Conformal subalgebra in the kernel of $v_{(0)}$

Put $q=(p-1)\mathrm{/}(2p)$. The field of $\omega$ is $\mathcal{T}(z)=:\alpha (z{)}^2:\mathrm{/}(4p)+q{\mathrm{\partial }}_z\alpha (z)$. The oscillator contraction $\alpha (z)\alpha (w)\sim 2p\mathrm{/}(z-w{)}^2$ gives $\mathcal{T}(z)\mathcal{T}(w)\sim (1\mathrm{/}2-12p{q}^2)\mathrm{/}(z-w{)}^4+2\mathcal{T}(w)\mathrm{/}(z-w{)}^2+{\mathrm{\partial }}_w\mathcal{T}(w)\mathrm{/}(z-w)$. Specifically the double quadratic contraction is $1\mathrm{/}2$ in fourth order; the two improvement terms give $q^2{\mathrm{\partial }}_z{\mathrm{\partial }}_w(2p\mathrm{/}(z-w{)}^2)=-12p{q}^2\mathrm{/}(z-w{)}^4$; the mixed cubic poles cancel and their remaining poles are $2q\mathrm{\partial }\alpha (w)\mathrm{/}(z-w{)}^2+q{\mathrm{\partial }}^2\alpha (w)\mathrm{/}(z-w)$. The quadratic single contractions give the remaining quadratic part and its derivative. Taking residues using [F2] proves the Virasoro relations and the displayed central charge. The modes are $L(n)=\frac{1}{4p}{\sum }_{j\in \mathbb{Z}}:{\alpha }_j{\alpha }_{n-j}:-q(n+1){\alpha }_n\mathrm{.}$ At $n=-1$ the improvement term vanishes. Pairing the terms with indices $j=m$ and $j=-m-1$ for $m\ge 0$ gives $L(-1)=\frac{1}{2p}({\alpha }_{-1}{\alpha }_0+{\sum }_{m\ge 1}{\alpha }_{-m-1}{\alpha }_m)\mathrm{.}$ On ${\mathcal{F}}_t$ this acts as $t{x}_1+{\sum }_{m\ge 1}m{x}_{m+1}{\mathrm{\partial }}_{x_m}$, precisely the translation $T$ in [F1] and [F2]. At $n=0$ the mode formula gives the weight operator in the Statement. On $V_L$ its vacuum-charge weights are $p{n}^2-(p-1)n$, nonnegative integers with equality only at $n=0$. Every weight space is finite because this quadratic bounds the charges, and the polynomial space in each remaining oscillator degree has the finite partition basis. Thus $V_L$ has all properties in [F3].

The same single/double contractions with an exponential of charge $t$ give $\mathcal{T}(z){\mathcal{E}}_t(w)\sim (p{t}^2-(p-1)t){\mathcal{E}}_t(w)\mathrm{/}(z-w{)}^2+{\mathrm{\partial }}_w{\mathcal{E}}_t(w)\mathrm{/}(z-w)$. The double quadratic contraction contributes $p{t}^2$, the improvement contributes $-(p-1)t$, and the single quadratic contraction contributes $t:\alpha {\mathcal{E}}_t:=\mathrm{\partial }{\mathcal{E}}_t$ by differentiating its explicit exponential formula. Thus the corresponding vacuum states are primary of the given weights. In particular $e^{\alpha }$ and $v$ have weight one, while $F$ has weight $2p-1$. For a weight-one primary $a$, the residues give $[L(n),Y(a,z)]={\mathrm{\partial }}_z(z^{n+1}Y(a,z))$, whose zero-mode residue is zero. Both screenings therefore commute with every $L(n)$. Since $Q\mathbf{1}=0$ and $\omega =L(-2)\mathbf{1}$, this implies $Q\omega =0$. The zero-mode identities in [F2] make $Q$ and $v_{(0)}$ derivations of every state product, with $Q$ restricting to $V_L$. By [F5] with $i=0$, $n=1$, $E=Q^{2}F\mathrm{\ne }0$; hence $H=QF\mathrm{\ne }0$, since otherwise $QH=E$ would vanish. The state $F=e^{-\alpha }$ is itself a nonzero constant polynomial in its labelled Fock space. Commuting $Q$ with Virasoro supplies the weight and positive-mode equations for $H,E$; together with this nonvanishing, all three states are primary in the stated nonzero sense. Moreover $QE=Q^{3}F$ would have charge $2\alpha$ and weight $2p-1$, but that charge has minimum weight $2p+2$. Hence $QE=0$.

Directly $Y(v,z)\omega$ is obtained by substituting $x_1+2z^{-1}$ and $x_2+2z^{-2}$ into $x_1^2\mathrm{/}(4p)+q{x}_2$, followed by multiplication by the creation exponential. Before that multiplication the result is $\omega +(x_1\mathrm{/}p)z^{-1}+z^{-2}$, while the exponential begins $1-(x_1\mathrm{/}p)z+\cdots$. Thus $v_{(0)}\omega =0$, $v_{(1)}\omega =v$ and $v_{(n)}\omega =0$ for $n\ge 2$. Also $Y(v,z)F$ begins with $z^2$, so $v_{(0)}F=0$. The explicit exponentials give $Qv=x_1{e}^{(1-1\mathrm{/}p)\alpha }=\frac{p}{p-1}T{e}^{(1-1\mathrm{/}p)\alpha }$. By [F2], $[Q,v_{(0)}]=(Qv{)}_{(0)}=0$, since a derivative has zero zero-mode residue. Hence $v_{(0)}H=v_{(0)}E=0$. The derivation property makes its kernel a vertex subalgebra containing all four generators, so it contains $W$. The $Q$ table of step 2.1 and its derivation property imply $QW\subset W$. Each generator is homogeneous; [F4] shows that products of homogeneous states have the mode-weight rule, so the span generated by all their products is graded. It inherits finite weight spaces, nonnegative weights, the one-dimensional vacuum space, the Virasoro vector and translation from $V_L$, so it has the properties in [F3]. Restricting the full module Jacobi identity of [F2] to this actual subalgebra supplies its ordinary module actions; no classification or endpoint claim is used.