Physical States and Type II Superstring Spectra

The previous pages built the ingredients of the RNS superstring: worldsheet fermions, NS and R spin structures, ghosts and superghosts, BRST cohomology, and vertex operators. We now use those ingredients to answer a concrete question:

What are the physical spacetime particles in the ten-dimensional superstring?

The answer is remarkably rigid. The open RNS string, after the GSO projection, contains the on-shell fields of ten-dimensional $N=1$ super-Yang—Mills theory. Tensoring left and right movers gives the closed-string theories: type IIA and type IIB. Their common NS—NS sector contains the metric, Kalb—Ramond two-form, and dilaton, while their R—R sectors differ by chirality: IIA has odd R—R potentials $C_1,C_3$, whereas IIB has even potentials $C_0,C_2,C_4$ with a self-dual five-form field strength.

The key ideas are simple, but the bookkeeping is unforgiving. We will keep the normal-ordering constants, the little-group representations, and the chirality choices visible at every step.

Conventions and oscillator numbers

We work in ten-dimensional flat space with mostly-plus signature. For open strings the physical mass is $M^2=-k^2$. In the RNS matter sector the oscillator numbers are

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

and

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

The normal-ordering constants are

$$a_{\rm NS}={1\over 2}, \qquad a_{\rm R}=0.$$

Thus the open-string mass formulae are

$$\boxed{\alpha' M^2=N_{\rm NS}-{1\over 2}\quad{\rm(NS)},} \qquad \boxed{\alpha' M^2=N_{\rm R}\quad{\rm(R)}.}$$

The difference between the two sectors is already dramatic. The NS vacuum has $M^2=-1/(2\alpha')$, while the Ramond ground state is massless before any oscillator is added. The superstring becomes a sensible supersymmetric theory only after a projection that removes the NS tachyon and chooses a Ramond chirality.

For closed strings we have independent left- and right-moving oscillator numbers. It is useful to introduce

$$\nu=\begin{cases} {1\over 2},& {\rm NS},\\ 0,& {\rm R}, \end{cases} \qquad \widetilde\nu=\begin{cases} {1\over 2},& \widetilde{\rm NS},\\ 0,& \widetilde{\rm R}. \end{cases}$$

The closed-string mass and level-matching conditions are

$$\boxed{\ {\alpha' M^2\over 4}=N-\nu=\widetilde N-\widetilde\nu\ }, \qquad \boxed{\ N-\nu=\widetilde N-\widetilde\nu\ }.$$

Equivalently,

$$\alpha' M^2=2\left(N+\widetilde N-\nu-\widetilde\nu\right), \qquad N-\widetilde N=\nu-\widetilde\nu.$$

For example, a massless NS—NS state has $N=\widetilde N=1/2$, a massless R—R state has $N=\widetilde N=0$, and a massless NS—R state has $N=1/2$, $\widetilde N=0$.

Physical-state conditions in old covariant language

For quick spectrum counting one often uses old covariant quantization and then translates the result into BRST language. The matter conditions are

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

with the additional mass-shell condition

$$(L_0-a)|\Phi\rangle=0.$$

Here $a=a_{\rm NS}=1/2$ in the NS sector and $a=a_{\rm R}=0$ in the R sector. In the Ramond sector there is also the zero-mode condition

$$G_0|\Phi\rangle=0.$$

Strictly speaking, after fixing conformal gauge the physical Hilbert space is BRST cohomology, not just the matter Hilbert space satisfying these equations. For the massless spectrum in flat space, however, the old covariant conditions give precisely the familiar transversality, gauge equivalence, and Dirac equations. BRST language refines the statement: null or pure-gauge states are BRST exact.

Let us see this explicitly for the first open-string states.

The open NS sector: vector bosons and the tachyon problem

The NS ground state is a spacetime scalar,

$$|0;k\rangle_{\rm NS},$$

with

$$N_{\rm NS}=0, \qquad \alpha'M^2=-{1\over 2}.$$

This is the open-string tachyon. The superstring projection must remove it.

