Closed Superstring Sectors and Type II Theories

The open superstring taught us how the NS and R sectors combine with the GSO projection to produce a supersymmetric spectrum. A closed superstring has two independent chiral halves, so the Hilbert space is a tensor product

$$\mathcal H_{\rm closed} = \mathcal H_L\otimes \mathcal H_R.$$

Each side may be NS or R. Before imposing the final GSO choice, this gives four sectors:

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

The first and last sectors are spacetime bosonic; the mixed sectors are spacetime fermionic. Type IIA and Type IIB differ only in how the two Ramond chiralities are chosen.

A closed NSR state is a tensor product of a left-moving state and a right-moving state. The four sectors are NS-NS, NS-R, R-NS, and R-R.

Mass formula and level matching

Let $a_{\rm NS}=1/2$ and $a_{\rm R}=0$. For a closed string, the physical conditions include

$$(L_0-a_L)|\Psi\rangle=0, \qquad (\widetilde L_0-a_R)|\Psi\rangle=0,$$

and the positive-mode constraints

$$L_n|\Psi\rangle=\widetilde L_n|\Psi\rangle=0, \qquad n>0,$$

together with the corresponding supercurrent constraints in the NS or R sector. Since

$$L_0=\frac{\alpha' k^2}{4}+N, \qquad \widetilde L_0=\frac{\alpha' k^2}{4}+\widetilde N,$$

we get

$$\frac{\alpha'M^2}{4}=N-a_L=\widetilde N-a_R.$$

Thus there are two requirements:

$$M^2=\frac{4}{\alpha'}(N-a_L), \qquad N-a_L=\widetilde N-a_R.$$

The second equation is level matching. It says that the left and right chiral excitations must carry equal worldsheet energy after the zero-point shifts are included.

For example, in the NS-NS sector the massless level is

$$N=\widetilde N=\frac12,$$

and a representative state is

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

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

The NS-NS polarization tensor $\epsilon_{\mu\nu}$ decomposes into irreducible spacetime fields:

$$\epsilon_{\mu\nu} = \epsilon_{(\mu\nu)}^{\rm traceless} + \epsilon_{[\mu\nu]} + \frac{1}{D-2}\,\Pi_{\mu\nu}\,\epsilon^\rho{}_{\rho},$$

where $\Pi_{\mu\nu}$ denotes the transverse projector. These three pieces are interpreted as

$$\epsilon_{(\mu\nu)}^{\rm traceless} \longleftrightarrow G_{\mu\nu},$$ $$\epsilon_{[\mu\nu]} \longleftrightarrow B_{\mu\nu},$$

and

$$\epsilon^\rho{}_{\rho} \longleftrightarrow \Phi.$$

So every Type II theory contains the same NS-NS massless fields:

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

Here $G_{\mu\nu}$ is the spacetime metric, $B_{\mu\nu}$ is the antisymmetric two-form gauge field, and $\Phi$ is the dilaton. At low energies these are the fields of the NS-NS sector of ten-dimensional supergravity.

The massless NS-NS tensor splits into a symmetric traceless graviton, an antisymmetric two-form, and a scalar dilaton.

The gauge redundancies are the linearized versions of spacetime gauge symmetries:

$$\delta G_{\mu\nu}=\partial_\mu\xi_\nu+\partial_\nu\xi_\mu,$$

and

$$\delta B_{\mu\nu}=\partial_\mu\Lambda_\nu-\partial_\nu\Lambda_\mu.$$

The first is linearized diffeomorphism invariance; the second is the gauge symmetry of a two-form potential.

The mixed sectors: gravitini and dilatini

The NS-R and R-NS sectors contain spacetime fermions. Schematically, a massless NS-R state has the form

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

and a massless R-NS state is obtained by exchanging left and right. These are vector-spinor states. The physical constraints remove unphysical components and split the result into gravitini and dilatini.

In field-theory language, the mixed sectors provide the fermionic superpartners of the NS-NS and R-R bosons. Type II strings have $32$ real spacetime supercharges, twice as many as the open Type I superstring.

The R-R sector: bispinors become forms

The R-R ground state is a tensor product of two spinor ground states,

$$|A\rangle_L\otimes|\widetilde B\rangle_R.$$

Thus a general R-R polarization is a bispinor $F_{A\widetilde B}$. Gamma matrices translate this bispinor into a sum of antisymmetric tensors:

$$F_{A\widetilde B} = \sum_p \frac{1}{p!} F_{\mu_1\cdots \mu_p} (\Gamma^{\mu_1\cdots \mu_p})_{A\widetilde B}.$$

The massless Dirac equations on the left and right imply the free equations for these differential forms,

$$dF=0, \qquad d{*F}=0,$$

with the appropriate self-duality condition in Type IIB.

The R-R sector naturally produces a polyform. The chirality of the two Ramond ground states determines which form degrees are allowed.

It is often cleaner to discuss R-R field strengths in the worldsheet vertex-operator language and R-R potentials in the spacetime effective action. D-branes couple electrically to the potentials $C_{p+1}$; this is why R-R forms will become central in the D-brane section.

Type IIA versus Type IIB

The NS projection is the same in Type IIA and Type IIB: it removes the NS tachyon and keeps the massless NS excitation. The distinction is the relative chirality of the two Ramond ground states.

