Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 53 (String Theory)

Use target-space signature $(-,+,\dots ,+)$ and let a dot between target vectors denote contraction with the Minkowski metric. Varying the string embedding map in the Polyakov action, with periodic variations around the closed string, givesThus the embedding equation of motion isThe independent worldsheet metric variation uses $\delta \sqrt{-\gamma }=-\frac{1}{2}\sqrt{-\gamma }{\gamma }_{ab}\delta {\gamma }^{ab}$. It sets the worldsheet stress tensor to zero:In conformal gauge, ${\gamma }_{ab}=e^{2\omega }\mathrm{diag}(-1,1)$, the Weyl factor cancels. Define worldsheet derivatives ${\partial }_{\pm }=({\partial }_{\tau }\pm {\partial }_{\sigma })/2$. The equations and Virasoro constraints reduce toEquivalently, $\ddot{X}-X^{\prime \prime }=0$, $\dot{X}\cdot X^{\prime }=0$ and ${\dot{X}}^2+X^{\prime 2}=0$. The free wave equation permits independent periodic left- and right-moving fields.

Take $0\le \sigma <2\pi$ and the uncompactified closed-string mode expansionThe Virasoro generators are the Fourier coefficients of the two quadratic worldsheet stress tensors. Their quantum expressions areThe canonical commutation relations are $[{\alpha }_m^{\mu },{\alpha }_n^{\nu }]=m{\eta }^{\mu \nu }{\delta }_{m+n,0}$, with an identical tilded algebra and commuting chiral sectors. Normal ordering places annihilation modes to the right. It gives $L_0={\alpha }^{\prime }{p}^2/4+N$, where $N={\sum }_{n>0}{\alpha }_{-n}\cdot {\alpha }_n$, and similarly on the other side. In old covariant string quantization, physical string states obeymodulo physical null string states. Negative generators are adjoint constraints on bras, not further annihilation conditions on the same ket. In critical bosonic string theory, $d=26$ and the string intercept is $a=1$. These closed-string physical-state Virasoro conditions implyThus the massless first closed-string level has $N=\tilde{N}=1$, and its general state isTo derive the polarization tensor constraints, the oscillator algebra gives $[L_n,{\alpha }_m^{\mu }]=-m{\alpha }_{n+m}^{\mu }$. Applying $L_1$ to this state replaces ${\alpha }_{-1}^{\mu }$ by ${\alpha }_0^{\mu }$; the independent tilded condition replaces the other oscillator by ${\tilde{\alpha }}_0^{\nu }$. All higher positive generators annihilate the state automatically. ThereforeThere is no condition that the tensor trace vanish: each chiral side has only one oscillator, so neither $L_2$ nor ${\tilde{L}}_2$ imposes a trace condition. The null-state quotient of a string also identifiesFor example, $L_{-1}({\xi }_{\nu }{\tilde{\alpha }}_{-1}^{\nu }|0,p⟩)$ is a momentum-longitudinal state with zero inner product against every transverse physical polarization. This explains the string-state gauge redundancy, rather than merely counting covariant components. For an ordinary nonzero null momentum, choose a frame adapted to $p$. The constraints remove one null direction in each index, and these null-state identifications remove the remaining longitudinal directions. The representatives are then ${\zeta }_{ij}$, $i,j=1,\dots ,24$.

Decompose that transverse matrix into its symmetric trace-free part, antisymmetric part and trace. They describe the graviton, Kalb–Ramond field and dilaton, respectively. Their particle polarization counts areThese are representations of the rotational part $SO(24)$ of the massless little group. The dilaton corresponds to the transverse trace; a covariant scalar polarization can be written proportional to ${\eta }_{\mu \nu }-p_{\mu }{\ell }_{\nu }-{\ell }_{\mu }{p}_{\nu }$, with ${\ell }^2=0$ and $p\cdot \ell =1$. It is transverse in both indices and has trace 24. The nonzero-momentum assumption is appropriate for propagating particle states; zero momentum requires separate treatment of gauge and background zero modes.

