# Do continuous motions of the vertices of convex polyhedra that maintain local convexity imply global convexity? (Reference request)

A convex polyhedron has all of its internal dihedral angles in $$(0, \pi)$$. However, if I start with an abstract polyhedron $$P$$, let's say a triangulated one, so I don't have to worry about planarity of faces, and I embed each vertex $$i\in V(P)$$ at some point of $$p_i \in \mathbb{E}^3$$, measuring the convexity of the dihedral angles is not enough to check that the polyhedron is convex because it could be self intersecting and have all of its dihedral angles between 0 and $$\pi$$.

On the other hand, assume I start with an embedded convex polyhedron $$P$$ and allow its vertex positions to vary continuously over some interval $$t\in I$$ so that the position of a vertex $$i$$ at time $$t$$ is given by a continuous function $$p_i(t)$$. Let $$P(t)$$ denote the family of polyhedra I obtain from the vertex position functions with $$P(0) = P$$. Suppose throughout this motion every dihedral angle takes a value within $$(0, \pi)$$, it seems obvious to me that $$P(t)$$ remains convex for all $$t$$.

I'm looking for a reference for where this appears as a theorem (if indeed I'm not missing something and its false). Bonus points for a projective version where one vertex is allowed to be at infinity.

EDIT: I am looking for this over every motion that maintains the dihedral angles in $$(0, \pi)$$, not just sufficiently small motions.

EDIT 2: It took me writing out nearly an entire proof of the theorem above to realize how Joe O'Rourke's reference is all you need. To reproduce it here: Alexandrov's observation is that the strictly convex polyhedra are an open set of $$\mathbb{R}^{3v}$$. Similarly, we may observe that the strictly non-convex polyhedra form an open set of $$\mathbb{R}^{3v}$$. To pass through the two requires passing from one open set into another, the boundary is a non-strictly convex polyhedron. Thus to pass from one to the other at least one dihedral angle must become $$0$$ or $$\pi$$.

EDIT 3: I think you also need to add that no edge length goes to 0. Otherwise you can sort of invert the polyhedron by inverting edges.

• Yes, I should have caught the need to avoid $|e| \to 0$. Jun 10 '20 at 15:52

Perhaps this quote from Alexandrov's book, p.147, suffices: Alexandrov, Alexandr D. Convex Polyhedra. Springer Science & Business Media, 2005.

• Thanks for the reference, though I'm don't quite see how this quite gets all the way to the result. My thinking is that Alexandrov's statement here applies only for a sufficiently small open neighborhood of $\mathbb{R}^{3v}$ and thus $t$ values close enough to 0, but not necessarily for the entire interval.
– John
Jun 10 '20 at 13:50
• I think maybe I was unclear that I'm interested in any motion maintaining the dihedral angle bounds, not just sufficiently small motions. I edited the question to clarify.
– John
Jun 10 '20 at 13:55
• Ok, I think I see how this proves it. Strict convexity is an open set of $\mathbb{R}^{3v}$ and strict non-convexity is an open set of $\mathbb{R}^{3v}$. Therefore, a continuous motion passing from one to the other must pass through a non-strictly convex state, and thus a dihedral angle has become $\pi$. Took me longer than it should have. Thanks!
– John
Jun 10 '20 at 14:32
• @John: I was out of touch for a while there, but I think Yes, strict convexity is an open set in that space. Jun 10 '20 at 14:37