It is known that outer automorphisms of semi-simple Lie algebras are automorphisms of their corresponding Dynkin diagrams. But would it be correct to say that for a semi-simple Lie algebra all outer automorphisms are exhausted by outer automorphisms of its simple components and outer automorphisms corresponding to exchanging the simple isomorphic components of the semi-simple algebra?
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6The answer is positive. If you have an automorphism of the Lie algebra, you get an induced action on the the Dynkin diagram. By composing with a permutation of the simple components you can thus assume that your automorphism is a product of automorphisms of the simple components. But as you said, such automorphisms arise from an automorphisms of the Dynkin diagram. – Ehud Meir Aug 13 '15 at 21:26
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1It could be useful to say how "outer automorphism" is defined. I expect inner automorphisms to be elements of the group of automorphisms generated by exponentials of all nilpotent elements $ad(x)$ (and Out being the quotient Aut/Inn)? Anyway when the field is not algebraically closed, it can be larger than the group of automorphisms of the Dynkin diagram, even for split semisimple Lie algebras. So maybe some hypothesis "finite-dimensional Lie algebra over an algebraically closed field of characteristic zero" is missing. – YCor Aug 16 '15 at 21:59