Timeline for Complete atomless Boolean algebras with abelian automorphism group
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 26, 2018 at 12:33 | comment | added | YCor | Indeed for a few seconds I first expected that the automorphism group could be cyclic, but the topological intuition makes it clear that it's much larger. | |
| Aug 26, 2018 at 8:59 | answer | added | YCor | timeline score: 4 | |
| Aug 25, 2018 at 23:47 | vote | accept | Iian Smythe | ||
| Aug 25, 2018 at 23:45 | comment | added | Iian Smythe | YCor, Oops, I should read comments with a bit more care. For some reason I read your comment as suggesting the automorphism group was cyclic, which of course you are not. | |
| Aug 25, 2018 at 23:15 | answer | added | Don Monk | timeline score: 8 | |
| Aug 25, 2018 at 22:15 | comment | added | Iian Smythe | I should mention, de Groot (in the same paper I mentioned in my last comment), gives an example of a 0-dim, I believe non-compact, subset of $\mathbb{R}$, with non-trivial homeomorphism group where every element is order two (and hence, is abelian). Maybe this can be adapted to what I want by taking the Cech-Stone compactification? | |
| Aug 25, 2018 at 21:08 | comment | added | YCor | What about $A\times A$ when $A$ has a trivial automorphism group (and maybe some additional assumption)? Here $(A,+)$ acts by automorphisms, namely $a\cdot (b,c)=((1-a)b+ac,(1-a)c+ab)$, and we could expect there's nothing more. | |
| Aug 25, 2018 at 19:38 | history | asked | Iian Smythe | CC BY-SA 4.0 |