Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 44 (String Theory)

Use target-space signature $(-,+,+,+)$ and conformal gauge $g_{ab}=e^{2\omega }\mathrm{diag}(-1,1)$. Dots and primes denote differentiation with respect to $\tau$ and $\sigma$. Varying the Polyakov action with respect to the string embedding map gives the wave equation; varying the worldsheet metric sets the worldsheet stress tensor to zero. Explicitly,With ${\sigma }^{\pm }=\tau \pm \sigma$ and ${\partial }_{\pm }=\frac{1}{2}({\partial }_{\tau }\pm {\partial }_{\sigma })$, the two Virasoro constraints are $({\partial }_{+}X{)}^2=({\partial }_{-}X{)}^2=0$.

A fixed spatial circle would sweep out a cylindrical string worldsheet. In static coordinates its string embedding map is $X=(t,R\cos \theta ,R\sin \theta ,0)$ and its induced worldsheet metric is $\mathrm{diag}(-1,R^2)$. The radial part of the Nambu–Goto equations of motion is proportional to $R^{-2}{\partial }_{\theta }^2(R\cos \theta ,R\sin \theta )=-(\cos \theta ,\sin \theta )/R$, which is nonzero. Thus a nonzero-radius cylinder cannot solve the equations. Motion of labels around this same cylinder is a worldsheet diffeomorphism and cannot alter this conclusion.

The familiar pulsating circular string makes the obstruction particularly explicit. Put $X^0=\kappa \tau$ and $X^1+i{X}^2=r(\tau )e^{i\sigma }$. The wave equation and Virasoro constraints reduce toA constant $r=R\ne 0$ violates the first equation. Even rotating the labels, $X^1+i{X}^2=R{e}^{i(\sigma +\Omega \tau )}$, cannot help: the wave equation requires ${\Omega }^2=1$, whereas $\dot{X}\cdot X^{\prime }=R^2\Omega$ requires $\Omega =0$.

Nor does a rigid rotation of the circular plane rescue the usual circular ansatz in three spatial dimensions. Write its spatial part as $A(\tau )\cos \sigma +B(\tau )\sin \sigma$, with $A^2=B^2=R^2$ and $A\cdot B=0$. The wave equation gives $\ddot{A}=-A$ and $\ddot{B}=-B$. The constant norms give $A\cdot \dot{A}=B\cdot \dot{B}=0$ and ${\dot{A}}^2={\dot{B}}^2=R^2$. The mixed Virasoro constraint, together with the derivative of $A\cdot B=0$, gives $\dot{A}\cdot B=A\cdot \dot{B}=0$. Differentiating $A\cdot B$ twice gives $\dot{A}\cdot \dot{B}=0$. This would require four mutually orthogonal nonzero spatial vectors in ${\mathbb{R}}^3$. This obstruction is specific to the stated spatial dimension; a rigid circular string in four spatial dimensions has more room.

To express the Virasoro constraints in modes, extend the string oscillators byDifferentiating the two chiral expansions gives ${\partial }_{-}{X}^{\mu }=\sqrt{{\alpha }^{\prime }/2}{\sum }_n{\alpha }_n^{\mu }{e}^{-in{\sigma }^{-}}$ and the corresponding expression with tildes for ${\partial }_{+}X$. Equating every Fourier coefficient of their squares to zero yieldsReality of the string embedding map means ${\alpha }_{-n}={\alpha }_n^{\ast }$ and ${\tilde{\alpha }}_{-n}={\tilde{\alpha }}_n^{\ast }$. In particular, define the classical string level operators by $N={\sum }_{n>0}{\alpha }_{-n}\cdot {\alpha }_n$ and $\tilde{N}={\sum }_{n>0}{\tilde{\alpha }}_{-n}\cdot {\tilde{\alpha }}_n$. The zero-mode Virasoro constraints becomeConsequently the classical bosonic string mass spectrum isThe equality of the two levels is closed-string level matching. There is no string intercept here: that shift comes from quantum normal ordering, not from these classical equations.

For the compact boson, closed-string periodicity permits a winding number $w\in \mathbb{Z}$, so $X^3(\sigma +2\pi )-X^3(\sigma )=2\pi wR$. Let $q$ be the momentum in this direction. Its zero-mode expansion and the corresponding chiral momenta areThus replace the equal left and right zero modes by ${\tilde{\alpha }}_0^3=\sqrt{{\alpha }^{\prime }/2}{p}_L^3$ and ${\alpha }_0^3=\sqrt{{\alpha }^{\prime }/2}{p}_R^3$; the nonzero string oscillators remain integer-moded. Classically $q$ is continuous. Quantizing the compact boson makes momentum wavefunctions single-valued around its circle, giving $q=n/R$ with $n\in \mathbb{Z}$.

