Exact chiral free-boson stress tensor and vertex OPE

For $m,n>0$ or $m,n<0$, the operators in [F1] commute. For $m>0,n=-r<0$, $[a_m,a_n]=[m{\mathrm{\partial }}_{x_m},x_r]=m{\delta }_{m,r}I$; reversing $m,n$ changes the sign. Thus $[a_m,a_n]=m{\delta }_{m,-n}I$ for all integer indices, including $a_0=0$. The current contraction is ${\sum }_{n\ge 1}n{z}^{-n-1}{w}^{n-1}=z^{-2}{\sum }_{n\ge 1}n(w\mathrm{/}z{)}^{n-1}=\frac{1}{(z-w{)}^2}\mathrm{.}$ The geometric-series identity is valid in the formal radial region.

The double contraction in $T(z)T(w)$ pairs the two $J$'s in two ways, giving $\frac{1}{2}(z-w{)}^{-4}$. The single contractions leave $\frac{:J(z)J(w):}{(z-w{)}^2}=\frac{2T(w)}{(z-w{)}^2}+\frac{\mathrm{\partial }T(w)}{z-w}+O(1),$ because the formal Taylor expansion gives $J(z)=J(w)+(z-w)\mathrm{\partial }J(w)+O((z-w{)}^2)$. Hence $T(z)T(w)\sim \frac{1\mathrm{/}2}{(z-w{)}^4}+\frac{2T(w)}{(z-w{)}^2}+\frac{\mathrm{\partial }T(w)}{z-w}\mathrm{.}$ Comparing with [F3] gives $c\mathrm{/}2=1\mathrm{/}2$, so $c=1$, in agreement with [F2].

Pairing each of the two currents in $T(z)=\frac{1}{2}:J(z)J(z):$ with $V_{\alpha }(w)$ gives the double-pole coefficient $\frac{1}{2}{\left(\frac{\alpha }{z-w}\right)}^2{V}_{\alpha }(w)=\frac{{\alpha }^2}{2}\frac{V_{\alpha }(w)}{(z-w{)}^2}\mathrm{.}$ The two single contractions give $\frac{\alpha :J(z)V_{\alpha }(w):}{z-w}=\frac{{\mathrm{\partial }}_w{V}_{\alpha }(w)}{z-w}+O(1),$ indeed, write $V_{\alpha }(z)=E_{\alpha }{z}^{\alpha p}{C}_{\alpha }(z)A_{\alpha }(z)$, with $C_{\alpha }$ and $A_{\alpha }$ its creation and annihilation exponentials. Differentiating the four factors gives ${\mathrm{\partial }}_z{V}_{\alpha }=\alpha E_{\alpha }{z}^{\alpha p}[(\frac{p}{z}+{\sum }_{n\ge 1}{a}_{-n}{z}^{n-1})C_{\alpha }{A}_{\alpha }+C_{\alpha }\left({\sum }_{n\ge 1}{a}_n{z}^{-n-1}\right)A_{\alpha }]=\alpha :J{V}_{\alpha }:,$ where the normal ordering places creation modes left and the zero mode to the right of $E_{\alpha }{z}^{\alpha p}$. The remaining terms are regular, so $T(z)V_{\alpha }(w)\sim \frac{{\alpha }^2}{2}\frac{V_{\alpha }(w)}{(z-w{)}^2}+\frac{{\mathrm{\partial }}_w{V}_{\alpha }(w)}{z-w}$, and $h(V_{\alpha })={\alpha }^2\mathrm{/}2$.

