If $S $ is infinite, locally finite graph which is not tree, $\tilde{S} $ is its universal cover, $p:\tilde{S}\rightarrow S $ is covering map,and $G $ is acting on $\tilde{S} $ with finite quotient $Y $, $q:\tilde{S}\rightarrow Y $ is the corresponding covering, can we find a group $H $ which acts on $S $ with quotient $Y$ such that the diagram commutes (i.e. $r\circ p=q $ where $r:S\rightarrow Y=S/H $ is natural map)?
Group acting on graph
1
$\begingroup$
group-theory
geometric-group-theory
-
1I don't understand where the "natural map" $r: S \to Y$ should come from without further assumptions. You may want to look at Bass's [Covering theory for graphs of groups](http://dx.doi.org/10.1016/0022-4049(93)90085-8). After all, $\tilde{S}$ *is* a locally finite tree. – 2011-08-29
-
0I'm mentioning this since this question [reminds me of this question on MO](http://mathoverflow.net/questions/73942/groups-acting-on-graph). For a good and elementary introduction to these ideas look at the appendix to Chapter 1 of [Hatcher's book](http://www.math.cornell.edu/~hatcher/AT/ATpage.html). Also, [the book by Dicks and Dunwoody](http://books.google.com/books?id=4sPiqRfOxD0C) might be a place to look at. – 2011-08-29
-
0@Theo: Here natural map $r:S\rightarrow Y=S/H$ is the quotient map by action of $H$. The question is whether there exist $H$ acting on $S$ with quotient map $r:S\rightarrow Y=S/H$ with diagram commutative. I am looking the ref. in comment 1; but couldn't handle the question. – 2011-08-29
-
0I understand that that's what you want. But why should $S$ even be a cover of $Y$ (you're asking for much more)? – 2011-08-29
-
0@Theo: Here I have assumed $S$ is infinite graph which is not tree and $Y$ is a ***finite*** quotient of $\tilde{S}$. therefore, I was expecting whether there is such map from $S$ to $Y$. – 2011-08-29