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A geodesic net is an embedding of a multigraph $(V,E)$ into a Riemannian manifold $(M,g)$, so that the vertices are mapped to points of $M$ and the edges to geodesics connecting them. Additionally, one imposes the following balancing condition: for every vertex, the outward unit tangent vectors sum to zero.

(Note that loops are allowed, as are edges with integer multiplicities---in the latter case that the multiplicity must be accounted for in the balancing conditions.)

For example in the round two-sphere $\mathbf{S}^2$, an equatorial geodesic is the simplest net. It has just one edge with length $l(e) = 2\pi$. The simplest one that is not a union of geodesics is the union of three half-circles, equal to the set \begin{equation} \{ (r,\theta,z) \mid \theta = 0, 2\pi/3,4\pi/3 \} \end{equation} in cylindrical polar coordinates. This has three edges, with total length $l(e_1) + l(e_2) + l(e_3) = 3\pi.$

Question. Is there a classification of geodesic nets available in $\mathbf{S}^2$? For example, is there a list of those that have $\sum_{e \in E} l(e) \leq 10 \pi$?

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    $\begingroup$ I guess you already did, but you can check out the work of Hass & Morgan "Geodesic nets on the 2-sphere" (and the articles cited there or the article which cite this work). On the other hand: consider 9 half-circles let them meet at one vertex on one side, and consider a small neighbourhood antipodal to that point. Since curvature there will play no big rôle, a simpler question would be to classify planar nets with 9 leaves. I'm not sure if this simpler question has been answered yet. $\endgroup$
    – ARG
    Commented Oct 29, 2021 at 12:59
  • $\begingroup$ Could you elaborate on the example with the nine half-circles? I don't completely follow the argument---how do you reduce to the plane, and what conclusion does one draw from the construction? $\endgroup$
    – Leo Moos
    Commented Oct 29, 2021 at 14:05
  • $\begingroup$ sure! Start with 9 half-circles (I'm just picking 9 because it's less than the $10\pi$ you mention) which meet at two antipodal points. There is some condition on the angles at the vertices but there is a lot of freedom. Erase an arbitrarily small neighbourhood U of one vertex, and consider what net could fit there (i.e. what other net could come inside U, instead of just all half-circles meeting boringly again at the same point). Because we can remove an arbitrarily small neighbourhood, my point is that we can (I mean we probably can, didn't check if this is rigorous) reduce part of $\endgroup$
    – ARG
    Commented Oct 29, 2021 at 15:31
  • $\begingroup$ your question to the following question: fix 9 points in the plane. How many "geodesic nets in the plane with 9 leaves"* can you fit in there? Although it sounds easy, there are not many bounds known for this problem. You can check this paper and references therein. Note that the length condition essentially disappear since we are in a ridiculously small neighbourhood. *by a net with 9 leaves: I mean that there are no angle conditions at the leaves; these are fixed; they are the points where the soap filament is attached if you wish $\endgroup$
    – ARG
    Commented Oct 29, 2021 at 15:39
  • $\begingroup$ by extension, every time you have a vertex with degree $n$ you can [probably] replace an infinitesimally small neighbourhood of the vertex with a planar graph. So as long as the planar question remains open, I doubt there are many chances to get the answer on the 2-sphere. $\endgroup$
    – ARG
    Commented Oct 29, 2021 at 15:42

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When one restricts attention to nets all of whose vertices are trivalent (which is the natural condition when one is studying area minimizing cones), then Lamerle (in 1864) supposedly found all but one configuration. This classification was completed (independent of knowledge of Lamerle's work) by in 1964 who found all 10 of them. Jean Taylor indendently found these nets again in 1976 who also showed that all but three of them are unstable for area (and the three that aren't are precisely the singularities observed in physical soap-films by Plateau).

Heppes, A., Isogonale sphärische Netze, Ann. Univ. Sci. Budap. Rolando Eötvös, Sect. Math. 7, 41-48 (1964). ZBL0127.37601.

Lamerle, E., Sur la stabilit ́e des syst`emes liquides en lames minces. M ́em. Acad. R. Belg. 35 (1864), 3–104.

Taylor, J., The structure of singularities in soap-bubble-like and soap-film-like minimal sur- faces. Ann. of Math. (2) 103 (1976), no. 3, 489–539

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  • $\begingroup$ That's fascinating, thank you for the references. Do you know anything about results when the degree-three hypothesis is dropped or relaxed? For example, say we go to the other extreme, and impose that the degrees all be even. $\endgroup$
    – Leo Moos
    Commented Oct 29, 2021 at 14:09
  • $\begingroup$ @LeoMoos when all degrees are 4 it's also easy (because it can only be a bunch of great circles which cross each other)... but otherwise... $\endgroup$
    – ARG
    Commented Oct 29, 2021 at 15:24
  • $\begingroup$ @ARG You're right about that of course. I guess I was hoping for something like: at the specific length $L < 10 \pi$ you will find the first (indecomposable) network that is not a union of geodesic half-circles. $\endgroup$
    – Leo Moos
    Commented Oct 31, 2021 at 15:33
  • $\begingroup$ @LeoMoos My guess would be that you find it before $10 \pi$ (inflated tetrahedron?). As RBega2 pointed out, note also that the minimal nets should have degree 3. In particular, if you start with a fairly simple graph (like two great circle) then look at (degree 3) perturbations of it to find the one which is minimal, you should get something. It could happen that the graph degenerates, so it might be better to start the (numerical) simulation with one degree 4 vertex which is fixed and the other replaced by 2 degree 3 vertices. Or maybe you need to start with more complicated graphs. $\endgroup$
    – ARG
    Commented Nov 1, 2021 at 19:13

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