Timeline for Is the group of integer points of a simple real linear algebraic group a maximal closed subgroup?
Current License: CC BY-SA 4.0
13 events
when toggle format | what | by | license | comment | |
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Oct 14, 2021 at 18:21 | vote | accept | Ian Gershon Teixeira | ||
Oct 14, 2021 at 13:12 | answer | added | David E Speyer | timeline score: 2 | |
Oct 14, 2021 at 12:50 | history | edited | YCor | CC BY-SA 4.0 |
formatting, added tag
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Oct 14, 2021 at 12:33 | comment | added | Ian Gershon Teixeira | I just meant maximal closed. In my head I say "almost dense" instead of maximal closed because I think it's more interesting to frame it as "this subgroup is so close to being dense that if you add any single new element and take the closure of the group generated that way you get every element of $G_\mathbb{R}$". So I use "maximal closed subgroup" and "almost dense subgroup" interchangeably and that was supposed to say almost dense | |
Oct 14, 2021 at 12:29 | history | edited | Ian Gershon Teixeira | CC BY-SA 4.0 |
deleted 23 characters in body
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Oct 14, 2021 at 9:05 | history | became hot network question | |||
Oct 14, 2021 at 3:24 | answer | added | Venkataramana | timeline score: 8 | |
Oct 14, 2021 at 2:59 | history | edited | LSpice | CC BY-SA 4.0 |
Proofreading; `\operatorname`
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Oct 14, 2021 at 2:57 | comment | added | LSpice | By "Integer points are in dense for $\operatorname{SO}_3$", do you mean "the integer points form a maximal closed subgroup of the real points when $G = \operatorname{SO}_3$" or something? \\ Also, the hypothesis 'simple' should probably go in the title; otherwise the additive group $G$ is a counterexample. | |
Oct 14, 2021 at 2:29 | comment | added | Ian Gershon Teixeira | haha oops I just fixed it with an edit | |
Oct 14, 2021 at 2:29 | history | edited | Ian Gershon Teixeira | CC BY-SA 4.0 |
added 21 characters in body
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Oct 14, 2021 at 1:22 | comment | added | Francois Ziegler | Where by “no” you mean “yes”? | |
Oct 14, 2021 at 1:05 | history | asked | Ian Gershon Teixeira | CC BY-SA 4.0 |