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Consider the closed bosonic string on a rectangular product of spatial circles with no background mixing between them. Let $Y^j$ have period $2\pi R_j$, take $0\le \sigma <2\pi$, and choose $Y_L(\tau +\sigma )$ to be the tilded oscillator sector. Closed-string topology permitsSingle-valued center-of-mass wavefunctions give quantized momentum $p^j=n_j/R_j$, $n_j\in \mathbb{Z}$. These are the momentum and winding modes. The oscillator modes remain integer moded; the zero modes are split asThe two intercept-one Virasoro constraints, with noncompact $k^2=-M^2$, giveThus the mass formula and compact-circle closed-string level matching condition areHere $N$ labels right movers and $\tilde{N}$ left movers; exchanging the labels changes the displayed sign convention, not the spectrum. Noncompact oscillator polarizations are still restricted by positive-mode Virasoro conditions and null-state identifications. T-duality exchanges $n_j,w_j$ and $R_j,{\alpha }^{\prime }/R_j$, preserving masses and the physical state conditions.

At generic radius, neutral $(N,\tilde{N})=(1,1)$ states with one noncompact oscillator and one compact oscillator give two massless vectors per circle. They are the two Abelian gauge fields associated with the compact metric and two-form components. Additional charged vectors can occur at oscillator level sum one. For one circle, set all other compact charges to zero. Take $n=w=\pm 1$ and $(N,\tilde{N})=(1,0)$, with a right-moving noncompact oscillator ${\alpha }_{-1}^{\mu }$; or take $n=-w=\pm 1$ and $(N,\tilde{N})=(0,1)$, with a left-moving noncompact oscillator ${\tilde{\alpha }}_{-1}^{\mu }$. These obey level matching. Their common squared mass isTherefore four extra charged vector particles become massless at $R=\sqrt{{\alpha }^{\prime }}$, the self-dual circle. Away from that radius they are massive. They are genuine vector states: the noncompact level-one oscillator carries a vector polarization, and the Virasoro constraint plus its null identification impose transversality and remove the gauge polarization at zero mass.

The resulting self-dual circle current algebra makes the gauge enhancement explicit. At the special radius, the left-moving charges above have $p_L=\pm 2/\sqrt{{\alpha }^{\prime }}$, $p_R=0$. With $Y_L(z)Y_L(w)\sim -{\alpha }^{\prime }\log (z-w)/2$, the currentshave conformal weight one. The operator products give $J^3{J}^3\sim 1/[2(z-w{)}^2]$, $J^3{J}^{\pm }\sim \pm J^{\pm }/(z-w)$ and $J^{+}{J}^{-}\sim 1/(z-w{)}^2+2J^3/(z-w)$. They form a level-one SU(2) current algebra; the right-moving copy supplies another SU(2). Pairing each chiral current with the opposite-sector noncompact derivative yields a weight-$(1,1)$ vector vertex. Thus the generic $U(1{)}_L\times U(1{)}_R$ becomes $SU(2{)}_L\times SU(2{)}_R$, with four charged generators in addition to the two Cartan vectors. Several self-dual circles give the corresponding product enhancement.

Compactification also creates scalar states and moduli; not every additional massless state is a vector. The mechanism just shown uses the intercept-one bosonic spectrum. A superstring projection changes the allowed oscillator states, so the same mass formula cannot simply be applied to a GSO-projected superstring without modification.

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 signature $(-,+,\dots ,+)$ and a closed-string spatial coordinate $0\le \sigma \le 2\pi$. Write $Y$ for the compact coordinate and $X^A$ for the noncompact coordinates. In conformal gauge, the wave equation factorizes, so every coordinate is a sum of a function of $\tau +\sigma$ and a function of $\tau -\sigma$. The closed-string boundary conditions areThus the derivatives of each chiral part are periodic and have integer Fourier modes; only their worldsheet zero modes can produce the winding displacement. Integrating these Fourier series gives the closed-string mode expansionThe oscillator reality conditions are ${\alpha }_m^{†}={\alpha }_{-m}$ and similarly for the tilded modes. The coefficient of $\tau$ follows by integrating the canonical momentum density $\dot{Y}/(2\pi {\alpha }^{\prime })$. Single-valued wavefunctions of the center coordinate $y\sim y+2\pi R$ have momentum $n/R$, with $n\in \mathbb{Z}$. Equivalently, the compact chiral worldsheet zero modes are ${\alpha }^{\prime }{p}_L(\tau +\sigma )/2$ and ${\alpha }^{\prime }{p}_R(\tau -\sigma )/2$, whereHere tildes denote the left-moving sector. This convention will also fix the sign in compact-circle closed-string level matching.

