Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 306 (String Theory)

The Euler-Lagrange equation of each component of the string embedding map is the wave equation $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^{\mu }=0$. Its left- and right-moving solutions are identified by the endpoint Neumann boundary conditions, so the resulting open-string mode expansion isIndeed, the Fourier cosine series makes ${\partial }_{\sigma }{X}^{\mu }$ vanish at $\sigma =0,\pi$. The canonical momentum density is ${\Pi }_{\mu }=(2\pi {\alpha }^{\prime }{)}^{-1}{\dot{X}}_{\mu }$. Imposing the equal-time canonical commutation relations $[X^{\mu }(\sigma ),{\Pi }_{\nu }({\sigma }^{\prime })]=i{\delta }_{\nu }^{\mu }\delta (\sigma -{\sigma }^{\prime })$ gives the covariant quantization of the bosonic stringwith ${\alpha }_0^{\mu }=\sqrt{2{\alpha }^{\prime }}{p}^{\mu }$.

An infinitesimal Lorentz transformation is $\delta X^{\mu }={\omega }^{\mu }{}_{\nu }{X}^{\nu }$ with ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$. Applying Noether theorem to this continuous symmetry gives the conserved Lorentz currentIts Noether charge is $M^{\mu \nu }={\int }_0^{\pi }d\sigma J_{\tau }^{\mu \nu }$. Substituting the open-string mode expansion and using orthogonality of the cosine modes yieldsThe first term is orbital angular momentum; the sum is the contribution of the string oscillators.

It is cleaner to use the action of the charge than to expand every double sum. The commutators from part a give, for $v^{\rho }=x^{\rho },p^{\rho }$, or any ${\alpha }_n^{\rho }$,Thus $M^{\mu \nu }$ acts on every mode in the vector representation of the Lorentz algebra. Apply the Jacobi identity to the action of $[M^{\mu \nu },M^{\kappa \lambda }]$ on every mode. The result agrees with the action ofThere is no extra scalar term: the orbital and oscillator pieces commute with one another, and direct use of their canonical commutation relations gives no central contribution. Hence the displayed operators obey precisely the standard Lorentz algebra.

The universal stress-tensor operator-product expansion in a two-dimensional conformal field theory isIts fourth-order coefficient defines the central charge $c$.

Differentiating $X(z)X(w)\sim -({\alpha }^{\prime }/2)\log (z-w)$ gives the contractions needed for Wick theorem. The free-boson part of $T$ supplies $1/[2(z-w{)}^4]$, corresponding to central charge one. The cross-contractions between $:\partial X\partial X:$ and ${\partial }^{2}X$ produce the required lower poles but no fourth-order scalar term. Finally,so the product of the two improvement terms contributes $3{\alpha }^{\prime }{q}^2/(z-w{)}^4$. Matching this with $c/[2(z-w{)}^4]$ in the stress-tensor operator-product expansion gives the linear dilaton conformal field theory

The first term of the action is the free boson conformal field theory. For the curvature coupling, insert the supplied first variation of $\sqrt{g}{R}^{(2)}$, integrate by parts twice and discard the boundary term. Its metric variation is the stress-tensor improvementin the conventions of the question. In a locally flat complex coordinate, the holomorphic component of the complete stress-energy tensor is thereforeas required. Thus the coupling $q\int \sqrt{g}X{R}^{(2)}$ makes $X$ a background-charge scalar field, equivalently a worldsheet coordinate in a linear dilaton conformal field theory.

Take $D$ embedding coordinates and give one coordinate the background charge $q$. The remaining $D-1$ free bosons contribute $D-1$, while the distinguished coordinate contributes $1+6{\alpha }^{\prime }{q}^2$, so the matter central charge isThe worldsheet ghosts contributes $-26$. Cancellation of the worldsheet Weyl anomaly therefore requireswhich is real for $D<26$ and produces a noncritical bosonic string.

In target-space language the dilaton is linear in this coordinate, so the local string coupling $g_s=e^{\Phi }$ changes exponentially. One end of the target direction is weakly coupled and the other is strongly coupled. Consequently string perturbation theory is trustworthy only in the weak-coupling region, rather than throughout the full background.

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 $(1,1)$.

The target metric is a coupling of the string nonlinear sigma model. Quantum consistency requires the gauge-fixed worldsheet theory to preserve Weyl invariance, so its sigma-model beta functions must vanish. With no B-field or varying dilaton, the metric beta function begins asTherefore vanishing of the worldsheet Weyl anomaly requiresto leading order in ${\alpha }^{\prime }$: the target-space metric must be Ricci-flat at this order.

The $p$-dependent part of the first-order action can be completed to a square:The Gaussian functional integral over $p$ therefore leavesafter the stipulated omission of its functional determinant. Identifying $X^{\mu }=(Y^i,Z)$ gives exactly $S_1[X]$, establishing the classical equivalence.

Integrating by parts makes the $Z$ dependence $-i\int Z{\partial }_{\alpha }{p}^{\alpha }$. Its functional integral imposes the constraint ${\partial }_{\alpha }{p}^{\alpha }=0$. On the simply connected plane, the Poincare lemma therefore permits the local and global parametrizationSubstitution into $S_2$ givesThis is the Buscher procedure for the translation isometry in $z$, and its T-duality replaces $G_{zz}=V$ by ${\tilde{G}}_{\tilde{z}\tilde{z}}=V^{-1}$.

The classical elimination in parts b and c discarded the field-dependent Gaussian functional integral determinant. At one loop that determinant is a local curvature coupling and supplies the Buscher rules dilaton shiftThe metric alone consequently need not obey ${\tilde{R}}_{\mu \nu }=0$. The relevant leading sigma-model beta function in the dual background instead containsThe new dilaton term cancels the failure of the dual metric to be Ricci-flat, so the complete metric-dilaton background remains conformal and physically T-dual to the original one.