$\frak{g}^{\bot}$ $=Z(\frak{g})$ if and only if $\frak{g}$ is reductive

Question

Let $\frak{g}$ be a finite dimensional Lie algebra over $k$, a field of characteristic $0$.

Recall that $\frak{g}$ is called reductive if the center $Z(\frak{g})$ is equal to the radical, rad$(\frak{g})$, the largest solvable ideal of $\frak{g}$.

Let $\kappa$ denote the killing form, and $\frak{g}^{\bot} = \{$ $x \in \frak{g}:$ $\forall y \in \frak{g}$, $\kappa(x,y) = 0 \}$ its radical.

Show $\frak{g}^{\bot}$ $=Z(\frak{g})$ if and only if $\frak{g}$ is reductive.

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Assume first that $\frak{g}$ is reductive.

By definition $Z(\frak{g}) \subset$ $\frak{g}^{\bot}$, so we need the converse inclusion. By Cartan's criterion for solvability $\frak{g}^{\bot}$ is solvable, hence, as it is an ideal, $\frak{g}^{\bot} \subset$ rad$(\frak{g})$ as required.

The other direction is what I'm having issues with:

I know that if $I$ is a solvable ideal of $\frak{g}$, the last term in its derived series is abelian. Also it is true that for any abelian ideal $J$, $\kappa(J, \frak{g}) = 0$ as for $j \in J, x \in \frak{g}$ $ad(j)ad(x)$ is nilpotent.

So if we assume $\frak{g}^{\bot}$ $=Z(\frak{g})$, we get that $J \subset Z(\frak{g})$, and moreover $I \cap Z(\frak{g}) \neq \{0\}$.

Questions:

1. Can I continue this attempt for the unsolved direction?

2. If not, can you provide a hint to continue?

Answer 1

Let $I$ be a solvable ideal, and $I,[I,I]=I^2,...,[I^{n-1},I^{n-1}]=I^n$ its derived series. Suppose $I^n\neq 0$, you have already shown that $I^n$ is in the center.

Remark that for every $x \in I^{n-1}, y,z\in g$, $(ad_x\circ ad_y)(z)=[x,[y,z]]$ since $x$ is an element of the ideal $I^{n-1}$ we deduce that $a=[x,[y,z]]\in I^{n-1}$.

We have $(ad_x\circ ad_y)^2(z)=(ad_x(ad_y(a))=[x,[y,a]]$ since $x$ and $[y,a]$ are in $I^{n-1}$, we deduce that $b=[x,[y,a]]\in I^n$. This implies that $(ad_x\circ ad_y)^3(z)=(ad_x(ad_y(b))=0$, we deduce that $ad_x\circ ad_y$ is nilpotent and $\kappa(x,y)=0$. This implies by hypothesis that $x\in Z(g)$. We deduce that $I^{n-1}\subset Z(g)$. We can apply a recursive argument to show that $I^{n-2},...,I^2,I \subset Z(g)$, this implies that $R(g)\subset Z(g)$.