Shortest Path Embeddings of Graphs on Surfaces

Abstract

The classical theorem of Fáry states that every planar graph can be represented by an embedding in which every edge is represented by a straight line segment. We consider generalizations of Fáry’s theorem to surfaces equipped with Riemannian metrics. In this setting, we require that every edge is drawn as a shortest path between its two endpoints and we call an embedding with this property a shortest path embedding. The main question addressed in this paper is whether given a closed surface S, there exists a Riemannian metric for which every topologically embeddable graph admits a shortest path embedding. This question is also motivated by various problems regarding crossing numbers on surfaces. We observe that the round metrics on the sphere and the projective plane have this property. We provide flat metrics on the torus and the Klein bottle which also have this property. Then we show that for the unit square flat metric on the Klein bottle there exists a graph without shortest path embeddings. We show, moreover, that for large g, there exist graphs G embeddable into the orientable surface of genus g, such that with large probability a random hyperbolic metric does not admit a shortest path embedding of G, where the probability measure is proportional to the Weil–Petersson volume on moduli space. Finally, we construct a hyperbolic metric on every orientable surface S of genus g, such that every graph embeddable into S can be embedded so that every edge is a concatenation of at most O(g) shortest paths.

Appendix A: Tutte’s Embedding Theorem in a Hyperbolic Setting

In this section, we explain the proof of the subsequent theorem, following the arguments of Y. Colin de Verdière [7].

Theorem 6.4

Let G be a graph embedded as a triangulation in a hyperbolic hexagon H endowed with the metric $m_H$. If there are no dividing edges in G, i.e., edges between two non-adjacent vertices on the boundary of H, then G can be embedded with geodesics, with the vertices on the boundary of H in the same positions as in the initial embedding.

As announced, the proof follows from a spring-like construction, i.e. we think of the edges of the graph G as springs with some arbitrary stiffness, the vertices which are not on the boundary are allowed to move and we prove that the equilibrium state for this physical system is an embedding of the graph.

For an embedding $\varphi :G \rightarrow H$, denote by $e_{ij}$ the map $[0,1] \rightarrow H$ representing the edge (i, j). Starting with an embedding $\varphi _0:G \rightarrow H$ and given assignments $c_{i,j}:E(G) \rightarrow \mathbb {R}^+$, we are interested in the map $\varphi :G \rightarrow H$ minimizing the energy functional

with fixed vertices on the boundary of H. This is the equilibrium state of the spring system with the $c_{i,j}$ coefficients specifying the stiffness of the springs. We claim that $\varphi$ is an embedding such that the edges are geodesics.

Step 1: Existence. The existence of $\varphi$ follows from classical compactness considerations, since an Arzelà-Ascoli argument proves the compactness of sets with bounded energy. Then an extremum of $E_{\varphi }$ corresponds to a $\varphi$ where all the arcs $e_{i,j}$ are geodesics. Furthermore, every vertex $\varphi (x)$ which is not on the boundary lies in the strict hyperbolic convex hull of its neighbors which are not mapped to the same point.

Step 2: Curvature considerations. Since $\varphi _0$ provides an embedding of G into H, G can be seen as a topological subspace of H. The corresponding simplicial complex will be denoted by X (it is of course homeomorphic to H) and its set of vertices, edges and triangles by V, E, and T. By extending $\varphi$ separately with a local homeomorphism in the interior of each non-degenerate triangle, we can extend it into a map $\Phi :X \rightarrow H$ agreeing with $\varphi$ on G.

Now, the map $\Phi :X \rightarrow H$ provides values for the angles of the non-degenerate triangles in X. For degenerate triangles, values of the angles are taken arbitrarily so that they sum to $\pi$ (therefore morally their hyperbolic area is zero). For an interior vertex v, let us define the curvature $K(v)= 2 \pi - \sum _i \alpha ^i_v$, where $\alpha ^i_v$ are the angles adjacent to v. For a vertex v on the interior of a geodesic boundary, we define it by $K(v)= \pi - \sum _i \alpha _v^i$, and on the six vertices of H, we take it to be $K(v)=\pi /2 - \sum _i \alpha _v^i$.

