Ten-Dimensional Spinors and Ramond Ground States

The Ramond sector of the NSR string is the first place where spacetime spinors appear. This is not put in by hand. It follows from one elementary fact: Ramond worldsheet fermions have zero modes, and those zero modes obey a spacetime Clifford algebra.

The previous page found

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

Therefore the Ramond ground state is not a single scalar state. It is a degenerate multiplet on which the gamma matrices act:

$$|0;k\rangle_{\rm R} \quad\rightsquigarrow\quad |u;k\rangle_{\rm R},$$

where $u$ is a spacetime spinor. The physical-state condition $G_0|u;k\rangle=0$ then becomes the massless Dirac equation,

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

This page supplies the spinor technology needed to understand this statement in ten dimensions.

The Ramond zero modes $\psi_0^\mu$ act on the degenerate ground states. After $\Gamma^\mu=\sqrt2\psi_0^\mu$, their algebra is exactly the spacetime Clifford algebra.

Clifford algebras and spinor modules

Let $D$ be the spacetime dimension, with mostly-plus metric

$$\eta_{\mu\nu}=\operatorname{diag}(-,+,\ldots,+).$$

The Lorentzian Clifford algebra is generated by matrices $\Gamma^\mu$ satisfying

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

A spinor is an element of a vector space on which these gamma matrices act. The group $\mathrm{Spin}(1,D-1)$ is generated inside the Clifford algebra by

$$\Sigma^{\mu\nu}=\frac14[\Gamma^\mu,\Gamma^\nu],$$

and spinors transform under these generators rather than as tensors under vector indices.

For the Ramond ground state, this is precisely what happens. The zero modes $\psi_0^\mu$ act on the degeneracy labels of the ground state, and the Lorentz generators acting on the spinor part are built from

$$S^{\mu\nu}_{\rm R} = \frac12[\psi_0^\mu,\psi_0^\nu] = \frac14[\Gamma^\mu,\Gamma^\nu].$$

So the Ramond sector produces spacetime spinors because the worldsheet fermion zero modes are gamma matrices.

Building spinors as a fermionic Fock space

The quickest way to count spinor components is to temporarily work in even Euclidean dimension $D=2K$. Pair the gamma matrices into fermionic creation and annihilation operators,

$$a_j=\frac12\bigl(\Gamma^{2j-1}+i\Gamma^{2j}\bigr), \qquad a_j^\dagger=\frac12\bigl(\Gamma^{2j-1}-i\Gamma^{2j}\bigr), \qquad j=1,\ldots,K.$$

The Clifford algebra implies

$$\{a_i,a_j^\dagger\}=\delta_{ij}, \qquad \{a_i,a_j\}=\{a_i^\dagger,a_j^\dagger\}=0.$$

Starting from a Clifford vacuum $|0\rangle$ satisfying

$$a_j|0\rangle=0,$$

one obtains all states by acting with the $K$ creation operators:

$$|0\rangle, \quad a_i^\dagger|0\rangle, \quad a_i^\dagger a_j^\dagger|0\rangle, \quad\ldots\quad, \quad a_1^\dagger\cdots a_K^\dagger|0\rangle.$$

Each $a_j^\dagger$ can appear at most once, so the complex Dirac spinor has

$$\dim_{\mathbb C}S_{\rm Dirac}=2^K=2^{D/2}$$

components.

Pairing gamma matrices into $K$ fermionic creation and annihilation operators gives a $2^K$-dimensional spinor module. Occupation number parity becomes chirality.

Chirality

In even dimension there is a chirality matrix. In Lorentzian $D=10$ we use

$$\Gamma_{11}=\Gamma^0\Gamma^1\cdots\Gamma^9$$

up to the standard phase convention chosen so that

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

The eigenvalues of $\Gamma_{11}$ split a Dirac spinor into two Weyl spinors,

$$S_{\rm Dirac}=S_+\oplus S_-, \qquad \Gamma_{11}u_\pm=\pm u_\pm.$$

In the Fock construction, chirality is essentially the parity of the number of creation operators:

$$\Gamma_{D+1}\sim (-1)^{N_F}, \qquad N_F=\sum_{j=1}^K a_j^\dagger a_j,$$

up to an overall convention. Thus even occupation-number states form one Weyl representation and odd occupation-number states form the other.

For even $D$, the chirality operator separates the spinor Fock space into even and odd occupation-number sectors. The overall label $+$ or $-$ is conventional.

A small caution: the labels $8_s$ and $8_c$, or $S_+$ and $S_-$, depend on a chirality convention. What is invariant is that there are two inequivalent chiral spinor representations and the GSO projection chooses one of them.

Ten-dimensional Majorana-Weyl spinors

For $D=10$, a complex Dirac spinor has

$$2^{10/2}=32$$

complex components. A Weyl projection halves this to

$$16$$

complex components.

Lorentzian ten-dimensional spinors also admit a Majorana reality condition. In a suitable basis, this can be written schematically as

$$u^c=u,$$

where charge conjugation is defined using a charge-conjugation matrix $C$. The detailed matrix conventions are not important for the spectrum; the important structural fact is that in $D=10$ one may impose Majorana and Weyl conditions simultaneously. A ten-dimensional Majorana-Weyl spinor has

$$16$$

real off-shell components.

