Vertex Modules, Rationality, and Logarithmic Regimes

For $N\ge 0$, the coefficient of $z_0^{-N-1}$ in the first term is the polynomial $(z_1-z_2{)}^N$ expanded in $z_2$. In the second term the same coefficient is $(z_1-z_2{)}^N$ expanded in $z_1$: the sign from $(-z_0{)}^{-N}$ cancels the sign in $(z_2-z_1{)}^N$. These are the same polynomial, so their difference is zero. The right side has only nonnegative powers of $z_0$.

Fix $j\ge 0$. The coefficient of $z_0^j$ in the first term is ${\sum }_{k\ge 0}\left(\genfrac{}{}{0px}{}{j+k}{j}\right)z_1^{-j-1-k}{z}_2^k$. The coefficient in the second delta term before its outer minus is $(-1{)}^{j+1}{\sum }_{k\ge 0}\left(\genfrac{}{}{0px}{}{j+k}{j}\right)z_1^k{z}_2^{-j-1-k}$, by expanding $(z_2-z_1{)}^{-j-1}$ in nonnegative powers of $z_1$. Thus the left-side coefficient is the sum of the first series and $(-1{)}^j$ times ${\sum }_{k\ge 0}\left(\genfrac{}{}{0px}{}{j+k}{j}\right)z_1^k{z}_2^{-j-1-k}$. On the right, the coefficient is $(-1{)}^j{\sum }_{n\in \mathbb{Z}}\left(\genfrac{}{}{0px}{}{n}{j}\right)z_1^{n-j}{z}_2^{-n-1}$. Splitting at $n=j$, using $n=-1-k$ for the negative indices and $\left(\genfrac{}{}{0px}{}{-1-k}{j}\right)=(-1{)}^j\left(\genfrac{}{}{0px}{}{j+k}{j}\right)$, gives exactly the same two series. Coefficients of every integer power of $z_0$ agree, proving the identity.

On every polynomial, differentiation and multiplication give $[a_m,a_n]=m{\delta }_{m,-n}I$. All sufficiently large positive modes kill that polynomial. Thus $a(z)$ is a field, and its commutator is ${\mathrm{\partial }}_w\delta (z,w)$, obtained by summing $m{z}^{-m-1}{w}^{m-1}$. Multiplying this distribution by $(z-w{)}^2$ gives zero (differentiate $(z-w)\delta (z,w)=0$), so the generator is local with itself. Directly $[T,a_n]=-n{a}_{n-1}$ and $T1=0$. Also $a(z)1={\sum }_{r\ge 1}{x}_r{z}^{r-1}=e^{zT}{x}_1$, since $T^j{x}_1=j!x_{j+1}$. The negative modes are the multiplication operators $x_r$, whose monomials span $P$. All hypotheses of [F2] hold.

The reconstruction has precisely the stated normal fields. Indeed the two binomial expansions in [F3], with index $-r$, give $a(z{)}_{(-r)}B(z)=:b_r(z)B(z):$: taking the residue separates the modes $a_{-r-j}$, with coefficient $\left(\genfrac{}{}{0px}{}{r+j-1}{j}\right)z^j$, from the modes $a_j$, with coefficient $(-1{)}^{r-1}\left(\genfrac{}{}{0px}{}{r+j-1}{j}\right)z^{-r-j}$. These are the divided derivatives of the negative and nonnegative parts of $a$. By [F2]'s vacuum-state transport, the state of that product is $a_{-r}v=x_{r}v$ if $B=Y(v,z)$. Induction on the number of factors gives the displayed monomial formula. The same-sign modes commute, so nesting these normal products introduces no extra contractions. This proves the candidate state fields coincide with the reconstructed ones.