The first excited NS state is

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

The mass formula gives

$$M^2=0.$$

The physical-state conditions impose the usual gauge-boson constraints. The condition $G_{1/2}|\zeta;k\rangle=0$ gives

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

There is also a null state generated by a gauge parameter. In BRST language it is exact; in spacetime language it is the gauge redundancy

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

Thus the first surviving NS state is not an arbitrary ten-vector. A massless vector in ten dimensions has $10-2=8$ physical polarizations. These transform as the vector representation $\mathbf 8_v$ of the massless little group $SO(8)$.

This is the first hint that the light-cone spectrum is the cleanest way to read off the physical degrees of freedom. The ten-dimensional Lorentz representation is covariant, but the physical polarizations are organized by $SO(8)$.

The massless NS vector loses two components through transversality and gauge equivalence, leaving eight transverse polarizations in $\mathbf 8_v$. The Ramond ground state obeys a massless Dirac equation; after a Weyl projection it also carries eight physical polarizations.

The open Ramond sector: spacetime spinors

The Ramond sector has integer-moded worldsheet fermions. The crucial new feature is the zero mode:

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

This is a Clifford algebra. We may represent it as

$$\psi_0^\mu={1\over\sqrt 2}\Gamma^\mu, \qquad \{\Gamma^\mu,\Gamma^\nu\}=2\eta^{\mu\nu}.$$

Therefore the Ramond ground states are spacetime spinors. We write them as

$$|u;k\rangle_{\rm R},$$

where $u$ is a ten-dimensional spinor. Since $a_{\rm R}=0$, the Ramond ground state is massless:

$$\alpha'M^2=0.$$

Now impose the zero-mode physical-state condition. On the ground state the oscillator part of $G_0$ drops out, and $G_0$ is proportional to $k\cdot\psi_0$. Thus

$$G_0|u;k\rangle=0 \quad\Longrightarrow\quad k_\mu\Gamma^\mu u=0.$$

This is the massless Dirac equation in ten dimensions.

A Dirac spinor in ten dimensions has $32$ complex components before reality conditions. In Lorentzian ten-dimensional string theory one can impose a Majorana condition and, independently for massless states, a Weyl chirality projection. A Majorana—Weyl spinor has $16$ real components off shell. The massless Dirac equation then removes half of them, leaving $8$ on-shell physical polarizations. These transform as either $\mathbf 8_s$ or $\mathbf 8_c$ of $SO(8)$, depending on the chirality convention.

This matches the open NS vector:

$$\mathbf 8_v \quad\longleftrightarrow\quad \mathbf 8_s\ \text{or}\ \mathbf 8_c.$$

The equality of dimensions is not an accident; after the GSO projection the open superstring has ten-dimensional spacetime supersymmetry.

The open-string GSO projection

The raw RNS spectrum is not yet a spacetime-supersymmetric theory. The NS sector contains a tachyon, and the Ramond sector contains too many spinor states unless a chirality is chosen. The GSO projection is the worldsheet projection that fixes both problems.

Let $(-1)^F$ be worldsheet fermion parity. In the NS sector one convention is

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

Every fermionic oscillator $\psi_{-r}^\mu$ flips the sign. Therefore

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

The projection

$${1+(-1)^F\over 2}$$

removes the NS tachyon and keeps the massless vector.

In the Ramond sector, the projection is a chirality projection. Let

$$\Gamma_{11}=\Gamma^0\Gamma^1\cdots\Gamma^9, \qquad \Gamma_{11}^2=1.$$

A choice of GSO projection keeps either

$$\Gamma_{11}u=+u \qquad\text{or}\qquad \Gamma_{11}u=-u.$$

For an oriented open superstring one may choose either chirality. The two choices are equivalent up to a convention for what is called $\mathbf 8_s$ and what is called $\mathbf 8_c$. The physical massless content is a ten-dimensional $N=1$ vector multiplet:

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

where $A_\mu$ has $8$ on-shell bosonic degrees of freedom and $\lambda$ is a Majorana—Weyl gaugino with $8$ on-shell fermionic degrees of freedom.

