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The author seems to get quite close to a proof that there must be 6 'defects' but then gives up for some reason. To get the result all you really need is the a

The author seems to get quite close to a proof that there must be 6 'defects' but then gives up for some reason. To get the result all you really need is the additional requirement that the vertices on the edge each have 4 neighbours. Once you have that then you can just glue it to its mirror image. This now functions as a triangulation of the sphere so it must have 12 more vertices with 5 neighbours than vertices with 7 neighbours, since it's composed of 2 equal halves each half must have exactly 6 more vertices with 5 neighbours. You can probably generalize this by letting vertices on the edge have one neighbour more or fewer as well, but things get a bit messy.

kmill 5y ago
The answer to the puzzle is in the comments on that page (Prof. Baez gives your argument in one of them, too). It would be interesting knowing why there are roughly six clusters of defects. The figure seems to technically have seven clusters, but it seems plausible that low-energy configurations tend to have at most six. One point of view is that in a triangulation, the degree d of a vertex corresponds to something like the curvature there, where the curvature is concentrated into a cone angle of 60 * d. The total cone angle defect (the sum of 360 - 60 * d over all vertices) is 720 degrees for a sphere, either by Euler characteristic or Gauss-Bonnet, so if the only vertex degrees you see are 5 and 6, there are exactly 12 vertices of degree 5. So, fancifully, it seems like the potential function makes the space "want" to be flat (i.e., have vertices of degree 6 for cone angle defect of 0), so it seems plausible that these defects, some kind of fundamental particle of curvature, would "want" to be concentrated in packets of total-defect-60-degrees, and that these would "want" to be relatively far apart from one another.

kmill 5y ago

The answer to the puzzle is in the comments on that page (Prof. Baez gives your argument in one of them, too). It would be interesting knowing why there are roughly six clusters of defects. The figure seems to technically have seven clusters, but it seems plausible that low-energy configurations tend to have at most six. One point of view is that in a triangulation, the degree d of a vertex corresponds to something like the curvature there, where the curvature is concentrated into a cone angle of 60 * d. The total cone angle defect (the sum of 360 - 60 * d over all vertices) is 720 degrees for a sphere, either by Euler characteristic or Gauss-Bonnet, so if the only vertex degrees you see are 5 and 6, there are exactly 12 vertices of degree 5. So, fancifully, it seems like the potential function makes the space "want" to be flat (i.e., have vertices of degree 6 for cone angle defect of 0), so it seems plausible that these defects, some kind of fundamental particle of curvature, would "want" to be concentrated in packets of total-defect-60-degrees, and that these would "want" to be relatively far apart from one another.