Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 306 (String Theory)

Define the worldsheet stress-energy tensor byUsing $\delta \sqrt{|h|}=-\sqrt{|h|}{h}_{ab}\delta h^{ab}/2$ in the Polyakov action givesBecause the worldsheet is two-dimensional,This classical tracelessness is the Noether identity for Weyl invariance.

The Nambu–Goto action isThe metric equation of the Polyakov action is $T_{ab}=0$, which saysThus $h_{ab}$ is conformal to the induced worldsheet metric, $h_{ab}=e^{2\omega }{\gamma }_{ab}$. Substitution into the Polyakov action cancels the conformal factor and yieldsThe two actions are therefore classically equivalent after the auxiliary worldsheet metric is eliminated.

For an infinitesimal Weyl transformation $\delta h^{ab}=-2\omega h^{ab}$, the definition of the stress tensor givesThe insertions are Weyl invariant by assumption, so varying the normalized path integral givesConsequently the stated condition $⟨h^{ab}{T}_{ab}{\varphi }_1\cdots {\varphi }_n⟩=0$ makes the correlation function Weyl invariant. Quantum failure of this condition is the worldsheet Weyl anomaly.

Changing the target metric by $\delta G_{\mu \nu }={ϵ}_{\mu \nu }$ changes the action bySince the path-integral weight is $e^{iS}$, the leading change is the insertionwhereA change of worldsheet metric within a gauge orbit changes only the redundant description and cannot alter gauge-invariant observables. The target-space metric is instead a physical coupling of the string nonlinear sigma model, so changing it changes the theory and its correlation functions.

The free embedding fields haveIn $T_X(z)T_X(w)$, the double contraction of the two derivatives in each normal-ordered stress tensor has two pairings. It givesThe stress-tensor operator-product expansion therefore has central charge $c=D$. Contour integration against the infinitesimal holomorphic vector field givesso in the notation of the question

The worldsheet ghosts form an anticommuting bc system of weights $(2,-1)$. Its double contractions give central chargeThereforeand hence

Adding the matter and ghost transformations givesFor the Laurent modes $L_n=\oint dz{z}^{n+1}{T}_{\mathrm{tot}}(z)/(2\pi i)$, the corresponding Virasoro algebra isAt the bosonic-string critical dimension of string theory, $D=26$, the total central charge and central term vanish. The total stress tensor then transforms as a genuine conformal tensor of weight two, and the quantum worldsheet gauge anomaly cancels.

The three unintegrated vertices fix the residual Möbius symmetry of the sphere and saturate its three holomorphic and antiholomorphic ghost zero modes. Moving a fixed insertion changes the amplitude by a BRST-exact operator, whose expectation with physical BRST-closed vertices vanishes. The remaining integrated positions are dummy variables. Thus the amplitude is independent of the chosen coordinates $(z_i,{\bar{z}}_i)$.

The zero mode of the free embedding field enforces momentum conservation, while contractions of normal-ordered exponentials give the Koba-Nielsen factor:Combining this with the supplied ghost correlator yields, up to the conventional overall normalization,Thereforewith the factor $(2\pi {)}^{26}$ absorbed into the amplitude normalization used by the question.

Fix $z_1=0$, $z_2=1$, and $z_3=\infty$, and write $z_4=z$. For the closed-string tachyon, $k_i^2=4/{\alpha }^{\prime }$. Hencewhere momentum conservation was used in the second identity. Taking the limit at infinity together with the ghost factor gives the Virasoro–Shapiro amplitudeChoosing a different set of three punctures to fix merely applies a Möbius transformation and relabels identical external tachyons. Exchanging the roles of the relevant punctures interchanges $s$ and $t$ without changing the integral, proving crossing symmetry under $s\leftrightarrow t$.

Varying the gauge-fixed Polyakov action gives the bulk wave equation and the boundary termIt vanishes in each target direction either through the Neumann boundary conditionor through the Dirichlet boundary conditionwhich fixes the endpoint positions.

Along the common D-brane worldvolume, the Neumann solution isIn each transverse direction, the endpoint conditions $X^I(0,\tau )=x_1^I$ and $X^I(\pi ,\tau )=x_2^I$ giveThe linear term is the classical stretch between the two parallel D-branes.

The zero-mode Virasoro constraint iswhere the string level operator isSince the string tension isthe mass formula is

For a string beginning and ending on the same brane, the $N=1$ states are massless. Oscillators polarized tangentially give a gauge field on the $(p+1)$-dimensional worldvolume, while transverse polarizations give scalar fields describing fluctuations of the brane position. Strings joining two separated branes have the additional stretching mass $T|x_2-x_1|$ and are charged under the two endpoint gauge groups. When the two branes coincide, these off-diagonal states also become massless and Coincident-D-brane gauge enhancement enlarges $U(1)\times U(1)$ to $U(2)$. The bosonic open string also retains its $N=0$ tachyon.