In a general background, these fields replace the flat-space free action by the string nonlinear sigma model. A convenient Euclidean convention isHere ${ϵ}^{ab}$ is the orientation density; reversing its convention changes the displayed sign of the two-form coupling. The metric tensor is $G_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }+\cdots$, $B$ is the antisymmetric Kalb–Ramond field, and $\Phi$ is the dilaton. Their worldsheet couplings reproduce the corresponding massless closed-string vertex operators when expanded about flat space. The Polyakov path integral integrates $e^{-S_E}$ over embedding fields and metrics modulo worldsheet diffeomorphisms and Weyl transformations; gauge fixing supplies ghost determinants and moduli integrations. Quantum consistency of a general background additionally requires vanishing sigma-model beta functions, not just insertion of arbitrary backgrounds into a free theory.

Finally, separate the constant dilaton mode, $\Phi ={\Phi }_0+\hat{\Phi }$. By the Gauss-Bonnet theorem, its action isFor a connected oriented surface of genus $g$, its Euler characteristic is $\chi =2-2g$. Defining the string coupling by $g_s=e^{{\Phi }_0}$ givesThus the string genus expansion is a topology sum with two extra powers of $g_s$ per handle. An amplitude with $n$ canonically normalized external closed-string vertices has the additional vertex normalization $g_s^n$, so its overall factor is $g_s^{2g-2+n}$. The curvature term, rather than a guessed coupling inserted at each internal interaction, supplies this dependence directly from the functional integral.

Distinguish target-space light-cone coordinates from the worldsheet $\pm$ labels. In target space take $X^{\pm }=(X^0\pm X^{25})/\sqrt{2}$, so the Minkowski metric is $d{s}^2=-2d{X}^{+}d{X}^{-}+{\sum }_i(d{X}^i{)}^2$. After conformal gauge has been chosen, the residual transformations of $\tau +\sigma$ and $\tau -\sigma$ may be used, on a patch with $p^{+}\ne 0$, to makeFor an open string on $0\le \sigma \le \pi$, the conventional normalization is $\kappa =2{\alpha }^{\prime }{p}^{+}$. For the closed-string convention of the preceding solution, $\kappa ={\alpha }^{\prime }{p}^{+}$. In either case, the two Virasoro constraints readSince ${\partial }_{\pm }{X}^{+}=\kappa /2$, they determine the longitudinal coordinate:Its oscillator content is fixed by the transverse coordinates, leaving $d-2$ independent fields. This is the light-cone gauge in string theory; it solves the constraints rather than imposing them afresh on each transverse state.

For Neumann boundary conditions at both endpoints, the open-string expansion isThe transverse string oscillators satisfy $[{\alpha }_m^i,{\alpha }_n^j]=m{\delta }^{ij}{\delta }_{m+n,0}$, $i,j=1,\dots ,24$, and ${\alpha }_{n>0}^i|0,p⟩=0$. Define $L_n^{\perp }=\frac{1}{2}{\sum }_m:{\alpha }_{n-m}^i{\alpha }_m^i:$ and ${\alpha }_0^{+}=\sqrt{2{\alpha }^{\prime }}{p}^{+}$. The solved longitudinal modes, including the zero-mode constraint, areThe critical bosonic string theory has 24 transverse bosons. Their regulated oscillator vacuum energy is $24\cdot \frac{1}{2}{\sum }_{n\ge 1}n=-1$, since the finite term of ${\sum }_{n\ge 1}n{e}^{-\epsilon n}={\epsilon }^{-2}-1/12+O({\epsilon }^2)$ is $-1/12$. Thus the normal-ordering constant of a string is $a=1$. A normalized transverse Fock space basis for general excited physical string states iswith finitely many nonzero occupation numbers. Arbitrary physical states are superpositions of this basis, with each momentum on its corresponding mass shell. The string level operator and open bosonic string mass spectrum giveAll spatial directions have free endpoints, so this string ends on a space-filling D25-brane. More generally, a D-brane has Neumann boundary conditions along its worldvolume and Dirichlet boundary conditions in its transverse directions; a D$p$-brane has $p$ spatial worldvolume directions. A stack may additionally supply Chan-Paton factors, without changing the oscillator construction above.

