Massless Closed-String Vertices and Gauge Invariance

The closed bosonic string has a universal massless level. In oscillator language it is

$$\epsilon_{\mu\nu}\,\alpha_{-1}^\mu\tilde\alpha_{-1}^\nu|0;k\rangle, \qquad k^2=0,$$

subject to physical-state constraints and gauge equivalences. In the CFT language the corresponding matter vertex operator is

$$V_\epsilon(k;z,\bar z) = \epsilon_{\mu\nu}:\partial X^\mu(z)\,\bar\partial X^\nu(\bar z) e^{ik\cdot X(z,\bar z)}:.$$

For an integrated closed-string vertex we use

$$\int d^2z\,V_\epsilon(k;z,\bar z),$$

while the unintegrated sphere vertex is

$$c(z)\tilde c(\bar z)\,V_\epsilon(k;z,\bar z).$$

The operator $V_\epsilon$ must be a $(1,1)$ primary. This single worldsheet statement contains the spacetime wave equation, the transversality conditions, and the gauge structure of the graviton, two-form, and dilaton.

The massless closed-string vertex factorizes into a left-moving oscillator, a right-moving oscillator, and a center-of-mass plane wave. The polarization tensor $\epsilon_{\mu\nu}$ carries the spacetime field content.

The primary-field test

We use the free-boson conventions

$$X^\mu(z,\bar z)X^\nu(w,\bar w) \sim -\frac{\alpha'}{2}\eta^{\mu\nu}\ln |z-w|^2, \qquad T(z)=-\frac{1}{\alpha'}:\partial X\cdot\partial X:.$$

The exponential has conformal weights

$$:e^{ik\cdot X}: \quad\Longrightarrow\quad (h,\bar h)=\left(\frac{\alpha' k^2}{4},\frac{\alpha' k^2}{4}\right).$$

Since $\partial X^\mu$ has weight $(1,0)$ and $\bar\partial X^\nu$ has weight $(0,1)$, the double-pole part of the OPE with $T$ says

$$h=1+\frac{\alpha' k^2}{4}.$$

Thus the condition $h=1$ gives the massless wave equation

$$k^2=0.$$

But being a primary is stronger than having the right dimension. A primary field $\phi$ satisfies

$$T(z)\phi(w,\bar w) \sim \frac{h\phi(w,\bar w)}{(z-w)^2} +\frac{\partial\phi(w,\bar w)}{z-w} +\text{regular},$$

with no higher-order singularity. For $V_\epsilon$, a double Wick contraction produces a third-order pole:

$$T(z)V_\epsilon(w,\bar w) \sim -\frac{i\alpha'}{2}\frac{k^\mu\epsilon_{\mu\nu}:\bar\partial X^\nu e^{ik\cdot X}:}{(z-w)^3} +\frac{hV_\epsilon(w,\bar w)}{(z-w)^2} +\frac{\partial V_\epsilon(w,\bar w)}{z-w} +\cdots .$$

The antiholomorphic OPE similarly contains a third-order pole proportional to $\epsilon_{\mu\nu}k^\nu$. Therefore the massless closed-string vertex is a $(1,1)$ primary exactly when

$$\boxed{ k^2=0, \qquad k^\mu\epsilon_{\mu\nu}=0, \qquad \epsilon_{\mu\nu}k^\nu=0. }$$

The double pole imposes the weight condition, while the absence of a third-order pole imposes transversality. The same statement in the antiholomorphic sector gives right transversality.

These conditions are the CFT version of the old covariant physical-state constraints

$$L_n|\text{phys}\rangle=\tilde L_n|\text{phys}\rangle=0\quad(n>0), \qquad (L_0-1)|\text{phys}\rangle=(\tilde L_0-1)|\text{phys}\rangle=0.$$

Decomposition into $G_{\mu\nu}$, $B_{\mu\nu}$, and $\Phi$

For a massless particle, the physical little group is $SO(D-2)$. We may choose a light-cone frame in which only transverse polarizations survive. Then

$$\epsilon_{\mu\nu}\quad\longrightarrow\quad \epsilon_{ij}, \qquad i,j=1,\ldots,D-2.$$

