Is there any example of a Lie Algebra, who has nontrivial radical but contains no abelian ideal?

Here, the radical of a Lie algebra means its maximal solvable ideal.

This question occurs in the proof of the theorem which states that "A Lie algebra is semsimple if and only if its killing form is nondegenerate." In the proof, it is written that "to prove that $L$ is semisimple, it will suffice to prove that every abelian ideal of $L$ lies in the radical of the killing form." So I am wondering what if $L$ contains no abelian ideal whiling being semisimple.

Is it possible?

Many thanks.

  • $\begingroup$ In what proof of that theorem? There are many proofs... Presumably you are reading a specific book? $\endgroup$ Mar 13 '11 at 19:06
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    $\begingroup$ Also, you did not write in the last sentence in your second paragraph what you meant. $\endgroup$ Mar 13 '11 at 19:08

If the radical $\mathfrak r$ of a Lie algebra $\mathfrak g$, then $\mathfrak r$ is a solvable Lie algebra. It follows that either $[\mathfrak r, \mathfrak r]$ is zero, so that $\mathfrak r$ is abelian, or $[\mathfrak r, \mathfrak r]$ is a non-trivial nilpotent ideal in $\mathfrak r$. In the last case, then the center of $[\mathfrak r,\mathfrak r]$, which is not trivial because of nilpotency, is an abelian ideal of $\mathfrak g$.

  • $\begingroup$ Yes, yes. It is so annoying to type \mathfrak r each time! :) $\endgroup$ Mar 13 '11 at 19:37
  • $\begingroup$ I am sorry for giving out so many problems on Lie algebra and keeping you doing the annoying typing. I hope someday I can do some favor for you :) BTW, you are an Argentinian? I love Argentina because of Maradona and Messi~ $\endgroup$ Mar 14 '11 at 20:26

Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{r}$ its radical, and $\mathfrak{a}$ the center of $\mathfrak{r}$ (the set of $x \in \mathfrak{r}$ s.t. for all $y \in \mathfrak{r}$, $[x,y]=0$). By the Jacobi identity, and since $\mathfrak{r}$ is an ideal of $\mathfrak{g}$, $\mathfrak{a}$ is an ideal of $\mathfrak{g}$.

Moreover a nontrivial solvable Lie algebra has nontrivial center. EDIT: This last sentence is indeed false (thank you Matt E). So it's easier to formulate it this way: by Jacobi, whenever $\mathfrak{a}$ and $\mathfrak{b}$ are ideals of a Lie algebra, $[\mathfrak{a},\mathfrak{b}]$ is also an ideal, so the last non-trivial $\mathcal{D}^n \mathfrak{r}$ is an abelian ideal of $\mathfrak{g}$.

  • $\begingroup$ ..and a solvable Lie algebra has non-trivial center. Remember: the question is: why is there a non-trivial abelian ideal! $\endgroup$ Mar 13 '11 at 19:21
  • $\begingroup$ Dear Plop, It's not true that a non-trivial solvable Lie algebra has a non-trivial centre. E.g. the unique (up to isomorphism) two-dimensional solvable Lie algebra has trivial centre. Regards, $\endgroup$
    – Matt E
    Mar 13 '11 at 19:27
  • $\begingroup$ @Matt: If I am not mistaken, over any field $k$ there are up to isomorphism exactly two $2$-dimensional Lie algebras, both of which are solvable. One of them of course has trivial bracket and thus has nontrivial center. The other, as you say, is solvable but has trivial center. $\endgroup$ Mar 13 '11 at 19:35
  • $\begingroup$ @Pete: Dear Pete, You're right of course; I meant to say the (unique up to isomorphism) non-abelian two-dimensional algebra. Best wishes, $\endgroup$
    – Matt E
    Mar 13 '11 at 20:05
  • $\begingroup$ I prefer Plop's use of the derived series since it generalizes to groups and non-connected Lie groups. The situation that the derived sub-thingy is nilpotent is pretty weird (and rare outside of Lie algebras and connected Lie groups). The fact that the second to last derived sub-thingy is abelian is just the definition of derived series terminating in the identity. $\endgroup$ Mar 14 '11 at 1:55

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