2

I understand that the dual graph $G^*$ is really defined for a plane graph $G$, i.e., a graph along with an embedding into $\mathbb{R}^2$, instead of just a planar graph, but I still have difficulty understanding why it is well defined.

Say $G$ is $K_3$ and $K_4$ connected by an edge $e$, embedded into the plane in the natural way ($K_4$ is embedded as a triangle with its center). In $G^*$ there is a loop $l$ at the vertex corresponding to the unbounded face of $G$. Should $l$ cross $e$? If so, should it surround $K_3$ or $K_4$? I believe the resulting "dual graphs" are isomorphic as graphs but different as plane graphs, although the latter two seem like the same if considered as graphs on sphere. There also seem to be some freedom when connecting the unbounded face with other faces. What is the exact algorithm to draw a dual graph?

Edit: I have come to the conclusion that dual graph is well-defined up to embedding into sphere, although I can't show this rigorously.

For convenience here is the part in Combinatorial Mathematics by Douglas West about dual.

enter image description here

n901
  • 460
  • 2
  • 8
  • @Morgan Rodgers But I have seen at different places the statement that double dual of a connected graph is itself, so the dual should be something more than a graph, otherwise it does not make sense to talk about double dual. – n901 Apr 14 '21 at 19:36
  • $G^{**}$ is itself as a graph, just not as a planar graph. – Mike Earnest Apr 14 '21 at 19:39
  • @Mike Earnest How do you define the dual of $G^*$ if it is given only as a graph? – n901 Apr 14 '21 at 19:42
  • @n901 Fair point! The only explanation I can think of is that you choose any embedding for $G^{}$ and use that to construct $G^{}$, then prove that any embedding for $G^{}$ results in the same graph $G^{**}$. But I do not know if this is actually true. – Mike Earnest Apr 14 '21 at 19:45
  • @MikeEarnest No, that is not true. – xxxxxxxxx Apr 14 '21 at 19:46
  • @Morgan Rodgers The text we use is Combinatorial Mathematics by West, where the statement appears at the beginning of chapter 9 (but it is left as an exercise). There are also threads on this forum such as https://math.stackexchange.com/questions/1585631/dual-of-the-dual-of-a-non-planar-graph – n901 Apr 14 '21 at 19:52

2 Answers2

2

The dual of a plane graph is a graph, and not itself a plane graph; any way you put the loop will give you an isomorphic graph, so it doesn't matter where from this point of view.

As far as whether the choices will give you equivalent embeddings on a sphere, this is not the case; in the example you give, you have 4 choices of where to put the loop; you can put it enclosing the vertex corresponding to the $K_{3}$, or you can put it enclosing the triangle corresponding to the $K_{4}$, or enclosing both, or enclosing neither. They are all different as plane graphs, they will pair up into two separate classes as embeddings on the sphere.

edit: After mentioning they are using West

In West's Introduction To Graph Theory, he gives the following definition of the graph dual:

Definition 6.1.7: The dual graph $G^*$ of a plane graph $G$ is a plane graph whose vertices correspond to the faces of $G$. The edges of $G^{*}$ correspond to the edges of $G$ as follows: If $e$ is an edge of $G$ with face $X$ on one side and face $Y$ on the other, then the endpoints of the dual edge $e^*$ joining the vertices $x$, $y$ in $G^*$ corresponding to the faces $X$ and $Y$. The order in the plane of the edges incident to $x \in V(G^{*})$ is the order of the edges bounding the face $X$ of $G$ in a walk around its boundary. (emphasis added).

This gives the notion you are looking for as to where to draw the loop; I think it suggests that the loop should contain one of $K_{3}$ or $K_{4}$, but not neither or both; however it doesn't resolve the last bit of ambiguity. This is addressed in remark 6.1.9 of the same book, but not adequately; I think this is a minor oversight made by West.

xxxxxxxxx
  • 13,302
  • 1
    It seems people generally agree that the loop should cross the connecting edge; in that case I don't see how is it possible to enclose both/neither. There are some extra choices for other egdes in $G^*$, but so far they all seem the same to me (as sphere graphs). – n901 Apr 14 '21 at 19:56
  • @n901 I made an edit given that you are using West. – xxxxxxxxx Apr 14 '21 at 20:08
  • It is a different book, and I am even more confused by this definition...Is there some kind of coherent orientation on edges? What if an edge appears twice in the walk? – n901 Apr 14 '21 at 20:17
  • It half resolves where to put the loop; since to traverse the outer face's boundary, you would go around the $K_{3}$, across the bridge, around the $K_{4}$, back across the bridge, the two endpoints of the loop should be separated, separating the $K_{3}$ and $K_{4}$ edges on either side. But it is ambiguous about what is enclosed by the loop. The "clarifying" comment I allude to above (but do not transcribe; sorry cat on my lap makes it hard to type) does not resolve it either. – xxxxxxxxx Apr 14 '21 at 20:25
  • @n901 I do think you may be correct though that under this definition, you have equivalence as spherical embeddings. But not 100% sure. – xxxxxxxxx Apr 14 '21 at 20:29
1
  1. Should l cross e?

Yes, l should cross e. Consider the simpler example of a graph A-B-C-D. In this case, if the self loop across B-C doesn't cross the B-C edge, you would not be able to replicate the original graph by taking the dual of the dual.

  1. If so should it surround $K_3$ or $K_4$.

Both are valid and isomorphic. For example, irrespective of which one it surrounds, i'm able to reconstruct the original graph from the dual of the corresponding dual graphs, so both are valid.

  1. Is there an algorithm?

See if this helps?

Rahul Madhavan
  • 2,789
  • 1
  • 11
  • 14
  • Of course they are isomorphic as graphs, but I was wondering whether they are the same as plane graphs, and now I think the answer is no, but maybe it becomes yes if we view them as sphere graphs. – n901 Apr 14 '21 at 19:29
  • Ah that's interesting I hadn't thought of that embedding (as embeddings on th surface of the sphere). Now that I think about it, they are isomorphic as sphere graphs - you can think of that loop as a greater circle on S2 – Rahul Madhavan Apr 14 '21 at 19:35