# Let ${{\mathfrak g}}$ be a finite-dim. Lie-algebra and ${\mathfrak r}$ a solvable Lie-ideal, does ${\mathfrak r}$ contain $Z({\mathfrak g})$?

Let $${{\mathfrak g}}$$ be a finite-dimensional Lie-algebra and $${{\mathfrak r}}$$ a solvable Lie-ideal, does $${{\mathfrak r}}$$ contain $${Z({\mathfrak g})}$$ ?

I know this is true for $${{\mathfrak r}} = Rad ~ {\mathfrak g}$$, but why is it true in general?

In particular, I am trying to understand Theorem 7 in the proof of Ado by Terence Tao(https://terrytao.wordpress.com/2011/05/10/ados-theorem/): "Let $${{\mathfrak g}}$$ be a finite-dimensional (abstract) Lie algebra, and let $${{\mathfrak r}}$$ be a solvable ideal in $${{\mathfrak g}}$$. Then $${[{\mathfrak g},{\mathfrak g}] \cap {\mathfrak r}}$$ is an (abstractly) nilpotent ideal of $${{\mathfrak g}}.$$"

I do not understand one of his last arguments: "To establish the claim in the abstract case, we simply use the adjoint representation, which effectively quotients out the centre $${Z({\mathfrak g})}$$ of $${{\mathfrak g}}$$ (which will also be an ideal of $${{\mathfrak r}})$$ to convert an finite-dimensional abstract Lie algebra into a concrete Lie algebra over a finite-dimensional space."

Thank you!

No, it need not hold in general. Consider an abelian Lie algebra $$L$$ of dimension $$n\ge 2$$ with basis $$(e_1,\ldots ,e_n)$$ and choose a $$1$$-dimensional subalgebra $$\mathfrak{r}$$ generated by $$e_1\in L$$. Then $$\mathfrak{r}$$ is a solvable ideal in $$L$$, which doesn't contain $$Z(L)=L$$.
Concerning Tao's proof, we just have $${\rm ad}(\mathfrak{g})\cong \mathfrak{g}/Z(\mathfrak{g})$$ by using $$\phi(L)\cong L/\ker(\phi)$$ for the representation $$\phi={\rm ad}$$.
• Thank you! But I have trouble with the part: When $([g,g] \cap r )/ Z(g)$ is nilpotent, then so is $[g,g] \cap r$. I know this wokrs if $g/ Z(g)$ is nilpotent, so is $g$. But if $([g,g] \cap r )$ does not contain $Z(g)$, how does the argument go? Apr 12 '21 at 19:04
• You mean Theorem $7$? There the first line says "Proof: Without loss of generality we may take ${{\mathfrak r}}$ to be the radical of ${{\mathfrak g}}$". Apr 12 '21 at 19:22