MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

At the moment I am polishing my lecture notes which in particular cover Alexandrov's existence theorem.

Denote by $\mathbf{P}_k$ the space of isometry classes of polyhedral metrics on the $\mathbb S^2$ with k-verteces and nonnegative curvature at each vertex.

In the proof of Alexandrov's existence theorem one has to show that any space in $\mathbf{P}_k$ can be continuously deformed in $\mathbf{P}_k$ into a realizable polyhedral metric; i.e. the one which is isometric to the surface of convex polyhedron.

The original Alexandrov's proof is simple but tedious. First he deforms the given polyhedral space into space with one degenerate vertex. Second he shows that any polyhedral metric near the degenerate space is realizable.

It seems that I found a simpler way to do this. I proof that $\mathbf{P}_k$ is path connected by constructing a path between given spaces in $\mathbf{P}_k$ using induction on $k$. The existence of one realizable space in $\mathbf{P}_k$ is not a problem.

Was it done before?

share|cite|improve this question
@Anton: Does your new proof imply an algorithm? Right now there really is no practical method to construct the polyhedron that his theorem guarantees exists. – Joseph O'Rourke May 18 '12 at 14:11
@Joseph, no I do not think so. In fact no new idea is involved, I only realized that the patching (which was also introduced by Alexandrov but much later) can be used to simplify the proof. – Anton Petrunin May 19 '12 at 11:04

I am not aware of an elementary argument along these lines. A high-powered argument to show a priori connectedness uses Teichmuller theory (the space of polyhedral metrics fibers over teichmuller/moduli space (depending on how you identify things), with contractible fiber, see for example Troyanov's paper on cone metrics in Ens. Mathematique in the late eighties; this can also be gotten from Schwartz-Christoffel). But this, while not difficult, is not elementary, so if you have a simple combinatorial proof, that's certainly a good thing.

share|cite|improve this answer
Thank you. My proof is elementary; all I do is cutting holes and patching them with one-parameter family of patches. – Anton Petrunin May 17 '12 at 12:38

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.