Ideals, Simple and Semisimple Lie Algebras

Lie Algebra Ideals

Lie Algebra Ideal and Quotients

A subspace $\mathfrak{a}$ of a Lie algebra $\mathfrak{g}$ is an ideal, denoted $\text{a}◃\text{g}$, if $[\text{g},\text{a}]\subseteq \text{a}$. That is, for any $x\in \text{g}$ and $y\in \text{a}$, we have $[x,y]\in \text{a}$. If $\text{a}◃\text{g}$, then the quotient $\text{g}/\text{a}$ is well-defined, and $[x+\text{a},y+\text{a}]:=[x,y]+\text{a}$.

e.g.

1. For any homomorphism $\phi :{\text{g}}_1\to {\text{g}}_2$, the kernel $\ker (\phi )$ is an ideal of ${\text{g}}_1$.
2. The center of $\text{g}$, $Z(\text{g})=\{x\in \text{g}\mid [x,y]=0\text{ for all }y\in \text{g}\}$, is an ideal.
3. The centralizer of a subset $S\subseteq \text{g}$, $C_{\text{g}}(S)=\{x\in \text{g}\mid [x,s]=0\text{ for all }s\in S\}$, is an ideal if $S$ is an ideal. (This is sufficient but not necessary)
4. The commutator algebra of $\text{g}$, $[\text{g},\text{g}]=\text{span}\{[x,y]\mid x,y\in \text{g}\}$, is an ideal.
5. For ideals $\text{a},\mathfrak{b}◃\text{g}$, their sum $\text{a}+\mathfrak{b}=\{x+y\mid x\in \text{a},y\in \mathfrak{b}\}$, and their Lie bracket $[\text{a},\mathfrak{b}]=\text{span}\{[x,y]\mid x\in \text{a},y\in \mathfrak{b}\}$ are also ideals.

Simple Lie Algebras

A Lie algebra $\text{g}$ is simple if it has no non-trivial ideals.

e.g. The Lie algebra ${\mathfrak{sl}}_2(F)={\text{span}}_F\{h,x,y\}$ with brackets $[h,x]=2x$, $[h,y]=-2y$, $[x,y]=h$ is simple provided $\text{char}(F)\ne 2$. Proof Let $\text{a}◃{\mathfrak{sl}}_2(F)$ be a non-zero ideal. By the commutation relation, we know that the adjoint map ${\mathrm{ad}}_h$ is diagonalizable with eigenvalues $2,0,-2$ corresponding to eigenvectors $x,h,y$. Take any non-zero element $u=ax+by+ch\in \text{a}$. If $b\ne 0$, apply ${\text{ad}}_x$ twice: ${\text{ad}}_x(u)=bh-2cy$ and ${\text{ad}}_x^2(u)=-2bx$. Since $\text{char}(F)\ne 2$ and $b\ne 0$, this implies $x\in \text{a}$. Similarly, if $a\ne 0$, applying ${\text{ad}}_y$ twice gives ${\text{ad}}_y^2(u)=2ay$, showing $y\in \text{a}$. Once $x$ or $y$ is in $\text{a}$, we have $[x,y]=h\in \text{a}$. Then $[h,x]=2x$ and $[h,y]=-2y$ show that if one of $x,y$ is in $\text{a}$, the other must be as well. If $a=b=0$, then $u=ch$ with $c\ne 0$, so $0\notin \text{a}$, yielding a contradiction. $\square$

Isomorphism Theorems for Lie Algebras

The following isomorphism theorems hold for Lie algebras, similar to those for groups and rings:

• Suppose $\phi :\text{g}\to \mathfrak{h}$ is a Lie algebra homomorphism, then $\text{g}/\ker \phi \cong \mathrm{im}\phi$.
• If $\text{a},\text{b}◃\text{g}$ and $\text{a}\subseteq \text{b}$ are ideals, then $(\text{g}/\text{a})/(\text{b}/\text{a})\cong \text{g}/\text{b}$.
• If $\text{a},\text{b}◃\text{g}$, then $\text{b}/(\text{a}\cap \text{b})\cong (\text{a}+\text{b})/\text{a}$.

Proposition

If $\text{a}◃\text{g}$ is an ideal, then $[\text{a},\text{a}]◃[\text{g},\text{g}]$, and $[\text{g},\text{g}]/[\text{a},\text{a}]\cong [\text{g}/\text{a},\text{g}/\text{a}]$.

Proof Suppose $x,y\in \text{g}$ and $z,w\in \text{a}$, let $u=[x,y]\in [\text{g},\text{g}]$, then $[u,[z,w]]+[z,[w,u]]+[w,[u,z]]=0.$So $[u,[z,w]]\in \text{a}$, hence $[\text{a},\text{a}]◃[\text{g},\text{g}]$. Now consider the quotient map $\pi :\text{g}\twoheadrightarrow \text{g}/\text{a}$. We have $\pi ([\text{g},\text{g}])=[\text{g}/\text{a},\text{g}/\text{a}]$ and $\pi ([\text{a},\text{a}])=\{0\}$. Thus, by the first isomorphism theorem, $[\text{g},\text{g}]/[\text{a},\text{a}]\cong [\text{g}/\text{a},\text{g}/\text{a}]$. $\square$

Solvability and Semi-Simplicity

Derived Series and Solvability

The derived series of a Lie algebra $\text{g}$ is a sequence of ideals defined recursively as follows: $D^0(\text{g})=\text{g},D^{k+1}(\text{g})=[D^k(\text{g}),D^k(\text{g})].$ A Lie algebra $\text{g}$ is solvable if $D^N(\text{g})=\{0\}$ for some integer $N$.

e.g. The Borel subalgebra ${\text{b}}_n(F)$ of ${\mathfrak{gl}}_n(F)$, consisting of all upper triangular matrices, is solvable.