With Chan—Paton factors included, $A_\mu$ and $\lambda$ become adjoint-valued fields. In the low-energy limit their interactions are those of ten-dimensional $N=1$ super-Yang—Mills theory.

Closed strings: tensoring left and right movers

A closed-string state is a tensor product of a left-moving state and a right-moving state. Before the final GSO choice, the four sectors are

$$\mathrm{NS}\otimes\widetilde{\mathrm{NS}}, \qquad \mathrm{NS}\otimes\widetilde{\mathrm R}, \qquad \mathrm R\otimes\widetilde{\mathrm{NS}}, \qquad \mathrm R\otimes\widetilde{\mathrm R}.$$

The tilde emphasizes that the right-moving GSO projection is independent of the left-moving one.

The massless states are obtained by the minimal oscillator numbers compatible with each sector:

$$\begin{array}{c|c|c|c} \text{sector} & N & \widetilde N & \text{massless state} \\ \hline \mathrm{NS}\otimes\widetilde{\mathrm{NS}} & {1\over2} & {1\over2} & \psi_{-1/2}^\mu\widetilde\psi_{-1/2}^\nu|0;k\rangle \\ \mathrm{NS}\otimes\widetilde{\mathrm R} & {1\over2} & 0 & \psi_{-1/2}^\mu|0;k\rangle\otimes|\widetilde u;k\rangle \\ \mathrm R\otimes\widetilde{\mathrm{NS}} & 0 & {1\over2} & |u;k\rangle\otimes\widetilde\psi_{-1/2}^\mu|0;k\rangle \\ \mathrm R\otimes\widetilde{\mathrm R} & 0 & 0 & |u;k\rangle\otimes|\widetilde u;k\rangle \end{array}$$

The little group organizes these states elegantly. After the GSO projection the NS massless state is $\mathbf 8_v$. The Ramond ground state is either $\mathbf 8_s$ or $\mathbf 8_c$. Thus the closed-string massless spectrum is a collection of tensor products of $SO(8)$ representations.

The NS—NS sector is common to type IIA and type IIB. The distinction between the two theories is the relative chirality of the left- and right-moving Ramond sectors.

The NS—NS sector: $G_{\mu\nu}$, $B_{\mu\nu}$, and $\Phi$

The massless NS—NS state has the form

$$\epsilon_{\mu\nu}\psi_{-1/2}^\mu\widetilde\psi_{-1/2}^\nu|0;k\rangle_{\rm NS\text{--}NS}.$$

The physical-state conditions imply

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

with gauge equivalences inherited from left and right null states. In light-cone gauge the polarization tensor is an $8\times8$ tensor under $SO(8)$:

$$\mathbf 8_v\otimes\mathbf 8_v.$$

Decompose this tensor into symmetric traceless, antisymmetric, and trace parts:

$$\mathbf 8_v\otimes\mathbf 8_v = \mathbf{35}_v\oplus\mathbf{28}\oplus\mathbf 1.$$

These are the on-shell degrees of freedom of

$$\boxed{ \mathbf{35}_v\leftrightarrow G_{\mu\nu}, \qquad \mathbf{28}\leftrightarrow B_{\mu\nu}, \qquad \mathbf 1\leftrightarrow \Phi. }$$

Here $G_{\mu\nu}$ is the spacetime metric, $B_{\mu\nu}$ is the Kalb—Ramond two-form, and $\Phi$ is the dilaton. This sector appears in both type IIA and type IIB.

The counting is worth internalizing. In $D$ spacetime dimensions, a massless graviton has

$${D(D-3)\over 2}$$

physical polarizations. For $D=10$ this is $35$. A massless two-form has

$${D-2\choose 2}={8\choose 2}=28$$

physical polarizations. The dilaton has one. Thus the NS—NS sector contains

$$35+28+1=64$$

bosonic degrees of freedom.

