The Killing form; factorising non-Abelian Lie groups

It could be fun to try and define a "dot product" on a Lie algebra.

You know, you might've already realised that the cross product is a Lie bracket of sorts -- you know, given its antisymmetry and the whole $a^\mu b^\nu - a^\nu b^\mu$ representation of the wedge product and all that. It's a short exercise to verify that the Lie algebra $\mathfrak{so}(3)$ of $SO(3)$ is the algebra of skew-symmetric matrices, and with the Lie bracket $XY-YX$ is isomorphic to $\mathbb{R}^3$ with the cross product.

Well, the dot product on $\mathbb{R}^3$ has an interesting connection to $SO(3)$ -- it is precisely the form that is invariant under the action of $SO(3)$. Well, but that's $SO(3)$ acting on $\mathbb{R}^3$ -- what is that action in the notation of $\mathfrak{so}(3)$? As it turns out (and you can work this out), it is precisely the adjoint map $\mathrm{Ad}_gX:=gXg^{-1}$ which corresponds to this "rotating $X$ by $g$". It's not really that unexpected, if you ask me -- conjugation is always the natural way to transform matrices in linear algebra when vectors are multiplied on the left.

So the "dot product" is an $\mathrm{Ad}$-invariant bilinear form. In fact, adding a symmetricity requirement allows us to just bother with norms (as a symmetric inner product can be determined from the norm, through the cosine rule). Conjugation basically allows you to determine the "contours" of this norm or inner product. The question is: can we determine the bilinear form -- up to scaling -- just from "$\mathrm{Ad}$-invariant symmetric bilinear form" alone?

This is equivalent to asking "is the orbit of some non-zero $X$ under conjugation by $G$ equal to $\mathfrak{g}$?" (so that the norm of that $X$ would suffice to determine all norms -- do you see why?) Well, this is equivalent to asking "is $X$ contained in some non-trivial ideal?" (prove that these are equivalent!), and this is equivalent to asking "does $\mathfrak{g}$ have any non-trivial ideals?" (do you see why?)

A Lie algebra without nontrivial ideals is called a simple Lie algebra. Our demonstration above shows that a simple Lie algebra has a unique $\mathrm{Ad}$-invariant symmetric bilinear form, determined by the value of $\langle X, X\rangle$ for some non-zero $X$.

Even before we actually derive what this form must look like, we can derive one important consequence of automorphism invariance: $\langle X, [X, Y]\rangle = 0$ (prove it!), i.e. the tangent to an automorphism curve is perpendicular to the position vector at every point. The understanding of the group as acting as a "rotation group" on its Lie algebra in the adjoint representation really makes sense!

Someone tell me if they know how one may "derive" the trace-form formula from this characterisation rather than pulling it out of the blue and then proving it is the unique $\mathrm{Ad}$-invariant symmetric bilinear form. Here's something I started to write:

Here's an idea for the base length (i.e. to define the scaling): $X$ has length 1 iff the length of $[X,Y]$ equals the length of $Y$ for all $Y$ perpendicular to $X$ -- equivalently: $\forall V\in\mathfrak{g}, |[X,[X,V]]|=|[X,V]|$. We need to check that this condition is well-defined, i.e. that:
  1. Given an $X$, $|[X,[X,U]]|=|[X,U]|$ for some $U$ not a multiple of $X$ implies that $|[X,[X,V]]|=|[X,V]|$ for all $V$.
  2. $X$ satisfying $|[X,[X,V]]|=|[X,V]|$ implies that all conjugates $gXg^{-1}$ of it satisfy it too. This is trivial from considering $V=gV'g^{-1}$ (since the identity is true for all $V$).
Is the first one even true outside $\mathfrak{so}(3)$ -- for all simple Lie algebras?

One may come up with the idea of defining a form $\langle X, Y\rangle = \mathrm{tr}[X,[Y,\cdot]]$ (example of some weak motivation -- the vector triple product $x\times(x\times v)$ has as eigenvectors the vectors $v$ perpendicular to $x$ and the eigenvalues depend on the length of $x$) and check that this is indeed an $\mathrm{Ad}$-invariant symmetric bilinear form, and is thus unique up to scaling for simple Lie algebras. This form is called the Killing form.

Factorisation of Lie groups

We have seen the classification of connected Abelian Lie groups: they are products of circles and lines. We wonder if such a classification is possible for more general Lie groups.

The natural way to "factorise" groups by taking quotients over normal subgroups -- we wonder if this means that all Lie groups can be written as direct products of simple Lie groups (groups that don't have a nontrivial connected normal subgroup -- can you see why "connected" matters?). Well, not really -- the quotients need not be subgroups at all, after all. Instead, the "factorisation" takes the form of what is known as a group extension. A group for which it is a direct product is called a reductive Lie group -- and its Lie algebra is the direct sum of simple Lie algebras, or a reductive Lie algebra.

It is more conventional in the literature to define a simple Lie algebra excluding the one-dimensional/abelian case. In this definition, direct sums of simple Lie algebras are semisimple Lie algebras, and reductive Lie algebras are direct sums of semisimple and abelian Lie algebras.

TBC: Cartan's criterion, solvability, nilpotency

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