The $bc$ Ghost System and Ghost Zero Modes

Conformal gauge fixes the worldsheet metric locally, but gauge fixing in a path integral is never free. The Faddeev-Popov determinant produced by fixing diffeomorphisms and Weyl transformations is represented by anticommuting fields. These are the reparameterization ghosts $b$ and $c$.

The $bc$ system has two jobs. Algebraically, it cancels the conformal anomaly of the matter fields in $D=26$. Geometrically, its zero modes encode the remaining conformal Killing symmetries and the moduli of punctured Riemann surfaces. This page develops the CFT of the ghosts; the next page uses it in the Polyakov path integral.

Why ghosts appear

Start with the Polyakov path integral schematically:

$$Z=\int\frac{\mathcal D h\,\mathcal D X}{\mathrm{Vol}(\mathrm{Diff}\times\mathrm{Weyl})} \,e^{-S_P[X,h]}.$$

Conformal gauge chooses a representative

$$h_{ab}=e^{2\omega}\widehat h_{ab}.$$

Gauge fixing introduces a Jacobian. As in ordinary gauge theory, this Jacobian may be written as a path integral over anticommuting fields. The infinitesimal conformal transformation parameter becomes the ghost $c$, while the antighost $b$ is naturally paired with deformations of the metric.

In a chiral sector, the action is the first-order system

$$S_{bc}=\frac{1}{2\pi}\int d^2z\, b\bar\partial c.$$

There is an antiholomorphic copy for closed strings,

$$S_{\widetilde b\widetilde c}=\frac{1}{2\pi}\int d^2z\,\widetilde b\partial\widetilde c.$$

For bosonic string reparameterization ghosts,

$$h_b=2, \qquad h_c=-1.$$

The negative conformal weight of $c$ is not a typo. It is why $c$ has zero modes on the sphere.

The anticommuting first-order system

It is useful to study a one-parameter family of anticommuting first-order systems,

$$h_b=\lambda, \qquad h_c=1-\lambda.$$

The basic OPE is

$$b(z)c(w)\sim\frac{1}{z-w}, \qquad c(z)b(w)\sim\frac{1}{z-w},$$

with no singular $bb$ or $cc$ OPEs. The fields are Grassmann odd, so their modes anticommute.

The $bc$ system is a first-order anticommuting CFT. For the bosonic string, $\lambda=2$, so $b$ has weight $2$ and $c$ has weight $-1$.

The stress tensor is

$$T_{bc}(z) = (1-\lambda):\partial b\,c:(z)-\lambda:b\partial c:(z).$$

It gives the primary-field OPEs

$$T_{bc}(z)b(w) \sim \frac{\lambda b(w)}{(z-w)^2} + \frac{\partial b(w)}{z-w},$$

and

$$T_{bc}(z)c(w) \sim \frac{(1-\lambda)c(w)}{(z-w)^2} + \frac{\partial c(w)}{z-w}.$$

The central charge is

$$c_{bc}=1-3(2\lambda-1)^2 =-2(6\lambda^2-6\lambda+1).$$

For reparameterization ghosts, $\lambda=2$, hence

$$c_{bc}=-26.$$

For $D$ free bosons, $c_X=D$, so the total central charge is

$$c_{\mathrm{tot}}=c_X+c_{bc}=D-26.$$

The condition of vanishing Weyl anomaly is therefore

$$D=26.$$

The anticommuting $bc$ central charge is $c_{bc}=1-3(2\lambda-1)^2$. The reparameterization ghost value $\lambda=2$ gives $c=-26$.

A useful side example: $\lambda=1/2$

At $\lambda=1/2$ the system has

$$h_b=h_c=\frac12, \qquad c_{bc}=1.$$

This is equivalent to a complex chiral fermion, or two real Majorana fermions. If

$$\psi=\frac{1}{\sqrt2}(\psi^1+i\psi^2), \qquad \bar\psi=\frac{1}{\sqrt2}(\psi^1-i\psi^2),$$

then

$$\psi^a(z)\psi^b(w)\sim\frac{\delta^{ab}}{z-w}, \qquad T_\psi(z)=-\frac12:\psi^a\partial\psi^a:(z), \qquad c_{\text{Majorana}}=\frac12.$$

On the Euclidean cylinder, fermions can be periodic or antiperiodic:

$$\psi(\sigma+2\pi)=+\psi(\sigma) \quad\text{Ramond}, \qquad \psi(\sigma+2\pi)=-\psi(\sigma) \quad\text{Neveu-Schwarz}.$$

This is only a preview, but it explains why the NSR superstring will require spin structures.

