3 The Killing form - describe solvable and semi-simple

TLDR

Any Lie algebra roughly can be described by a solvable part and a semi-simple part.

Let $g$ be a finite dimensional complex Lie algebra. Then $g/\mathrm{rad}g$ is semi-simple. It can be proved later that there is a Lie subalgebra $h\cong g/\mathrm{rad}g$ s.t. $g\cong \mathrm{rad}g\oplus h$. Essentially, $g=\mathrm{rad}g:g/\mathrm{rad}g$.

So we use $\mathrm{rad}B$ to “detect” whether a Lie algebra is solvable / nilpotent. Two main results are proved here.

Basic setup

Simple and semi-simple

simple

A finite dimensional complex Lie algebra is called simple if it has no non-trivial ideal. One-dimensional Lie algebras are called trivial simple Lie algebra.

Remark. If $g$ is a non-trivial simple Lie algebra, then $[g,g]=g$.

semi-simple

A finite dimensional complex Lie algebra is called semi-simple if its radical is $0$, where the radical is the maximal solvable ideal.

Symmetric pairing and Killing form

symmetric pairing

Let $g$ be a finite dimensional complex Lie algebra. If $(\rho ,V)$ is a finite dimensional complex representation of $g$, we can find a symmetric pairing $(X,Y{)}_{\rho }=\mathrm{Tr}(\rho (X)\rho (Y))$. Specially, the pairing $(\cdot ,\cdot {)}_{\mathrm{ad}}$ is called the Killing form on $g$, which we denote by $B$ or $B_g$.

Definition

For $U\subseteq g$, $U^{\perp }=\{x\in g:B(u,x)=0\}$. Specially, define $\mathrm{rad}B:=g^{\perp }=\{x\in g:B(x,g)=0\}$. If $\mathrm{rad}B=0$, we say $B$ is non-degenerate.

By $\mathrm{tr}(AB)=\mathrm{tr}(BA)$, we can prove the following property:

invariant property

The Killing form satisfies “invariant property”: $B([Z,X],Y)+B([X,[Z,Y]])=0$. In fact, for any representation $\rho :g\to \mathrm{End}(V)$, we have $B_{\rho }([Z,X],Y)+B_{\rho }([X,[Z,Y]])=0$.

By the above property we can prove the following lemma:

Lemma

Let $g$ be a finite dimensional complex Lie algebra.

• If $a\subseteq g$ is an ideal, then $a^{\perp }:=\{x\in g:B(a,x)=0\}$ is an ideal and $B_a=B_g{|}_a$.
• If $g=a\oplus b$ where $a,b$ are ideals, then $B_g=B_a\oplus B_b$.

$\mathrm{rad}g$ and $\mathrm{nil}g$

Lemma

Let $g$ be a finite dimensional complex Lie algebra. Then $g$ has a unique maximal solvable ideal and $g$ has a unique maximal nilpotent ideal.

\begin{proof}i) Claim: if $a,b\subset g$ are solvable ideals, then $a+b$ is also a solvable ideal. Consider $0\to a\to a+b\to (a+b)/a\to 0$. Since the canonical projection $f:a+b\to (a+b)/a$ is a homomorphism, there is a $n$ s.t. $(a+b{)}^n\subseteq a$ and so $a+b$ is solvable. Thus let $\mathrm{rad}g$ be a solvable ideal of maximal dimension and it is what we want.

ii) Similarly, claim that if $a,b\subseteq g$ are nilpotent ideals of $g$, then $a+b$ is also a nilpotent ideal. By i) we know $a+b$ is solvable. Note that the solvable Lie subalgebra $a+b$ has a maximal nilpotent ideal and the maximal nilpotent ideal of $a+b$ contains both $a$ and $b$, so $a+b$ is nilpotent. Thus let $\mathrm{nil}g$ be a nilpotent ideal of maximal dimension and it is what we want. \end{proof}

Definition

For a finite dimensional complex Lie algebra $g$, denote the unique maximal solvable ideal as $\mathrm{rad}g$ and denote the unique maximal nilpotent ideal as $\mathrm{nil}g$.

