Timeline for Finite vertex-transitive graphs that look like infinite vertex-transitive graphs
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Mar 30, 2013 at 21:22 | history | edited | HJRW | CC BY-SA 3.0 |
Rolled back after mistaken rewrite.
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| Mar 30, 2013 at 21:20 | comment | added | HJRW | R W - many thanks for pointing out my silly mistake. Of course you're right. I'm increasingly intrigued by this problem. Anyway, I'd better roll back my answer. | |
| Mar 30, 2013 at 20:17 | comment | added | R W | Oh - are you claiming that Schreier graphs are transitive? They are not! | |
| Mar 30, 2013 at 20:12 | comment | added | R W | There is something I don't understand in this argument. Any, say 4-regular graph, can be edge-labelled to become a Schreier graph of the free group $F_2$. Do you claim it's embeddable into a homogeneous graph? | |
| Mar 30, 2013 at 18:04 | history | edited | HJRW | CC BY-SA 3.0 |
Completely rewritten.
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| Mar 30, 2013 at 12:02 | comment | added | HJRW | Brendan, this is an answer to the question in the the final paragraph 'are there any other graphs G where such a family exists?'. (Note that the set of Cayley graphs of residually finite groups is much larger than the set of Cayley graphs of $\mathbb{Z}^d$.) Since the first question is stronger than the famous open problem 'Is every finitely generated group sophic' (as pointed out by R W), it may well be impossible to answer. | |
| Mar 30, 2013 at 9:34 | comment | added | Brendan McKay | I don't see an answer to the question here. I only see an argument that a particular way of deriving an answer from the group doesn't always work. | |
| Mar 30, 2013 at 8:59 | vote | accept | Jon Schneider | ||
| Mar 30, 2013 at 6:39 | history | edited | HJRW | CC BY-SA 3.0 |
Removed one implication.
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| Mar 30, 2013 at 6:30 | comment | added | HJRW | Sorry, Boris, of course you're right. I was implicitly assuming that the approximating graphs are Cayley graphs. | |
| Mar 29, 2013 at 21:28 | comment | added | Boris Bukh | I am sorry, but I remain unconvinced. We see the loops, but how do we write from knowing the loops the relations? The edges are not labelled with the names of generators. So, we do not know that two "parallel-looking" edges correspond to the same generator. | |
| Mar 29, 2013 at 21:14 | history | edited | HJRW | CC BY-SA 3.0 |
Added missing hypothesis.
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| Mar 29, 2013 at 20:59 | comment | added | HJRW | Not when $G$ is finitely presentable. If the ball $B_n$ is large enough that it contains loops corresponding to all relations of $G$, then any finite group with $B_n$ embedded in its Cayley graph is necessarily a quotient, since it satisfies all the relations of $G$. When $G$ is finitely generated but not finitely presentable, yes, it's conceivable. | |
| Mar 29, 2013 at 20:49 | comment | added | Boris Bukh | For the original question, isn't it conceivable that an infinite graph has symmetry group $G$, but for every $d$ there is a group $G_d$ and a set of generators such that the Cayley graph looks like $G$ in a neighborhood of size $d$, but these groups $G_d$ have nothing whatsoever to do with the original $G$? | |
| Mar 29, 2013 at 20:17 | history | answered | HJRW | CC BY-SA 3.0 |