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…)

Advertisements

OK, now we’ve gotten some of the basic facts about the root space decomposition down. So, as usual ${\mathfrak{g}}$ is a semisimple Lie algebra and ${\mathfrak{h}}$ a Cartan subalgebra; we have the decomposition ${\mathfrak{g} = \mathfrak{h} \oplus \bigoplus_{\alpha \in \Phi} \mathfrak{g}_{\alpha}}$, where ${\Phi \subset \mathfrak{h}^{\vee}}$ is the root system. For each ${\alpha \in \Phi}$, we can choose a pair of vectors ${X_{\alpha} \in \mathfrak{g}_{\alpha}< Y_{\alpha} \in \mathfrak{g}_{-\alpha}, H_{\alpha} \in \mathfrak{h}}$. Then ${X_{\alpha}, Y_{\alpha}, H_{\alpha}}$ generate a subalgebra ${\mathfrak{s}_{\alpha} \subset \mathfrak{g}}$ which is isomorphic to ${\mathfrak{sl}_2}$. Here ${\alpha(H_{\alpha})=2}$ and ${H_{\alpha}}$ is a multiple of ${T_{\alpha}}$, which in turn is the dual to ${\alpha}$ under the Killing form that identifies ${\mathfrak{h} \simeq \mathfrak{h}^{\vee}}$.

That was a lightning review of where we are; if you’ve missed something, check back at this post.

The notation ${\mathfrak{s}_{\alpha}}$ suggests that the algebra should only depend on ${\alpha}$ and not on the particular choice of ${X_{\alpha}, Y_{\alpha}}$ (but ${H_{\alpha}}$ is uniquely determined from ${\alpha(H_{\alpha})=2}$ and ${H_{\alpha} \in \mathbb{C} T_{\alpha}}$). Indeed, this is the case, and it follows from

Proposition 1 When ${\alpha \in \Phi}$, ${\mathfrak{g}_{\alpha}}$ is one-dimensional.

Choose any ${\mathfrak{s}_{\alpha}}$ coming from suitable ${X_{\alpha}, Y_{\alpha}}$ and ${H_{\alpha}}$. We have a representation of ${\mathfrak{s}_{\alpha}}$ on $\displaystyle V := \bigoplus_{\mathbb{Z} \alpha} \mathfrak{g}_{\alpha}$

(recall ${\mathfrak{g}_0 = \mathfrak{h}}$) and we can apply the representation theory of ${\mathfrak{sl}_2}$ to it. (more…)

Let ${\mathfrak{g}}$ be a semisimple Lie algebra over ${\mathbb{C}}$ and ${\mathfrak{h}}$ a Cartan subalgebra.

Given ${\alpha \in \mathfrak{h}^{\vee}}$, we can define a subspace of ${\mathfrak{g}}$ $\displaystyle \mathfrak{g}_{\alpha} = \{ x \in \mathfrak{g}: (\mathrm{ad} H)x = \alpha(H) x , \ \forall H \in \mathfrak{h} \}.$

The nonzero ${\alpha}$ that occur with ${\mathfrak{g}_{\alpha} \neq 0}$ are called roots, and they form a set ${\Phi}$. Because ${\mathfrak{h}}$ acts on ${\mathfrak{g}}$ by commuting diagonalizable operators (by semisimplicity of the elements of ${\mathfrak{h}}$), it follows by simultaneous diagonalization, that $\displaystyle \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_{\alpha \in \Phi} \mathfrak{g}_{\alpha}.$

Recall that ${\mathfrak{g}_0 = \mathfrak{h}}$, because a Cartan subalgebra is maximal abelian.

This is called the root space decomposition. A simple but important property is that ${[\mathfrak{g}_{\alpha}, \mathfrak{g}_{\beta}] \subset \mathfrak{g}_{\alpha + \beta}}$; this is checked because the ${\mathrm{ad} H}$ are derivations.

The root space decomposition is highly useful in studying simple representations of ${\mathfrak{g}}$.

I shall collect here a few facts about it.

Proposition 1 ${\mathfrak{g}_{\alpha}, \mathfrak{g}_{\beta}}$ are orthogonal under the Killing form unless ${\alpha + \beta = 0}$.

This follows by a familiar argument, in view of ${[\mathfrak{g}_{\alpha}, \mathfrak{g}_{\beta}] \subset \mathfrak{g}_{\alpha + \beta}}$. (more…)