If $k^a$, $a=0,1,2$, is the noncompact momentum and $M^2=-k^a{k}_a$, the two zero-mode Virasoro constraints giveAveraging and subtracting gives the momentum and winding modes and compact-circle closed-string level matching formulasThe latter becomes $\tilde{N}-N+nw=0$ on imposing quantum momentum quantization. The sign follows from assigning tildes to the left mover ${\sigma }^{+}$.

An explicit winding-supported circular string, valid classically for every $R>0$, isThe noncompact projection is a circle of constant radius. Every coordinate obeys the wave equation. Its compact winding is $w=1$, and a direct calculation givesThus both Virasoro constraints hold and the induced worldsheet metric is $2R^2\mathrm{diag}(-1,1)$, not a degenerate cylinder. The compact motion cancels the mixed stress that spoiled rotation of labels in flat space. Here $q=-R/{\alpha }^{\prime }$, $N=0$ and $\tilde{N}=R^2/{\alpha }^{\prime }$ satisfy the classical compact-circle closed-string level matching condition. Requiring this particular classical momentum to equal an individual quantum eigenvalue would impose an additional radius-dependent condition; that is not required to construct the classical solution.

Use the free-boson operator product expansionWrite $E_p=:e^{ip\cdot X}:$ and $ϵ=z-w$. The needed Wick contractions are $\partial X^{\mu }(z)\partial X^{\nu }(w)\sim -{\alpha }^{\prime }{\eta }^{\mu \nu }/(2{ϵ}^2)$ and $\partial X^{\mu }(z)E_p(w)\sim -i{\alpha }^{\prime }{p}^{\mu }{E}_p(w)/(2ϵ)$. For $V_{\zeta }={\zeta }_{\mu }:\partial X^{\mu }{E}_p:$, the cross double contraction in the worldsheet stress tensor has a cubic pole: either of its two derivatives can contract with the derivative in $V_{\zeta }$, while the other contracts with $E_p$. Including the factor $-1/{\alpha }^{\prime }$ givesA primary operator has no cubic pole. Hence the first vertex is primary precisely whenThe antiholomorphic conformal weight is $\bar{h}={\alpha }^{\prime }{p}^2/4$. Primarity alone puts no mass-shell condition on $p$.

For $W_{\zeta }={\zeta }_{\mu \nu }:\partial X^{\mu }\bar{\partial }{X}^{\nu }{E}_p:$, the holomorphic cubic pole is $-i{\alpha }^{\prime }{p}^{\mu }{\zeta }_{\mu \nu }:\bar{\partial }{X}^{\nu }{E}_p:/(2{ϵ}^3)$. The antiholomorphic operator product expansion gives the analogous pole involving $p^{\nu }{\zeta }_{\mu \nu }$. Thus the transversality of derivative-exponential primary vertices requiresThere is no trace-free requirement for primarity. A physical integrated string vertex operator must additionally have conformal weights $(1,1)$; for this vertex that gives $p^2=0$.

Now keep the fermionic signs in the real chiral Majorana fermion calculation. Contracting the external $\psi (w)$ through $:\psi (z)\partial \psi (z):$ givesMultiplying by $-1/2$ and Taylor expanding $\psi (z)$ at $w$ producesThis proves that $\psi$ is a primary operator with weight $h_{\psi }=1/2$. To find the central charge, in $T(z)T(w)$ the two double-contraction pairings contribute, before the overall factor $1/4$,The single Wick contractions give $2T(w)/{ϵ}^2+\partial T(w)/ϵ$, soComparing the leading term with $c/(2{ϵ}^4)$ proves $c_{\psi }=1/2$, as in the free chiral Majorana fermion conformal field theory.

With four embedding bosons and the ordinary reparameterization bc ghost system, cancellation of the worldsheet Weyl anomaly in each chirality requiresThus there are 44 real left-moving and 44 real right-moving fermions, equivalently 44 two-dimensional nonchiral Majorana fermions. These are internal free fermions in a four-dimensional bosonic string. Adding internal matter to the bosonic string does not introduce local worldsheet supersymmetry or its superghosts.

For the requested polynomial string vertex operators, work in the unprojected Neveu–Schwarz sector. A full model also chooses spin structures and a consistent projection, which the central-charge condition alone does not specify. If the holomorphic and antiholomorphic oscillator contributions to the conformal weights are ${\ell }_L,{\ell }_R$, physicality requiresTherefore only equal levels at most one can contribute to the requested nonpositive bosonic string mass spectrum.

