# Levi Decomposition and relations with nilradicals

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?

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 $${\rm nil}(L)$$.