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Dec 8, 2020 at 14:41 comment added user166831 The point is that you can get a "Jordan decomposition" in $\mathfrak{g}$, as in Erdmann and Wildon 9.15 or in Premet's answer. However, to prove that it maps to the Jordan decomposition under every representation (as in E&W 9.16), you seem to need complete reducibility. The argument before E&W 9.16 fails because we don't know that $d$ lies in $L$.
Aug 22, 2018 at 10:51 comment added Alexander Premet For us V=g and the Jordan decomposition of any element in gl(g) is unique. We know that for D=ad x the derivations D_s and D_n of g are inner (since Der(g)=ad(g)). This gives us commuting inner derivations ad(x_s) and ad(x_n) and hence unique commuting x_s,x_n in g such that x=x_s+x_n. This is all we need really.
Aug 22, 2018 at 10:43 history edited Alexander Premet CC BY-SA 4.0
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Aug 22, 2018 at 7:30 comment added spin I am not sure if I am missing something, but in the first edition of the book by Karin Erdmann and Mark Wildon they tried to do something like in your answer to avoid Weyl's complete reducibility theorem, but their approach had some problems (see the answer by Jim Humphreys).
Aug 22, 2018 at 7:27 comment added spin Let $x = x_s' + x_n'$ be the Jordan decomposition of $x$ in $\mathfrak{gl}(V)$. So by your argument, the restriction of $\operatorname{ad}_{\mathfrak{gl}(V)} x_s'$ to $\mathfrak{g}$ equals $\operatorname{ad}_{\mathfrak{g}} x_s$, where $D_s = \operatorname{ad} x_s$ as in your answer. So $x_s - x_s'$ lies in the centralizer of $\mathfrak{g}$ in $\mathfrak{gl}(V)$, but I do not see how we get $x_s' \in \mathfrak{g}$.
Aug 22, 2018 at 4:30 history edited Alexander Premet CC BY-SA 4.0
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Aug 21, 2018 at 20:00 review Late answers
Aug 21, 2018 at 20:05
Aug 21, 2018 at 19:40 history answered Alexander Premet CC BY-SA 4.0