Let $L$ be a Lie algebra (with field $F$). Consider the killing form
$$\begin{align*} k:&L \times L \to F \\
&(x,y) \to Tr(ad(x)ad(y))
\end{align*}$$
We know that $L$ is semisimple if and only if its killing form is non degenerate.
Now If $L \subseteq \mathfrak{gl}(V)$, I can define
$$\begin{align*} k’:&L \times L \to F \\
&(x,y) \to Tr(xy)
\end{align*} $$
I want to prove: $L$ is semisimple if and only if $k’$ is non degenerate. One implication is trivial:
Suppose $L$ semisimple. If $x\in L^{\bot}$(respect $k’$) and $y\in L$ then $Tr(xy)=0$, so in particular this is true for $y \in [L^\bot,L^\bot]$. So, by Cartan’s criterion, $L^\bot$ is solvable and so it is zero since $RadL=0$.
I don’t know how to complete the other implication. My idea is to replicate the proof of the general case (Humphreys pg 22).
Some ideas?
Best Answer
The trace form of the Lie algebra $\mathfrak{gl}_n(F)$ is nondegenerate over a field $F$ of characteristic zero. However, $\mathfrak{gl}_n(F)$ is not semisimple. So the converse is not true.
However, it is known, see Bourbaki's book on Lie algebras, that a Lie algebra $L$ having a nondegenerate trace form for some representation is reductive. For references see this post:
Killing form of a reductive symmetric Lie algebra
Also, if you require that all trace forms $B_{\rho}$ of $L$ are nondegenerate, then $L$ is semisimple, See this post:
Semisimplicity of Lie algebras and non-degeneracy of associated bilinear forms of representations