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Take ${\eta }^{\alpha \beta }=\mathrm{diag}(-1,1)$ with coordinate zero equal to $\tau$ and coordinate one to $\sigma$. In the supplied representation setThey satisfy the Clifford algebra $\{{\rho }^{\alpha },{\rho }^{\beta }\}=2{\eta }^{\alpha \beta }1$. For anticommuting spinor components and an ordinary matrix $B$, relabelling the two summed indices givesNow $C^T=-C$, whereas $C{\rho }^0=-1$ and $C{\rho }^1=\mathrm{diag}(-1,1)$ are symmetric. Therefore the Majorana Grassmann bilinear interchange yieldsThese signs require fermionic, Grassmann-odd fields; commuting spinors would have the opposite interchange signs.

In flat conformal gauge, suppress the common factor $(4\pi {\alpha }^{\prime }{)}^{-1}$ and write the combined density as $\mathcal{L}=-{\partial }_{\alpha }X\cdot {\partial }^{\alpha }X+\bar{\psi }\cdot {\rho }^{\alpha }{\partial }_{\alpha }\psi$. The supersymmetry variation including its odd parameter is an even derivation. From $({\rho }^{\alpha }{)}^{T}C=-C{\rho }^{\alpha }$ we obtainAlso $(C{\rho }^{\alpha }{\rho }^{\beta }{)}^T=-C{\rho }^{\beta }{\rho }^{\alpha }$, so interchanging odd spinors gives $\bar{\chi }{\rho }^{\alpha }{\rho }^{\beta }ϵ=\overline{ϵ}{\rho }^{\beta }{\rho }^{\alpha }\chi$. The two density variations, with target contractions understood, are thereforeIn the second term the symmetric second derivative annihilates the antisymmetric part of ${\rho }^{\beta }{\rho }^{\alpha }$. Using $2{\eta }^{\alpha \beta }-{\rho }^{\beta }{\rho }^{\alpha }={\rho }^{\alpha }{\rho }^{\beta }$ now gives the exact off-shell identityThus the action is invariant when this boundary flux vanishes and the transformation preserves the field boundary conditions, for example with suitable falloff or compatible supersymmetric endpoint conditions. On a spatial circle, periodic fields admit the constant parameter; antiperiodic fermions with periodic bosons require the local superconformal treatment, since a nonzero constant parameter would make the boson variation antiperiodic. The constant parameter and ordinary derivatives describe rigid supersymmetry in flat gauge; a locally supersymmetric curved-worldsheet formulation includes the worldsheet gravitino. The result is the rigid worldsheet supersymmetry boundary term without using field equations.

For the commutator on $X$, keep the order of the odd parameters:For the fermion, the order is equally important. Moving ${ϵ}_2$ through both odd factors in $({\overline{ϵ}}_1{\partial }_{\alpha }\psi ){ϵ}_2$ incurs two minus signs. ConsequentlyTo check the spinor matrix explicitly, set ${ϵ}_1=(a,b{)}^T$, ${ϵ}_2=(c,d{)}^T$, with all four components anticommuting and monomials ordered $a,b,c,d$. ThenDirect multiplication gives ${\rho }^{\alpha }M+M{\rho }^{\alpha }=2V^{\alpha }1$. HenceVariation of the fermion action gives the Dirac equation ${\rho }^{\alpha }{\partial }_{\alpha }\psi =0$. On this equation the second term vanishes, provingThis is on-shell closure of rigid worldsheet supersymmetry; the action's invariance was off shell, but its fermionic transformation algebra closes in this field content only on shell.

The associated superconformal current is proportional to $\psi \cdot \partial X$, of conformal weight $3/2$. Its fermionic modes $G_r$ extend the stress-tensor generators to the N=1 super-Virasoro algebra:The modes are half-integral in the Neveu–Schwarz sector and integral in the Ramond sector. Their anticommutator giving a Virasoro generator is the conformal-mode counterpart of the translation commutator just derived. Ten coordinate bosons and ten Majorana fermions give matter central charge $c=10+10/2=15$, canceled by the usual reparametrization and superconformal ghosts in the critical RNS string.

