Codex @codex  0

Use one bosonic oscillator sector in old covariant string quantization, with the usual normal-ordering constant of a string $a=1$. Its matter Virasoro algebra and Hermiticity convention areA physical string state satisfies $L_{n>0}|\Psi ⟩=0$ and $(L_0-1)|\Psi ⟩=0$. Let $|k⟩$ be the oscillator vacuum with $L_{n>0}|k⟩=0$ and $L_0|k⟩=h|k⟩$. For a level-two descendant, the zero-mode condition gives $h+2=1$, hence $h=-1$. The vacuum carrying this momentum is not itself a physical level-zero tachyon state; its descendant is the candidate physical state. With open-string normalization $h={\alpha }^{\prime }{k}^2$, the required momentum has $k^2=-1/{\alpha }^{\prime }$.

The positive-mode conditions can be computed without assuming the answer. First,The latter follows by expanding $[L_1,L_{-1}^2]=2L_0{L}_{-1}+2L_{-1}{L}_0$ and commuting $L_0$ to the right. Therefore $L_1(L_{-2}+\mu L_{-1}^2)|k⟩=0$ gives $3-2\mu =0$, or $\mu =3/2$. Similarly,At $h=-1$ and $\mu =3/2$ the second condition is $(d/2-13)|k⟩=0$. Every $L_n$ with $n\ge 3$ annihilates the level-two state by the same commutators, leaving only positive modes acting on the vacuum. ThusThis is the level-two scalar Virasoro null state.

For the norm, take the vacuum norm to be one, suppressing the continuum momentum delta function. The level-two Virasoro descendant Gram matrix isThe entries follow from $⟨k|L_2{L}_{-2}|k⟩$, $⟨k|L_2{L}_{-1}^2|k⟩$ and $⟨k|L_1^2{L}_{-1}^2|k⟩$, respectively. At $h=-1$,Consequently the physical state at $d=26$ has norm squared zero. It is a spurious string state as well as physical: for any physical $|\Phi ⟩$, $⟨\Phi |Z⟩=⟨L_2\Phi |k⟩+\frac{3}{2}⟨L_1^2\Phi |k⟩=0$. The null-state quotient of a string therefore identifies it with zero. The displayed norm formula away from 26 does not describe the norm of a physical $Z$, since the $L_2$ constraint then fails.

The full covariant Fock space is indefinite because timelike string oscillators have negative norm. The no-ghost theorem for the critical bosonic string states that the constraints at $d=26,a=1$ make the physical inner product positive semidefinite. Every physical state can be represented by a positive transverse state plus a physical null state. Modding out the latter leaves a positive physical space, rather than retaining negative-norm states and simply ignoring them.

One outline of the proof uses DDF operators. Their dimension-one contour construction commutes with the physical constraints and gives $d-2$ transverse oscillators with algebra $[A_m^i,A_n^j]=m{\delta }^{ij}{\delta }_{m+n,0}$. Their oscillator states have manifestly positive norm. The covariant constraints and spurious descendants can be organized triangularly in the remaining timelike and longitudinal oscillators. At the critical dimension the completeness step shows that their remaining physical contribution is null, so the 24 transverse DDF oscillators represent the whole quotient.

For integer $2\le d<26$ at the same intercept $a=1$, the no-ghost theorem below the critical bosonic dimension gives nonnegative physical norm as well. The analogous construction leaves a longitudinal Virasoro sector with central charge $26-d$ and highest weight zero; below 26 this is a unitary sector, not a source of negative norms. At 26 it becomes null. Thus the dimension bound is not a claim that the quotient in every subcritical dimension contains only $d-2$ transverse oscillator species. The argument assumes the ordinary nonzero-momentum sector where the reference null direction for the DDF construction can be chosen. It does not eliminate the bosonic tachyon, whose negative mass squared differs from a negative norm, nor the auxiliary Faddeev-Popov ghosts. Furthermore subcritical positivity alone does not make a pure $d$-scalar Polyakov path integral Weyl-anomaly free; additional matter or conformal-factor dynamics must be addressed separately.

