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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.

Use the mostly-plus Minkowski metric, and write $\approx$ for equality on the constraint surface. Assume that the mechanical constraints are locally independent. They are first-class constraints whenThus their Poisson brackets vanish on the constraint surface, and their Hamiltonian flows preserve that surface. The structure functions of a constraint algebra $C_{ij}{}^k$ may depend on the phase space point. The finite real span of the constraints is a Lie algebra if it closes with constant structure coefficients, in a suitable choice of generators. The Jacobi identity then gives the usual conditions on the structure constants of a Lie algebra. With general structure functions the finite real span need not close, even though the Poisson bracket of all smooth functions is itself a Lie bracket.

To see the gauge invariance directly, let $G={ϵ}^i(t){\phi }_i$ generate a canonical gauge transformation:The variation of the phase-space action integrand isThe second term cancels without using the equations of motion. Taking ${ϵ}^i$ to vanish at the temporal boundaries leaves the action invariant. Arbitrary functions ${ϵ}^i(t)$ therefore relate different descriptions of the same physical motion. This reasoning also works with structure functions; constant structure coefficients are only needed for the finite-dimensional Lie algebra claim.

For a closed string, choose $0\le \sigma <2\pi$ and periodic fields. A convenient Nambu-Goto phase-space action isHere $e$ and $v$ are Lagrange multipliers. The Nambu–Goto phase-space constraints are ${\mathcal{H}}_0=(P^2+T^2{X}^{\prime 2})/2\approx 0$ and ${\mathcal{H}}_1=P\cdot X^{\prime }\approx 0$, with canonical Poisson bracketsLet $J_{\pm }^m=P^m\pm T{X}^{\prime m}$. Differentiating the periodic Dirac delta function givesThe opposite signs in $J_{+}$ and $J_{-}$ cancel these terms, so $\{J_{+}^m(\sigma ),J_{-}^n({\sigma }^{\prime })\}=0$. Replace the original constraints by the equivalent chiral densitiesTheir mixed Poisson brackets vanish. Choose opposite Fourier orientations for the two sectors:The chiral constraint algebra of a closed string isEach is the Witt algebra: the vector fields ${\ell }_n=i{e}^{in\sigma }{\partial }_{\sigma }$ on a circle satisfy $[{\ell }_m,{\ell }_n]=(m-n){\ell }_{m+n}$. Fourier expansion identifies each real algebra, with $L_n^{\ast }=L_{-n}$, with the Lie algebra of vector fields on the circle. The two commuting copies give ${\mathrm{Diff}}_1\oplus {\mathrm{Diff}}_1$, not a quantum central extension.

For an open string, allowed boundary conditions must remove the endpoint term in the variation of the action, consistently with the allowed endpoint variations. The spatial boundary term isIt expresses the open-string endpoint momentum flux. In the temporal gauge for a string $X^0=t$, take a boundary-adapted parametrization with $v=0$ at the ends. Fixing $X^i(t,a)=0$ gives $\delta X^i(t,a)=0$, a Dirichlet boundary condition. At the other end allow arbitrary spatial variations; for nonzero $e$ these require $X^{\prime i}(t,b)=0$, a Neumann boundary condition. Also $X^{\prime 0}=0$ in this temporal gauge for a string, so $X^{\prime m}(t,b)=0$. The constraint at this free-end string boundary condition reduces to $P^2=0$. Hamilton's equation $\dot{X}=eP+v{X}^{\prime }$ consequently gives ${\dot{X}}^2=0$ there. Since ${\dot{X}}^0=1$, the free endpoint has spatial speed one. This is the null motion of a free string endpoint.

A straight rotating string with one fixed endpoint supplies the required solution in at least two spatial dimensions. Set $e=1/T$, $v=0$, andTake $P_m=T{\dot{X}}_m$. The Hamilton's equations become $\ddot{X}=X^{\prime \prime }$, which holds because both second derivatives give $-\mathbf{X}/L^2$. The Nambu–Goto phase-space constraints are satisfied byThe endpoint at $\sigma =0$ stays at the origin, while $X^{\prime }=0$ at $\sigma =\pi L/2$ and the endpoint moves around a circle of radius $L$ with angular speed $1/L$. At each time the whole string lies on a straight radial segment. Its spatial proper length isThe velocity is everywhere perpendicular to the segment, so this also equals the sum of local rest-frame lengths. The induced worldsheet metric becomes degenerate at the null free endpoint, as expected for the limiting free-end solution.

Work away from $x=0$, where the inverse-square potential is singular. Variation of the phase-space action givesThe last equation is the constraint imposed by the Lagrange multiplier $e$. The canonical momentum conjugate to $t$ is $-E$, not $E$. Consequently the nonzero canonical Poisson brackets areIn particular,With the convention $\delta z=\alpha \{z,Q\}$, the Noether charge generating $\delta t=\alpha$ is $Q=-E$. Equivalently, the conserved physical energy is $E$; a convention that calls the energy the time-translation charge absorbs this minus sign into its parameter.

For the dilation charge,on the constraint surface. Its infinitesimal canonical transformations areFor constant $\alpha$, extending this by ${\delta }_{D}e=-\alpha e$ preserves the action: the kinetic terms are invariant and $E-H$ scales oppositely to $e$. Thus the dilation acts on time and position with their nonrelativistic relative scaling. Its Poisson bracket with the energy is

Put $C=E-H$. Direct variation of the Hamiltonian givessoThe supplied transformations are generated by the special conformal charge of inverse-square mechanicsChooseThen $\delta (eC)=0$ even for a time-dependent parameter. Expanding the kinetic-term variation, keeping the terms in $\dot{\beta }$, givesHenceFor a constant parameter only the boundary term remains, identifying $K$ as the Noether charge. Independently, the equations of motion giveUsing the same canonical Poisson bracket convention,Together, $\{D,E\}=-E$, $\{D,K\}=K$ and $\{E,K\}=2D$ form the sl2R Lie algebra of conformal mechanics. These are time translation, dilation and special conformal transformation. A useful normalization check is the Casimir elementon the constraint surface.

For the extended canonical variables with $\{t,E\}=-1$ and $\{x,p\}=1$, the charge $K=t^{2}E-txp+x^2/2$ generates a special conformal transformation. Its time derivative is $2te(E-H)$, so it is conserved on the constraint surface. With a time-dependent parameter $\beta$, the multiplier varies as $\delta e=-2te\beta$ and the action variation is $\dot{\beta }K-d(\beta x^2/2)/d\lambda$.