First recall some definitions :
Let $B$ be a Killing form on Lie algebra $\mathfrak{g}$ over ${\bf R}$ such that $B(X,Y)\doteq Tr(ad_Xad_Y)$.
$\mathfrak{g}$ is semisimple if $B$ is non-degenerate.
Define $\partial \mathfrak{g} \doteq \{ D | D[X,Y]=[DX,Y] + [X,DY] \}$ Clearly it contains $ad(\mathfrak{g})\doteq \{ ad_X | X\in \mathfrak{g}\}$.
My question is about proof of $\partial \mathfrak{g}=ad(\mathfrak{g})$ :
I am reading Helgason's book. We can show that $ ad(\mathfrak{g}) = \mathfrak{g}$ easily.
I cannot proceed the proof since $B$ is defined on $\mathfrak{g}$ only.
How can I complete the proof ? Thank you in advance.
Here's a proof from Humphrey's excellent book:
First of all, show that $I=ad(\mathfrak g)\subset\partial\mathfrak g$ is an ideal of the Lie algebra $\partial\mathfrak g$: if $x\in\mathfrak g$ and $\delta\in\partial\mathfrak g$, then $$[\delta,ad(x)]=\cdots\in I$$ This implies that the Killing form on $\mathfrak g$ coincides with that induced by the Killing form on $\partial\mathfrak g$. Since $\mathfrak g$'s Killing form is non degenerate by semisimplicity, we have $$I\oplus I^{\perp}=\partial\mathfrak g$$ with respect to the Killing form on the derivations. Now consider $\delta\in I^{\perp}$. For any $x\in\mathfrak g$, since both $I$ and $I^{\perp}$ are ideals in $\partial\mathfrak g$, $$[\delta,ad(x)]\in I\cap I^{\perp}=\lbrace 0\rbrace$$ equals $ad(\delta(x))$ by the calculation above, so for all $x\in\mathfrak g,~\delta(x)\in\ker(ad)=\lbrace 0\rbrace$. That is, $\delta=0$ and $I^{\perp}=\lbrace 0\rbrace$, so that $ad(\mathfrak g)=\partial\mathfrak g$.