Admissible triplet action and the Jordan top of the cyclic logarithmic module

Write $q=(p-1)\mathrm{/}(2p)$. The field of $\omega$ is $\mathcal{T}(z)=:\alpha (z{)}^2:\mathrm{/}(4p)+q{\mathrm{\partial }}_z\alpha (z)$. The contraction $\alpha (z)\alpha (w)\sim 2p\mathrm{/}(z-w{)}^2$, obtained directly from the oscillator commutator, gives the singular product $\mathcal{T}(z)\mathcal{T}(w)\sim (1\mathrm{/}2-12p{q}^2)\mathrm{/}(z-w{)}^4+2\mathcal{T}(w)\mathrm{/}(z-w{)}^2+{\mathrm{\partial }}_w\mathcal{T}(w)\mathrm{/}(z-w)$. Here the double contraction of the quadratic terms 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, leaving $2q\mathrm{\partial }\alpha (w)\mathrm{/}(z-w{)}^2+q{\mathrm{\partial }}^2\alpha (w)\mathrm{/}(z-w)$. The remaining quadratic single contractions give $2:\alpha (w{)}^2:\mathrm{/}(4p)$ and its derivative. Taking mode residues proves the Virasoro relations with central charge $1-24p{q}^2=1-6(p-1{)}^2\mathrm{/}p$. In particular $L(n)=\frac{1}{4p}{\sum }_{j\in \mathbb{Z}}:{\alpha }_j{\alpha }_{n-j}:-q(n+1){\alpha }_n\mathrm{.}$ The formula for $n=-1$ acts on $f{e}^{t\alpha }$ as ${\sum }_{k\ge 1}k{x}_{k+1}{\mathrm{\partial }}_{x_k}f+t{x}_{1}f$, the translation in [F2]. The $n=0$ formula is $L(0)={\sum }_{k\ge 1}k{x}_k{\mathrm{\partial }}_{x_k}+p{t}^2-(p-1)t$. Thus it is diagonal with the displayed weights on both spaces. In $V_L$ these weights are nonnegative integers: $p{n}^2-(p-1)n\ge 0$ for every integer $n$, with equality only at $n=0$. Its weight spaces are finite, since the quadratic bounds the possible charges and the polynomial space in each oscillator degree is finite.

The same contraction calculation gives $\mathcal{T}(z){\mathcal{E}}_t(w)\sim (p{t}^2-(p-1)t){\mathcal{E}}_t(w)\mathrm{/}(z-w{)}^2+{\mathrm{\partial }}_w{\mathcal{E}}_t(w)\mathrm{/}(z-w)$. Indeed the quadratic double contraction contributes $p{t}^2$, the improvement contributes $-(p-1)t$, and the quadratic single contraction contributes $t:\alpha {\mathcal{E}}_t:=\mathrm{\partial }{\mathcal{E}}_t$, as follows by differentiating its explicit exponential formula. In particular $v=e^{-\alpha \mathrm{/}p}$ is primary of weight one. To check the screening kernel on the generators directly, apply its exponential to $\omega$: the annihilation exponential substitutes $x_1+2z^{-1}$ and $x_2+2z^{-2}$ into $x_1^2\mathrm{/}(4p)+q{x}_2$. The result before its creation exponential is $\omega +(x_1\mathrm{/}p)z^{-1}+z^{-2}$; the creation exponential begins $1-(x_1\mathrm{/}p)z+\cdots$. Hence $v_{(1)}\omega =v$, $v_{(0)}\omega =0$ and $v_{(n)}\omega =0$ for $n\ge 2$. Also $Y(v,z)F$ starts with $z^2$, so $v_{(0)}F=0$. Finally $Qv=x_1{e}^{(1-1\mathrm{/}p)\alpha }=\frac{p}{p-1}T{e}^{(1-1\mathrm{/}p)\alpha }$. The zero-mode commutator formula from the adjoint Jacobi identity gives $[Q,v_{(0)}]=(Qv{)}_{(0)}=0$, since the zero mode of a derivative is zero. Therefore $v_{(0)}H=v_{(0)}E=0$. The same commutator formula makes $v_{(0)}$ a derivation of every state product, so its kernel is a vertex subalgebra containing $\omega ,F,H,E$; it contains $W(p)$.

