Timeline for Question about maximal compact subgroups of Lie groups
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 4, 2022 at 18:55 | history | became hot network question | |||
| Nov 4, 2022 at 16:03 | vote | accept | Mira | ||
| Nov 4, 2022 at 14:20 | comment | added | YCor | @LSpice no, I missed the assumption (I edited to streamline) | |
| Nov 4, 2022 at 14:19 | history | edited | YCor | CC BY-SA 4.0 |
simplified formulation of assumption
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| Nov 4, 2022 at 14:18 | comment | added | LSpice | @YCor, right, but, if the centre is positive-dimensional, then $G_{\mathbb C}$ is not semisimple, which is assumed here. Or am I misunderstanding? (Anyway I guess one can pull back to $Z(G)^\circ \times G_\text{der}$ and make essentially the same argument there.) | |
| Nov 4, 2022 at 14:16 | comment | added | YCor | @LSpice Still, if the center is positive-dimensional the Killing form is not negative-definite on $\mathfrak{g}$. | |
| Nov 4, 2022 at 14:15 | comment | added | LSpice | @YCor, ah, thanks! I didn't notice. It's too late to change my comment, but I have changed the wording in my answer accordingly. | |
| Nov 4, 2022 at 14:14 | comment | added | YCor | @LSpice ah, the formulation "maximal on which..." is ambiguous. | |
| Nov 4, 2022 at 14:14 | comment | added | LSpice | @YCor, I'm not sure what you mean. I did not claim that $\mathfrak g$ is maximal, only that it is maximal for the Killing form being negative definite. Is that false? (As to the centre, I agree that one has to worry about it, but, since $G$ is semisimple, there is not much centre.) | |
| Nov 4, 2022 at 14:13 | comment | added | YCor | @LSpice For $\mathfrak{g}$ not simple, it is not true that $\mathfrak{g}$ is maximal in $\mathfrak{g}_\mathbf{C}$. For $\mathfrak{g}$ semisimple the larger subalgebras correspond to non-compact subgroups anyway, so it's fine. In the general case one should deal with the center too. | |
| Nov 4, 2022 at 14:12 | answer | added | LSpice | timeline score: 5 | |
| Nov 4, 2022 at 14:11 | comment | added | YCor | @LSpice maximal compact subgroups of connected Lie groups are connected (the inclusion is even a homotopy equivalence). | |
| Nov 4, 2022 at 14:05 | comment | added | LSpice | $\mathfrak g$ is a maximal subalgebra of its complexification on which the Killing form is negative definite, so at least $G$ is a maximal connected compact subgroup. | |
| Nov 4, 2022 at 10:55 | history | asked | Mira | CC BY-SA 4.0 |