A periodic layout for the Petersen graph

Question

The Petersen graph is a well-known graph with genus $1$, which means it can be drawn without crossings on a torus. Here is one possible embedding of this type. Topologically, we can think of a torus as a square with opposite sides identified. So every time we draw something on a torus, we can tile the plane with that drawing, and get a periodic picture in the plane.

The Petersen graph has $10$ vertices, and there is a simple $\sqrt{10} \times \sqrt{10}$ square that can tile an infinite grid, as shown in the diagram below:

My question. Is there a nice periodic embedding of the Petersen graph that puts its $10$ vertices down at the points labeled $0$ through $9$ in the diagram above, and connects every point by an edge to some instance of its three neighbors?

The meaning of "nice" is complicated; there are a bunch of mathematical considerations that I am prepared to trade off for an aesthetically pleasing result.

• We should definitely be able to draw this in such a way that none of the edges cross. That's what it means for the Petersen graph to have genus $1$.
• It would be nice for all the edges to be straight line segments. (This is definitely possible for some embeddings; it may no longer be possible when we put the vertices at these specific points in the grid, but it probably is.)
• It would also be nice to have some kind of symmetry or hint of symmetry - for example, different vertices of the embedding having similarly drawn edges out of them, or different faces of the embedding being congruent, aside from what is required by the periodicity.

To illustrate the question a bit more, here is a "near miss". It is a straight-line embedding, and has $180^\circ$ symmetry (almost $90^\circ$ symmetry, in fact) about every point labeled $0$ or $5$, but it is not quite an embedding of the Petersen graph. To be the Petersen graph, it would need an edge between $0$ and $5$.

Answer 1

Here is a non-crossing embedding using only straight line segments of length $1$ and $\sqrt{2}$:

(One nice way to see that this embedding is of a Petersen graph is to note the relatively clean $0-4-3-7-6-0$ and $9-2-5-8-1-9$ cycles, then observe that points on one of the cycles connect to alternating points in the opposite cycle, which easily bijects with this standard view of the Petersen graph.)

It is not very nice, but there are good reasons to think that no solutions with a nontrivial symmetry exist.

The problem is that any reflectional symmetry that does not agree with the existing red grid will, when combined with the tiling across the torus, produce several symmetry constraints that must be obeyed simultaneously, which seems likely to be very difficult. (For instance, being reflectionally symmetric about any one vertical line implies doing so across all vertical lines.)

180-degree rotational symmetries can fare better in this respect, but they cannot be centered around a point without forcing that point to have even degree. I don't see immediate obstructions to central symmetry about an edge or a face of one of the small squares, but running a SAT solver on those constraints with all edges of length at most $5$ didn't turn up any solutions (at least, if my code works properly).

Answer 2

To add on to the excellent accepted answer, here are some diagrams of the embedding it contains, and two more embeddings that I found after seeing it. For the sake of space, these are given as icons; click on them to see a full-size image.

RavenclawPrefect's original embedding (with short, straight edges):

An embedding with near-$180^\circ$ symmetry (but with longer edges):

An embedding with near-$90^\circ$ symmetry (but with curved edges):

The idea behind the two symmetric embeddings is as follows. We see from the original embedding that the Petersen graph ends up having $5$ faces on the torus, of lengths $5, 5, 5, 6, 9$. I'm pretty sure this should be a constant across all embeddings, because the Petersen graph is $3$-connected, and $3$-connected planar graphs definitely have a unique embedding on the sphere. Anyway, if we delete an edge between the faces of length $5$ and $6$, we get another face of length $9$, and a graph with faces of length $5,5,9,9$ could have an embedding with $180^\circ$ rotational symmetry that exchanges two faces of the same length. This is what happens in the second embedding: there is a symmetry that all edges except edge $05$ respect.

In the third embedding, we delete some more edges: the edge separating the faces of length $5$, and the edge separating the faces of length $9$. At that point, we get two faces: one of length $8$ and one of length $16$. Now it is reasonable to have a $90^\circ$ rotational symmetry which preserves each of those faces. When we put the deleted edges back, we get an embedding of the Petersen graph in which every edge except $06, 38, 79$ respects the $90^\circ$ symmetry (and every edge except $38$ respects the $180^\circ$ symmetry). Unfortunately, with this approach, there seems to be no way to avoid making one of the edges curved.

The more standard way to draw a toroidal embedding of a graph is to just draw one of the $\sqrt{10} \times \sqrt{10}$ squares that repeat in the previous diagrams. Here are the corresponding representations of the three embeddings above, rotated so that the squares are upright:

(In all three cases, I shifted around the repeating part to try to minimize the number of edges that have to cross from one side of the square to the other.)

My original motivation for this question was not just to find a nice toroidal embedding of the Petersen graph, but to use it to draw some Petersen-graph-inspired pictures. So here are three repeated tilings of the plane obtained by coloring each face of the embedding a different color. (Only one tile is shown in the post; click on the image to see it repeated several times horizontally and vertically.)

I am by no means convinced that this exhausts the limit of what can be done with periodic layouts for the Petersen graph. New ideas are always welcome!