Mode expansions and oscillator algebra

For general $\lambda$, the plane mode expansions are

$$b(z)=\sum_{n\in\mathbb Z} b_n z^{-n-\lambda}, \qquad c(z)=\sum_{n\in\mathbb Z} c_n z^{-n+\lambda-1}.$$

Equivalently,

$$b_n=\oint_0\frac{dz}{2\pi i}z^{n+\lambda-1}b(z), \qquad c_n=\oint_0\frac{dz}{2\pi i}z^{n-\lambda}c(z).$$

The OPE gives

$$\{b_m,c_n\}=\delta_{m+n,0}, \qquad \{b_m,b_n\}=\{c_m,c_n\}=0.$$

The Virasoro modes follow from

$$T_{bc}(z)=\sum_{m\in\mathbb Z}L_m^{bc}z^{-m-2}.$$

A convenient normal-ordered expression is

$$L_m^{bc} = \sum_{n\in\mathbb Z} \left((\lambda-1)m-n\right):b_{m+n}c_{-n}:.$$

For reparameterization ghosts, $\lambda=2$, so

$$b(z)=\sum_{n\in\mathbb Z} b_n z^{-n-2}, \qquad c(z)=\sum_{n\in\mathbb Z} c_n z^{-n+1},$$

and

$$L_m^{gh}=\sum_{n\in\mathbb Z}(m-n):b_{m+n}c_{-n}:.$$

The singular OPE $b(z)c(w)\sim1/(z-w)$ is equivalent to the mode algebra $\{b_m,c_n\}=\delta_{m+n,0}$. For $\lambda=2$, the three modes $c_{-1},c_0,c_1$ are associated with global conformal transformations on the sphere.

Ghost number

The chiral ghost-number current is often written as

$$j_{gh}(z)=-:b c:(z).$$

With this convention,

$$N_{gh}(c)=+1, \qquad N_{gh}(b)=-1.$$

Ghost number is useful because string amplitudes must soak up ghost zero modes. On compact worldsheets, the ghost-number bookkeeping is tied to the geometry of conformal Killing vectors and moduli.

Ghost vacua

Because the $c$ field has weight $-1$ in the bosonic string, its vacuum structure is slightly unusual. On the open-string cylinder one often uses two ground states,

$$|\downarrow\rangle, \qquad |\uparrow\rangle=c_0|\downarrow\rangle,$$

with

$$b_0|\downarrow\rangle=0, \qquad b_0|\uparrow\rangle=|\downarrow\rangle.$$

This two-state zero-mode system is the oscillator version of a path-integral fact: if a Grassmann zero mode appears and nothing absorbs it, the integral vanishes.

The $b_0,c_0$ zero modes generate a two-state ghost-vacuum structure. This is the simplest avatar of the general ghost zero-mode rule in string amplitudes.

The sphere and three $c$ zero modes

On the Riemann sphere the holomorphic conformal Killing vectors are

$$\epsilon(z)=a+bz+cz^2.$$

They generate the $SL(2,\mathbb C)$ transformations

$$z\mapsto\frac{az+b}{cz+d}.$$

The corresponding $c$-ghost zero-mode part is

$$c(z)=c_1+c_0 z+c_{-1}z^2+\cdots,$$

where the indexing follows the plane expansion $c(z)=\sum_n c_n z^{-n+1}$. A chiral sphere correlator therefore vanishes unless it contains three $c$ insertions.

