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Jan 26, 2023 at 14:32 comment added Callum On a Lie algebra level, a compact Lie algebra $\mathfrak{g}$ has maximal dimension of a maximal abelian equal to $\mathrm{rank}(\mathfrak{g})$. For a non-compact simple Lie algebra you can often find the largest abelian subalgebra as the nilradical of one of the maximal parabolic subalgebras. I think this will work for $G$ complex or split for all the classical cases plus $E_6$ and $E_7$ but breaks down in some of the real forms and for $E_8$, $F_4$ and $G_2$.
Jan 25, 2023 at 15:43 comment added Yushi MuGiwara @YCor I really appreciate your help, it was extremely helpful.
Jan 25, 2023 at 7:47 comment added YCor If $G$ has a finite center, then $d(G)=c(G)$. The inequality $c(G)\le d(G)$ is trivial, and to get the reverse one, we can suppose $G$ has trivial center hence is algebraic, and pass to the Zariski closure. In general, if the center has $\mathbf{Q}$-rank $r$ one deduces $d(G)=c(G)+r$. So everything boils down to computing $c(G)$.
Jan 24, 2023 at 19:42 review Close votes
Feb 4, 2023 at 3:02
Jan 24, 2023 at 19:41 history undeleted Asaf Karagila
Jan 24, 2023 at 18:28 history deleted Yushi MuGiwara via Vote
Jan 24, 2023 at 18:28 history edited Yushi MuGiwara CC BY-SA 4.0
added 67 characters in body
Jan 24, 2023 at 18:25 review Close votes
Jan 24, 2023 at 18:34
Jan 24, 2023 at 18:21 history undeleted Yushi MuGiwara
Jan 20, 2023 at 16:31 history deleted Yushi MuGiwara via Vote
Jan 20, 2023 at 15:59 comment added Yushi MuGiwara I know that we have equality in the case of nilpotent Lie group ... to be specific I am looking for the case where G is semi-simple.
Jan 20, 2023 at 15:51 comment added YCor This inequality is trivial. Aren't you asking whether equality holds?
Jan 20, 2023 at 15:50 history edited YCor CC BY-SA 4.0
formatting, changed tag, clarified title
Jan 20, 2023 at 15:41 history asked Yushi MuGiwara CC BY-SA 4.0