Since $Y(\omega ,z)=\frac{1}{2}:a(z{)}^2:$, its coefficient is the quadratic sum stated. For a fixed $n$ and polynomial, only finitely many terms survive: modes of large positive index kill the polynomial after normal ordering, and the sum of two negative indices is fixed. The oscillator commutator gives $[L(n),a_m]=-m{a}_{n+m}$ by the two terms $k=-m$ and $n-k=-m$ in that sum. Directly the sums at $n=0,-1$ give $L(0)={\sum }_{r\ge 1}{a}_{-r}{a}_r$ and $L(-1)={\sum }_{r\ge 1}{a}_{-r-1}{a}_r=T$, since $a_0=0$.

Let $C_{m,n}=[L(m),L(n)]-(m-n)L(m+n)$. Step 3.1 and the operator commutator identity show that $C_{m,n}$ commutes with every $a_r$. Any operator commuting with all multiplication operators and all positive derivatives is scalar: its value on $1$ is killed by every partial derivative and hence is constant, and commuting with multiplication determines its value on every monomial. Each $L(n)$ has degree $-n$, so this scalar is zero unless $m+n=0$. For $m>0$, $L(m)1=L(0)1=0$ and $L(-m)1=\frac{1}{2}{\sum }_{r=1}^{m-1}{x}_r{x}_{m-r}$. Applying $L(m)$ with step 3.1 gives $C_{m,-m}1=\frac{1}{2}{\sum }_{r=1}^{m-1}r(m-r)1=(m^3-m)1\mathrm{/}12$. Antisymmetry covers negative $m$, and $m=0$ is zero. Thus $[L(m),L(n)]=(m-n)L(m+n)+(m^3-m){\delta }_{m,-n}I\mathrm{/}12$.

The explicit Euler operator in step 3.1 has weight $d$ on $P_d$. For fixed $d$, only $x_1,\dots ,x_d$ can occur, with $m_r\le d\mathrm{/}r$, so its monomial basis is finite. There are no negative degrees, and degree zero consists of constants. Along with $L(-1)=T$ and the Virasoro relations of step 4.1, these verify every requirement in [F4]. No choice principle or analytic completion is used.

The polynomial formulas give $[a_m,a_n]=m{\delta }_{m,-n}I$ on $F(E,A)$: the zero mode commutes with every mode because it acts on $E$ alone. Every positive mode of index exceeding the degree of a polynomial kills it. In a fixed coefficient of a normal product applied to a degree-$d$ vector, all nonnegative modes act first. Their positive indices have bounded total sum, at most $d$; there are finitely many such choices, including zero indices. The total exponent then bounds the sum of the remaining negative indices, so there are finitely many remaining choices too. Every coefficient is therefore a genuine endomorphism. Counting degrees in a term of $b_r^F(z)$ gives degree change $-n$ at exponent $-n-r$. For a weight-$w$ monomial, $w=\sum r_i$, so its coefficient of $z^{-k-1}$ has degree change $w-k-1$. This proves the displayed shift and lower truncation; $Y_F(1,z)=I$ is the vacuum action.

The delta identity [F3] implies, for each scalar Laurent monomial $z^r{w}^s(z-w{)}^{-K}$ with $K\ge 0$, the corresponding identity $D_{12}{\iota }_{z,w}G-D_{21}{\iota }_{w,z}G=D_{20}{\iota }_{w,z_0}G(w+z_0,w)$. Indeed multiplying either of the first two delta sums by its expansion of $(z-w{)}^{-K}$ shifts the delta summation index and gives $z_0^{-K}$ times that delta sum. Also $D_{20}z=D_{20}(w+z_0)$, directly from its delta sum; integer powers give substitution of $z^r$, using the binomial inverse of $w+z_0$ for negative $r$. Multiplying [F3] by $z^r{w}^s{z}_0^{-K}$ therefore gives exactly the asserted identity, with the prescribed expansions.

