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I am reading from the paper here. Let $X$ be a graph and $G$ a finite group which acts on it (as per the definition in the paper, the group action is by default faithful). Now, we say that an edge $e=\pmatrix{d_1 & d_2}$ (where $d_i$ denote darts) is invertible by $G$ if there is some $g\in G$ such that $g(d_1)=d_2$ and $g(d_2)=d_1$. Suppose no edge is invertible by $G$. On page 3 of the paper, the author then says that the factor graph $X/G$ is well defined (the first line of the proof of Theorem 3). Why is well definiteness requiring the absence of invertible edges is my question.

The factor graph, according to my understanding, consists of vertices given by the orbits of the vertices of $X$ (i.e. all the $[v]=\{g\cdot v:g\in G\}$) and the edges given by the orbits of the edges of $X$ (i.e. all the $[\pmatrix{d_1 & d_2}]=\{g\cdot \pmatrix{d_1 & d_2}:g\in G\}$) where two vertices $[u],[v]$ are incident on an edge $[\pmatrix{d_1 & d_2}]$ if there exists $x\in [u],y\in[v],\pmatrix{d_i & d_j}\in[\pmatrix{d_1 & d_2}]$ such that $u,v$ are incident on $\pmatrix{d_i & d_j}$. This is not the exact definition given in the paper but I am convinced that this is what the author really means.

Such a factor graph will be defined even if there are invertible edges, then why the emphasis on non invertiblity?

  • One possibility is that you want the quotient map $q : X \to X/G$, defined by $q(v) = [v]$, to be a graph homomorphism. A graph homomorphism must map adjacent vertices to adjacent (and thus distinct!) vertices - which is impossible if two adjacent vertices belong to the same orbit. – Brandon du Preez Sep 29 '22 at 11:39
  • Invertibility is a stronger condition then vertices being in the same orbit –  Sep 29 '22 at 12:16
  • ok, now I think I got it. The problem here is in the inverse operation of edges. For non-invertible edges, you get that the quotient class of an edge and its inverse are distinct. For an invertible edge, instead, the two class are the same, so the quotient graph would present a tail, not an edge anymore – Exodd Sep 29 '22 at 12:18
  • Notice that a tail is NOT a loop, since even loops have distinct inverse loops – Exodd Sep 29 '22 at 12:20

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