4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Oh crap..it is the complete opposite. Don't know why I did that. I will correct this. $\endgroup$ – B. Pasternak Mar 27 '17 at 15:21
  • $\begingroup$ @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. $\endgroup$ – B. Pasternak Mar 27 '17 at 15:25
  • $\begingroup$ Why does $L'$ decompose into a direct sum of irreducible $ad L$-submodules? $\endgroup$ – JDZ Jul 10 '18 at 16:43
  • $\begingroup$ @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 $W\oplus W'=V$). $\endgroup$ – B. Pasternak Oct 24 '18 at 13:21
0
$\begingroup$

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.

$\endgroup$

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.