Timeline for Commutator of radical and Levi factor in Lie algebra
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 25, 2016 at 21:17 | comment | added | Robin Goodfellow | @YCor -- Indeed! In retrospect, I probably should have checked what I wrote against an example. I have redone the answer. Thank you. | |
| Nov 25, 2016 at 21:14 | history | edited | Robin Goodfellow | CC BY-SA 3.0 |
Major corrections to fix silly fallacies
|
| Nov 25, 2016 at 18:45 | comment | added | YCor | In general it's just $\subset$. For instance, if $S$ is $SL_2$, $R$ is the 3-dimensional Heisenberg Lie algebra (with standard action), then $[S,R]=R$, so $[R,[S,R]]$ is the 1-dimensional center, but $[S,[R,R]]=0$. So the "Jacobi identity for subalgebras" here is a proper inclusion. | |
| Nov 25, 2016 at 17:40 | comment | added | Robin Goodfellow | @YCor -- Perhaps I am missing something obvious, but I do not see why that would be true. | |
| Nov 25, 2016 at 17:21 | history | edited | Robin Goodfellow | CC BY-SA 3.0 |
added 194 characters in body
|
| Nov 25, 2016 at 1:34 | comment | added | YCor | I guess "this fact" refers to the fact $[S,R]$ is an ideal, not just the equality $[S,R]=[S,[S,R]]$. | |
| Nov 24, 2016 at 23:04 | history | answered | Robin Goodfellow | CC BY-SA 3.0 |