Suppose I am given an arbitrary 3-dimensional convex polyhedron $P\subset\Bbb R^3$ that contains the origin.
I can "blow it up" to a spherical polyhedron by projecting all edges and vertices (away from the origin) to the unit sphere (centered at the origin):
What about the other direction?
Question: Given a spherical polyhedron, is there a "convex polyhedron" whose projection is exactly the given spherical polyhedron? And how to construct it explicitly?
For me, a spherical polyedron is a tiling of the 2-sphere where the edges are great circle arcs.
And I know that there is always a convex polyhedron with the same combinatorics as the given spherical polyhedron, but I ask specifically about a convex polyhedron that projects to the given spherical polyhedron.
Best Answer
The answer is negative: not every spherical polyhedron comes from a convex one. The argument is basically the same I gave over here: the spherical polyhedron can have more degrees of freedom than the convex polyhedron.
The following is a copy of a part of the answer behind the link:
Now consider the $7$-sided prism (the argument can certainly be improved, but as presented here, we need an $n$-prism with $n\ge 7$). This prism has $14$ vertices, and by the argument presented above, the spherical 7-prism has $2\times 14=28$ degrees of freedom.
However, a convex polyhedron has as many degrees of freedom as its dual (because they determine each other uniquely). The dual of the 7-prism is the 7-sided bipyramid, which has $9$ vertices. And the position of these vertices determine the bipyramid uniquely. Each vertex has three dregrees of freedom, and so the 7-prism has at most $3\times 9=27$ degrees of freedom.
In other words, the projection of the convex prism to the spherical one (which is continuous) cannot be surjective, given the larger dimension of the realization space of the image.