Timeline for Does every linear group admit a subgroup of dimension 1?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 26, 2013 at 9:07 | vote | accept | Tomasz Lenarcik | ||
| Sep 26, 2013 at 8:26 | history | edited | Tomasz Lenarcik | CC BY-SA 3.0 |
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| Sep 26, 2013 at 2:29 | comment | added | Marguax | It is not necessary to assume $k = \overline{k}$ if allowing any characteristic; separably closed is enough. In char. 0 "real closed" is enough; e.g., the conclusion holds for $k = \mathbf{R}$. By the way, why do you highlight points of infinite order? That doesn't seem germane, since every nontrivial torus over a field $k$ not algebraic over a finite field (even such a $k$-torus with no nontrivial proper $k$-subtori) contains a $k$-point generating a Zariski-dense subgroup, so one cannot produce a 1-dimensional smooth closed $k$-subgroup by using any old point of infinite order. | |
| Sep 26, 2013 at 0:21 | comment | added | Jim Humphreys | As indicated in the answers, your language is too loose to be clear: the meaning of "linear group defined over some field" needs to be made more precise, at which point the question is easy to answer. | |
| Sep 25, 2013 at 23:36 | answer | added | David E Speyer | timeline score: 11 | |
| Sep 25, 2013 at 23:25 | answer | added | abz | timeline score: 7 | |
| Sep 25, 2013 at 21:30 | history | edited | Tomasz Lenarcik | CC BY-SA 3.0 |
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| Sep 25, 2013 at 21:04 | answer | added | Daniel Loughran | timeline score: 11 | |
| Sep 25, 2013 at 20:53 | review | First posts | |||
| Sep 25, 2013 at 21:06 | |||||
| Sep 25, 2013 at 20:35 | history | asked | Tomasz Lenarcik | CC BY-SA 3.0 |