Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 50 / 2

Take $0\le \sigma <2\pi$. Variation of the conformal-gauge action, with periodic variations, gives $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^{\mu }=0$. Hence $X^{\mu }=X_R^{\mu }(\tau -\sigma )+X_L^{\mu }(\tau +\sigma )$. In an uncompactified direction, periodicity allows equal linear slopes and independent periodic parts. Expanding the periodic parts in a Fourier series yields the closed-string mode expansionHere ${\alpha }_m$ labels the right-moving sector and ${\tilde{\alpha }}_m$ the left-moving sector; exchanging the names is harmless if done consistently. Integration of the canonical momentum density $(2\pi {\alpha }^{\prime }{)}^{-1}{\dot{X}}^{\mu }$ over the string gives $p^{\mu }$, fixing the zero-mode coefficient. Reality gives ${\alpha }_m^{†}={\alpha }_{-m}$ and the analogous relation for tildes, with ${\alpha }_0={\tilde{\alpha }}_0=\sqrt{{\alpha }^{\prime }/2}p$.

The closed-string physical-state Virasoro conditions arewhere $L_0={\alpha }^{\prime }{p}^2/4+N$ and ${\tilde{L}}_0={\alpha }^{\prime }{p}^2/4+\tilde{N}$. Thus closed-string level matching gives $N=\tilde{N}$, and $M^2=-p^2=4(N-1)/{\alpha }^{\prime }$. The massless level is $N=\tilde{N}=1$. Its general state isThe string oscillator commutators give $[L_m,{\alpha }_n^{\mu }]=-n{\alpha }_{m+n}^{\mu }$: the two terms obtained by commuting through the quadratic generator coincide and cancel its factor $1/2$. Therefore $L_1$ replaces ${\alpha }_{-1}^{\mu }$ by ${\alpha }_0^{\mu }$, while $L_{m\ge 2}$ produces positive string oscillators that annihilate this level. The same holds independently for tildes. The complete polarization conditions areNull string states identify ${\zeta }_{\mu \nu }$ with ${\zeta }_{\mu \nu }+p_{\mu }{\lambda }_{\nu }+{\tilde{\lambda }}_{\mu }{p}_{\nu }$, with transverse $\lambda ,\tilde{\lambda }$. At nonzero momentum this leaves $24\cdot 24=576$ physical polarizations. Their symmetric traceless, antisymmetric and transverse-trace parts are the graviton, Kalb–Ramond field and dilaton.

The original PDF's condition ${\alpha }^{\prime }{p}^2=4$ is inconsistent with calling the displayed excited states massless. It is the on-shell condition of the string oscillator vacuum as a physical tachyon. For the level-one state, that value would give $L_0={\tilde{L}}_0=2$, violating the physical conditions. The ket $|0;p⟩$ used as the basis for a massless excitation must instead carry the excitation's momentum $p^2=0$; by itself it is then not an on-shell physical vacuum state.

For the compact coordinate, $X^1(\tau ,\sigma +2\pi )=X^1(\tau ,\sigma )+2\pi wR$, $w\in \mathbb{Z}$. Its momentum is $n/R$, $n\in \mathbb{Z}$, since the centre-of-mass wavefunction is single-valued on the circle. Thus its zero modes becomewithThese are momentum and winding modes. Let $k$ be the 25-dimensional noncompact momentum and $M^2=-k^2$. The two zero-mode constraints now giveTheir difference and average give, respectively,This also displays T-duality $R\leftrightarrow {\alpha }^{\prime }/R$, $n\leftrightarrow w$, with one chiral momentum reversed.

At a generic radius, the massless solutions are only $n=w=0$, $N=\tilde{N}=1$. Their fields in 25 dimensions are a graviton (275 polarizations), antisymmetric two-form (253), dilaton (one), two abelian vectors from $G_{a1}$ and $B_{a1}$ (23 each), and the radius scalar (one). Hence the generic massless spectrum has $275+253+1+46+1=576$ polarizations, with gauge group $U(1{)}_L\times U(1{)}_R$. Here generic excludes isolated radii where exceptional massless ground states of a bosonic circle occur. Indeed $N=\tilde{N}=0$ forces $nw=0$, and becomes massless at $R=|n|\sqrt{{\alpha }^{\prime }}/2$ or $R=2\sqrt{{\alpha }^{\prime }}/|w|$. These additional thresholds matter in the bosonic theory.

At the self-dual circle $R=\sqrt{{\alpha }^{\prime }}$, masslessness and level matching readSince $N,\tilde{N}\ge 0$, their sum is at most two. For sum two, $n=w=0$ and level matching gives the ordinary $(1,1)$ family. For sum one the four possibilities areThey have one right-moving ${\alpha }_{-1}$ or one left-moving ${\tilde{\alpha }}_{-1}$ excitation, as their levels specify. Each family has 24 physical polarizations: a 25-dimensional vector with 23 polarizations and one compact-oscillator scalar. Thus these families add 96 states.

For sum zero, level matching forces $nw=0$, while the mass equation requires $n^2+w^2=4$. It gives four additional scalar states with charges $(\pm 2,0)$ and $(0,\pm 2)$, without any string oscillators. They must not be omitted. The massless spectrum at the bosonic self-dual circle therefore has 100 additional polarizations, 676 in total. The four new vector fields enhance the gauge group to $SU(2{)}_L\times SU(2{)}_R$. Including the original radius scalar, the nine non-dilaton scalars form its $(3,3)$ representation. Equivalently, the full count is $275+253+1+6\cdot 23+9=676$.