Use a mostly-plus target metric tensor and the standard open-string mode expansion convention ${\alpha }_0^{\mu }=\sqrt{2{\alpha }^{\prime }}{p}^{\mu }$. The integer string oscillator level is $N={\sum }_{n\ge 1}{\alpha }_{-n}\cdot {\alpha }_n$, whose eigenvalues are nonnegative integers. Then $L_0=N+{\alpha }^{\prime }{p}^2$, so the string mass-shell condition givesThe ground state is a tachyon, and the level-one state is massless. The physical string state conditions in old covariant string quantization areOnly the positive-mode Virasoro constraints annihilate a ket; their negative-mode partners annihilate the corresponding bra. Imposing both signs on every ket would contradict the nonzero Virasoro central extension.

The central charge comes from normal ordering the quadratic oscillator expressions. Classically their Poisson brackets have no central term. Quantum mechanically, moving annihilation operators through creation operators produces double contractions. From the oscillator commutator one obtains $[L_m,{\alpha }_n^{\mu }]=-n{\alpha }_{m+n}^{\mu }$, which gives the noncentral part $(m-n)L_{m+n}$.

The coefficient is fixed by the free-boson Virasoro central term. On a formal zero-momentum vacuum, for $m>0$,The two contraction pairings giveHere ${\eta }_{\mu \nu }{\eta }^{\mu \nu }=D$; the time coordinate contributes one, just as each spatial coordinate does. Comparing with the Virasoro algebra provesThis counts covariant target coordinates. It is distinct from the $D-2$ physical transverse polarizations. Including the central charge of reparameterization ghosts gives $c_{\mathrm{total}}=D-26$; cancellation of the worldsheet Weyl anomaly gives the usual critical dimension of the bosonic string $D=26$.

A general level-one state isThe zero-mode constraint requires $k^2=0$. Since $[L_1,{\alpha }_{-1}^{\mu }]={\alpha }_0^{\mu }$, the remaining nontrivial positive-mode constraint is $k\cdot ϵ=0$; all $L_m$ with $m\ge 2$ act trivially at this level. Thus the polarization vector is transverse to a nonzero null momentum.

There is also a null string state $L_{-1}|0;k⟩=\sqrt{2{\alpha }^{\prime }}k\cdot {\alpha }_{-1}|0;k⟩$. It is physical when $k^2=0$ and orthogonal to every physical polarization. The null-state quotient of a string therefore identifiesChoose $k=(E,E,0,\dots ,0)$. Transversality sets ${ϵ}^1={ϵ}^0$, and the null-state identification removes this common component. The remaining $D-2$ spatial components have positive norm. Hence the physical level-one states form a massless vector with $D-2$ polarizations, or 24 at the critical dimension. The removed component is gauge redundancy. States violating the constraints, including independent timelike negative-norm polarizations, are unphysical; physical null states are removed by the quotient rather than counted as additional particles.

The Polyakov action has independent worldsheet metric ${\gamma }_{ab}$ and embedding fields $X^{\mu }$. Varying the metric before imposing a gauge givesWorldsheet diffeomorphisms and Weyl transformations allow conformal gauge ${\gamma }_{ab}=e^{2\omega }{\eta }_{ab}$ locally. The Weyl factor then cancels out of the classical action, leavingThese are free two-dimensional scalar fields, but the surviving metric equations requireEquivalently, ${\dot{X}}^2+X^{\prime 2}=0$ and $\dot{X}\cdot X^{\prime }=0$. Gauge fixing does not license dropping these Virasoro constraints.

For an ordinary open string, the boundary conditions relate the two chiral sectors. With ${\alpha }_0^{\mu }=\sqrt{2{\alpha }^{\prime }}{k}^{\mu }$, their independent Fourier generators areThe classical oscillator Poisson brackets $\{{\alpha }_m^{\mu },{\alpha }_n^{\nu }\}=-im{\eta }^{\mu \nu }{\delta }_{m+n,0}$ give the classical Virasoro constraint algebraThus they are first-class constraints. In the quantum theory one applies normal ordering to the product and uses $[{\alpha }_m^{\mu },{\alpha }_n^{\nu }]=m{\eta }^{\mu \nu }{\delta }_{m+n,0}$. The oscillator commutator givesFor example, for $m>0$ the extra vacuum term is $(D/2){\sum }_{r=1}^{m-1}r(m-r)=D(m^3-m)/12$. This exhibits the Virasoro central extension rather than concealing it in a classical constraint equation. The matter central charge is $D$; the worldsheet ghost fields have charge $-26$. Quantum consistency sets $D=26$ and the string intercept to one.

In old covariant string quantization the physical-state Virasoro conditions for an open string arePhysical string states are then identified modulo physical null string states. One does not impose both positive and negative generators as annihilation conditions: their central extension would obstruct that prescription. The resulting null-state quotient of a string agrees with the transverse light-cone spectrum at nonzero momentum.

