Timeline for A question on linear groups
Current License: CC BY-SA 3.0
6 events
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Jul 3, 2016 at 21:40 | comment | added | YCor | I've downvoted your answer because it contains a mistake and you didn't edit it to fix/clarify it. You should say explicitly what you prove, namely that $\mathfrak{so}_n(\mathbf{R)})$ is a maximal subalgebra, and the consequence that $\mathrm{SO}(n)$ is maximal among closed subgroups of $\mathrm{SL}_n(\mathbf{R})$, showing in particular that the closure of $\langle \mathrm{SO}(n),t\rangle$ contains $\mathrm{SL}_n$ whenever $t$ is not a scalar multiple of an isometry. | |
Jun 30, 2016 at 16:06 | comment | added | YCor | No, I didn't. When I say "the set of $x$ in the Lie algebra...", I mean, in the Lie algebra $\mathfrak{gl}_n$ (but indeed I didn't justify that it's a Lie subalgebra, which requires a little argument). | |
Jun 30, 2016 at 15:30 | comment | added | Venkataramana | @Ycor; actually, your comments also assume $G$ has a Lie algebra so you have also assumed that $G$ is an analytic subgroup. So your proof is just as incomplete! | |
Jun 30, 2016 at 15:05 | comment | added | YCor | Your proof is incomplete, since it assumes implicitly $G$ closed (or, at least, an analytic subgroup). There's some further stuff in between. | |
Jun 30, 2016 at 14:22 | history | edited | Venkataramana | CC BY-SA 3.0 |
added 77 characters in body
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Jun 30, 2016 at 14:15 | history | answered | Venkataramana | CC BY-SA 3.0 |