The actual two-dimensional triplet zero-mode plane modulo the endpoint

Put $d=(e^{\delta \alpha }{)}_{(-1)}t$ and $f=F_{(2p-2)}b$. The exponential formula gives $d=S_N(\delta x)w$ and $f=S_N(-x)w$: their initial powers are $u^{-N}$ and $u^{-3p+2}$, respectively, so the indicated coefficients select oscillator degree $N$ in each case. At $x_k=-2p$ the generating series for $S_j(ax)$ becomes $(1-u{)}^{2pa}$. Hence $S_N(\delta x)$ evaluates to $(-1{)}^N\left(\genfrac{}{}{0px}{}{2p-2}{N}\right)=C_0$, and $S_N(-x)$ evaluates to $\left(\genfrac{}{}{0px}{}{2p+N-1}{N}\right)=A_p$. Thus $Qd=C_{0}b$ and $Qf=A_{p}b$. It follows directly that $Q{w}^{\mathrm{\prime }}=2\nu (C_0{A}_p-A_p{C}_0)b=0$. Applying [F4] proves $w^{\mathrm{\prime }}\in U({\mathrm{V}\mathrm{i}\mathrm{r}}_{-})w\subset N_2$.

Directly $z=Qt=S_N(x)e^{\alpha \mathrm{/}2}\mathrm{\ne }0$. The ordinary $F_{(2p-2)}$ has degree shift zero. Its action on $z$ lands at charge $-\alpha \mathrm{/}2$ and degree $N$, hence is a scalar multiple of $t$. To compute it, project its exponential field on $z$ to oscillator degree zero: the annihilation exponential replaces $x_k$ in $S_N(x)$ by $2p{u}^{-k}$ and the initial power is $u^{-p}$. The coefficient of $u^{-2p+1}$ is $S_N(2p,2p,\dots )=A_p$. Therefore $F_{(2p-2)}z=A_{p}t$. Also $F_{(2p-2)}t=0$ and $F_{(2p-2)}d=0$, since their possible output charges $-3\alpha \mathrm{/}2$ and $(\beta -1)\alpha$ have minimum degrees $4p-2$ and $p+1$, strictly greater than $N$. Finally $Qb=0$ and $Qz=0$ by charge and degree: their possible output charges have vacuum degrees $4p-2$ and $p+1$. Using [F3] gives $Gt=z+\nu d$, $Gz=\nu C_{0}b$, and $G{w}^{\mathrm{\prime }}=Q{w}^{\mathrm{\prime }}=0$, since $w^{\mathrm{\prime }}$ is in the second sector.

Write $F_0=F_{(2p-2)}$, $H_0={\tilde{H}}_{(2p-2)}$, and $E_0={\tilde{E}}_{(2p-2)}$ in this step. By [F3] and step 2.1, $H_{0}t=[G,F_0]t=-A_{p}t,H_{0}z=G(A_{p}t)-F_0(\nu C_{0}b)=A_{p}z+\nu A_{p}d-\nu C_{0}f=A_{p}z-\frac{1}{2}{w}^{\mathrm{\prime }}\mathrm{.}$ Taking the second commutator on $t$ gives $E_{0}t=G^2{F}_{0}t-2G{F}_{0}Gt+F_0{G}^{2}t=-2A_p(z+\nu d)+2\nu C_{0}f=-2A_{p}z+w^{\mathrm{\prime }}\mathrm{.}$ Since $w^{\mathrm{\prime }}\in N_2\subset M_1$ and $A_p\mathrm{\ne }0$, this proves $z\in M_1$. On the second sector $H_0$ is ordinary, and its zero-mode derivation expression gives $H_{0}b=[Q,F_0]b=Qf=A_{p}b$. Thus the other second commutator yields $E_{0}z=G(A_{p}z-w^{\mathrm{\prime }}\mathrm{/}2)-H_0(\nu C_{0}b)=0$. The deformation of $F$ is zero by [F3]. This proves every asserted generator action before taking classes.

The vectors $t$ and $z$ are in the first summand, with different charges $-\alpha \mathrm{/}2$ and $\alpha \mathrm{/}2$; $z\mathrm{\ne }0$ by step 2.1. Since $N_2$ is in the second summand, their classes are linearly independent. They both have degree $N$, and [F2] makes $\tilde{L}(0)$ the scalar $h+N$ on them. Reducing the identities in step 3.1 modulo $N_2$ proves the three matrices. Their span is stable, contains $\bar{t}$, and also contains $\bar{z}=-o(E)\bar{t}\mathrm{/}(2A_p)$. Hence it is exactly the two-dimensional subspace generated under the four indicated zero modes.