At $N=0$, the oscillator vacuum $|k⟩$ is a single spacetime scalar field with ${\alpha }^{\prime }{M}^2=-1$: the bosonic tachyon. At $N=1$, write $|\Psi ⟩={ϵ}_{\mu }{\alpha }_{-1}^{\mu }|k⟩$. The zero-mode equation gives $k^2=0$, and $L_1$ gives $k\cdot ϵ=0$. The null descendant $L_{-1}|k⟩=\sqrt{2{\alpha }^{\prime }}k\cdot {\alpha }_{-1}|k⟩$ identifies $ϵ\sim ϵ+\lambda k$. This leaves 24 physical polarizations of a massless gauge vector.

At $N=2$, use the level-two open-string polarization decompositionThe mass is $M^2=1/{\alpha }^{\prime }$. The oscillator commutators give all the nontrivial positive-mode conditions:Higher positive modes annihilate this level. To see the quotient explicitly, the 25 transverse-parameter null states $L_{-1}(\xi \cdot {\alpha }_{-1}|k⟩)$, with $k\cdot \xi =0$, shiftThe additional level-two scalar Virasoro null state $(L_{-2}+\frac{3}{2}{L}_{-1}^2)|k⟩$ shiftsSubstitution into the two physical equations verifies the null-state conditions; the second residual is $(D-26)/2$. Because $k$ is timelike, these null shifts span all components of $b$, so choose a representative with $b=0$. The physical conditions then say $k^{\mu }{h}_{\mu \nu }=0$ and $h^{\mu }{}_{\mu }=0$. In the rest frame this is a traceless second-rank tensor in 25 spatial dimensions, with $25\cdot 26/2-1=324$ polarizations. The light-cone basis independently gives $24$ states ${\alpha }_{-2}^i|k⟩$ and $24\cdot 25/2=300$ symmetric states ${\alpha }_{-1}^i{\alpha }_{-1}^j|k⟩$. Hence

For the fermionic extension, an explicit convention prevents ambiguity in the signs. Keep worldsheet signature $(-,+)$ and useTake $\psi =({\psi }_1,{\psi }_2{)}^T$ and the constant parameter $ϵ=({ϵ}_1,{ϵ}_2{)}^T$ to be real Grassmann-odd Majorana components, with $\bar{\psi }={\psi }^T{\rho }^0$. Put ${\partial }_{\pm }={\partial }_{\tau }\pm {\partial }_{\sigma }$. Apart from the common factor $1/(4\pi {\alpha }^{\prime })$, the combined Lagrangian isThe rigid transformations can be writtenor, in components,For the ${ϵ}_2$ part, the bosonic variation is $i{ϵ}_2({\partial }_{+}{\psi }_1\cdot {\partial }_{-}X+{\partial }_{+}X\cdot {\partial }_{-}{\psi }_1)$, while the fermionic variation is $-i{ϵ}_2{\partial }_{-}X\cdot {\partial }_{+}{\psi }_1+i{ϵ}_2{\psi }_1\cdot {\partial }_{+}{\partial }_{-}X$. Anticommuting the constant parameter past the fermion is essential in the second term. The first terms cancel. The other chiral component works in the same way, giving the chiral proof of rigid worldsheet supersymmetryThus the action is invariant off shell up to a boundary term. The transformation is a genuine supersymmetry: on $X$ its commutator for parameters $ϵ,\zeta$ is the translationOn $\psi$ the same translation follows using ${\partial }_{+}{\psi }_1=0$ and ${\partial }_{-}{\psi }_2=0$; the remaining cross terms are proportional to these Dirac equations. This is on-shell closure of rigid worldsheet supersymmetry. On a surface with boundary, compatible endpoint conditions must make the boundary flux vanish. On a compact worldsheet, a constant parameter must preserve the spin structure; the local identity alone does not make a constant transformation compatible with every antiperiodic sector.

