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Take ${\partial }_{\pm }=({\partial }_{\tau }\pm {\partial }_{\sigma })/2$ on the strip $0\le \sigma \le \pi$. Varying the Grassmann-odd worldsheet Majorana fermions gives ${\partial }_{-}{\psi }_1=0$, ${\partial }_{+}{\psi }_2=0$ and, up to the orientation of the strip boundary, the surface termAt each endpoint it vanishes if ${\psi }_1=\eta {\psi }_2$ with $\eta =\pm 1$, with the same relation on variations. A common sign can be absorbed by redefining one chiral field, so set ${\eta }_0=+1$. Since ${\psi }_1=f(\tau +\sigma )$ and ${\psi }_2=f(\tau -\sigma )$, the second endpoint imposes $f(u+2\pi )={\eta }_{\pi }f(u)$. Equal endpoint signs give the periodic Ramond sector, and opposite endpoint signs give the antiperiodic Neveu–Schwarz sector:These gluing choices preserve the Neumann-endpoint worldsheet supersymmetry conditions and give integer or half-integer worldsheet Majorana fermion modes, respectively.

First fix a normalization issue in the original PDF. With its stated canonical string oscillator algebra, the standard superconformal current mode iswithout the printed factor $1/2$. To check the Ramond supercurrent zero-mode square directly, pair positive and negative integer modes:The zero-mode Clifford relations give $({\alpha }_0\cdot {\psi }_0{)}^2={\alpha }_0^2/2$. Different $Q_m$ anticommute, and so do $Q_m$ with the zero-mode term. The two same-sign terms in $Q_m^2$ vanish by antisymmetry of the fermions and symmetry of the bosonic product. The cross terms giveThe second term comes from commuting ${\alpha }_m$ through ${\alpha }_{-m}$, so the frequency factor $m$ is essential. ThereforeHere $L_0^R$ uses vacuum-annihilating Ramond normal ordering; its string intercept is zero. A plane-to-cylinder central shift must be included consistently if a different zero-mode convention is used.

For the mode $F_r=G_r/2$ actually printed, the algebra instead gives $\{F_0,F_0\}=L_0^R/2$. Already the oscillator-free term gives $\{F_0,F_0\}={\alpha }_0^2/4$, whereas $2L_0^R={\alpha }_0^2$. Thus the requested identity is not valid with both that prefactor and the supplied string oscillator algebra. Removing the prefactor gives the identity just derived. Renaming $L_0^R/4$ as $L_0$ would fix this isolated equality but would spoil the usual Virasoro normalization and the subsequent mass formula. We use the standard normalized modes $G_r$ in the remaining physical constraints; their vanishing is unaffected by a common nonzero rescaling.

The RNS string physical-state conditions in the two sectors areThe last condition is the fermionic zero-mode constraint; its square implies the Ramond mass-shell condition. Physical null string states are quotiented. Matter central charge is $3d/2$, while the reparameterization worldsheet ghost fields and superconformal ghosts contribute $-26+11=-15$. Thus $d=10$ is the critical dimension of the RNS superstring, with $a_{NS}=1/2$ and $a_R=0$.

To solve the constraints, choose $X^{\pm }=(X^0\pm X^9)/\sqrt{2}$, so ${\eta }_{+-}=-1$, and work on a patch with ${\alpha }_0^{+}=\sqrt{2{\alpha }^{\prime }}{p}^{+}\ne 0$. In light-cone gauge in string theory fix all nonzero ${\alpha }_n^{+}$ and all ${\psi }_r^{+}$ to zero. For the Ramond sector, the printed restriction to nonzero fermion modes is not by itself complete: one must also reduce the longitudinal fermionic zero modes using the zero-mode supersymmetry constraint. The usual reduced light-cone description sets the remaining ${\psi }_0^{+}$ to zero as well and retains the transverse Clifford algebra. Covariantly this reduction is the Dirac equation on the ground-state particle polarization; it is not an extra unconstrained canonical string oscillator.

With this reduction the string oscillator generators have the formRepeated $i$ runs over the eight transverse directions. The coefficient in $L_n^{\perp }$ comes from the fermion worldsheet stress tensor $-\psi \partial \psi /2$; for $n=0$, positive and negative $r$ combine to give the occupation energy $r{\psi }_{-r}{\psi }_r$. Thus the light-cone super-Virasoro constraints are explicitly solved byHere $a$ is $1/2$ in the NS sector and zero in the R sector. The zero-mode equation says $2p^{+}{p}^{-}=p_i^2+(N-a)/{\alpha }^{\prime }$. All longitudinal excitations are consequently determined by transverse bilinears. The critical dimension and string intercepts are necessary for the resulting quantum theory to preserve the target Lorentz algebra.

