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If $L$ is a Lie algebra, $\text{Rad}(L)$ denotes its largest solvable ideal. Then $L$ is reductive if $\text{Rad}(L) = Z(L)$ (the center of $L$).

An exercise in Humphreys asks: $L$ is reductive if and only if $L$ is a completely reducible $\text{ad}(L)$-module. Doesn't this latter criterion amount to saying that $L$ is a semisimple Lie algebra? I'm under this impression for the following reason. $\text{ad}(L)$ acts on $L$ by $\text{ad}(x)y := [x, y]$, right? This is exactly how $L$ acts on $L$ too, so I don't see why $L$ wouldn't have the same structure as an $L$-module as it does an $\text{ad}(L)$-module. The issue is that not all reductive Lie algebras are semisimple --- for example, if $L$ is (nonzero and) abelian then it is reductive but not semisimple.

So here's my "proof" that if $L$ is reductive, then $L$ is semisimple. We know $\text{Rad}(L) = Z(L)$, so $\text{ad}(L) \simeq L/Z(L) = L/\text{rad}(L)$ which is semisimple. Therefore every module for $\text{ad}(L)$ is completely reducible (Weyl's Theorem). In particular $L$ is completely reducible as a module for $\text{ad}(L)\simeq L/Z(L)$. But $Z(L)$ annihilates $L$, so the submodule structure of $L$ as an $L/Z(L)$-module and an $L$-module are exactly the same. Therefore $L$ is completely reducible.

Am I being too liberal with "$\simeq$"? Or maybe throwing around annihilators too sloppily?

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    $\begingroup$ Completely reducible does not imply semisimple (just think of an abelian Lie algebra). $\endgroup$ Apr 6, 2014 at 18:54
  • $\begingroup$ Ah, that makes a lot of sense! So my claim that $L$ has the submodules as an $L$-module as it does an $\text{ad}(L)$-module is correct? $\endgroup$
    – Ehsaan
    Apr 6, 2014 at 18:59
  • $\begingroup$ If the lie algebra is reductive, why $Rad(L)=Z(L)$? @Tobias Kildetoft using a def of reductive lie algebra that defines it as a finite dim lie algebra that is completely reducible as an L-module. $\endgroup$
    – user864806
    May 27, 2022 at 17:11

1 Answer 1

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L being a completely reducible $ad(L)$-module means that $L=L_1 \oplus \dots \oplus L_t$ where each $L_i$ is a irreducible $ad(L)$-submodule. The fact that $L_i$ is a $ad(L)$-submodule translates into $L_i$ being an ideal, but irreducibility does not imply simplicity of the ideal. Such implication is true only when $dim L_i > 1$. So that is how I see the difference of semisimple and reductive Lie algebra: the first one can be written as sum of simple ideals, while the second can be written as sum of ideals, but not all simple. There might be some ideals of dimension $1$. The sum of those ideals of dimension $1$ is exactly the center and the radical of the algebra.

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