Nilradical is contained Killing radical.

Question

I have a question about Lie algebras.

The exercise is to show that $\mathfrak{n}\subset\text{rad}(K)$, where $\mathfrak{n}$ is the nilradical and $\text{rad}(K)$ is the radical of the Killing form $K$ and that they do not coincide in general.

The hint was to use Engel's theorem, that is $x\in\mathfrak{n}$ iff $\text{ad}(x)$ is nilpotent, that is $\text{ad}(x)^n=0$ for sufficiently high $n$. This implies that $\text{tr}(\text{ad}(x))=0$ as all eigenvalues are $0$, but I am not able to deduce from this that $\text{tr}(\text{ad}(x)\circ \text{ad}(y))=0$ or (which suffice) $\text{tr}((\text{ad}(x)\circ \text{ad}(y))^n)=0$ for all $y\in\mathfrak{g}$ and some $n$.

Answer 1

You can show that $\operatorname{ad}(y) \operatorname{ad}(x)$ is nilpotent when $x$ lies in the nilradical. Write down a high power of $\operatorname{ad}(y) \operatorname{ad}(x)$: $$(\operatorname{ad}(y) \operatorname{ad}(x))^n = \operatorname{ad}(y) \operatorname{ad}(x) \cdots \operatorname{ad}(y) \operatorname{ad}(x)\operatorname{ad}(y) \operatorname{ad}(x) \,.$$ Now we use that $\operatorname{ad}(x)\operatorname{ad}(y) - \operatorname{ad}(y)\operatorname{ad}(x) = \operatorname{ad}([x, y])$ in order to move the right-most factor $\operatorname{ad}(y)$ one position to the left: $$\begin{align*} &\operatorname{ad}(y) \operatorname{ad}(x) \cdots \operatorname{ad}(y) \operatorname{ad}(x) \operatorname{ad}(y) \color{red}{ \operatorname{ad}(x) \operatorname{ad}(y)} \operatorname{ad}(x) \\ & \quad = \operatorname{ad}(y) \operatorname{ad}(x) \cdots \operatorname{ad}(y) \operatorname{ad}(x) \operatorname{ad}(y) \color{red}{\operatorname{ad}(y)\operatorname{ad}(x)} \operatorname{ad}(x) \\ & \qquad + \operatorname{ad}(y) \operatorname{ad}(x) \cdots \operatorname{ad}(y) \operatorname{ad}(x) \operatorname{ad}(y) \color{red}{\operatorname{ad}([x, y])} \operatorname{ad}(x) \end{align*}$$

We can keep moving factors $\operatorname{ad}(y)$ to the left, until we obtain a finite sum of products of the form $$\operatorname{ad}(y)^k \cdot (\text{product of }n\text{ factors }\operatorname{ad}(x) \text{ or }\operatorname{ad}([x,y])\text{ in any order}) \,.$$ It suffices now to show that such a product is zero when $n$ is large enough. This is because $\operatorname{ad}(\mathfrak n)^n(\mathfrak g) \subset \operatorname{ad}(\mathfrak n)^{n-1}(\mathfrak n)$ is zero when $n$ is large enough, because $\mathfrak n$ is nilpotent.

---

If in the argument you do not start with the right-most factors $\operatorname{ad}(y)$, you will in the end also get factors of the form $\operatorname{ad}([[x,y], y]), \operatorname{ad}([[[x,y], y], y]), \ldots$ but that is no problem as those brackets still lie in the nilradical.