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Consider the vertices $v_i$ of a subdivided icosahedron $J$. In my case, each vertex $v_i$ has an ordered tuple of nodes denoting the edges of $J$ starting in $v_i$. All vertices have 6 edges, except for the vertices of the 'original' icosahedron which only have 5. This could be described by a map $N:J \rightarrow J^6 \cup J^5, v \mapsto (n_i)_i$.

Is it possible to define an $N$ such that $N(g(v)) = (g(n_i))_i$ for all rotations $g$ of the icosahedral symmetry group and for all $v \in J$? (simply put: is it possible to make the link relationships "look the same" for all vertices of the original icosahedron?)

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  • $\begingroup$ Do I understand this correctly: it is clear which vertices are contained in $N(v)$, just not the order in which they do? $\endgroup$
    – M. Winter
    Commented Jun 6, 2020 at 15:11
  • $\begingroup$ I would say this is impossible, because there are symmetries of $J$ that fix a vertex, but not its neighbors. $\endgroup$
    – M. Winter
    Commented Jun 6, 2020 at 15:14
  • $\begingroup$ @m-winter Yes the vertices in $N(v)$ are fixed but the order is not. Regarding your second comment: do you mean rotations about an axis going through the vertex? $\endgroup$
    – user1479670
    Commented Jun 8, 2020 at 7:20
  • $\begingroup$ Yes; that is one example of such a symmetry. $\endgroup$
    – M. Winter
    Commented Jun 8, 2020 at 7:52

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