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The Polyakov action introduces an independent worldsheet metric $h_{ab}$:Its redundancies are worldsheet diffeomorphisms and Weyl transformations. A Polyakov path integral over all metrics without dividing by this gauge volume would count physically equivalent configurations repeatedly. Fixing conformal gauge therefore needs the Faddeev-Popov determinant, not merely the substitution of a flat metric into the action.

After using the Weyl transformation to remove the trace of a metric variation, the infinitesimal worldsheet diffeomorphism ghost operator isThe Faddeev-Popov determinant of this operator is represented by anticommuting fields: a vector Faddeev-Popov ghost $c^a$ and a symmetric traceless Faddeev-Popov antighost field $b^{ab}$. In Euclidean coordinates their worldsheet ghost action is, up to a consistent overall normalization,The two chiral copies of this bc system have conformal weights $(2,-1)$ and anticommutators $\{b_m,c_n\}={\delta }_{m+n,0}$. They are auxiliary fields for the gauge determinant, not additional spacetime particles. Their use is different from the unwanted timelike negative-norm string states of covariant matter quantization.

The chiral holomorphic stress-energy tensor of the bc system isDouble contractions in its operator product expansion give the central charge of reparameterization ghosts $c_{\mathrm{gh}}=-26$. Equivalently, a fermionic bc system with antighost weight $\lambda$ has $c=1-3(2\lambda -1{)}^2$, and $\lambda =2$ gives $-26$. The free coordinate matter contributes $c_{\mathrm{m}}=D$. Thus the total worldsheet Weyl anomaly vanishes for the flat critical bosonic string theory precisely whenThis statement assumes no additional matter or compensating Liouville field theory.

BRST symmetry expresses the gauge symmetry after gauge fixing by replacing its infinitesimal parameter with the Grassmann variable $c$. In one chiral conformal patch, its graded action has the formthere is an analogous barred copy for a closed string. The Grassmann parity makes $s^{2}X=0$: the term from $sc$ cancels the term from applying $s$ to $\partial X$, while the term with $c^2$ vanishes. The gauge-fixed chiral transformations are on shell when their conservation equations are used. At the quantum level, BRST nilpotence also tests the anomaly.

For example, an open string BRST charge can be written using the matter Virasoro generators asUse the ghost oscillator vacuum $b_{n\ge 0}|0{⟩}_{mathrmgh}=c_{n>0}|0{⟩}_{mathrmgh}=0$ and the corresponding oscillator normal ordering. In this convention the last term is $-c_0$ in the critical theory. With a differently shifted ghost $L_0$, the displayed intercept must be shifted consistently; one must not count the same shift twice. Squaring $Q_B$ shows why both critical conditions are needed. Its anomalous terms, in this convention, have coefficientsThe independent cubic and linear terms vanish at $D=26$, $a=1$. The equivalent chiral BRST current is $j_B=c(T_{mathrmm}+T_{mathrmgh}/2)+(3/2){\partial }^{2}c$, with its contour integral defining the charge.

The graded identity $\{Q_B,b_n\}=L_n^{\mathrm{tot}}$ recovers the gauge constraints. In particular, on $|\psi ⟩\otimes |0{⟩}_{mathrmgh}$, BRST-closed imposes $L_{n>0}^{\mathrm{m}}|\psi ⟩=0$ and $(L_0^{\mathrm{m}}-1)|\psi ⟩=0$. Physical states are classes in BRST cohomology,at the appropriate ghost number: one for the usual open-string vertex and two for the closed-string vertex, with the closed-string zero-mode and closed-string level matching conditions also imposed. Since $Q_B^2=0$, an exact shift preserves closure. BRST Ward identities make exact insertions decouple from physical amplitudes and establish gauge independence. Together with the no-ghost theorem for the critical bosonic string, this explains how the auxiliary ghosts enforce gauge invariance while the physical spectrum retains positive norm.

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 $a=1,D=26$.