TITLE: Connecting Lemma in the Alexandrov's existence theorem.
QUESTION [12 upvotes]: 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?

REPLY [6 votes]: 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.