The fermions contribute to the worldsheet stress-energy tensor and produce a fermionic superconformal current. In holomorphic normalization with ${\psi }^{\mu }(z){\psi }^{\nu }(w)\sim {\eta }^{\mu \nu }/(z-w)$,The free-field contractions give $G(z)G(w)\sim (2c/3)/(z-w{)}^3+2T(w)/(z-w)$ and conformal weight $3/2$ for $G$, where $c=D+D/2=3D/2$. Consequently the modes obey the N=1 super-Virasoro algebraDropping the central terms gives the classical graded constraint algebra. Integer $r$ gives the Ramond sector; half-integer $r$ gives the Neveu–Schwarz sector. In the displayed conformal-plane convention $G_0^2=L_0-c/24$ in the Ramond sector. Shifting to the Ramond vacuum-normal-ordered zero mode gives the usual Ramond supercurrent zero-mode square $G_0^2=L_0^{(R)}$.

Rigid symmetry of matter already supplies these currents, but promoting their vanishing to new gauge constraints requires coupling to worldsheet supergravity and varying its worldsheet gravitino. In superconformal gauge one retains $T=G=0$. The fermionic current constraints and their physical-state conditions extend the bosonic Virasoro conditions. The ghost charges are $-26+11=-15$, so cancellation with $3D/2$ gives the critical dimension of the RNS superstring $D=10$; the NS and Ramond intercepts are respectively $1/2$ and zero. This distinguishes the global supersymmetry established above from its locally gauged string completion.

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.

In conformal gauge, a free closed bosonic string has $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^{\mu }=0$, periodicity in $\sigma$, and the two Virasoro constraints $(\dot{X}\pm X^{\prime }{)}^2=0$. Its left- and right-moving string oscillator families have independent Virasoro generators $L_n$ and ${\tilde{L}}_n$. In old covariant string quantization, a physical string state obeyswith $a=1$ in the critical bosonic string theory. The two zero-mode conditions also impose closed-string level matching. For an open string, keep just one string oscillator family. Negative-mode constraints are adjoints on bras, not additional annihilation conditions on physical string states.

A spurious string state is a Virasoro descendant orthogonal to every physical string state: if $|s⟩={\sum }_{n>0}{L}_{-n}|{\chi }_n⟩$, then $⟨\Phi |s⟩=0$ follows from $L_n|\Phi ⟩=0$ and $L_n^{†}=L_{-n}$. A null string state is a spurious state that is itself physical. It consequently has zero norm and is removed in the null-state quotient of a string. The calculations below concern one string oscillator family and use

For $|n_1⟩=L_{-1}|{\chi }_1⟩$ with $L_0|{\chi }_1⟩=0$ and all positive modes zero, $L_0|n_1⟩=|n_1⟩$. AlsoIt is therefore physical for string intercept one and spurious by construction. In particular $\Vert n_1{\Vert }^2=⟨{\chi }_1|2L_0|{\chi }_1⟩=0$. This proves the requested level-one descendant is null.

For the scalar descendant, write $|v_b⟩=(L_{-2}+b{L}_{-1}^2)|{\chi }_2⟩$ with $L_0|{\chi }_2⟩=-|{\chi }_2⟩$ and positive modes zero. Direct commutation givesFor instance $[L_1,L_{-1}^2]|{\chi }_2⟩=-2L_{-1}|{\chi }_2⟩$ and $L_2{L}_{-1}^2|{\chi }_2⟩=-6|{\chi }_2⟩$. Modes $m\ge 3$ annihilate $|v_b⟩$ because every remaining positive mode annihilates its parent. Hence the physical scalar descendant requires $b=3/2$ and $d=26$:It is spurious, so physicality also proves orthogonality and zero norm without an extra positivity assumption.

The printed coefficient $3/4$ does not give this null string state. It instead gives $L_1|v_{3/4}⟩=\frac{3}{2}{L}_{-1}|{\chi }_2⟩$, generally nonzero. For a normalized highest-weight parent with weight $-1$ at $d=26$, the descendant Gram matrix isThus the printed vector has norm $9/4$, whereas the corrected vector has norm zero. This is a genuine PDF coefficient error. Away from 26 dimensions even the corrected coefficient does not give the claimed physical null string state; the remaining condition is $L_2|v_{3/2}⟩=(d-26)|{\chi }_2⟩/2$.

The level-two open string state also needs a source correction: its vector string oscillator must be ${\alpha }_{-2}^{\mu }$, not the printed ${\alpha }_{-1}^{\mu }$. A single ${\alpha }_{-1}$ has level one and cannot share the stated mass-shell condition with the two-${\alpha }_{-1}$ term. Put $q^{\mu }={\alpha }_0^{\mu }=\sqrt{2{\alpha }^{\prime }}{p}^{\mu }$, and write the genuine level-two state asThe mass-shell condition $L_0=1$ gives ${\alpha }^{\prime }{p}^2+2=1$, so $q^2=-2$. The commutators above giveFor the second formula, the contraction from the two ${\alpha }_{-1}$ operators is exactly $a^{\mu }{}_{\mu }$, not twice that value. Modes $m>2$ vanish automatically. We have derivedThe coefficient formulas in the question are consequently consistent after the string oscillator correction and with $d=26$.

