Timeline for A generously vertex transitive graph which is not Cayley?
Current License: CC BY-SA 4.0
21 events
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| Nov 28, 2019 at 7:49 | comment | added | YCor | @user334639 Oh, indeed my comment I forgot to write what I meant... the claim is that the group $G$ (which is a locally compact topological group under the pointwise convergence topology) is unimodular. And it also works for the closure in $G$ of every generously vertex-transitive subgroup. | |
| Nov 28, 2019 at 1:12 | comment | added | user334639 | @YCor Thanks for the comment and for editing the question! I'm not quite sure what you are trying to prove in your comment. Are you justifying that GVT implies unimodular? | |
| Nov 26, 2019 at 9:49 | history | edited | YCor |
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| Nov 26, 2019 at 9:20 | comment | added | YCor | Remark: if $X$ is such a graph and $G$ is its automorphism group (with a fixed left Haar measure), any two vertex stabilizers are conjugate by some element $g$ whose square fixes a vertex. In particular, $g$ belongs to a compact subgroup and hence $\Delta(g)=1$; thus all vertex stabilizers have the same Haar measure. Thus $G$ is unimodular. (While there exist many vertex-transitive finite degree connected graph with non-unimodular automorphism group.) | |
| Nov 26, 2019 at 8:10 | history | edited | YCor | CC BY-SA 4.0 |
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| Nov 26, 2019 at 8:06 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
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| Jul 31, 2015 at 21:31 | vote | accept | user334639 | ||
| Jul 31, 2015 at 17:43 | answer | added | Adam P. Goucher | timeline score: 5 | |
| Jul 31, 2015 at 14:11 | history | edited | user334639 | CC BY-SA 3.0 |
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| Jul 31, 2015 at 13:52 | history | edited | user334639 | CC BY-SA 3.0 |
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| Jul 31, 2015 at 4:50 | answer | added | Igor Rivin | timeline score: 3 | |
| Jul 31, 2015 at 3:33 | comment | added | user334639 | Thank you all for feedback. Question has been edited to remove some of these obscurities. | |
| Jul 31, 2015 at 3:31 | history | edited | user334639 | CC BY-SA 3.0 |
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| Jul 30, 2015 at 9:59 | comment | added | nvcleemp | I think the OP means "For the last fact..." | |
| Jul 30, 2015 at 1:39 | comment | added | Gerry Myerson | "For the last claim...." But no claims are made. What do you mean? | |
| Jul 29, 2015 at 22:27 | comment | added | Chris Godsil | A permutation group $G$ on a set $V$ is generously transitive if, for each pair of points from $F$, there is an element of $G$ that swaps them. The literature I am aware of focusses on the finite case. | |
| Jul 29, 2015 at 18:39 | comment | added | Dave Witte Morris | Perhaps you want to assume the graph has finite valence? Otherwise, the first question is answered by the complement of @ErikRijcken's example. | |
| Jul 29, 2015 at 14:05 | history | edited | user334639 | CC BY-SA 3.0 |
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| Jul 29, 2015 at 13:49 | comment | added | Erik Rijcken | Do you assume your graph to be connected? Otherwise, to answer your first question, you could take infinitely many copies of the Petersen graph. | |
| Jul 29, 2015 at 13:03 | review | First posts | |||
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| Jul 29, 2015 at 12:59 | history | asked | user334639 | CC BY-SA 3.0 |