To see T-duality of D-brane boundary conditions directly, decompose the compact coordinate as $X=X_L(\tau +\sigma )+X_R(\tau -\sigma )$ and define its dual by $\tilde{X}=X_L-X_R$. Differentiation givesTherefore the original Neumann boundary condition ${\partial }_{\sigma }X=0$ becomes ${\partial }_{\tau }\tilde{X}=0$ at each endpoint. The dual endpoint coordinate is time-independent: the dual condition is Dirichlet. Conversely a fixed endpoint in $X$ becomes Neumann in the dual coordinate. For a circle, the T-duality radius is $\tilde{R}={\alpha }^{\prime }/R$. A wrapped Neumann brane loses one spatial dimension in the dual picture, for example D25 becomes D24. The two dual endpoint constants need not be identical; their positions encode the original Wilson-line data.

For the separated-brane problem, use two static parallel D$p$-branes and no relative angle or background field. Let their displacement in transverse directions be $d^I$, with ${\sum }_I(d^I{)}^2=d^2$. A Dirichlet coordinate contains the classical termThe spatial gradient contributes to the worldsheet HamiltonianParallel branes have integer modes in their NN and DD directions, so their oscillator vacuum energy is still $-1$. The zero-mode Virasoro constraint is consequentlyWith string tension $T=1/(2\pi {\alpha }^{\prime })$, the Open string stretched between D-branes hasThe classical stretching energy is $Td$, while the quantum ground-state energy, when real, isThe vacuum correction is an additive term in energy squared, not a constant subtraction from the classical stretching energy. It follows that the bosonic stretched-string tachyon threshold isBelow this threshold the lowest state is a tachyon: small-momentum modes have imaginary frequency and the background is unstable, rather than having an ordinary real negative ground-state energy. It is massless at the threshold. The assumption of parallel branes is needed for the integer moding used here; relative angles change the vacuum contribution.

The Polyakov action is invariant under worldsheet diffeomorphisms and Weyl transformations, ${\gamma }_{ab}\mapsto e^{2\omega }{\gamma }_{ab}$. A two-dimensional metric has three local components: two coordinate transformations and one Weyl function remove these locally. In conformal gauge the factor $\sqrt{-\gamma }{\gamma }^{ab}$ is independent of the conformal scale, and on a flat coordinate patch the matter action isThus it is the action of $d$ free scalar fields, with their target-space contractions using the Minkowski metric. On a Euclidean sphere there are no unpunctured conformal moduli: a metric can be put into the conformal class of the round metric, and in stereographic patches the scalar action has this free-field form. This does not mean that the sphere has a globally flat metric. A spherical worldsheet here is understood after Wick rotation; it has no nonsingular global Lorentzian metric. The residual Möbius transformations and their zero modes still have to be divided out or fixed by vertex positions.

To derive the worldsheet ghost action, consider an infinitesimal diffeomorphism generated by $v^a$ and a Weyl variation $\omega$. The metric variation separates asIntegration over the Weyl parameter removes the trace gauge condition. The trace-free part produces the Faddeev-Popov determinant of an operator mapping vectors to symmetric traceless tensors. This is trace elimination in string Faddeev-Popov gauge fixing. The stated Grassmann Gaussian integral exponentiates this determinant with an odd vector $c^a$ and an odd symmetric trace-free antighost $b_{ab}$.

