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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.)

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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 '11 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 '11 at 2:30
up vote 3 down vote accepted

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.

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There are many ways of proving this, using all sorts of different methods. In the second edition my book "Lie groups, Lie algebras, and representations" (following Brocker and tom Dieck) I use the mapping degree theorem. If we fix a single maximal torus $T$ and consider the conjugation map $\Phi:T \times (K/T) \rightarrow K$ be given by $\Phi(t,[x])=xtx^{-1}$. If we can show that this map has nonzero mapping degree, we can conclude it is surjective, which is just what we are trying to show. In Section 11.5, I show that $\Phi$ has mapping degree equal to the order of the Weyl group. This approach also gives a proof of the Weyl integral formula by the same computation.

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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.

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Maybe the OP forgot to mention that the subgroup is connected? – José Figueroa-O'Farrill Apr 26 '11 at 2:17
    
Thanks, I added "connected". Hope that now it is the same as maximal torus:) – Fedor Petrov Apr 26 '11 at 7:00

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