Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the study of groups acting on graphs, one often starts with hypothesis "Let $X$ be a graph, and $Aut(X)$ acting on $X$ without inversion.."

But, is it true that $Aut(X)$ always acts on $X$ without inversion?

Here $Aut(X)$ is the set of bijections between all vertices of $X$ which preserve adjcency, and a group $G$ acts without inversion means there is no oriented edge $e$ of $X$ such that $g.e = \bar{e}$, $g\in G$; $\bar{e}$ is the edge $e$ with reverse orientation.

Some people say that, "if $G$ is acting on $X$ with inversion, then taking barycentric subdivision, we will have action of group $G$ without inversion.."

But, by taking barycentric division of $X$, is it not true that its automorphism group may also change?

What would be the precise way to consider the action of full automorphism group on the graph and study its quotient graph?

share|cite|improve this question
Yes, the automorphism group of the barycentric subdivision $X'$ of $X$ may be larger, but the automorphism group of $X$ will embed in an obvious way. So if $G$ acts on $X$ then we have a homomorphism $G \to Aut(X)$ and composing this with the homomorphism $Aut(X) \hookrightarrow Aut(X')$ we get an action of $G$ on $X'$ and you should check that $G$ acts without edge inversion on $X'$. I'm pretty sure that this appeared first in Serre's lectures Arbres, amalgames et $SL_2$ (translated to English as Trees). – t.b. Jul 18 '11 at 9:04
The point is of course that $Aut(X)$ acts on $X'$ without edge inversion. – t.b. Jul 18 '11 at 9:20
By the way: you mean "there is no oriented edge $e$ of $X$ such that $g.e = \bar{e}$". For an example in which you get a larger automorphism group by barycentric subdivision, take a cyclic graph. – t.b. Jul 18 '11 at 9:23
@Theo: it is the meaning of action without inversion; the condition should be for all elements of group. – William Serre Jul 18 '11 at 9:25
Taking $X$ to be the connected tree on two vertices shows that $\operatorname{Aut}(X)$ does not necesarily act without inversions... I guess you knew this? – Mariano Suárez-Alvarez Jul 19 '11 at 6:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.