The Superconformal Algebra and NS/R Sectors

The RNS matter theory is a free two-dimensional supersymmetric field theory, but it is not merely a collection of free fields. Its real organizing principle is the superconformal algebra. This algebra packages the stress tensor $T$ and its fermionic partner $T_F$ into one structure, and it is the algebra whose constraints define physical superstring states.

We focus first on one holomorphic sector. The antiholomorphic sector is identical, with tildes. For open strings the doubling trick leaves a single chiral algebra, while for closed strings the left-moving and right-moving algebras are independent.

Throughout this page we use the common CFT normalization $\alpha'=2$, so that

$$X^\mu(z,\bar z)X^\nu(w,\bar w) \sim -\eta^{\mu\nu}\ln |z-w|^2,$$

and

$$\psi^\mu(z)\psi^\nu(w)\sim {\eta^{\mu\nu}\over z-w}.$$

The holomorphic currents are

$$T(z) =-{1\over2}:\partial X^\mu\partial X_\mu: -{1\over2}:\psi^\mu\partial\psi_\mu:,$$ $$T_F(z)=i:\psi^\mu\partial X_\mu:.$$

The current $T$ has conformal weight $2$, while $T_F$ has conformal weight $3/2$. The pair $(T,T_F)$ is the local current multiplet of $N=1$ worldsheet supersymmetry.

The superconformal OPEs

The stress tensor OPE with a primary field of weight $h$ is

$$T(z)\phi(w)\sim {h\phi(w)\over (z-w)^2}+{\partial\phi(w)\over z-w}.$$

Applying this to $T_F$ gives

$$T(z)T_F(w) \sim {3\over2}{T_F(w)\over (z-w)^2}+{\partial T_F(w)\over z-w}.$$

Thus $T_F$ is a primary field of weight $3/2$. The stress tensor with itself gives the Virasoro OPE

$$T(z)T(w) \sim {c/2\over (z-w)^4}+{2T(w)\over (z-w)^2}+{\partial T(w)\over z-w}.$$

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

$$c_X=D, \qquad c_\psi={D\over2}, \qquad c={3D\over2}.$$

It is often convenient in superconformal field theory to define

$$\hat c={2c\over3}.$$

For the free RNS matter system, this simply gives

$$\hat c=D.$$

The most important OPE is the supercurrent with itself:

$$T_F(z)T_F(w) \sim {2c/3\over (z-w)^3}+{2T(w)\over z-w} ={\hat c\over (z-w)^3}+{2T(w)\over z-w}.$$

This formula is the local CFT version of the slogan

$$\text{two supersymmetries} = \text{one translation}.$$

Indeed, the simple pole contains the stress tensor, and the stress tensor generates conformal translations. The cubic pole is the quantum central extension.

The singular OPEs of $T$ and $T_F$ are equivalent, by contour integration, to the super-Virasoro algebra of the modes $L_n$ and $G_r$.

A quick check of the central term is useful. The bosonic stress tensor contributes $D$ to $c$. A real holomorphic fermion contributes $1/2$, so $D$ such fermions contribute $D/2$. The matter central charge is therefore $3D/2$. This is not yet the full criticality condition of the superstring, because the reparametrization ghosts and superconformal ghosts must also be included. That will lead to $D=10$.

Modes of the currents

The Laurent modes of the stress tensor are defined by

$$T(z)=\sum_{n\in\mathbb Z}{L_n\over z^{n+2}}, \qquad L_n=\oint {dz\over2\pi i}\,z^{n+1}T(z).$$

The supercurrent has weight $3/2$, so its modes are

$$T_F(z)=\sum_r {G_r\over z^{r+3/2}}, \qquad G_r=\oint {dz\over2\pi i}\,z^{r+1/2}T_F(z).$$

The allowed values of $r$ are not fixed by the local OPE. They are fixed by the global boundary condition of the worldsheet fermion around the spatial circle. This is where the Ramond and Neveu—Schwarz sectors enter.