The transverse matrix decomposes into irreducible $SO(D-2)$ pieces:

$$\boxed{ \epsilon_{ij} = \left(\epsilon_{(ij)}-\frac{\delta_{ij}}{D-2}\epsilon^k{}_k\right) + \epsilon_{[ij]} + \frac{\delta_{ij}}{D-2}\epsilon^k{}_k. }$$

The three terms are interpreted as

$$\begin{array}{ccl} \text{symmetric traceless} &:& \text{graviton } h_{\mu\nu},\\ \text{antisymmetric} &:& \text{Kalb-Ramond two-form } B_{\mu\nu},\\ \text{trace} &:& \text{dilaton } \Phi. \end{array}$$

The tensor product of left and right transverse oscillators decomposes into the graviton, antisymmetric two-form, and scalar dilaton.

For the critical bosonic string, $D=26$ and the transverse little group is $SO(24)$. The number of physical polarizations is

$$\#h_{ij}=\frac{24\cdot25}{2}-1=299, \qquad \#B_{ij}=\frac{24\cdot23}{2}=276, \qquad \#\Phi=1.$$

Their sum is $299+276+1=576=24^2$, exactly the number of entries of a transverse $24\times24$ matrix.

Gauge redundancy from total derivatives

The polarization tensor is not unique. It has the equivalence

$$\boxed{ \epsilon_{\mu\nu} \sim \epsilon_{\mu\nu}+k_\mu\xi_\nu+\tilde\xi_\mu k_\nu. }$$

This is not an extra assumption. It follows directly from the fact that the corresponding change in the integrated vertex is a total derivative. Indeed,

$$k_\mu\partial X^\mu e^{ik\cdot X} = -i\partial\left(e^{ik\cdot X}\right),$$

so

$$\begin{aligned} \delta \int d^2z\,V_\epsilon &= \int d^2z\, (k_\mu\xi_\nu)\partial X^\mu\bar\partial X^\nu e^{ik\cdot X} \\ &=-i\int d^2z\, \partial\left(\xi_\nu\bar\partial X^\nu e^{ik\cdot X}\right), \end{aligned}$$

up to the free equation of motion $\partial\bar\partial X^\nu=0$ and contact terms. On a closed worldsheet with no boundary this integral vanishes. The same argument applies to the shift $\tilde\xi_\mu k_\nu$ using $\bar\partial$.

A polarization shift proportional to $k_\mu$ or $k_\nu$ changes the integrated vertex by a total derivative. On a closed worldsheet this decouples from amplitudes.

In spacetime terms, the symmetric and antisymmetric pieces become the familiar gauge symmetries

$$h_{\mu\nu} \sim h_{\mu\nu}+\partial_\mu\xi_\nu+\partial_\nu\xi_\mu,$$

and

$$B_{\mu\nu} \sim B_{\mu\nu}+\partial_\mu\Lambda_\nu-\partial_\nu\Lambda_\mu.$$

The gauge-invariant field strength of the two-form is

$$H_{\mu\nu\rho}=3\partial_{[\mu}B_{\nu\rho]}.$$

The dilaton is the gauge-invariant scalar trace component. It will play a special role on the next page: its expectation value controls the string coupling.

The graviton gauge symmetry is linearized diffeomorphism invariance. The two-form has its own one-form gauge parameter. The dilaton is a scalar and controls the string coupling in perturbation theory.

Linearized field equations

The primary-field conditions are equivalent to the free spacetime equations of motion in a convenient gauge. For the graviton, the linearized Einstein equation in de Donder gauge reduces to

$$k^2 h_{\mu\nu}=0, \qquad k^\mu h_{\mu\nu}=0,$$

with residual gauge transformations generated by $\xi_\mu$. For the two-form, the free equation and gauge condition may be written

$$k^2B_{\mu\nu}=0, \qquad k^\mu B_{\mu\nu}=0, \qquad B\sim B+d\Lambda.$$

Thus the worldsheet primary condition is not merely a kinematic trick. It is the first appearance of a recurring principle:

physical spacetime equations arise from the requirement that vertex operators be allowed operators in a conformal field theory.

