RNS Local Central-Charge Cancellation

For one scalar, the only $u^{-4}$ contribution to $T_X(z)T_X(w)$ is its two pairings of $\mathrm{\partial }X$ with $\mathrm{\partial }X$, each $-{\alpha }^{\mathrm{\prime }}\mathrm{/}(2u^2)$ by [F2]. Their product with the prefactor ${\alpha }^{\mathrm{\prime }-2}$ is $1\mathrm{/}(2u^4)$, so this scalar contributes $c=1$. For one Majorana field, the contraction $\psi (z)\psi (w)=u^{-1}$ and its derivatives give a double-contraction coefficient $1\mathrm{/}(4u^4)$ after the factor $(-1\mathrm{/}2{)}^2$, so it contributes $c=1\mathrm{/}2$. Independent fields have no cross contractions; summing the $D$ copies gives $c_m=D+D\mathrm{/}2=3D\mathrm{/}2$.

For a fermionic first-order pair of weights $(\lambda ,1-\lambda )$ with $T=(1-\lambda ):\mathrm{\partial }bc:-\lambda :b\mathrm{\partial }c:$, put $A=:\mathrm{\partial }bc:$ and $B=:b\mathrm{\partial }c:$. The CAR in [F3] give the radial contractions $b(z)c(w)=c(z)b(w)=u^{-1}$; differentiating gives $\mathrm{\partial }b(z)c(w)=-u^{-2}$, $b(z)\mathrm{\partial }c(w)=u^{-2}$, $\mathrm{\partial }b(z)\mathrm{\partial }c(w)=-2u^{-3}$, and $\mathrm{\partial }c(z)b(w)=-u^{-2}$. Pairing the two cross contractions in each normal-ordered product, with the fermionic Wick sign, gives the double-contraction coefficients $-1,-2,-2,-1$ times $u^{-4}$ for $AA,AB,BA,BB$. Weighting them by $(1-\lambda {)}^2$, $-\lambda (1-\lambda )$ twice, and ${\lambda }^2$ yields $-(1-\lambda {)}^2-{\lambda }^2+4\lambda (1-\lambda )=-(6{\lambda }^2-6\lambda +1)$. At $\lambda =2$, this is the explicit $T_{bc}$ declared in [F2], and the coefficient is $-13$, so $c_{bc}\mathrm{/}2=-13$ and $c_{bc}=-26$.

Contracting the two factors of $G_m=i\sqrt{2\mathrm{/}{\alpha }^{\mathrm{\prime }}}\psi \cdot \mathrm{\partial }X$ once in each field gives $D{u}^{-3}$; the single contractions give $2T_m(w)u^{-1}$. Hence $G_m(z)G_m(w)\sim D{u}^{-3}+2T_m(w)u^{-1}$. For a weight-$3\mathrm{/}2$ current, if the leading OPE coefficient is $\kappa (z-w{)}^{-3}$, it contributes $\frac{\kappa }{2}(r^2-\frac{1}{4}){\delta }_{r+s,0}$ to $\{G_r,G_s\}$: expand $z^{r+1\mathrm{/}2}$ to second order about $w$ in the two contour integrals. Matching [F5]'s coefficient $\frac{c}{3}(r^2-\frac{1}{4})$ gives $\kappa =2c\mathrm{/}3$. Here $\kappa =D$, so $c_m=3D\mathrm{/}2$, agreeing with step 1.1.

For the commuting pair, radial exchange in [F4]'s ordered convention gives $\gamma (z)\beta (w)=u^{-1}$ and $\beta (z)\gamma (w)=-u^{-1}$. Thus the four double-contraction coefficients for $AA,AB,BA,BB$ in the reordered first-order stress tensor have signs $+1,+2,+2,+1$ times $u^{-4}$. Reordering [F4] gives $T_{\beta \gamma }=-\frac{3}{2}:\beta \mathrm{\partial }\gamma :-\frac{1}{2}:\mathrm{\partial }\beta \gamma :$, the same bilinear form with $\lambda =3\mathrm{/}2$ and commuting Wick signs. Its coefficient is $(1-\lambda {)}^2+{\lambda }^2-4\lambda (1-\lambda )=6{\lambda }^2-6\lambda +1=11\mathrm{/}2$, so $c_{\beta \gamma }=11$.

The total stress tensor is the sum of independent matter, $bc$, and $\beta \gamma$ stresses, so there are no cross-contraction central terms. Adding the charges from steps 1.1, 1.2 and 2.2 gives $c_{tot}=3D\mathrm{/}2-26+11=3D\mathrm{/}2-15$. Solving $c_{tot}=0$ gives $D=10$, and substituting $D=10$ gives zero; since $D$ is an integer, this is exactly the stated iff. This is a local chiral cancellation only.