The Lie algebra of a maximal torus on $G$ is the maximal abelian Lie subalgebra of $\mathfrak{g}$.

Let $T$ be a maximal torus of the lie group $G$. I want to show that the lie algebra $\mathfrak{t}$ of T is a maximal abelian Lie subalgebra of $\mathfrak{g}$.

It's straightforward that $\mathfrak{t}$ is a abelian subalgebra of $\mathfrak{g}$ (once $ad=0$).

So, let $\mathfrak{a}$ be a abelian Lie subalgebra of $\mathfrak{g}$ containing $\mathfrak{t}$. Then we have a Lie subgroup $A$ of $G$ such that $\mathfrak{a}$ is the Lie algebra corresponding to $A$. As $A$ is connected it is generated by $\exp[\mathfrak{a}]$. Thus it is abelian. Also, as $\mathfrak{t}\subset \mathfrak{a}$, we have that $T\subset A$.

How do I proceed from this point? Is there any way to show that $A$ is a torus?

• Do you mean to assume $G$ is compact? – Eric Wofsey Jun 26 '18 at 20:23

This statement is only correct if $G$ is compact (consider $G=\mathbb{R}$ for instance!). Given that assumption, let $B=\overline{A}$. Since $A$ is abelian and connected, so is $B$, and therefore $B$ is a torus since it is compact. Since $T\subseteq A\subseteq B$, maximality of $T$ implies $T=B$ and hence $T=A$. Thus $\mathfrak{t}=\mathfrak{a}$, as desired.