The frequency-weighted Neveu–Schwarz level operator isThe PDF's displayed number operator omits the fermion frequency weights and the $r=1/2$ term. With the stated anticommutators that cannot be a level operator: it would assign no level to the first fermionic excitation. The bosonic term already includes its frequency through $[{\alpha }_n,{\alpha }_{-n}]=n$, whereas the fermionic term must include its frequency explicitly.

One independent way to obtain the string intercept is the transverse zero-point energy. Each periodic boson contributes $\frac{1}{2}{\sum }_{n\ge 1}n=-1/24$. Each antiperiodic real fermion contributes $-\frac{1}{2}{\sum }_{r\ge 1/2}r=-1/48$, with the minus sign due to fermionic normal ordering. For example a common exponential cutoff givesSubtracting the local divergent terms leaves $8(-1/24-1/48)=-1/2$, so the NS string intercept is $a=1/2$. In the Ramond sector the integer fermions instead contribute $+1/24$ per component and cancel the bosonic contribution, giving $a_R=0$.

The unprojected NS sector oscillator vacuum has $N=0$ andIt is a spacetime scalar tachyon. The first excitation ${\psi }_{-1/2}^i|0;p⟩$ has $N=1/2$ and is massless, with eight vector particle polarizations. Covariantly its particle polarization obeys $p\cdot ϵ=0$ modulo the longitudinal null particle polarization, leaving the same eight degrees of freedom.

The R sector oscillator vacuum is also massless, but it is degenerate because $\{{\psi }_0^{\mu },{\psi }_0^{\nu }\}={\eta }^{\mu \nu }$. The operators ${\Gamma }^{\mu }=\sqrt{2}{\psi }_0^{\mu }$ give the target Clifford algebra, and $G_0|u;p⟩=0$ becomes $p_{\mu }{\Gamma }^{\mu }u=0$. At nonzero null momentum this Dirac equation leaves sixteen unprojected on-shell spinor states. Equivalently the eight transverse zero modes give the sixteen-dimensional spin representation $8_s\oplus 8_c$ of $SO(8)$. Thus the Ramond ground state describes spacetime fermions, not a single scalar oscillator vacuum.

To define the GSO projection explicitly, let $f_{NS}$ count excited NS fermions and give its vacuum even parity. ThenIt retains odd fermion number, removes the tachyon, and retains the eight massless vector particle polarizations. In the Ramond sector let ${\Gamma }_{\ast }$ be transverse chirality and $f_R$ count nonzero-mode fermions. Choose one consistent chirality sign $s=\pm 1$ and useEach operator squares to itself. At the ground level $P_R$ retains one eight-dimensional spinor chirality, so the retained spectrum has eight massless bosonic and eight massless fermionic states. The parity convention can instead assign the NS vacuum odd parity and write a plus projector; the retained states are identical.

Equality continues at every mass level. Let $q$ count ${\alpha }^{\prime }{M}^2$, suppressing common momentum and endpoint-label factors, and set $D(q)={\prod }_{n\ge 1}(1-q^n{)}^8$. The projected string oscillator generating functions areThe Ramond factor eight follows because either chirality projector selects eight ground states for every nonzero-oscillator parity. The Jacobi abstruse identity makes the bracket equal to $16q^{1/2}{\prod }_{n\ge 1}(1+q^n{)}^8$, soThis proves level-by-level equality of bosonic and fermionic multiplicities. The appropriate spacetime supercharge from an RNS spin field relates the two sectors; the projection ensures that it acts within the retained state space. These sector and chirality conventions agree with the primary superstring lecture notes. Equal total counts alone are not the definition of the projection; they follow from the stated parity selection and the string oscillator identity.

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.

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.

A Majorana spinor equals its transform under charge conjugation, so its spinor components are not independent of their conjugates. In a real two-dimensional gamma matrix representation it can be taken real, with anticommuting Grassmann variables as components. Each target index labels a separate worldsheet Majorana fermion; it is not itself a target-space spinor.

Use worldsheet signature $(-,+)$ and the real gamma matricesTake $\bar{\psi }={\psi }^{T}C$ with $C={\gamma }^0$. Define the chirality matrix with the chosen worldsheet orientation byIt squares to one and anticommutes with each gamma matrix. Chirality is its eigenvalue, selected by the chiral projectors. Its overall sign is conventional.

Writing $\psi =({\psi }_{+},{\psi }_{-}{)}^T$, the fermion equation of motion ${\gamma }^{\mu }{\partial }_{\mu }\psi =0$ becomesHenceThe first profile travels toward increasing $\sigma$, and the second toward decreasing $\sigma$. Thus the choice of chirality orientation realizes the requested worldsheet chirality and propagation direction correspondence; reversing the orientation interchanges the labels.

