Ghosts, Superghosts, and BRST Quantization

Gauge fixing removes the worldsheet metric and gravitino as local propagating variables, but their constraints remain. The ghost fields encode those constraints in an odd operator $Q_{\mathrm B}$. Its square vanishes when the quantum constraint algebra is anomaly free, and physical states are its closed states modulo its exact states.

This page derives the two ghost systems, their charge and picture gradings, and the BRST operator with mutually consistent signs and coefficients. A central result is that the RNS matter theory must have $c_m=15$. Ten flat spacetime dimensions are one realization of this condition; a suitable internal superconformal field theory can supply part of the matter central charge.

We use the shared complex measure and ghost normalization, the canonical matter supercurrent, and the plane NS/R algebra. The preceding page introduced spin fields and explained why supersymmetry transformations must track superghost pictures.

Gauge fixing produces two first-order systems

Diffeomorphisms and Weyl transformations permit a locally flat conformal metric; local supersymmetry and super-Weyl symmetry similarly permit a vanishing gravitino locally. On a general surface, moduli, odd moduli, and zero modes remain. Setting the gravitino to zero in a coordinate patch is not a global elimination of this information.

The gauge parameters determine the ghost weights. A holomorphic vector-field parameter has weight $-1$, giving a ghost $c$ of weight $-1$ and its conjugate antighost $b$ of weight $2$. A local supersymmetry parameter has weight $-1/2$, giving a ghost $\gamma$ of weight $-1/2$ and antighost $\beta$ of weight $3/2$. Ghost statistics are opposite to the corresponding gauge-parameter statistics: $b,c$ anticommute, whereas $\beta,\gamma$ commute.

For one chiral sector the first-order action is

$$S_{\mathrm{gh}}=\frac1{2\pi}\int d^2z\, \bigl(b\bar\partial c+\beta\bar\partial\gamma\bigr).$$

The closed string has an independent antiholomorphic sector. An open-string boundary identifies the two sectors in the doubling construction.

Ordered propagators and their signs

The shared measure gives $\bar\partial_z[1/(z-w)]=2\pi\delta_z^{(2)}(z-w)$. This identity inverts the first-order kinetic operator. For the commuting system, varying the path integral with respect to $\beta(z)$ gives

$$\bar\partial_z\langle\gamma(z)\beta(w)\rangle =2\pi\delta_z^{(2)}(z-w).$$

Thus $\gamma(z)\beta(w)\sim1/(z-w)$. Reversing the ordered fields changes the sign of the first-order kernel; commuting statistics supply no compensating minus sign. For the anticommuting pair, fermionic exchange does supply that sign. The four ordered products are consequently

$$\begin{aligned} b(z)c(w)&\sim\frac1{z-w}, &c(z)b(w)&\sim\frac1{z-w},\\ \beta(z)\gamma(w)&\sim-\frac1{z-w}, &\gamma(z)\beta(w)&\sim\frac1{z-w}. \end{aligned}$$

The relative signs must also be used in the stress tensors and BRST operator. Changing only one OPE sign would change the generated translations or the constraint algebra.

The plane modes and brackets are

$$\begin{aligned} b(z)&=\sum_{n\in\mathbb Z}b_nz^{-n-2}, &c(z)&=\sum_{n\in\mathbb Z}c_nz^{-n+1},\\ \beta(z)&=\sum_r\beta_rz^{-r-3/2}, &\gamma(z)&=\sum_r\gamma_rz^{-r+1/2}, \end{aligned}$$ $$\{b_m,c_n\}=\delta_{m+n,0}, \qquad [\beta_r,\gamma_s]=-\delta_{r+s,0}.$$

Here $r\in\mathbb Z+1/2$ in NS and $r\in\mathbb Z$ in R. These are the cylinder spin structures, even though the displayed expansions are on the plane. The supercurrent and superghosts have matching spin structures, so their product in the BRST current is single-valued.

Stress tensors and the ghost anomaly

For a first-order pair $B,C$ of weights $\lambda,1-\lambda$, the stress tensor is

$$T_{BC}=(1-\lambda):\partial B\,C: -\lambda:B\partial C:.$$

With the appropriate ordered OPEs above, it generates

$$T(z)B(w)\sim\frac{\lambda B(w)}{(z-w)^2} +\frac{\partial B(w)}{z-w},$$

and the corresponding formula with weight $1-\lambda$ for $C$. For example, the two terms of $T_{\beta\gamma}$ contribute respectively $-\tfrac12\partial\beta/(z-w)$ and $\tfrac32[\beta/(z-w)^2+\partial\beta/(z-w)]$. Their sum has the required weight $3/2$ and unit translation coefficient.

Double contractions in $T(z)T(w)$ give

$$\begin{aligned} c_{BC}^{\mathrm{anticommuting}} &=-2(6\lambda^2-6\lambda+1),\\ c_{BC}^{\mathrm{commuting}} &=+2(6\lambda^2-6\lambda+1). \end{aligned}$$

The statistics reverse the sign of the closed contraction. In the RNS case,

$$\begin{aligned} T_{bc}&=-:\partial b\,c:-2:b\partial c:, &c_{bc}&=-26,\\ T_{\beta\gamma}&=-\frac12:\partial\beta\,\gamma: -\frac32:\beta\partial\gamma:, &c_{\beta\gamma}&=11. \end{aligned}$$

The total ghost central charge is $c_{\mathrm{gh}}=-15$. Therefore a matter superconformal theory can cancel it when $c_m=15$. For $D$ free flat coordinate multiplets, $c_m=D+D/2$, giving $D=10$. The statement concerns the complete matter theory, not just the number of noncompact coordinates.

The arrows add the two ghost central charges: $-26+11=-15$. Their horizontal displacements use a common central-charge scale. The condition is $c_m=15$; the conclusion $D=10$ assumes that all matter consists of free flat coordinate multiplets. Internal matter sectors must also be included in $c_m$.

