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Use the critical ten-dimensional RNS string and light-cone gauge in string theory, leaving eight transverse bosons and eight transverse fermions. The bosonic oscillators ${\alpha }_{-m}^i$ have integer $m>0$. In the Neveu–Schwarz sector the fermionic oscillators $b_{-r}^i$ have positive half-integer $r$, while in the Ramond sector the fermionic oscillators $d_{-m}^i$ have integer $m$, including zero modes. Their anticommutators are $\{b_r^i,b_s^j\}={\delta }^{ij}{\delta }_{r+s,0}$ and $\{d_m^i,d_n^j\}={\delta }^{ij}{\delta }_{m+n,0}$. The level and mass formulas areBefore projection the NS ground state is a tachyon of mass squared $-1/(2{\alpha }^{\prime })$. Its oscillator tower describes spacetime bosons. The Ramond zero-mode Clifford algebra $\{\sqrt{2}{d}_0^i,\sqrt{2}{d}_0^j\}=2{\delta }^{ij}$ makes the Ramond vacuum a 16-dimensional spinor of $SO(8)$, split into the two chiral eight-dimensional spinors. The Ramond tower describes spacetime fermions; there is no Ramond tachyon.

Choose the supersymmetric GSO projection. In the NS sector it keeps states with an odd number of fermionic creation operators, eliminating the zero-oscillator tachyon. In the Ramond sector it fixes the combined parity of the zero-mode chirality and the nonzero-mode fermion number. Thus the Ramond vacuum retains one chirality; adding an odd number of nonzero-mode fermions requires the opposite zero-mode chirality. The projected towers both have ${\alpha }^{\prime }{M}^2=0,1,2,\dots$. Here the counts refer to oscillator polarizations for a single endpoint multiplicity; any common Chan-Paton multiplicity multiplies both counts.

At the first retained mass level, $M^2=0$, the NS states are $b_{-1/2}^i|0;k⟩$ for $i=1,\dots ,8$, furnishing the eight polarizations of the ten-dimensional gauge vector. The Ramond ground state is one chiral spinor with eight physical polarizations. HenceThese form the massless supersymmetric vector multiplet.

At the next mass level, ${\alpha }^{\prime }{M}^2=1$, the NS level is $3/2$. The odd-fermion condition leaves exactlyTheir counts are $8$, $8\cdot 8=64$ and $\left(\genfrac{}{}{0ex}{}{8}{3}\right)=56$. Anticommutation requires distinct indices in the third family. There are no other partitions of $3/2$ with odd fermion number, so these are all 128 bosons. They assemble into the massive little-group $SO(9)$ representations of dimensions 44 and 84: a traceless symmetric two-tensor and a three-form.

At Ramond level one, the only possibilities are ${\alpha }_{-1}^i|s⟩$ and $d_{-1}^i|s^{\prime }⟩$. The spinors $|s⟩$ and $|s^{\prime }⟩$ each have eight components and have opposite zero-mode chirality because the fermionic oscillator reverses the GSO parity. Consequently there are $64+64=128$ fermions. They form a gamma-traceless vector-spinor of the massive little group: a vector-spinor has $9\cdot 16$ components and its gamma trace removes 16. Therefore the first massive level of the open RNS string hasThis is the open-string mass normalization; the same chiral oscillator count used in a closed string would correspond to ${\alpha }^{\prime }{M}^2=4$.