Write the annihilation exponential of $V_{\alpha }(z)$ as $e^{A_{\alpha }(z)}$, $A_{\alpha }(z)=-\alpha {\sum }_{n\ge 1}{a}_n{z}^{-n}\mathrm{/}n$, and the creation exponential of $V_{\beta }(w)$ as $e^{B_{\beta }(w)}$, $B_{\beta }(w)=\beta {\sum }_{n\ge 1}{a}_{-n}{w}^n\mathrm{/}n$. Their commutator is central: $[A_{\alpha }(z),B_{\beta }(w)]=-\alpha \beta {\sum }_{n\ge 1}\frac{(w\mathrm{/}z{)}^n}{n}=\alpha \beta \log (1-w\mathrm{/}z)\mathrm{.}$ The formal Baker–Campbell–Hausdorff identity therefore contributes $(1-w\mathrm{/}z{)}^{\alpha \beta }$. Commuting $z^{\alpha p}$ past $E_{\beta }$ contributes $z^{\alpha \beta }$, and $E_{\alpha }{E}_{\beta }=E_{\alpha +\beta }$. The product of scalar factors is $z^{\alpha \beta }(1-w\mathrm{/}z{)}^{\alpha \beta }=(z-w{)}^{\alpha \beta }$; after extracting these scalars, the remaining zero-mode factor is $E_{\alpha +\beta }{z}^{\alpha p}{w}^{\beta p}$, and the oscillator factors are the normally ordered creation and annihilation exponentials. More explicitly, define the bilocal normally ordered factor $N_{\alpha ,\beta }(z,w):=E_{\alpha +\beta }{z}^{\alpha p}{w}^{\beta p}\exp \text{\hspace{0.17em}⁣}({\sum }_{n\ge 1}\frac{a_{-n}}{n}(\alpha z^n+\beta w^n))\exp \text{\hspace{0.17em}⁣}(-{\sum }_{n\ge 1}\frac{a_n}{n}(\alpha z^{-n}+\beta w^{-n}))\mathrm{.}$ The exact radial identity is $V_{\alpha }(z)V_{\beta }(w)=(z-w{)}^{\alpha \beta }{N}_{\alpha ,\beta }(z,w)$, where the scalar power is expanded as $z^{\alpha \beta }(1-w\mathrm{/}z{)}^{\alpha \beta }$ on the chosen branch. For the local expansion set $s=z-w$, expand every $(w+s{)}^q$ binomially in $s\mathrm{/}w$, and put $D_k(w):=\frac{1}{k!}({\mathrm{\partial }}_z^k{N}_{\alpha ,\beta })(w,w)$ for integers $k\ge 0$. These are formal coefficient fields: on a fixed input polynomial the annihilation exponential is finite, and at each fixed oscillator degree the creation coefficients and their derivatives are finite sums. Consequently the Taylor expansion is well defined coefficientwise and gives the full generalized-power OPE $V_{\alpha }(w+s)V_{\beta }(w)=s^{\alpha \beta }{\sum }_{k\ge 0}{s}^k{D}_k(w),D_0(w)=V_{\alpha +\beta }(w)\mathrm{.}$ This local re-expansion uses $\mathrm{\mid }s\mathrm{\mid }<\mathrm{\mid }w\mathrm{\mid }$ as its binomial-expansion convention, distinct from the original radial expansion; neither is an assertion of numerical convergence of the field series. The leading fused term alone is $s^{\alpha \beta }{V}_{\alpha +\beta }(w)$; the $k\ge 1$ fields retain all regular descendants.

For each pair $i<j$, commuting the annihilation exponential of the outer field $V_{{\alpha }_i}(z_i)$ through the creation exponential of $V_{{\alpha }_j}(z_j)$ contributes $(1-z_j\mathrm{/}{z}_i{)}^{{\alpha }_i{\alpha }_j}$ by the central-commutator calculation of step 1.4. Moving the outer zero-mode factor past the later charge shift contributes $z_i^{{\alpha }_i{\alpha }_j}$. Their product is $(z_i-z_j{)}^{{\alpha }_i{\alpha }_j}$ in the radial region $\mathrm{\mid }{z}_j\mathrm{/}{z}_i\mathrm{\mid }<1$. After all such pairwise commutations, $\epsilon$ kills every positive oscillator-degree term and every nonzero total charge. Therefore $C({\alpha }_1,z_1;\dots ;{\alpha }_N,z_N)=\{\begin{array}{ll}{\displaystyle \prod _{1\le i<j\le N}(z_i-z_j{)}^{{\alpha }_i{\alpha }_j},}& {\sum }_i{\alpha }_i=0,\\ 0,& {\sum }_i{\alpha }_i\mathrm{\ne }0.\end{array}$ The powers use the fixed radial branches. If any ${\alpha }_i=0$, its field is the identity and every factor involving it equals $1$, reducing the formula to the $N-1$-field formula. The neutrality condition is for the declared algebraic vacuum functional, not a full scalar path integral.

If $\alpha =0$, the charge shift is the identity, $V_0=I$, the weight ${\alpha }^2\mathrm{/}2$ is zero, and both OPEs reduce to the identity-field cases; if $\alpha \beta =0$, the scalar vertex factor has no branch because its exponent is zero. If $N=1$, the functional gives $1$ for ${\alpha }_1=0$ and $0$ otherwise, matching the empty product and charge-neutrality rule. These are the zero-charge and one-insertion boundary cases. The example asserts no compactification, winding sectors, or physical path-integral measure.