Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 49 / 3

Independent metric and Hamiltonian parametrization. The Lorentzian Polyakov action isThe worldsheet metric is independent of the embedding. Its equation sets the traceless worldsheet stress tensor to zero, making the metric locally conformal to the induced worldsheet metric; substitution recovers the Nambu–Goto action. Eliminating momentum in the Nambu-Goto phase-space action instead gives $(\dot{X}-u{X}^{\prime }{)}^2/(2e)-e{T}^2{X}^{\prime 2}/2$. On the positive-lapse branch, choose the independent metric, up to a positive Weyl transformation, asIts determinant is $-e^2{T}^2$. Multiplication by $-T/2$ recovers the reduced density, including the mixed term $-u\dot{X}\cdot X^{\prime }/e$. This proves the phase-space identification of the Polyakov metric.

Conformal gauge and residual coordinate freedom. In the Hamiltonian formulation, conformal gauge is $e=1/T,u=0$. The equations become $\ddot{X}-X^{\prime \prime }=0$, while ${\dot{X}}^2+X^{\prime 2}=0$ and $\dot{X}\cdot X^{\prime }=0$ remain as constraints. In the metric formulation, conformal gauge is ${\gamma }_{ab}=e^{2\omega }{\eta }_{ab}$, or simply ${\eta }_{ab}$ after fixing Weyl invariance. The metric parametrization proves these choices equivalent. Gauge-fixing the metric does not discard the Virasoro constraints.

With ${\sigma }^{\pm }=t\pm \sigma$, independent reparameterizations ${\sigma }^{\pm }\mapsto f_{\pm }({\sigma }^{\pm })$ multiply the flat metric by a conformal factor. A compensating Weyl transformation restores its chosen representative. Infinitesimally the Conformal Killing equation is ${\partial }_{-}{c}^{+}=0,{\partial }_{+}{c}^{-}=0$. These are the residual conformal transformations of two-dimensional Minkowski spacetime, with the periodicity conditions appropriate to a closed string.

Ghost determinant and anomaly. For gauge conditions $G_A(\varphi )=0$, the Faddeev-Popov determinant is the determinant of the linearized variationInsert this determinant in the gauge-fixed path integral and exponentiate it with anticommuting ghosts and antighosts: the additional action is proportional to $\int b^A{M}_{AB}{c}^B$. In metric gauge fixing, the trace is removed by Weyl invariance, and a worldsheet diffeomorphism changes the trace-free metric throughConsequently $c^a$ is a vector ghost and $b^{ab}$ a symmetric trace-free antighost. A conventional normalization of the worldsheet ghost action isA rescaling of the antighost changes only this overall kinetic normalization. In conformal gauge, after normalizing the chiral fields, the kinetic terms areThe closed-string ghosts are periodic, with independent left and right systems. Their conformal weights are $2$ for $b$ and $-1$ for $c$, so the supplied bc ghost system expression gives $c_{\mathrm{gh}}=-2(6\cdot 2^2-6\cdot 2+1)=-26$. Each free embedding boson has central charge one per chirality. HenceThis central charge of reparameterization ghosts cancels the string conformal anomaly in each chirality separately. Having two sectors does not double the required target dimension.

Channel variables, poles and the vector state. With particles 1,2 incoming and 3,4 outgoing, the Mandelstam variables are $s=-(p_1+p_2{)}^2$ and $t=-(p_1-p_3{)}^2$. They measure squared center-of-mass energy and momentum transfer. In the specified units ${\alpha }^{\prime }=1$, the external tachyon mass squared is $-1$, so $s+t+u=-4$.

The Gamma function residue at $-n$ is $(-1{)}^n/n!$. At $s=n-1$, its argument $-1-s$ varies with the opposite sign to $s$. The Gamma function recurrence givesThus the Veneziano amplitude pole residues areFor generic fixed $t$ the simple poles are $-1,0,1,\dots$. A zero of the residue product at an exceptional negative integer $t$ makes the corresponding pole removable. Fixed $t$ must be away from its own channel poles; fixing it at a channel singularity does not give an ordinary finite meromorphic function of $s$.

At $s=0$, the residue is $-(t+2)=-(t-u)/2$. This is proportional to $(p_1-p_2)\cdot (p_3-p_4)$, the contraction of conserved scalar-pair vector currents. Each current is orthogonal to the exchanged momentum because the external masses in its pair are equal. Alternatively, with scattering-angle cosine $z$, one has $t=-2(s/4+1)(1-z)$; at this pole $t+2=2z$, a pure spin-one angular dependence. ThereforeThe massless vector pole of the Veneziano amplitude is the massless open-string vector seen in the two-tachyon exchange channel.