Screening coefficient and nonsection obstruction for the actual triplet projection

Since $e^{\alpha }$ is primary of weight one by [F3], its zero mode $Q$ commutes with every $L(n)$: taking the $z^{-1}$ coefficient of $[L(n),Y(e^{\alpha },z)]=z^{n+1}{\mathrm{\partial }}_{z}Y(e^{\alpha },z)+(n+1)z^{n}Y(e^{\alpha },z)$ gives zero. It annihilates the vacuum. Thus $Q\omega =0$. Also $QF=H$, $QH=E$, and $QE=Q^{3}F=0$: the latter has charge $2\alpha$ and weight $2p-1$, whereas that charge has minimum weight $2p+2$. By the zero-mode derivation identity from [F2], $Q$ therefore maps the four-generator algebra $W(p)$ into itself. Directly, $Qv=x_{1}B=\nu TB$ and $QB=0$, because the leading power in $Y(e^{\alpha },z)B$ is $z^{2p-2}$. The translation identity gives $(TB{)}_{(n)}=-n{B}_{(n-1)}$, so $[Q,v_{(n)}]=-\nu n{B}_{(n-1)}$ and $[Q,B_{(-1)}]=0$ on both the algebra and its module.

For $a\in W(p)$ put $c_n=(-1{)}^{n+1}\mathrm{/}n$. The difference between $[Q,\tilde{Y}(a,z)]$ and $\tilde{Y}(Qa,z)$ is ${\sum }_{n\ge 1}{c}_n{z}^{-n}Y([Q,v_{(n)}]a,z)=\nu {\sum }_{n\ge 1}(-1{)}^n{z}^{-n}Y(B_{(n-1)}a,z)$. On the other hand [F2] with $j=-1$ gives $[B_{(-1)},Y(a,z)]={\sum }_{i\ge 0}(-1{)}^i{z}^{-i-1}Y(B_{(i)}a,z)$. The two displayed sums cancel after $i=n-1$. Every deformation correction has a state in the sector-one ideal; its field and $B_{(-1)}$ have zero composite in either order on this explicit module, since both raise the sector index by one. Therefore $[G,\tilde{Y}(a,z)]=\tilde{Y}(Qa,z)\mathrm{.}$ All sums here are finite on the fixed state $a$, so no completion of the module is used. In particular $[G,F_{(k)}]={\tilde{H}}_{(k)}$ and $[G,{\tilde{H}}_{(k)}]={\tilde{E}}_{(k)}$, since the deformation of $F$ is zero.

The state $B$ has weight zero, hence $B_{(-1)}$ and $Q$ preserve the undeformed weight. The vector $Q^{2}t$ would have degree $p-1$ and charge $3\alpha \mathrm{/}2$, whose vacuum degree is $p+1$; it is zero. Moreover $B_{(-1)}^2=0$ by the sector rule. Using step 1.1 gives $G^{2}t=2\nu Q{B}_{(-1)}t$. The latter vector has charge $(3\mathrm{/}2-1\mathrm{/}p)\alpha$ and degree $p-1$, exactly the vacuum degree of that charge, so it is a scalar multiple of $b$. To compute its scalar, take ${\mathrm{Res}}_x{\mathrm{Res}}_y{y}^{-1}Y(e^{\alpha },x)Y(B,y)t$ and project the output to oscillator degree zero. The explicit exponential contraction in [F1] leaves the Laurent polynomial $x^{-p}{y}^{-(p-1)}(x-y{)}^{2p-2}\mathrm{.}$ The coefficient selected is $[x^{-1}{y}^0]$ before multiplication by $y^{-1}$. Expanding the last factor as ${\sum }_{j=0}^{2p-2}(-1{)}^j\left(\genfrac{}{}{0px}{}{2p-2}{j}\right)x^{2p-2-j}{y}^j$, only $j=p-1$ contributes. Thus $Q{B}_{(-1)}t=(-1{)}^{p-1}\left(\genfrac{}{}{0px}{}{2p-2}{p-1}\right)b$, proving $G^{2}t=C_{p}b$.

Put $m=3p-3$. The mode $F_{(m)}$ lowers degree by $p-1$. The vector $t$ has degree $p-1$ and charge $-\alpha \mathrm{/}2$, and $Gt$ has that same degree with possible charges $\alpha \mathrm{/}2$ and $(1\mathrm{/}2-1\mathrm{/}p)\alpha$. After applying $F_{(m)}$ their possible charges are $-3\alpha \mathrm{/}2$, $-\alpha \mathrm{/}2$ and $(-1\mathrm{/}2-1\mathrm{/}p)\alpha$. None is a degree-zero charge: [F3] shows that those are only $\alpha \mathrm{/}2$ and $(1\mathrm{/}2-1\mathrm{/}p)\alpha$. Hence $F_{(m)}t=F_{(m)}Gt=0$. Applying the double commutator from step 2.1 gives ${\tilde{E}}_{(m)}t=F_{(m)}{G}^{2}t$. The leading term of $Y(F,z)b$ is $z^{-3p+2}w$, with coefficient one in the specified trivial cocycle. Step 2.2 now yields ${\tilde{E}}_{(m)}t=C_{p}w$. Since $C_p\mathrm{\ne }0$, $w\in M_1$ and thus $N_2\subset M_1$. On the second sector all corrections vanish, so this copy of $N_2$ has precisely its undeformed action.

Every correction maps the first summand into the second and annihilates the second. Thus projection to the first intertwines the deformed action with the undeformed action. Its image on the cyclic module $M_1$ is exactly the undeformed cyclic space $\mathrm{\Pi }(1)$; $K=M_1\cap V_{L+\alpha \mathrm{/}2-\alpha \mathrm{/}p}$, and step 3.1 gives $N_2\subset K$. The explicit degree decomposition restricts to each cyclic space and to $K$: as in [F3], projection onto a finite set of eigencomponents of any vector is a polynomial in the weight operator. Consequently the image, kernel and inclusion are admissible with the inherited grading, and the displayed sequence with kernel $K$ is short exact.

In the undeformed first sector the same double commutator with $Q$ gives $E_{(m)}t=Q^2{F}_{(m)}t-2Q{F}_{(m)}Qt+F_{(m)}{Q}^{2}t=0$. Indeed the first and last terms vanish by steps 2.2 and 3.1, while $F_{(m)}Qt$ would have degree zero and charge $-\alpha \mathrm{/}2$, so it too vanishes. Suppose there were an admissible section $s$ of $\pi$. Since it preserves degree and $\pi s(t)=t$, write $s(t)=t+k$ with $k$ in the second sector of degree $p-1$. The second-sector vacuum degrees $p{n}^2-n$ show that its only possible charges are $(1\mathrm{/}2-1\mathrm{/}p)\alpha$ and $(3\mathrm{/}2-1\mathrm{/}p)\alpha$: all $n\le -1$ give degree at least $p+1$ and all $n\ge 2$ give degree at least $4p-2$. The undeformed $E_{(m)}$ has charge $\alpha$ and lowers degree by $p-1$. On these two possible charges its degree-zero output would have charge $(3\mathrm{/}2-1\mathrm{/}p)\alpha$ or $(5\mathrm{/}2-1\mathrm{/}p)\alpha$, both absent in degree zero. Corrections annihilate this sector; hence ${\tilde{E}}_{(m)}k=0$. Intertwining would now give $0=s(E_{(m)}t)={\tilde{E}}_{(m)}s(t)=C_{p}w\mathrm{\ne }0$, a contradiction. This proves absence of a section for the actual projection, independently of the unresolved equality $K=N_2$.