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I came across an alternative definition of reductive Lie algebra as follows:

$\mathfrak{g}$ is said to be reductive of all abelian ideals of it are contained in its center $Z(\mathfrak{g})$ and $Z(\mathfrak{g})\cap [\mathfrak{g},\mathfrak{g}]=0$.

My question is, does this definition coincide with the usual one with $Z(\mathfrak{g})=Rad(\mathfrak{g})$? In particular, how can one show that $\mathfrak{g}=Z(\mathfrak{g})\oplus \mathfrak{s}$ where the latter one is semisimple and in fact is indeed $[\mathfrak{g}, \mathfrak{g}]$? Thank you in advance!

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2 Answers 2

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Suppose that $\mathfrak{g}$ is reductive. Then its adjoint representation is semisimple, so that the subrepresentation $[\mathfrak{g}, \mathfrak{g}]$ has a complement $W$ by Weyl's theorem, i.e., we have $$ \mathfrak{g}=[\mathfrak{g}, \mathfrak{g}]\oplus W. $$ Because of $[\mathfrak{g},W]\subseteq [\mathfrak{g}, \mathfrak{g}] \cap W=0$ we have $W\subseteq Z(\mathfrak{g})$, so that $[\mathfrak{g}, \mathfrak{g}]+Z(\mathfrak{g})=\mathfrak{g}$. This sum is in fact direct, since the subrepresentation $[\mathfrak{g}, \mathfrak{g}]$ is semisimple, so that we have a complement $U$ to $[\mathfrak{g}, \mathfrak{g}]\cap Z(\mathfrak{g})$, and $$ [\mathfrak{g}, \mathfrak{g}]=[\mathfrak{g}, [\mathfrak{g}, \mathfrak{g}] +Z(\mathfrak{g})] =[\mathfrak{g}, [\mathfrak{g}, \mathfrak{g}]]=[\mathfrak{g},U]\subseteq U, $$ hence $[\mathfrak{g}, \mathfrak{g}]\cap Z(\mathfrak{g})=0$. It follows that $\mathfrak{g}=[\mathfrak{g}, \mathfrak{g}] \oplus Z(\mathfrak{g})$, with $[\mathfrak{g}, \mathfrak{g}]$ semisimple, and $Rad(\mathfrak{g})=Z(\mathfrak{g})$.

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  • $\begingroup$ Why is [g,g] semisimple? $\endgroup$
    – JDZ
    Dec 31, 2016 at 23:33
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    $\begingroup$ @JDZ See the first sentence above: "First recall that a Lie algebra is semisimple if and only if it does not contain nonzero abelian ideals". The definition given by the OP then implies that $[g,g]$ is semisimple. $\endgroup$ Jan 1, 2017 at 16:37
  • $\begingroup$ @Dietrich Burde how did you induce that Rad(g)=Z(g) in the end of the proof? $\endgroup$
    – user864806
    May 28, 2022 at 9:28
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    $\begingroup$ @Bestmat See below: "so that semisimplicity of $[\mathfrak{g}, \mathfrak{g}]$ implies $[\mathfrak{g}, \mathfrak{r}]=0$, that is, the radical is central. " $\endgroup$ May 28, 2022 at 9:45
  • $\begingroup$ Thanks ,but if $g=r\oplus s$ (I am not sure if the symbol written means direct sum) then why $[g,g]=[g,r]\oplus s$? And $[g,r]=0$ since it is contained in $r$ while $[g,g]$ is semisimple having no solveble ideals right? @Dietrich Burde $\endgroup$
    – user864806
    May 28, 2022 at 11:36
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First recall that a Lie algebra is semisimple if and only if it does not contain nonzero abelian ideals. This is because for non-semisimple Lie algebras the last nonzero term in the series of commutator subalgebras of its radical is an abelian ideal.

In your formulation, since all abelian ideals are contained in $\mathfrak{z}$ and $\mathfrak{z}\cap [\mathfrak{g}, \mathfrak{g}]=0$ we conclude that $[\mathfrak{g}, \mathfrak{g}]$ is semisimple.

Some authors call a Lie algebra $\mathfrak{g}$ reductive if $[\mathfrak{g}, \mathfrak{g}]$ is semisimple. This is equivalent to the other more common definition. To see this let $$\mathfrak{g}=\mathfrak{r}\rtimes\mathfrak{s}$$ be a Levi decomposition of $\mathfrak{g}$. Then $$[\mathfrak{g}, \mathfrak{g}]=[\mathfrak{g}, \mathfrak{r}]\rtimes \mathfrak{s},$$ so that semisimplicity of $[\mathfrak{g}, \mathfrak{g}]$ implies $[\mathfrak{g}, \mathfrak{r}]=0$, that is, the radical is central. Therefore $$\mathfrak{g}=\mathfrak{r}\oplus\mathfrak{s}.$$

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    $\begingroup$ Could you please explain your second paragraph? Why is an abelian ideal of $[\mathfrak{g},\mathfrak{g}]$ an ideal of $\mathfrak{g}$? Thanks! $\endgroup$ Apr 12, 2022 at 5:25

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