# If a Lie algebra $L$ is a completely reducible $\mathrm{ad}\,L$-module, then $L$ is reductive.

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.

• Oh crap..it is the complete opposite. Don't know why I did that. I will correct this. 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. 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 $W\oplus W'=V$). 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 $L$. Apr 2 '20 at 0:54

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.