The RNS Spectrum and the GSO Projection

The previous page isolated the two possible spin structures of the RNS fermions. The Neveu—Schwarz sector has half-integer modes and no fermion zero modes. The Ramond sector has integer modes, including zero modes satisfying a Clifford algebra. We now turn these facts into a spacetime spectrum.

This is the first place where the superstring looks dramatically better than the bosonic string. Before one more projection, the NS sector still contains a tachyon. After the GSO projection, the tachyon disappears, the Ramond spinor has definite chirality, and the massless open-string spectrum becomes precisely the field content of ten-dimensional $N=1$ super Yang—Mills theory:

$$\text{massless open superstring}: \qquad A_\mu \oplus \lambda,$$

where $A_\mu$ is a gauge boson and $\lambda$ is a Majorana—Weyl gaugino. In light-cone language this is the equality

$$8_v \oplus 8_s \qquad\text{or}\qquad 8_v \oplus 8_c,$$

depending on the chirality convention.

We will mostly discuss the open superstring, because it is the cleanest place to see the mechanism. Closed-string spectra are obtained by taking left-moving and right-moving tensor products; the type IIA and type IIB theories differ by the relative chiralities chosen in the two Ramond sectors.

Physical-state conditions and conventions

For the spectrum it is useful to restore $\alpha'$ explicitly. The open-string mass formulas will be written as

$$\alpha' M^2=N-a,$$

where $N$ is the oscillator number and $a$ is the intercept. Equivalently, writing the light-cone zero-point energy as $E_0=-a$,

$$\alpha'M^2=N+E_0.$$

In old covariant language, a physical state obeys the positive-mode constraints

$$L_n|\Phi\rangle=0, \qquad n>0,$$

and

$$G_r|\Phi\rangle=0, \qquad r>0,$$

with $r\in\mathbb Z+1/2$ in the NS sector and $r\in\mathbb Z$ in the R sector. The mass-shell condition is the $L_0$ constraint. In the Ramond sector there is also the zero-mode constraint $G_0|\Phi\rangle=0$, which will become the spacetime Dirac equation.

In light-cone gauge the constraints have already been solved. Only the transverse oscillators remain:

$$i=1,\ldots,8,$$

for a ten-dimensional critical superstring. The open-string transverse number operators are

$$N_{\rm NS} = \sum_{n=1}^{\infty}\alpha_{-n}^i\alpha_n^i + \sum_{r=1/2}^{\infty} r\psi_{-r}^i\psi_r^i,$$

and

$$N_{\rm R} = \sum_{n=1}^{\infty}\alpha_{-n}^i\alpha_n^i + \sum_{n=1}^{\infty} n\psi_{-n}^i\psi_n^i.$$

The Ramond zero modes $\psi_0^i$ are not part of $N_{\rm R}$. They act inside the degenerate Ramond ground-state space.

In light-cone gauge the Virasoro and super-Virasoro constraints remove the longitudinal oscillators. The physical RNS Fock space is built from eight transverse bosons and eight transverse fermions.

The central result of this page is the pair of mass formulas

$$\boxed{\alpha'M^2=N_{\rm NS}-{1\over2}} \qquad\text{and}\qquad \boxed{\alpha'M^2=N_{\rm R}}.$$

Let us derive the intercepts and then read off the states.

Normal-ordering constants

The zero-point energy is easiest to compute in light-cone gauge. Each transverse boson contributes the usual integer-mode vacuum energy

$$E_0^X={1\over2}\sum_{n=1}^{\infty}n ={1\over2}\zeta(-1) =-{1\over24}.$$

The sign for a fermionic oscillator is opposite after normal ordering. For an NS fermion the modes are half-integer, so

$$E_0^\psi({\rm NS}) =-{1\over2}\sum_{r=1/2}^{\infty}r =-{1\over2}\zeta\left(-1,{1\over2}\right) =-{1\over48}.$$

Thus one transverse boson-fermion pair contributes

$$E_0^{X+\psi}({\rm NS}) =-{1\over24}-{1\over48} =-{1\over16}.$$

For $D-2$ transverse pairs,

$$E_0({\rm NS})=-{D-2\over16}.$$

In the critical theory $D=10$, hence

$$E_0({\rm NS})=-{1\over2}, \qquad a_{\rm NS}={1\over2}.$$

In the Ramond sector the fermion modes are integer-moded. The zero modes carry no oscillator frequency, and the nonzero modes give

$$E_0^\psi({\rm R}) =-{1\over2}\sum_{n=1}^{\infty}n =+{1\over24}.$$

This cancels the bosonic contribution:

$$E_0^{X+\psi}({\rm R}) =-{1\over24}+{1\over24}=0.$$

Therefore

$$E_0({\rm R})=0, \qquad a_{\rm R}=0.$$

This cancellation is the first concrete signal that worldsheet supersymmetry has softened the vacuum energy. In the NS sector the boundary condition breaks the cancellation by half a unit; in the R sector the cancellation is exact.

