Abelian, Nilpotent and Soluable Lie Algebras

Definition 1. A Lie algebra $L$ is abelian if $[L,L]=0$, which means for all $x,y\in L,\ [x,y]=0$.

Notation:
$$
L^1=L,\qquad L^{n+1}=[L^n,L]\qquad (n\geq 1)
$$

Lemma 1. If $I,J$ are ideals of $L$, so is $[I,J]$.

Proof: Let $x\in I,\ y\in J,\ z\in L$, then
$$
[[x,y],z]=[x,[y,z]]-[y,[x,z]]\in [I,J].\quad \square
$$

Proposition 1. (i) $L^n$ is an ideal of $L$,

(ii) Also
$$
L=L^1\supseteq L^2\supseteq L^3\supseteq \cdots
$$

Proof: (i) Follows from the above lemma.

(ii)
$$
L^{n+1}=[L^n,L]\subseteq L^n.\quad \square
$$

Definition 2. A Lie algebra $L$ is nilpotent if $L^n=0$ for some $n\geq 1$,

Remark: Every subalgebra and every quotient algebra of a nilpotent Lie algebra are nilpotent.

Notation:
$$
L^{(0)}=L,\qquad L^{(n+1)}=[L^{(n)},L^{(n)}]\qquad (n\geq 0).
$$

Proposition 2. (i) $L^{(n)}$ is an ideal of $L$.

(ii)
$$
L=L^{(0)}\supseteq L^{(1)}\supseteq L^{(2)}\supseteq \cdots
$$

Proof: Similar to above. $\square$

Definition 3. A Lie algebra $L$ is soluble if $L^{(n)}=0$ for some $n\geq 0$.

Proposition 3. (i)
$$
[L^m,L^n]\subseteq L^{m+n}\qquad \text{for }m,n\geq 1,
$$

(ii)
$$
L^{(n)}\subseteq L^{2^n}\qquad \text{for }n\geq 0,
$$

(iii) nilpotent implies soluble.

Proof: (i) We prove by induction on $n$. Base case $n=1$ is clear. Inductive step:

Suppose (i) holds for $n\leq r$,
$$
[L^m,L^{r+1}]
=[L^m,[L^r,L]]
=[ [L^m,L],L^r ]+[ [L^m,L^r],L ]
$$
by Jacobi identity. Hence
$$
[L^m,L^{r+1}]
\subseteq [L^{m+1},L^r]+[L^{m+r},L]
\subseteq L^{m+r+1}.
$$
(ii) We prove by induction on $n$. Base case $n=0$ is clear. Inductive step:

Suppose (ii) holds for $n\leq r$,
$$
L^{(r+1)}=[L^{(r)},L^{(r)}]
\subseteq [L^{2^r},L^{2^r}]
\subseteq L^{2^r+2^r}=L^{2^{r+1}}.
$$

(iii) Suppose $L$ is nilpotent. Then $L^n=0$ for some $n$. We can pick $k$ s.t. $2^k\geq n$.

$$
L^{(k)}\subseteq L^{2^k}\subseteq L^n=0.
$$
Hence $L^{(k)}=0$. So $L$ is soluble. $\square$

Proposition 4. Let $I$ be an ideal of $L$. If $I$ and $L/I$ are soluble, so is $L$.

Proof: We have $(L/I)^{(n)}=0$ for some $n$. Then
$$
L^{(n)}\subseteq I.
$$

Also, $I^{(m)}=0$, for some $m$. Hence
$$
L^{(n+m)}=(L^{(n)})^{(m)}\subseteq I^{(m)}=0.
$$
So $L$ is soluble. $\square$

Proposition 5. (i) Every finite dimensional Lie algebra $L$ contains a unique maximal soluble ideal $R$.

(ii) Moreover, $L/R$ contains no non-zero soluble ideal.

Proof: (i) Let $I,J$ be soluble ideals of $L$. Then $I+J$ is an ideal of $L$.

Also, $I$ is a soluble ideal of $I+J$. And

$$
(I+J)/I \cong J/(I\cap J),
$$
which is soluble. Therefore, by proposition 4, $I+J$ is soluble.

We have thus shown that the sum of two soluble ideals is soluble, it follows that $L$ has a unique maximal soluble ideal $R$. $\square$

(ii) Let $I/R$ be a soluble ideal of $L/R$. Then $I$ is a soluble ideal of $L$. By part (i), we have

$$
I\subseteq R.
$$
Therefore,
$$
I=R,
$$
and hence
$$
I/R=0.\quad \square
$$
Definition 4. $R$ is called the soluble radical of $L$.

Definition 5. A Lie algebra $L$ is semisimple if $R=0$. In other words, $L$ is semisimple if and only if $L$ has no non-zero soluble ideal.

Definition 6. $L$ is simple if $L$ has no proper ideal.

Example 1. Suppose $L$ has $\dim L=1$. Then $L$ has a basis $\{e\}$. Since $[e,e]=0$, we have
$$
L^2=0.
$$
So $L$ is abelian. Also, $L$ is simple. $L$ is called the trivial simple Lie algebra.

Proposition 6. Every non-trivial simple Lie algebra is semisimple.

Proof: Suppose $L$ is simple but not semisimple. Then
$$
R\neq 0.
$$

Since $L$ is simple and $R$ is an ideal of $L$, we must have
$$
R=L.
$$

Because $R$ is soluble, there exists some $n\geq 0$ such that
$$
L^{(n)}=0.
$$

Then
$$
L^{(1)}\neq L,
$$
for otherwise we would have
$$
L^{(n)}=L\neq 0
$$
for all $n$, which is a contradiction.

Since $L^{(1)}$ is an ideal of $L$ and $L$ is simple, it follows that
$$
L^{(1)}=0.
$$

That is,
$$
[L,L]=0.
$$

Therefore, every subspace of $L$ is an ideal of $L$.

But $L$ is simple, so necessarily
$$
\dim L=1.
$$

This contradicts the assumption that $L$ is non-trivial. Hence $L$ must be semisimple. $\square$

Reference

This series of articles on Lie algebras is based on Carter R., Lie Algebras of Finite and Affine Type (Cambridge University Press, 2005), as well as the class notes for Introduction to Lie Algebras by Chenwei Ruan at BIMSA.

The cover image of this article was taken in Wellington, New Zealand.