Why the Killing form? - MathOverflow

Why the Killing form?

2010-07-19T23:03:32Z

I'm teaching a short summer course on algebraic groups and it's time to talk about the Killing form on the Lie algebra. The students are all undergrads of varying levels of inexperience, and I try to make everything seem like it has a point (going back to the basic goals of "what is an algebraic group" and "what does this have to do with representation theory"). I am having a hard time justifying the Killing form from anything like first principles: it is useful, and I can prove theorems explaining why it is useful, but I can't think of an explanation of why it is reasonable to invent. The ideal answer to this question will be a "naive" explanation. Other interesting answers (which I would appreciate for myself) can be more sophisticated.

Answer by Steve Huntsman for Why the Killing form?

2010-07-19T23:09:33Z

A one-sentence answer is that the Killing form provides the appropriate generalized inner product on a Lie algebra. A slightly longer answer is that the Killing form gives structural information about (e.g.) solvability and semisimplicity via the Cartan criteria and (e.g.) allows Levi decompositions to be effected in practice and the compact real form of the semisimple part of a Lie group to be constructed via Weyl’s so-called "unitary trick".

Answer by Ben Webster for Why the Killing form?

2010-07-19T23:51:56Z

For me, the important property of the Killing form is its naturality with respect to ideals (an illuminating fact to prove). Then, suddenly, its connection to semi-simplicity becomes quite clear: could there be an abelian ideal? Well, it would have to be in the radical of the Killing form.

Answer by Akhil Mathew for Why the Killing form?

2010-07-20T00:30:16Z

I always thought of the Killing form as the natural way to introduce an invariant inner product (that was nontrivial) on a Lie algebra, hence the ideal tools for proving theorems in the semisimple case. Indeed, the trace form is a reasonable invariant bilinear form on $\mathfrak{gl}_n$, and the adjoint map is the first choice of a map from a Lie algebra into $\mathfrak{gl}_n$ that one would think of.

Answer by David Jordan for Why the Killing form?

2010-07-20T01:55:54Z

Hi Ryan,

I presume given your description of the students that they know finite groups pretty well, and have seen the averaging idempotent $e=\frac{1}{|G|}\sum_{g\in G} g$, and how this can be used to construct an invariant inner product on any representation of a finite group. Perhaps you can convince them that compact groups admit the same sort of averaging idempotents via integral, and so perhaps you can construct the invariant inner product on finite dimensional representations of a compact group in more or less direct analogy with finite groups. Then you can derive the properties the Killing form should satisfy on the Lie algebra by setting g=e^tX, and taking derivatives of the axioms of the group's inner product?

This is the closest connection I can think of to finite group theory, which is hopefully well-understood by, or at least familiar to, your students.

What do you think? -david

Answer by james-parson for Why the Killing form?

2010-07-20T03:13:18Z

You might find Thomas Hawkin's book "Emergence of the Theory of Lie Groups" an interesting place to look for first-principles explanations of Lie-theory facts. He explains how Killing, Cartan, and Weyl first came up with the structure theory for semi-simple Lie algebras. (See Section 6.2, in particular, for a detailed discussion of Cartan's contributions to Killing-style structure theory---including his introduction and use of the Killing form.)

According to Hawkins, one of Killing's insights in his structure theory for a Lie algebra $\mathfrak{g}$ was to consider the characteristic polynomial $$ {\rm det} (t I - {\rm ad}(X)) = t^n -\psi_1(X)t^{n-1} + \psi_2(X)t^{n-2} + \cdots + (-1)^n\psi_n(X) $$ as a function of $X$. (The start of the structure theory was to consider those $X$---regular elements---such that the eigenvalue $0$ has minimal multiplicity.) In general the coefficients $\psi_i(X)$ are polynomial functions on $\mathfrak{g}$ that are invariants for the adjoint action of $\mathfrak{g}$ on itself.

Consider, in particular, a simple Lie algebra $\mathfrak{g}$. Killing observed that the coefficient $\psi_1(X)$, which is linear a linear functional on $X$, must vanish identically, since its kernel is an ideal. Cartan considered the coefficient $\psi_2(X)$, which is a quadratic form on $X$. (The value $\psi_2(X)$ is essentially the sum of the squares of the eigenvalues---roots---of $X$, since $\psi_1(X)=0$.) The bilinear form associated to this quadratic form is the usual Killing form. The invariance of $\psi_2$ under the adjoint action translates to the "associativity" property of the Killing form. Cartan observed (in essence) that for $\mathfrak{g}$ simple, the kernel of the the associated bilinear form is either $0$ or $\mathfrak{g}$ (by invariance + simplicity), and he managed to prove that the kernel is always $0$, starting his repair of the faults in Killing's structure theory. One can look at Hawkins' book for the details of the story, stripped of modern efficiencies.

It is tempting to think that the simpler structure theory of finite-dimensional associative algebras (where non-degeneracy of the trace form also characterizes semi-simplicity) may have inspired Cartan. It seems (again according to Hawkins) that Molien introduced this form for associative algebras (as a bilinear form---as opposed to Cartan's quadratic form) independently in the same year (1893) that Cartan published his thesis.

Answer by Simon Pepin Lehalleur for Why the Killing form?

2010-07-27T15:00:05Z

A less algebraic answer, but one that really helped me to understand the role of the Killing form, is that it induces the unique G-invariant riemannian metric on symmetric spaces $G(\mathbb{R})/K$ (K maximal compact subgroup), another fact which was very dear to Cartan as well...