On the infinitesimal rigidity of weakly convex polyhedra

The main motivation here is a question: whether any polyhedron which can be subdivided into convex pieces without adding a vertex, and which has the same vertices as a convex polyhedron, is infinitesimally rigid. We prove that it is indeed the case for two classes of polyhedra: those obtained from a convex polyhedron by “denting” at most two edges at a common vertex, and suspensions with a natural subdivision. 1 A question on the rigidity of polyhedra A question. The rigidity of Euclidean polyhedra has a long and interesting history. Legendre [LegII] and Cauchy [Cau13] proved that convex polyhedra are rigid: if there is a continuous map between the surfaces of two convex polyhedra that is a congruence when restricted to each face, then the map is a congruence between the polyhedra (see [Sab04]). However the rigidity of non-convex polyhedra remained an open question until the first example of flexible (non-convex) polyhedra were discovered [Con77]. We say that a polyhedral surface is weakly strictly convex if for every vertex pi there is a (support) plane that intersects the surface at exactly pi. If it is also true that every edge e of the triangulated surface has a (support) plane that intersects the surface at exactly e, we say the surface is strongly strictly convex. If there is an edge such that the internal dihedral angle is greater than 180◦, we say that edge is a non-convex edge of the surface. In addition to being rigid, strongly strictly convex polyhedra with all faces triangles are infinitesimally rigid: there is no non-trivial first-order deformation that is an infinitesimal congruence on each triangular face. This point, which was first proved by Dehn [Deh16], is important in Alexandrov’s subsequent theory concerning the induced metrics on convex polyhedra (and from there on convex bodies, see [Ale58]). Alexandrov also showed that Dehn’s Theorem can be extended to the case when the polyhedral surface is weakly strictly convex, as well as being convex. In other words, vertices of the subdivision can only be vertices of the convex set, and they cannot appear in the interior of faces, for example. If vertices of a convex polyhedral surface do lie in the interior of a face, then the surface is rigid, but not infinitesimally rigid. This shows that the underlying framework is what determines infinitesimal rigidity, rather than simply the surface as a space. Our main motivation here is a question concerning the infinitesimal rigidity of a class of frameworks determined by polyhedra which are weakly strictly convex. Question 1.1. Let P ⊂ E3 be a polyhedral surface, with vertices p1, · · · ,pn, such that: i.) P is weakly convex; ii.) P is decomposable, i.e., it can be written as the union of non-overlapping convex polyhedra, such that any two intersect in a common face, without adding any new vertices. Is P then necessarily infinitesimally rigid ? ∗(visiting Cambridge University until August 2006) Research supported in part by NSF Grant No. DMS-0209595