What is the semidirect product we use in Levi Decomposition

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.

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

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

$$\operatorname{Rad}(\mathfrak{g})$$ 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 $$\operatorname{Rad}(\mathfrak{s})=\operatorname{Rad}\left(\mathfrak{g}/\operatorname{Rad}(\mathfrak{g})\right) = \operatorname{Rad}(\mathfrak{g})/\operatorname{Rad}(\mathfrak{g}) = \{0\}.$$

• That's all correct but not an answer to the OP's question. Commented Jun 14 at 3:20
• Then I didn't get it. Is he looking for the embedding? Commented Jun 14 at 3:24
• 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?
– Albi
Commented Jun 14 at 3:26
• @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. Commented Jun 14 at 3:53
• 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.
– Albi
Commented Jun 14 at 4:04