With zero fermion insertions, level zero gives the tachyon vertex operator $E_p$, a scalar with $M^2=-4/{\alpha }^{\prime }$. Level one gives the massless closed-string vertex operator ${\zeta }_{\mu \nu }:\partial X^{\mu }\bar{\partial }{X}^{\nu }{E}_p:$, with the transverse conditions above and the string-state gauge redundancy that removes longitudinal polarizations. The physical transverse polarization tensor decomposes into a symmetric trace-free graviton, an antisymmetric Kalb–Ramond field, and a scalar dilaton. In four spacetime dimensions these have respectively two, one, and one physical polarizations; the two-form can be dualized to a scalar.

With two fermion insertions there are two possibilities. One fermion in each chirality givesThese are $44^2=1936$ scalar tachyons before any projection. There is no spacetime derivative or spacetime polarization to impose a vector transversality condition.

Two different fermions in the same chirality form an antisymmetric fermion bilinear, a weight-one current $J^{ij}=:{\psi }^i{\psi }^j:$; equal-species undifferentiated bilinears vanish. Closed-string level matching then requires a weight-one bosonic derivative in the opposite chirality. The resulting string vertex operators areThey have $M^2=0$ and are gauge vectors, with $p\cdot A^{ij}=0$ and $A_{\mu }^{ij}\sim A_{\mu }^{ij}+p_{\mu }{\lambda }^{ij}$, and likewise for $\tilde{A}$. Each chirality supplies $\left(\genfrac{}{}{0ex}{}{44}{2}\right)=946$ gauge vectors. The fermion bilinears generate the two $\mathfrak{so}(44)$ current algebras associated with the Special orthogonal Lie algebra, giving the unprojected gauge group $SO(44{)}_L\times SO(44{)}_R$. The tempting vertex $:{\psi }^i{\psi }^j{E}_p:$ alone fails closed-string level matching because its weights are $(1,0)$ before the common momentum contribution. Additional derivatives raise the levels above the nonpositive-mass range; a product of a left and a right bilinear has four fermion insertions and is outside this requested sector. This completes the zero- and two-fermion vertices of an internally fermionized bosonic string; a specified global projection can remove some of the listed states.

Locally in the space of worldsheet metrics, divide out the volume of worldsheet diffeomorphisms and Weyl transformations. With stabilizer zero modes omitted, the inverse Faddeev-Popov determinant at a metric $g$ isThis is the linearized orbit integral: the delta functional is the product of the metric-component delta functions, and its Jacobian is the inverse determinant of the gauge-orbit map. The omitted zero modes and any metric moduli must be handled separately; this local formula is not an integration over the physical moduli themselves.

The trace elimination in string Faddeev-Popov gauge fixing is essential. In two dimensions, split the variation asThe second term is trace-free. Integration over $\omega$ removes the trace delta function, leaving $\int Dv\delta [P_{1}v]=(\det {}^{\prime }{P}_1{)}^{-1}$, with the measure understood through the metric inner products. Equivalently, a Fourier representation of the metric delta functional introduces a commuting symmetric tensor ${\beta }^{ab}$. Integration over $\omega$ enforces $g_{ab}{\beta }^{ab}=0$, and the remaining exponent is proportional to $\int \sqrt{g}{\beta }^{ab}(P_{1}v{)}_{ab}$. This gives the same inverse Faddeev-Popov determinant and explains why its tensor variable must be trace-free.

For the Polyakov path integral, insert the gauge-fixing identitywhere $\hat{g}(m)$ is a representative depending on the worldsheet moduli $m$. Gauge invariance of the action, measure, and Faddeev-Popov determinant lets us change variables along the gauge orbit. Its volume cancels the original gauge-group denominator, leaving an integral over $X$, the moduli, and the determinant evaluated at $\hat{g}$. This is how the determinant enters the numerator of the gauge-fixed integral, despite the inverse determinant occurring in the orbit integral above.