The R—R sector: spinor bilinears and differential forms

The R—R ground states are tensor products of left and right spacetime spinors. A useful representation-theoretic fact is that spinor bilinears in ten dimensions are equivalent to differential forms. Schematically,

$$S\otimes S' \sim \sum_p C_{\mu_1\cdots\mu_p}\Gamma^{\mu_1\cdots\mu_p}.$$

The allowed values of $p$ are controlled by the relative chiralities of $S$ and $S'$.

In type IIA the left and right Ramond ground states have opposite ten-dimensional chirality. The R—R potentials are odd-degree forms:

$$\boxed{\text{type IIA: } C_1,\ C_3.}$$

Their magnetic dual potentials are $C_5$ and $C_7$ in the democratic description, but the minimal two-derivative formulation may use $C_1$ and $C_3$.

In type IIB the left and right Ramond ground states have the same ten-dimensional chirality. The R—R potentials are even-degree forms:

$$\boxed{\text{type IIB: } C_0,\ C_2,\ C_4.}$$

The field strength of $C_4$ is constrained by self-duality,

$$\widetilde F_5=*\widetilde F_5.$$

At the level of $SO(8)$ physical polarizations this becomes

$$\begin{aligned} \text{IIA:}\qquad \mathbf 8_s\otimes\mathbf 8_c&=\mathbf 8_v\oplus\mathbf{56}_v,\\ \text{IIB:}\qquad \mathbf 8_s\otimes\mathbf 8_s&=\mathbf 1\oplus\mathbf{28}\oplus\mathbf{35}_+. \end{aligned}$$

The dimensions are again $64$ in either theory:

$$8+56=64, \qquad 1+28+35=64.$$

Thus the total number of massless bosonic degrees of freedom in either type II theory is

$$64_{\rm NS\text{--}NS}+64_{\rm R\text{--}R}=128.$$

This number will match the fermions.

The mixed sectors: gravitini and dilatini

The NS—R and R—NS sectors are spacetime fermions. A typical NS—R state is

$$\psi_{-1/2}^\mu|0;k\rangle_{\rm NS}\otimes|\widetilde u;k\rangle_{\rm R}.$$

In light-cone language this is a tensor product

$$\mathbf 8_v\otimes\mathbf 8_s \quad\text{or}\quad \mathbf 8_v\otimes\mathbf 8_c.$$

The decomposition is

$$\mathbf 8_v\otimes\mathbf 8_s=\mathbf{56}_c\oplus\mathbf 8_c, \qquad \mathbf 8_v\otimes\mathbf 8_c=\mathbf{56}_s\oplus\mathbf 8_s.$$

The $\mathbf{56}$ representation is the on-shell content of a gravitino; the $\mathbf 8$ is the on-shell content of a dilatino. Because there are two mixed sectors, closed type II strings have two gravitini and two dilatini.

The relative chirality distinguishes the theories:

$$\boxed{ \begin{array}{ccl} \text{type IIA} &:& \text{two gravitini of opposite ten-dimensional chirality},\\ \text{type IIB} &:& \text{two gravitini of the same ten-dimensional chirality}. \end{array}}$$

This is the origin of the names:

$$\text{IIA is non-chiral } N=(1,1), \qquad \text{IIB is chiral } N=(2,0).$$

Each mixed sector contains $64$ fermionic degrees of freedom, so the total is

$$64_{\rm NS\text{--}R}+64_{\rm R\text{--}NS}=128.$$

Therefore each type II theory has

$$\boxed{128\ \text{bosonic} + 128\ \text{fermionic massless degrees of freedom}.}$$

This is the massless spectrum of ten-dimensional type II supergravity.

Type IIA versus type IIB in one table

The entire distinction is encoded in the relative chirality of the two Ramond sectors. Let $R_+$ and $R_-$ denote Ramond ground states of positive and negative ten-dimensional chirality. Then, schematically,

$$\begin{array}{ccl} \text{IIA} &:& (R_+,\widetilde R_-) \quad\text{or equivalently}\quad (R_-,\widetilde R_+),\\ \text{IIB} &:& (R_+,\widetilde R_+) \quad\text{or equivalently}\quad (R_-,\widetilde R_-). \end{array}$$

A parity transformation flips one chirality, so type IIA is parity invariant in ten dimensions, while type IIB is chiral. The chirality of IIB is visible in the self-duality of $\widetilde F_5$ and in the fact that its two supersymmetry generators have the same chirality.

