 Solution

First-class reduction. A regular set of independent constraints is first class when their Poisson brackets vanish on the constraint surface, equivalently locally $\{{\phi }_i,{\phi }_j\}=C_{ij}{}^k{\phi }_k$. The coefficients can be functions on phase space. With the generator $G={\sum }_i{\xi }^i{\phi }_i$, an infinitesimal canonical gauge transformation isEach independent first-class condition removes one phase-space dimension, and quotienting its independent gauge orbit removes another. Thus the regular physical phase space hasThis regular first-class phase-space reduction requires independent constraints and gauge directions. The count can fail at singular strata or for reducible constraints; first-class closure alone does not guarantee independence. For a concrete counterexample to counting redundant equations, take two canonical pairs and ${\phi }_1=p_1$, ${\phi }_2=2p_1$. Both equations are first class, but there is only one independent condition and one gauge direction. The surviving pair $(q^2,p_2)$ has dimension two, whereas blindly substituting $N=n=2$ gives zero.

Oscillator symplectic form. The center-of-mass pair $(x^m,p_m)$ describes translations and total momentum. The nonzero Fourier coefficients ${\alpha }_k^m$ describe the standing-wave modes compatible with free-end Neumann boundary conditions; their reality condition is ${\alpha }_{-k}={\alpha }_k^{\ast }$, and ${\alpha }_0^m=\sqrt{2{\alpha }^{\prime }}{p}^m$ in the standard normalization. For $k>0$, regarding ${\alpha }_k^m$ as a complex coordinate makes its conjugate momentum $i{\alpha }_{-k,m}/k$. Equivalently write ${\alpha }_k=(q_k+i{s}_k)/\sqrt{2}$. Up to a total derivative, its kinetic term is $-q_k\cdot {\dot{s}}_k/k$, a real canonical form. Inverting this oscillator symplectic form of an open string givesThe zero oscillator commutes with nonzero oscillators but is not independent of $p$: $\{x^m,{\alpha }_0^n\}=\sqrt{2{\alpha }^{\prime }}{\eta }^{mn}$. Brackets between the center pair and independent nonzero oscillators vanish.

The quadratic constraints obeyFor example, the two terms in the bracket with $L_n=\frac{1}{2}{\sum }_j{\alpha }_j\cdot {\alpha }_{n-j}$ give equal contributions after relabelling $j$. Applying this identity to both factors of $L_m$ givesThis classical Virasoro constraint algebra has no central term and closes on the constraints, hence is first class. For $G={\sum }_n{\xi }_{-n}{L}_n$, the oscillator gauge transformation isReality is respected when ${\xi }_{-n}={\xi }_n^{\ast }$.

Light-cone reduction and mass. Choose light-cone coordinates $X^{\pm }=(X^0\pm X^{D-1})/\sqrt{2}$; a vector square is $-2{\alpha }^{+}{\alpha }^{-}+{\mathbf{\alpha }}^2$, with $D-2$ transverse components. On the proposed gauge slice ${\alpha }_{k\ne 0}^{+}=0$, the variation becomesFor ${\alpha }_0^{+}\ne 0$, every nonzero-mode gauge parameter has an invertible coefficient, so the conditions locally fix the corresponding gauge freedom. Globally this is the usual patch in which $X^{+}$ is an admissible worldsheet clock; it is not a claim about strings for which that coordinate has turning points. The zero-mode reparameterization is left over.

On that slice the nonzero Virasoro constraints are linear in the longitudinal oscillators:No nonzero longitudinal oscillator remains independent. The residual action isThis residual mass-shell action in light-cone string gauge displays the remaining zero-mode constraint. At the quantum level normal ordering replaces its oscillator term by $N-a$. One can further set $x^{+}$ equal to time and solve for the light-cone Hamiltonian $p^{-}=[{\mathbf{p}}^2+(N-a)/{\alpha }^{\prime }]/(2p^{+})$.

Canonical quantization gives the transverse relationsHere $\hslash =1$, ${\alpha }_{k>0}$ annihilate the oscillator vacuum, and ${\alpha }_{-k}$ create excitations. Define $a_k^I={\alpha }_k^I/\sqrt{k}$ for $k>0$. The string level operator isThus it counts oscillator number weighted by mode number, and has nonnegative integer eigenvalues. The residual mass-shell condition givesThe constant $a$ is the normal-ordering constant of a string, or intercept, not an extra classical tension. In the usual Lorentz-invariant critical bosonic string, level one carries the $D-2$ transverse polarizations of a massless vector. A massive vector would need $D-1$ polarizations; the longitudinal one is not present. Lorentz consistency therefore requires that this vector level be massless, giving $a=1$. Equivalently the regularized transverse zero-point value $a=(D-2)/24$ gives $D=26$. This critical-vector assumption in the string intercept argument concerns the usual critical quantum theory. In $D=3$ the lone transverse level-one polarization transforms trivially under the transverse rotation group; a massive scalar interpretation is not excluded by counting. Thus the stated masslessness conclusion is not a consequence of polarization counting for arbitrary $D$. An arbitrary intercept in a transverse oscillator model also need not define the usual covariant critical theory.

The full self-dual massless spectrum. In the closed-string sector, $N$ and $\tilde{N}$ are the independent nonnegative integer oscillator levels of the two chiral sectors. The integer $n$ quantizes center momentum around the circle, $p_{\mathrm{circle}}=n/R$, and $w$ counts how many times the string winds it. At the self-dual circle $R=\sqrt{{\alpha }^{\prime }}$, zero mass requiresBoth levels are nonnegative, so their sum is at most two. Exhausting these possibilities gives the massless spectrum at the bosonic self-dual circle:

The last family is easy to miss because the uncompactified ground state is a tachyon; its positive compact energy cancels that negative contribution at these charges. There are no other possibilities: at level sum one, $n^2+w^2=2$ forces both charges to be $\pm 1$; at sum zero, $n^2+w^2=4$ gives exactly the four listed pairs. With $d=D-2$ transverse oscillators the number of independent massless polarizations is $d^2+4d+4=D^2$, hence 676 in the critical $D=26$ bosonic theory. The circle oscillator is included among the $d$ components; it should not be discarded when interpreting the lower-dimensional scalar states.