7

I'm currently self-studying Lie Algebras and I came across the definition of the Killing Form. As I understand it, the Killing Form gives you an inner product with which you can visualize the roots of a Lie Algebra. Two questions here:

  1. The definition of the Killing Form seems very random. Is there a natural reason why someone would choose this particular inner product with which to visualize the fundamental roots? Is there really no simpler inner product to choose?

  2. What deeper insight does the root system give you about the Lie Algebra? As an example, I've attached a screenshot of a sample root system below. My issue is that it's so many layers thick with abstraction (each point is an "eigenvalue of the action of Cartan Subalgebra under the adjoint map" -- gosh, even saying that makes my head spin!) that I can't get a grip of what the diagram is saying morally.

To sum up, where I'm at right now is this: "the eigenvalues of the adjoint map form a nice picture if we arrange them according to this seemingly random inner product (the Killing Form)." But why are the eigenvalues of the adjoint map significant, and why is their arrangement in the below diagram significant? I feel like I am missing the big picture. Any suggestions would be appreciated!

Root System of

Adithya Chakravarthy
  • 2,582
  • That screenshot is not the root system of $\mathfrak{sl}_4(\mathbb C)$ in any way I recognise. The root system is of type $A_3$, which in its full beauty looks like this: https://en.wikipedia.org/wiki/Root_system#/media/File:A3vzome.jpg, and whose Dynkin diagram is just a line with three dots. – Torsten Schoeneberg Aug 09 '20 at 23:21
  • 1
    As for the insight the root system gives us into the Lie algebra, one way to understand this is: First, understand how one can understand the structure of all of $\mathfrak{sl}_n$ from the diagonal and the "one non-zero entry off the diagonal" matrices. Then, this game generalises this to other classical (and the exceptional simple) Lie algebras: The Cartan subalgebra plays the role of the diagonal, and the root spaces play the role of those "one non-zero entry off the diagonal" matrices. Cf. my answer to https://math.stackexchange.com/q/2095754/96384 – Torsten Schoeneberg Aug 09 '20 at 23:28
  • Thanks for the comment - I've corrected the $\mathfrak{sl_4}$ mistake above. The example in the link you have was very helpful. I'm still looking for some intuition on why the root diagram gives you insight on the Lie Algebra, so I'll keep the question open. – Adithya Chakravarthy Aug 10 '20 at 02:09
  • By "root diagram" do you mean the Dynkin diagram of a root system? But that picture in your post still is neither a root system nor the Dynkin diagram of one. I just don't know how it is supposed to relate to any root system. What is its source? – Torsten Schoeneberg Aug 10 '20 at 02:36
  • 1
    This is from "Representation Theory: A First Course" by Fulton and Harris (page 213). They refer to this picture as a way to visualize roots. It's definitely not a Dynkin Diagram - perhaps it's a root lattice? – Adithya Chakravarthy Aug 10 '20 at 02:49
  • Honestly I have no idea what it is, and unfortunately I do not have acces to the book right now. I am sure though that Fulton and Harris know what they're doing. Any chance you can reproduce more context to that diagram? – Torsten Schoeneberg Aug 10 '20 at 21:09

1 Answers1

5

Let me try to explain your first point, about the origin and importance of the Killing form. If I have a break from my work I can try to go into the second point, or someone more of an expert in Lie algebras than I can do it first.

If $\mathfrak g$ is a simple Lie algebra, then there is a unique non-degenerate bilinear form on the adjoint representation of $\mathfrak g$. This is a general fact about simple modules, and simply comes from the fact that the adjoint representation is self-dual, so there is a unique map $V\otimes V\to k$. (I chose $V$ and $k$ here because this is a general statement about self-dual simple modules over some object and a field $k$, be they Lie algebras, algebraic groups, etc.)

It turns out that the map is symmetric (i.e., comes from a map from the symmetric square of the adjoint, rather than the exterior square). So the reason for the definition in one sense is that the Killing form is unique, and that is it.

If one takes a step back, and looks at the theory of finite-dimensional $k$-algebras, then one encounters (nowadays, certainly not in 1910) the idea of a symmetric algebra. This is a $k$-algebra with a symmetric bilinear form satsifying $(ab,c)=(a,bc)$. The Killing form satisfies this relation also. So the Killing form is trying to turn the Lie algebra into a symmetric algebra. Now normally symmetric algebras are associative, but we'll not worry about this.

What do symmetric bilinear forms look like? They are often called symmetrizing trace forms, and we start to see the first connections with the definition of a Killing form. It turns out that this is the usual way to define symmetrizing trace forms, they come from trace maps. Indeed, the symmetrizing form on a matrix algebra is simply the trace map.

So not only is the Killing form the only way to define it, it is the standard way to define such a map.

David A. Craven
  • 11,561
  • Thanks for the thorough reply! That helps motivate the Killing Form. As a clarification, when you say "a bilinear form on the adjoint representation of $\mathfrak{g}$", what do you mean exactly? I understand what the adjoint map $\text{ad}_X$ is, but what does "a bilinear form on the adjoint representation" mean? Is the Killing Form a map on some space other than $\mathfrak{g}$? Thanks for the help! – Adithya Chakravarthy Aug 09 '20 at 22:12
  • 1
    The adjoint representation is just the Lie algebra acting upon itself. Associated to any module is (potentially) a trace form, as its just a homomorphism to the field from $V\otimes V^$. If $V\cong V^$, then this is a map from $V\otimes V$ to the field. So in order to set it up 'properly', I needed to specify which representation I was using. The adjoint representation is the only non-trivial representation you obviously have. – David A. Craven Aug 09 '20 at 22:14
  • 2
    @chaad It's also the right place to think about roots. The adjoint representation gives us a nice module. Restrict that module to the Cartan subalgebra. How does it break up into indecomposable (irreducible) summands? You now see the roots. – David A. Craven Aug 09 '20 at 22:16