Cancellation of central charges is necessary. To establish BRST nilpotence one must also construct the nonlinear ghost terms correctly; we will calculate the remaining obstruction below.

Background charge changes weights and central charge

Bosonization is particularly useful for ghost vacua. Begin with a normalized scalar $Y(z)Y(w)\sim-\log(z-w)$ and an improved stress tensor

$$T_Y=-\frac12:(\partial Y)^2:+A\partial^2Y.$$

The improvement changes the coupling to worldsheet curvature. Since

$$\partial Y(z)e^{aY(w)}\sim-\frac{a}{z-w}e^{aY(w)}, \qquad \partial^2Y(z)\partial^2Y(w)\sim\frac6{(z-w)^4},$$

the weight and central charge become

$$h(e^{aY})=-\frac{a^2}{2}+Aa, \qquad c_Y=1+12A^2.$$

The ordinary quadratic stress contributes $1/2$ to the fourth-order pole of $TT$, while the two improvement terms contribute $6A^2$. Their mixed contractions affect the lower poles, not the central coefficient.

Reparametrization-ghost bosonization

For a scalar $\sigma$ with this same logarithmic OPE, choose $A=3i/2$ and write, with consistent cocycles,

$$b\sim e^{-i\sigma}, \qquad c\sim e^{i\sigma}, \qquad T_\sigma=-\frac12:(\partial\sigma)^2: +\frac{3i}{2}\partial^2\sigma.$$

The weight formula gives $h_b=2$ and $h_c=-1$, and $c_\sigma=1+12(3i/2)^2=-26$. The imaginary improvement is part of this nonunitary ghost description; it is not a real spacetime dilaton background.

The background charge also explains anomalous charge selection. For a normalized scalar with improvement $A\partial^2Y$, the holomorphic genus-$g$ selection rule is $\sum a_i=2A(1-g)$, with the corresponding curvature coupling and zero-mode prescription. For $a_i=iq_i$ in the $bc$ description this becomes $\sum q_i=3(1-g)$. On the sphere, the three conformal Killing vectors are paired with three $c$ zero modes. On the disk, after doubling, three boundary positions are fixed by three $c$ insertions. Integrating over moduli requires the appropriate $b$ insertions as well.

The related linear-dilaton mechanism

A real $A$ for a spacelike matter scalar raises its central charge. In a bosonic string with another matter sector of central charge $c_\perp$, anomaly cancellation would require

$$c_\perp+1+12A^2-26=0.$$

This is a different theory from the RNS ghost system. If $Y=\sqrt{2/\alpha'}X$ is a normalized spacetime coordinate, its linear dilaton is $\Phi=A Y$ and the string coupling varies as $e^{A Y}$. Exercise 2 derives the shifted conformal weights and the associated tachyon mass parameter. It also identifies why this background is not equivalent to flat space with a constant coupling.

The sigma-model and linear-dilaton discussion develops its spacetime interpretation. The Weyl-anomaly discussion explains how a conformal factor can acquire dynamics when the anomaly does not cancel. Neither connection makes the bosonized ghost scalar into an extra physical target coordinate.

Superghost bosonization needs an extra fermion pair

Trying $\beta=e^{-\phi}$ and $\gamma=e^\phi$ fails: their exponential product begins as $z-w$, whereas the first-order system requires a pole. Their weights would also be wrong. Introduce an anticommuting pair of weights $h_\eta=1$, $h_\xi=0$, with

$$\eta(z)\xi(w)\sim\frac1{z-w}, \qquad \xi(z)\eta(w)\sim\frac1{z-w},$$

and use

$$\beta=e^{-\phi}\partial\xi, \qquad \gamma=\eta e^\phi, \qquad \phi(z)\phi(w)\sim-\log(z-w).$$

The cocycles assign $e^{\pm\phi}$ odd exchange parity, so each product defining $\beta$ or $\gamma$ is even. Keeping this ordering gives

$$\begin{aligned} \beta(z)\gamma(w)&\sim (z-w)\left[-\frac1{(z-w)^2}\right] =-\frac1{z-w},\\ \gamma(z)\beta(w)&\sim (z-w)\left[\frac1{(z-w)^2}\right] =\frac1{z-w}. \end{aligned}$$

Thus the derivative of $\xi$ repairs both the pole and the conformal weight. Matching the stress tensor gives

$$T_{\beta\gamma}=T_\phi+T_{\eta\xi}, \qquad T_\phi=-\frac12:(\partial\phi)^2:-\partial^2\phi, \qquad T_{\eta\xi}=-:\eta\partial\xi:.$$

The scalar has $A=-1$, hence $c_\phi=13$, while the fermion pair has $c_{\eta\xi}=-2$. Their sum is 11. The exponential weight is

$$h(e^{q\phi})=-\frac12q(q+2).$$

In particular, $e^{-\phi}$ has weight $1/2$, $e^{-\phi/2}$ and $e^{-3\phi/2}$ have weight $3/8$, and $e^\phi$ has weight $-3/2$. Multiplication by $\partial\xi$ or $\eta$ then gives $h_\beta=3/2$ and $h_\gamma=-1/2$. This construction and its different vacua are developed in Friedan, Martinec, and Shenker.

Ghost vacua specify which modes annihilate a state

The $bc$ identity vacuum satisfies

$$b_n|0\rangle_{bc}=0\quad(n\geq-1), \qquad c_n|0\rangle_{bc}=0\quad(n\geq2).$$

These conditions follow by requiring $b(z)|0\rangle$ and $c(z)|0\rangle$ to be regular at the origin. The often-used state $c_1|0\rangle_{bc}$ instead has $L_0=-1$ and ghost number one. The identity vacuum and this state must not be assigned the same zero-point weight.