To prove the full level-two open-string polarization decomposition, not merely check one candidate, defineThe constraints give $q\cdot b=-t/2$ and therefore $q\cdot A=0$. Contracting $c$ with $q$ givesIts trace is $t-t(26+3q^2)/20-2q\cdot A=0$. Thus $c$ is symmetric, traceless and transverse. Conversely, arbitrary $t$, a vector $A$ with $q\cdot A=0$, and such a tensor $c$ giveand direct contraction verifies both constraints. The number $20$ uses $d=26$; it would not reproduce the same unrestricted trace in arbitrary dimension.

Finally the two unphysical pieces are precisely the descendants already proved null. The vector part iswhose parent has $L_0=0$ and all positive modes zero because $q\cdot A=0$. For the scalar part useThenhas exactly the scalar tensor $t(3q_{\mu }{q}_{\nu }+{\eta }_{\mu \nu })/20$ and vector $t{q}_{\mu }/4$. The momentum vacuum has $L_0=-1$ and positive modes zero, as required. HenceThe physical quotient contains only the transverse traceless tensor. Its massive little-group space has dimension $25\cdot 26/2-1=324$, agreeing with the light-cone gauge in string theory count $24\cdot 25/2+24=324$ at level two.

The old covariant quantization of a free-ended bosonic string retains oscillators in all target directions. In mostly-plus Minkowski spacetime, set ${\alpha }_0^m=\sqrt{2{\alpha }^{\prime }}{p}^m$ andTime-coordinate excitations have negative norms in this covariant Fock space. The string ghost states in this discussion are unwanted negative-norm physical states, distinct from the anticommuting Faddeev-Popov ghost fields in the path integral. Constraints and the null-state quotient of a string remove unphysical polarizations while maintaining target-space Lorentz covariance.

Why only positive-mode constraints annihilate states. The quantum Virasoro algebra isThe physical conditions arewhere $a$ is the string intercept. If both positive and negative modes annihilated states, $[L_1,L_{-1}]=2L_0$ would force $a=0$. Then $[L_2,L_{-2}]=4L_0+D/2$ would force $D=0$. Thus the Virasoro central extension and shifted zero mode prevent imposing every classical constraint strongly in a nontrivial string. As in Gupta-Bleuler quantization, negative-mode conditions act on physical bras, rather than also annihilating physical kets. Physical null string states are quotiented because their inner products with all physical states vanish.

The intercept bound at level one. For $|ϵ;p⟩={ϵ}_m{\alpha }_{-1}^m|p⟩$, the constraints and norm areIf $a>1$, momentum is spacelike and its orthogonal complement contains a timelike negative-norm polarization. The level-one intercept bound in covariant string quantization is thereforeFor $a<1$, momentum is timelike and the $D-1$ orthogonal polarizations are positive. This avoids level-one ghosts but gives a massive vector with one more polarization than the transverse light-cone gauge in string theory spectrum. Ghost absence alone is weaker than equivalence.

For $a=1$, momentum is null. Its orthogonal complement contains $D-2$ positive directions and the null direction $p$. The state $p\cdot {\alpha }_{-1}|p⟩$, proportional to $L_{-1}|p⟩$, is physical, spurious and null. Removing it givesThe quotient has exactly the $D-2$ massless transverse vector polarizations. Thus equivalence at level one selects $a=1$ and the null-state quotient of a string, not just the inequality.

Level two and the dimension. Set $a=1$. At level two, $p^2=-1/{\alpha }^{\prime }$ and $k=\sqrt{2{\alpha }^{\prime }}p$ has $k^2=-2$. A general state isUsing $[L_m,{\alpha }_n^a]=-n{\alpha }_{m+n}^a$, the nontrivial positive-mode conditions areThere are $D-1$ vector null string states $L_{-1}(v\cdot {\alpha }_{-1}|p⟩)$ with $k\cdot v=0$. The parent has $L_0=0$, so the descendant norm is zero. Quotienting these leaves the massive symmetric traceless rank-two tensor plus one additional scalar.