It is often useful to summarize the GSO choices as follows.

Type IIA and type IIB are the two maximally supersymmetric oriented closed superstrings. Type 0A and type 0B are non-supersymmetric but modular-consistent alternatives with an NS—NS tachyon and no spacetime fermions.

Type 0 strings as a useful contrast

The GSO projection is not uniquely fixed by modular invariance alone. There are non-supersymmetric oriented closed-string theories, called type 0A and type 0B, obtained by a diagonal projection that removes all mixed NS—R and R—NS sectors. Since those mixed sectors carry spacetime fermions, type 0 strings have no spacetime fermions.

They do keep an NS—NS tachyon. The schematic sectors are

$$\begin{array}{ccl} \text{type 0A} &:& (\mathrm{NS}_+,\widetilde{\mathrm{NS}}_+) \oplus (\mathrm{NS}_-,\widetilde{\mathrm{NS}}_-) \oplus (R_+,\widetilde R_-) \oplus (R_-,\widetilde R_+),\\ \text{type 0B} &:& (\mathrm{NS}_+,\widetilde{\mathrm{NS}}_+) \oplus (\mathrm{NS}_-,\widetilde{\mathrm{NS}}_-) \oplus (R_+,\widetilde R_+) \oplus (R_-,\widetilde R_-). \end{array}$$

Here $\mathrm{NS}_+$ denotes the GSO-even NS sector, whose first state is the massless vector in the open-string case or the massless NS—NS fields in the closed-string case. The $\mathrm{NS}_-$ sector contains the tachyon.

The R—R spectrum of type 0 theories is doubled relative to type II: type 0A has doubled odd R—R potentials, while type 0B has doubled even R—R potentials. These theories are unstable in flat space because of the tachyon, but they are useful conceptual laboratories. They show that the absence of tachyons and the presence of spacetime supersymmetry are additional dynamical virtues, not automatic consequences of writing a modular-invariant RNS partition function.

How the covariant fields know about gauge invariance

The little-group counting gives the physical degrees of freedom, but the covariant fields are more useful in spacetime effective actions. The map between the two descriptions works because the physical-state constraints and BRST equivalences impose gauge invariance.

For the open string,

$$\zeta_\mu\psi_{-1/2}^\mu|0;k\rangle$$

obeys

$$k\cdot\zeta=0, \qquad \zeta_\mu\sim\zeta_\mu+k_\mu\lambda.$$

This is exactly the momentum-space gauge invariance of a Maxwell field.

For the closed NS—NS field,

$$\epsilon_{\mu\nu}\psi_{-1/2}^\mu\widetilde\psi_{-1/2}^\nu|0;k\rangle$$

has two independent gauge equivalences, one from the left movers and one from the right movers. Decomposing

$$\epsilon_{\mu\nu}=h_{\mu\nu}+b_{\mu\nu}+{1\over 10}\eta_{\mu\nu}\epsilon^\rho{}_{\rho}$$

identifies

$$\begin{aligned} h_{\mu\nu}&=h_{\nu\mu},\\ b_{\mu\nu}&=-b_{\nu\mu},\\ \Phi&\sim \epsilon^\rho{}_{\rho}. \end{aligned}$$

The corresponding gauge transformations are the linearized forms of

$$\delta h_{\mu\nu}=\partial_\mu\xi_\nu+\partial_\nu\xi_\mu, \qquad \delta b_{\mu\nu}=\partial_\mu\Lambda_\nu-\partial_\nu\Lambda_\mu.$$

Thus the string spectrum does not merely contain particles with the right number of components; it contains the gauge redundancies required for gravity and antisymmetric tensor gauge theory.

The same philosophy applies to R—R fields. The R—R vertex operators describe gauge-invariant field strengths most naturally, while the low-energy supergravity action is often written in terms of potentials $C_p$ with gauge transformations

$$C_p\mapsto C_p+d\Lambda_{p-1}$$

and, in the presence of $B_2$, modified gauge-invariant field strengths.

