The level-one affine sl2 current relation

The affine simple-root triple at node $0$ has raising operator $e_0=f_1$, lowering operator $f_0=e_{-1}$ and Cartan element $h_0=c-h$. On the vacuum, the raising operator vanishes and $(c-h)\mathbf{1}=\mathbf{1}$, since $c$ acts by level $1$ and $h\mathbf{1}=0$. Integrability in [F2] makes this a finite-dimensional ${\mathfrak{s}\mathfrak{l}}_2$ string, so its highest-weight-one vector is killed by $f_0^2$, giving $e_{-1}^2\mathbf{1}=0$. The finite root has $(\alpha ,\alpha )=2$, so $m_{\alpha }=2\mathrm{\ell }\mathrm{/}(\alpha ,\alpha )=1$ and $N_1=2$. The displayed invariant form gives $B(e,e)=B(f,f)=B(e,h)=B(f,h)=0$ by invariance, so the $B$-dual basis to $(e,f,h)$ is $(f,e,h\mathrm{/}2)$. The adjoint Casimir is therefore ${\mathrm{\Omega }}_B=\mathrm{ad}(e)\mathrm{ad}(f)+\mathrm{ad}(f)\mathrm{ad}(e)+\frac{1}{2}\mathrm{ad}(h{)}^2$. On $e$ it acts by $[e,[f,e]]+[f,[e,e]]+\frac{1}{2}[h,[h,e]]=2e+0+2e=4e$. By the normalized convention $h^{\vee }={\gamma }_B\mathrm{/}2$ where ${\mathrm{\Omega }}_B={\gamma }_{B}I$ in the adjoint representation, $h^{\vee }=2$. Since $\dim {\mathfrak{s}\mathfrak{l}}_2=3$, [F5] gives $c_{\mathrm{V}\mathrm{i}\mathrm{r}}=3\mathrm{/}(1+2)=1$.

The root current modes commute by [F3], so its normal product equals its coefficientwise product. The vertex-algebra iterate identity sends the zero state $e_{-1}^2\mathbf{1}$ to the field $:J^e(z)J^e(z):=0$, and the module iterate identity gives the same zero product on every weak module. For $v\in M$, choose $k$ such that $e_{n}v=0$ for all $n>k$ by lower truncation. In the coefficient of $z^{-2(k+1)}$ in $J_M^e(z{)}^{2}v$, any term with an index above $k$ vanishes after commuting that mode to the right; if both indices are at most $k$ and sum to $2k$, both equal $k$. The coefficient is therefore $e_k^{2}v$, and the zero product forces it to vanish. The bound applies to this truncation index $k$; [F4] proves local nilpotence for arbitrary fixed modes. This is an algebraic current-mode identity; it asserts no Hilbert-space positivity or analytic CFT construction.

To check the limit of the coefficient conclusion, on the level-one VOA put $E=f_2$, $F=e_{-2}$ and $H=2c-h_0$, where $h_0$ here denotes the horizontal zero mode of $h$. The affine bracket gives $[E,F]=H$, $[H,E]=2E$ and $[H,F]=-2F$. The vacuum obeys $E\mathbf{1}=0$ and $H\mathbf{1}=21$, so $EF\mathbf{1}=21$ and $E{F}^2\mathbf{1}=(HF+FH)\mathbf{1}=2F\mathbf{1}$. Hence $E^2{F}^2\mathbf{1}=41\mathrm{\ne }0$, and therefore $e_{-2}^2\mathbf{1}\mathrm{\ne }0$. Thus the cutoff condition in step 1.2 is essential.