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Let $L$ be a Lie algebra over a algebraically closed field with characteristic zero. Call $L$ reductive if $\mathrm{Rad}\,L=Z(L)$.

Suppose that $L$ is a completely reducible $\mathrm{ad}\,L$-module. I am trying to prove that $L$ is reductive.

Note first that an $\mathrm{ad}\,L$-submodule of $L$ is the same as an ideal of $L$.

Since $Z(L)$ is an ideal of $L$ and $L$ is a completely reducible $\mathrm{ad}\,L$-module, there exists some $\mathrm{ad}\,L$-submodule $L^\prime$ of $L$ such that $L=Z(L)\oplus L^\prime$. This $L^\prime$ then decomposes as a direct sum of irreducible $\mathrm{ad}\,L$-submodules, i.e. as a direct sum of non-zero ideals. It is clear that each summand is either simple or one dimensional. Then $[LL]$ is a direct sum of simple ideals, so $[LL]$ is semisimple. Since $[LL]$ is an irreducible $\mathrm{ad}\,L$-submodule we have that $L=[LL]\oplus M$ for some $\mathrm{ad}\,L$-submodule $M$ of $L$. Since $M\cong L/[LL]$, $M$ is abelian.

Edit: as Dietrich Burde pointed out, I wasn't thinking when I wrote that $Z(M)=0$. In fact, $Z(M)=M$. Since the center is solvable we also have that $\mathrm{Rad}\,M=M$, which proves the statement. I think it is okay like this.

B. Pasternak
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  • Oh crap..it is the complete opposite. Don't know why I did that. I will correct this. – B. Pasternak Mar 27 '17 at 15:21
  • @DietrichBurde I was actually also a little bit unsure as to why $Z([LL]\oplus M)=Z([LL])\oplus Z(M)$ would hold. For the radical it is true since there is one inclusion because $\mathrm{Rad}([LL])\oplus\mathrm{Rad},M$ is solvable, and the other is there because the canonical images of $\mathrm{Rad}([LL])\oplus M)$ in $[LL]$ and $M$ are solvable. – B. Pasternak Mar 27 '17 at 15:25
  • Why does $L'$ decompose into a direct sum of irreducible $ad L$-submodules? – JDZ Jul 10 '18 at 16:43
  • @JDZ A module $V$ is completely reducible if it is a direct sum of irreducible modules, or equivalently, if every submodule has a complement (i.e. for every submodule W there exists a submodule W' such that $W\oplus W'=V$). – B. Pasternak Oct 24 '18 at 13:21
  • How do you know an $ad(L)$-submodule of $L$ is the same as an ideal of $L$? We don't even know the action of $ad(L)$ on $L$. – user5826 Apr 02 '20 at 00:54
  • If you know where in Humphrey's book this question comes from, I think this comment would answer itself. The action of ad(L) on L is determined by the inclusion ad(L)\subset gl(L). Perhaps easier just to say that its action is determined by the iso ad(L)\cong L/Z(L). – B. Pasternak Apr 02 '20 at 14:31

1 Answers1

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Assume $L$ is a completely reducible $\textrm{ad }L$-module.
$\quad$ We claim that every abelian ideal $I\subseteq L$ is contained in $Z(L)$. If $I\subseteq L$ is an abelian ideal, then $I$ is an $\textrm{ad }L$-submodule of $L$, which posseses a complement $J$ (cf. Humphreys' Exercise II.6.2). Then $[IL]=[I,I\oplus J]\subseteq[II]+[IJ]\subseteq I\cap J=0$, so $I\subseteq Z(L)$.
$\quad$ Since $L$ is completely reducible, $L=\bigoplus L_{i}$ for some irreducible $\textrm{ad }L$-submodules $L_{i}$. For each $i,$ consider the ideals $(\textrm{Rad }L)\cap L_{i}\subseteq L_{i}$. Since $L_{i}$ is irreducible, either $(\textrm{Rad }L)\cap L_{i}=L_{i}$ or $(\textrm{Rad }L)\cap L_{i}=0$.
$\quad$ If $(\textrm{Rad }L)\cap L_{i}=L_{i}$, then $L_{i}\subseteq\textrm{Rad }L$ implies $L_{i}$ is solvable. It follows that $L_{i}$ is abelian because either $[L_{i}L_{i}]=L_{i}$ or $[L_{i}L_{i}]=0$, but $[L_{i}L_{i}]=L_{i}$ is contradictory to $L_{i}$ being solvable. Hence $L_{i}\subseteq Z(L)$ in this case, by the above claim.
$\quad$ If $(\textrm{Rad }L)\cap L_{i}=0$, then for any $x\in\textrm{Rad }L$ and $y\in L_{i}$, we have $[xy_{i}]=0$.
$\quad$ Now let $x\in\textrm{Rad }L$ and $y\in L$. Write $y=\sum y_{i}$ for $y_{i}\in L_{i}$. Then $[xy]=\sum[xy_{i}]=0$ since each summand $[xy_{i}]\in(\textrm{Rad }L)\cap L_{i}$, which is either $0$ or contained in $Z(L)$. This proves $\textrm{Rad }L\subseteq Z(L)$, and therefore $L$ is reductive.

JDZ
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