Summary of the massless type II spectra

The massless bosonic fields are

$$\boxed{ \begin{array}{c|c|c} \text{theory} & \text{NS--NS} & \text{R--R} \\ \hline \text{type IIA} & G_{\mu\nu},\ B_{\mu\nu},\ \Phi & C_1,\ C_3 \\ \text{type IIB} & G_{\mu\nu},\ B_{\mu\nu},\ \Phi & C_0,\ C_2,\ C_4^+ \end{array}}$$

The fermions are

$$\boxed{ \begin{array}{c|c} \text{theory} & \text{massless fermions} \\ \hline \text{type IIA} & \text{two gravitini of opposite chirality plus two dilatini} \\ \text{type IIB} & \text{two gravitini of the same chirality plus two dilatini of the opposite chirality} \end{array}}$$

Both theories have $32$ real supercharges and $128+128$ on-shell massless degrees of freedom. The difference between them is not the number of states but their chirality and R—R form content. This difference will become physically decisive when we study T-duality and D-branes: T-duality along one circle flips one Ramond chirality, and therefore exchanges type IIA with type IIB.

Exercises

Exercise 1. The massless vector from the NS sector

Consider the open-string NS state

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

Use the mass formula and the physical-state condition $G_{1/2}|\zeta;k\rangle=0$ to show that this state is a massless transverse vector. Explain why it has eight physical polarizations in ten dimensions.

Solution

The state has oscillator number

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

because one fermionic oscillator $\psi_{-1/2}^\mu$ has level $1/2$. The NS mass formula gives

$$\alpha'M^2=N_{\rm NS}-{1\over2}=0.$$

Thus $k^2=0$.

The supercurrent mode satisfies schematically

$$G_{1/2}\sim \sum_n \alpha_n\cdot\psi_{1/2-n}.$$

When acting on the one-oscillator state, the relevant term contracts $\alpha_0^\mu\propto k^\mu$ with $\psi_{-1/2}^\mu$, giving

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

The physical-state condition therefore requires

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

There is also a gauge equivalence $\zeta_\mu\sim\zeta_\mu+\lambda k_\mu$. A ten-vector has $10$ components; transversality removes one and gauge equivalence removes another, leaving $8$ physical polarizations. These transform as $\mathbf 8_v$ of the massless little group $SO(8)$.

Exercise 2. Ramond zero modes and the Dirac equation

Using

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

show that Ramond ground states transform as spacetime spinors. Then show that the physical-state condition $G_0|u;k\rangle=0$ gives the massless Dirac equation.

Solution

The zero modes obey a Clifford algebra. If

$$\psi_0^\mu={1\over\sqrt2}\Gamma^\mu,$$

then

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

A Hilbert space carrying this algebra is a spinor representation of the ten-dimensional Lorentz group. Therefore the Ramond ground state can be written as $|u;k\rangle_{\rm R}$, where $u$ is a spacetime spinor.

On the Ramond ground state the oscillator terms in $G_0$ vanish. The zero-mode contribution is proportional to

$$k_\mu\psi_0^\mu={1\over\sqrt2}k_\mu\Gamma^\mu.$$

Thus

$$G_0|u;k\rangle=0 \quad\Longrightarrow\quad k_\mu\Gamma^\mu u=0.$$

This is the massless Dirac equation. After imposing a Majorana—Weyl condition and the massless Dirac equation, the state has eight physical polarizations.

Exercise 3. Counting the NS—NS fields

Show that

$$\mathbf 8_v\otimes\mathbf 8_v = \mathbf{35}_v\oplus\mathbf{28}\oplus\mathbf 1.$$

Identify the corresponding spacetime fields.