The normalization of the operator and antighost can be chosen to match the printed convention. For example, in a flat patch take $P=-P_1/2$. It differs from the orbit operator only by a field-independent determinant factor. Tracelessness and symmetry give $b_{ab}(Pc{)}^{ab}=-b_{ab}{\partial }^a{c}^b$. Integration by parts, followed by exchanging two odd factors, gives $\int b_{ab}{\partial }^a{c}^b=\int c^a{\partial }^b{b}_{ab}$. Consequently the determinant exponent is $+(2\pi {)}^{-1}\int c^a{\partial }^b{b}_{ab}=i{I}_{\mathrm{gh}}$ withOverall constant normalization does not affect the determinant's dependence on the fields. The Faddeev-Popov ghost fields are vectors, while the antighost fields are symmetric trace-free tensors; they are not extra target-space string excitations.

Now use ${\sigma }^{\pm }=\tau \pm \sigma$, ${\partial }_{\pm }=({\partial }_{\tau }\pm {\partial }_{\sigma })/2$ and $c^{\pm }=c^0\pm c^1$. In these coordinates ${\eta }_{+-}={\eta }_{-+}=-1/2$ and ${\partial }^{\pm }=-2{\partial }_{\mp }$. Tracelessness implies $b_{+-}=0$, so directlyThereforeFor a Cartesian check, $b_{11}=b_{00}$ and $b_{++}=(b_{00}+b_{01})/2$, $b_{--}=(b_{00}-b_{01})/2$ reproduce the same result.

The original PDF repeats ${\delta }_{\eta }{c}^{+}$ for its final ghost transformation. The second occurrence must be ${\delta }_{\eta }{c}^{-}$; otherwise the displayed rules give two incompatible transformations for one ghost and none for the other. Write the corrected transformation as ${\delta }_{\eta }\Phi =\eta s\Phi$, with $\eta$ on the left. Then the odd left-acting BRST differential satisfies the graded Leibniz rule $s(AB)=(sA)B+(-1{)}^{|A|}A(sB)$. Its local rules areThe last two expressions are $s{b}_{\pm \pm }=i{\theta }_{\pm \pm }/{\alpha }^{\prime }$ with the matter-plus-ghost worldsheet stress tensor in the question.

Here is an off-shell check of the action variation, including the Grassmann signs. Temporarily set $c=c^{+}$, $f=c^{-}$, $b=b_{++}$, $h=b_{--}$, $u={\partial }_{+}X$ and $v={\partial }_{-}X$. Apart from the common factor $1/\pi$, the total Lagrangian is $\mathcal{L}=u\cdot v/{\alpha }^{\prime }+i(c{\partial }_{-}b+f{\partial }_{+}h)$. Its matter variation isThe matter terms in $sb$ and $sh$ give the ghost-action contribution ${\alpha }^{\prime -1}[c{\partial }_{-}(u^2)+f{\partial }_{+}(v^2)]$. Combining these terms with the last two matter terms produces ${\alpha }^{\prime -1}[{\partial }_{-}(c{u}^2)+{\partial }_{+}(f{v}^2)]$.

The remaining contribution from the $c,b$ sector iswhere subscripts denote derivatives. In the last equality the terms $c_{-}{c}_{+}b$ and $c_{+}{c}_{-}b$ cancel by anticommutation. The $f,h$ sector has the same identity with $+\leftrightarrow -$. ThuswhereHence the total action is BRST invariant up to a boundary term, which vanishes on a closed worldsheet or under the usual compatible boundary and time-end conditions. This calculation requires no equations of motion.

Nilpotence needs a qualification for these particular chiral rules. For either ghost, $s(c\partial c)=(c\partial c)\partial c-c\partial (c\partial c)=0$, because $c^2=(\partial c{)}^2=0$. Thus $s^2{c}^{\pm }=0$ algebraically. For the embedding, however, the graded calculation givesMixed second derivatives of $X$ cancel because $c^{+}{c}^{-}=-c^{-}{c}^{+}$. The remaining terms vanish on the ghost equations of motion, ${\partial }_{-}{c}^{+}={\partial }_{+}{c}^{-}=0$, which follow by varying the antighosts in the chiral ghost action. They need not vanish off shell. An explicit counterexample uses independent Grassmann variables $\xi ,\rho$: take $c^{+}=\xi$, $c^{-}=\rho {\sigma }^{+}$ and one coordinate $X={\sigma }^{-}$. Then $s^{2}X=-\xi \rho \ne 0$.