Main theorems

solvable iff $[g,g]\subseteq \mathrm{rad}B$

Let $g$ be a finite dimensional complex Lie algebra. Then:

• $g$ is solvable iff $[g,g]\subseteq \mathrm{rad}B:=g^{\perp }$, i.e. $B(g,[g,g])=0$.
• In general, $[g,g{]}^{\perp }=\mathrm{rad}g$. As a consequence, $\mathrm{rad}B\subseteq \mathrm{rad}g$.

\begin{proof}i) Suppose $g$ is solvable. To show $B(g,[g,g])=0$, it suffices to show for any $X,Y,Z\in g$, $\mathrm{Tr}(\mathrm{ad}[X,Y]\mathrm{ad}Z)=0$. By $g$ solvable, one can choose a set of basis such that any $\mathrm{ad}X$ is upper triangular matrices. So $\mathrm{ad}[X,Y]=[\mathrm{ad}X,\mathrm{ad}Y]$ is a strictly triangular matrix. Then $\mathrm{Tr}(\mathrm{ad}[X,Y]\mathrm{ad}Z)=0$.

Now suppose $B(g,[g,g])=0$. We want to show $g$ is solvable. Recall $g$ solvable $⟺$ $[g,g]$ nilpotent $⟺$ $\mathrm{ad}X$ nilpotent for all $X\in [g,g]$. So it is enough to show each $\mathrm{ad}y$ $(y\in [g,g])$ is a nilpotent element in $\mathrm{End}([g,g])$. We will actually argue that $\mathrm{ad}y\in \mathrm{End}(g)$ is nilpotent. By Jordan-Chevalley Decomposition, there exist $s,n$ such that $\mathrm{ad}y=s+n$. Additional, there is a polynomial ${\alpha }_0(t)$ with $0$ constant term such that $s={\alpha }_0(\mathrm{ad}y)$.

One need to argue that for $y\in [g,g]$, the associated $s=0$. We will indeed argue that $\mathrm{Tr}(\bar{s}s)=0$. Note that

$$\begin{array}{rl}\mathrm{Tr}(\bar{s}s)& =\mathrm{Tr}(\bar{s}y)\\ & =\mathrm{Tr}(\bar{s}\mathrm{ad}(\sum _i[z_i,w_i]))\\ & =\sum _i\mathrm{Tr}(\bar{s}[\mathrm{ad}{z}_i,\mathrm{ad}{w}_i])\\ & =\sum _i\mathrm{Tr}([\bar{s},\mathrm{ad}{z}_i]\mathrm{ad}{w}_i)\\ & =\sum _i\mathrm{Tr}(\mathrm{ad}\bar{s}\mathrm{ad}{z}_i\mathrm{ad}{w}_i).\end{array}$$

Notice that if $\mathrm{ad}y=s+n$, then $\mathrm{ad}(\mathrm{ad}y)=\mathrm{ad}s+\mathrm{ad}n$ is the Jordan decomposition of $\mathrm{ad}(\mathrm{ad}y)$. Thus $\mathrm{ad}s=\alpha (\mathrm{ad}(\mathrm{ad}y))$ where $\alpha$ is a polynomial with $0$ constant term. Notice that there exists a polynomial $\beta (t)$ with $0$ constant term such that $\mathrm{ad}\bar{s}=\beta (\mathrm{ad}s)$. Thus

$$\mathrm{ad}\bar{s}=\beta (\mathrm{ad}s)=\beta (\alpha (\mathrm{ad}\mathrm{ad}y)):=\gamma (\mathrm{adad}y),$$

where $\gamma$ is also a polynomial with zero constant.

At this moment,

$$\mathrm{Tr}(\bar{s}s)=\sum _i\sum _j{c}_j\mathrm{Tr}((\mathrm{adad}y{)}^j(\mathrm{ad}{z}_i)(\mathrm{ad}{w}_i)).$$

Since $(\mathrm{adad}y{)}^j(\mathrm{ad}{z}_i)=\mathrm{ad}(\mathrm{ad}{y}^{j}z)$, we have

$$\mathrm{Tr}((\mathrm{adad}y{)}^j(\mathrm{ad}{z}_i)(\mathrm{ad}{w}_i))=\mathrm{Tr}(\mathrm{ad}(\mathrm{ad}{y}^{j}z)\mathrm{ad}{w}_i)=B(\mathrm{ad}{y}^j{z}_i,w_i).$$

The assumption of $B([g,g],g)=0$ and $\mathrm{ad}{y}^j{z}_i\in [g,g]$ yields $B(\mathrm{ad}{y}^j{z}_i,w_i)=0$ and so $\mathrm{Tr}(\bar{s}s)=0$.

ii) Denote that $b^{\prime }=[g,g{]}^{\perp }$ and $b=\mathrm{rad}g$. One direction is easy: $b^{\prime }{\perp }_{B_g}[g,g]\Rightarrow b^{\prime }{\perp }_{B_b}[b^{\prime },b^{\prime }]\Rightarrow b^{\prime }$ is solvable $\Rightarrow b^{\prime }\subseteq \mathrm{rad}g$.

