<p>You need to specify that $G$ is a simple graph (else the trigonal <a href="http://en.wikipedia.org/wiki/Hosohedron">hosohedron</a> is a counterexample) but then the answer is yes. What can go wrong? If a dual face is not a simple cycle then one of its edges is repeated, giving a self-loop in the primal, or a vertex but not an edge is repeated, in which case this face and vertex form a configuration of the same type in the primal. If two dual faces meet at more than one edge then there is a multiple adjacency in the primal. If the dual has a self-loop or multiple adjacency then we get a face with a repeated edge or two dual faces meeting at more than one edge in the primal. If two dual faces meet at two vertices but do not share an edge between these two vertices then a configuration of the same type happens in the primal. And if two dual faces meet in an edge and a third vertex then in the primal we have three faces sharing the same two vertices and they can't all three contain the edge connecting these two vertices.</p>

CC BY-SA 3.0
2012-06-17T05:36:52.703Z
2012-06-17T07:28:09.630Z
2012-06-17T07:28:09.630Z
David Eppstein
<p>A polyhedral embedding of a graph on a surface is an embedding without edge crossings such that all the faces are bounded by simple cycles, and any two faces share a common vertex, share a common edge, or do not intersect at all. Moreover, all the faces are disks.</p>

<p>Given a graph $G$ with a polyhedral embedding on a surface $S$, one can define a dual graph $G'$ embedded in the same surface $S$, by creating a vertex in $G'$ for each face of $G$, and adding an edge between two vertices of $G'$ for each edge that the corresponding two faces in $G$ share.</p>

<p>I have a feeling that this will be simple, but I don't know how to prove it, so here it goes anyways. <strong>Question</strong>: The dual embedding that one obtains for $G'$ is also polyhedral?</p>

<co.combinatorics><topological-graph-theory>
Is the dual of a polyhedral embedding a polyhedral embedding?
a06e
# Is the dual of a polyhedral embedding a polyhedral embedding?

(Asked by: username_0 on 2012-06-17)

A polyhedral embedding of a graph on a surface is an embedding without edge crossings such that all the faces are bounded by simple cycles, and any two faces share a common vertex, share a common edge, or do not intersect at all. Moreover, all the faces are disks.

Given a graph $G$ with a polyhedral embedding on a surface $S$, one can define a dual graph $G'$ embedded in the same surface $S$, by creating a vertex in $G'$ for each face of $G$, and adding an edge between two vertices of $G'$ for each edge that the corresponding two faces in $G$ share.

I have a feeling that this will be simple, but I don't know how to prove it, so here it goes anyways. **Question**: The dual embedding that one obtains for $G'$ is also polyhedral?

**Tags:** co.combinatorics, topological-graph-theory


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## 1 answers

### Answer by username_1 on 2012-06-17. Score: 7 | Accepted answer

You need to specify that $G$ is a simple graph (else the trigonal [hosohedron](http://en.wikipedia.org/wiki/Hosohedron) is a counterexample) but then the answer is yes. What can go wrong? If a dual face is not a simple cycle then one of its edges is repeated, giving a self-loop in the primal, or a vertex but not an edge is repeated, in which case this face and vertex form a configuration of the same type in the primal. If two dual faces meet at more than one edge then there is a multiple adjacency in the primal. If the dual has a self-loop or multiple adjacency then we get a face with a repeated edge or two dual faces meeting at more than one edge in the primal. If two dual faces meet at two vertices but do not share an edge between these two vertices then a configuration of the same type happens in the primal. And if two dual faces meet in an edge and a third vertex then in the primal we have three faces sharing the same two vertices and they can't all three contain the edge connecting these two vertices.


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