# A reductive lie algebra $L$ satisfies that $[L,L]$ is semisimple

Let $$L$$ be a reductive Lie algebra (i.e., $$\mathrm{Z}(L) = \operatorname{Rad}(L)$$). Then $$\mathrm{ad} \colon L \to \mathfrak{gl}(V)$$ is completely reducible, $$L = [L, L] \oplus \operatorname{Z}(L)$$ and $$[L, L]$$ is semisimple.

I was able to prove that $$\mathrm{ad}$$ is a completely reducible representation, and if I assume the second part then $$[L, L] \cong L/\mathrm{Z}(L)$$ which is semisimple.

My attempt of the second part:

We have that since $$L/\mathrm{Z}(L)$$ is semisimple, so $$[L, L]/\mathrm{Z}(L) \cong [L/\mathrm{Z}(L), L/\mathrm{Z}(L)] = L/\mathrm{Z}(L) \,.$$ And therefore for every $$z \in L$$ there exist $$y \in [L, L]$$ and $$c \in \mathrm{Z}(L)$$ s.t. $$z = y + c$$. What I’m stuck with is proving that $$[L, L] \cap \mathrm{Z}(L) = \{ 0 \}$$.

Any help would be appreciated.

• In this post we show why $[L,L]\cap Z(L)=0$ then. Nov 23, 2022 at 21:39
• @DietrichBurde He seems to be using the fact that $[L,L]$ is semisimple which makes the claim circular. Do you know a proof that $[L,L]$ is semisimple that is independent of the composition? Nov 23, 2022 at 21:46
• No, it is only used that the subrepresentation $[L,L]$ of the adjoint representation is semisimple, because we already know that the adjoint representation is semisimple. Nov 24, 2022 at 9:26
• And "semisimple" for a representation means "completely reducible". Nov 24, 2022 at 9:45
• See also this post, and this one. They also show that the commutator subalgebra $[L,L]$ is a semisimple Lie algebra. Nov 24, 2022 at 10:58

By complete reducibility, the subrepresentation $$Z(L)$$ has a complement, say

$$L \simeq A \oplus Z(L)$$

(as $$L$$-modules). In particular, $$A$$ is a subalgebra. $$A \simeq L/Z(L)$$ is semisimple by the assumption $$Z(L)=rad(L)$$, hence it is perfect (i.e. $$[A,A]=A$$).

But then $$[L,L]=[A+Z(L), A+Z(L)]=[A,A]=A$$, i.e. $$A$$ was the derived subalgebra all along.

Let $$L$$ be a finite-dimensional Lie algebra whose adjoint representation is completely reducible. This means that we have a decomposition $$L = L_1 ⊕ \dotsb ⊕ L_n \tag{\ast}$$ into irreducible subrepresentations $$L_j$$.

Each $$L_j$$ is thus an ideal of $$L$$, and $$(\ast)$$ is a decomposition of $$L$$ into ideals. This implies that the Lie bracket of $$L$$ is calculated summand-wise with respect to the decomposition $$(\ast)$$; i.e., we have $$[x_1 + \dotsb + x_n,\; y_1 + \dotsb + y_n] = [x_1, y_1] + \dotsb + [x_n, y_n]$$ for all $$x_j, y_j ∈ L_j$$. This has the following consequences.

• We have $$[L, L] = [L_1, L_1] ⊕ \dotsb ⊕ [L_n, L_n]$$.

• We have $$\mathrm{Z}(L) = \mathrm{Z}(L_1) ⊕ \dotsb ⊕ \mathrm{Z}(L_n)$$.

• The $$L$$-subrepresentations of $$L_j$$ are precisely the $$L_j$$-subrepresentations of $$L_j$$. But $$L_j$$ is irreducible as an $$L$$-representations, so it is irreducible as an $$L_j$$-representation. This means that $$L_j$$ is either simple as a Lie algebra, or one-dimensional and abelian.

We may rearrange the summands $$L_1, \dotsc, L_n$$ in such a way that $$L_1, \dotsc, L_m$$ are simple and $$L_{m + 1}, \dotsc, L_n$$ are one-dimensional and abelian. We find that \begin{align*} [L, L] &= [L_1, L_1] ⊕ \dotsb ⊕ [L_m, L_m] ⊕ [L_{m+1}, L_{m+1}] ⊕ \dotsb ⊕ [L_n, L_n] \\ &= L_1 ⊕ \dotsb ⊕ L_m ⊕ 0 ⊕ \dotsb ⊕ 0 \\ &= L_1 ⊕ \dotsb ⊕ L_m \end{align*} is a sum of simple Lie algebras, and therefore semi-simple. We also find that \begin{align*} \mathrm{Z}(L) &= \mathrm{Z}(L_1) ⊕ \dotsb ⊕ \mathrm{Z}(L_m) ⊕ \mathrm{Z}(L_{m+1}) ⊕ \dotsb ⊕ \mathrm{Z}(L_n) \\ &= 0 ⊕ \dotsb ⊕ 0 ⊕ L_{m+1} ⊕ \dotsb ⊕ L_n \\ &= L_{m+1} ⊕ \dotsb ⊕ L_n \,, \end{align*} and therefore $$L = L_1 ⊕ \dotsb ⊕ L_m ⊕ L_{m+1} ⊕ \dotsb ⊕ L_n = [L, L] ⊕ \mathrm{Z}(L) \,.$$