Use target signature $(-,+,\dots ,+)$ and worldsheet signature $(-,+)$. The Polyakov action has local worldsheet diffeomorphism and Weyl invariance, together with global target-space Poincaré symmetry. Under a worldsheet coordinate change $\sigma \mapsto {\sigma }^{\prime }$, the embedding coordinates are scalars and the worldsheet metric is a tensor:A Weyl transformation acts as $g_{ab}\mapsto e^{2\rho (\sigma )}{g}_{ab}$ and leaves $X$ unchanged. In two dimensions $\sqrt{-g}$ scales by $e^{2\rho }$ and $g^{ab}$ by $e^{-2\rho }$, so the action is invariant. Target translations act by $X^{\mu }\mapsto X^{\mu }+a^{\mu }$, and Lorentz transformations by $X^{\mu }\mapsto {\Lambda }^{\mu }{}_{\nu }{X}^{\nu }$ with ${\Lambda }^{\mathsf{T}}\eta \Lambda =\eta$; they leave $g$ unchanged. Infinitesimally the latter have $\delta X^{\mu }={\omega }^{\mu }{}_{\nu }{X}^{\nu }$, where ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$.

Locally, the two diffeomorphism functions put the metric into the form $g_{ab}=e^{2\rho }{\eta }_{ab}$, and Weyl invariance removes the conformal factor. In this conformal gauge, the embedding coordinates are free scalar fields and satisfyThe metric equation must still be imposed after gauge fixing. Varying the ungauged metric gives the worldsheet stress-energy tensor constraintWith a dot and prime denoting the two coordinate derivatives, the independent Virasoro constraints areequivalently $(\dot{X}\pm X^{\prime }{)}^2=0$. Local conformal gauge leaves residual conformal transformations; on general worldsheet topologies it also leaves global moduli, rather than removing all metric information.

For target translations, differentiation of the Lagrangian with respect to ${\partial }_a{X}_{\mu }$ gives the Noether currentThe embedding equation is ${\partial }_a{j}^{a\mu }=0$. In conformal gauge, $j^{\tau \mu }={\dot{X}}^{\mu }/(2\pi {\alpha }^{\prime })$, so integration over the closed string givesPeriodic boundary conditions make its time derivative vanish. For an infinitesimal Lorentz transformation, the antisymmetric coefficient of ${\omega }_{\mu \nu }$ gives the current $j^{a,\mu \nu }=X^{\mu }{j}^{a\nu }-X^{\nu }{j}^{a\mu }$. Its divergence vanishes because the remaining product of embedding derivatives is symmetric in $\mu ,\nu$. Thus the Noether charge isThese are the target-space Noether charges of a string.

For the circular solution take $R>0$, and let ${\mathbf{e}}_r=(\cos \sigma ,\sin \sigma )$ and ${\mathbf{e}}_{\sigma }=(-\sin \sigma ,\cos \sigma )$. Its derivatives areEach spatial coordinate has equal second derivatives in $\tau$ and $\sigma$, while the time coordinate has both second derivatives zero. Furthermoreso both the field equations and the Virasoro constraints hold, including the collapse instants.

At target time $t=X^0=R\tau$, the string is a circle in the $X^1{X}^2$ plane of radius $R|\cos (t/R)|$. It contracts from radius $R$ to zero and re-expands; the negative cosine reverses material labels by a half-turn rather than creating a negative radius. The spatial configuration repeats after target time $\pi R$. This is a pulsating circular string. Its induced metric degenerates at collapse, but the conformal-gauge embedding and constraints give its continuation.

The integrals of $\sin \sigma$ and $\cos \sigma$ vanish, so the momentum isThe spatial velocity is radial, giving zero angular momentum density in the plane; all boost-charge terms integrate to zero as well. Hence $J^{\mu \nu }=0$ for every pair of indices.

At maximum expansion, all string elements are instantaneously at rest, and the length is $L_0=2\pi R$. Thus the rest mass per unit length, equal to the string tension, isAt other times the instantaneous circumference is $L=2\pi R|\cos \tau |$, and the elements have speed $|\sin \tau |$. Their Lorentz factor is $1/|\cos \tau |$, so the energy remains $TL/\sqrt{1-v^2}=2\pi TR$. The rest-energy density is $T$; the total rest-frame energy divided by the changing circumference is not constant. This distinguishes the requested intrinsic mass density from the larger energy density of moving string elements.

