maximal tori cover compact Lie group - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T22:55:32Z http://mathoverflow.net/feeds/question/62985 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62985/maximal-tori-cover-compact-lie-group maximal tori cover compact Lie group Fedor Petrov 2011-04-26T00:08:41Z 2011-04-26T06:58:35Z <p>Let $G$ be a compact connected Lie group, $T$ be some maximal torus in $G$ (that is, inclusion-maximal connected abelian subgroup). Then the union of tori $gTg^{-1}$, $g\in G$, is the whole $G$. This is well-known (4.21 in Adams book). My question is rather methodological: is there any proof without use of algebraic topology? Adams presents A. Weil's proof, which uses some kind of Lefschetz fixed points theorem. (Yes, I am sorry but my motivation is mostly that I teach second year students this stuff.) </p> http://mathoverflow.net/questions/62985/maximal-tori-cover-compact-lie-group/62989#62989 Answer by Faisal for maximal tori cover compact Lie group Faisal 2011-04-26T00:43:43Z 2011-04-26T00:43:43Z <p>There are proofs that avoid algebraic topology: see for example Chapter 16 in Bump's <em>Lie Groups</em> or IV.5 in Knapp's <em>Lie Groups Beyond an Introduction</em>.</p> http://mathoverflow.net/questions/62985/maximal-tori-cover-compact-lie-group/62995#62995 Answer by Richard Borcherds for maximal tori cover compact Lie group Richard Borcherds 2011-04-26T02:04:48Z 2011-04-26T02:04:48Z <p>The claim in the question that maximal tori are the same as inclusion-maximal abelian subgroups is not correct. For example, the diagonal matrices with +1 or -1 on the diagonal form a maximal abelian subgroup of SO(n) that is not a torus. </p>