Is it, that every edge in the dual counts twice?

I've drawn a 3-regular graph on a double torus, i.e. I've drawn it on the fundamental polygon of the double torus (no double torus at hand), in the following way:

It has 6 faces which are all octagons. Corresponding faces are colored accordingly. Now I'm interested in its dual graph. Therefore I connected faces that share an edge and found the following graph:

This is a 4-regular graph, but I expected an 8-regular one, since I started out with octagons.

Is it, that every edge in the dual counts twice?

Like a multiedge, so that my dual graph is not simple...

Solution 1:

Here is an embedding of the dual graph in the double torus:

I'm using the same fundamental polygon as you. So there is only one yellow vertex in this picture (placed at all $8$ corners of the octagon, which are the same point), and only one red, blue, purple, and cyan vertex (each one placed at the midpoints of two opposite sides of the octagon, which are also the same point).

We see two edges between each pair of vertices, and we see the $16$ triangular faces we expected to get (since we started with a $16$-vertex $3$-regular graph).

Here is a version of the diagram above with the $24$ distinct edges labeled by numbers $1,\dots,24$ (some edges appear in the diagram twice, because they're along opposite sides of the octagon, which are identified). This helps us see that even though some of the triangular faces have all the same vertices (e.g. there are two faces with the red, yellow, and blue vertices) they are different faces, because they have different edges.

If you are not interested in the embedding, only in the graph structure, you can get the same graph by doubling every edge of your octahedral graph.