On the infinitesimal rigidity of polyhedra with vertices in convex position

Let $P \subset \R^3$ be a polyhedron. It was conjectured that if $P$ is weakly convex (i. e. its vertices lie on the boundary of a strictly convex domain) and decomposable (i. e. $P$ can be triangulated without adding new vertices), then it is infinitesimally rigid. We prove this conjecture under a weak additional assumption of codecomposability. The proof relies on a result of independent interest concerning the Hilbert-Einstein function of a triangulated convex polyhedron. We determine the signature of the Hessian of that function with respect to deformations of the interior edges. In particular, if there are no interior vertices, then the Hessian is negative definite.Comment: 12 page

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

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 o