Consider a Lie algebra $\mathfrak{g}$ that can be view as a subset of some $\mathfrak{gl}(V)$.

We can decompose $\mathfrak{g}$ using the Levi decomposition, so we write it as the semidirect decomposition of its radical $\mathfrak{r}$ (the maximal solvable ideal) and a semisimple Lie algebra $\mathfrak{s}$. As vector space we have: $$\mathfrak{g}=\mathfrak{s}\oplus \mathfrak{r}$$ Now the solvable ideal can be decomposed as a sum of its nilradical $\mathfrak{n}$ (the maximal nilpotent ideal) and a complement, let's call it $\mathfrak{a}$.

In the end, we have

$$\mathfrak{g}=\mathfrak{s}\oplus \mathfrak{n}\oplus\mathfrak{a}$$

Obviuosly we have $[\mathfrak{s},\mathfrak{s}]=\mathfrak{s}$, and $[\mathfrak{s},\mathfrak{r}],[\mathfrak{r},\mathfrak{r}]\subset\mathfrak{r}$.

The question is the following: can $\mathfrak{n}$ be considered as and ideal (and hence as the nilradical) of $\mathfrak{g}$ ? If not, is $\mathfrak{n}$ unique?

Can be alwayse chosen the complement $\mathfrak{a}$ to be abelian?


1 Answer 1


Let $L$ be a Lie algebra and ${\rm rad}(L)$ be the solvable radical and ${\rm nil}(L)$ the nilpotent radical, i.e., the maximal solvable resp. nilpotent ideal in $L$. Then $$ {\rm nil}({\rm rad}(L))={\rm nil}(L), $$ because ${\rm nil}(L)$ is also a nilpotent ideal in ${\rm rad}(L)$. Hence the nilradical of the solvable radical of $L$ coincides with the nilradical of $L$ itself.

In particular, your $\mathfrak{n}$ is just the nilradical of $L$ and hence an ideal. And indeed, ${\rm rad}(L)/{\rm nil}(L)$ is always abelian.

Note however that the short exact sequence $$ 0\rightarrow {\rm nil}(L)\rightarrow L\rightarrow L/{\rm nil}(L)\rightarrow 0 $$ does not split in general, i.e., there isn't always a subalgebra complementary to ${\rm nil}(L)$.


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