Superconformal Algebra and NS/R Sectors

The NSR matter theory is not just a free boson CFT plus some extra fermions. The stress tensor $T$ and supercurrent $G$ form a closed chiral algebra, and the allowed boundary conditions of the fermions split the Hilbert space into two sectors with very different spacetime interpretations.

The whole story begins with the two OPEs

$$X^\mu(z,\bar z)X^\nu(w,\bar w) \sim -\frac{\alpha'}{2}\eta^{\mu\nu}\ln |z-w|^2, \qquad \psi^\mu(z)\psi^\nu(w) \sim \frac{\eta^{\mu\nu}}{z-w}.$$

With the normalization used in the previous page, the matter currents are

$$T(z) = -\frac{1}{\alpha'}:\partial X^\mu\partial X_\mu: -\frac12:\psi^\mu\partial\psi_\mu:, \qquad G(z)=i\sqrt{\frac{2}{\alpha'}}:\psi^\mu\partial X_\mu:.$$

The stress tensor generates conformal transformations; the supercurrent generates the residual worldsheet supersymmetry left after superconformal gauge fixing.

The chiral NSR matter CFT is controlled by the stress tensor $T(z)$ and its spin-$3/2$ superpartner $G(z)$.

The matter superconformal OPEs

The singular OPEs are

$$T(z)T(w) \sim \frac{c/2}{(z-w)^4} + \frac{2T(w)}{(z-w)^2} + \frac{\partial T(w)}{z-w},$$ $$T(z)G(w) \sim \frac{\frac32G(w)}{(z-w)^2} + \frac{\partial G(w)}{z-w},$$

and

$$G(z)G(w) \sim \frac{2c/3}{(z-w)^3} + \frac{2T(w)}{z-w}.$$

For $D$ free bosons and $D$ free Majorana fermions,

$$c_{\rm matter}=D+\frac{D}{2}=\frac{3D}{2}.$$

The second OPE says that $G$ is a primary field of weight $3/2$. The third OPE is the supersymmetric analogue of the $T(z)T(w)$ OPE: the anticommutator of two supersymmetry generators produces a translation, represented here by the stress tensor.

Super-Virasoro modes

Expand the currents as

$$T(z)=\sum_{n\in\mathbb Z}L_n z^{-n-2}, \qquad G(z)=\sum_r G_r z^{-r-3/2}.$$

The allowed values of $r$ depend on the fermion spin structure:

$$r\in\mathbb Z+\frac12 \quad\text{in the NS sector}, \qquad r\in\mathbb Z \quad\text{in the R sector}.$$

Contour manipulations then give the matter super-Virasoro algebra

$$[L_m,L_n] =(m-n)L_{m+n} +\frac{c}{12}(m^3-m)\delta_{m+n,0},$$ $$[L_m,G_r] =\left(\frac{m}{2}-r\right)G_{m+r},$$ $$\{G_r,G_s\} = 2L_{r+s} +\frac{c}{3}\left(r^2-\frac14\right)\delta_{r+s,0}.$$

The super-Virasoro algebra is the mode algebra of $T(z)$ and $G(z)$. The same formula applies in the NS and R sectors, but the allowed labels $r$ are different.

In the full string theory, the matter algebra is accompanied by the $bc$ ghosts and the commuting $\beta\gamma$ superghosts. The critical dimension $D=10$ follows from cancellation of the total central charge. For the spectrum, however, the key point is simpler: physical states are annihilated by the positive modes of both $T$ and $G$.

NS and R boundary conditions

A worldsheet fermion can be antiperiodic or periodic around the spatial circle of the cylinder. For a chiral fermion,

$$\psi^\mu(\sigma+2\pi)=\eta\,\psi^\mu(\sigma),$$

with

$$\eta=-1 \quad\text{for the Neveu--Schwarz sector}, \qquad \eta=+1 \quad\text{for the Ramond sector}.$$

The NS sector is antiperiodic around the cylinder; the R sector is periodic. For closed strings the left- and right-moving sectors can be chosen independently.

For open strings, the boundary conditions at $\sigma=0,\pi$ relate the left- and right-moving fermions, so there is one NS sector and one R sector. For closed strings, left and right movers are independent, giving

$$\text{NS-NS}, \qquad \text{NS-R}, \qquad \text{R-NS}, \qquad \text{R-R}.$$

The first contains familiar bosons such as the graviton, the mixed sectors contain spacetime fermions, and the R-R sector contains antisymmetric tensor gauge potentials.

Fermion modes

On the plane the holomorphic fermion has the mode expansion

$$\psi^\mu(z)=\sum_r \psi_r^\mu z^{-r-1/2},$$

with oscillator algebra

$$\{\psi_r^\mu,\psi_s^\nu\} = \eta^{\mu\nu}\delta_{r+s,0}.$$

Thus

$$\begin{array}{c|c|c} \text{sector} & r & \text{zero mode?}\\ \hline \text{NS} & \mathbb Z+\frac12 & \text{no}\\ \text{R} & \mathbb Z & \text{yes, }\psi_0^\mu \end{array}$$

The absence or presence of the zero mode $\psi_0^\mu$ is the crucial difference between the two sectors. The Ramond zero modes generate spacetime spinors.

The NS vacuum is comparatively simple: it is annihilated by all positive half-integer modes. The R vacuum is degenerate because the zero modes do not annihilate the state. Instead,

$$\{\psi_0^\mu,\psi_0^\nu\}=\eta^{\mu\nu}.$$

Equivalently,

$$\Gamma^\mu=\sqrt2\,\psi_0^\mu \quad\Longrightarrow\quad \{\Gamma^\mu,\Gamma^\nu\}=2\eta^{\mu\nu}.$$

So the Ramond ground states furnish a representation of the spacetime Clifford algebra. This is the first place where spacetime spinors enter the NSR formalism.

