Timeline for Does Aut(G) → Out(G) always split for a compact, connected Lie group G?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 6, 2020 at 16:36 | comment | added | LSpice | @YCor, good idea. I have done so. | |
| Dec 6, 2020 at 16:36 | history | edited | LSpice | CC BY-SA 4.0 |
The answer is 'yes'
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| Dec 6, 2020 at 15:38 | comment | added | YCor | Maybe it would help to start saying the answer is "yes"? | |
| Dec 5, 2020 at 19:28 | comment | added | LSpice | @anniemarieheart's question points up some of the difficulties that can arise. | |
| Dec 5, 2020 at 19:27 | comment | added | LSpice | For this argument to work, we need some appropriate notion of a pinning. The usual notion is for a complex group, where, instead of considering real rays in root groups, we just consider individual non-$0$ root vectors; and this works just as well for any real Lie group that is quasisplit, i.e., contains a Borel subgroup (where now we require $X_{\overline\alpha} = \overline{X_\alpha}$). Without a Borel subgroup, or some appropriate substitute (such as a maximal torus in a compact group), I don't know how to construct an appropriate lifting (but I don't know that it doesn't exist). | |
| Dec 5, 2020 at 19:26 | vote | accept | Ben Heidenreich | ||
| Dec 5, 2020 at 19:23 | comment | added | Ben Heidenreich | So then the sequence splits whether or not $G$ is compact, correct? | |
| Dec 5, 2020 at 18:56 | comment | added | LSpice | I think that this question has appeared here before, but I can't find it right now. | |
| Dec 5, 2020 at 18:54 | history | answered | LSpice | CC BY-SA 4.0 |