The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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September 3, 2012 - Posted by | Algebra, Lie Algebras

5 Comments »

  1. [...] first and most important structural result using the Killing form regards its “radical”. We never really defined this before, but it’s not hard: [...]

    Pingback by The Radical of the Killing Form « The Unapologetic Mathematician | September 6, 2012 | Reply

  2. [...] go back to our explicit example of and look at its Killing form. We first recall our usual [...]

    Pingback by Back to the Example « The Unapologetic Mathematician | September 7, 2012 | Reply

  3. [...] no way to spread another simple ideal across two or more summands in this decomposition. And the Killing form of a summand is the restriction of the Killing form of to that summand, as we expect for any ideal [...]

    Pingback by Decomposition of Semisimple Lie Algebras « The Unapologetic Mathematician | September 8, 2012 | Reply

  4. [...] makes an ideal, so the Killing form latex I$ is the restriction of of the Killing form of . Then we can define to be the subspace [...]

    Pingback by All Derivations of Semisimple Lie Algebras are Inner « The Unapologetic Mathematician | September 11, 2012 | Reply

  5. [...] is, a bilinear form is invariant if and only if it is associative, in the sense that the Killing form is: Share this:StumbleUponDiggRedditLike this:LikeBe the first to like [...]

    Pingback by The Submodule of Invariants « The Unapologetic Mathematician | September 21, 2012 | Reply


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