For the rigid worldsheet supersymmetry variation, let $ϵ$ be a constant Grassmann-odd Majorana spinor. With the displayed conventions,The minus sign follows from $({\gamma }^{\mu }{)}^{T}C=-C{\gamma }^{\mu }$. A second identity from the Majorana Grassmann bilinear interchange is $\bar{\chi }{\gamma }^{\nu }{\gamma }^{\mu }ϵ=\overline{ϵ}{\gamma }^{\mu }{\gamma }^{\nu }\chi$. Keeping these Grassmann signs is essential.

The bosonic kinetic variation is $2i{\partial }_{\mu }{X}_a\overline{ϵ}{\partial }^{\mu }{\psi }^a$. The two fermionic kinetic variations, using the two identities above, give the full resultFor the second equality use ${\gamma }^{\nu }{\gamma }^{\mu }=2{\eta }^{\mu \nu }-{\gamma }^{\mu }{\gamma }^{\nu }$ and symmetry of the second embedding derivative. No field equation has been used. Thus the variation is an off-shell total derivative:This is the rigid worldsheet supersymmetry boundary term. It vanishes when the boundary flux vanishes, for example on a closed worldsheet with compatible fields and variations. On a spatial circle, a constant supersymmetry parameter must also respect the chosen spin structure. Periodic fermions allow this rigid transformation; antiperiodic fermions and a constant parameter would give an antiperiodic $\delta X$ and fail to preserve a periodic bosonic embedding. The local action identity remains valid, but global symmetry requires compatible boundary data. This is the spin-structure obstruction to constant worldsheet supersymmetry.

For the spectrum, assume the usual critical RNS string, physical-state constraints, a nonzero null target momentum, and the supersymmetric GSO projection. Analyze just one chiral sector. Its matter central charge is $d+d/2$; the bc system and superconformal ghosts contribute $-26+11$. Therefore $3d/2-15=0$ gives the critical dimension of the RNS superstring $d=10$. Light-cone gauge in string theory leaves eight transverse bosonic and eight transverse fermionic string oscillator directions.

In the Neveu–Schwarz sector, the normal-ordering intercept is $a_{\mathrm{NS}}=1/2$. The massless level is $N=1/2$ and has statesThe GSO projection retains these eight transverse vector polarizations while removing the tachyonic ground state. They are spacetime bosons, although the creating string oscillator is a worldsheet fermion. In covariant language transversality and the longitudinal null-state quotient leave $10-2=8$ polarizations.

In the Ramond sector, bosonic and fermionic zero-point energies cancel, so $a_{\mathrm{R}}=0$ and the ground states are massless. The eight transverse fermion zero modes obeyThis Ramond zero-mode Clifford algebra acts on a sixteen-dimensional ground-state space. Its two chirality subspaces each have dimension eight. The GSO projection keeps one, an $8_s$ or $8_c$ spinor of $\mathrm{SO}(8)$, the rotation subgroup of the massless little group. These are spacetime fermions. Equivalently a ten-dimensional Majorana-Weyl spinor has sixteen real components before the massless Dirac equation reduces the physical polarization count to eight.

Consequently the massless chiral RNS spectrum after GSO projection satisfiesThe GSO projection is essential: without it, the transverse Ramond sector would retain both eight-dimensional spinor chiralities. These are counts for one chiral sector, not the tensor-product counts of the full closed-string spectrum.

Use $x^{\pm }=(x^0\pm x^{D-1})/\sqrt{2}$, transverse coordinates $x^i$, $i=1,\dots ,d=D-2$, and the Minkowski metricThe relativistic particle phase-space action becomesIn the light cone gauge $x^{+}=t$, solve the mass-shell condition for $p_{+}$, assuming $p_{-}\ne 0$. The reduced phase-space action is $\int dt({\dot{x}}^{-}{p}_{-}+{\dot{x}}^i{p}_i-H)$, withThe last equality selects the future-directed momentum sector and makes positivity transparent. With $\hslash =1$ and $p_a=-i{\partial }_a$, the Schrödinger equation isThe inverse acts only on Fourier modes with nonzero $p_{-}$. Multiplication by $2p_{-}$ gives $2{\partial }_{-}{\partial }_{+}\Psi =({\Delta }_{\perp }-{\mu }^2)\Psi$. ThereforeThe light-cone Hamiltonian thus gives the same Klein-Gordon equation as covariant quantization.