On the Euclidean cylinder $w=\tau_E+i\sigma$, with $\sigma\sim\sigma+2\pi$, the bosons must be periodic:

$$X^\mu(w+2\pi i,\bar w-2\pi i)=X^\mu(w,\bar w).$$

Fermions may instead be periodic or antiperiodic:

$$\psi^\mu_{\rm cyl}(w+2\pi i)=+\psi^\mu_{\rm cyl}(w) \qquad\text{Ramond, R},$$ $$\psi^\mu_{\rm cyl}(w+2\pi i)=-\psi^\mu_{\rm cyl}(w) \qquad\text{Neveu--Schwarz, NS}.$$

Equivalently, with

$$\nu= \begin{cases} 0, & \text{R},\\ {1\over2}, & \text{NS}, \end{cases}$$

the plane expansion is

$$\psi^\mu(z)=\sum_{r\in\mathbb Z+\nu}{\psi_r^\mu\over z^{r+1/2}}.$$

Thus

$$\begin{aligned} \text{NS}:\quad & r\in\mathbb Z+{1\over2}, \\ \text{R}:\quad & r\in\mathbb Z. \end{aligned}$$

The bosonic modes are the same as before:

$$\partial X^\mu(z)=-i\sum_{n\in\mathbb Z}{\alpha_n^\mu\over z^{n+1}}.$$

The oscillator algebra is

$$[\alpha_m^\mu,\alpha_n^\nu]=m\eta^{\mu\nu}\delta_{m+n,0},$$ $$\{\psi_r^\mu,\psi_s^\nu\}=\eta^{\mu\nu}\delta_{r+s,0}.$$

The Ramond sector has fermion zero modes $\psi_0^\mu$. The NS sector does not. This one fact has enormous consequences: Ramond ground states transform as spacetime spinors after quantization, while NS ground states are spacetime bosons.

NS fermions have half-integer modes and no zero mode. R fermions have integer modes, including the Clifford zero modes $\psi_0^\mu$.

The current modes may be written in terms of oscillators as

$$G_r=\sum_{n\in\mathbb Z}\alpha_n\cdot\psi_{r-n},$$

and

$$L_m={1\over2}\sum_{n\in\mathbb Z}:\alpha_{m-n}\cdot\alpha_n: +{1\over2}\sum_r\left(r+{m\over2}\right):\psi_{-r}\cdot\psi_{m+r}:.$$

The colons denote normal ordering. For $L_0$, the oscillator part takes the familiar form

$$L_0={1\over2}\alpha_0^2+N+ a,$$

where $N$ counts oscillator excitations and $a$ is the sector-dependent normal-ordering constant. The detailed values of $a$ are the next ingredient in the spectrum.

The super-Virasoro algebra

Contour integration of the OPEs gives the $N=1$ super-Virasoro algebra:

$$[L_m,L_n]=(m-n)L_{m+n}+{c\over12}m(m^2-1)\delta_{m+n,0},$$ $$[L_m,G_r]=\left({m\over2}-r\right)G_{m+r},$$ $$\{G_r,G_s\}=2L_{r+s}+{c\over3}\left(r^2-{1\over4}\right)\delta_{r+s,0}.$$

Equivalently, using $\hat c=2c/3$,

$$\{G_r,G_s\}=2L_{r+s}+{\hat c\over2}\left(r^2-{1\over4}\right)\delta_{r+s,0}.$$

This algebra has two versions:

$$\begin{array}{c|c|c} \text{sector} & r & \text{fermion boundary condition} \\ \hline \text{NS} & r\in\mathbb Z+{1\over2} & \psi(\sigma+2\pi)=-\psi(\sigma) \\ \text{R} & r\in\mathbb Z & \psi(\sigma+2\pi)=+\psi(\sigma) \end{array}$$

The same local OPEs produce both algebras. What changes is the spin structure of the fermion bundle on the cylinder.

What the cylinder-to-plane map does to fermions

The conformal map from the cylinder to the plane is

$$z=e^w.$$

A primary field of weight $h=1/2$ transforms as

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

Therefore

$$\psi_{\rm plane}(z)=z^{-1/2}\psi_{\rm cyl}(w).$$

The extra factor $z^{-1/2}$ reverses the naive periodicity when one encircles the origin. Going once around the origin corresponds to $w\to w+2\pi i$, and $z^{-1/2}\to -z^{-1/2}$. Hence:

• NS fermions are antiperiodic on the cylinder but single-valued on the plane.
• R fermions are periodic on the cylinder but change sign around the origin on the plane.

This is why the NS vacuum is represented by the identity operator on the plane, while the Ramond vacuum is created by a spin field. A spin field is an operator that inserts a branch cut for $\psi^\mu$. Its detailed construction by bosonization will be one of the main tools for deriving spacetime supersymmetry.

The square-root Jacobian in $z=e^w$ makes the NS sector single-valued on the plane, while the R sector requires a branch cut ending on a spin-field insertion.