Solution

The tensor product of two eight-dimensional vector representations is the space of $8\times8$ tensors $T_{ij}$. It decomposes into symmetric traceless, antisymmetric, and trace parts:

$$T_{ij}=T_{(ij)}^{\rm traceless}+T_{[ij]}+{1\over8}\delta_{ij}T^k{}_k.$$

The dimensions are

$$\dim({\rm symmetric})={8\cdot9\over2}=36,$$

so the symmetric traceless part has dimension $36-1=35$. The antisymmetric part has dimension

$${8\cdot7\over2}=28,$$

and the trace has dimension $1$. Therefore

$$8\times8=35+28+1.$$

The symmetric traceless part is the graviton $G_{\mu\nu}$, the antisymmetric part is the two-form $B_{\mu\nu}$, and the trace is the dilaton $\Phi$.

Exercise 4. R—R forms in IIA and IIB

Use the rule that a product of two ten-dimensional spinors of the same chirality gives even-degree forms, while a product of spinors of opposite chirality gives odd-degree forms. Explain why IIA has R—R potentials $C_1,C_3$, while IIB has $C_0,C_2,C_4^+$.

Solution

R—R ground states are spinor bilinears. Gamma matrices with $p$ antisymmetrized indices map spinors of one chirality to spinors of either the same or the opposite chirality depending on $p$. Even $p$ preserves ten-dimensional chirality, while odd $p$ flips it.

Equivalently, the bispinor expansion has the schematic form

$$S_+\otimes S_+\sim \text{even forms}, \qquad S_+\otimes S_-\sim \text{odd forms}.$$

Type IIB has left and right Ramond ground states of the same chirality, so its R—R potentials are even-degree forms. In the minimal formulation these are

$$C_0, \qquad C_2, \qquad C_4.$$

The five-form field strength of $C_4$ is self-dual, so we write $C_4^+$ as a reminder of the self-duality constraint on $F_5$.

Type IIA has left and right Ramond ground states of opposite chirality, so its R—R potentials are odd-degree forms. The minimal potentials are

$$C_1, \qquad C_3.$$

Their magnetic duals may be included in the democratic formulation, but they are not independent degrees of freedom.

Exercise 5. Degree-of-freedom matching in type II strings

Show that the massless bosonic and fermionic degrees of freedom in type II string theory both equal $128$.

Solution

The NS—NS bosons contribute

$$35+28+1=64.$$

For type IIA the R—R degrees are

$$\mathbf 8_v\oplus\mathbf{56}_v,$$

which contribute

$$8+56=64.$$

For type IIB the R—R degrees are

$$\mathbf1\oplus\mathbf{28}\oplus\mathbf{35}_+,$$

which contribute

$$1+28+35=64.$$

Thus either theory has

$$64+64=128$$

massless bosonic degrees of freedom.

For fermions, each mixed sector contains

$$\mathbf 8_v\otimes\mathbf 8_s \quad\text{or}\quad \mathbf 8_v\otimes\mathbf 8_c,$$

which decomposes as

$$\mathbf{56}\oplus\mathbf8.$$

This has $64$ states. There are two mixed sectors, NS—R and R—NS, so the fermionic count is

$$64+64=128.$$

Therefore type II strings have $128$ bosonic and $128$ fermionic massless degrees of freedom.

Exercise 6. Why type 0 strings are not supersymmetric

Type 0 theories keep NS—NS and R—R sectors but remove NS—R and R—NS sectors. Explain why this removes spacetime fermions and why a tachyon remains.

Solution

Spacetime statistics in the RNS string are tied to the number of Ramond factors. NS—NS and R—R sectors are spacetime bosonic. The mixed sectors NS—R and R—NS are spacetime fermionic because exactly one side carries a Ramond spinor.

If a projection removes the mixed sectors, no spacetime fermions remain. Without spacetime fermions there cannot be spacetime supersymmetry, since supersymmetry generators map bosons to fermions.

The type 0 diagonal GSO projection also keeps the sector

$$\mathrm{NS}_-\otimes\widetilde{\mathrm{NS}}_-.$$

The lowest state in this sector is the NS—NS tachyon. Its mass is negative because each NS vacuum contributes $-1/2$ to the left or right intercept. Therefore type 0 strings are non-supersymmetric and unstable in flat ten-dimensional spacetime.