TITLE: Constructing a convex polyhedron from a spherical polyhedron
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: It depends on whether you require the vertices to stay fixed, or whether you consider isomorphs (with the same structure but different vertex positions on the sphere) to be the "same" polyhedron.
If you want them to stay put, then there is a much simpler proof than the one given in another answer, though it amounts to the same thing:

For a convex polyhedron, all vertices of a given face must be
coplanar,
this condition is not necessary for a spherical polyhedron, and spherical polyhedra with skew faces exist
therefore there are spherical polyhedra which have no
corresponding convex polyhedron.

But if you consider all isomorphs to be essentially the same polyhedron, then it always has a corresponding convex polyhedron, known as its canonical form.