An abstract Jordan decomposition of an element of a Lie algebra L is a decomposition of the form a = a$_{s}$ + a$_{n}$, where

(a) ad a$_{s}$ is a diagonalizable (equivalently semisimple) endomorphism of L.

(b) ad a$_{n}$ is a nilpotent endomorphism.

(c) [a$_{s}$, a$_{n}$] = 0 .

This note defines the abstract Jordan decomposition in an arbitrary Lie algebra. Abstract Jordan decomposition in a Lie algebra is unique when it exists iff its centre is zero. It seems that the abstract Jordan decomposition maybe not exist even when its centre is zero, who can show me an example?

The same question is at here with no answer.

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    $\begingroup$ it is better if you define what abstract Jordan decomposition is. The note you have referred to talks about Jordan decomposition in complex semi simple Lie algebras $\endgroup$ – Venkataramana Nov 2 '19 at 6:51
  • $\begingroup$ OK, I rewrite the definition 12.1 in the note. $\endgroup$ – Strongart Nov 3 '19 at 4:50

In $\mathfrak g=\left\{\begin{pmatrix}x&x&y\\0&x&z\\0&0&0\end{pmatrix}:x,y,z\in \mathbf R\right\}$, $\mathrm{ad}$-semisimple and $\mathrm{ad}$-nilpotent elements all have $x=0$; so they don’t span.

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  • $\begingroup$ About the ad-semisimple case, choose the basis e$_{11}$+e$_{12}$+e$_{22}$,e$_{13}$ and e$_{23}$? $\endgroup$ – Strongart Nov 4 '19 at 5:03
  • $\begingroup$ @Strongart Well, take any basis (that doesn’t affect diagonalizability and nilpotency), but yes, using this one, the above matrix $a$ has $$\mathrm{ad}(a)=\begin{pmatrix}0&0&0\\-y-z&x&x\\-z&0&x\end{pmatrix}$$ which is nilpotent iff $x=0$, and not semisimple if $x\ne0$: $\det(\lambda I-\mathrm{ad}(a))=\lambda(\lambda -x)^2$, and eigenvalue $x$ has only a 1-dimensional eigenspace ($\mathbf Re_{13}$). $\endgroup$ – Francois Ziegler Nov 4 '19 at 5:37
  • $\begingroup$ I see, even when x = 0, it is also not ad-semisimple, it has no ad-semisimple part. $\endgroup$ – Strongart Nov 4 '19 at 14:11

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