The NS sector before projection

The NS ground state is annihilated by all positive modes:

$$\alpha_n^i|0;k\rangle_{\rm NS}=0, \qquad n>0,$$

and

$$\psi_r^i|0;k\rangle_{\rm NS}=0, \qquad r>0.$$

It has oscillator number $N_{\rm NS}=0$, so the mass formula gives

$$\alpha'M^2=-{1\over2}.$$

Thus the unprojected NS ground state is tachyonic. This tachyon is less severe than the bosonic open-string tachyon, whose mass is $\alpha'M^2=-1$, but it is still an instability of the perturbative vacuum.

The first excited NS state is obtained by acting with one fermionic oscillator:

$$|\zeta;k\rangle_{\rm NS} =\zeta_i \psi_{-1/2}^i|0;k\rangle_{\rm NS}.$$

It has

$$N_{\rm NS}={1\over2},$$

and therefore

$$M^2=0.$$

In covariant notation this state is written

$$|\zeta;k\rangle_{\rm NS} =\zeta_\mu \psi_{-1/2}^\mu|0;k\rangle_{\rm NS}.$$

The $G_{1/2}$ constraint imposes transversality. Since

$$G_{1/2}=\sum_n \alpha_n\cdot\psi_{1/2-n},$$

the only term that acts nontrivially on the one-fermion state is the zero-mode piece $\alpha_0\cdot\psi_{1/2}$. This gives

$$G_{1/2}\left(\zeta_\mu \psi_{-1/2}^\mu|0;k\rangle_{\rm NS}\right) \propto (k\cdot \zeta)|0;k\rangle_{\rm NS},$$

so physical states obey

$$k\cdot\zeta=0.$$

The null-state equivalence generated by the constraints identifies

$$\zeta_\mu \sim \zeta_\mu+\lambda k_\mu.$$

Thus the first NS excitation is a massless vector. In the light-cone little group $SO(8)$ it has eight polarizations and transforms as $8_v$.

The first few unprojected open-string NS levels are therefore

$$\begin{array}{c|c|c|c} N_{\rm NS} & \text{representative state} & \alpha'M^2 & \text{interpretation} \\ \hline 0 & |0;k\rangle_{\rm NS} & -1/2 & \text{tachyon} \\ 1/2 & \psi_{-1/2}^i|0;k\rangle_{\rm NS} & 0 & \text{massless vector }8_v \\ 1 & \alpha_{-1}^i|0;k\rangle_{\rm NS},\; \psi_{-1/2}^i\psi_{-1/2}^j|0;k\rangle_{\rm NS} & 1/2 & \text{unprojected massive states} \end{array}$$

The tachyon and the mismatch between integer and half-integer levels tell us that this cannot yet be the final superstring spectrum.

The Ramond sector before projection

The Ramond ground state is defined by

$$\alpha_n^\mu|A;k\rangle_{\rm R}=0, \qquad n>0,$$

and

$$\psi_n^\mu|A;k\rangle_{\rm R}=0, \qquad n>0.$$

Here $A$ is not a decorative label. It is a spacetime spinor index, because the zero modes obey

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

Equivalently, define

$$\Gamma^\mu=\sqrt2\, \psi_0^\mu.$$

Then

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

The Ramond ground states therefore form a representation of the ten-dimensional Clifford algebra. In covariant language the ground state is a spacetime spinor:

$$|A;k\rangle_{\rm R}, \qquad A=1,\ldots,32$$

for a real ten-dimensional Majorana spinor before imposing the massless equation and chirality.

The R-sector mass formula is

$$\alpha'M^2=N_{\rm R}.$$

Since the ground state has $N_{\rm R}=0$, it is massless:

$$M^2=0.$$

The remaining zero-mode superconformal constraint is $G_0|\Phi\rangle=0$. On a Ramond ground state,

$$G_0 \sim \alpha_0\cdot\psi_0,$$

so it becomes

$$k_\mu\Gamma^\mu u(k)=0.$$

This is precisely the massless Dirac equation in ten dimensions. Thus the Ramond ground state is not merely a collection of worldsheet fermion zero modes; it is a spacetime fermion.

