Timeline for Does this poset have a unique minimal element?
Current License: CC BY-SA 3.0
26 events
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| Jan 20, 2013 at 10:29 | comment | added | Pietro Majer | Dear Andrew, don't worry, I especially like the cute green color of accepted answers, either mine or other people's! You did the right thing. | |
| Jan 20, 2013 at 1:01 | comment | added | ARupinski | On reading through, everything seems to make sense with your fix in place. I wish it were possible to accept two answers in situations like this where two posters both come up with beautiful answers. | |
| Jan 20, 2013 at 1:00 | vote | accept | ARupinski | ||
| Jan 20, 2013 at 1:00 | |||||
| Jan 17, 2013 at 22:47 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Jan 16, 2013 at 3:19 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Jan 16, 2013 at 2:13 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Jan 15, 2013 at 23:22 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Jan 15, 2013 at 21:40 | comment | added | Pietro Majer | I happened to spare a whole day in the weekend, and I get back to your problem, producing the above proof. | |
| Jan 15, 2013 at 21:25 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Jan 8, 2013 at 17:09 | comment | added | Pietro Majer | Thank you, then if you think it may be useful I'll leave it. While there isn't yet an accepted answer, there are more chances to get another. | |
| Jan 5, 2013 at 19:56 | comment | added | ARupinski | @Pietro: I will unaccept it if you really prefer, but I think the ideas are nevertheless useful for getting to a final proof or at least avoiding some cases. In particular, as you note, for any minimal element, removal of a leaf leads to a unique non-trivial automorphism of order 2 on the resulting sub-tree has led me to start thinking about trees with this latter property and how one might do something with them to build up to a proof of the original conjecture. | |
| Dec 30, 2012 at 1:58 | comment | added | Pietro Majer | @ARupinski: if you un-accept my answer I will be able to delete it. | |
| Dec 30, 2012 at 1:53 | comment | added | Pietro Majer | Well, this evening I tried to combine the preceding lemmas into a real proof. But I always meet some case they do not cover. They may be useful to get a proof, or even to find a counterexample. In short, the situation is conveniently depicted by withdrawing the preceding answer. My excuses, gentlemen. | |
| Dec 29, 2012 at 10:11 | comment | added | Pietro Majer | Brendan, I understand that my rough hints are sufficient for ARupinski to get a complete proof, but certainly what I wrote is far from a proof. As I see it now, there should be no surprise, but who knows, so I'd like to fix everything before I eventually leave the thing. | |
| Dec 29, 2012 at 1:31 | comment | added | Brendan McKay | I'm confused. Is the proof now complete? | |
| Dec 28, 2012 at 9:06 | comment | added | Pietro Majer | A more precise statement: Assume (S,o) is a rooted tree such that, removing any of its leaves, one always get isomorphic r-trees. Then, the degree of any vertex only depends on its height. (e.g. grandgrandfather Adam has 3 sons; each of Adam's son has 7 sons; each of 21 Adam's grandsons has 5 sons and so on). | |
| Dec 28, 2012 at 2:43 | vote | accept | ARupinski | ||
| Jan 5, 2013 at 19:56 | |||||
| Dec 27, 2012 at 7:57 | comment | added | Pietro Majer | Thank you... maybe the above point 2 is sufficient to conclude. The relevant fact should be: (S,o) is a rooted tree such that, removing any of its leaves, one always get isomorphic r. trees. This shold imply : if (S,o) had more than one leaf, then it has a nontrivial automorphism, which can be extended to the whole T, a contradiction, so one should conclude that S has only one leaf, that immediately leads to $E_ 7$ | |
| Dec 26, 2012 at 23:57 | comment | added | ARupinski | @Pietro: thanks for the answer... I am digesting the details now, but so far everything seems to fit nicely. | |
| Dec 26, 2012 at 16:24 | comment | added | Pietro Majer | An useful lemma should be the following. If a tree S has a vertex y which is fixed by no nontrivial automorphism of S, then S has at most one nontrivial automorphism (which is necessarily an involution). It holds both in the rooted and unrooted case. Do you agree? The proof is not difficult (in case I'll write it down). This applies to S:=T\{x}, T :=a minimal tree (either in the rooted or unrooted case) and x := a leaf of T; y:= the father of the leaf x. | |
| Dec 26, 2012 at 11:25 | comment | added | Pietro Majer | Yes, sorry for the mess. The idea is that for a minimal object T (both in the case of AFT and of AFRT) and a leaf x, any nontrivial automorphism ϕ of T\{x} must have some special properties, otherwise one can use it to make a nontrivial automorphism ψ of T. For instance, it can't fix the father of x, and can't have certain invariant sets. Also, ϕ is unique, and it is an involution. I'll try to state these small bricks more clearly. | |
| Dec 26, 2012 at 5:19 | comment | added | Dima Pasechnik | IMHO the claim of 1 looks very plausible. So it's matter of fixing the proof there. In 2, the reduction to an involutory automorphism of T\{x} looks incomplete - what's wrong about freezing a part of T? Otherwise, very interesting! | |
| Dec 25, 2012 at 20:51 | comment | added | Pietro Majer | warning: there is a gap in part 1, since T\C may have non trivial automorphisms, so that it may fail to be in RATF. I think the proof could be fixed, so I will not delete it for the moment. | |
| Dec 25, 2012 at 20:48 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Dec 25, 2012 at 18:22 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Dec 25, 2012 at 18:15 | history | answered | Pietro Majer | CC BY-SA 3.0 |