For the superghosts, let $|q\rangle$ be created by $e^{q\phi}(0)$ from the regular $\eta\xi$ vacuum. The exponential products imply

$$\beta(z)|q\rangle=O(z^q), \qquad \gamma(z)|q\rangle=O(z^{-q}).$$

Comparing with the mode expansions derives the annihilation conditions

$$\begin{aligned} \beta_r|q\rangle&=0&&\text{for }r\geq-q-\frac12,\\ \gamma_r|q\rangle&=0&&\text{for }r\geq q+\frac32. \end{aligned}$$

Integer $q$ is compatible with the NS mode lattice; half-integer $q$ is compatible with R. Three important choices are

Superghost vacuum	Weight	Annihilators
NS, $q=-1$	$1/2$	$\beta_r,\gamma_r$ for $r\geq1/2$
R, $q=-1/2$	$3/8$	$\beta_r$ for $r\geq0$; $\gamma_r$ for $r\geq1$
R, $q=-3/2$	$3/8$	$\beta_r$ for $r\geq1$; $\gamma_r$ for $r\geq0$

The Ramond choices exchange the role of the commuting zero modes. In picture $-1/2$, $\gamma_0$ generates a tower and $\beta_0$ annihilates the vacuum; in picture $-3/2$ their roles reverse. This is why stating a picture is part of specifying a ghost Fock representation.

The total weight of a familiar unintegrated NS representative $c e^{-\phi}V_m$ is $h_m-1/2$. Its mass-shell condition is therefore $h_m=1/2$. For $c e^{-\phi/2}V_m^{\mathrm R}$ it is $h_m-5/8$, giving $h_m=5/8$. These are plane weights, consistent with the physical conditions derived earlier. The Ramond matter vacuum already carries its $5/8$ weight; it is not an additional intercept to be added a second time.

Ghost number, picture, and the small Hilbert space

The bosonized variables permit two distinct gradings. We extend ghost number so that $\eta$ has ghost number $+1$, $\xi$ has $-1$, and $e^{q\phi}$ has zero. Picture assigns $q$ to $e^{q\phi}$, $-1$ to $\eta$, and $+1$ to $\xi$. The corresponding currents are

$$j_{\mathrm{gh}}=-:bc:-:\xi\eta:, \qquad j_P=-\partial\phi+:\xi\eta:.$$

Their charges are

Operator	Ghost number	Picture
$b,c$	$-1,+1$	$0,0$
$\eta,\xi$	$+1,-1$	$-1,+1$
$e^{q\phi}$	$0$	$q$
$\beta,\gamma$	$-1,+1$	$0,0$

For example, $P(\beta)=-1+1=0$ and $P(\gamma)=-1+1=0$. Derivatives do not change these charges. This extension agrees with the original ghost charges on $b,c,\beta,\gamma$ but also specifies how the enlarged bosonized space is graded; the two gradings must not be collapsed into one. See Fuchs and Kroyter, section 8.1, for the charge assignments.

The original $\beta\gamma$ fields involve $\partial\xi$, not the zero mode $\xi_0$. Including that zero mode enlarges the state space. The small Hilbert space is

$$\eta_0|\Psi\rangle=0, \qquad \{\eta_0,\xi_0\}=1.$$

The large Hilbert space also contains states with $\xi_0$ dependence. Physical states and their gauge parameters will both be chosen in the small space. This restriction is essential when interpreting picture changing as a BRST commutator.

Constructing the BRST charge

The matter generators obey the canonical positive-closure OPE

$$G_m(z)G_m(w)\sim \frac{2c_m/3}{(z-w)^3}+\frac{2T_m(w)}{z-w}.$$

For free matter, the shared normalization is $G_m=i\sqrt{2/\alpha'}:\psi\cdot\partial X:$; the factor of $i$ is fixed by this positive closure. The BRST operator begins with $cT_m+\gamma G_m$, but these terms alone do not square to zero. The $GG$ algebra generates a Virasoro transformation, which must be canceled by a ghost interaction.

A useful contour-current representative is

$$\begin{aligned} j_0={}&:cT_m:+:\gamma G_m: +:bc\partial c:+:cT_{\beta\gamma}: -:b\gamma^2:,\\ Q_{\mathrm B}={}&\oint\frac{dz}{2\pi i}\,j_0(z). \end{aligned}$$

All free-ghost products are plane normal ordered with the OPEs established above. The contour is counterclockwise. This is a definition of a representative for the charge; we do not assume that this particular local current is a primary.

Why the cubic term has a minus sign

Consider the simple graded algebra $\{G,G\}=2L$, $[L,G]=0$, with $\{b,c\}=1$. Let the odd matter constraint anticommute with $b,c$, and treat $\gamma$ as commuting. For

$$Q=cL+\gamma G+a\,b\gamma^2,$$

one finds

$$Q^2=\gamma^2L+a\gamma^2L=(1+a)\gamma^2L.$$

Therefore $a=-1$. The complete local algebra also has conformal-transformation terms; these generate $bc\partial c$ and $cT_{\beta\gamma}$. The finite example isolates the required sign but is not by itself a proof of quantum string nilpotence.

Current representatives and the ghost supercurrent

Define

$$\begin{aligned} T_{\mathrm{gh}}&=T_{bc}+T_{\beta\gamma},\\ G_{\mathrm{gh}}&=-2:b\gamma:+:c\partial\beta: +\frac32:\partial c\,\beta:. \end{aligned}$$

A compact charge expression is $c(T_m+T_{\mathrm{gh}}/2)+\gamma(G_m+G_{\mathrm{gh}}/2)$. When its ghost factors are expanded and fully normal ordered, the mixed derivative terms are

$$\frac34:\partial c\,\beta\gamma: +\frac14:c\partial\beta\,\gamma: -\frac34:c\beta\partial\gamma:.$$

Their difference from $:cT_{\beta\gamma}:$ is $\tfrac34\partial(:c\beta\gamma:)$. Its contour integral vanishes. Other total-derivative improvements can be used to change the transformation properties of a local current, but the same charge results. A local current’s higher poles must therefore be quoted together with its chosen representative; equality of contour charges does not mean equality of all local OPEs. The expanded charge is also given in Polchinski, section 10.5.

