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Inner automorphisms of a Lie algebra are typically defined as automorphisms generated by elements of the form $exp (ad_X)$ where $X$ is nilpotent. Is $exp (ad_X)$ not inner for $X$ having a non-trivial Jordan-Chevalley decomposition?

This question is important to classification questions, which are often up to inner automorphism; classifying subalgebras up to inner automorphism, for instance. Certainly one needs to understand the definition of inner automorphism if one hopes to undertake such a classification.

wellington
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    Worth having a look at this – Callum Aug 04 '23 at 22:00
  • Thank you, Mariano and Callum. – wellington Aug 05 '23 at 02:22
  • I don't quite see why $exp (ad_{S+N})$ wouldn't make sense as an automorphism, where $S$ is semisimple and $N$ is nilpotent with $[S,N]=0$? I'm not sure what I'm missing: $exp (ad_{S+N}) = Ad_{exp(S+N)}=Ad_{exp(S) exp(N)}$, which is just conjugation by $exp(S) exp(N)$? – wellington Aug 05 '23 at 02:27
  • If S is semisimple, one can still compute $exp(S)$: Since $S$ is semisimple (diagonalizable assuming algebraic closure) we have an invertible matrix $G$ such that $S=GDG^{-1}$, where $D=diag(d_1,…,d_m)$. Then

    $exp(S)=exp(GDG^{-1})=G ~exp(D)~G^{-1}= G ~diag(exp(d_1), …, exp(d_m))~ G^{-1}.$

     – wellington Aug 06 '23 at 04:33
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    @Wellington if we assume algebraic closure (or at least if we assume we are working over $\mathbb{C}$) then the different definitions of inner automorphism are equivalent anyway. Note also your definition is slightly off. It is elements generated by exponentials of nilpotent elements rather than only the pure exponentials (which aren't closed under multiplication). In the split case this recovers all the elements you are looking for (with no convergence issues) – Callum Aug 06 '23 at 16:25
  • Thanks @Callum. I believe it's clear now. I appreciate it. – wellington Aug 06 '23 at 16:52

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