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 ${[\mathfrak{g}_{\alpha}, \mathfrak{g}_{\beta}] \subset \mathfrak{g}_{\alpha + \beta}}$.

As a result, we find that if ${\alpha \in \Phi}$, then ${-\alpha \in \Phi}$ too, because there has to be something with root ${-\alpha}$ to dualize ${\mathfrak{g}_{\alpha}}$ under the Killing form.

We now consider the commutators ${[\mathfrak{g}_{\alpha}, \mathfrak{g}_{-\alpha}] \subset \mathfrak{h}}$.

Proposition 2 We have ${[\mathfrak{g}_{\alpha}, \mathfrak{g}_{-\alpha}] = \mathbb{C} T_{\alpha}}$, where ${T_{\alpha}}$ is the dual to ${\alpha}$ under the Killing form.

The definition of ${T_{\alpha}}$ is that $\displaystyle B(T_{\alpha}, H) = \alpha(H), \ \forall H \in \mathfrak{h}.$

This turns out to be relatively easy to prove. Let ${X \in \mathfrak{g}_{\alpha}, Y \in \mathfrak{g}_{-\alpha}}$; then I claim $\displaystyle [X,Y] = B(X,,Y) T_{\alpha} .$

To see this, we will show that the two quantities above (which are in ${\mathfrak{h}}$, on which ${B}$ is nondegenerate) have the same “inner products” (via ${B}$) with any other vector ${H}$ in ${\mathfrak{h}}$. But by invariance and symmetry of the Killing form: $\displaystyle B(H, [X,Y]) = B([H,X], Y) = \alpha(H) B(X,Y) = B( H, T_{\alpha} B(X,Y)).$

This proves the claim.

Proposition 3 We have ${\alpha(T_\alpha) = B(T_{\alpha}, T_{\alpha}) \neq 0}$.

Choose nonzero ${X \in \mathfrak{g}_{\alpha}, Y \in \mathfrak{g}_{\alpha}}$ with ${B(X,Y) \neq 0}$. Suppose ${\alpha(T_{\alpha})=0}$. Now the subspace ${\mathbb{C} T_{\alpha} \oplus \mathbb{C} X \oplus \mathbb{C} Y}$ is a subalgebra ${\mathfrak{s}}$ by the commutation relations. Moreover ${[T_{\alpha},X]=[T_{\alpha},Y] = 0}$ by hypothesis. Therefore, this is actually a solvable Lie algebra, and its adjoint action on ${\mathfrak{g}}$ can be made to happen by upper-traingular matrices. In particular ${T_{\alpha} \in [\mathfrak{s}, \mathfrak{s}]}$ must have ${\mathrm{ad} T_{\alpha}}$ nilpotent on ${\mathfrak{g}}$. However, ${\mathrm{ad} T_{\alpha}}$ is also semisimple, which means ${\mathrm{ad} T_{\alpha} = 0}$, contradiction.

Therefore ${\alpha(T_{\alpha}) \neq 0}$. So, choose ${H_{\alpha}}$ an appropriate multiple of ${T_{\alpha}}$ such that $\displaystyle \alpha(H_{\alpha}) = 2.$

Choose ${X_{\alpha} \in \mathfrak{g}_{\alpha}, Y_{\alpha} \in \mathfrak{g}_{-\alpha}}$ scaled appropriately so that ${[ X_{\alpha},Y_{\alpha}] = H_{\alpha}}$. Then ${X_{\alpha}, Y_{\alpha}, H_{\alpha}}$ satisfy: $\displaystyle [H_{\alpha}, X_{\alpha}] = 2 X_{\alpha}, \ [H_{\alpha}, Y_{\alpha}] = -2 Y_{\alpha}, \ [ X_{\alpha},Y_{\alpha}] = H_{\alpha}.$

These are precisely the relations for ${\mathfrak{sl}_2}$. Indeed, this explains a remark I made way back when about ${\mathfrak{sl}_2}$ being a highly important example. Let the subalgebra of ${\mathfrak{g}}$ so defined be denoted ${\mathfrak{s}_\alpha}$; ${\mathfrak{s}_\alpha}$ is isomorphic to ${\mathfrak{sl}_2}$.

Proposition 4 When ${\alpha \neq 0}$, ${\mathfrak{g}_{\alpha}}$ is one-dimensional.

We will prove this next time by a more careful look at the representation theory of ${\mathfrak{sl}_2}$.