Several direct residues test the full relative normalization:

$$\begin{aligned} \{Q_{\mathrm B},b\}&=T_m+T_{\mathrm{gh}}, &[Q_{\mathrm B},\beta]&=G_m+G_{\mathrm{gh}},\\ \{Q_{\mathrm B},c\}&=c\partial c-\gamma^2, &[Q_{\mathrm B},\gamma]&=c\partial\gamma-\frac12(\partial c)\gamma. \end{aligned}$$

For example, contracting $cT_{\beta\gamma}$ with $\beta$ gives $c\partial\beta+\tfrac32(\partial c)\beta$, and contracting $-b\gamma^2$ with $\beta$ gives $-2b\gamma$. The resulting ghost terms are exactly $G_{\mathrm{gh}}$. This check tests the derivative coefficients as well as the cubic sign.

The quantum obstruction to nilpotence

The terms proportional to $T_m$, $G_m$, and the nonlinear ghost monomials cancel using the noncentral super-Virasoro algebra and the ghost brackets. Equivalently, the cubic ghost terms implement the graded structure constants of the constraints. Central contractions leave a possible quantum obstruction.

For an odd contour charge,

$$Q_{\mathrm B}^2=\frac12\oint\frac{dw}{2\pi i} \mathop{\mathrm{Res}}_{z=w}j_0(z)j_0(w).$$

The matter central terms are especially transparent. In $cT_m(z)cT_m(w)$, Taylor expansion of $c(z)c(w)$ to third order gives an integrated residue $-c_m c\partial^3c/12$. In $\gamma G_m(z)\gamma G_m(w)$, expanding $\gamma(z)$ to second order gives $c_m\gamma\partial^2\gamma/3$.

The free-ghost double contractions give the following coefficients, after integrating by parts. The cross term in the last row includes both orderings.

Pair of current terms	Coefficient of $c\partial^3c$	Coefficient of $\gamma\partial^2\gamma$
$cT_m$ with itself	$-c_m/12$	$0$
$\gamma G_m$ with itself	$0$	$c_m/3$
$bc\partial c$ with itself	$13/6$	$0$
$cT_{\beta\gamma}$ with itself	$-11/12$	$0$
$cT_{\beta\gamma}$ with $-b\gamma^2$	$0$	$-5$

For instance, $13/6-11/12=15/12$, exactly the shift required in the first column. Summing and including the factor $1/2$ gives

$$Q_{\mathrm B}^2=(c_m-15)\oint\frac{dz}{2\pi i} \left[-\frac1{24}:c\partial^3c: +\frac16:\gamma\partial^2\gamma:\right].$$

This formula concerns the integrated residue; total derivatives in the local $j_0j_0$ product have already been removed. It exhibits both independent ghost structures in the anomaly. With the stated plane normalization, $c_m=15$ makes the charge nilpotent. An incorrect cubic interaction would leave noncentral terms even at that value.

Converting the two residual terms to modes gives a useful check:

$$\begin{aligned} Q_{\mathrm B}^2={}&\frac{c_m-15}{24} \sum_{n\in\mathbb Z}(n^3-n)c_{-n}c_n\\ &+\frac{c_m-15}{6} \sum_r\left(r^2-\frac14\right)\gamma_{-r}\gamma_r. \end{aligned}$$

The sums run over both signs of the mode number; they must not be compared with positive-mode sums without their factors of two. Plane normal ordering is already included. Reexpressing the charge using creation and annihilation operators relative to a particular ghost vacuum produces the corresponding vacuum constants; it does not permit appending an independent intercept to this OPE result.

Physical cohomology requires a specified state space

At criticality, let $\mathcal C_P^g$ denote the chosen small-space chain space at ghost number $g$ and fixed picture $P$. The charge has ghost number one and picture zero:

$$Q_{\mathrm B}:\mathcal C_P^g\longrightarrow\mathcal C_P^{g+1}, \qquad Q_{\mathrm B}^2=0.$$

The open-string physical cohomology is taken at ghost number one:

$$H_P^1(Q_{\mathrm B})= \frac{\ker(Q_{\mathrm B}:\mathcal C_P^1\to\mathcal C_P^2)} {\operatorname{im}(Q_{\mathrm B}:\mathcal C_P^0\to\mathcal C_P^1)}.$$

Gauge parameters belong to the same small-space, momentum, picture, and boundary-condition sector. Other ghost-number cohomologies have distinct roles; they are not all propagating one-particle states.

For the usual open-string relative cohomology, one further imposes $b_0|\Psi\rangle=0$ and $L_0^{\mathrm{tot}}|\Psi\rangle=0$. These conditions are compatible because $\{Q_{\mathrm B},b_0\}=L_0^{\mathrm{tot}}$. In the canonical Ramond representation, a standard relative choice also uses $\beta_0|\Psi\rangle=G_0^{\mathrm{tot}}|\Psi\rangle=0$. Changing Ramond picture changes the zero-mode representation, so these restrictions cannot be copied blindly between pictures.

For closed strings, the physical degree is normally two in total, and the standard semirelative conditions include $b_0^-|\Psi\rangle=0$ and $L_0^-|\Psi\rangle=0$, with $b_0^-=b_0-\widetilde b_0$ and $L_0^-=L_0-\widetilde L_0$. Left and right pictures and any Ramond restrictions must also be specified. We will mainly use open-string representatives below.

