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A naive question from a physicist, so forgive the lack of rigor. Consider a Lie group, acting on its Lie algebra by the adjoint action. Does every orbit go through the Cartan subalgebra? Alternatively, for which Lie groups is this true? (I view the statement as a generalization of the elementary fact that every Hermitian matrix can be diagonalized by a unitary transformation.) It would be nice if it held at least for real, compact, and semi-simple Lie algebras. Thank you for any hint or reference to literature!

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The group acts on the algebra by Lie algebra automorphisms. Now the elements of the Cartan subalgebra are semisimple, and there are orbits consisting of non-semisimple elements. –  Mariano Suárez-Alvarez Jan 27 '11 at 18:04
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Thank you, I now understand that the claim cannot hold generally for complex Lie algebras. How about the real ones? As I am a physicist, compact and semi-simple would be enough. It is certainly true for su(n) (diagonalization of a Hermitian matrix by a unitary transformation) as well as so(n) (block-diagonalization of an antisymmetric matrix by an orthogonal transformation). Can one somehow prove it also for the symplectic or even the exceptional algebras? –  Tomáš Brauner Jan 28 '11 at 15:39
    
its true for semi-simple complex algebras. –  Eric O. Korman Jan 30 '11 at 22:45

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If G is a compact and connected Lie Group,that must be true!!About this idea,you can read this book(Differential Geometry,Lie Groups and Symmetric Spaces,author:Helgason) Chapter V,6th section.

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Thanks for the detailed reference! I will have a look. –  Tomáš Brauner Feb 28 '12 at 19:10

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