Talk:Gain graph

Group action?

Does a gain graph satisfy at least some of the axioms of a group action? So, for example, if g,h,k are group elements, with k=gh, and if v a vertex, then, if the edge kv exists, and the edge hv exists and if the edge g(hv) exists, then will kv be the same vertex as g(hv) ? If so, then it seems to me that a gain graph is like a group action

$G\times V\to V\,$

except that it is only a partial function on the set of pairs $G\times V$. Is that correct?

If this holds, it seems like one could legitimately call a grain graph a "partial group action", right?

If this does not hold, then it is not at all clear (to me) why one is justified in calling the edge labels "group elements"; one could just as easily call then "foobar elements", and that would be a meaningless label unless the axioms of a foobar were to be made to hold ... linas (talk) 21:23, 24 March 2008 (UTC)

The group elements don't label vertices, they label edges. Thus, there's no such edge as kv. An edge has two end vertices, so you might describe an edge as uv, where u and v are vertices. This edge will have a gain (group element), say g. However, uv isn't the best notation for an edge because there could be many edges uv, each with a different gain. I like the notation e:uv which distinguishes e from another edge f:uv with the same endpoints.

The multiplication takes place between consecutive edges in a path (or walk) in the graph. Suppose you have an edge uv with gain g, and an edge vw with gain h. Then the path uvw composed of those two edges has gain gh=k. I hope this answers your question.

Obviously, the article needs a diagram or two, but I don't have time to make any now. It probably also needs an example explained in detail. Zaslav (talk) 06:08, 25 March 2008 (UTC)

Ahh, I think the intent of kv was that v was the initial vertex, and k was the group elt starting at v, which would be appropriate if one was thinking of group actions. The appropriate response would be: there is at most one edge labelled by group elt k, and so there is at most one v that is the initial vertex of k. There is no other vertex w with kw, unless v=w. 67.198.37.16 (talk) 16:59, 16 August 2016 (UTC)

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