Recovering the matter constraints

In familiar canonical representatives, BRST closure reproduces the positive-mode conditions $L_n^m|v\rangle=0$ for $n>0$ and $G_r^m|v\rangle=0$ for $r>0$. In NS the matter plane condition is $(L_0^m-1/2)|v\rangle=0$. In R it is $(L_0^m-5/8)|v\rangle=0$, together with $G_0^m|v\rangle=0$ at $c_m=15$. The latter implies the mass-shell equation through $(G_0^m)^2=L_0^m-5/8$; the converse does not follow in a covariant indefinite state space.

These conditions apply to the matter part of the representative, not indiscriminately to a general ghost-excited BRST state. The resulting quotient reproduces the transverse physical spectrum under the hypotheses of the no-ghost theorem. Nilpotence alone does not establish positive norm: a badly chosen matter theory can satisfy a central-charge equation without satisfying the theorem’s unitarity assumptions.

Picture changing acts between complexes

In the large space define the even operator

$$\mathcal X(z)=\{Q_{\mathrm B},\xi(z)\}.$$

With the displayed ghost ordering, its local expression is

$$\mathcal X=c\partial\xi+e^\phi G_m +:b\partial\eta\,e^{2\phi}: +\partial(:b\eta e^{2\phi}:).$$

The last two terms follow from $-b\gamma^2=-:b\eta\partial\eta e^{2\phi}:$: contracting $\eta$ with $\xi$ gives the first simple-pole term, and contracting $\partial\eta$ gives a double pole whose residue differentiates the remaining fields. Each term has weight zero, ghost number zero, and picture $+1$. In particular, $c\partial\xi$ carries picture $+1$ without any $\phi$ exponential.

Nilpotence and the graded Jacobi identity imply

$$[Q_{\mathrm B},\mathcal X]=0, \qquad [\eta_0,\mathcal X]=0.$$

The second statement uses $\{\eta_0,Q_{\mathrm B}\}=0$ and $\{\eta_0,\xi\}=1$. For a map of states at fixed $L_0$, use its weight-zero mode

$$\mathcal X_0=\oint\frac{dz}{2\pi iz}\,\mathcal X(z) =\{Q_{\mathrm B},\xi_0\}.$$

It maps small-space closed states to closed states and exact states to exact states, while raising picture by one. It also preserves the usual $b_0$ relative condition: from $[b_0,\mathcal X(z)]=z\partial\xi(z)$ one obtains $[b_0,\mathcal X_0]=\oint dz\,\partial\xi/(2\pi i)=0$. Ramond zero-mode restrictions require their own compatibility analysis. A local vertex map using $\mathcal X(z)V(w)$ instead requires a defined finite-part OPE prescription; it should not be conflated with unrestricted pointwise multiplication.

This state map need not be trivial even though $\mathcal X_0=\{Q_{\mathrm B},\xi_0\}$. For a closed small-space state $V$, the would-be gauge parameter $\xi_0V$ is generally not in the small space: $\eta_0(\xi_0V)=V$. Exactness using a forbidden gauge parameter is not exactness in the physical complex.

Horizontal arrows are the nilpotent BRST differential. Vertical arrows are the zero-mode maps $\mathcal X_0$, which raise picture and commute with the differential; no inverse is assumed. At ghost number one, exact states form a subspace of closed states, and the physical space is their quotient. The auxiliary operator $\xi_0$ leaves the small Hilbert space, so its appearance in $\mathcal X_0=\{Q_{\mathrm B},\xi_0\}$ does not make this map trivial there. The regions are schematic vector spaces, not dimension counts.

A local inverse-picture representative is $\mathcal Y=c\partial\xi e^{-2\phi}$, of weight and ghost number zero and picture $-1$. Suitable regulated products implement inverse maps on appropriate physical classes; this is not a claim of invertibility on every off-shell state or at every momentum. The Ramond calculation on the preceding page gives a concrete nonzero-momentum isomorphism and shows why its momentum-slash argument fails at zero momentum.

Colliding picture-changing insertions can have singular OPEs. Normal ordering one product does not by itself settle spurious singularities or the global choice of insertions over supermoduli space. Those require an amplitude prescription, such as the constructions discussed by Sen and Witten. The local identities here are necessary ingredients, not a substitute for that global analysis.

BRST descent relates local vertices and integrated insertions

Write $[A,B\}=AB-(-1)^{|A||B|}BA$ for the graded bracket. For a BRST-closed unintegrated open-string vertex $\mathcal V$ of ghost number one, define

$$U=[b_{-1},\mathcal V\}.$$

Since $\{Q_{\mathrm B},b_{-1}\}=L_{-1}$, the graded Jacobi identity gives

$$[Q_{\mathrm B},U\}=\partial\mathcal V.$$

Thus an integrated boundary insertion $\int U$ is invariant up to endpoint or moduli-boundary contributions. The argument uses parity correctly whether $\mathcal V$ is even or odd.

For a representative of the form $\mathcal V=cU$ with $b(z)U(w)$ regular, the descent simply removes $c$. That shortcut is not universal. For example, let $U_F$ be a matter superconformal primary of weight $1/2$, and set $U_B=G_{-1/2}U_F$. The picture-zero BRST-closed representative is

$$\mathcal V^{(0)}=cU_B+\gamma U_F.$$

The extra superghost term is required by BRST closure. Nevertheless $[b_{-1},\mathcal V^{(0)}\}=U_B$, because $b$ has no pole with $\gamma U_F$. Exercise 5 derives this expression and its pure-gauge longitudinal specialization.

A closed-string integrated vertex uses two-dimensional descent, with both $b$ and $\widetilde b$. The open boundary formula should not be substituted for that construction. Higher-genus moduli insertions and boundary terms are also part of amplitude calculations, even when every external local vertex is BRST closed.

Picture selection in amplitudes

The $A=-1$ background charge of $\phi$ gives total holomorphic picture $2g-2$ for a genus-$g$ small-space amplitude, with the prescribed ghost and moduli insertions. A closed amplitude obeys this separately on the left and right. The doubled disk requires total picture $-2$.