Then we will argue $\mathrm{rad}g\subseteq [g,g{]}^{\perp }$. For any $Z\in \mathrm{rad}g$ and $X,Y\in g$, we need to verify $B(Z,[X,Y])=0$. When $X,Y\in b$, there is nothing to prove. Let’s assume $X,Y\notin b$ for instance and compute $B([Y,Z],X)=\mathrm{Tr}(\mathrm{ad}[Y,Z]\mathrm{ad}X)$. Put $b_1=b\oplus \mathbb{C}X$, then $b_1$ is solvable. And we have $B([Y,Z],X)=B_{b_1}([Y,Z],X)$ because

$$\mathrm{ad}X=\left[\begin{array}{ccc}\ast & 0& \ast \\ 0& 0& \ast \\ 0& 0& \ast \end{array}\right]\text{ and }\mathrm{ad}[Y,Z]=\left[\begin{array}{ccc}\ast & \ast & \ast \\ 0& 0& 0\\ 0& 0& 0\end{array}\right],$$

where $g=b\oplus \mathbb{C}X\oplus U$. Since $b_1$ is solvable, one can choose a basis of $b_1$ such that $b_1$‘s action on $b$ is expressed by upper triangular matrix. Thus w.r.t. basis $\{v_1,\cdots ,v_l,X\}$, $\mathrm{ad}b\subseteq \left[\begin{array}{cc}{A}_b& \ast \\ 0& 0\end{array}\right]$ and $\mathrm{ad}X=\left[\begin{array}{cc}A& 0\\ 0& \ast \end{array}\right]$ where $A_b$ and $A$ are upper triangular matrices. Since $b_2=b\oplus \mathbb{C}Y$ is solvable, $[b_2,b_2]$ is nilpotent. So $\mathrm{ad}[Y,Z]\in \mathrm{End}(b_2)$ is nilpotent. Then $\mathrm{ad}[Y,Z]$ acting on $b$ is nilpotent. So the diagonal entries of $A_b$ are all zero. Thus $B(Z,[X,Y])=0$. \end{proof}

semi-simple iff $\mathrm{rad}B=0$

Let $g$ be a finite dimensional complex Lie algebra. Then $g$ is semi-simple iff $B$ is non-degenerate. As a consequence, $g$ is semi-simple iff it is a direct sum of non-trivial simple Lie algebra.

\begin{proof}If $g$ is semi-simple, then $\mathrm{rad}g=0$. By last theorem, we know $[g,g{]}^{\perp }=\mathrm{rad}g$. Thus $\mathrm{rad}B=g^{\perp }\subseteq [g,g{]}^{\perp }=0$ and so $B$ is non-degenerate.

On the other hand, suppose $g$ is not semi-simple, and we will show $\mathrm{rad}B\ne 0$, i.e. there is a non-zero $v\in g$ s.t. $v\perp g$. Since $\mathrm{rad}g\ne 0$ is the maximal solvable ideal, $(\mathrm{rad}g{)}^k=0$ for some $k$. Then $a=\mathrm{rad}{g}^{k-1}$ is an abelian ideal. Write $g=a\oplus u$ where $u$ is a subspace, then elements of $\mathrm{ad}a$ are expressed as $\left[\begin{array}{cc}0& \ast \\ & 0\end{array}\right]$ while for $y\in u$, $\mathrm{ad}y$ is expressed by $\left[\begin{array}{cc}\ast & \ast \\ 0& \ast \end{array}\right]$. So $a\perp u$ also $a\perp a$. Thus $a\perp g$ and $a\subseteq \mathrm{rad}B$. Therefore $\mathrm{rad}B\ne 0$.

Furthermore, if $g={\oplus }_{i=1}^k{a}_i$ where $a_i$ are non-trivial simple Lie algebras. Then $B_g={\oplus }_{i=1}^k{B}_{a_i}$ where $B_{a_i}$ are non-degenerate by ^c72e16. So $B_g$ is non-degenerate and $g$ is semi-simple. On the other hand, suppose $g$ is semi-simple. Let $a$ be the minimal nonzero ideal. Since $a\cap a^{\perp }$ is either $a$ or $0$. If $a\cap a^{\perp }=a$, then $a{\perp }_{B_g}a$ and $a{\perp }_{B_a}a$ and $a$ is solvable, which contradict $g$ semi-simple. Thus $g=a\oplus a^{\perp }$. Then by doing induction on dimension, $g$ is a direct sum of non-trivial simple Lie algebras. \end{proof}