For the massive two-form field, take $\mu \ne 0$. Apply ${\partial }^n$ to its field equation. Antisymmetry of $F_{mnp}$ makes ${\partial }^n{\partial }^m{F}_{mnp}=0$, soExpanding $F=dA$, the other divergence terms vanish by this condition, leaving $({\square }_D-{\mu }^2)A_{np}=0$. The light-cone decomposition of a massive two-form makes its dependent components explicit. The divergence equation isTaking $n=-$ and $n=i$, respectively, givesThe $n=+$ equation follows from these expressions: the two terms containing ${\partial }_{+}{\partial }_i{A}_{-i}$ cancel and ${\partial }_i{\partial }_j{A}_{ji}=0$. Consequently $A_{-i}$ and $A_{ij}$ are independent, each satisfying the Klein-Gordon equation with mass $\mu$. The number of independent particle polarizations isThis is the exterior square of the vector representation of the massive little group $SO(D-1)$. In the analogous Proca equation, ${\partial }^m{A}_m=0$ determines $A_{+}$ from $A_{-}$ and $A_i$, leaving $D-1$ components. A massive field has no gauge freedom that would justify setting these longitudinal components to zero. If $\mu =0$, instead use the two-form gauge field symmetry $A\mapsto A+d\Lambda$: the light-cone gauge for a two-form removes $A_{-m}$, leaving $\left(\genfrac{}{}{0ex}{}{D-2}{2}\right)$ transverse particle polarizations. The massive and massless counts are different.

In the closed-string mode expansion, $x^m,p_m$ are center-of-mass canonical variables, while ${\alpha }_k^i,{\tilde{\alpha }}_k^i$ are independent left- and right-moving transverse string oscillators. Their complex conjugates are ${\alpha }_{-k}^i,{\tilde{\alpha }}_{-k}^i$. The two zero-mode Lagrange multipliers impose the remaining mass-shell condition and closed-string level matching. The string level operators areTheir quantum definitions use normal ordering. The symplectic terms in the phase-space action givewith all brackets between distinct sectors zero. The nonzero-index string oscillators obey ${\alpha }_n^{i†}={\alpha }_{-n}^i$ and similarly for the right-moving sector. Define the momentum-labelled oscillator vacuum byFor $k>0$, $a_k^i={\alpha }_k^i/\sqrt{k}$ has $[a_k^i,a_l^{j†}]={\delta }_{kl}{\delta }^{ij}$. HenceStarting with $N|0;p⟩=0$, a finite product with $r_{ki}$ creation operators of mode $k$ has eigenvalue ${\sum }_{k,i}k{r}_{ki}$. The Fock space is generated by these products; both level operators have nonnegative integer eigenvalues. This establishes the integer string oscillator level property. Subtracting their physical zero-mode constraints enforces $N=\tilde{N}$.

There is a distinction between the displayed classical zero modes and their quantum constraints. With the normal-ordering constant of a string $a$, these areAt the massless first closed-string level, the states areTheir transverse polarization tensor splits into a symmetric trace-free part, an antisymmetric part, and its trace. These are the graviton, Kalb–Ramond field, and dilaton, with respective particle polarization counts $d(d+1)/2-1$, $d(d-1)/2$, and one. They have the transverse little group representations of massless particles. In a Lorentz-consistent bosonic string theory, the first chiral level is a massless vector, not a massive vector with one missing physical polarization; the closed-string products are therefore massless. This fixes $a=1$. Equivalently, regularized transverse zero-point energy gives $a=(D-2)/24$, and Lorentz consistency fixes the critical dimension of the bosonic string $D=26$.

It follows that the bosonic string mass spectrum isThe ground state has $M_0^2=-8\pi T$ and is a tachyon; level one is massless; for $N\ge 2$ the mass is $M_N=\sqrt{8\pi T(N-1)}$. The masslessness claim uses the consistent quantum theory, rather than an unshifted reading of the classical $L_0$.

A massive two-form at closed-string level two is present. To see it without confusing it with the level-one massless Kalb–Ramond field, the level-two states in one chiral sector areThey have $d+d(d+1)/2=(D-1)D/2-1$ components and assemble into the symmetric traceless square $S_0^{2}V$ of the massive little group vector space $V={\mathbb{R}}^{D-1}$. The full closed-string level is $S_0^{2}V\otimes S_0^{2}V$. For two symmetric trace-free matrices $S,\tilde{S}$, the mapis an equivariant map onto antisymmetric matrices. To verify surjectivity, take $S$ diagonal with distinct entries in positions $a,b$ and $\tilde{S}$ with only its symmetric $ab$ entry nonzero. Their commutator gives the $ab$ antisymmetric basis element. Finite-dimensional representations of the compact little group are completely reducible, so this quotient representation is also a subrepresentation. It has exactly $\left(\genfrac{}{}{0ex}{}{D-1}{2}\right)$ particle polarizations and is described by the massive field equation with ${\mu }^2=8\pi T$. At $D=26$ this gives 300 particle polarizations, consisting in light-cone coordinates of 24 components $A_{-i}$ and 276 components $A_{ij}$.