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It's known that if $L\subset gl(V)$, with $V$ finite dimensional, is a semisimple Lie algebra, then the abstract and usual Jordan decompositions in $L$ coincide. Is it possible to provide a counter-example if $L$ isn't semisimple?

Remark: The underlying field is algebraically closed of characteristic $0$ .

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The underlying field should be algebraically closed of characteristic 0 (otherwise the discussion gets more complicated). –  Jim Humphreys Apr 15 '11 at 22:17
You're right. I'll edit my post. Thank you! –  user14312 Apr 15 '11 at 22:54
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2 Answers 2

up vote 14 down vote accepted

I'm tempted to amplify Ben's precise short answer by emphasizing how subtle the notion of abstract (or intrinsic) Jordan decomposition really is. You start with a matrix Lie algebra which is (1) required to be isomorphic to its image under the adjoint representation (in other words, centerless). Then it makes sense to interchange the Jordan decompositions in the two settings. But these might not agree for every matrix realization, unless you also require (2) that the given Lie algebra satisfy a complete reducibility theorem for finite dimensional representations. Then the proof I first saw in Bourbaki, which goes back to Richard Brauer's early work influenced by Weyl, takes over. (It needs characteristic 0 and at first an algebraically closed field to apply Weyl's complete reducibility theorem directly.)

Ben's example already fails (1), as does a general linear Lie algebra, whereas a centerless solvable Lie algebra only fails (2).

The somewhat elaborate-looking Bourbaki argument tempts people to take shortcuts (even in one published textbook), but as far as I can see the more sophisticated proof is really needed.

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Consider the subalgebra $\left(\begin{matrix} 0& a\\ 0& 0\end{matrix}\right)$ in $\mathfrak{gl}(2)$. This is abelian, so in abstract JD, every element is semi-simple, but these are nilpotent linear operators.

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