Summary

The massless closed-string vertex operator

$$\epsilon_{\mu\nu}:\partial X^\mu\bar\partial X^\nu e^{ik\cdot X}:$$

is physical when it is a $(1,1)$ primary modulo total derivatives. This gives

$$k^2=0, \qquad k^\mu\epsilon_{\mu\nu}=0, \qquad \epsilon_{\mu\nu}k^\nu=0, \qquad \epsilon_{\mu\nu}\sim\epsilon_{\mu\nu}+k_\mu\xi_\nu+\tilde\xi_\mu k_\nu.$$

After decomposing the transverse polarization tensor, the massless closed string contains

$$G_{\mu\nu}, \qquad B_{\mu\nu}, \qquad \Phi.$$

The next page asks what spacetime action reproduces their low-energy interactions.

Exercises

Exercise 1: The third-order pole

Using

$$\partial X^\mu(z)\partial X^\nu(w) \sim -\frac{\alpha'}{2}\frac{\eta^{\mu\nu}}{(z-w)^2}, \qquad \partial X^\mu(z)e^{ik\cdot X(w)} \sim -\frac{i\alpha'}{2}\frac{k^\mu}{z-w}e^{ik\cdot X(w)},$$

show that the third-order pole in $T(z)V_\epsilon(w,\bar w)$ is proportional to $k^\mu\epsilon_{\mu\nu}$.

Solution

The stress tensor is $T(z)=-\alpha'^{-1}:\partial X_\rho\partial X^\rho:$. The third-order pole comes from a double contraction: one $\partial X$ in $T$ contracts with $\partial X^\mu$ in the vertex, while the other contracts with the exponential. There are two identical Wick pairings, so

$$T(z)V_\epsilon(w,\bar w) \supset -\frac{1}{\alpha'}2 \left(-\frac{\alpha'}{2}\frac{\eta^{\rho\mu}}{(z-w)^2}\right) \left(-\frac{i\alpha'}{2}\frac{k_\rho}{z-w}\right) \epsilon_{\mu\nu}:\bar\partial X^\nu e^{ik\cdot X}:.$$

Hence

$$T(z)V_\epsilon(w,\bar w) \supset -\frac{i\alpha'}{2}\frac{k^\mu\epsilon_{\mu\nu}:\bar\partial X^\nu e^{ik\cdot X}:}{(z-w)^3}.$$

A primary field has no third-order pole, so $k^\mu\epsilon_{\mu\nu}=0$.

Exercise 2: Count the massless polarizations

Let $n=D-2$. Show that

$$\left[\frac{n(n+1)}{2}-1\right]+\frac{n(n-1)}{2}+1=n^2.$$

Interpret the three terms.

Solution

The first term is the number of symmetric traceless components of an $n\times n$ matrix. The second term is the number of antisymmetric components. The last term is the trace. Adding gives

$$\frac{n(n+1)}{2}-1+\frac{n(n-1)}{2}+1 = \frac{n^2+n+n^2-n}{2}=n^2.$$

This equals the dimension of the product of left and right transverse vector polarizations.

Exercise 3: Gauge shift as a total derivative

Show that the shift $\epsilon_{\mu\nu}\to\epsilon_{\mu\nu}+k_\mu\xi_\nu$ changes the integrated vertex by a total derivative on the worldsheet.

Solution

The change in the vertex is

$$\delta V=(k_\mu\xi_\nu)\partial X^\mu\bar\partial X^\nu e^{ik\cdot X}.$$

Since

$$k_\mu\partial X^\mu e^{ik\cdot X}=-i\partial(e^{ik\cdot X}),$$

we get

$$\delta V=-i\xi_\nu\partial(e^{ik\cdot X})\bar\partial X^\nu.$$

Using the equation of motion $\partial\bar\partial X^\nu=0$ away from contact terms,

$$\delta V=-i\partial\left(\xi_\nu\bar\partial X^\nu e^{ik\cdot X}\right).$$

Thus $\int d^2z\,\delta V$ vanishes on a closed worldsheet without boundary.

Exercise 4: Gauge invariance of $H=dB$

Verify that $H_{\mu\nu\rho}=3\partial_{[\mu}B_{\nu\rho]}$ is invariant under $B_{\mu\nu}\to B_{\mu\nu}+\partial_\mu\Lambda_\nu-\partial_\nu\Lambda_\mu$.

Solution

In differential-form notation, the transformation is $B\to B+d\Lambda$. Therefore

$$H=dB\quad\longrightarrow\quad dB+d^2\Lambda=dB,$$

because $d^2=0$. In components, all terms contain commuting second derivatives antisymmetrized over two indices, so they cancel.