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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?

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    $\begingroup$ @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. $\endgroup$
    – Joseph O'Rourke
    Commented May 18, 2012 at 14:11
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    $\begingroup$ @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. $\endgroup$
    – Anton Petrunin
    Commented May 19, 2012 at 11:04

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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.

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  • $\begingroup$ Thank you. My proof is elementary; all I do is cutting holes and patching them with one-parameter family of patches. $\endgroup$
    – Anton Petrunin
    Commented May 17, 2012 at 12:38

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