# 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

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