For closed strings, the left-moving and right-moving fermions may independently be NS or R. Thus, before projections, the four closed-string sectors are

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

The NS—NS and R—R sectors contain spacetime bosons. The mixed sectors contain spacetime fermions. Later, the GSO projection will correlate these sectors in a way that removes tachyons and produces spacetime supersymmetry.

Highest-weight states in the NS sector

A superconformal primary state $|\phi\rangle$ in the NS sector obeys

$$L_n|\phi\rangle=0\quad(n>0), \qquad G_r|\phi\rangle=0\quad(r>0),$$

and

$$L_0|\phi\rangle=h|\phi\rangle.$$

Since the NS values of $r$ are half-integers, the lowest lowering supercharge is $G_{-1/2}$. The commutator

$$[L_0,G_r]=-rG_r$$

implies

$$L_0G_{-1/2}|\phi\rangle= \left(h+{1\over2}\right)G_{-1/2}|\phi\rangle.$$

Thus $G_{-1/2}|\phi\rangle$ is the superpartner of $|\phi\rangle$, and its conformal weight is $h+1/2$.

The algebra also gives

$$\{G_{-1/2},G_{-1/2}\}=2L_{-1},$$

or

$$G_{-1/2}^2=L_{-1}.$$

This is the radial-quantization version of supersymmetry squaring to a translation.

A useful norm identity follows from

$$\{G_{1/2},G_{-1/2}\}=2L_0.$$

For a highest-weight state,

$$\|G_{-1/2}|\phi\rangle\|^2 =\langle\phi|G_{1/2}G_{-1/2}|\phi\rangle =2h\langle\phi|\phi\rangle.$$

In a unitary theory $h\ge0$. The identity has $h=0$, so its would-be superpartner is null:

$$G_{-1/2}|0\rangle=0.$$

This is why the NS vacuum is special. Ordinary NS primaries with $h>0$ come in supermultiplets.

The Ramond zero mode and the shifted ground-state energy

In the Ramond sector, $r\in\mathbb Z$, so the algebra contains $G_0$. Setting $r=s=0$ in the super-Virasoro algebra gives

$$\{G_0,G_0\}=2L_0-{c\over12}.$$

Equivalently,

$$G_0^2=L_0-{c\over24}.$$

Using $\hat c=2c/3$, this may also be written as

$$G_0^2=L_0-{\hat c\over16}.$$

Therefore the natural Ramond ground-state weight in a unitary superconformal theory is

$$h_R={c\over24}={\hat c\over16}.$$

For the free matter theory with $c=3D/2$, this gives

$$h_R={D\over16}.$$

This number will reappear when we construct spin fields. In $D$ free fermions, the Ramond vacuum is created by a spin field of precisely this conformal weight.

The matter fermion zero modes themselves obey

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

Up to a conventional factor of $\sqrt2$, this is the Clifford algebra of spacetime gamma matrices. This is the microscopic reason the Ramond sector carries spacetime spinor indices. The next page turns this statement into the actual superstring spectrum.

Summary

The RNS matter theory is governed by a local $N=1$ superconformal algebra. The free-field currents

$$T=-{1\over2}:\partial X\cdot\partial X:-{1\over2}:\psi\cdot\partial\psi:, \qquad T_F=i:\psi\cdot\partial X:$$

have central charge

$$c={3D\over2}, \qquad \hat c=D.$$

The global spin structure of $\psi$ separates the theory into NS and R sectors:

$$\psi_{\rm NS}: r\in\mathbb Z+{1\over2}, \qquad \psi_{\rm R}: r\in\mathbb Z.$$

The NS sector has no fermion zero modes and starts from a bosonic ground state. The R sector has fermion zero modes forming a Clifford algebra, and its ground states carry spacetime spinor quantum numbers. These facts are the raw material for the GSO projection and spacetime supersymmetry.

Exercises

Exercise 1: central charge of the RNS matter system

Use the known central charges of a free holomorphic boson and a free holomorphic Majorana fermion to derive

$$c={3D\over2}, \qquad \hat c=D.$$

Solution

A free holomorphic scalar has central charge $1$. For $D$ target-space coordinates $X^\mu$, the bosons contribute

$$c_X=D.$$

A free holomorphic Majorana fermion has central charge $1/2$. For $D$ fermions $\psi^\mu$, the fermions contribute

$$c_\psi={D\over2}.$$

Therefore the matter central charge is

$$c=c_X+c_\psi=D+{D\over2}={3D\over2}.$$

By definition,

$$\hat c={2c\over3},$$

so

$$\hat c={2\over3}{3D\over2}=D.$$

Exercise 2: mode quantization from the cylinder boundary condition

Let

$$\psi_{\rm cyl}(w)=\sum_r \psi_r e^{-rw}.$$

Show that periodic fermions have $r\in\mathbb Z$, while antiperiodic fermions have $r\in\mathbb Z+1/2$.

