MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question

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.]

share|cite|improve this answer
It is easy to see what is meant by a non-strictly convex polyhedron: some edges have dihedral angle of $\pi.$ – Igor Rivin Feb 26 '12 at 20:23
@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. – Joseph O'Rourke Feb 26 '12 at 20:37
@Joseph: yes, exactly. – Igor Rivin Feb 26 '12 at 21:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.