0
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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 $\phi={\rm ad}$.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.