I'm trying to understand root system of semisimple Lie algebra but having trouble following one of the step in the note which explain why each root spaces are 1-dimensional. According to the note, given a semisimple Lie algebra $\mathfrak{g}$ we can decompose this into
\begin{equation}
\mathfrak{g} = \mathfrak{h} \oplus (\bigoplus_{\alpha \in \mathfrak{h}^*-{0}}\mathfrak{g}_{\alpha})
\end{equation}
where $\mathfrak{g}_\alpha := \{x \in \mathfrak{g} | [h,x] = \alpha(h)x, \forall h \in \mathfrak{h}\}$ and $\mathfrak{h} := \{x \in \mathfrak{g} | [\tilde{h},x] = 0\}$ where $\tilde{h}$ is regular and semisimple. Now the note said that, pick $X_{\alpha} \in \mathfrak{g}_{\alpha}$, $Y_{\alpha} \in \mathfrak{g}_{-\alpha}$ and $H_{\alpha} \in \mathfrak{h}_{\alpha} := [\mathfrak{g}_{\alpha}, \mathfrak{g}_{-\alpha}] \subset \mathfrak{h}$ (which was proven in the note earlier that $\mathfrak{h}_{\alpha}$ is 1-dimensional), these will generates a subalgebra isomorphic to $\mathfrak{sl}(2)$ and we will consider the adjoint representation of $\mathfrak{sl}(2)$.
So far everything seems alright, but the following is what confused me: Suppose that the dimension of $\mathfrak{g}_{-\alpha}$ is greater than 1 then we can find another $Z \in \mathfrak{g}_{-\alpha}$ such that $[X_{\alpha}, Z] = 0$ because $\mathfrak{h}_{\alpha}$ is 1-dimensional.
I don't understand why such a claim is true. I can only see that if $\mathfrak{h}_{\alpha}$ was 1-dimensional then $[X_{\alpha}, Z]$ should be proportional to $H_{\alpha}$. Could someone explain to me the step in obtaining the claim or else could someone guide me how to complete the proof from this point? (or just an explanation in general why root spaces are 1-dimensional because I tried to read from other places but the proof seems to be much more complicated than this one).
The note I'm referring to is http://stacky.net/files/written/LieGroups/LieGroups.pdf lecture 13, page 67-68.
Best Answer
If $\dim {\mathfrak g}_{-\alpha}>1$, then the map $[-,X_\alpha]: {\mathfrak g}_{-\alpha}\to {\mathfrak h}_\alpha$ can't be injective since ${\mathfrak h}_\alpha$ is $1$-dimensional as you already know. Explicitly, if $Z\in{\mathfrak g}_{-\alpha}$ then $Z^{\prime} := Z - [Z,X_{\alpha}]X_{-\alpha}$ satisfies $[Z,X_\alpha]=0$.
Alternatively, the proof I know and like is the following: Consider $${\mathfrak g}(\alpha) := {\mathbb k}\cdot X_{-\alpha}\oplus {\mathfrak h}_{\alpha}\oplus\bigoplus\limits_{n\geq 1} {\mathfrak g}_{n\alpha}.$$ It is naturally a module over the Lie subalgebra ${\mathbb k}\cdot\{X_{-\alpha},H_\alpha,X_\alpha\}\cong{\mathfrak sl}_2({\mathbb k})$ of ${\mathfrak g}$. Hence, since $H_\alpha = [X_\alpha,X_{-\alpha}]$, the trace of the adjoint action $[H_\alpha,-]$ on ${\mathfrak g}(\alpha)$ must be $0$, but on the other hand equals $$-2 + 0 + 2\cdot \dim {\mathfrak g}_\alpha + 4\cdot\dim {\mathfrak g}_{2\alpha} + ...$$ It follows that $\dim{\mathfrak g}_\alpha=1$ and $n\alpha\notin\Phi$ for all $n>1$.
In the infinite-dimensional setting of Kac-Moody algebras things break down very early, since it is not true in general that $[{\mathfrak h}_\alpha,{\mathfrak g}_\alpha] = [[{\mathfrak g}_\alpha,{\mathfrak g}_{-\alpha}],{\mathfrak g}_\alpha]\neq 0$, preventing you from constructing an ${\mathfrak sl}_2$-triple from any root (it only works for the real roots).