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 $\phi={\rm ad}$.

  • $\begingroup$ 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? $\endgroup$
    – Sarah
    Apr 12 at 19:04
  • 1
    $\begingroup$ You mean Theorem $7$? There the first line says "Proof: Without loss of generality we may take ${{\mathfrak r}}$ to be the radical of ${{\mathfrak g}}$". $\endgroup$ Apr 12 at 19:22

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