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Consider given a finite collection of points along the boundary of the unit disk $D \subset \mathbf{R}^2$: \begin{equation} p_1,\dots,p_{2n} \in \partial D. \end{equation} We assume that these are all distinct from one another, and that there is an even number of them in total. The collection $P = \{ p_1,\dots,p_{2n}\}$ may be balanced (meaning $p_1 + \cdots + p_{2n} = 0$) or unbalanced.

Is there a classification of all geodesic nets (with even-degree vertices) in $D$ spanning the boundary $P$? What function $N(P)$ describes the number of solutions?

A geodesic net in $D$ (with boundary $P$) is a smooth embedding $f$ of a graph $G = (V,E)$ into $\bar{D}$ so that every point of $P$ is the image of a vertex of $G$, and so that $f(G)$ is a critical point for the length-functional \begin{equation} \lVert f(G) \rVert := \sum_{e \in E} \mathrm{length}(f(e)). \end{equation}

This means that for every one-parameter family of diffeomorphisms $(\Phi_t)$ with $\Phi_0 = \mathrm{id}$ fixing $P$, one has \begin{equation} \frac{\mathrm{d}}{\mathrm{d} t} \lVert \Phi_{t\#}(f(G)) \rVert = 0 \quad \text{ at $t = 0$}. \end{equation}

Additionally we make the simplifying assumption that vertices in $P$ have degree one, and that all vertices in the interior have even degree. Without this there are configurations with infinitely many spanning geodesic net—see the example below.

Example. Take $2n = 6$ equally spaced vertices $p_j = \mathrm{e}^{j \frac{\mathrm{i}\pi}{3}}$. The standard cone $C$ bounded by $P$ can then be modified to yield an uncountable family of solutions. Pick any radius $r \in (0,1)$, and split off triple junctions from each ray. The resulting net looks like a regular hexagon with vertices at $r p_j$, with rays shooting out of the vertex in the direction of $p_j$, intersecting $\partial D$ exactly in the set $P$. Therefore this set $P$ bounds uncountably many geodesic nets.

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  • $\begingroup$ Do you know about the classification of stationary 1varifolds by Allard and almgren? $\endgroup$ Mar 23, 2023 at 15:07
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    $\begingroup$ @OtisChodosh Sure, but I think if one works with one-varifolds then there's a ton of different solutions. Say if $n = 3$ and $p_j = \exp(j \frac{\mathrm{i} \pi}{3})$ then you have the cone over $P$, but you can also split off triple junctions at any radius $r \in (0,1)$. (The resulting varifolds look like a regular hexagon, with rays shooting out of every vertex.) So in this case there's uncountably many solutions. $\endgroup$
    – Leo Moos
    Mar 23, 2023 at 15:19

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