# Connecting Lemma in the Alexandrov's existence theorem.

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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@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