The Lie groups book I'm reading (Knapp, Lie Groups Beyond an Introduction, page 255) goes to some trouble to prove that every element of a compact Lie group is contained in a maximal torus. Why isn't this obvious from the fact that every element is contained in a maximal abelian subgroup (since they are closed under nested union), which is therefore closed so compact (if not, close it and get a bigger abelian subgroup).

In my book, this is a corollary to the result that, given $T$ a maximal Torus of a compact Lie group, $G$, every element of $G$ is conjugate to an element of $T$. This result takes three pages pages to prove via induction despite the fact that it seems to follow trivially, as above, from the immediately preceding result in the text that any two maximal tori of a compact Lie group are conjugate.


I think the issue is connectedness. I suppose it's not clear that $g\in G$ is contained in a maximal compact abelian connected subgroup.

  • 3
    $\begingroup$ I agree. If connectedness followed for formal reasons then the exponential map would always be surjective, which isn't true. $\endgroup$ – Qiaochu Yuan Dec 19 '14 at 8:58

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