Integrable highest weight modules for rank one gcm

The negative algebra is generated by its sole generator $f$, so every bracket of length at least two is zero, beginning with $[f,f]=0$ and inducting on bracket length. Hence F5 spans $L_A(m)$ by powers of $f$ on $v_0$. Put $v_k=f^k{v}_0$. By F1 and F2, exactly the powers through $m$ are nonzero. F4 gives $h{v}_k=(m-2k)v_k$ by induction from the highest label. These $m+1$ eigenvalues are distinct, so the vectors are independent: Lagrange polynomials in $h$ isolate each coefficient in a finite relation. Thus they form a basis.

The formula for $f$ is the definition of $v_k$. Starting with $e{v}_0=0$, the recurrence $e{f}^{k+1}{v}_0=fe{f}^k{v}_0+h{f}^k{v}_0$ gives the coefficient recurrence $c_{k+1}=c_k+m-2k$, $c_0=0$. Summing yields $c_k=km-k(k-1)=k(m-k+1)$, proving the displayed formula for $e$. It includes $k=0$ and at $k=m+1$ gives zero consistently with $v_{m+1}=0$.

The formulas directly check the brackets on every basis vector. Whenever the corresponding shifted vector is present, $[h,e]v_k=2k(m-k+1)v_{k-1}=2e{v}_k$ and $[h,f]v_k=-2v_{k+1}=-2f{v}_k$; at the omitted endpoints both sides vanish. For $0<k<m$, $[e,f]v_k=((k+1)(m-k)-k(m-k+1))v_k=(m-2k)v_k$. At $k=0<m$, this is $m{v}_0$; at $k=m>0$, it is $-m{v}_m$. For $m=0$ all three operators act as zero and all relations hold. Both $e^{m+1}$ and $f^{m+1}$ vanish. If a submodule is nonzero, interpolation in $h$ gives some $v_k$ in it; then $e^k{v}_k=({\prod }_{j=1}^{k}j(m-j+1))v_0\mathrm{\ne }0$, with empty product $1$ for $k=0$. Applying $f$ then supplies every basis vector, proving simplicity directly. No AC is used.