Choose a noncompact light-cone pair and impose light-cone gauge in string theory $X^{+}=x^{+}+{\alpha }^{\prime }{p}^{+}\tau$, with $p^{+}\ne 0$. The Virasoro constraints $({\partial }_{\pm }X{)}^2=0$ then solve for ${\partial }_{\pm }{X}^{-}$ in terms of the transverse coordinates. For example, with ${\partial }_{\pm }={\partial }_{\tau }\pm {\partial }_{\sigma }$, one has ${\partial }_{\pm }{X}^{-}=({\partial }_{\pm }{X}^i{)}^2/(2{\alpha }^{\prime }{p}^{+})$. Thus there are two independent transverse oscillator Fock spaces, including $Y$ among the transverse coordinates. Quantization gives $[{\alpha }_m^i,{\alpha }_k^j]=m{\delta }^{ij}{\delta }_{m+k,0}$, and the same relation for the tilded oscillators, with the two sectors commuting. Lorentz consistency gives the critical dimension of the bosonic string $D=26$ and string intercept one in each sector. Accordinglycount 24 transverse oscillator species. The two zero-mode constraints give the lower-dimensional bosonic string mass spectrumTaking their difference and average yieldsEvery choice of transverse string oscillators and integer charges satisfying this closed-string level matching is a physical string state in light-cone gauge. The eliminated longitudinal oscillators do not supply additional polarizations.

To classify all massless states, positivity of the two compact terms implies $N+\tilde{N}\le 2$. When the sum is two, masslessness forces $n=w=0$, and closed-string level matching then forces $N=\tilde{N}=1$. The states ${\alpha }_{-1}^i{\tilde{\alpha }}_{-1}^j|0⟩$ have $24^2=576$ polarizations. In 25 noncompact dimensions, their decomposition gives a graviton, a Kalb–Ramond field, a dilaton, two Abelian gauge vectors from the metric and two-form with one compact index, and the radius scalar field. Their physical polarization counts areThe first two numbers are the symmetric traceless second-rank tensor and antisymmetric second-rank tensor of the massless little group $SO(23)$.

When $N+\tilde{N}=1$, closed-string level matching requires $|nw|=1$. The compact contribution obeysand equality occurs precisely at $R=\sqrt{{\alpha }^{\prime }}$. The charges $n=w=\pm 1$ have $(N,\tilde{N})=(1,0)$; the charges $n=-w=\pm 1$ have $(N,\tilde{N})=(0,1)$. Each charge has 24 polarizations supplied by its single excited sector. At a general radius these have

Finally, $N=\tilde{N}=0$ requires $nw=0$. The exceptional massless ground states of a bosonic circle occur atEach charge is a scalar field. These special radii must be included if “general radius” is to mean a complete classification, rather than only a generic radius. At a generic radius the massless spectrum is the neutral 576-state spectrum above; at exceptional radii one adds exactly the states just listed. The neutral ground state remains tachyonic.

The mapsends $p_L$ to $p_L$ and $p_R$ to $-p_R$. Reversing all right-moving compact string oscillators, ${\alpha }_m^Y\mapsto -{\alpha }_m^Y$, preserves their commutators and oscillator number; the left-moving oscillators are unchanged. Both the mass formula and compact-circle closed-string level matching are therefore unchanged, with a one-to-one map of the complete oscillator basis. This proves T-duality of the full spectrum, including winding, massive and tachyonic states.