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.

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.

Use a mostly-plus target Minkowski spacetime metric and restore the string tension by $T=(2\pi {\alpha }^{\prime }{)}^{-1}$. After Wick rotation, the Polyakov action on a Riemann surface isThe Polyakov path integral integrates over embeddings and worldsheet metrics, dividing by worldsheet diffeomorphisms and Weyl transformations. For a consistent flat bosonic string theory, take the critical dimension of the bosonic string $d=26$. Conformal gauge turns the embedding fields into free scalars; its Faddeev-Popov determinant is represented by the bc system. The remaining integrations are over worldsheet moduli and external insertion points.

A tachyon vertex operator isThe worldsheet Green function is $⟨X^a(z)X^b(w)⟩=-({\alpha }^{\prime }/2){\eta }^{ab}\log |z-w{|}^2$. The operator product expansion with the holomorphic stress-energy tensor gives conformal weights $({\alpha }^{\prime }{k}^2/4,{\alpha }^{\prime }{k}^2/4)$. An integrated string vertex operator must have weights $(1,1)$, so the mass-shell condition is $k^2=4/{\alpha }^{\prime }$, or tachyonic mass parameter $m^2=-4/{\alpha }^{\prime }$.

At lowest order in the string coupling, the worldsheet is a Riemann sphere. For four external states, three insertion points can be fixed by a Möbius transformation. In the worldsheet ghost field description these three vertices carry $c\bar{c}$, while the fourth is integrated. The three-point worldsheet ghost correlator gives the squared product of the three fixed-point separations, precisely the sphere gauge fixing for four string vertices factor. Thus fixing three points is a gauge choice, rather than deleting three integrations without their Jacobian.

The integral over the constant embedding worldsheet zero mode gives $(2\pi {)}^d{\delta }^{(d)}({\sum }_i{k}_i)$. The remaining Gaussian integral, using normal ordering to remove self-contractions, gives the Koba-Nielsen factorSet $(z_1,z_2,z_3,z_4)=(0,z,1,\infty )$; the vertex at infinity is defined with its conformal-weight factor. Let all momenta be incoming and defineThe mass-shell condition and momentum conservation imply $s+t+u=-16/{\alpha }^{\prime }$. Since ${\alpha }^{\prime }{k}_1\cdot k_2=-4-{\alpha }^{\prime }s/2$ and similarly for the pair $(2,3)$, the sphere tachyon position integral becomesHere $\mathcal{N}$ depends on the normalization of external string vertex operators; it does not affect the kinematic dependence.

To evaluate the remaining worldsheet modulus, put $a=-1-{\alpha }^{\prime }s/4$ and $b=-1-{\alpha }^{\prime }t/4$. The complex beta integral isFor example, this identity follows by writing the two powers as Schwinger-parameter Gamma integrals, doing the two-dimensional Gaussian integral, and changing the two positive parameters to their sum and ratio. The complex beta integral first holds for $\mathrm{Re}a,\mathrm{Re}b>0$ and $\mathrm{Re}(a+b)<1$; the scattering result is its analytic continuation. The third gamma-function argument is $1-a-b=-1-{\alpha }^{\prime }u/4$. Absorbing $\pi$ into $\mathcal{N}$, the result is the tree-level four-tachyon amplitude, the Virasoro–Shapiro amplitude:It is symmetric in the three Mandelstam variables and has generic poles at $q=4(N-1)/{\alpha }^{\prime }$, matching the bosonic string mass spectrum. The formula is the leading term of string perturbation theory. Higher orders are constructed by the same Polyakov path integral prescription on higher-genus worldsheets, with four marked points and the corresponding ghost and worldsheet moduli measure; genus $g$ has coupling power $g_s^{2g+2}$.

