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The unprojected level-one states form a transverse vector representation and an exterior square. Fermionic anticommutation makes the two-fermion states antisymmetric. Together they restrict from an antisymmetric rank-two tensor of $SO(D-1)$, not a massive spin-two field.

Solving the massive subsidiary conditions. In light-cone coordinates, use $d{s}^2=-2d{X}^{+}d{X}^{-}+{\delta }_{IJ}d{X}^{I}d{X}^J$ and write $d=D-2$. The divergence condition readsThus, when ${\partial }_{-}$ is invertible,First apply this with $n=-$, then with $n=I$, and finally with $n=+$; symmetry supplies the mixed components already determined. The trace condition becomesThe independent components are $h_{--}$, $h_{-I}$ and the trace-free part of $h_{IJ}$. Under transverse rotations they form a scalar representation, a vector representation, and a symmetric traceless rank-two tensor. ThereforeThis is the light-cone decomposition of a massive spin-two field for $D\ge 3$. In $D=2$, the trace and divergence give $h_{+-}=h_{++}=0$ and ${\partial }_{+}{h}_{--}=0$. The massive wave equation then forces $h_{--}=0$, so there are no polarizations, consistent with the zero value of the printed count. The massive particle little group is $SO(D-1)=SO(d+1)$, and its symmetric traceless square branches asThe mixed components with the extra direction give the vector; one independent trace combination gives the scalar. These are exactly the polarizations of a massive spin-two field. The remaining independent components retain the massive Klein-Gordon equation.

Transverse bosonic modes and mass levels. The variables ${\alpha }_k^I$ are Fourier amplitudes of the physical transverse open-string mode expansion. Classically, reality requires ${\alpha }_{-k}^I=({\alpha }_k^I{)}^{\ast }$. The symplectic term in the action fixes their quantum commutators:With string tension convention ${\alpha }^{\prime }=(2\pi T{)}^{-1}$, the zero mode of the constraint givesThe longitudinal nonzero modes have already been removed in light-cone gauge in string theory. Quantum normal ordering introduces the string intercept $a$, giving the open bosonic string mass spectrumEach bosonic occupation number is a nonnegative integer, so $N$ is a nonnegative integer weighted by oscillator frequency.

Suppressing the common momentum label, the lowest light-cone levels of an open bosonic string areThe oscillator vacuum is annihilated by every positive ${\alpha }_k^I$. At level one there are only $d$ vector polarizations. For a Lorentz-consistent vector, these are the transverse polarizations of a massless particle, transforming under the rotation part of its massless particle little group. A massive vector would need $D-1$ polarizations, including a scalar under $SO(d)$ that is absent here. Thus the first bosonic vector level must be massless, fixing $a=1$.

At level two the commuting creation operators give a symmetric square. Its scalar trace and symmetric traceless square, together with the mode-two vector, are the massive-spin-two decomposition above. In the consistent bosonic theory they form one massive spin-two field with $m^2=1/{\alpha }^{\prime }$. The covariant equations describe its propagation while eliminating the redundant components. For $D=26$ the transverse counts are $300+24=324$, the symmetric traceless rank-two tensor dimension of $SO(25)$.

Half-integer fermionic modes. The Neveu–Schwarz sector has antiperiodic worldsheet Majorana fermions. Its Neveu–Schwarz fermionic oscillators obeyThe oscillator vacuum satisfies ${\alpha }_k^I|0;p⟩=0$ for $k>0$ and $b_r^I|0;p⟩=0$ for $r>0$. A negative fermion mode is a fermionic creation operator for a transverse worldsheet excitation. The Neveu–Schwarz level operator and mass condition areThe smallest positive frequency is $1/2$, so the only first-excited states areThe same vector-polarization argument requires them to be massless in Lorentz-consistent quantization, fixing $a_{\mathrm{NS}}=1/2$.

At $N=1$, ${\alpha }_{-1}^I|0;p⟩$ is a vector, while $b_{-1/2}^I{b}_{-1/2}^J|0;p⟩$ is the exterior square: interchanging the indices changes the sign and equal indices give zero. Together they branch from an antisymmetric tensor:Thus the Neveu–Schwarz level-one massive tensor has $(D-1)(D-2)/2$ polarizations and mass squared $1/(2{\alpha }^{\prime })$. It differs from spin two because the two-fermion tensor is antisymmetric and has neither the symmetric trace-free representation nor its scalar trace. At $D=10$, the count is $8+28=36$, compared with $44$ for a massive spin-two field. The specified states are before the GSO projection; the usual tachyon-removing GSO projection also removes this integer level.

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.

The level-two scalar in covariant string quantization can be chosen, for every $D$, asOn its three displayed structures, $L_1$ gives respectively $2,-4,2$ times $k\cdot {\alpha }_{-1}|p⟩$, and $L_2$ gives $D,-2,-4$ times the vacuum. The coefficients make both combinations vanish. The first two structures have norms $2D,8$ and cross inner product $-4$; the mode-two structure has norm $-4$ and is orthogonal to them. For $B=(D+4)/10$, $C=(D-1)/5$, this yieldsAbove 26 this is a physical negative-norm string state. Below 26 it is an extra positive-norm scalar, which cannot be discarded just to force the ordinary light-cone state count. At 26 it becomes null and can be removed. Thus level-two ghost absence gives $D\le 26$, whereas equivalence to the ordinary transverse spectrum requires $D=26$.

The critical scalar is also a Virasoro descendant. For $L_0|p⟩=-|p⟩$, the level-two scalar Virasoro null state candidateobeysAt $D=26$ it is physical and null, with $|S_{26}⟩=2|Z⟩$. At other dimensions it is not physical, so its norm must not be treated as a physical ghost test; the already-physical $|S_D⟩$ gives the correct test.

Finally, the covariant level-two oscillator space has dimension $D(D+3)/2$. The $D$ conditions from $L_1$ and one from $L_2$ leave $(D-1)(D+2)/2$ physical components. Quotienting $D-1$ vector null states and the critical scalar null state leavesthe massive spin-two field count and the level-two light-cone gauge in string theory count. The vector null descendants change the components of $\zeta$ orthogonal to $k$, and the critical scalar null descendant changes its component along $k$. Thus $\zeta$ can be set to zero. A representative then has $\zeta =0$, $k^m{ϵ}_{mn}=0$ and ${ϵ}^m{}_m=0$. The two approaches consequently agree on the massless level-one vector and massive level-two spin-two tensor for $a=1,D=26$.