Timeline for Does this poset have a unique minimal element?
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| S Apr 7, 2019 at 13:32 | history | suggested | Ilmari Karonen | CC BY-SA 4.0 |
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| Apr 7, 2019 at 13:06 | review | Suggested edits | |||
| S Apr 7, 2019 at 13:32 | |||||
| Feb 11, 2013 at 15:56 | comment | added | Michał Kukieła | @Tom: unique minimal element is NOT the same as the least element (consider integers with standard order and one additional element A with A<0 and no other comparabilities; A is the unique minimal element, but there is no least one). Nevertheless, I agree that in some cases it seems somewhat strange to talk about "the unique minimal element", when "the least element" does the job. | |
| Jan 20, 2013 at 22:16 | comment | added | ARupinski | There also seem to be examples with infinite degree vertices, so perhaps total classification might be out of reach, but classification of those with only finite degree vertices might be doable. I am definitely going to think about this some more. | |
| Jan 20, 2013 at 22:15 | comment | added | ARupinski | @Pietro: I was actually thinking about that exact thing a few days ago. I was able to come up with some infinite families, but I didn't think about it enough to come up with a conjecture as to what a complete list would look like. The first infinite family I was able to come up with consists of taking the doubly infinite path and attaching 3 single leafs to nodes which are not evenly spaced; the other families I had thought of were variations of this with multiple infinite branches. | |
| Jan 20, 2013 at 16:23 | comment | added | Pietro Majer | The problem being solved by now, I wonder if there is a characterization of mimimal elements for analogous sets of possibly infinite automorphism free trees (e.g. assuming finite branching, or not). Clearly, there are new minimal elements. For instance, think of a tree consisting of a doubly infinite path indexed by $\mathbb{Z}$ ( $x$ and $y$ are adjacent if and only $|x-y|=1$), plus another single leaf, and a two-steps path, attached by their endpoints at some different points in $\mathbb{Z}$. | |
| Jan 20, 2013 at 1:00 | vote | accept | ARupinski | ||
| Jan 20, 2013 at 1:00 | vote | accept | ARupinski | ||
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| Jan 20, 2013 at 0:54 | vote | accept | ARupinski | ||
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| Jan 16, 2013 at 0:21 | comment | added | ARupinski | But for what its worth, I agree that for local rings you are probably right as far as largest would suffice (if people actually thought about what 'largest' is constrained to mean in context). | |
| Jan 16, 2013 at 0:17 | comment | added | ARupinski | For other instances such as local rings, I might venture to guess that the usage might have something to do with largest not always being equivalent to unique maximal. For example, a given group may have several 'largest' proper subgroups all of the same size (and so not unique, think $G = (\mathbb{Z}/2\mathbb{Z})^n$ with $n\geq 2$) and likewise it may have maximal subgroups which are not largest in size among all subgroups (for example the Monster has a maximal subgroup with only 1640 elements, far smaller than many of its non-maximal subgroups). | |
| Jan 16, 2013 at 0:11 | comment | added | ARupinski | @Tom: interesting meta-question. At least in this case I suppose describing it this way I meant to make the focus of the present question whether colloquially 'all roads lead to Rome' or, translated into this problem, 'all paths in $\mathcal{AFT}$ lead to $E_7$'. | |
| Jan 15, 2013 at 23:54 | comment | added | Tom Leinster | Kind of off topic, but: why do people so often say "unique minimal element" when "least element" is shorter and (I would say) more vivid? They're the same, at least under the axiom of choice. But undeniably, the definition of local ring is phrased in terms of a "unique maximal (proper) ideal" waaay more often than "largest (proper) ideal". Why? | |
| Jan 14, 2013 at 21:43 | answer | added | Ilhee Kim | timeline score: 8 | |
| Dec 28, 2012 at 2:43 | vote | accept | ARupinski | ||
| Jan 5, 2013 at 19:56 | |||||
| Dec 27, 2012 at 0:11 | comment | added | ARupinski | We conclude that some branch of $N$ must have a sub-path from $N$ of length at least 3. Again by considering automorphisms fixing $N$, some other branch must have a sub-path from $N$ of length at least 2 (otherwise all other branches from $N$ have length 1 leading to an automorphism of the tree), and finally since every branch has a sub-path from $N$ of length at least 1, we find that $E_7$ is a subgraph of every element of $\mathcal{AFT}$. | |
| Dec 27, 2012 at 0:08 | comment | added | ARupinski | @Per: unfortunately, this set is empty. Any element of $\mathcal{AFT}$ contains a node $N$ of degree at least 3. From $N$, consider all branches emanating from it; if each has maximal length 2 to any leaf then it is a star graph (a hub node with $k$ spokes for some $k\geq 1$) with one of its spokes' terminus equal to $N$. Clearly any star branch with 3 or more spokes will admit an automorphism fixing the hub; but there are only 2 possible star branches with fewer than 3 spokes; thus any choice of these for the $\geq 3$ branches of $N$ leads to an automorphism of the tree fixing $N$. | |
| Dec 25, 2012 at 18:15 | answer | added | Pietro Majer | timeline score: 7 | |
| Dec 25, 2012 at 16:15 | comment | added | Dima Pasechnik | Computer says that, up to 22 vertices, there is no other minimal element in AFT. There is apparently a well-established counting "mechanics" for AFT, see oeis.org/A000220 Perhaps an answer to Q3 can be dug up there. | |
| Dec 25, 2012 at 15:24 | comment | added | Per Alexandersson | Consider all trees which do not contain $E_7$ as sub-tree. This set should not be too large (finite), or am I too quick? If you can show that there is not another minimal element among these, then maybe this leads to something? | |
| Dec 25, 2012 at 2:58 | comment | added | Daniel Litt | Just wanted to say--this is a great question. | |
| Dec 24, 2012 at 18:40 | history | edited | ARupinski | CC BY-SA 3.0 |
updated definition
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| Dec 24, 2012 at 18:39 | comment | added | ARupinski | @Joel: Point taken on these two smaller trees. I exclude them mainly because I am considering this poset in the context of deleting leaves; since the sub-poset consisting of the empty tree and the 1-vertex tree is totally disconnected from the larger trees it isn't very interesting in terms of deletion sequences. I will update the formulation of $\mathcal{AFT}$ to reflect this. | |
| Dec 24, 2012 at 17:08 | comment | added | Gerhard Paseman | Show that every large member has E_7 as a subtree, and then show that every large member is not minimal. Large can probably be taken to mean 10 or more vertices. Finish up with a detailed examination of not large members. Gerhard "Sketches Of Proof Sketches Sketched" Paseman, 2012.12.24 | |
| Dec 24, 2012 at 17:08 | comment | added | Joel David Hamkins | Why don't you regard the one-point tree in AFT? Or the empty tree? | |
| Dec 24, 2012 at 16:35 | history | asked | ARupinski | CC BY-SA 3.0 |