The Lorentz generators acting on the zero-mode spinor are

$$S^{\mu\nu}_0 =-{i\over2}[\psi_0^\mu,\psi_0^\nu] =-{i\over4}[\Gamma^\mu,\Gamma^\nu].$$

Together with the orbital and oscillator pieces, they furnish the usual $SO(9,1)$ Lorentz action. This is how spacetime spin enters the RNS formalism even though the fundamental field $\psi^\mu$ was only a worldsheet spinor.

The Ramond zero modes are gamma matrices. The Ramond ground state is a spacetime spinor; the zero-mode constraint gives the massless Dirac equation, and the GSO projection selects a definite chirality.

Counting physical fermion polarizations

The counting is worth spelling out because it is the simplest check of spacetime supersymmetry.

A ten-dimensional Dirac spinor has $32$ real components. The massless Dirac equation halves the number of independent on-shell components. A chirality projection halves it again at the covariant level, but for a massless spinor the cleanest physical statement is made in the light-cone little group.

For a massless particle in ten dimensions, the little group is

$$SO(8).$$

The physical polarizations of a gauge boson transform as the vector representation

$$8_v.$$

The Ramond ground state, after choosing a definite chirality, supplies one of the two inequivalent spinor representations

$$8_s \qquad\text{or}\qquad 8_c.$$

Thus the massless open superstring contains exactly eight bosonic and eight fermionic physical polarizations. The equality is not an accident; it is the first visible form of ten-dimensional spacetime supersymmetry.

The GSO projection

The projection introduced by Gliozzi, Scherk, and Olive is a projection on worldsheet fermion number. Its consequences are much deeper than its definition suggests. It removes the NS tachyon, fixes the chirality of the Ramond spinor, makes the one-loop theory modular invariant, and produces spacetime supersymmetry.

In the NS sector define an oscillator fermion parity $(-1)^{F_{\rm osc}}$ by saying that each fermionic oscillator $\psi_{-r}^i$ changes the sign. The convention appropriate for the open superstring is

$$(-1)^{F_{\rm NS}} =-(-1)^{F_{\rm osc}}.$$

Therefore

$$(-1)^{F_{\rm NS}}|0;k\rangle_{\rm NS}=-|0;k\rangle_{\rm NS},$$

while

$$(-1)^{F_{\rm NS}}\psi_{-1/2}^i|0;k\rangle_{\rm NS} =+\psi_{-1/2}^i|0;k\rangle_{\rm NS}.$$

The GSO projection keeps

$$(-1)^{F_{\rm NS}}=+1.$$

Hence the tachyon is removed and the massless vector is kept.

In the Ramond sector the zero modes make fermion parity more subtle. The operator $(-1)^F$ includes the spacetime chirality matrix. Schematically,

$$(-1)^{F_{\rm R}} =\Gamma^{11}(-1)^{F_{\rm osc}},$$

up to a conventional sign. The GSO projection keeps one chirality:

$$P_\pm={1\over2}\left(1\pm(-1)^{F_{\rm R}}\right).$$

The two choices correspond to the two possible ten-dimensional Majorana—Weyl chiralities. For a single oriented open string, choosing one or the other amounts to choosing whether the massless gaugino transforms as $8_s$ or $8_c$ under the transverse little group. For closed strings, the relative choice between left and right movers is physical: equal chiralities give type IIB, opposite chiralities give type IIA.

The unprojected RNS spectrum has an NS tachyon and a nonchiral Ramond ground state. The GSO projection keeps odd NS fermion number and one Ramond chirality, producing the massless open-superstring vector multiplet.

After the projection, the open-string massless states are

$$\begin{array}{c|c|c|c} \text{sector} & \text{state} & SO(8)\text{ representation} & \text{spacetime field} \\ \hline \text{NS} & \psi_{-1/2}^i|0;k\rangle_{\rm NS} & 8_v & A_\mu \\ \text{R} & |s;k\rangle_{\rm R}^{\rm GSO} & 8_s\text{ or }8_c & \lambda \end{array}$$

The resulting low-energy theory on a stack of open strings is ten-dimensional $N=1$ super Yang—Mills. Chan—Paton factors will turn the single vector multiplet into an adjoint-valued vector multiplet.

Why the projection is physically forced

At this stage it may look as if the GSO projection is a clever surgery performed on the spectrum. That is too weak a view. Several independent demands point to the same projection.

