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Let $Y$ be the compact coordinate, identified modulo $2\pi R$. The closed string may wrap the circle:Single-valued momentum eigenfunctions $e^{ipY}$ require $p=n/R$, $n\in \mathbb{Z}$. These are the momentum and winding modes. Write the chiral zero-mode momenta asThe left-moving string oscillators will be tilded: $N_L=\tilde{N}$, $N_R=N$. This fixes the sign in compact-circle closed-string level matching.

Define the lower-dimensional mass by $M^2=-p_{\mathrm{noncompact}}^2$. The two closed-string physical-state Virasoro conditions, with bosonic string intercept one, areTheir average and difference derive the complete free bosonic string mass spectrum on the circle:The allowed states have nonnegative integer oscillator levels and obey the Virasoro constraints within each chiral sector. Compactification has not removed oscillator polarizations; it has added quantized compact momentum and allowed winding while changing the lower-dimensional interpretation of those polarizations.

At large $R$, the momentum and winding modes with $w=0$ have closely spaced momentum energies $|n|/R$, whereas nonzero winding costs $|w|R/{\alpha }^{\prime }$. At small $R$ their roles reverse. The precise equality is T-duality:It leaves $p_L$ unchanged and reverses $p_R$, and leaves the mass formula and closed-string level matching unchanged. On the string embedding map it reverses the right-moving part of $Y$. Therefore the small-radius spectrum is the large-radius spectrum of the dual theory, rather than a spectrum in which all charged states simply become heavy.

At a generic radius, the neutral level $(N_L,N_R)=(1,1)$ contains the lower-dimensional graviton, Kalb–Ramond field and dilaton, two gauge bosons from the mixed components of the metric and two-form, and the radius modulus of a string compactification. The two vector charges are the Cartan charges of $U(1{)}_L\times U(1{)}_R$.

Consider instead the following charged states:Every state satisfies compact-circle closed-string level matching. The single noncompact string oscillator supplies a vector polarization. At the massless point, the level-one Virasoro constraint imposes transverse polarization and its physical null string state identifies polarizations differing by the noncompact momentum. Thus these are genuine lower-dimensional vector states. Their squared mass isThus four additional charged massless spin-one particles appear atFor $(N_L,N_R)=(0,1)$ the charges have $p_R=0$, $p_L=\pm 2/\sqrt{{\alpha }^{\prime }}$; for $(1,0)$ they have $p_L=0$, $p_R=\pm 2/\sqrt{{\alpha }^{\prime }}$. This identifies the two pairs of charged roots.

The enhanced gauge group can be derived rather than just inferred from counting. At the self-dual circle, the chiral compact boson satisfies $Y_L(z)Y_L(u)\sim -({\alpha }^{\prime }/2)\log (z-u)$. The self-dual circle current algebra has currentsThe exponentials have conformal weight ${\alpha }^{\prime }{p}_L^2/4=1$. Free-boson contractions give the operator product expansionsThey form the level-one SU(2) Lie algebra, and there is an independent right-moving copy. The string vertex operators $J_L^a\bar{\partial }{X}^{\mu }$ and $\partial X^{\mu }{J}_R^a$ give the six massless vectors. ConsequentlyChanging $R$ gives the four charged vectors the mass found above, consistent with the radius deformation as a Higgs mechanism.

For completeness, the enhanced spectrum also has scalar fields. Using a compact rather than noncompact oscillator in the four singly excited charged families gives four scalars. At $(N_L,N_R)=(0,0)$, closed-string level matching requires $nw=0$; the mass equation then gives the four charges $(\pm 2,0),(0,\pm 2)$. These exceptional massless ground states of a bosonic circle give another four scalars. Together with the radius scalar they form nine scalars in the $(3,3)$ representation, in addition to the neutral dilaton. The ever-present neutral bosonic tachyon still has $M^2=-4/{\alpha }^{\prime }$. The enhancement establishes the additional massless vectors, not stability of the entire bosonic vacuum.

Use target-space signature $(-,+,+,+)$ and conformal gauge $g_{ab}=e^{2\omega }\mathrm{diag}(-1,1)$. Dots and primes denote differentiation with respect to $\tau$ and $\sigma$. Varying the Polyakov action with respect to the string embedding map gives the wave equation; varying the worldsheet metric sets the worldsheet stress tensor to zero. Explicitly,With ${\sigma }^{\pm }=\tau \pm \sigma$ and ${\partial }_{\pm }=\frac{1}{2}({\partial }_{\tau }\pm {\partial }_{\sigma })$, the two Virasoro constraints are $({\partial }_{+}X{)}^2=({\partial }_{-}X{)}^2=0$.

A fixed spatial circle would sweep out a cylindrical string worldsheet. In static coordinates its string embedding map is $X=(t,R\cos \theta ,R\sin \theta ,0)$ and its induced worldsheet metric is $\mathrm{diag}(-1,R^2)$. The radial part of the Nambu–Goto equations of motion is proportional to $R^{-2}{\partial }_{\theta }^2(R\cos \theta ,R\sin \theta )=-(\cos \theta ,\sin \theta )/R$, which is nonzero. Thus a nonzero-radius cylinder cannot solve the equations. Motion of labels around this same cylinder is a worldsheet diffeomorphism and cannot alter this conclusion.

The familiar pulsating circular string makes the obstruction particularly explicit. Put $X^0=\kappa \tau$ and $X^1+i{X}^2=r(\tau )e^{i\sigma }$. The wave equation and Virasoro constraints reduce toA constant $r=R\ne 0$ violates the first equation. Even rotating the labels, $X^1+i{X}^2=R{e}^{i(\sigma +\Omega \tau )}$, cannot help: the wave equation requires ${\Omega }^2=1$, whereas $\dot{X}\cdot X^{\prime }=R^2\Omega$ requires $\Omega =0$.