Semi-Simple Lie Algebras

A Lie algebra $\text{g}$ is semi-simple if its only solvable ideal is $\{0\}$.

Properties of Solvable Lie Algebras

Let $\text{g}$ be a Lie algebra.

1. If $\text{g}$ is solvable, so are its subalgebras and homomorphic images.
2. If $\text{a}◃\text{g}$ is a solvable ideal and the quotient $\text{g}/\text{a}$ is solvable, then $\text{g}$ is solvable.
3. If $\text{a},\mathfrak{b}◃\text{g}$ are solvable ideals, then their sum $\text{a}+\mathfrak{b}$ is also a solvable ideal.

Proof

1. Suppose $\text{h}<\text{g}$ is a subalgebra. Then $D^k(\text{h})\subseteq D^k(\text{g})$ for all $k$. If $\text{g}$ is solvable, then $D^N(\text{g})=\{0\}$ for some $N$, so $D^N(\text{h})\subseteq D^N(\text{g})=\{0\}$, hence $\text{h}$ is solvable. If $\phi :\text{g}\to \mathfrak{k}$ is a homomorphism, then $\phi (D^k(\text{g}))=D^k(\phi (\text{g}))$ for all $k$. If $\text{g}$ is solvable, then $D^N(\text{g})=\{0\}$ for some $N$, so $D^N(\phi (\text{g}))=\phi (D^N(\text{g}))=\{0\}$, hence $\phi (\text{g})$ is solvable.
2. Consider the quotient map $\pi :\text{g}\twoheadrightarrow \text{g}/\text{a}$. We have $\pi (D^k(\text{g}))=D^k(\text{g}/\text{a})$ for all $k$. If $\text{a}$ and $\text{g}/\text{a}$ are solvable, then $D^M(\text{a})=\{0\}$ and $D^N(\text{g}/\text{a})=\{0\}$ for some $M,N$. From the proposition, we know that $D^{M+N}(\text{g})/D^{M+N}(\text{a})\cong D^{M+N}(\text{g}/\text{a})=\{0\}$, so $D^{M+N}(\text{g})=D^{M+N}(\text{a})=\{0\}$, $\text{g}$ is solvable.
3. We note that $\text{a}\hookrightarrow \text{a}+\text{b}\twoheadrightarrow (\text{a}+\text{b})/\text{a}\cong \text{b}/(\text{a}\cap \text{b})$, so by part 2, $\text{a}+\text{b}$ is solvable.

$\square$

This implies there exists a unique maximal solvable ideal in $\text{g}$:

Radical of a Lie Algebra

For any Lie algebra $\text{g}$, there exists a unique maximal solvable ideal in $\text{g}$, called the radical of $\text{g}$, denoted $\mathrm{Rad}(\text{g})$. The quotient algebra $\text{g}/\text{Rad}(\text{g})$ is semi-simple.

Nilpotent Lie Algebras

Lower Central Series and Nilpotency

The lower central series of a Lie algebra $\text{g}$ is defined recursively: $C^0(\text{g})=\text{g},C^{k+1}(\text{g})=[\text{g},C^k(\text{g})]$ A Lie algebra $\text{g}$ is nilpotent if $C^N(\text{g})=\{0\}$ for some integer $N$.

e.g.

• The Lie algebra of strictly upper triangular matrices is nilpotent. However, the Borel subalgebra of all upper triangular matrices is solvable but not nilpotent.
• The affine Lie algebra ${\mathfrak{aff}}_1=\mathrm{span}\{H,X\}$ with $[H,X]=X$ is solvable but not nilpotent.

Proposition

Nilpotency implies solvability.

Proof: $D^k(\text{g})\subseteq C^k(\text{g})$ for all $k\ge 1$. $\square$

Properties of Nilpotent Lie Algebras

Let $\text{g}$ be a Lie algebra.

1. Subalgebras and homomorphic images of a nilpotent Lie algebra are nilpotent.
2. If $\text{g}/Z(\text{g})$ is nilpotent, then $\text{g}$ is nilpotent.
3. If $\text{g}$ is nilpotent and non-zero, then its center $Z(\text{g})$ is non-zero.

Proof (1) is clear, we will show (2) and (3). (2) Let $\pi :\text{g}\twoheadrightarrow \text{g}/Z(\text{g})$ be the quotient map. Suppose $C^n(\text{g}/Z(\text{g}))=(0)$. Then $\pi (C^n(\text{g}))=C^n(\text{g}/Z(\text{g}))=(0)$. So $C^n(\text{g})\subseteq Z(\text{g})$. Thus, $C^{n+1}(\text{g})=[\text{g},C^n(\text{g})]\subseteq [\text{g},Z(\text{g})]=\{0\}$. Hence, $\text{g}$ is nilpotent. (3) Suppose $\text{g}$ is nilpotent, non-zero and $Z(\text{g})=\{0\}$. Assume that $C^n(\text{g})=(0)$. This means $C^{n-1}(\text{g})\subseteq Z(\text{g})$, so $C^{n-1}(\text{g})=(0)$. Repeating this argument, we get $\text{g}=C^0(\text{g})=(0)$, a contradiction. Hence, $Z(\text{g})\ne \{0\}$. $\square$