I’m keeping the same notation as all the previous posts here on semisimple Lie algebras.

Consider the real vector space

$\displaystyle E = \sum_{\alpha \in \Phi} \mathbb{R} \alpha \subset \mathfrak{h}^{\vee}.$

I claim that the form ${(\cdot, \cdot)}$ (obtained by the isomorphism ${\mathfrak{h}^{\vee} \rightarrow \mathfrak{h}}$ induced by the Killing form and the Killing form itself) is actually an inner product making ${E}$ into a euclidean space. To see this, we will check that ${(\alpha, \alpha) > 0}$ for all ${\alpha}$. Indeed:

$\displaystyle (\alpha, \alpha) = B(T_{\alpha}, T_{\alpha})$

where ${B}$ is the Killing form, by definition.

Now

$\displaystyle B(T_{\alpha}, T_{\alpha}) = \mathrm{Tr}_{\mathfrak{g}} ( \mathrm{ad} T_{\alpha}^2) = \sum_{\beta \in \Phi} \mathrm{Tr}_{\mathfrak{g}_{\beta}} ( \mathrm{ad} T_{\alpha}^2) .$

Now ${T_{\alpha}}$ acts by the scalar ${\beta(T_{\alpha}) = (\beta, \alpha)}$ on ${\mathfrak{g}_{\beta}}$, so after dividing by ${(\alpha, \alpha)^2}$, this becomes

$\displaystyle (\alpha, \alpha)^{-1} = \sum_{\beta \in \Phi} \left( \frac{ (\beta, \alpha)}{(\alpha, \alpha ) } \right)^2.$

But as we showed yesterday, ${\frac{ (\beta, \alpha)}{(\alpha, \alpha )} \in \mathbb{Q}}$, so the sum in question is actually positive. This proves one half of:

Proposition 1 ${E}$ is a euclidean space and ${\mathfrak{h}^{\vee} = E \oplus iE}$.

It is clear that ${\mathfrak{h}^{\vee} = E + iE}$; indeed, the roots span ${\mathfrak{h}^{\vee}}$, because if ${H \in \mathfrak{h}}$ were orthogonal to all the roots, then ${ \mathrm{ad} H}$ would act by zero on ${\mathfrak{g}}$. (I probably should have said this before explicitly.)

I claim now that if we can write

$\displaystyle \alpha = c_1 \alpha_1 + \dots + c_n \alpha_n$

for ${\alpha_1, \dots, \alpha_n, \alpha \in \Phi}$ and the vectors ${\alpha_1, \dots, \alpha_n}$ linearly independent, then each ${c_i \in \mathbb{Q}}$. This will prove the final claim of the proposition, that the sum is direct. In fact, we get equations of the form

$\displaystyle \frac{(\alpha, \alpha_i)}{(\alpha_i, \alpha_i )} = \sum_{j=1}^n \frac{ (\alpha_j, \alpha_i)}{(\alpha_i, \alpha_i)} c_j, \quad 1 \leq i \leq n$

from which we can conclude, since all coefficients are rational, that ${c_i \in \mathbb{Q}}$.

${\Phi}$ as a root system

I now will go into some abstract generality on root systems. First, I’ve got to say a few things on reflections in a euclidean vector space ${E}$—that is, a real one with an inenr product ${(\cdot, \cdot)}$. Given a nonzero vector ${v \in E}$, the reflection around ${v}$ is defined by

$\displaystyle s_v(w) := w - 2\frac{ (v,w)}{(w,w)} v .$

A reflection is characterized by fixing the hyperplane perpendicular to ${v}$ and sending ${v}$ onto ${-v}$. Conversely, any endomorphism of ${E}$ that does that is the reflection.

A root system in ${E}$ is a finite set ${\Phi \subset E}$ not containing zero that spans ${E}$. We require:

1. The only multiples of any ${\alpha \in \Phi}$ are ${\pm \alpha}$

2. ${\Phi}$ is invariant under the reflections ${s_{\alpha}, \alpha \in \Phi}$.

3. If ${\alpha, \beta \in \Phi}$, then ${2\frac{ (\alpha, \beta)}{(\alpha, \alpha)} }$ is an integer.

I will develop some of the theory of root systems in the sequel.   John Armstrong at The Unapologetic Mathematician has been doing a series on this, starting here.     Incidentally, if you haven’t seen his blog, you ought to check it out.  It’s a free tour up the slopes of Mount Bourbaki, designed to show the interested nonmathematician (or aspiring mathematician) that staggering precipice one initially sees when encountering the subject conceals a gentle and mostly walkable path up the other side of the mountain.  (OK, so my recent reading of Climbing Mount Improbable may have made me prone to abusing this particular metaphor.)  I’m going to overlap with him slightly, but probably no more than I absolutely have to in order to do the basic representation theory.

First, we need a bit of motivation:

Proposition 2 If ${\Phi \subset E}$ is the set of roots of a semisimple Lie algebra ${\mathfrak{g}}$ with a fixed Cartan subalgebra ${\mathfrak{h}}$ and ${E}$ is defined as above, then ${\Phi}$ is a root system.

It is clear from the definition that ${\Phi}$ spans ${E}$; finiteness follows because ${\mathfrak{g}}$ is finite-dimensional. We proved 3 and 1 yesterday and actually computed ${2\frac{ (\alpha, \beta)}{(\alpha, \alpha)} }$ in terms of maximal strings. The one to check is 2: if ${\alpha, \beta \in \Phi}$, we must prove ${s_{\alpha}(\beta) \in \Phi}$.

To see this, consider the maximal string of roots ${\beta + p\alpha, \dots, \beta+q\alpha}$ as in yesterday; then ${q+p = - 2\frac{ (\alpha, \beta)}{(\alpha, \alpha)}}$. Here ${\beta}$ is ${q}$ units down from ${\beta+q\alpha}$; by this logic, ${\beta + (p+q)\alpha}$ which is ${q}$ units up from the bottom should belong to ${\Phi}$. Now ${s_{\alpha}(\beta) = \beta + (q+p) \alpha}$ though, so this proves the proposition.

I’ve been slightly disingenuous here. The choice of a root system seems to depend very heavily on the choice of the Cartan subalgebra, which is not unique. However, it is a theorem (which I may or may not give a proof of soon) that Cartan subalgebras are unique up to conjugation.

Anway, it turns out that you can recover a Lie algebra from a root system, so the classification of simple Lie algebras becomes reduced to the classification of simple root systems, which was successfully done—there are four “classical” infinite families and five exceptional cases.