The area of a geodesic hyperbolic triangle is $\pi$ minus the sum of its angles. Summing over all the triangles of $\Phi (X)$, we obtain $|T| \pi - \sum _v \sum _i \alpha ^i_v = \sum _{t \in T} \mathrm{Area}(t)$. With Euler’s formula and double counting, this gives $\sum _{t \in T} \mathrm{Area}(t)=\pi + \sum _v K(v)$. Since the boundary is fixed, $\Phi$ has degree one and is thus surjective, therefore the sum of the areas of the triangles is at least the area of the hexagon, which is $\pi$ since it is right-angled. Therefore $\sum _v K(v) \ge 0$.

Step 3: Punctual degeneracies. In this step we investigate which subcomplexes of X can be mapped to a single point. We show that no triangle can be mapped to a single point, and that a set of edges mapped to a single point forms a path subgraph in G.

Let $X_1$ be a maximal connected subcomplex of X which is mapped to a point x by $\Phi$. This subcomplex has to be simply connected, otherwise the region inside could be mapped to x as well which would reduce the value of $E_{\varphi }$. Since the boundary edges are fixed by $\varphi$, $X_1$ does not contain any edge on the boundary or triangle adjacent to the boundary.

For every vertex v in $\Phi ^{-1}(x)$, $\Phi (v)=\varphi (v)$ lies in the strict convex hull of its neighbors which are not mapped to x, as was observed in Step 1. Therefore the angles of the non-degenerate triangles adjacent to v sum up to at least $2 \pi$. Indeed the angular opening at $\varphi (v)$ has to be at least $\pi$ by the convexity hypothesis, but if a map $\mathbb {S}^1 \rightarrow \mathbb {S}^1$ is not surjective then every point in the image has at least two pre-images, in which case this angular opening of at least $\pi$ amounts to at least $2 \pi$ in the sum of angles around v. This shows that $K(x):=\sum _{v \in \Phi ^{-1}(x)} K(v)$ is non-positive. Since the boundary edges are fixed, we also have $K(v)\le 0$ for the vertices on the boundary.

Summing over all the values of x, we obtain that $\sum _v K(v) \le 0$, and thus this sum is zero by the previous paragraph, and each of the K(x) is also zero.

Fig. 10

Any triangulation inducing a linear degeneracy would require either multiple edges (top) or a dividing edge (bottom)

From that we infer that $X_1$ contains no triangle: if it did, there would be at least 3 preimages of x for which the angles of the adjacent non-degenerate triangles would sum up to at least $2\pi$. Summing them into K(x) we would obtain a nonzero value. Similarly, $X_1$ can only be a linear subgraph of G, and every triangle adjacent to a $X_1$ not reduced to a point is degenerate.

Step 4: Linear degeneracies. Now that we showed that triangles cannot be mapped to points, we show that triangles are not mapped to lines either, or equivalently that edges are not mapped to points.

Let $X_2$ be a maximal connected subcomplex of X such that the image of the triangles of $X_2$ by $\varphi$ are degenerate. Let us assume that $X_2$ is non-empty. Then the image $\Phi (X_2)$ is an arc of a geodesic of H: indeed if there was a broken line in $\Phi (X_2)$, around the breaking points there would be non-degenerate triangles adjacent to a $X_1$ not reduced to a point, which is absurd by the previous paragraph.

If this geodesic is not a boundary geodesic of H, two of the points on the boundary of $X_2$ are mapped to the endpoints of the arc of geodesic, and all the other vertices have their adjacent edges within $X_2$ because of the convexity condition. Therefore, there must be two arcs connecting the two boundary points, as in the top of Fig. 10, which is impossible in the simplicial complex X.

If this geodesic is on the boundary of H, then by the same convexity argument, two vertices of $\partial X$ must map to the endpoints of this arc of geodesic, and the other vertices have all their edges within $X_2$. Therefore there is a dividing edge connecting these two vertices, as in the bottom of Fig. 10, which is a contradiction.

Step 5: Conclusion. Since $X_2$ is empty, no triangle in the image of $\Phi$ is degenerate. Furthermore, all the $X_1$ are reduced to a single point and thus K(v) is zero for all the vertices v. The only remaining possible pathology is if all the triangles adjacent to a non-boundary vertex v are mapped to a half-plane around $\Phi (v)$. By the convexity constraint, this can only happen if the edges adjacent to v are aligned, but this would yield degenerate triangles. Therefore $\Phi$ is a local homeomorphism of degree 1, hence it is a global homeomorphism and $\varphi$ is an embedding.