Timeline for Exotic group topologies on the affine group $ax+b$
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 15, 2015 at 10:25 | vote | accept | hosain | ||
| Jul 9, 2015 at 12:34 | history | edited | YCor | CC BY-SA 3.0 |
Changed title and removed general-topology tag
|
| Jul 9, 2015 at 12:31 | answer | added | YCor | timeline score: 3 | |
| Jul 9, 2015 at 12:03 | history | edited | Tomasz Kania | CC BY-SA 3.0 |
edited title
|
| Jul 9, 2015 at 11:46 | answer | added | Dave Witte Morris | timeline score: 3 | |
| Jul 9, 2015 at 11:06 | history | reopened |
YCor Yemon Choi Felipe Voloch Stefan Kohl♦ Daniel Moskovich |
||
| Jul 8, 2015 at 20:16 | review | Reopen votes | |||
| Jul 9, 2015 at 11:06 | |||||
| Jul 8, 2015 at 19:54 | comment | added | YCor | Here $\hat{\mathbf{Q}}$ cannot work because the its automorphism group is not transitive on nonzero elements (because it's isomorphic to the automorphism group of $\mathbf{Q}$, namely $\mathbf{Q}^*$, which is countable hence can't be transitive on an uncountable set). But I'm not sure about $\hat{\mathbf{Q}}^{\mathbf{N}}$, which is abstractly isomorphic to $\mathbf{R}$ and has an uncountable automorphism group. | |
| Jul 8, 2015 at 19:48 | comment | added | YCor | Indeed the Pontryagin dual of the discrete group $\mathbf{Q}$ is a compact group whose underlying discrete group is isomorphic to the underlying discrete group of $\mathbf{R}$. It's not clear if this can be extended to the semidirect structure, but the question is reasonable and even if one can argue about its interest, it's not ambiguous at all and the "put on hold as unclear" is not justified. | |
| Jul 8, 2015 at 18:23 | history | closed |
R W Joonas Ilmavirta András Bátkai Alex Degtyarev Andreas Thom |
Needs details or clarity | |
| Jul 8, 2015 at 18:07 | comment | added | Will Brian | It is possible to put a compact group topology on $\mathbb{R}$ (Halmos, "Comment on the real line", 1944). But any such topology is weird, in the sense that it cannot look very much like the usual topology on $\mathbb{R}$. For example, $[0,1)$ is not Haar measurable in any compact group topology on $\mathbb{R}$ (if it were, then you get a contradiction by the same argument that shows Vitali sets are non-measurable). | |
| Jul 8, 2015 at 17:37 | review | Close votes | |||
| Jul 8, 2015 at 18:28 | |||||
| S Jul 8, 2015 at 17:19 | history | suggested | Michael Albanese | CC BY-SA 3.0 |
Added MathJax and a tag.
|
| Jul 8, 2015 at 17:14 | comment | added | Alain Valette | Have you noticed that $K$ is isomorphic (as a group) to the additive group of the real line? Can you put a compact group topology on $\mathbb{R}$? | |
| Jul 8, 2015 at 17:06 | review | Suggested edits | |||
| S Jul 8, 2015 at 17:19 | |||||
| Jul 8, 2015 at 17:01 | history | asked | hosain | CC BY-SA 3.0 |