Simple root string in an integrable kac moody module

Write $e=e_i,f=f_i,h=h_i$ and initially $a=\eta (h_i)$. F2 gives $h{f}^{k}v=(a-2k)f^{k}v$. From $ev=0$ and $e{f}^{k+1}v=fe{f}^{k}v+h{f}^{k}v$, induction gives $e{f}^{k}v=k(a-k+1)f^{k-1}v$. Let $N\ge 1$ be the least exponent with $f^{N}v=0$, supplied by F1. Then $f^{N-1}v\mathrm{\ne }0$ and $0=e{f}^{N}v=N(a-N+1)f^{N-1}v$ forces $a=N-1=m\in {\mathbb{Z}}_{\ge 0}$. Thus precisely the powers from $0$ through $m$ are nonzero.

The span $S$ of these powers is invariant under $e,f,h$ by the formulas in 1.1, and contains $v$. Conversely every vector in the list is obtained by applying a power of $f$ to $v$. Hence $S$ is exactly the cyclic simple-root module. Full Cartan commutation in F2 gives weight $\eta -k{\alpha }_i$ to $f^{k}v$. These weights are distinct, since ${\alpha }_i(h_i)=2$, so the nonzero vectors are independent. Their raising coefficients are $k(m-k+1)$, nonzero for $1\le k\le m$ and zero at the highest endpoint.

Directly using F3, $s_i(\eta -k{\alpha }_i)=\eta -k{\alpha }_i-(m-2k){\alpha }_i=\eta -(m-k){\alpha }_i$. Within $S$, the explicit map $f^{k}v\mapsto f^{m-k}v$ between each pair of one-dimensional weight spaces is a linear isomorphism. F4 additionally identifies the corresponding full ambient weight spaces, without a finite-multiplicity assumption. At $k=0$ and $k=m$ this exchanges the endpoints. If $m=0$, the sole vector is killed by both $e,f$ and the reflection fixes its weight. Zero $v$ is excluded; no assumption about other weight strings or the whole coset intersection was made. Only finite strings and F4's choice-free maps are used.