The root-string property

The subspace $V={⨁}_{k\in \mathbb{Z}}{\mathfrak{g}}_{\beta +k\alpha }$ is finite-dimensional. By [L2], ${\mathrm{ad}}_{e_{\alpha }}$ maps its $k$-th summand into its $(k+1)$-st summand and ${\mathrm{ad}}_{f_{\alpha }}$ maps it into its $(k-1)$-st summand, including any case in which the target is ${\mathfrak{g}}_0=\mathfrak{h}$; ${\mathrm{ad}}_{h_{\alpha }}$ preserves every summand. Thus $V$ is a finite-dimensional module over the copy of ${\mathfrak{s}\mathfrak{l}}_2$ in [L1]. On ${\mathfrak{g}}_{\beta +k\alpha }$, $h_{\alpha }$ has eigenvalue $\beta (h_{\alpha })+2k$ by [L5], and these eigenvalues are distinct as $k$ varies. Hence the nonzero summands ${\mathfrak{g}}_{\beta +k\alpha }$ are exactly the $h_{\alpha }$-weight spaces of $V$.

Decompose $V$ into irreducibles as in [L3]. If $W$ is one irreducible summand, its $h_{\alpha }$-weights are $m,m-2,\dots ,-m$ with $m\ge 0$ integer. Thus the indices $k$ for which $W\cap {\mathfrak{g}}_{\beta +k\alpha }\mathrm{\ne }0$ form an interval of integers $\{a_W,a_W+1,\dots ,b_W\}$ determined by $\beta (h_{\alpha })+2a_W=-m$ and $\beta (h_{\alpha })+2b_W=m$. Consequently $a_W+b_W=-\beta (h_{\alpha })$ for every irreducible summand $W$.

The intervals in step 2.1 all have centre $-\beta (h_{\alpha })\mathrm{/}2$, so they are nested and their finite union is the interval $\{a,a+1,\dots ,b\}$ with $a+b=-\beta (h_{\alpha })$. This union is exactly $\{k:{\mathfrak{g}}_{\beta +k\alpha }\mathrm{\ne }0\}$ by step 1.1. It contains $0$, because ${\mathfrak{g}}_{\beta }\mathrm{\ne }0$ for $\beta \in \mathrm{\Phi }\cup \{0\}$ and ${\mathfrak{g}}_0=\mathfrak{h}\mathrm{\ne }0$; hence $a\le 0\le b$. Put $p=-a\ge 0$ and $q=b\ge 0$. Then the index set is $\{-p,\dots ,q\}$ and $p-q=-(a+b)=\beta (h_{\alpha })\in \mathbb{Z}$.