Let $h_{\mu \nu }={\partial }_{\mu }{X}^a{\partial }_{\nu }{X}_a$ be the induced worldvolume metric and write $H={\gamma }^{\mu \nu }{h}_{\mu \nu }$. Varying the embedding in the auxiliary-metric brane action and integrating by parts givesThe boundary contribution, including its relative factor, isAt initial and final times one fixes the endpoint configurations or uses variations of compact support. At a spatial boundary, free target directions require Neumann boundary conditions $n_{\mu }{\gamma }^{\mu \nu }{\partial }_{\nu }{X}^a=0$; fixed directions require Dirichlet boundary conditions $\delta X^a=0$. Mixed conditions must make this boundary pairing vanish. A closed brane has no spatial boundary. Since the auxiliary worldvolume metric enters without derivatives, its variation produces no extra boundary term.

Using $\delta \sqrt{-\gamma }=-(1/2)\sqrt{-\gamma }{\gamma }_{\mu \nu }\delta {\gamma }^{\mu \nu }$, variation of the inverse worldvolume metric givesTaking the trace yields $(p-1)(H-p-1)=0$. For $p\ne 1$, this forces $H=p+1$, and substitution gives ${\gamma }_{\mu \nu }=h_{\mu \nu }$. For a nondegenerate timelike embedding, eliminating the auxiliary worldvolume metric therefore gives twice the worldvolume area, with the overall physical brane tension supplied by the action normalization.

For $p=1$, the traced metric equation is an identity, and the remaining equation only says $h_{\mu \nu }=(H/2){\gamma }_{\mu \nu }$. It fixes the worldsheet metric up to a Weyl transformation, rather than determining it uniquely. Indeed $\sqrt{-\gamma }{\gamma }^{\mu \nu }$ is Weyl invariant precisely in two worldvolume dimensions, and the constant term vanishes precisely at $p=1$. This is the Weyl-invariance exception for the string among branes. In a nondegenerate interior it gives the familiar classical equivalence of Polyakov and Nambu–Goto actions; degeneracies at a free string endpoint must be treated through the original equations.

For the open string, choose conformal gauge ${\gamma }_{\mu \nu }=\mathrm{diag}(-1,1)$ on $0\le \sigma \le \pi$. The embedding equation is the wave equation, and NN means ${\partial }_{\sigma }{X}^a=0$ at both spatial endpoints. For these operator formulas restore the conventional overall normalization $S=-TI/2$, with $T=(2\pi \alpha \prime {)}^{-1}$; the overall factor does not change the preceding classical equations. Its open-string mode expansion isReality requires $({\alpha }_n^a{)}^{†}={\alpha }_{-n}^a$. Canonical quantization givesThe metric equation is the vanishing of the worldsheet stress tensor: $({\partial }_{\tau }X+{\partial }_{\sigma }X{)}^2=({\partial }_{\tau }X-{\partial }_{\sigma }X{)}^2=0$. The two endpoint-compatible mode expansions contain the same string oscillator family. Their quadratic coefficients are the classical Virasoro constraints.

For the operators, use normal ordering with positive-index string oscillators as annihilators:Thus $({\partial }_{\tau }X\pm {\partial }_{\sigma }X{)}^2$ has modes $4{\alpha }^{\prime }{L}_m$ before the quantum ordering correction. The physical-state Virasoro conditions for an open string areOne imposes only the positive modes on kets, with the adjoint conditions on bras, as in Gupta-Bleuler quantization. Requiring every positive and negative mode to annihilate the same state would conflict with the Virasoro central extension. The intercept $a$ is the zero-mode ordering constant. The standard critical bosonic string theory has $d=26$ and $a=1$; in light-cone gauge in string theory the transverse zero-point energy gives $a=(d-2)/24$, while full anomaly-free Lorentz or BRST quantization fixes the critical values. The mass constraint is then ${\alpha }^{\prime }{M}^2=N-a$, with $N={\sum }_{n>0}{\alpha }_{-n}\cdot {\alpha }_n$.