These totals include explicit picture-changing operators. With all external NS vertices at picture $-1$ and all R vertices at $-1/2$, the number of picture-raising insertions needed in one chiral sector is

$$n_{\mathcal X}=2g-2+n_{\mathrm{NS}}+\frac12n_{\mathrm R}.$$

This formula is a charge count, not a proof that the resulting amplitude is nonzero or that arbitrary insertion locations are acceptable. Spin-structure compatibility, zero-mode saturation, BRST invariance, and avoidance or treatment of spurious singularities still matter. Integrating a vertex changes its ghost number through descent; it does not automatically change its picture.

Common pitfalls

A sign convention must be changed throughout the calculation. The ordered $\beta\gamma$ OPEs, stress tensor, mode bracket, supercurrent, and BRST cubic term form one consistent set. A correct central-charge sum cannot rescue an inconsistent charge.

Picture is not generally the exponent of $\phi$. That shortcut works for representatives without $\eta$ or $\xi$. The fields $\beta,\gamma$ are picture neutral, while $c\partial\xi$ has picture $+1$ and $c\partial\xi e^{-2\phi}$ has picture $-1$.

Cohomology depends on its domain. The small-space condition, ghost degree, picture, zero-mode restrictions, and boundary conditions are part of the definition. A large-space primitive or a local inverse-picture expression does not automatically establish equivalence in every physical state space.

Exercises

Exercise 1: stress tensors test the ordered OPEs

For $T=(1-\lambda):\partial B\,C:-\lambda:B\partial C:$, derive the double-pole and simple-pole coefficients in $T(z)B(w)$ for both fermionic and commuting first-order systems. Compute the fourth-order coefficient in $TT$. Then take an anticommuting pair of weights $(\lambda,1-\lambda)$ and a commuting pair of weights $(\lambda-1/2,3/2-\lambda)$. Determine their total central charge and explain the RNS choice. What fails if the commuting ordered OPE signs are reversed without changing $T$?

Solution

For the commuting pair, use $B(z)C(w)\sim-1/(z-w)$ and $C(z)B(w)\sim1/(z-w)$. Contracting the first term of $T$ with $B(w)$ gives $(1-\lambda)\partial B/(z-w)$. The second term contracts $\partial C(z)$ with $B(w)$ and gives

$$\lambda\left[\frac{B(w)}{(z-w)^2} +\frac{\partial B(w)}{z-w}\right].$$

The sum is $\lambda B/(z-w)^2+\partial B/(z-w)$. For fermions, $C(z)B(w)$ still has a positive pole, but contracting across the other odd field contributes the signs needed for the same result. In both cases $T$ generates the desired conformal transformation.

Put $a=1-\lambda$ and $d=-\lambda$. The commuting double contractions of $a:\partial B C:+d:B\partial C:$ with itself give

$$\frac{a^2+d^2+4ad}{(z-w)^4} =\frac{6\lambda^2-6\lambda+1}{(z-w)^4}.$$

For fermions the coefficient has the opposite sign. Multiplying by two gives the central charges in the text. The two systems in the problem therefore have

$$\begin{aligned} c_{\mathrm{pair}} &=\bigl[1-3(2\lambda-1)^2\bigr] +\bigl[-1+3(2\lambda-2)^2\bigr]\\ &=9-12\lambda. \end{aligned}$$

The RNS ghost weights correspond to $\lambda=2$, giving $c_{\mathrm{pair}}=-15$.

Reversing both commuting OPE signs while retaining this stress tensor multiplies the entire $TB$ singular part by $-1$. The result has both weight $-\lambda$ and translation coefficient $-1$, rather than merely an alternative name for the same normalization. One would have to change the stress tensor and dependent formulas consistently.

Exercise 2: the linear-dilaton mass shift

Consider a closed bosonic string with $d$ other free coordinates, including one time coordinate, and one spacelike normalized scalar $Y$ with improvement $A\partial^2Y$. Let $c_\perp=d$ and take real $A$. Derive the anomaly-cancellation condition and the weights of

$$\mathcal V=e^{(A+i\ell)Y}e^{i\boldsymbol k\cdot\boldsymbol Y_\perp},$$

where $Y(z)Y(w)\sim-\log(z-w)$ and $Y_\perp^\mu(z)Y_\perp^\nu(w)\sim-\eta^{\mu\nu}\log(z-w)$. Impose the matched closed-string ground-state conditions $h=\bar h=1$ and determine the effective mass parameter $\mu^2\equiv\alpha'M_{\mathrm{eff}}^2/2$. Explain its relation to the usual flat-space tachyon and the limitation of interpreting this as a translation-invariant scattering problem.

Solution

The scalar contributes $1+12A^2$ and the $bc$ ghosts contribute $-26$, so

$$A^2=\frac{25-d}{12}.$$

A real spacelike slope in this construction therefore requires $d\leq25$. Substituting $a=A+i\ell$ into the improved weight gives

$$-\frac12(A+i\ell)^2+A(A+i\ell) =\frac12(A^2+\ell^2).$$

The terms linear in $\ell$ cancel. Including the other coordinates,

$$h=\frac12(\boldsymbol k^2+\ell^2+A^2)=1.$$

Writing $\boldsymbol k^2+\ell^2=-\mu^2$ therefore gives

$$\mu^2=A^2-2=\frac{1-d}{12}, \qquad M_{\mathrm{eff}}^2=\frac2{\alpha'}(A^2-2).$$

At $d=25$, the slope vanishes and $M_{\mathrm{eff}}^2=-4/\alpha'$, the closed bosonic tachyon mass. For $1<d<25$, the shifted parameter is still negative; at $d=1$ it reaches zero. The factor $e^{AY}$ in the vertex implements the background-charge shift of the momentum basis, so these are not ordinary plane waves of the unweighted field.

