3
$\begingroup$

I saw this question recently, which asks for a "geometric" example where a certain automorphism doesn't exist. Since some counterexamples are well known, I thought it should be a simple task to convert a counterexample into a "geometric" version by looking at cayley graphs.

After thinking about it for a little while, I wasn't able to come up with a relationship between the cayley graph $\Gamma(G)$ and the automorphism group $\text{Aut}(G)$. Since $G$ acts naturally on $\Gamma(G)$ and $\text{Aut}(G)$ acts naturally on $G$, I (perhaps naively) thought that there should be a kind of "higher order" relationship between $\Gamma(G)$ and $\text{Aut}(G)$. After all, once we fix an "origin" in $\Gamma(G)$, we can identify the vertices with $G$.

It seems there is some precedent for special cases of this question (here and here), but I can't seem to find any more general descriptions.

The question, then:

Is there a relationship between $\text{Aut}(G)$ and $\Gamma(G)$? Is there a more natural geometric object that $\text{Aut}(G)$ acts on than the obvious $\Gamma(\text{Aut}(G))$? I would be interested more broadly in ways of understanding $\text{Aut}(\text{Sym}(X))$ in terms of $X$ for any object $X$ in some category, not necessarily Cayley Graphs.

Thanks in advance!

$\endgroup$
6
  • $\begingroup$ You seem to assume that "the automorphism group acts on the Cayley graph" which is wrong. The error is helped by using the notation $\Gamma(G)$ and call it "the" Cayley graph. The Cayley graph depends on a choice of generating subset $S$ of $G$ and should be denoted $\Gamma(G,S)$. It admits no action of $\mathrm{Aut}(G)$ in general (only if $S$ is invariant under automorphisms, which is quite rare) . $\endgroup$
    – YCor
    Oct 14, 2020 at 8:35
  • $\begingroup$ I actually don't think that $\text{Aut}(G)$ acts in any obvious way on any Cayley graph. That's a big part of this question. And yes, a Cayley graph depends on a choice of generators, but this seems to be a common abuse of language. (though I admit, since I'm not necessarily interested in coarse geometry, it becomes much more problematic and I should have been more specific). The question is "what CAN we say about the automorphisms of $G$ in relation to things that $G$ acts on?" $\endgroup$ Oct 14, 2020 at 8:44
  • 1
    $\begingroup$ It does, as I said, when the subset is invariant under automorphisms. For instance, if $X$ is a set of cardinal $\neq 6$, it can be shown that the set of transpositions in $S(X)$ is automorphism-invariant. Hence the corresponding Cayley graph inherits an $\mathrm{Aut}(S(X))$-action. This can be used to eventually show that every automorphism of $S(X)$ is inner. $\endgroup$
    – YCor
    Oct 14, 2020 at 8:47
  • 1
    $\begingroup$ Concerning your question, it's typically non-trivial and well-studied for something such as the automorphism group of a free group. There's especially a big literature about spaces with $\mathrm{Out}(F_n)$-actions (outer space, Culler–Vogtmann), the case of $\mathrm{Aut}$ is possibly treated somewhere too. Also for surface groups $\mathrm{Out}$ is mapping class group and acts on Teichmüller space; I'm not sure about $\mathrm{Aut}$. For $\mathrm{GL}_n(\mathbf{Z})$ there a whole theory of cutting cusps in symmetric spaces. $\endgroup$
    – YCor
    Oct 14, 2020 at 8:56
  • 1
    $\begingroup$ To get an action of $Aut(F_n)$ you can see auter-space for example in arxiv.org/pdf/1610.08545.pdf $\endgroup$
    – 1123581321
    Oct 14, 2020 at 9:02

0

You must log in to answer this question.