$\text{sl}(2,k)$ is not nilpotent

Question

Why is $L=\text{sl}(2,k)$, $k$ a field, not nilpotent as a Lie algebra (of traceless $2\times 2$ matrices)?

We have to show that its centralised series $C^0 L=L, C^{i+1}L=[L,C^i L]$ doesn't stabilise as $0$.

Answer 1

Let$$H=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\ X=\begin{pmatrix}0&1\\0&0\end{pmatrix}\text{, and }Y=\begin{pmatrix}0&0\\1&0\end{pmatrix}.$$Then $(H,X,Y)$ is a basis of $\mathfrak{sl}(2,k)$. Furthermore, $[H,X]=2X$, $[H,Y]=-2Y$, and $[X,Y]=H$. But then $\bigl[\mathfrak{sl}(2,k),\mathfrak{sl}(2,k)\bigr]=\mathfrak{sl}(2,k)$. Therefore, $\mathfrak{sl}(2,k)$ is not nilpotent. To be more precise,$$(\forall n\in\mathbb{Z}_+):\mathcal{C}^n\bigl(\mathfrak{sl}(2,k)\bigr)=\mathfrak{sl}(2,k)$$

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I will add here somehting from the comments. What was written above is valid if the characteristic of $k$ is different from $2$. However, if it is equal to $2$, then $\mathfrak{sl}(2,k)$ is nilpotent. In fact, it follows then from the computations made above that then $\mathcal{C}^1\bigl(\mathfrak{sl}(2,k)\bigr)=kH$. But, using again the fact that we are working in characteristic $2$,$$kH=\{\text{scalar matrices}\},$$since $H=\operatorname{Id}$. Therefore $\mathcal{C}^2\bigl(\mathfrak{sl}(2,k)\bigr)=\{0\}$.

Answer 2

By Engel's theorem, $L$ is nilpotent if and only if all adjoint operators $\operatorname{ad}(x)$ are nilpotent. Hence, if $2\neq 0$, $L$ is not nilpotent, since $\operatorname{ad}(h)=\operatorname{diag}(2,-2,0)$ is not nilpotent in this case. If $2=0$ in $K$, then all adjoint operators are nilpotent, hence $L$ is nilpotent.