Writing a Lie algebra as a sum of two ideals

idealslie-algebraslinear algebrasemisimple-lie-algebras

Let $\mathfrak{g}$ be a semisimple Lie algebra and let Der$(\mathfrak{g})$ be the Lie algebra of derivations of $\mathfrak{g}$. Let $ad$ be the adjoint representation defined by $ad(X)Y=[X,Y]$ Let $I=$Im$(ad)$, then $I$ is an ideal in Der($\mathfrak{g})$ and $I$ is semisimple. Define $I^\perp=\{Y\in\text{Der}(\mathfrak{g}):\kappa_{\text{Der}(\mathfrak{g})}(X,Y)=0\,\forall X\in I\}$ with $\kappa_{\text{Der}(\mathfrak{g})}$ the Killing form on Der$(\mathfrak{g})$. Then $I^\perp$ is an ideal of Der$(\mathfrak{g})$ as well and $I\cap I^\perp=\{0\}$ (follows from Cartan's criterium). Does it follow directly that Der($\mathfrak{g})=I\oplus I^\perp$ and that $[I,I^\perp]=\{0\}$?

Best Answer

Yes, we have the following result:

Proposition: Let $L$ be a finite-dimensional Lie algebra over a field $K$ of characteristic zero and $I$ be a semisimple ideal in $L$. Then $L=I\oplus I^{\perp}$ and $I^{\perp}$ is the centralizer of $I$ in $L$.

The proof follows directly from the fact that the Killing form restricted to $I$ is nondegenerate.

Now if $\mathfrak{g}$ is semisimple, we have $Z(\mathfrak{g})=0$, so that $I=ad(\mathfrak{g})\cong \mathfrak{g}$ is a semisimple ideal in the Lie algebra $L=Der(\mathfrak{g})$. Hence we have $$ Der(\mathfrak{g})=I\oplus I^{\perp}. $$

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