All weight spaces are 1 dimensional implies representation is irreducible for $\mathfrak{sl}(2,\mathbb{C})$? I'm following Kirillov's book "An Introduction to Lie Groups and Lie Algebras".
In Theorem 6.44 of Chapter 6 (page 124) he mentions the following statement:
For roots $\alpha, \beta \neq \alpha$, the subspace
$V = \bigoplus\limits_{k \in \mathbb{Z}} \mathfrak{g}_{\beta + k \alpha}$
is an irreducible representation of $\mathfrak{sl}(2,\mathbb{C})_\alpha$.
For some context,
$\mathfrak{g}_{\beta + k \alpha}$ is the root space associated with the root $\beta + k \alpha$,
$\mathfrak{sl}(2,\mathbb{C})_\alpha$ is the subalgebra (isomorphic to $\mathfrak{sl}(2,\mathbb{C})$) generated by $e \in \mathfrak{g}_\alpha$, $f \in \mathfrak{g}_{-\alpha}$ and $h=[e,f]$,
The representation being referred to here is the adjoint representation restricted to the $\mathfrak{sl}(2,\mathbb{C})_\alpha$ subalgebra.
In the proof for the statement, the author just states that it follows immediately from the fact that dim $\mathfrak{g}_{\beta + k \alpha} = 1$. This is not obvious to me at all. Is there some theorem that connects the dimensions of the weight spaces in a representation of $\mathfrak{sl}(2,\mathbb{C})$ to the reducibility of the representation? Any help would be much appreciated.
Edit:
You are not allowed to use (at least not directly) the fact that if $\alpha$, $\beta$, $\alpha + \beta$ are roots, then $[\mathfrak{g}_\alpha, \mathfrak{g}_\beta] = \mathfrak{g}_{\alpha + \beta}$ (More specifically $\text{ad}_{e_\alpha} e_\beta \neq 0$ ($e_\alpha \in \mathfrak{g}_\alpha$, $e_\beta \in \mathfrak{g}_\beta$) if $\mathfrak{g}_{\alpha + \beta} \neq \{0\}$)
This is because the book uses the irreducibility of the above representation to prove this fact. So we'd end up in a circular argument.
 A: I feel like in the phrasing of my question, I've ended up asking two questions in one post. The question in the title has been answered to be false in general with the counter example $V(1) \oplus V(0)$ thanks to @JyrkiLahtonen. More explicitly the representation
\begin{equation}
h=
\begin{pmatrix}
0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1
\end{pmatrix}
,\;
e=
\begin{pmatrix}
0 & 0 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{pmatrix}
,\;
f=
\begin{pmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 1 & 0
\end{pmatrix}
\end{equation}
has weights $0,1,-1$ with 1 dimensional weight spaces but it is reducible.
$\\$
As for the theorem in Kirillov's book, I have come up with the following proof:
Let $e_\alpha \in \mathfrak{g}_\alpha$ and $e_{-\alpha} \in \mathfrak{g}_{-\alpha}$
Choose some $e_\beta \in \mathfrak{g}_\beta$ and consider the subspace
$S_1 = \text{span} \{..., (\text{ad}_{e_{-\alpha}})^2 e_\beta, (\text{ad}_{e_{-\alpha}}) e_\beta, e_\beta, (\text{ad}_{e_\alpha}) e_\beta, (\text{ad}_{e_\alpha})^2 e_\beta, ...\}$
Now $S_1$ is an irreducible representation of $\mathfrak{sl}(2,\mathbb{C})_\alpha$
Further $S_1$ is a subspace of $V$
Due to the fact that $\mathfrak{g}_{\beta + k \alpha}$ is 1 dimensional, $S_1 = \bigoplus\limits_{\;\:\,k \in \mathbb{Z}\\-m \leq k \leq n} \mathfrak{g}_{\beta + k \alpha}$
Where $n$ is such that $(\text{ad}_{e_\alpha})^n e_\beta \neq 0$ but $(\text{ad}_{e_\alpha})^{n+1} e_\beta = 0$
and $m$ is such that $(\text{ad}_{e_{-\alpha}})^m e_\beta \neq 0$ but $(\text{ad}_{e_{-\alpha}})^{m+1} e_\beta = 0$
So then $V = (\bigoplus\limits_{k \in \mathbb{Z}\\k>m} \mathfrak{g}_{\beta - k \alpha}) \oplus S_1 \oplus (\bigoplus\limits_{k \in \mathbb{Z}\\k>n} \mathfrak{g}_{\beta + k \alpha})$
Let $S' = \bigoplus\limits_{k \in \mathbb{Z}\\k>n} \mathfrak{g}_{\beta + k \alpha}$
and $S'' = \bigoplus\limits_{k \in \mathbb{Z}\\k>m} \mathfrak{g}_{\beta - k \alpha}$
We need to show that $S'=0$ and $S''=0$. We will show $S'=0$. Similar arguments can be used to show $S''=0$.
Suppose $S' \neq 0$, there must exist some $l>n$ such that $\mathfrak{g}_{\beta + l \alpha} \neq 0$
Let $l$ be the smallest integer greater than $n$ such that $\mathfrak{g}_{\beta + l \alpha} \neq 0$
Choose some $e_{\beta + l \alpha} \in \mathfrak{g}_{\beta + l \alpha}$ and consider the subspace
$S_2 = \{e_{\beta + l \alpha}, (\text{ad}_{e_\alpha}) e_{\beta + l \alpha}, (\text{ad}_{e_\alpha})^2 e_{\beta + l \alpha}, ...\}$
Again $S_2$ is an irreducible representation of $\mathfrak{sl}(2,\mathbb{C})_\alpha$
Now $S_1$ is an irrep of $\mathfrak{sl}(2,\mathbb{C})_\alpha$ with the highest weight $\beta + n \alpha$
and $S_2$ is an irrep of $\mathfrak{sl}(2,\mathbb{C})_\alpha$ with the lowest weight $\beta + l \alpha$
But considering the fact that any irreducible representation of $\mathfrak{sl}(2,\mathbb{C})_\alpha$ must have weights of the form:
$\{...,-\alpha,0,\alpha,...\}$ for odd dim
$\{...,-\alpha/2,\alpha/2,...\}$ for even dim
We can't have $\beta + n \alpha$ be the highest weight of one irrep and $\beta + l \alpha$ be the lowest weight of another irrep when $l>n$. Hence the proof.
