Lifting of a spherical graph

Let us be given a topological graph $G$ on the unit sphere in $\mathbb{R}^3$ whose edges are minor arcs of great circles. We suppose that the graph is $3$-vertex-connected and that a pair of edges may only share a vertex incident to both edges.

We say that a polyhedron $P$ is a lifting of $G$ iff $G$ is obtained by radial projection of the edge-skeleton of $P$ onto the unit sphere.

My question is whether there is a simple criterion to check if a given spherical graph possesses a lifting. Moreover, I would like to compute that lifting (if it exists). (Computationally, I am interested in some spherical graphs with about $10$ to $20$ vertices)

The analogue planar question is classical and well-understood. That is, a planar graph, i.e., a graph with vertices in $\mathbb{R}^2$ and straight edges, can be lifted into a polyhedron, iff it is possible to find an equilibrium configuration of forces, see here.

The condition is similar to the equilibrium condition on the plane, only this time the sum of the forces at each vertex must not be zero but collinear with the radius-vector of the vertex: $$\sum_j w_{ij}(p_i-p_j) = c_i p_i$$ Given a collection of weights $w_{ij}=w_{ji}$ satisfying the above condition, we can construct a dual polyhedron with faces $F_i$ orthogonal to the radius-vectors $p_i$. Namely, note that $$\sum_j w_{ij} (p_i \times p_j) = p_i \times \sum_j w_{ij} p_j = 0$$ Therefore for every $i$ the vectors $w_{ij} (p_i \times p_j)$ form a closed polygonal line in a plane orthogonal to $p_i$. This will be our face $F_i$. These faces can be fitted together to form a polyhedron. The polar dual of this polyhedron has the desired combinatorics, and its vertices are multiples of $p_i$.
If all weights $w_{ij}$ were positive, the resulting polyhedron is convex. Otherwise self-intersections can happen. Is this bad or good, depends on your situation.
• Yes, exactly. Positive weights correspond to dihedral angles smaller than $\pi$ (although there is no simple formula to express the angles in terms of weights). – Ivan Izmestiev Nov 11 '16 at 9:21