At the self-dual circle the four single-oscillator charge families give $4\cdot 24=96$ new massless states. Each family decomposes into a 25-dimensional gauge vector with 23 polarizations and one scalar field. There are also four massless ground states, with charges $(\pm 2,0)$ and $(0,\pm 2)$. Thus the massless spectrum at the bosonic self-dual circle hasThe gauge interpretation follows directly from the chiral compact-boson operators. With $⟨Y_L(z)Y_L(0)⟩=-{\alpha }^{\prime }\log z/2$, the exponential $e^{ik{Y}_L}$ has conformal weight ${\alpha }^{\prime }{k}^2/4$. At the self-dual radius,have weight one and satisfy $J^3(z)J^3(0)\sim 1/(2z^2)$, $J^3(z)J^{\pm }(0)\sim \pm J^{\pm }(0)/z$, and $J^{+}(z)J^{-}(0)\sim z^{-2}+2J^3(0)/z$. These are the level-one $SU(2)$ current relations; the right sector supplies another copy. Multiplying these currents by a noncompact oscillator produces six gauge vectors in total. The nine scalar string vertex operators $J_L^a{J}_R^b$ form the $(3,3)$ representation, accounting for the radius scalar field and eight new scalar fields. The remaining neutral fields are the graviton, two-form and dilaton.

A radius deformation as a Higgs mechanism is the expectation value of the Cartan–Cartan scalar string vertex operator $J_L^3{J}_R^3\propto \partial Y_L\bar{\partial }{Y}_R$. Its stabilizer is $U(1{)}_L\times U(1{)}_R$, soThe four compact single-oscillator scalar field polarizations become longitudinal polarizations of the four massive vectors, exactly as in the Higgs mechanism. The other four extra scalar fields have masses $4/R^2-4/{\alpha }^{\prime }$ or $4R^2/{\alpha }^{\prime 2}-4/{\alpha }^{\prime }$ and need not remain massless; one pair becomes tachyonic on either side of the self-dual point. Together with the original bosonic tachyon, this means that the gauge-symmetry interpretation is formal and does not assert a stable bosonic-string vacuum.

Use the mostly-positive target Minkowski spacetime metric and a closed string coordinate $0\le \sigma <2\pi$. In conformal gauge, variation of the Polyakov action gives $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^{\mu }=0$. The compact coordinate, denoted $X^{25}$, can change by $2\pi Rw$ around the closed string, with $w\in \mathbb{Z}$. Write $w^{\mu }=w{\delta }^{\mu ,25}$. Integrating the two chiral wave solutions and expanding their periodic parts gives the closed-string mode expansionReality imposes ${\alpha }_{-n}=({\alpha }_n{)}^{\ast }$ and the analogous condition in the other chirality. The classical compact momentum is continuous; quantization of a wavefunction on the circle subsequently gives $p^{25}=n/R$ with $n\in \mathbb{Z}$. The momentum and winding modes are most conveniently packaged asFor noncompact directions both zero modes are $\sqrt{{\alpha }^{\prime }/2}{p}^{\mu }$. Thus the formula includes both independent string oscillator families and all compact topological sectors.

Define the worldsheet stress-energy tensor by ${\theta }_{ab}=-2(-\gamma {)}^{-1/2}\delta I/\delta {\gamma }^{ab}$. Its normalization and components areAn overall rescaling of the named worldsheet stress tensor does not alter its vanishing. Variation of the independent worldsheet metric imposes ${\theta }_{ab}=0$, equivalently $(\dot{X}+X^{\prime }{)}^2=(\dot{X}-X^{\prime }{)}^2=0$. These are the Virasoro constraints that restrict the otherwise general wave solution.

The conjugate momentum density is $P_{\mu }={\dot{X}}_{\mu }/(2\pi {\alpha }^{\prime })$. The gauge-fixed Hamiltonian generating $\tau$ translation is thereforewith the analogous tilded generators. In particular, before imposing the Virasoro constraints,The target momentum includes its timelike component, so this is not a positive particle-energy formula. On the classical constraint surface the reparameterization Hamiltonian vanishes; target energy is the separate charge $p^0$. The difference $L_0-{\tilde{L}}_0=0$ imposes the residual spatial constraint.

