadrilaterals) admit finite deformations? 2. Delaunay unfolding. Can the boundary of a convex polytop be unfolded into the plane without self-overlap by cutting the surface along adges of the Delaunay tesselation T of the boundary? (Note that the Delaunay tesselation is determined by the metric). In particular if all the faces of a polytop are acute triangles can it be unfolded by cutting alonfg the edges of the polytope? 3. Koenigs nets. are nets of planar quadrilateral with the combinatorics of the square grid such that the intersection points of the diagonals of four neighboring quads are coplanar. They possess remarkable transformations and rigidity properties to be studied. A.4 Dolbilin, Nikolay Here I select among them the most interesting topics. B. Point 1.. Given a convex polyhedron, whether there is at least one its \edge" unfolding homeomorphic to a disc? This problem has become of especial interest because of a quite new result. In 2006 I suggested to prove (or disprove) the so-called Anti-Durer¨ conjecture: Let P be a convex polyhedron and c(P ) minimal number of non-overlapping components in edge unfoldings of P , then C = sup c(P ) over all convex polyhedra is 1. The well-known conjecture ("the Durer¨ conjecture") says that in a set of convex polyhedra C = sup c(P ) = 1. I, in person, cannot say what answer is more probable: C equals either 1 or 1. But the C seems to me not to be able to be in between. I offered to Alexei Tarasov and Alexei Glazyrin to prove the analogue the Anti-Durer¨ conjecture in the set of concave polyhedra. They proved it recently. B. Point 4. The problem of finding the maximum of the volume bounded by a closed surface, which is isometric (or submetric) to a given polyhedral sphere, could be called the isometric problem. In contrast to the classical isoperimetric problem, in the isometric problem the maximal volume is being searched for in a "subspace" of all pairwise isometric surfaces of the whole space of all closed surfaces with same area. If the organizers are going to run a series of talks by participants, I could offer to the organizers to give one or two talks: 1. The new results on the Minkowski theorem on convex polyhedra and its applications. 2. Rigidity of zonohedra. 5 The last talk is going to be a survey of results on rigidity of surfaces with center symmet- rical faces obtained by M.Stan'ko, M.Stogrin, and N.Dolbilin. A typical result here sounds as follows: assume a polyhedral sphere with centrosymmetrical faces, which is immersed into space, is rigid (non-convexity and self-intersection under the immersion are allowed). A.5 Fernandez, Silvia Here is an open problem that may be of interest to some of the participants of the workshop. Let K be a strictly convex domain (i.e., a bounded subset of the plane such that if x and y are boundary points of K then the open segment xy is contained in the interior of K) with smooth boundary B(K). Consider a unit rigid rod (segment) R with endpoints p and q. Let R travel counter-clockwise along K in such a way that p and q are on the boundary of K at all times. More precisely, R rotates continuously and counter-clockwise 360 degrees, in such a way that p and q are always points of B(K). Of course, this is impossible sometimes. For example, if K is too small then R may not even fit inside K. And even if R fits inside K, it can suddenly get stuck. (This happens when the width of K in a certain direction is less than 1.) Now, assuming that R can complete a full turn around K, follow the trajectory of its end points along B(K). When K is \round" and \big" enough p and q simply move counter-clockwise along B(K). But interestingly enough, for some sets K, the points p and q may change directions while moving along B(K). My questions are: 1. Under which conditions on K, the points p and q change direction along the bound- ary of K a finite number of times? 2. Is it true that for every K the points p and q change direction along the boundary of K a finite number of times? 3. Is it possible to find, for every integer n > 0 a set K such that p changes direction at least n times while moving along the boundary of K? A.6 Fillastre, Francois A famous theorem of A.D.
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