Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 306 / 3

A local operator $\mathcal{O}(w,\bar{w})$ is a primary operator of conformal weights $(h,\tilde{h})$ when its operator product expansions with the holomorphic and antiholomorphic stress tensors arewith no more singular poles. Equivalently, under a local conformal transformation it transforms covariantly with holomorphic exponent $h$ and antiholomorphic exponent $\tilde{h}$.

LetThe exponential is a primary operator with conformal weights $({\alpha }^{\prime }{k}^2/4,{\alpha }^{\prime }{k}^2/4)$. In the $T(z)V(w,\bar{w})$ OPE, contracting one derivative in $T$ with $\partial X^{\mu }$ and the other with the exponential produces a third-order pole proportional to $k^{\mu }{ϵ}_{\mu \nu }$. Its antiholomorphic counterpart is proportional to ${ϵ}_{\mu \nu }{k}^{\nu }$. Therefore $V$ is primary exactly when its polarization tensor is transverse in both indices,The remaining second-order poles give

The matter part of the massless closed-string vertex operator isIts symmetric trace-free polarization is the graviton, while its antisymmetric polarization is the Kalb–Ramond field, or B-field. The mass-shell condition $k^2=0$ and transversality make $V_{ϵ}$ a primary operator of conformal weights $(1,1)$, as required for an integrated string vertex operator.

The graviton polarization has the linearized gauge redundancywhile the antisymmetric polarization obeysIn either case the change in the integrated vertex is a worldsheet total derivative, hence vanishes on a closed worldsheet; in covariant language it is BRST-exact. This string-state gauge redundancy is the vertex-operator form of linearized target-space diffeomorphism or two-form gauge invariance.

For $E_{\mu \nu }={\eta }_{\mu \nu }$, transversality would require $k^{\mu }{\eta }_{\mu \nu }=k_{\nu }=0$. Hence the operator has a third-order stress-tensor pole and is not a primary operator for any nonzero $k$.

Forthe two transversality conditions coincide and reduce toIf $k^2=0$, this says $k\cdot \xi =-1$, with arbitrary additional component transverse to $k$. If $k^2\ne 0$, it forcesso $E_{\mu \nu }={\eta }_{\mu \nu }-k_{\mu }{k}_{\nu }/k^2$ is the transverse projection operator. Once this condition removes the third-order poles, the conformal weights are again $(1+{\alpha }^{\prime }{k}^2/4,1+{\alpha }^{\prime }{k}^2/4)$; in the null case they are $(1,1)$.