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}.