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The circle packing theorem (Koebe–Andreev–Thurston theorem) states that every finite planar graph is the nerve of some disk packing in the plane, where the nerve of a packing $P$ is a graph $G=(V,E)$, where there is an edge $(u,w) \in E(G)$ when $\partial P_{u} \cap \partial P_{w} \neq \varnothing$, for disks $P_{u},P_{w} \in P$.

More succintly, for every connected simple planar graph $G$ there is a circle packing in the plane whose intersection graph is (isomorphic to) $G$.

I am interested in the possibility of extending this result to dimension three, where we would be considering a homogeneous connected simplicial $3$-complex as the nerve of a sphere packing $P$.

Has there been any research done to suggest that such a theorem exists in three dimensions, or a counterexample that this is not the case?

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    $\begingroup$ Similar to mathoverflow.net/questions/85547/… $\endgroup$
    – Lee Mosher
    Commented May 7, 2012 at 18:27
  • $\begingroup$ Wait… do you want to allow balls with intersecting interiors? $\endgroup$
    – Zsbán Ambrus
    Commented May 7, 2012 at 20:00
  • $\begingroup$ @Zsban: That would not be a generalization of the "standard" circle packing theorem, but a generalization of the generalized circle packing theorem... $\endgroup$
    – Igor Rivin
    Commented May 7, 2012 at 20:32
  • $\begingroup$ @Zsban: By the standard definition of sphere "packing" we would only have that $\partial P_{u} \cap \partial P_{w} \neq \varnothing$, not that $\text{int}(P_{u}) \cap \text{int}(P_{w}) \neq \varnothing$. $\endgroup$
    – Samuel Reid
    Commented May 7, 2012 at 21:01

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@Lee's comment is correct, and the answer to the question he cites give an almost complete picture, but you might also want to look at the following:

Combinatorial scalar curvature and rigidity of ball packings D. Cooper and I. Rivin Math Res Letters, 1996

(note: all the results in the papers are correct, but one of the proofs is wrong, and fixed by Dave Glickenstein a few years later).

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