Solution

Under a spatial loop on the Euclidean cylinder,

$$w\to w+2\pi i.$$

A single mode transforms as

$$e^{-r(w+2\pi i)}=e^{-rw}e^{-2\pi i r}.$$

For a periodic fermion we require

$$e^{-2\pi i r}=1,$$

which gives

$$r\in\mathbb Z.$$

For an antiperiodic fermion we require

$$e^{-2\pi i r}=-1,$$

which gives

$$r\in\mathbb Z+{1\over2}.$$

Thus periodic fermions are Ramond fermions, and antiperiodic fermions are Neveu—Schwarz fermions.

Exercise 3: why the NS sector is single-valued on the plane

Use

$$\psi_{\rm plane}(z)=z^{-1/2}\psi_{\rm cyl}(w), \qquad z=e^w,$$

to show that the NS fermion is single-valued on the plane, whereas the R fermion changes sign around $z=0$.

Solution

Taking $z$ once around the origin corresponds to

$$w\to w+2\pi i.$$

The factor $z^{-1/2}$ changes sign under this continuation:

$$z^{-1/2}\to -z^{-1/2}.$$

In the NS sector, the cylinder fermion is antiperiodic:

$$\psi_{\rm cyl}(w+2\pi i)=-\psi_{\rm cyl}(w).$$

The two minus signs cancel, so

$$\psi_{\rm plane}(e^{2\pi i}z)=\psi_{\rm plane}(z).$$

Thus the NS fermion is single-valued on the plane.

In the R sector, the cylinder fermion is periodic:

$$\psi_{\rm cyl}(w+2\pi i)=\psi_{\rm cyl}(w).$$

Only the Jacobian contributes a minus sign, so

$$\psi_{\rm plane}(e^{2\pi i}z)=-\psi_{\rm plane}(z).$$

Thus the R fermion has a branch cut on the plane. The operator that creates this branch cut is the Ramond spin field.

Exercise 4: the weight of a superpartner

Let $|\phi\rangle$ be an NS superconformal primary with

$$L_0|\phi\rangle=h|\phi\rangle.$$

Use

$$[L_0,G_r]=-rG_r$$

to show that $G_{-1/2}|\phi\rangle$ has weight $h+1/2$.

Solution

Compute

$$L_0G_{-1/2}|\phi\rangle =G_{-1/2}L_0|\phi\rangle+[L_0,G_{-1/2}]|\phi\rangle.$$

The first term is

$$G_{-1/2}L_0|\phi\rangle=hG_{-1/2}|\phi\rangle.$$

The commutator gives

$$[L_0,G_{-1/2}]={1\over2}G_{-1/2}.$$

Therefore

$$L_0G_{-1/2}|\phi\rangle =\left(h+{1\over2}\right)G_{-1/2}|\phi\rangle.$$

So $G_{-1/2}|\phi\rangle$ has conformal weight $h+1/2$.

Exercise 5: supersymmetry squares to translation

Use the super-Virasoro algebra to prove

$$G_{-1/2}^2=L_{-1}$$

in the NS sector. Then use the Ramond zero-mode relation to show that

$$G_0^2=L_0-{c\over24}.$$

Solution

In the NS sector, set $r=s=-1/2$ in

$$\{G_r,G_s\}=2L_{r+s}+{c\over3}\left(r^2-{1\over4}\right)\delta_{r+s,0}.$$

Since $r+s=-1$, the Kronecker delta vanishes. Hence

$$\{G_{-1/2},G_{-1/2}\}=2L_{-1}.$$

For an odd operator $G$, $\{G,G\}=2G^2$, so

$$G_{-1/2}^2=L_{-1}.$$

In the R sector, set $r=s=0$. Then $r+s=0$, so the central term contributes:

$$\{G_0,G_0\}=2L_0+{c\over3}\left(-{1\over4}\right) =2L_0-{c\over12}.$$

Therefore

$$2G_0^2=2L_0-{c\over12},$$

and hence

$$G_0^2=L_0-{c\over24}.$$