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Let L be a finite-dimensional complex semisimple lie algebra, then ad(L)=Der(L). (Der is short for derivation). In order to show that ad(L)=Der(L), the book says that it only need to show that the prependicular space to ad(L) is zero. This is where I don't quite understand. The book I am reading is Introduction to Lie Algebras by Karin Erdmann, and the proof is on page 85. I am sorry that I cannot find a link for this book.

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1 Answer 1

First of all, $\text{ad}(L)$ is an ideal of $\text{Der}(L)$ which can be obtained from the following observation;

$$[\delta, \text{ad}(x)]=\text{ad}(\delta x) \;\;\; \text{for} \;\; x \in L, \; \delta \in \text{Der}(L).$$

Since $L$ is semisimple, $Z(L)=0$ (its center), therefore, $L \to \text{ad}(L)$ is an isomorphism of Lie algebras, so $S=\text{ad}(L)$ has a non-degenerate Killing form.

If $D=\text{Der}(L)$, then by the above observation, we have $[D,S] \subset S.$ This implies that $\kappa_S$ is the restriction to $S \times S$ of the Killing form $\kappa_D.$ In particular, if $I=S^{\perp}$ is the subspace of $D$ orthogonal to $S$ under $\kappa_D,$ then the non-degeneracy of the Killing form of $\kappa_S$ forces $I \cap S=0.$ Both $I$ and $S$ are ideals of $D$ so we obtain $[I,S]=0.$ If $\delta \in I,$ this forces $\text{ad}(\delta x)=0$ by the above observation, for all $x \in L,$ and because $\text{ad}$ is one-to-one, we obtain $\delta x=0$ for every $x\in L.$ Hence, $\delta=0.$

In conclusion, $I=0$ and $\text{Der}(L)=\text{ad}(L).$

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Why does I=0 imply that Der(L)=ad(L)? –  Yuan Nov 1 '11 at 14:34
    
Because non-degeneracy of the Killing form implies that its radical is trivial, then $\text{dim}(I)=\text{dim}(D)-\text{dim}(S),$ or $D=S \oplus I,$ since we have shown that $I \cap S=0.$ –  Ehsan M. Kermani Nov 1 '11 at 19:30

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