TITLE: Can you determine whether a graph is the 1-skeleton of a polytope?
QUESTION [21 upvotes]: How do I test whether a given undirected graph is the 1-skeleton of a polytope?
How can I tell the dimension of a given 1-skeleton?

REPLY [3 votes]: Not an answer, but potentially useful:

A matrix whose rows form an orthogonal basis of an eigenspace of a graph's adjacency matrix has columns that serve as coordinate vectors of an harmonious geometric realization of the graph.

("harmonious" == automorphisms of the graph induce rigid isometries the realization)
When the graph has a high degree of symmetry, these realizations --which I call "spectral"[*]-- have a great visual appeal; in general, though, these realizations are jumbles of points in one-dimensional space. In most cases, multiple vertices (and edges) are collapsed to single points, so that the realizations aren't faithful.
If a graph happens to admit a faithful spectral realization, you might be able to tease out a cell structure (which may not be unique), though I've not investigated this. I'll note that, even for polyhedra, there's no guarantee that the "faces" of a spectral realization are bounded by planar cycles of edges.
[*] More precisely, the realizations as described here are orthogonal projections of realizations I call "spectral". (See my still-drafty note, "Spectral Realizations of Graphs", the bulk of which is dedicated to a gallery of spectral realizations of the uniform polyhedra.)