For the antighosts, use the permitted stress-tensor identity $s{\theta }_{ab}=0$ on the classical solution space. Then $s^2{b}_{\pm \pm }=(i/{\alpha }^{\prime })s{\theta }_{\pm \pm }=0$. Indeed direct variation of the displayed stress tensors giveswhich vanish using the ghost equations and the free matter wave equation. Therefore two successive BRST transformations annihilate every field on shell. This is chiral conformal-gauge BRST nilpotence is on shell, not an unrestricted off-shell identity for arbitrary ghosts. Quantum BRST nilpotence additionally requires cancellation of the worldsheet anomaly, giving $d=26$ in the critical bosonic theory.

Use ${\partial }_{\pm }=({\partial }_{\tau }\pm {\partial }_{\sigma })/2$ and keep each constant odd supersymmetry parameter on the left. The transformation of a fermion may be written $\delta {\psi }_1^{\mu }={ϵ}_1{\partial }_{+}{X}^{\mu }$ or ${\partial }_{+}{X}^{\mu }{ϵ}_1$, since the bosonic derivative commutes with the parameter. The variation $\delta$ itself is even, so it obeys the ordinary product rule; the signs below arise when moving an odd parameter past an odd field.

First consider the ${ϵ}_1$ transformation. Suppress target indices and put $\mathcal{L}=2X_{+}\cdot X_{-}+i{\psi }_1\cdot {\partial }_{-}{\psi }_1+i{\psi }_2\cdot {\partial }_{+}{\psi }_2$. Its bosonic part varies asFor the first fermion, moving ${ϵ}_1$ to the left givesThe second fermion does not vary under ${ϵ}_1$. Since ${\partial }_{-}{X}_{+}={\partial }_{+}{X}_{-}$, these terms combine to ${\delta }_1\mathcal{L}=-i{ϵ}_1{\partial }_{+}({\psi }_1\cdot X_{-})$. The other parameter gives the same cancellation with $+\leftrightarrow -$ and $1\leftrightarrow 2$. ThereforeThis proves worldsheet supersymmetry invariance up to the rigid worldsheet supersymmetry boundary term, without using field equations.

For an open string, the boundary term is part of the invariance statement. For example, a Neumann endpoint has ${\partial }_{+}X={\partial }_{-}X$ and compatible fermion gluing ${\psi }_1=\kappa {\psi }_2$, $\kappa =\pm 1$. The spatial boundary flux vanishes when ${ϵ}_2=\kappa {ϵ}_1$. Both endpoints must admit the same preserved parameter. Opposite gluing signs, giving the Neveu–Schwarz sector, do not preserve a nonzero constant parameter at both ends. Thus the local bulk identity and the existence of a global rigid charge on a particular open-string sector are distinct. The local superconformal symmetry remains the basis of the string constraints; this is consistent with the spin-structure obstruction to constant worldsheet supersymmetry.

Now write $A_{\pm }^{\mu }={\partial }_{\pm }{X}^{\mu }$ and $J_{+}^1={\psi }_1\cdot A_{+}$, $J_{-}^2={\psi }_2\cdot A_{-}$. The other two components vanish. The action implies ${\partial }_{-}{\psi }_1=0$, ${\partial }_{+}{\psi }_2=0$ and ${\partial }_{+}{\partial }_{-}X=0$, so these supercurrents are chiral and conserved locally. At equal time, the given canonical commutation relations implyWith $q_{ϵ}={\pi }^{-1}\int d\sigma ({ϵ}_1{\psi }_1\cdot A_{+}+{ϵ}_2{\psi }_2\cdot A_{-})$, contraction with this commutator givesThe parameter-weighted charge is even. Since the parameters anticommute with the fermions,Using the PDF's normalization $\{{\psi }_i^{\mu }(\sigma ),{\psi }_j^{\nu }({\sigma }^{\prime })\}=\pi {\eta }^{\mu \nu }{\delta }_{ij}\delta (\sigma -{\sigma }^{\prime })$ yieldsThus the charge generates all the specified supersymmetry transformations. Dropping the factor $\pi$ in the fermion anticommutator, as happens in the converted TeX, would fail this check. The original PDF has the correct factor.

