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I don't know how one can tackle the following kind of question, so any hint is welcome. I formulate a precise question in order to fix ideas, but it is to be considered as an example out of a more general class.

Example: Can one embed Petersen's graph in $\mathbb R^4$ in such a way that all edges are mapped to segments (of not necessarily equal lenght) and each pair of adjacent segments forms an angle of $2\pi/3$ ?

The question can be generalized in the following way, to give a wider class of questions:

  • instead of Petersen's graph, consider a given $k$-regular graph (we had $k=3$ above)
  • instead of $\mathbb R^4$ consider $\mathbb R^n, n>k$, or embeddings into the round sphere $S^n$ where the edges of the graph are mapped into geodesics
  • ask that the angles formed between any pair of adjacent edges are equal to the angle $VOV'$ made at which two vertices $V,V'$ of a regular euclidean $(k-1)$-simplex are seen from its barycenter $O$

I think that $1$-skeletons of spherical regular polytopes give such kind of graphs. I am interested in any partial answer which introduces essentially more ingredients than just constructions based on symmetry groups, or in obstructions which show negative answers.

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Assume $\Gamma$ is a 3-regular graph in $\mathbb{R}^n$ and the angles at each vertex are $\tfrac23\pi$. Then it is easy to see that the distance to the origin can not have a local maximum on $\Gamma$.

Therefore $\Gamma$ has to be infinite; in particular it can not be Petersen's graph.

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  • $\begingroup$ Thanks, I didn't notice this criterion. If there will not be more answers (for example on $S^n$ or for other $k$) within 2 weeks, I will accept this one. $\endgroup$
    – Mircea
    Commented Jul 9, 2012 at 14:33
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    $\begingroup$ Yet an other reason: the Petersen's graph has cycles of length 5; any 5-gon in $\mathbb R^k$ has sum of angles at most $3\cdot\pi$, so at least one of the angles has to be $\le \tfrac35\cdot\pi<\tfrac23\cdot\pi$. $\endgroup$
    – Anton Petrunin
    Commented Jul 9, 2012 at 17:11

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