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The statement:

If $G$ is a semisimple Lie group and $\mathfrak{g}$ the corresponding Lie algebra, then the general statement is that every element of $\mathfrak{g}$ is conjugate, via the adjoint action of $G$, to an element in any fixed Cartan subalgebra.

seems to be well known in the field of Lie groups, see https://math.stackexchange.com/a/2169798/7266 .

I was wondering, if there is a (canonical) reference of a book that proves this theorem?

Fabian
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  • The answer to your question is "no." What examples of noncompact simple Lie algebras do you know? Try to disprove the claim. – Moishe Kohan Apr 03 '23 at 13:05
  • @MoisheKohan: I guess the problem is due to Jordan blocks. That is what I found confusing when reading the answer linke in the OP. Is there a version of the statement that is true, allowing for Jordan blocks? Is there a good reference? – Fabian Apr 03 '23 at 19:58
  • I do not understand your question. But did you work out a counter example and for which Lie algebra? – Moishe Kohan Apr 03 '23 at 20:08
  • @MoisheKohan: I guess the problem is that I do not understand enough about the subject. I was thinking about the group $SL(2,\mathbb{C})$ and the element $\pmatrix{ 0 & 1 \ 0 & 0 }$ of the Lie-algebra. – Fabian Apr 03 '23 at 20:34
  • Yes, that's the simplest example. Do you understand why? All you need is linear algebra. – Moishe Kohan Apr 03 '23 at 20:49
  • @Moishe Kohan: there is only a single eigenvector to the eigenvalue 0. So there is no way to diagonalize this matrix with elements in SL. Still I am wondering, what the correct statement in the linked post is. – Fabian Apr 03 '23 at 20:57
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    Most likely, he assumed compactness. – Moishe Kohan Apr 03 '23 at 21:01

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