Split $a^F=a^{-}+a^{+}$ into $n<0$ and $n\ge 0$. The only mixed commutator is $[a^{+}(z),a^{-}(w)]={\sum }_{n\ge 1}n{z}^{-n-1}{w}^{n-1}={\iota }_{z,w}(z-w{)}^{-2}I$; same-sign parts commute. Differentiating gives the scalar contraction $K_{r,s}(z-w)=(-1{)}^{r-1}(r+s-1)!(z-w{)}^{-r-s}\mathrm{/}((r-1)!(s-1)!)$ for $b_r^F(z),b_s^F(w)$. For $u={\prod }_{i=1}^p{x}_{r_i}$ and $v={\prod }_{j=1}^q{x}_{s_j}$, let a partial matching $J$ pair distinct left factors with distinct right factors, and write $I(J),O(J)$ for its unmatched left and right indices. Define the common expression $G_F(z,w)={\sum }_J({\prod }_{(i,j)\in J}{K}_{r_i,s_j}(z-w)):{\prod }_{i\in I(J)}{b}_{r_i}^F(z){\prod }_{j\in O(J)}{b}_{s_j}^F(w):$. This is a finite sum over matchings. Reorder the annihilators of the left product across the creators of the right product. At each crossing use $XY=YX+[X,Y]$; every commutator is the displayed scalar and commutes with all remaining factors. Each choice of disjoint crossings contributes exactly one partial matching, including the empty matching. This proves $Y_F(u,z)Y_F(v,w)={\iota }_{z,w}{G}_F(z,w)$. Reversing the ordered products gives $Y_F(v,w)Y_F(u,z)={\iota }_{w,z}{G}_F(z,w)$: differentiation of $(w-z{)}^{-2}$ gives the same rational $K_{r_i,s_j}(z-w)$ after swapping variables, and the unmatched normal product is unchanged.

The same calculation on the vacuum space computes every state coefficient $u_{(n)}v$. Apply its Wick expression to $1$ and set $w=0$. This is legitimate because $Y(v,w)1$ has only nonnegative powers of $w$. All nonnegative oscillator modes in the normal product kill $1$, and the negative part of $b_r(z)$ is ${\sum }_{k\ge 0}\left(\genfrac{}{}{0px}{}{r+k-1}{k}\right)x_{r+k}{z}^k$. Consequently $Y(u,z_0)v$ is the state-valued Laurent series ${\sum }_J({\prod }_{(i,j)\in J}{K}_{r_i,s_j}(z_0))({\prod }_{i\in I(J)}{\sum }_{k\ge 0}\left(\genfrac{}{}{0px}{}{r_i+k-1}{k}\right)x_{r_i+k}{z}_0^k)({\prod }_{j\in O(J)}{x}_{s_j})$. Every coefficient is a finite sum, since the contraction pole order is bounded and the required sum of the $k$'s is fixed. Apply $Y_F(-,w)$ coefficientwise. The formal Taylor identity $b_r^F(w+z_0)={\sum }_{k\ge 0}\left(\genfrac{}{}{0px}{}{r+k-1}{k}\right)z_0^k{b}_{r+k}^F(w)$ follows by differentiating the binomial expansion of each power of $w$. The normal parts obey that same Taylor identity. Hence the resulting iterate is precisely $Y_F(Y(u,z_0)v,w)={\iota }_{w,z_0}{G}_F(w+z_0,w)$. This proves the complete iterate formula, including negative products, rather than only the singular OPE.

Apply step 1.2 termwise to the common expression in step 2.1. To justify every infinite-series operation, fix an input polynomial degree and project the output to degrees at most $D$. Each unmatched normal product then has only finitely many modes in each variable: the positive indices are bounded by the input degree, and the creation indices are bounded by the output degree plus the input degree. It is therefore a Laurent polynomial after this projection, multiplied by the finite-order contraction poles. Step 1.2 applies without an infinite rearrangement. Since $D$ is arbitrary and every vector is a finite sum of homogeneous polynomials, the identity holds coefficientwise on all of $F(E,A)$. Substitute the two ordered products from step 2.1 and the iterate from step 3.1. The resulting equality is exactly the full module Jacobi identity in [F4], for all monomial states and thus all states by linearity. Along with step 1.1 this proves the module assertion. All calculations are algebraic and choice-free.