To calculate the supercurrent bracket, put $A={\partial }_{+}X$ and $J={\psi }_1\cdot A$. The equal-time bosonic relations giveBosonic and fermionic fields commute. Expanding the two current products, commuting the bosonic factors once and using the fermion anticommutator gives the noncentral partFor an ordinary smooth bilocal coefficient, the distributional derivative identity isFor the fermion bilinear, its coincident classical value, or its coincident normal-ordered value, is zero by antisymmetry against the symmetric target metric. The second term is ${\psi }_1\cdot {\partial }_{\sigma }{\psi }_1$. Since ${\partial }_{-}{\psi }_1=0$, we have ${\partial }_{\sigma }{\psi }_1={\partial }_{+}{\psi }_1$. It follows that the noncentral current bracket isThis is the canonical supercurrent bracket of free RNS matter. The other chirality similarly gives ${\theta }_{--}=({\partial }_{-}X{)}^2+(i/2){\psi }_2\cdot {\partial }_{-}{\psi }_2$ and ${\theta }_{+-}=0$ on shell.

The printed current identity is the classical, or noncentral, part of the algebra. Fully quantized coincident products require normal ordering; the matter algebra then also has a central contact term proportional to ${\delta }^{\prime \prime }$ in the current anticommutator. The 10 bosons and 10 real fermions contribute matter central charge $c=10+10/2=15$. The central term cannot be discarded in a literal quantum operator identity, although it is irrelevant to identifying the local noncentral stress tensor above.

The other brackets close on these same currents. The stress-tensor commutator contains $\theta {\delta }^{\prime }$ and ${\theta }^{\prime }\delta$ and, quantum mechanically, a central ${\delta }^{\prime \prime \prime }$ term. The stress-tensor–supercurrent commutator contains $J{\delta }^{\prime }$ and $J^{\prime }\delta$, expressing the conformal weight $3/2$ of $J$. The supercurrent anticommutator contains $\theta \delta$ and its central ${\delta }^{\prime \prime }$ term. In a conventional normalization of Fourier modes these form the N=1 super-Virasoro algebra:Here $C_{mn}$ and $D_{rs}$ are central terms supported at zero total mode number; their detailed coefficients are not needed for this request. The indices $r$ are half-integers in the Neveu–Schwarz sector and integers in the Ramond sector. Opposite chiralities commute or anticommute as appropriate.

For the closed RNS string, impose a copy of the constraints in each chirality. In the Neveu–Schwarz sector, the physical conditions areIn the Ramond sector, use vacuum-normal-ordered generators with zero Ramond intercept. ThenThe Ramond zero-mode equation is essential: its square gives the mass-shell condition in this convention, $G_0^2=L_0$, and on the oscillator ground sector it is the spacetime Dirac constraint. In a convention whose unshifted plane generator has $G_0^2=L_0-c/24$, the zero-mode equation instead sets that shifted combination to zero; these are the same condition, not competing intercepts.

For either choice, the corresponding tilded constraints hold on the other side. Both chiral mass-shell conditions and closed-string level matching are required:The state sectors are NS–NS, NS–R, R–NS and R–R. Quotienting physical null states removes unphysical longitudinal polarizations; equivalently, use the appropriate BRST cohomology including the superconformal ghosts. A consistent superstring also specifies the GSO projection. It is an additional sector selection, not a replacement for these local super-Virasoro constraints. Physical states satisfy both chiral positive-mode and zero-mode constraints, level matching, and the null-state quotient; Ramond states additionally obey the fermionic zero-mode constraint.