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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.)

edited Apr 26 2011 at 6:58

asked Apr 26 2011 at 0:08

Fedor Petrov
6,789●1●12●38

	You are teaching abstract Lie groups to second year students?! It is not difficult to do "by hand" for the classical compact groups, since this follows from basic linear algebra; namely the normal forms for orthogonal and unitary matrices. – José Figueroa-O'Farrill Apr 26 2011 at 0:51
	As Richard has pointed out in the answer below, you may need to edit and write "inclusion-connected maximal abelian subgroup". – Somnath Basu Apr 26 2011 at 2:30

Faisal · Accepted Answer · 2011-04-26 00:43:43Z UTC

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There are proofs that avoid algebraic topology: see for example Chapter 16 in Bump's Lie Groups or IV.5 in Knapp's Lie Groups Beyond an Introduction.

answered Apr 26 2011 at 0:43

Faisal
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Richard Borcherds · Answer 2 · 2011-04-26 02:04:48Z UTC

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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.

answered Apr 26 2011 at 2:04

Richard Borcherds
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	Maybe the OP forgot to mention that the subgroup is connected? – José Figueroa-O'Farrill Apr 26 2011 at 2:17
	Thanks, I added "connected". Hope that now it is the same as maximal torus:) – Fedor Petrov Apr 26 2011 at 7:00

maximal tori cover compact Lie group

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