Timeline for Does every finite poset have a rigid endomorphism?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 21, 2020 at 16:17 | comment | added | lambda | @Pierre-YvesGaillard The boolean lattice $B_n$ consists of subsets of $\{1, \dots, n\}$ and its automorphism group is $S_n$ acting in the obvious way. A maximal chain looks like $\{X_0, X_1, \dots, X_n\}$ where $X_{i+1}$ is obtained by adding a single element to $X_i$. Any element of $S_n$ that fixes both $X_i$ and $X_{i+1}$ must therefore fix this new element, so an automorphism that fixes the whole chain is the identity. | |
| Oct 21, 2020 at 15:49 | comment | added | Pierre-Yves Gaillard | @lambda - Thanks a lot! Looks very interesting! If I understand correctly you're saying that if $C$ is a maximal chain in a boolean lattice, then $C$ isn't fixed by any nontrivial automorphism. Could you give me an argument or a reference? | |
| Oct 21, 2020 at 14:24 | comment | added | lambda | To pedantically correct myself, I meant any nonempty chain. | |
| Oct 21, 2020 at 14:09 | comment | added | lambda | Any chain is the image of an endomorphism, so if you can find a chain which isn't fixed by any nontrivial automorphism you're done. (I have no intuition for whether this should be expected to exist in general, but many highly symmetric posets I can think of have this property, eg. any maximal chain in a boolean lattice.) | |
| Oct 21, 2020 at 11:57 | history | edited | Pierre-Yves Gaillard | CC BY-SA 4.0 |
edit clearly indicated
|
| Oct 11, 2020 at 14:02 | comment | added | Pierre-Yves Gaillard | @SamHopkins- Thanks! No I don't have a counterexample for infinite posets, but I'm very interested in the infinite case as well. I thought it was natural to concentrate first on the finite case. | |
| Oct 11, 2020 at 13:49 | comment | added | YCor | Note that the question is whether the monoid $M$ of endomorphisms of every finite poset satisfies the purely monoid-wise property: $$\exists g\in M:\forall h,k\in M: (hk=kh=1 \text{ and } hg=gh) \Rightarrow h=1.$$ | |
| Oct 11, 2020 at 13:47 | comment | added | Sam Hopkins | This looks like a very interesting question (although the wall of text is a bit intimidating- I wish MO had a way of putting stuff inside spoiler tags). But one naive question: do you have a counter-example for infinite posets (where I think everything still makes perfect sense)? | |
| Oct 11, 2020 at 13:44 | comment | added | YCor | related (while distinct) to mathoverflow.net/questions/358057/…, mathoverflow.net/questions/359660/… | |
| Oct 11, 2020 at 13:44 | history | edited | YCor |
edited tags
|
|
| Oct 11, 2020 at 13:00 | history | asked | Pierre-Yves Gaillard | CC BY-SA 4.0 |