The standard normalization is

$$\langle c(z_1)c(z_2)c(z_3)\rangle =z_{12}z_{13}z_{23}, \qquad z_{ij}=z_i-z_j.$$

For closed strings, the antiholomorphic ghosts give

$$\langle c\widetilde c(z_1)c\widetilde c(z_2)c\widetilde c(z_3)\rangle =|z_{12}z_{13}z_{23}|^2.$$

This is why sphere amplitudes are often written with three unintegrated closed-string vertices,

$$c\widetilde c V_1, \qquad c\widetilde c V_2, \qquad c\widetilde c V_3,$$

and the remaining vertices integrated over the sphere.

The sphere has three holomorphic conformal Killing vectors. Three $c$ insertions soak up the corresponding zero modes and are usually attached to three unintegrated vertex operators.

What to remember

The $bc$ ghosts are the CFT representation of the Faddeev-Popov determinant for conformal gauge. For the bosonic string,

$$h_b=2, \qquad h_c=-1, \qquad c_{bc}=-26.$$

Together with $D$ free bosons, the total central charge is $D-26$, so flat bosonic string theory is Weyl invariant only at $D=26$. In amplitudes, ghost zero modes determine how many unintegrated vertex operators must be inserted. On the sphere, the magic number is three.

Exercises

Exercise 1. Weights from the ghost stress tensor

Verify that

$$T_{bc}=(1-\lambda):\partial b\,c:-\lambda:b\partial c:$$

makes $b$ a primary of weight $\lambda$ and $c$ a primary of weight $1-\lambda$.

Solution

Using $b(z)c(w)\sim1/(z-w)$, contract the stress tensor with $b(w)$ and $c(w)$. The singular terms are

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

and

$$T(z)c(w) \sim \frac{(1-\lambda)c(w)}{(z-w)^2} + \frac{\partial c(w)}{z-w}.$$

These are precisely the OPEs of primaries with weights $\lambda$ and $1-\lambda$.

Exercise 2. Central charge checks

Compute $c_{bc}$ for $\lambda=2$, $\lambda=1/2$, and $\lambda=1$.

Solution

Using

$$c_{bc}=1-3(2\lambda-1)^2,$$

we find

$$\lambda=2: \quad c=1-3(3)^2=-26,$$ $$\lambda=\frac12: \quad c=1,$$

and

$$\lambda=1: \quad c=1-3=-2.$$

Exercise 3. Mode algebra from the OPE

Use the contour definitions of $b_m$ and $c_n$ to derive $\{b_m,c_n\}=\delta_{m+n,0}$.

Solution

Compute

$$\{b_m,c_n\} = \oint_0\frac{dw}{2\pi i}w^{n-\lambda} \oint_w\frac{dz}{2\pi i}z^{m+\lambda-1}\frac{1}{z-w}.$$

The inner contour gives $w^{m+\lambda-1}$. Therefore

$$\{b_m,c_n\} = \oint_0\frac{dw}{2\pi i}w^{m+n-1} = \delta_{m+n,0}.$$

Exercise 4. Three $c$ ghosts on the sphere

Use conformal invariance to determine the $z_i$ dependence of $\langle c(z_1)c(z_2)c(z_3)\rangle$.

Solution

For three primary fields of weights $h_1,h_2,h_3$,

$$\langle\phi_1(z_1)\phi_2(z_2)\phi_3(z_3)\rangle = \frac{C}{z_{12}^{h_1+h_2-h_3}z_{13}^{h_1+h_3-h_2}z_{23}^{h_2+h_3-h_1}}.$$

Here all three fields are $c$ ghosts with $h=-1$. Thus each exponent equals $-1$, and

$$\langle c(z_1)c(z_2)c(z_3)\rangle = C z_{12}z_{13}z_{23}.$$

The standard normalization sets $C=1$.

Exercise 5. Why three, not two?

Explain why a sphere amplitude with only two $c$ insertions vanishes in the chiral ghost path integral.

Solution

The sphere has three holomorphic conformal Killing vectors, corresponding to three $c$ zero modes. A Grassmann integral over zero modes vanishes unless every zero mode appears once in the integrand. Two $c$ insertions can absorb at most two zero modes, leaving one unabsorbed. Hence the correlator vanishes.