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Find a 2-connected planar graph whose drawings are all topologically isomorphic, but whose planar embeddings are not all equivalent.

I think $K_{2,3}$ might be an example, but I'm not sure how to show this at all. Anything would be welcome.

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This sounds like homework, so I will give you a hint. The graph you are looking for must be 2-connected but not 3-connected. Thus, it must have two vertices whose removal disconnects the graph. Try drawing different graphs like this -- you will need something more complicated than $K_{2,3}$ to get the job done. –  Jim Belk Feb 8 '11 at 1:37
    
Like K_{2,4} or something? –  user6584 Feb 8 '11 at 9:11

2 Answers 2

The planar 3-connected graphs, which are the vertex-edge graphs of 3-dimensional convex polyhedra have essentially only one way to be embedded in the plane, by a theorem of Hassler Whitney. Though such embeddings may look different because the number of sides of the infinite face in different drawings may differ, the different embeddings are isomorphic. What can be accomplished using two vertices joined by paths of different lengths when the graph might be embedded in the plane in different ways?

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I think this might be K_{2,4} and I'm more or less sure. If anyone could confirm or say whether this is correct or not, would be nice. Just see that there's an automorphism that doesn't translate to a topological isomorphism.

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