A fermionic Gaussian integral exponentiates $\det {}^{\prime }{P}_1$ using an anticommuting vector $c^a$ and an anticommuting symmetric trace-free tensor $b^{ab}$:The constant $C$ can be absorbed into $b$. In local complex conformal gauge, trace-freeness sets $b_{z\bar{z}}=0$. The only surviving components of the worldsheet diffeomorphism ghost operator are proportional to $\bar{\partial }{c}^z$ and $\partial c^{\bar{z}}$. Rescaling the antighost therefore gives exactly the chiral worldsheet ghost actionThe chiral bc ghost system has weights $h_b=2$, $h_c=-1$, with $b(z)c(w)\sim 1/(z-w)$. Its worldsheet stress tensor is $T_{\mathrm{gh}}=-2:b\partial c:-:(\partial b)c:$. Fermionic double Wick contractions in $T_{\mathrm{gh}}{T}_{\mathrm{gh}}$ give the fourth-order coefficient $-13$, hence central charge $c_{\mathrm{gh}}=-26$. More generally this calculation gives $c=1-3(2h_b-1{)}^2$. Each embedding scalar contributes one, so the total chiral central charge is $D-26$. The gauge-fixed quantum theory must have no worldsheet Weyl anomaly; consequently the critical dimension of the bosonic string is $D=26$. Zero-mode insertions for Conformal Killing vector fields and moduli do not change this local anomaly calculation.

For the Veneziano amplitude, first examine an ordered beta-function term. A pole occurs when $-{\alpha }^{\prime }s-1=-n$, with $n=0,1,2,\dots$, so the exchanged masses follow the open bosonic string mass spectrumThere is a tachyon at level zero, a massless vector at level one, then an infinite tower with equally spaced squared masses. In oscillator quantization these levels are $N=n$; the poles alone do not determine all oscillator degeneracies or all states that might fail to couple to the chosen external particles.

The Veneziano amplitude pole residue gives considerably more information than a pole position. Near $s_n=(n-1)/{\alpha }^{\prime }$, use $\Gamma (-n+ϵ)\sim (-1{)}^n/(n!ϵ)$ and $ϵ=-{\alpha }^{\prime }(s-s_n)$. The finite gamma function ratio is a product of $n$ linear factors. Thus, for generic $t$,The empty product at $n=0$ is one. This degree-$n$ angular polynomial decomposes into spin partial waves; it reveals the spins that couple, up to maximal spin $n$ in the ordered channel, and the corresponding products of three-point couplings. For example, the level-one linear residue encodes vector exchange. A single external-state amplitude need not separate all states with the same mass and spin.

There is a qualification specific to the sum printed in the question. For four identical on-shell tachyons, $s+t+u=-4/{\alpha }^{\prime }$. At ${\alpha }^{\prime }s=n-1$, put $P_n(t)={\prod }_{j=1}^n({\alpha }^{\prime }t+j+1)$. Then ${\alpha }^{\prime }u=-n-3-{\alpha }^{\prime }t$, and reversing the product gives $P_n(u)=(-1{)}^n{P}_n(t)$. The $t,u$ beta-function term is regular at a generic $s$-channel pole. The residue of the full symmetric amplitude is therefore proportional toThis is odd-level pole cancellation in the symmetric Veneziano amplitude. The fully symmetrized identical-tachyon amplitude has only even-level poles at generic angle, including the tachyon but not the massless vector. This is an external-state selection rule, not removal of odd-level states from the string theory. Ordered amplitudes, or suitable amplitudes with color factors and different external states, retain those exchanges. Thus interpreting every candidate gamma function pole as a pole of the printed sum would give a false inference about that amplitude.

At high energy with $x=-t/s\in (0,1)$ fixed, the gamma function reflection formula followed by the Stirling formula gives the fixed-angle softness of an open-string amplitude:up to oscillatory factors and possible additional cancellations. The bracket is positive at a genuinely nonforward fixed angle. This is the envelope away from resonance poles, or the corresponding analytically continued asymptotic regime; it is not a uniform bound through real-axis poles. String scattering is exponentially soft at fixed angle, unlike the typical power-law behavior of point-particle interactions. The finite string length distributes a hard collision over an extended object. This is evidence of improved ultraviolet behavior, not by itself a proof that every loop amplitude is finite. The fixed-angle limit also differs from the fixed-$t$ Regge limit.

Choose a Euclidean string worldsheet and the usual normalization of the Polyakov action. For a closed oriented worldsheet $\Sigma$, the string nonlinear sigma model isHere ${\epsilon }^{12}=1$ is the alternating density, not the tensor with components $1/\sqrt{g}$. The imaginary coefficient in the Kalb–Ramond field term is the Euclidean continuation; in Lorentzian signature this term is real. $G$ is the spacetime metric and $\Phi$ the dilaton.

All three terms have worldsheet diffeomorphism invariance, and are covariant under spacetime coordinate transformations when $X$, $G$, $B$ and $\Phi$ are transformed together. Under a Weyl transformation $g_{ab}\mapsto e^{2\omega }{g}_{ab}$, $\sqrt{g}{g}^{ab}$ is invariant in two dimensions, so the metric coupling has classical Weyl invariance. The antisymmetric coupling is independent of the worldsheet metric, so it too has classical Weyl invariance.

