Recall that the reflections ${s_{\alpha} \in Aut(E)}$ for ${\alpha \in \Phi}$ preserve ${\Phi}$. They generate a group ${W}$ called the Weyl group. Moreover, since ${\Phi}$ spans ${E}$, the map ${W \rightarrow S_{\Phi}}$, the symmetric group on ${\Phi}$, is injective. So ${W}$ is a finite group of orthogonal isomorphisms of ${E}$, i.e. leaving invariant the bilinear form ${(\cdot, \cdot)}$.

Everything here actually makes sense for root systems in general, but we are restricting ourselves to the case of a root system associated to semisimple Lie algebra and a Cartan subalgebra.  The only difference is that one has to prove the result on maximal strings (which was proved in the case of Lie algebras here), though it can be done for root systems in general.

Now choose a base ${\Delta}$ for ${\Phi}$ and a corresponding partition ${\Phi = \Phi^+ \cup \Phi^-}$; the ${s_{\delta}}$ for ${\delta \in \Delta}$ are called simple reflections.

The goal we are aiming for is the following theorem, which gives a large amount of information about the Weyl group.

Theorem 1 ${W}$ acts simply transitively on Weyl chambers and on bases. Any root can be moved by an element of the Weyl group into a given base. ${W}$ is generated by the simple reflections. (more…)

So, suppose given a root system ${\Phi}$ in a euclidean space ${E}$, which arises from a semisimple Lie algebra and a Cartan subalgebra as before. The first goal of this post is to discuss the “splitting”

$\displaystyle \Phi = \Phi^+ \cup \Phi^-$

(disjoint union) in a particular way, into positive and negative roots, and the basis decomposition into simple roots. Here ${\Phi^- = - \Phi^+}$.

To do this, choose ${v \in E}$ such that ${(v, \alpha) \neq 0}$ for ${\alpha \in \Phi}$. Then define ${\Phi^+}$ to be those roots ${\alpha}$ with ${(v,\alpha)>0}$ and ${\Phi^-}$ those with ${(v,\alpha) < 0}$. This was easy. We talked about positive and negative roots before using a real-valued linear functional, which here is given by an inner product anyway.

Bases

OK. Next, I claim it is possible to choose a linearly independent set ${\Delta \subset \Phi^+}$ such that every root is a combination

$\displaystyle \alpha = \sum k_i \delta_i, \quad \delta_i \in \Delta, \ k_i \in \mathbb{Z}$

with all the ${k_i \geq 0}$ or all the ${k_i \leq 0}$.

Then ${\Delta}$ will be called a base. It is not unique, but I will show how to construct this below. (more…)

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 ${\mathfrak{h}^{\vee} = E \oplus iE}$. (more…)