In the quantum theory the same quadratic expressions require normal ordering: $[{\alpha }_m^{\mu },{\alpha }_n^{\nu }]=m{\eta }^{\mu \nu }{\delta }_{m+n,0}$ makes different orderings differ by divergent constants. Their regularized finite remainder is the normal-ordering constant of a string, or string intercept. The Virasoro central extension also appears. In the critical 26-dimensional bosonic string theory, the 24 transverse string oscillators give one-unit string intercept in each chirality. For example their regulated chiral vacuum energy is $24{\sum }_{k>0}k/2=-1$ using zeta function regularization. The physical zero-mode constraints are $(L_0-1)|\Psi ⟩=({\tilde{L}}_0-1)|\Psi ⟩=0$.

Let $M$ be the mass seen in the remaining 25 noncompact dimensions. Substituting the compact zero modes in these two constraints givesTaking their mean and difference yieldsThis sign of compact-circle closed-string level matching follows from assigning tildes to the left-moving sector. Under T-duality, set $R^{\prime }={\alpha }^{\prime }/R$, $n^{\prime }=w$, $w^{\prime }=n$. Then $p_L^{\prime }=p_L$ and $p_R^{\prime }=-p_R$, so reversing the compact right-moving string oscillators implements the transformation while leaving $N,\tilde{N}$ unchanged. Both the mass equation and closed-string level matching are invariant. This proves invariance of the full closed string spectrum, including its string oscillator degeneracies, rather than only the zero-mode contribution.

A D-brane with $p$ spatial worldvolume directions, called a D$p$-brane, is a $(p+1)$-dimensional timelike submanifold on which open string endpoints have Neumann boundary conditions tangentially and Dirichlet boundary conditions transversely. Apply T-duality to the compact coordinate by $\tilde{X}=X_L-X_R$. The differential relationsinterchange the two endpoint conditions. Consequently a D$p$-brane wrapping the original circle becomes a D$(p-1)$-brane localized on the dual circle. Its compact Wilson line specifies that position. This is T-duality of D-brane boundary conditions.

The separated-brane configuration must instead have the circle transverse to the D-branes. Two parallel D-branes wrapping the same circle cannot be separated along that worldvolume direction. Thus the latter part describes a new orientation, with both endpoints subject to compact Dirichlet boundary conditions. If their positions differ by $L$, an oriented Open string stretched between D-branes has displacementOn $0\le \sigma \le \pi$, its compact classical solution has a linear part $y_1+L_w\sigma /\pi$, plus sine modes vanishing at the endpoints. This adds $L_w^2/(4{\pi }^2{\alpha }^{\prime })$ to the Virasoro generator zero mode. Thus, for momentum $\mathbf{p}$ along the D-branes,The second equation is the quantum rest-energy squared of the string oscillator ground state; the smallest one uses the shortest allowed compact separation $|L_w|$. Classically the straight string energy is $T|L_w|=|L_w|/(2\pi {\alpha }^{\prime })$. The quantum ground state is a tachyon if $|L_w|<2\pi \sqrt{{\alpha }^{\prime }}$, so in that regime it is an instability rather than a state with real rest energy. Dropping the string intercept would describe the classical stretched string, not its quantum ground state.

The transverse-circle T-duality changes these two D-branes to D$(p+1)$-branes wrapping the dual circle. Their positions become the eigenvalues $A_i=y_i/(2\pi {\alpha }^{\prime })$ of a constant gauge connection. The Wilson loop phases are $e^{i2\pi R^{\prime }{A}_i}=e^{i{y}_i/R}$. A string connecting D-brane $1$ to D-brane $2$ has dual compact momentumIts momentum contribution to $E^2$ is exactly the former stretching term, including every winding sector. Qualitatively the two geometrically coincident wrapped D-branes are distinguished by their Wilson lines, and their connecting strings have shifted momentum quantization. This is how brane separation becomes a Wilson line.

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.