Apply [F1] with the displayed $E,A_{\lambda }$. This already gives a genuine module with all three-variable Jacobi coefficients verified, not merely a representation of the current-mode Lie algebra. Formula [F2] and the action definition give $Y_F(\omega ,z)=\frac{1}{2}:a^F(z{)}^2:$, so its modes are the displayed quadratic sums, finite on each polynomial vector by the same normal-order bounds as [F1]. At $n=0$ this gives $L_F(0)={\sum }_{r\ge 1}{a}_{-r}{a}_r+a_0^2\mathrm{/}2=D\otimes I+A_{\lambda }^2\mathrm{/}2$. Since $N^2=0$, $A_{\lambda }^2\mathrm{/}2=({\lambda }^2\mathrm{/}2)I+\lambda N$.

The conformal modes also satisfy the central-charge-one Virasoro relations directly. The oscillator commutator gives $[L_F(n),a_m]=-m{a}_{n+m}$. Thus $C_{m,n}=[L_F(m),L_F(n)]-(m-n)L_F(m+n)$ commutes with every multiplication and derivative mode and is determined by its action on the common positive-mode kernel $E$. Its degree is $-m-n$; an operator commuting with all positive modes must send $E$ into $E$, so $C_{m,n}{\mathrm{\mid }}_E=0$ if $m+n\mathrm{\ne }0$. For $m>0$, $L_F(m)E=0$ and $L_F(-m)e=x_m{A}_{\lambda }e+\frac{1}{2}{\sum }_{r=1}^{m-1}{x}_r{x}_{m-r}e$. Applying $L_F(m)$ gives $m{A}_{\lambda }^{2}e+\frac{1}{2}{\sum }_{r=1}^{m-1}r(m-r)e$. Subtracting $2m{L}_F(0)e=m{A}_{\lambda }^{2}e$ leaves $(m^3-m)e\mathrm{/}12$. Antisymmetry and the $m=0$ case complete the Virasoro relation; commuting with negative multiplication modes extends it from $E$ to every polynomial vector.

The degree shift in [F1] is precisely [F3]'s admissible shift, so $F_{\lambda }={⨁}_{d\ge 0}{P}_d\otimes E$ is admissible. On its degree-$d$ subspace, step 1.1 gives $(L_F(0)-d-{\lambda }^2\mathrm{/}2{)}^2=0$. Distinct degrees have distinct eigenvalues. If $h\mathrm{\ne }d+{\lambda }^2\mathrm{/}2$, then $(d+{\lambda }^2\mathrm{/}2-h)I+\lambda N$ is invertible on that degree, with inverse $c^{-1}I-\lambda c^{-2}N$ where $c=d+{\lambda }^2\mathrm{/}2-h$. Since vectors have only finitely many degree components, the generalized eigenspaces are exactly the stated degree spaces. They have finite dimension $2\dim P_d$ by [F2], and their real parts are bounded below by $\mathrm{Re}({\lambda }^2\mathrm{/}2)$. This verifies grading restriction in [F3].

If $\lambda \mathrm{\ne }0$, on the lowest space $E$ the vector $e_1$ satisfies $(L_F(0)-{\lambda }^2\mathrm{/}2)e_1=\lambda e_0\mathrm{\ne }0$ and its second iterate is zero. Thus $L_F(0)$ is not semisimple, and the module is logarithmic by [F4]. If $\lambda =0$, step 1.1 gives $L_F(0)=D$, diagonal on the displayed grading, so the module is ordinary by [F4] and not logarithmic. At $\lambda =1$ the explicit formulas become $a_0=I+N$ and $L_F(0){\mathrm{\mid }}_E=I\mathrm{/}2+N$. No unproved module-twist, reconstruction-for-modules, classification, or rationality theorem is used.