The spacetime coupling varies as $g_s(Y)\propto e^{AY}$. A real slope has a strong-coupling end, breaks translation invariance along $Y$, and changes the appropriate state inner product. The mass parameter derived from the conformal weight should therefore not be interpreted as establishing a globally controlled flat-space scattering vacuum. This is a bosonic matter example, separate from the superghost $\phi$ and from the RNS condition $c_m=15$.

Exercise 3: Ramond ghost zero modes and paired pictures

Derive the annihilation thresholds for $|q\rangle$ from the local OPEs. Realize the Ramond zero-mode algebra by $\gamma_0=x$, $\beta_0=-d/dx$. Identify representative wavefunctions for the $q=-1/2$ and $q=-3/2$ vacua. Determine which picture pairs with a picture-$q$ state in a sphere two-point function, and check that the two paired exponential weights agree. Why is this not an ordinary positive-norm oscillator Hilbert space?

Solution

The first possible power in $\beta(z)|q\rangle$ is $z^q$. A mode $\beta_r$ would give $z^{-r-3/2}$, so modes with $-r-3/2<q$ must annihilate the state. On the unit-spaced mode lattice this is $r\geq-q-1/2$. Similarly, the first power for $\gamma$ is $z^{-q}$, giving $r\geq q+3/2$ as the annihilation range.

In R, the proposed zero-mode realization satisfies $[\beta_0,\gamma_0]=-1$. The $q=-1/2$ vacuum is annihilated by $\beta_0$, represented by the constant function $1$; successive $\gamma_0$ actions generate polynomials in $x$. The $q=-3/2$ vacuum is annihilated by $\gamma_0$, represented distributionally by $\delta(x)$; successive $\beta_0$ actions generate its derivatives. These formal realizations explain the exchanged creator/annihilator roles.

The superghost background charge on the sphere requires $q+q'=-2$, hence

$$q'=-q-2, \qquad h(q')=-\frac12(-q-2)(-q)=h(q).$$

The NS picture $-1$ pairs with itself, while the R pictures $-1/2$ and $-3/2$ pair with one another. Appropriate $bc$ insertions and the small-space prescription are still required for a full nonzero inner product.

The constant function and delta distribution are not two normalizable ground states of a positive-definite harmonic oscillator. Ghost Fock spaces carry an indefinite or distributional pairing and include fields of negative conformal weight. Physical positivity concerns the final BRST quotient under the no-ghost theorem’s assumptions, not the unconstrained ghost sector.

Exercise 4: detect the quantum anomaly beyond global modes

Convert the integrated obstruction in the text into mode sums, retaining sums over positive and negative indices. Let $\delta c=c_m-15$. Compute $[Q_{\mathrm B}^2,b_n]$ and $[Q_{\mathrm B}^2,\beta_r]$. Explain why testing only the global NS generators cannot detect the central-charge obstruction. Finally, verify the necessity of the negative cubic sign in the finite model $L=1$, $G^2=1$ with $\{b,c\}=1$.

Solution

In the $c$ expansion, three derivatives on $z^{-n+1}$ give $-(n^3-n)z^{-n-2}$. The contour condition pairs indices $n$ and $-n$. Two derivatives on the $\gamma$ mode give $(r^2-1/4)z^{-r-3/2}$. Thus

$$Q_{\mathrm B}^2= \frac{\delta c}{24}\sum_n(n^3-n)c_{-n}c_n +\frac{\delta c}{6}\sum_r(r^2-1/4)\gamma_{-r}\gamma_r.$$

Use $[c_{-m}c_m,b_n]=c_{-m}\delta_{m+n,0}-\delta_{-m+n,0}c_m$. Since $m^3-m$ is odd in $m$, the two contributions add with a minus sign:

$$[Q_{\mathrm B}^2,b_n] =-\frac{\delta c}{12}(n^3-n)c_n.$$

For the commuting pair, $[\gamma_s,\beta_r]=\delta_{r+s,0}$ and $r^2-1/4$ is even, giving

$$[Q_{\mathrm B}^2,\beta_r] =\frac{\delta c}{3}(r^2-1/4)\gamma_r.$$

The global NS modes have $n=0,\pm1$ and $r=\pm1/2$, where both polynomials vanish. They can therefore close even in a noncritical theory. Modes such as $n=2$ or $r=3/2$ detect the obstruction. In R, $r=0$ is itself sensitive to the shifted central polynomial.

For the finite sign test, choose odd $G$ that anticommutes with both ghost operators, and let $\gamma$ be a commuting number. The terms $(\gamma G)^2=\gamma^2$ and $\{c,a b\gamma^2\}=a\gamma^2$ are the only nonzero contributions to $Q^2$. Therefore $a=-1$ cancels them, whereas $a=+1$ gives $2\gamma^2$. This verifies a structural sign independently of central charges, but does not replace the full mode or OPE calculation.

Exercise 5: a picture-zero vertex and its exact longitudinal part

Let $U_F$ be an odd matter superconformal primary of weight $1/2$ and $U_B=G_{-1/2}U_F$. Derive the BRST closure of $\mathcal V^{(0)}=cU_B+\gamma U_F$ and its descent. For an open massless vector use $W_p=:e^{2ip\cdot X_L}:$, $p^2=0$, $U_F=\zeta\cdot\psi W_p$ and $p\cdot\zeta=0$. Compute $U_B$ and show that the choice $\zeta=p$ is BRST exact with a small-space picture-zero gauge parameter.

