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.

Now {\mathbb{C} H_{\alpha}} belongs to the representation {\mathfrak{s}_{\alpha} \subset V}—that is, the adjoint representation of {\mathfrak{s}_{\alpha}}, which is irreducible because {\mathfrak{s}_{\alpha}} is simple! So by Weyl’s theorem, we have a decomposition of {\mathfrak{s}_{\alpha}}-modules

\displaystyle V = \mathfrak{s}_{\alpha} \oplus W .

The space {\ker(\alpha) \subset \mathfrak{h} \subset V} is also a {\mathfrak{s}_{\alpha}} submodule of {V}, disjoint from {\mathfrak{s}_{\alpha}}; indeed it is the zero representation. We can expand this decomposition

\displaystyle V = \mathfrak{s}_{\alpha} \oplus \ker(\alpha) \oplus W'.

I claim {W' = 0}. Indeed, all weights of the action of {H_{\alpha}} on {V} are even because {\alpha(H_{\alpha})=2}, so any irreducible submodule of {V} (and a fortiori of {W'}) must have an even highest weight, and thus a nonzero vector with weight zero. But all the vectors of weight zero have been accounted for in {\mathfrak{s}_{\alpha} \oplus \ker(\alpha)}. So {W' = 0}.

In particular,

\displaystyle \bigoplus_{\mathbb{Z} \alpha} \mathfrak{g}_{\alpha} = \mathbb{C} Y_{\alpha} \oplus \mathfrak{h} \oplus \mathbb{C} X_{\alpha}. \ (*)

The proposition thus follows.

From this we see in particular that the only integral multiples of {\alpha} that are roots are {\pm \alpha}. More strongly:

Proposition 2 If {\alpha \in \Phi}, the only multiples of {\alpha} that are roots are {\pm \alpha}.


Indeed, we can consider the representation of {\mathfrak{s}_{\alpha}} on {\bigoplus_{c \in \mathbb{C}} \mathfrak{g}_{c \alpha}}. On this {H_{\alpha}} must act with integral weights only by the representation theory of {\mathfrak{sl}_2}, so in particular, only multiples of {\frac{1}{2} \alpha} can occur. But if {\frac{1}{2}\alpha} were a root, then {\alpha} as an integral multiple of {\frac{1}{2} \alpha} would not allowed to be a root. If {\frac{k}{2} \alpha} with {k} odd occurred, then {k} would be a weight for {H_{\alpha}}—and so would {1}, meaning {\mathfrak{g}_{\frac{1}{2}\alpha} \neq 0}, which we’ve just proved can’t happen.

We saw a trick in the previous arguments: consider the action of {\mathfrak{s}_{\alpha}} on a suitable subalgebra of {\mathfrak{g}} and apply the representation theory of {\mathfrak{sl}_2} to it.

Now let {\alpha, \beta \in \Phi}. We want to see for which {k} is {\beta + k\alpha} a weight.

Let {k} range over an interval {S=[p,q]} such that {\beta + k \alpha \in \Phi} for {k \in S}, but {\beta + (q+1)\alpha, \beta + (p+1)\alpha \notin \Phi}. Consider the submodule

\displaystyle W = \bigoplus_{k \in S} \mathfrak{g}_{\beta + k \alpha}.

Now {\mathfrak{s}_{\alpha}} preserves this. Moreover, {H_\alpha} acts on this space with trace

\displaystyle \sum_{k = p}^q \beta(H_{\alpha}) + 2k = (p-q+1) \beta(H_{\alpha}) + 2(p-q+1) \frac{p+q}{2}

This trace is also zero though—because {X_{\alpha}, Y_{\alpha}} preserve {W} and {H_{\alpha} = [X_{\alpha}, Y_{\alpha}]}. As a result:

\displaystyle \beta(H_{\alpha}) = 2 \frac{ B(T_{\beta}, T_{\alpha})}{B(T_{\alpha}, T_{\alpha})} = -(p+q)

since {H_{\alpha} = \alpha(T_{\alpha})^{-1} T_{\alpha} }.

Now define a bilinear form {(\cdot, \cdot)} on {\mathfrak{h}^{\vee}} by {(\alpha, \beta) := B(T_{\alpha}, T_{\beta})}. With this, we find

Proposition 3

If {\alpha, \beta \in \Phi},\displaystyle \frac{2(\beta, \alpha)}{(\alpha, \alpha)} \in \mathbb{Z}.

In particular, it is equal to {-(p+q)} where {p,q} are above. 


This will be important in the sequel, as this is one of the axioms of a root system. We will show that the roots form a root system.

If we had any two distinct sets {S=[p,q], S'=[p',q']} as above (Helgason calls them “maximal strings”), one would have to be disjoint from the other, which contradicts {p+q = p' + q'}. So:

Corollary 4 The set of {k \in \mathbb{Z}} such that {\beta+ k \alpha \in \mathbb{Z}} is an interval.