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So using the Levi Decomposition for any lie algebra $\mathfrak{g}$, there exists a semisimple subalgebra $\mathfrak{s}$ such that:

$Rad(\mathfrak{g})$$\ltimes$$\mathfrak{s}$=$\mathfrak{g}$

However in the proofs I've seen of this, I never get what the semidirect product is? In other words, what is the derivation we are using here? Could someone explain this to me, thaks in advance.

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  • $\begingroup$ Levi and Lie (not Levi and lie). $\endgroup$ Commented Jun 14 at 8:01

2 Answers 2

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$Rad(\mathfrak g)$ is an ideal; the only derivation in sight, and the only way that $\mathfrak{s}$ can act on this, is the adjoint action.

Formally, take $$\pi: \mathfrak s \rightarrow Der(Rad(\mathfrak g))\\ s \mapsto ad(s)_{\lvert Rad(\mathfrak g)}$$

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$\operatorname{Rad}(\mathfrak{g})$ is an ideal of $\mathfrak{g}$ and $\mathfrak{s}\cong \mathfrak{g}/\operatorname{Rad}(\mathfrak{g})$ is semisimple with the multiplication $$ [X+\operatorname{Rad}(\mathfrak{g}),Y+\operatorname{Rad}(\mathfrak{g})]=[X,Y]+\operatorname{Rad}(\mathfrak{g}). $$ The semisimplicity of $\mathfrak{s}$ results from $$ \operatorname{Rad}(\mathfrak{s})=\operatorname{Rad}\left(\mathfrak{g}/\operatorname{Rad}(\mathfrak{g})\right) = \operatorname{Rad}(\mathfrak{g})/\operatorname{Rad}(\mathfrak{g}) = \{0\}. $$

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  • $\begingroup$ That's all correct but not an answer to the OP's question. $\endgroup$ Commented Jun 14 at 3:20
  • $\begingroup$ Then I didn't get it. Is he looking for the embedding? $\endgroup$ Commented Jun 14 at 3:24
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    $\begingroup$ What I am asking is when you have a semidirect product, it is always with respect to a specific derivation, right? In this case what is that derivatoin? $\endgroup$
    – Albi
    Commented Jun 14 at 3:26
  • $\begingroup$ @Albi Do you mean $$[(X,R),(Y,S))]=([X,Y],[R,S]+X.S-Y.R)=([X,Y],[R,S]+[X,S]-[Y,R])$$? It is simply the inherited multiplication from $\mathfrak{g}.$ It is basically what I wrote. $\endgroup$ Commented Jun 14 at 3:53
  • $\begingroup$ Ohhh ok, thank you so much! I thought it could be a more complicated action for $X.S$ and $Y.R$ than the adjoint one. $\endgroup$
    – Albi
    Commented Jun 14 at 4:04

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