Cylinder-plane map and spin fields

Let

$$z=e^{-iw}, \qquad w=\sigma+i\tau_E.$$

A primary field of weight $h$ transforms as

$$\phi_{\rm cyl}(w)=\left(\frac{dz}{dw}\right)^h\phi_{\rm plane}(z).$$

For a fermion, $h=1/2$, so the square root matters:

$$\psi_{\rm cyl}(w)=\left(\frac{dz}{dw}\right)^{1/2}\psi_{\rm plane}(z).$$

Because $\psi$ has weight $1/2$, the cylinder-plane map includes a square root. On the plane the Ramond sector is naturally described by a branch cut ending on a spin-field insertion.

The Ramond ground state is therefore created by a spin field. The detailed construction by bosonization will come later; for now the important fact is that R-sector operators change the boundary condition of $\psi$ around the insertion.

Normal-ordering constants

In light-cone gauge there are $D-2$ transverse bosons and $D-2$ transverse fermions. The open NS zero-point energy is

$$E_0^{\rm NS} =-(D-2)\left(\frac{1}{24}+\frac{1}{48}\right) =-\frac{D-2}{16}.$$

For $D=10$ this is

$$E_0^{\rm NS}=-\frac12.$$

The R-sector fermions are periodic. Their zero-point energy cancels the bosonic contribution, so

$$E_0^{\rm R}=0.$$

Hence the open-string mass formulas in the critical NSR string are

$$\boxed{\alpha'M^2_{\rm NS}=N-\frac12,} \qquad \boxed{\alpha'M^2_{\rm R}=N.}$$

The NS and R sectors have different zero-point energies. In $D=10$, $a_{\rm NS}=1/2$ and $a_{\rm R}=0$ for the open superstring.

Before the GSO projection, the NS sector therefore has a tachyonic ground state, while the R ground state is massless. The next page builds these states explicitly.

Exercises

Exercise 1. The weight of the supercurrent

Using $h(\psi)=1/2$ and $h(\partial X)=1$, explain why the supercurrent $G=i\sqrt{2/\alpha'}:\psi\cdot\partial X:$ has conformal weight $3/2$.

Solution

For a normal-ordered product of free primary fields, the leading conformal weight is the sum of the weights unless singular contractions create additional anomalous terms. Here

$$h(\psi^\mu)+h(\partial X_\mu)=\frac12+1=\frac32.$$

A direct OPE computation gives

$$T(z)G(w) \sim \frac{\frac32G(w)}{(z-w)^2} + \frac{\partial G(w)}{z-w},$$

confirming that $G$ is a primary field of weight $3/2$.

Exercise 2. Fermion modes from boundary conditions

Show that antiperiodic boundary conditions imply $r\in\mathbb Z+1/2$, while periodic boundary conditions imply $r\in\mathbb Z$.

Solution

Write a chiral fermion on the cylinder as

$$\psi(\sigma,\tau)=\sum_r \psi_r(\tau)e^{-ir\sigma}.$$

Then

$$\psi(\sigma+2\pi,\tau) =\sum_r e^{-2\pi i r}\psi_r(\tau)e^{-ir\sigma}.$$

Periodic boundary conditions require $e^{-2\pi i r}=1$, so $r\in\mathbb Z$. Antiperiodic boundary conditions require $e^{-2\pi i r}=-1$, so $r\in\mathbb Z+1/2$.

Exercise 3. Super-Virasoro action on $G_r$

Use the fact that $G$ is a primary field of weight $3/2$ to derive

$$[L_m,G_r]=\left(\frac{m}{2}-r\right)G_{m+r}.$$

Solution

For a primary field $\phi$ of weight $h$ with modes $\phi_r$, the Virasoro action is

$$[L_m,\phi_r]=\big((h-1)m-r\big)\phi_{m+r}.$$

Setting $h=3/2$ and $\phi_r=G_r$ gives

$$[L_m,G_r] =\left(\frac{m}{2}-r\right)G_{m+r}.$$

Exercise 4. NS zero-point energy in $D=10$

Using $\zeta(-1)=-1/12$ and $\zeta(-1,1/2)=1/24$, compute the open NS zero-point energy in $D=10$.

Solution

For one transverse boson,

$$E_0^{\rm boson}=\frac12\sum_{n=1}^{\infty}n =\frac12\zeta(-1) =-\frac{1}{24}.$$

For one transverse NS fermion,

$$E_0^{\rm fermion} =-\frac12\sum_{r=1/2}^{\infty}r =-\frac12\zeta\left(-1,\frac12\right) =-\frac{1}{48}.$$

One transverse boson-fermion pair contributes

$$-\frac{1}{24}-\frac{1}{48}=-\frac{1}{16}.$$

In $D=10$ there are $D-2=8$ transverse pairs, hence

$$E_0^{\rm NS}=8\left(-\frac{1}{16}\right)=-\frac12.$$

Exercise 5. Ramond zero modes

Show that the Ramond zero modes generate the spacetime Clifford algebra.

Solution

The R-sector fermion modes obey

$$\{\psi_r^\mu,\psi_s^\nu\}=\eta^{\mu\nu}\delta_{r+s,0}.$$

Taking $r=s=0$ gives

$$\{\psi_0^\mu,\psi_0^\nu\}=\eta^{\mu\nu}.$$

Define $\Gamma^\mu=\sqrt2\psi_0^\mu$. Then

$$\{\Gamma^\mu,\Gamma^\nu\} =2\{\psi_0^\mu,\psi_0^\nu\} =2\eta^{\mu\nu}.$$

This is the Clifford algebra in spacetime. Therefore the Ramond ground states transform as spacetime spinors.