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Let $G$ be a semisimple Lie group. Denote $d(G)$ as the maximal integer $p$ such that $\mathbb{Z}^p$ is isomorphic to a discrete subgroup of $G$ and $c(G)$ is the maximal integer $q$ such that $\mathbb{R}^q$ is isomorphic to a closed subgroup of $G$.

is there a way to compute $d(G)$ and $c(G)$?

if you have any related references, please share them whit me.

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    $\begingroup$ This inequality is trivial. Aren't you asking whether equality holds? $\endgroup$
    – YCor
    Commented Jan 20, 2023 at 15:51
  • $\begingroup$ 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. $\endgroup$
    – Yushi MuGiwara
    Commented Jan 20, 2023 at 15:59
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    $\begingroup$ 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 $c(G)$. $\endgroup$
    – YCor
    Commented Jan 25, 2023 at 7:47
  • $\begingroup$ @YCor I really appreciate your help, it was extremely helpful. $\endgroup$
    – Yushi MuGiwara
    Commented Jan 25, 2023 at 15:43
  • $\begingroup$ 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 $G_2$. $\endgroup$
    – Callum
    Commented Jan 26, 2023 at 14:32

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