For a massless particle, the Dirac equation halves the physical polarizations:

$$k_\mu\Gamma^\mu u=0 \quad\Longrightarrow\quad 8\ \text{on-shell fermionic polarizations}.$$

This is the number that must match the $8$ transverse polarizations of a massless vector in ten dimensions.

In $D=10$, the Majorana and Weyl conditions are compatible. A Majorana-Weyl spinor has $16$ real components off shell, and the massless Dirac equation leaves $8$ on-shell polarizations.

The massless little group and $SO(8)$ spinors

For a massless particle in $D=10$, choose a frame with momentum along the light-cone direction. The subgroup preserving this momentum acts on the eight transverse directions. The physical little group is

$$SO(8).$$

The transverse vector polarizations of a massless gauge boson form the representation

$$8_v.$$

The two chiral spinor representations are

$$8_s, \qquad 8_c.$$

A useful way to label $SO(8)$ spinor weights is by four signs,

$$(s_1,s_2,s_3,s_4), \qquad s_i=\pm\frac12.$$

There are $2^4=16$ sign choices. They split into two sets of eight: one with an even number of minus signs and one with an odd number of minus signs. These are the two chiral spinor representations $8_s$ and $8_c$.

The ten-dimensional massless little group is $SO(8)$. The physical open-string vector gives $8_v$, while a chiral Ramond ground state gives either $8_s$ or $8_c$.

The special feature of $SO(8)$ is triality: the three eight-dimensional representations $8_v$, $8_s$, and $8_c$ are permuted by outer automorphisms. This is why the equality of bosonic and fermionic degrees of freedom in the open superstring can be so economical.

Back to the Ramond ground state

The open R ground state before the GSO projection is a ten-dimensional spinor subject to

$$k_\mu\Gamma^\mu u=0.$$

After imposing the GSO projection on the next page, one keeps a single ten-dimensional Majorana-Weyl chirality. On shell this leaves exactly

$$8$$

fermionic polarizations, matching the $8$ polarizations of the NS massless vector.

This is the key representation-theoretic fact behind the ten-dimensional open-superstring vector multiplet.

Exercises

Exercise 1. Ramond zero modes generate gamma matrices

Starting from

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

show that $\Gamma^\mu=\sqrt2\psi_0^\mu$ obeys the spacetime Clifford algebra.

Solution

Compute

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

This is exactly the Clifford algebra for mostly-plus Lorentzian signature.

Exercise 2. Dimension of a Dirac spinor in even dimension

Use the fermionic oscillator construction with $K$ creation operators to show that a complex Dirac spinor in $D=2K$ dimensions has dimension $2^K$.

Solution

Each creation operator $a_j^\dagger$ is fermionic, so it can either be absent or present. Thus a basis is labeled by $K$ occupation numbers,

$$n_j=0,1, \qquad j=1,\ldots,K.$$

The number of basis states is therefore

$$2\times 2\times\cdots\times 2=2^K.$$

Since $D=2K$, this is $2^{D/2}$.

Exercise 3. Chirality from occupation-number parity

Show that the spinor Fock space splits into two equal halves according to the parity of the number of creation operators.

Solution

The full Fock space has states with occupation numbers $n_j=0,1$. The operator

$$(-1)^{N_F}, \qquad N_F=\sum_{j=1}^K n_j,$$

has eigenvalue $+1$ on even occupation number and $-1$ on odd occupation number. Since multiplying by any gamma matrix changes the occupation number by one, this parity operator anticommutes with all gamma matrices, just like chirality.

The even and odd subspaces have equal dimension because

$$(1+1)^K=(1-1)^K+2\sum_{p\,\text{even}}\binom Kp$$

implies, for $K>0$,

$$\sum_{p\,\text{even}}\binom Kp = \sum_{p\,\text{odd}}\binom Kp =2^{K-1}.$$

Thus chirality splits the Dirac spinor into two Weyl spinors of equal dimension.

Exercise 4. Ten-dimensional Majorana-Weyl counting

Explain why a ten-dimensional Majorana-Weyl spinor has $16$ real off-shell components and $8$ massless on-shell polarizations.

Solution

A complex Dirac spinor in $D=10$ has

$$2^{10/2}=32$$

complex components. The Weyl condition halves this to $16$ complex components. In $D=10$ one can also impose a Majorana reality condition compatible with chirality, turning the $16$ complex components into $16$ real components.

For a massless spinor, the Dirac equation

$$k_\mu\Gamma^\mu u=0$$

removes half the independent components. Therefore the physical on-shell polarizations are

$$\frac{16}{2}=8.$$

Exercise 5. Count $SO(8)$ chiral spinor weights

Use the weights $(s_1,s_2,s_3,s_4)$ with $s_i=\pm1/2$ to show that each chiral spinor representation of $SO(8)$ has dimension $8$.

Solution

There are four independent signs, so there are

$$2^4=16$$

weights. Chirality separates these into two classes: even number of minus signs and odd number of minus signs. The number of weights with an even number of minus signs is

$$\binom40+\binom42+\binom44=1+6+1=8.$$

The number with an odd number of minus signs is

$$\binom41+\binom43=4+4=8.$$

Thus the two chiral spinor representations are both eight-dimensional.