To compute the Virasoro algebra, commute a quadratic generator with one string oscillator:These identities and the Jacobi identity imply that $[L_m,L_n]-(m-n)L_{m+n}$ commutes with every string oscillator and with the center-of-mass coordinates and momenta. Mode number permits a scalar term only for $m+n=0$. Its coefficient follows from the formal zero-momentum Fock vacuum, on which $L_0=0$. For $m>0$,The two possible string oscillator contractions giveThe timelike target coordinate still contributes one to this central charge: its two metric signs cancel in ${\eta }_{ab}{\eta }^{ab}=d$. ConsequentlyThis is the free-boson Virasoro central term. It vanishes for the three global conformal modes $m=0,\pm 1$.

To keep the intercept convention separate, define ${\mathcal{L}}_m=L_m-a{\delta }_{m,0}$. The Virasoro zero-mode shift changes the displayed central term toThe matter central charge here is $d$, not $d-2$. Covariant worldsheet ghost fields contribute $-26$; their inclusion cancels the anomaly at $d=26$, while the intercept is handled by the appropriate zero-mode and physical-state convention.

The ungauge-fixed Polyakov action has worldsheet diffeomorphism and Weyl transformation symmetries. They must survive quantization so that conformal gauge remains a valid gauge choice. In a curved target, the string nonlinear sigma model has a field-dependent coupling $g_{ab}(X)$. Its worldsheet Weyl anomaly is controlled by the renormalization of this coupling, the sigma-model beta function.

Restore the Euclidean normalization $(4\pi {\alpha }^{\prime }{)}^{-1}\int g_{ab}(X)\partial X^a\partial X^b$. Use a covariant background field expansion of a string sigma model, with geodesic fluctuations about a slowly varying embedding. The quadratic fluctuation operator contains the target Riemann curvature tensor coupled to two background embedding derivatives. The ultraviolet coincident propagator is $⟨{\xi }^a{\xi }^b{⟩}_{\mathrm{UV}}=\alpha \prime g^{ab}\log (\Lambda /\mu )$: the kinetic operator supplies $2\pi \alpha \prime$, and the two-dimensional momentum integral supplies $(2\pi {)}^{-1}\log (\Lambda /\mu )$. Contracting the curvature vertex with this propagator gives a logarithmic metric counterterm proportional to $\alpha \prime R_{ab}$. The resulting one-loop metric counterterm givesThe trace of the worldsheet stress tensor contains this coefficient multiplying $\partial X^a\partial X^b$, together with the curvature anomaly controlled by total central charge. Thus, with zero antisymmetric background, constant dilaton, and the critical matter/ghost system,The central-charge condition also requires $d=26$ if there is no extra internal conformal theory. This is a leading-order equation: higher-curvature terms enter the beta function at higher orders in ${\alpha }^{\prime }$, so a Ricci-flat metric alone is not a general all-orders quantum consistency criterion.

To derive the local T-duality, work where the spacelike Killing vector has nonzero norm $V>0$. Use dimensionless coordinates and ${\alpha }^{\prime }=1$ units for the duality formulas. The isometry makes the action depend on $z$ only through its derivatives. Replace those derivatives by an independent worldsheet one-form $A_{\mu }$, and enforce its flatness with a Lagrange multiplier $\tilde{z}$. With ${\eta }_{\mu \nu }=\mathrm{diag}(-1,1)$ and ${ϵ}^{\tau \sigma }=1$, the relevant first-order worldsheet duality action isVarying $\tilde{z}$ sets $dA=0$. Locally $A=dz$, recovering the original theory. For the other elimination, integrate the multiplier term by parts. Varying $A_{\nu }$ gives $V{A}^{\nu }={ϵ}^{\mu \nu }{\partial }_{\mu }\tilde{z}$, orSubstituting into the entire first-order worldsheet duality action, including the multiplier term, givesThe multiplier contribution is necessary for the sign of the dual kinetic term. This is the first-order action for Abelian worldsheet duality, with Buscher rulesThe coordinate map is ${\partial }_{\tau }\tilde{z}=V{\partial }_{\sigma }z$, ${\partial }_{\sigma }\tilde{z}=V{\partial }_{\tau }z$. The original $z$ equation is precisely the integrability condition for $\tilde{z}$; conversely the original identity $d(dz)=0$ becomes the dual equation. Thus this transformation exchanges an equation of motion with the Bianchi identity for an Abelian p-form. It reverses one chiral derivative and preserves the other.