Dehn's theorem states that any simplicial strictly convex polyedron P in Euclidean 3-space is infinitesimally rigid (that is, any non-trivial first order deformation of P induces a variation of its edges lengths). But many authors write « convex polyhedra are infinitesimally rigid »... Are the conditions « simplicial » and « strictly » necessary ?
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The way the Alexandrov phrases the theorem, in his book Convex Polyhedra (p.421) is:
Theorem. A closed convex polyhedron with infinitesimally rigid faces is infinitesimally rigid.
When the polyhedron is simplicial, i.e., all faces are triangles, then all faces are infinitesimally rigid. The 1-skeleton of a cube is not rigid, because the squares can deform to rhombi. But if the cube faces are rigid squares, then this nonsimplicial polyhedron is rigid. So it depends on what you consider a polyhedron—Is it built out of sticks or out of plates?
I am not sure what is meant by a non-«strictly convex polyhedron». [Igor clarifies below.]
answered Feb 26, 2012 at 19:33
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$\begingroup$ It is easy to see what is meant by a non-strictly convex polyhedron: some edges have dihedral angle of $\pi.$ $\endgroup$ Feb 26, 2012 at 20:23
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$\begingroup$ @Igor: Oh, I see. Thanks. So that is not a relevant consideration. For example, adding diagonals to a cube face both rigidifies it and makes it non-strictly convex along those new edges. $\endgroup$ Feb 26, 2012 at 20:37
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1$\begingroup$ Has this also been proven in higher dimensions? For example, if all 3-dimensional faces of a 4-polytope are infinitesimally rigid, is the polytope itself infinitesimally rigid? $\endgroup$ Mar 4 at 14:22