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Here's my idea for motivating the Killing form: the only notion we have of magnitudes and angles in a Lie algebra comes from conjugations, as they can be understood to be the "natural" transformations on the Lie algebra. So it's natural to ask for a norm map that satisfies $\forall g\in G$,

$$\|X\|=\|\mathrm{Ad}_gX\|$$ And hopefully we can then use symmetry to pin down a bilinear form. The idea is that we can already compare two vectors in the same line, and this condition creates contours that are precisely the orbits of conjugation, which means allowing us to compare vectors in the same ideal.

So in a simple Lie algebra, the bilinear form would then be completely determined up to scaling.

Am I on a sensible track? I guess what I'm asking is:

  1. Am I right to believe that "bilinear, symmetric and automorphism-invariant" uniquely determine the Killing form (up to scaling) for simple Lie algebras?
  2. If so, how can I prove the $\mathrm{tr}(\mathrm{ad}(x)\mathrm{ad}(y))$ formula from this characterisation?
  3. How might I extend this intuition to non-simple Lie algebras? I think I can "see" why the "semisimple equivalent to non-degenerate" property is true, though.

(See here for examples of the kind of intuition, motivation I'm looking for. Based on advice there, I'm splitting my "intuition for Lie algebras" questions.)

Abhimanyu Pallavi Sudhir
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  • For non-semisimple Lie algebras the Killing form is usually zero, e.g., for all nilpotent Lie algebras. Then it is "highly degenerate", so certainly not an inner product. – Dietrich Burde Sep 25 '19 at 13:22
  • @DietrichBurde Sorry, I changed it to "bilinear form". That's not really the kind of intuition I'm trying to get at, though -- I don't need help motivating a bilinear form. – Abhimanyu Pallavi Sudhir Sep 25 '19 at 14:23
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    Yes, you don't need help motivating a bilinear form, but you have asked how the intuition might extend to non-simple Lie algebras. And there you have the zero Killing form, which must fit into your intuition. For simple Lie algebras it looks right. It is unique up to scalar. – Dietrich Burde Sep 25 '19 at 14:26
  • (1) Yes. (2) Check that $(x, y) \mapsto \operatorname{tr}(\operatorname{ad}(x) \operatorname{ad}(y))$ satisfies the characterization you described in (1). I'm not posting this as an answer, because I don't quite understand what you mean in (3) by "this intuition" in "The idea is that we can already compare two vectors in the same line, and this condition creates contours that are precisely the orbits of conjugation, which means allowing us to compare vectors in the same ideal." – Travis Willse Sep 25 '19 at 14:49
  • Like I said, it follows from the definition that it is bilinear, symmetric, and invariant under the adjoint action, so if you have the uniqueness guaranteed by (1), there's nothing more to check. Are you asking how to prove the uniqueness you mention in (1)? – Travis Willse Sep 25 '19 at 16:43
  • @Travis Right, thanks -- I got confused for some reason. – Abhimanyu Pallavi Sudhir Sep 25 '19 at 16:44
  • You're welcome. – Travis Willse Sep 25 '19 at 16:44

1 Answers1

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One way to look at the Killing form is as a tool for examining ideals of $\mathfrak{g}$. Let's start not with the Killing form, but with any bilinear invariant form $B(-, -): \mathfrak{g} \times \mathfrak{g} \to \mathbb{C}$. Invariance here means either of the following equivalent things:

  1. $B([x, y], z) = B(x, [y, z])$ for all $x, y, z \in \mathfrak{g}$.
  2. The map $B: \mathfrak{g} \otimes \mathfrak{g} \to \mathbb{C}$ is a homomorphism of $\mathfrak{g}$-modules, where $\mathbb{C}$ is the trivial module on which $\mathfrak{g}$ acts by zero.

The radical of an invariant bilinear form $$ \operatorname{Rad}(B) = \{ x \in \mathfrak{g} \mid B(x, y) = 0 \text{ for all } y \in \mathfrak{g}\}$$ is an ideal of $\mathfrak{g}$. For a simple Lie algebra $\mathfrak{g}$:

  1. Every invariant form is either zero, or nondegenerate.
  2. Since a nondegenerate invariant form is the same as an isomorphism of representations $\mathfrak{g} \to \mathfrak{g}^*$, if $\mathfrak{g}$ is simple then any two nondegenerate forms are scalar multiples of each other.
  3. If $B$ is nonzero and invariant, then so is the form $C$ defined by switching arguments: $C(x, y) = B(y, z)$, so $B$ and $C$ must be multiples of each other. From this we can deduce that $B$ must be either symmetric of skew-symmetric. Some playing around with invariance and skew-symmetry will show that $B$ cannot be skew-symmetric, so it must be symmetric.

Hence for a simple Lie algebra, every nonzero invariant form is symmetric, and all such forms are scalar multiples of each other. So it is enough to find any nonzero invariant form.

Now, we need a way to manufacture invariant bilinear forms. In fact, there is such a form for every representation of $\mathfrak{g}$. Given a representation $\rho: \mathfrak{g} \to \mathfrak{gl}(V)$, define the trace form for $V$ to be the bilinear form $$ B_V(x, y) = \operatorname{tr}(\rho_V(x) \rho_V(y)).$$ Every trace form is symmetric and invariant. Of course, this might not help much, since given a random Lie algebra $\mathfrak{g}$, cooking up nontrivial representations is a nontrivial task. However, we get the adjoint representation $\operatorname{ad}: \mathfrak{g} \to \mathfrak{gl}(\mathfrak{g})$ for free, and we can define the associated trace form $$ B_\mathfrak{g}(x, y) = \operatorname{tr}((\operatorname{ad} x)(\operatorname{ad} y)).$$ All that we need to do is to show that the Killing form $B_\mathfrak{g}$ is nonzero. I think the fastest way to do this is by Cartan's Criterion for solvability (to apply Cartan's criterion, we view $\mathfrak{g}$ as embedded inside $\mathfrak{gl}(\mathfrak{g})$ via the homomorphism $\operatorname{ad}$).

As for extending this intuition to semisimple Lie algebras, I'd suggest you read through the (not long) proofs of Lie's theorem and Cartan's criterion, which deal with this more general case.

Joppy
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