First, the NS tachyon must be absent if the vacuum is to be supersymmetric. Keeping $(-1)^{F_{\rm NS}}=+1$ removes the tachyon and preserves the massless vector.

Second, the Ramond sector must have a definite chirality in ten dimensions. A nonchiral open-string Ramond ground state would not pair correctly with the NS vector to form a minimal supersymmetric multiplet.

Third, the projection is required by one-loop consistency. On the torus or annulus, summing over spin structures with the right signs is what produces modular-invariant amplitudes. Later, when we write theta-function partition functions, the same projection will appear as the identity behind spacetime supersymmetry.

Fourth, spacetime supercharges are built from spin fields. The GSO projection is exactly the mutual-locality condition that allows those spin fields to define good operators in the worldsheet CFT.

So the projection is not merely a way to delete an unwanted particle. It is part of the definition of the consistent superstring.

Closed strings in one paragraph

For closed RNS strings, the left-moving and right-moving sectors are independent. Before imposing projections, the four possibilities are

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

The mass formula is

$${\alpha'M^2\over4}=N_L-a_L=N_R-a_R,$$

with level matching. The GSO projection acts separately on left and right movers. The NS—NS sector gives the graviton, $B$-field, and dilaton. The mixed sectors give gravitini and dilatini. The R—R sector gives differential-form gauge potentials. The detailed type IIA/type IIB spectra will be worked out later, after vertex operators and ghosts are in place.

Summary

The open RNS string has two sectors:

$$\boxed{\alpha'M^2=N_{\rm NS}-{1\over2}} \qquad\text{and}\qquad \boxed{\alpha'M^2=N_{\rm R}}.$$

The NS ground state is a tachyon, while the first NS excitation is a massless vector. The Ramond ground state is massless because the R-sector zero-point energy vanishes. Its zero modes satisfy a spacetime Clifford algebra, so the ground state is a spacetime spinor, and $G_0=0$ becomes the massless Dirac equation.

The GSO projection keeps

$$(-1)^{F_{\rm NS}}=+1$$

in the NS sector, removing the tachyon, and keeps one chirality in the R sector. The massless spectrum after projection is

$$8_v \oplus 8_s \qquad\text{or}\qquad 8_v \oplus 8_c,$$

which is the on-shell field content of a ten-dimensional $N=1$ vector multiplet.

Exercises

Exercise 1: the NS zero-point energy

Use zeta-function regularization to show that one transverse boson plus one transverse NS fermion contributes $-1/16$ to the light-cone zero-point energy. Then derive $a_{\rm NS}=1/2$ in $D=10$.

Solution

For one transverse boson,

$$E_0^X={1\over2}\sum_{n=1}^{\infty}n ={1\over2}\zeta(-1) =-{1\over24}.$$

For one real NS fermion,

$$E_0^\psi({\rm NS}) =-{1\over2}\sum_{r=1/2}^{\infty}r =-{1\over2}\zeta\left(-1,{1\over2}\right).$$

Since

$$\zeta\left(-1,{1\over2}\right)={1\over24},$$

we get

$$E_0^\psi({\rm NS})=-{1\over48}.$$

Therefore one transverse boson-fermion pair contributes

$$E_0^{X+\psi}({\rm NS}) =-{1\over24}-{1\over48} =-{1\over16}.$$

There are $D-2$ transverse pairs, so

$$E_0({\rm NS})=-{D-2\over16}.$$

For $D=10$ this gives

$$E_0({\rm NS})=-{8\over16}=-{1\over2}.$$

Since $a=-E_0$, we find

$$a_{\rm NS}={1\over2}.$$

Exercise 2: the R zero-point energy

Show that the bosonic and fermionic zero-point energies cancel in the Ramond sector.

Solution

For one transverse boson,

$$E_0^X=-{1\over24}.$$

For one real Ramond fermion, the nonzero modes are integer-moded. The zero mode has no oscillator frequency and is not included in the vacuum energy. The nonzero-mode contribution is

$$E_0^\psi({\rm R}) =-{1\over2}\sum_{n=1}^{\infty}n =-{1\over2}\zeta(-1) =+{1\over24}.$$

Thus

$$E_0^{X+\psi}({\rm R}) =-{1\over24}+{1\over24}=0.$$

The cancellation occurs pair by pair, so

$$E_0({\rm R})=0$$

for any number of transverse pairs. In the open superstring this gives

$$\alpha'M^2=N_{\rm R}.$$

Exercise 3: transversality of the NS massless vector

Consider the covariant NS state

$$|\zeta;k\rangle=\zeta_\mu\psi_{-1/2}^\mu|0;k\rangle_{\rm NS}.$$

Use $G_{1/2}|\zeta;k\rangle=0$ to show that $k\cdot\zeta=0$.