In a common convention,

$$\text{Type IIA}:\qquad S_L^+\otimes S_R^-,$$

while

$$\text{Type IIB}:\qquad S_L^+\otimes S_R^+.$$

Equivalently, Type IIA has opposite left/right Ramond chiralities, while Type IIB has the same left/right Ramond chirality. Reversing every chirality label changes notation but not the physics.

The relative chirality of the two Ramond sectors distinguishes Type IIA from Type IIB.

The consequences are substantial:

$$\text{Type IIA is nonchiral}, \qquad \text{Type IIB is chiral}.$$

The R-R potentials also have different degrees:

$$\text{Type IIA}:\qquad C_1, C_3 \quad \text{or democratically} \quad C_1,C_3,C_5,C_7,C_9,$$

whereas

$$\text{Type IIB}: \qquad C_0, C_2,C_4^+ \quad \text{or democratically} \quad C_0,C_2,C_4,C_6,C_8.$$

The superscript $+$ on $C_4^+$ is a reminder that its field strength is self-dual:

$$F_5=*F_5.$$

Type IIA and Type IIB share the same NS-NS fields but differ in fermion chiralities and R-R form degrees.

Strong-coupling preview: Type IIA and eleven dimensions

The nonchiral structure of Type IIA is not an accident. At strong coupling, Type IIA develops an extra circular dimension. The relation is schematically

$$R_{11}\sim g_s\sqrt{\alpha'},$$

so the eleventh dimension is invisible at weak coupling and opens up as $g_s$ grows. This is the first hint of M-theory. In this course the main use of the statement is conceptual: Type IIA is naturally connected to an eleven-dimensional theory, while Type IIB is chiral and behaves differently.

The strong-coupling limit of Type IIA is naturally interpreted as an eleven-dimensional limit with circle radius proportional to $g_s\sqrt{\alpha'}$.

Exercises

Exercise 1. Level matching in the NS-R sector

Show that a massless NS-R state must have $N=1/2$ on the NS side and $\widetilde N=0$ on the R side, assuming $a_{\rm NS}=1/2$ and $a_{\rm R}=0$.

Solution

The closed-string mass formula is

$$\frac{\alpha'M^2}{4}=N-a_L=\widetilde N-a_R.$$

For an NS-R state, $a_L=1/2$ and $a_R=0$. Masslessness means $M^2=0$, so

$$N-\frac12=0, \qquad \widetilde N-0=0.$$

Thus

$$N=\frac12, \qquad \widetilde N=0.$$

The left state has one NS fermion oscillator $\psi_{-1/2}^\mu$, while the right state is a Ramond ground state.

Exercise 2. Counting the NS-NS decomposition

In light-cone gauge, the NS-NS massless polarizations transform as $8_v\otimes 8_v$ under the little group $SO(8)$. Decompose this tensor product into symmetric traceless, antisymmetric, and trace parts, and count dimensions.

Solution

The tensor product has dimension

$$8\times 8=64.$$

The symmetric part has dimension

$$\frac{8\cdot 9}{2}=36.$$

Removing one trace gives the symmetric traceless part of dimension

$$36-1=35.$$

The antisymmetric part has dimension

$$\frac{8\cdot 7}{2}=28.$$

The trace has dimension $1$. Hence

$$8_v\otimes 8_v=35\oplus 28\oplus 1.$$

These are the graviton, $B$-field, and dilaton polarizations.

Exercise 3. Why Type IIB is chiral

Explain why taking the same Ramond chirality on the left and right gives a chiral ten-dimensional theory, while taking opposite chiralities gives a nonchiral theory.

Solution

The spacetime supersymmetry charges come from the Ramond sectors. If the two Ramond sectors have the same ten-dimensional chirality, then the two supersymmetry generators have the same chirality. This is a chiral theory: it treats the two ten-dimensional spinor chiralities asymmetrically.

If the two Ramond sectors have opposite chirality, then the supersymmetry generators come in opposite-chirality pairs. The theory is nonchiral. This is the Type IIA case.

Exercise 4. R-R potentials and D-brane dimensions

Using the fact that a D$p$-brane couples electrically to $C_{p+1}$, determine which D-brane dimensions are allowed in Type IIA and Type IIB from the R-R potential degrees listed above.

Solution

Type IIA has odd-degree R-R potentials,

$$C_1,C_3,C_5,C_7,C_9.$$

Since a D$p$-brane couples to $C_{p+1}$, we get

$$p+1=1,3,5,7,9,$$

so

$$p=0,2,4,6,8.$$

Type IIB has even-degree R-R potentials,

$$C_0,C_2,C_4,C_6,C_8.$$

The electrically coupled D-branes have

$$p+1=2,4,6,8,$$

so

$$p=1,3,5,7,$$

and the $C_0$ potential is naturally associated with the D$(-1)$ instanton.

Exercise 5. The self-dual five-form

Why can Type IIB not have a standard covariant action with an unconstrained five-form field strength $F_5$?

Solution

The Type IIB five-form field strength satisfies the self-duality constraint

$$F_5=*F_5.$$

A conventional kinetic term would have the schematic form

$$\int F_5\wedge *F_5.$$

But if $F_5$ is self-dual, the equations obtained from an unconstrained variation would double the degrees of freedom. The usual supergravity treatment either imposes self-duality after deriving equations from a pseudo-action, or uses a more elaborate formalism that incorporates the constraint directly.