# The Unapologetic Mathematician

## The Killing Form

We can now define a symmetric bilinear form $\kappa$ on our Lie algebra $L$ by the formula

$\displaystyle\kappa(x,y)=\mathrm{Tr}(\mathrm{ad}(x)\mathrm{ad}(y))$

It’s symmetric because the cyclic property of the trace lets us swap $\mathrm{ad}(x)$ and $\mathrm{ad}(y)$ and get the same value. It also satisfies another identity which is referred to as “associativity”, though it may not appear like the familiar version of that property at first:

\displaystyle\begin{aligned}\kappa([x,y],z)&=\mathrm{Tr}(\mathrm{ad}([x,y])\mathrm{ad}(z))\\&=\mathrm{Tr}([\mathrm{ad}(x),\mathrm{ad}(y)]\mathrm{ad}(z))\\&=\mathrm{Tr}(\mathrm{ad}(x)[\mathrm{ad}(y),\mathrm{ad}(z)])\\&=\mathrm{Tr}(\mathrm{ad}(x)\mathrm{ad}([y,z]))\\&=\kappa(x,[y,z])\end{aligned}

Where we have used the trace identity from last time.

This is called the Killing form, named for Wilhelm Killing and not nearly so coincidentally as the Poynting vector. It will be very useful to study the structures of Lie algebras.

First, though, we want to show that the definition is well-behaved. Specifically, if $I\subseteq L$ is an ideal, then we can define $\kappa_I$ to be the Killing form of $I$. It turns out that $\kappa_I$ is just the same as $\kappa$, but restricted to take its arguments in $I$ instead of all of $L$.

A lemma: if $W\subseteq V$ is any subspace of a vector space and $\phi:V\to V$ has its image contained in $W$, then the trace of $\phi$ over $V$ is the same as its trace over $W$. Indeed, take any basis of $W$ and extend it to one of $V$; the matrix of $\phi$ with respect to this basis has zeroes for all the rows that do not correspond to the basis of $W$, so the trace may as well just be taken over $W$.

Now the fact that $I$ is an ideal means that for any $x,y\in I$ the mapping $\mathrm{ad}(x)\mathrm{ad}(y)$ is an endomorphism of $L$ sending all of $L$ into $I$. Thus its trace over $I$ is the same as its trace over all of $L$, and the Killing form on $I$ applied to $x,y\in I$ is the same as the Killing form on $L$ applied to the same two elements.

September 3, 2012 Posted by | Algebra, Lie Algebras | 5 Comments

## Cartan’s Criterion

It’s obvious that if $[L,L]$ is nilpotent then $L$ will be solvable. And Engel’s theorem tells us that if each $x\in[L,L]$ is ad-nilpotent, then $[L,L]$ is itself nilpotent. We can now combine this with our trace criterion to get a convenient way of identifying solvable Lie algebras.

If $L\subseteq\mathfrak{gl}(V)$ is a linear Lie algebra and $\mathrm{Tr}(xy)=0$ for all $x\in[L,L]$ and $y\in L$, then $L$ is solvable. We’d obviously like to use the trace criterion to show this, but we need a little massaging first.

The catch is that our $M$ consists of all the $x\in\mathfrak{gl}(V)$ such that $\mathrm{ad}(x)$ sends $L$ to $[L,L]$. Clearly $L\subseteq M$, but it may not be all of $M$; our hypothesis states that $\mathrm{Tr}(xy)=0$ for all $y\in L$, but the criterion needs it to hold for all $y\in M$.

To get there, we use the following calculation, which is a useful lemma in its own right:

\displaystyle\begin{aligned}\mathrm{Tr}([x,y]z)&=\mathrm{Tr}(xyz-yxz)\\&=\mathrm{Tr}(xyz)-\mathrm{Tr}(yxz)\\&=\mathrm{Tr}(xyz)-\mathrm{Tr}(xzy)\\&=\mathrm{Tr}(xyz-xzy)\\&=\mathrm{Tr}(x[y,z])\end{aligned}

Now, if $x,y\in L$ — so $[x,y]\in[L,L]$ — and $z\in M$ then

$\displaystyle\mathrm{Tr}([x,y]z)=\mathrm{Tr}(x[y,z])=\mathrm{Tr}([y,z]x)$

But since $z\in M$ we know that $[y,z]\in[L,L]\subseteq L$, which means that the hypothesis kicks in: $[y,z]\in[L,L]$ and $x\in L$ so $\mathrm{Tr}([y,z]x)=0$.

Then we know that all $x\in[L,L]$ are nilpotent endomorphisms, which makes them ad-nilpotent. Engel’s theorem tells us that $[L,L]$ is nilpotent, which means $L$ is solvable.

We can also extend this out to abstract Lie algebras: if $L$ is any Lie algebra such that $\mathrm{Tr}(\mathrm{ad}(x)\mathrm{ad}(y))=0$ for all $x\in[L,L]$ and $y\in L$, then $L$ is solvable. Indeed, we can apply the linear version to the image $\mathrm{ad}(L)\subseteq\mathfrak{gl}(L)$ to see that this algebra is solvable. The kernel $\mathrm{Ker}(\mathrm{ad})$ is just the center $Z(L)$, which is abelian and thus automatically solvable. The image $\mathrm{ad}(L)$ is thus the solvable quotient of $L$ by a solvable kernel, so we know that $L$ itself is solvable.

September 1, 2012 Posted by | Algebra, Lie Algebras | 3 Comments