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.

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    $\begingroup$ In my head it's pretty much "the adjoint action is natural, and traces are natural, and bilinear forms are useful." Is that not good enough? $\endgroup$ – Qiaochu Yuan Jul 20 '10 at 0:40
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    $\begingroup$ I just feel like I'm not doing them much of a service if I slap down the trace form and tell them it might be useful before deducing a few things that are not obviously related to traces. For example, if I say that bilinear forms (especially inner products) are useful, then they should have some reason to believe that two vectors in the Lie algebra having a trace-free product means they're orthogonal is appropriate. $\endgroup$ – Ryan Reich Jul 20 '10 at 0:52
  • $\begingroup$ An unsubstantiated notion is that if I do make this definition, and if this form is nondegenerate, then the sum of products that constitutes the Casimir operator has trace equal to dim(V) in any representation V. In particular, it is nonzero, and I think they can see the value in having a natural nonzero $\mathfrak{g}$-endomorphism of any representation. This is still slightly too long a daisy chain, though. $\endgroup$ – Ryan Reich Jul 20 '10 at 0:57
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    $\begingroup$ Ryan, tell them Killing form was introduced by Cartan (in thesis), to characterize solvable and especially ss Lie algebras (Bourbaki LIE I, 5.5 Cor. 1 and 6.1 Thm. 1). If one aims to characterize semisimple ones (reasonable, since Lie had proved earlier that the "classical" Lie algebras were simple, and ss must have been "in the air") then experience with finite group rep'ns (which one really should have before studying Lie algebras) suggests to look for "natural" invariant bilinear form to make complements. Why that trace form? Well, the analogue proves semisimplicity of matrix algebras... $\endgroup$ – BCnrd Jul 20 '10 at 2:17
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    $\begingroup$ I don't remember the details, but here's what went through my mind when I first struggled with this stuff: If you're teaching Lie algebras (and Lie groups?) to undergraduates, you'd better be focusing on matrix groups, so they should view a Lie algebra as a subspace of complex or real matrices that is closed under Lie bracket. Then a natural question is whether there is a natural inner product on the subspace. The first obvious thing to try is $<A, B> = tr A^tB$. The question then is whether this works or not and, if it doesn't, how to fix it. Or how to generalize it to non-matrix Lie algebras $\endgroup$ – Deane Yang Jul 20 '10 at 4:45

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

  • $\begingroup$ This is a nice idea! I did actually spend a lecture on finite groups that I can draw on for motivation, and I made a point of talking up the averaging procedure in the course of proving Maschke's theorem. In a better-structured course I would have done the inner product at that time, and then what you say would be perfect. Perhaps I can still rig it... $\endgroup$ – Ryan Reich Jul 20 '10 at 2:03
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    $\begingroup$ This seems to be a motivatation for an invariant form on a general representation of $g$, which is a worthy goal, but not for the Killing form per se, which has more to do with the structure of the Lie algebra itself and also has properties not shared by all invariant forms. $\endgroup$ – Victor Protsak Jul 20 '10 at 6:31
  • $\begingroup$ Are any of the Killing form's extra properties useful? $\endgroup$ – Allen Knutson Jul 20 '10 at 9:24
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    $\begingroup$ Allen, to state the obvious, Cartan's criterion is extremely useful and fails for other invariant forms. $\endgroup$ – Victor Protsak Jul 20 '10 at 16:17

You might find Thomas Hawkins' 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 a linear functional on $\mathfrak{g}$, 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.

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    $\begingroup$ That is absolutely fascinating. I will have to read that book. $\endgroup$ – Ryan Reich Jul 20 '10 at 3:27
  • $\begingroup$ James, I wondered what Hawkins' book would say on this (but was too lazy to walk over to the bookshelf); very nice. That's quite a coincidence about the case of associative algebras. So much for my attempt to derive history by pure thought. :) $\endgroup$ – BCnrd Jul 20 '10 at 3:38

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...


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.


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.


It's the essentially unique (Ad) $G$ invariant "inner product"... ok, maybe not positive or negative definite, so just "pairing" ... on $\mathfrak g$. So whatever "geometry" we imagine might live on $\mathfrak g$ is expressed in terms of the Killing form.


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".


Since you have already explained to them the process of exponentiation of the Lie algebra to go to a Lie group and the linearization of a Lie group to get to the Lie algebra, you can borrow some ideas from the following article to build some very easy connections to basic differential geometry concepts, such as the tangent plane and vectors that belong to the tangent plane. The wonderful article is

After explaining to them how the group is a manifold (with elements being points on the manifold) and how a Lie algebra is the tangent plane at the point that corresponds to the identity element (with elements of the Lie algebra being tangent vectors), you can just view the Killing form as the analogue/generalization of the inner product at the tangent space; in particular, with the inner product between vectors that they are probably most familiar with, we sum over 1 index (i.e. $\vec{v}\cdot \vec{u}=\sum_i v_i u^i$), the Killing form is the analogue of this in the case where our local tangent "vectors" are matrices: $$g_{\alpha\beta}=tr(A_\alpha A_\beta)=\sum_i\sum_j\left(A_\alpha \right)^i_j \left(A_\beta\right)^j_i$$ with $A_\alpha$ being a generator (labelled by the index $\alpha$) of the Lie algebra, in the adjoint representation (although you can give them the definition in any representation for simple Lie agebras).

  • $\begingroup$ "you can just view the Killing form as the analogue/generalization of the inner product at the tangent space" - the inner product on a tangent space to an abstract manifold? Even in case you think of $G$ as a concrete closed subgroup $\subseteq\mathrm{GL}_n(\mathbb{R})$, the Killing form on $\mathfrak{g}=T_1G$ is not in general that induced from the standard dot product (a.k.a. trace form) on $\mathrm{Mat}_n(\mathbb{R})$, as the latter is always (positive) definite. $\endgroup$ – Qfwfq Oct 12 '18 at 23:42
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    $\begingroup$ Now that I re-read, I realize you were making the analogy with the ordinary dot product on $\mathbb{R}^n$ being "summing over one index" while the Killing becomes "summing over two indices" because "the vectors are matrices". Personally I find this extrinsic viewpoint quite confusing, but maybe other people or the students can make sense out of it... $\endgroup$ – Qfwfq Oct 12 '18 at 23:48
  • $\begingroup$ @Qfwfq Yes, the analogy is the latter which you commented on. It is only an analogy meant to motivate undergrads. At that level, I can personally not find any intuitive way to think about it. Cheers! $\endgroup$ – TheQuantumMan Oct 12 '18 at 23:58

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