Timeline for Finite group representation as $\mathrm{Aut}(\Gamma)$ action $H^1(\Gamma,\mathbb{Z})$ of graph?
Current License: CC BY-SA 4.0
11 events
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| Jul 20, 2018 at 17:46 | vote | accept | CommunityBot | ||
| Jul 14, 2018 at 21:42 | comment | added | YCor | @LeeMosher Oops, I had a mistaken argument in mind. Indeed, the graph consisting of 2 vertices with 2 edges joining them (or its barycentric subdivision if one wants to forbid multi-edges) has an automorphism group with elements of order 3. Thanks for the correction and mentioning this theorem. | |
| Jul 14, 2018 at 14:06 | comment | added | Lee Mosher | @YCor: It's a theorem, proved independently by Culler, by Zimmerman, and by Khramstov, that every finite subgroup of $\text{Out}(F_n)$ can be realized by an action on a graph. For example the order 3 cyclic group acts on the rank 2 theta graph. | |
| Jul 14, 2018 at 9:06 | comment | added | YCor | However, for $n=2$ it seems that the automorphism group of a leafless finite connected graph with fundamental group $F_2$ can only be of order 1, 2, 4 or 8, and hence we miss cyclic subgroups of order 3 (adding leaves can increase the order of the automorphism group but will not change the action on homology). In general, this suggests that "most" finite subgroups of $\mathrm{Out}(F_n)$ can't be realized by an action on a graph. | |
| Jul 14, 2018 at 8:56 | comment | added | YCor | More precisely the linked paper gives finite cyclic subgroups which do not lift, for every $n\ge 3$. For $n\le 2$ the quotient homomorphism is an isomorphism and hence lifting holds. | |
| Jul 13, 2018 at 19:01 | comment | added | Gregory Arone | Anyhow, I edited. | |
| Jul 13, 2018 at 19:00 | history | edited | Gregory Arone | CC BY-SA 4.0 |
added 9 characters in body
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| Jul 13, 2018 at 18:59 | comment | added | Lee Mosher | I'll just throw in that this terminology, however imprecise, is nonetheless pretty well established in group theory. | |
| Jul 13, 2018 at 18:56 | comment | added | YCor | What you mean is correct; it's just that it sounded weird to me to call this an "action by outer automorphisms" (note that what you say in your comment does not make use of the group structure of $G$). Possibly by outer action of $G$ on a group $H$ one can just mean a homomorphism $G\to\mathrm{Out}(H)$. | |
| Jul 13, 2018 at 18:48 | comment | added | YCor | How do you define "an action by outer automorphisms" for a group action on another group? | |
| Jul 13, 2018 at 18:22 | history | answered | Gregory Arone | CC BY-SA 4.0 |