Solution

The superconformal algebra gives $h(U_B)=1$ and

$$G(z)U_F(w)\sim\frac{U_B(w)}{z-w}, \qquad G(z)U_B(w)\sim\frac{U_F(w)}{(z-w)^2} +\frac{\partial U_F(w)}{z-w}.$$

Writing $\delta=[Q_{\mathrm B},\,\cdot\,\}$, their transformations are

$$\begin{aligned} \delta U_F&=c\partial U_F+\frac12(\partial c)U_F+\gamma U_B,\\ \delta U_B&=c\partial U_B+(\partial c)U_B +\gamma\partial U_F+(\partial\gamma)U_F. \end{aligned}$$

Combining these with $\delta c=c\partial c-\gamma^2$ and $\delta\gamma=c\partial\gamma-\tfrac12(\partial c)\gamma$, and respecting the odd parity of $c$ and $U_F$, gives

$$\begin{aligned} \delta(cU_B)&=-\gamma^2U_B-c\gamma\partial U_F-c(\partial\gamma)U_F,\\ \delta(\gamma U_F)&=+\gamma^2U_B+c\gamma\partial U_F+c(\partial\gamma)U_F. \end{aligned}$$

They cancel. The $b_{-1}$ residue of the second term vanishes, while that of $cU_B$ is $U_B$. Hence $\delta U_B=\partial\mathcal V^{(0)}$, in agreement with graded descent.

For the vector, contracting $G_m=i\sqrt{2/\alpha'}\psi\cdot\partial X_L$ with $U_F$ gives

$$U_B=\left[i\sqrt{\frac2{\alpha'}}\,\zeta\cdot\partial X_L +\sqrt{2\alpha'}(p\cdot\psi)(\zeta\cdot\psi)\right]W_p.$$

The potential double pole is proportional to $p\cdot\zeta$ and vanishes. At $\zeta=p$, the fermion bilinear vanishes because its two factors coincide. Meanwhile $W_p$ is a weight-zero primary and

$$[Q_{\mathrm B},W_p] =c\partial W_p+\sqrt{2\alpha'}\,\gamma(p\cdot\psi)W_p =\sqrt{2\alpha'}\,\mathcal V_{\zeta=p}^{(0)}.$$

Thus the longitudinal state is exact with gauge parameter $W_p/\sqrt{2\alpha'}$, which has ghost number and picture zero and no $\xi_0$ dependence. Its integrated descendant is proportional to $\partial W_p$, a boundary total derivative. Removing only the $c$ term from the unintegrated expression would miss the superghost contribution essential for exactness and closure.

Exercise 6: small-space picture changing and mixed amplitudes

Use the graded Jacobi identity to prove that $\mathcal X_0$ maps small-space cohomology to small-space cohomology. Explain why its large-space primitive does not trivialize that map. Then count the picture-raising insertions for a disk amplitude with three NS and two R external states, all in canonical pictures, and for a sphere amplitude with two NS–NS and two R–R states. Give distributions that absorb the required insertions into external vertices. Finally count the holomorphic insertions for four canonical NS vertices at genus two, and state what none of these counts establishes by itself.

Solution

At criticality, $[Q_{\mathrm B},\{Q_{\mathrm B},\xi\}]=[Q_{\mathrm B}^2,\xi]=0$. Also

$$[\eta_0,\{Q_{\mathrm B},\xi\}] =-[Q_{\mathrm B},\{\eta_0,\xi\}]=-[Q_{\mathrm B},1]=0,$$

where $\{\eta_0,Q_{\mathrm B}\}=0$ was used. These local identities imply the corresponding identities for $\mathcal X_0$. It preserves the small-space condition and commutes with its differential. If $V=Q_{\mathrm B}\Lambda$ with an allowed small-space $\Lambda$, then $\mathcal X_0V=Q_{\mathrm B}(\mathcal X_0\Lambda)$, so the map is well defined on cohomology.

For a closed small-space state $V$, the formal large-space primitive $\xi_0V$ fails the gauge-parameter condition because $\eta_0(\xi_0V)=V$. It therefore does not prove that $\mathcal X_0V$ is trivial in the small space. Moreover, a commuting chain map need not be invertible; this proof says nothing about kernels at exceptional momentum or singular operator collisions.

On the disk, three NS vertices and two R vertices have total canonical picture $-3-1=-4$. Two picture-raising insertions bring it to $-2$. One possible distribution has two NS vertices in picture zero, one in picture $-1$, and the two R vertices in picture $-1/2$. The total is $0+0-1-1/2-1/2=-2$.

For the sphere example, each chiral sector has two NS pictures $-1$ and two R pictures $-1/2$, totaling $-3$. One picture-raising insertion is needed on each side. One can place an NS–NS vertex in picture $(0,0)$, leave the other in $(-1,-1)$, and retain both R–R vertices in $(-1/2,-1/2)$. Left and right totals are then separately $-2$. The $bc$ fixing of three positions and integration of the fourth is an independent ghost-number/descent issue.

At genus two the required holomorphic total is $2g-2=2$. Four canonical NS vertices contribute $-4$, so six picture-raising insertions are required in that sector. For a closed amplitude the right-moving count is imposed separately.

These are necessary charge counts. They neither construct a global supermoduli prescription nor prove a nonzero amplitude: spin structures, zero modes, kinematics, integration boundaries, and spurious singularities remain to be handled. Absorbing insertions into vertices is legitimate only within an appropriate regulated amplitude prescription.

References

• Friedan, Daniel, Emil Martinec, and Stephen Shenker. “Conformal invariance, supersymmetry and string theory.” Nuclear Physics B 271 (1986): 93–165, especially sections 4–6. Author-hosted full paper.
• Fuchs, Ehud, and Michael Kroyter. “Analytical solutions of open string field theory.” Physics Reports 502 (2011): 89–149, section 8.1. doi:10.1016/j.physrep.2011.01.003; author-accessible article.
• Polchinski, Joseph. String Theory, Volume II: Superstring Theory and Beyond. Cambridge: Cambridge University Press, 1998, sections 10.1 and 10.5. Publisher-distributed preview.
• Sen, Ashoke, and Edward Witten. “Filling the Gaps With PCO’s.” Journal of High Energy Physics 2015, no. 9 (2015): 004. arXiv:1504.00609.