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.