Apply [F3] to $\mathcal{A},I,v$ and $S=W(p)$. It supplies the full module Jacobi identity on $\mathcal{M}$, not merely the mode commutation relations. Step 2.1 gives $\mathrm{\Delta }(z)\omega =\omega +z^{-1}v$, whence $\tilde{L}(n)=L(n)+v_{(n)}$ and $\tilde{L}(0)=L(0)+v_{(0)}$. The screening $v_{(0)}$ has weight zero and square zero on $\mathcal{M}$, since it sends sector zero to sector one and annihilates sector one. For the two module charge strings, subtracting $h$ from the weight of the charge vacuum gives respectively $p{n}^2+n$ and $p{n}^2-n$, for $n\in \mathbb{Z}$. These are nonnegative integers, vanishing only at $n=0$. Thus $\mathcal{M}={⨁}_{d\ge 0}\mathcal{M}(d)$, with each degree finite and $L(0)=(h+d)I$ on it. Since $v_{(0)}$ preserves degree, these are exactly the generalized $\tilde{L}(0)$-weight spaces. Counting weights in the explicit fields gives $a_{(k)}\mathcal{M}(d)\subset \mathcal{M}(d+\mathrm{wt}a-k-1)$ for homogeneous $a\in V_L$: the charge shift, oscillator multiplications and derivatives have exactly these weight changes. In a correction $v_{(n)}a$ has weight $\mathrm{wt}a-n$ and the extra factor $z^{-n}$ increases the mode index by $n$, leaving the same shift. This is the admissible grading for the deformed fields.

Put $u=e^{\alpha \mathrm{/}2}$ and $w=e^{\alpha \mathrm{/}2-\alpha \mathrm{/}p}$. The screening exponential on $u$ starts with $z^{-1}w$, so $v_{(0)}u=w$; it annihilates $w$ by the sector rule. Both have original weight $h$. Therefore $\tilde{L}(0)u=hu+w$ and $\tilde{L}(0)w=hw$. Since the cyclic submodule containing $u$ is stable under $\tilde{L}(0)$, it also contains $w$. The degree-zero space of $\mathcal{M}$ is precisely $\mathbb{C}u\oplus \mathbb{C}w$ by step 3.1, so this is exactly the degree-zero space of $P_{+}^{p-1}$. Every invariant subspace of the direct sum of finite generalized eigenspaces of $\tilde{L}(0)$ decomposes as its intersections with them: for each vector, projection onto any one of its finitely many generalized eigencomponents is a polynomial in $\tilde{L}(0)$, obtained by the coprime factors $(X-(h+d){)}^2$. Thus the cyclic submodule inherits the admissible grading, finite generalized weight spaces and lower bound. The nonzero nilpotent action on its degree-zero space proves it logarithmic in [F4].

The correction to $F$ vanishes because $v_{(n)}F=0$ for all $n\ge 0$. The leading term of $Y_{\mathcal{M}}(F,z)u$ is $z^{-p}{e}^{-\alpha \mathrm{/}2}$, so ${\tilde{F}}_{(p-1)}u=e^{-\alpha \mathrm{/}2}$ and $M_1\subset P_{+}^{p-1}$. It inherits the same grading by the polynomial-projection argument in step 4.1. To check semisimplicity on $M_1$, the screening exponential on its cyclic vector has leading power $z^1$, giving $v_{(0)}{e}^{-\alpha \mathrm{/}2}=0$. For $a\in W(p)$ the ordinary module zero-mode commutator gives $[v_{(0)},Y_{\mathcal{M}}(a,z)]=Y_{\mathcal{M}}(v_{(0)}a,z)=0$. Every deformation correction has state in $I$, which is also annihilated by the adjoint screening zero mode; hence this commutation holds for the deformed fields too. Therefore $v_{(0)}$ vanishes on the entire cyclic space $M_1$. On it $\tilde{L}(0)=L(0)$, and the explicit oscillator-weight operator is diagonal. This proves the claimed semisimplicity and locates the Jordan block in $P_{+}^{p-1}$ rather than $M_1$.