Nor does a rigid rotation of the circular plane rescue the usual circular ansatz in three spatial dimensions. Write its spatial part as $A(\tau )\cos \sigma +B(\tau )\sin \sigma$, with $A^2=B^2=R^2$ and $A\cdot B=0$. The wave equation gives $\ddot{A}=-A$ and $\ddot{B}=-B$. The constant norms give $A\cdot \dot{A}=B\cdot \dot{B}=0$ and ${\dot{A}}^2={\dot{B}}^2=R^2$. The mixed Virasoro constraint, together with the derivative of $A\cdot B=0$, gives $\dot{A}\cdot B=A\cdot \dot{B}=0$. Differentiating $A\cdot B$ twice gives $\dot{A}\cdot \dot{B}=0$. This would require four mutually orthogonal nonzero spatial vectors in ${\mathbb{R}}^3$. This obstruction is specific to the stated spatial dimension; a rigid circular string in four spatial dimensions has more room.

To express the Virasoro constraints in modes, extend the string oscillators byDifferentiating the two chiral expansions gives ${\partial }_{-}{X}^{\mu }=\sqrt{{\alpha }^{\prime }/2}{\sum }_n{\alpha }_n^{\mu }{e}^{-in{\sigma }^{-}}$ and the corresponding expression with tildes for ${\partial }_{+}X$. Equating every Fourier coefficient of their squares to zero yieldsReality of the string embedding map means ${\alpha }_{-n}={\alpha }_n^{\ast }$ and ${\tilde{\alpha }}_{-n}={\tilde{\alpha }}_n^{\ast }$. In particular, define the classical string level operators by $N={\sum }_{n>0}{\alpha }_{-n}\cdot {\alpha }_n$ and $\tilde{N}={\sum }_{n>0}{\tilde{\alpha }}_{-n}\cdot {\tilde{\alpha }}_n$. The zero-mode Virasoro constraints becomeConsequently the classical bosonic string mass spectrum isThe equality of the two levels is closed-string level matching. There is no string intercept here: that shift comes from quantum normal ordering, not from these classical equations.

For the compact boson, closed-string periodicity permits a winding number $w\in \mathbb{Z}$, so $X^3(\sigma +2\pi )-X^3(\sigma )=2\pi wR$. Let $q$ be the momentum in this direction. Its zero-mode expansion and the corresponding chiral momenta areThus replace the equal left and right zero modes by ${\tilde{\alpha }}_0^3=\sqrt{{\alpha }^{\prime }/2}{p}_L^3$ and ${\alpha }_0^3=\sqrt{{\alpha }^{\prime }/2}{p}_R^3$; the nonzero string oscillators remain integer-moded. Classically $q$ is continuous. Quantizing the compact boson makes momentum wavefunctions single-valued around its circle, giving $q=n/R$ with $n\in \mathbb{Z}$.

If $k^a$, $a=0,1,2$, is the noncompact momentum and $M^2=-k^a{k}_a$, the two zero-mode Virasoro constraints giveAveraging and subtracting gives the momentum and winding modes and compact-circle closed-string level matching formulasThe latter becomes $\tilde{N}-N+nw=0$ on imposing quantum momentum quantization. The sign follows from assigning tildes to the left mover ${\sigma }^{+}$.

An explicit winding-supported circular string, valid classically for every $R>0$, isThe noncompact projection is a circle of constant radius. Every coordinate obeys the wave equation. Its compact winding is $w=1$, and a direct calculation givesThus both Virasoro constraints hold and the induced worldsheet metric is $2R^2\mathrm{diag}(-1,1)$, not a degenerate cylinder. The compact motion cancels the mixed stress that spoiled rotation of labels in flat space. Here $q=-R/{\alpha }^{\prime }$, $N=0$ and $\tilde{N}=R^2/{\alpha }^{\prime }$ satisfy the classical compact-circle closed-string level matching condition. Requiring this particular classical momentum to equal an individual quantum eigenvalue would impose an additional radius-dependent condition; that is not required to construct the classical solution.

For a target circle of radius $R$, maps may wind, so the boundary condition isSingle-valued target-space wavefunctions quantize the center-of-mass momentum as $p_Y=n/R$, with $n\in \mathbb{Z}$. The compact boson expansion becomesEquivalently, its left- and right-moving zero-mode momenta areThe integers $n$ and $w$ are the momentum and winding modes.

For a compact boson $X\sim X+2\pi R$, a closed spatial circuit of the worldsheet may wind around the target circle:Single-valued target-space wavefunctions quantize the zero-mode momentum as $p=n/R$, $n\in \mathbb{Z}$. In the conventionthe Hamiltonian and worldsheet momentum becomeThe $nw$ term is the zero-mode contribution to closed-string level matching.

At $R=\sqrt{{\alpha }^{\prime }}$, a chiral compact boson has operator product expansion $Y_L(z)Y_L(w)\sim -{\alpha }^{\prime }\log (z-w)/2$. The currents $J^3=i\partial Y_L/\sqrt{{\alpha }^{\prime }}$ and $J^{\pm }=:e^{\pm 2i{Y}_L/\sqrt{{\alpha }^{\prime }}}:$ have conformal weight one. Contractions give $J^3(z)J^3(w)\sim 1/[2(z-w{)}^2]$, $J^3(z)J^{\pm }(w)\sim \pm J^{\pm }(w)/(z-w)$ and $J^{+}(z)J^{-}(w)\sim 1/(z-w{)}^2+2J^3(w)/(z-w)$. These are the level-one SU(2) Lie algebra currents. The right-moving copy supplies a second factor. Combining a current with a noncompact coordinate derivative gives a massless string vertex operator for each of the six vectors; four are additional to the generic two Cartan vectors.