The curvature coupling requires care. An infinitesimal Weyl transformation gives ${\delta }_{\omega }(\sqrt{g}{R}^{(2)})=-2\sqrt{g}{\Delta }_g\omega$. Integration by parts on the closed worldsheet therefore gives the Weyl variation of a nonconstant dilaton coupling:A constant dilaton coupling is topological and Weyl invariant; a general varying dilaton coupling is not separately classically Weyl invariant. Its classical variation participates in canceling the quantum worldsheet Weyl anomaly, producing the improved background equations below.

To relate these background couplings to the flat-space bosonic string mass spectrum, expand $G_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }(X)$ and express $h$ and $B$ as superpositions of plane waves. The perturbation of the action is a sum of integrated string vertex operators of the formThis is the vertex of ${\alpha }_{-1}^{\mu }{\tilde{\alpha }}_{-1}^{\nu }|p⟩$ in the massless first closed-string level. Its conformal weights are $(1+{\alpha }^{\prime }{p}^2/4,1+{\alpha }^{\prime }{p}^2/4)$, so marginality gives $p^2=0$. The primary operator conditions give transversality, and the string-state gauge redundancy removes longitudinal components. The symmetric trace-free polarization tensor is the graviton and perturbs $G$; its antisymmetric part is the Kalb–Ramond field and perturbs $B$. The scalar physical polarization is the dilaton. At the level of background fields its vertex involves the metric trace and the curvature improvement, so an arbitrary off-shell metric trace should not simply be identified with the physical dilaton without this distinction.

The spacetime gauge invariance of the antisymmetric coupling isIndeed its action is $S_B=i(2\pi {\alpha }^{\prime }{)}^{-1}{\int }_{\Sigma }{X}^{\ast }B$. Its variation isfor a closed worldsheet, by Stokes theorem. In components this is the total derivative ${\partial }_a({\epsilon }^{ab}{\Lambda }_{\nu }(X){\partial }_b{X}^{\nu })$: the term containing ${\partial }_a{\partial }_b{X}^{\nu }$ vanishes by antisymmetry. Thus only the Kalb-Ramond field strength $H$ can enter the local spacetime equations. With boundaries, a compensating boundary gauge-field transformation is needed.

Separate the dilaton into a constant ${\Phi }_0$ and a varying part. The Gauss-Bonnet theorem givesfor a connected closed orientable genus-$g$ worldsheet. Its Euclidean path-integral weight containsConsequently the sum over worldsheet topologies is the string genus expansion ${\sum }_{g\ge 0}{g}_s^{2g-2}{\mathcal{A}}_g$, with each coefficient integrating over that genus's worldsheet moduli. Each extra handle costs $g_s^2$, which is the loop-counting factor of string perturbation theory. With the usual normalization of $n$ external closed-string vertices, the connected amplitude carries $g_s^{2g-2+n}$. This dilaton Euler-characteristic weighting is distinct from the expansion in ${\alpha }^{\prime }$ of a fixed-worldsheet string nonlinear sigma model.

Finally quantize that two-dimensional theory. The background fields are its couplings. The background field expansion of a string sigma model, $X=x+\xi$ in target-space normal coordinates, contains curvature vertices quadratic in the fluctuation $\xi$. Contracting their two fluctuation indices produces the Ricci tensor multiplying $\partial x^{\mu }\bar{\partial }{x}^{\nu }$. Logarithmic short-distance contractions renormalize $G$, $B$, and $\Phi$, giving functional sigma-model beta functions. Gauge independence requires vanishing of the worldsheet Weyl anomaly, including the classical improvement from the varying dilaton, not merely an arbitrary choice of renormalization scale.

For clarity, in conventional leading-order normalization the improved coefficients areHere $H^2=H_{\mu \nu \rho }{H}^{\mu \nu \rho }$. The constant term in the dilaton coefficient is the matter-plus-ghost central charge deficit. At $D=26$, setting these coefficients to zero gives the leading spacetime field equationsFor the last equation, take the trace of the first, $R-H^2/4+2{\nabla }^2\Phi =0$, and combine it with ${\overline{\beta }}^{\Phi }=0$; this verifies the coefficients and signs. These are equivalently the equations from the leading string-frame massless effective actionHigher worldsheet orders give higher-derivative ${\alpha }^{\prime }$ corrections, whereas additional handles give string-loop corrections controlled by $g_s$. Thus quantum consistency of a two-dimensional string nonlinear sigma model determines the dynamics of its spacetime background fields. These leading coefficient conventions agree with the primary string-theory lectures' background-field calculation.