4
$\begingroup$

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?

$\endgroup$
4
  • 1
    $\begingroup$ Similar to mathoverflow.net/questions/85547/… $\endgroup$
    – Lee Mosher
    May 7, 2012 at 18:27
  • $\begingroup$ Wait… do you want to allow balls with intersecting interiors? $\endgroup$ 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
    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 $\text{int}(P_{u}) \cap \text{int}(P_{w}) \neq \varnothing$. $\endgroup$ May 7, 2012 at 21:01

1 Answer 1

3
$\begingroup$

@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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.