Abstract
rigid, then the whole surface is rigid. It had been suspected that any (connected) polyhedral surface, convex or not, (with its triangular faces, say, held rigid) was rigid, but this has turned out to be false (see Connelly [7]). W e extend Cauchy’s theorem to show that any convex polyhedral surface, no matter how it is triangulated, is rigid. Note that vertices are allowed in the relative interior of the natural faces and edges. Here we say a triangulated surface in three-space is rigid if any continuous deformation -of the surface that keeps the distance fixed between any pair of points in each triangle (a part of the triangulation of the surface), keeps the distance fixed between any pair of points on the surface (and thus extends to an isometry of three-space). A natural face of a convex polyhedral surface is the two dimensional intersection of a support plane with the surface (see Fig. 3). Similarly natural edges and vertices are one- and zero-dimensional intersections, respectively. Thus in Cauchy’s theorem it is insisted that under a deformation of the surface the distance is fixed between any pair of points of a natural face and not just some triangle of a triangulation. Alexandrov [l] showed that if a convex polyhedral surface is triangulated with no vertices in the relative interior of a natural face (but they are allowed in natural edges), then the surface is rigid. This theorem is the basic result for the results of this paper. The author is grateful for a description of Alexandrov’s proof by Asimow and Roth [5]. Whiteley [ 151 also has a somewhat
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