Solution

The supercurrent mode is

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

For $r=1/2$,

$$G_{1/2}=\alpha_0\cdot\psi_{1/2}+\sum_{n\ne0}\alpha_n\cdot\psi_{1/2-n}.$$

On the state $\zeta_\mu\psi_{-1/2}^\mu|0;k\rangle_{\rm NS}$, all terms except the zero-mode momentum term either contain annihilators or create states at a different oscillator structure. Using

$$\{\psi_{1/2}^\mu,\psi_{-1/2}^\nu\}=\eta^{\mu\nu},$$

we find

$$G_{1/2}|\zeta;k\rangle \propto \alpha_0\cdot\zeta\, |0;k\rangle_{\rm NS}.$$

Since $\alpha_0^\mu$ is proportional to $k^\mu$, the physical-state condition gives

$$k\cdot\zeta=0.$$

This is the transversality condition for a massless vector polarization.

Exercise 4: the Dirac equation from $G_0$

Show that the Ramond zero-mode constraint $G_0|u;k\rangle_{\rm R}=0$ gives the massless Dirac equation

$$k_\mu\Gamma^\mu u(k)=0.$$

Solution

The supercurrent mode is

$$G_0=\sum_{n\in\mathbb Z}\alpha_n\cdot\psi_{-n}.$$

On a Ramond ground state, the positive modes annihilate the state, while the negative modes have no corresponding excitations to contract with. The only term acting within the ground-state space is

$$G_0\sim \alpha_0\cdot\psi_0.$$

The bosonic zero mode is proportional to momentum, and the fermion zero modes are gamma matrices:

$$\Gamma^\mu=\sqrt2\,\psi_0^\mu.$$

Therefore the constraint $G_0|u;k\rangle_{\rm R}=0$ becomes

$$k_\mu\Gamma^\mu u(k)=0,$$

up to an irrelevant normalization. This is the massless Dirac equation.

Exercise 5: how the NS GSO projection removes the tachyon

Let

$$(-1)^{F_{\rm NS}}=-(-1)^{F_{\rm osc}},$$

where each $\psi_{-r}$ oscillator flips $(-1)^{F_{\rm osc}}$. Show that the NS tachyon is removed while the massless vector is kept if one imposes $(-1)^{F_{\rm NS}}=+1$.

Solution

The NS ground state has no fermionic oscillator excitations, so

$$(-1)^{F_{\rm osc}}|0;k\rangle_{\rm NS}=+|0;k\rangle_{\rm NS}.$$

Because of the extra minus sign in the definition,

$$(-1)^{F_{\rm NS}}|0;k\rangle_{\rm NS} =-|0;k\rangle_{\rm NS}.$$

Thus the tachyonic ground state has eigenvalue $-1$ and is removed by the projection to $+1$.

The massless vector has one fermionic oscillator:

$$\psi_{-1/2}^i|0;k\rangle_{\rm NS}.$$

The oscillator parity gives a minus sign, and the extra NS convention gives another minus sign:

$$(-1)^{F_{\rm NS}}\psi_{-1/2}^i|0;k\rangle_{\rm NS} =+\psi_{-1/2}^i|0;k\rangle_{\rm NS}.$$

Therefore the massless vector survives.

Exercise 6: matching massless bosonic and fermionic degrees of freedom

Explain why the GSO-projected open superstring has eight massless bosonic and eight massless fermionic physical polarizations.

Solution

For a massless particle in ten dimensions, the little group is $SO(8)$.

The GSO-projected NS massless state is

$$\psi_{-1/2}^i|0;k\rangle_{\rm NS}, \qquad i=1,\ldots,8.$$

It transforms as the vector representation $8_v$ of $SO(8)$, so it has eight physical bosonic polarizations.

The GSO-projected Ramond ground state has a definite chirality. In light-cone gauge its physical polarizations form one of the two spinor representations of $SO(8)$:

$$8_s \qquad\text{or}\qquad 8_c.$$

Each has dimension eight. Hence the massless fermion has eight physical polarizations.

Thus the massless open-string spectrum is

$$8_v \oplus 8_s$$

or

$$8_v \oplus 8_c,$$

with equal bosonic and fermionic on-shell degrees of freedom. This is exactly what is required for a ten-dimensional $N=1$ vector multiplet.