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.

  • 1
    $\begingroup$ In this post we show why $[L,L]\cap Z(L)=0$ then. $\endgroup$ Nov 23, 2022 at 21:39
  • $\begingroup$ @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? $\endgroup$
    – Math101
    Nov 23, 2022 at 21:46
  • $\begingroup$ 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. $\endgroup$ Nov 24, 2022 at 9:26
  • $\begingroup$ And "semisimple" for a representation means "completely reducible". $\endgroup$ Nov 24, 2022 at 9:45
  • 1
    $\begingroup$ See also this post, and this one. They also show that the commutator subalgebra $[L,L]$ is a semisimple Lie algebra. $\endgroup$ Nov 24, 2022 at 10:58

2 Answers 2


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 $$ L = L_1 ⊕ \dotsb ⊕ L_m ⊕ L_{m+1} ⊕ \dotsb ⊕ L_n = [L, L] ⊕ \mathrm{Z}(L) \,. $$


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