= Solution
Let $Y$ be the compact coordinate, identified modulo $2\pi R$. The <closed string> may wrap the circle:
$$
Y(\tau,\sigma+2\pi)=Y(\tau,\sigma)+2\pi wR,\qquad w\in\mathbb Z.
$$
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 as
$$
p_L=\frac nR+\frac{wR}{\alpha'},\qquad
p_R=\frac nR-\frac{wR}{\alpha'}.
$$
The left-moving <string oscillators> will be tilded: $N_L=\widetilde N$, $N_R=N$. 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, are
$$
\frac{\alpha'}4(-M^2+p_L^2)+N_L-1=0,
\qquad
\frac{\alpha'}4(-M^2+p_R^2)+N_R-1=0.
$$
Their average and difference derive the complete free <bosonic string mass spectrum> on the circle:
$$
\boxed{M^2=\frac{n^2}{R^2}+\frac{w^2R^2}{\alpha'^2}
+\frac2{\alpha'}(N_L+N_R-2)},
\qquad
\boxed{N_L-N_R+nw=0}.
$$
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 $|w|R/\alpha'$. At small $R$ their roles reverse. The precise equality is <T-duality>:
$$
\boxed{R'=\frac{\alpha'}R,\qquad n'=w,\qquad w'=n}.
$$
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:
$$
(N_L,N_R)=(0,1),\quad (n,w)=(1,1),(-1,-1),
$$
$$
(N_L,N_R)=(1,0),\quad (n,w)=(1,-1),(-1,1).
$$
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> is
$$
M^2=\frac1{R^2}+\frac{R^2}{\alpha'^2}-\frac2{\alpha'}
=\left(\frac1R-\frac R{\alpha'}\right)^2.
$$
Thus \b[four additional charged massless spin-one particles appear at]
$$
\boxed{R=\sqrt{\alpha'}}.
$$
For $(N_L,N_R)=(0,1)$ the charges have $p_R=0$, $p_L=\pm2/\sqrt{\alpha'}$; for $(1,0)$ they have $p_L=0$, $p_R=\pm2/\sqrt{\alpha'}$. 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 $Y_L(z)Y_L(u)\sim-(\alpha'/2)\log(z-u)$. The <self-dual circle current algebra> has currents
$$
J_L^3=\frac{i}{\sqrt{\alpha'}}\partial Y_L,\qquad
J_L^\pm=:\!\exp\left(\pm\frac{2iY_L}{\sqrt{\alpha'}}\right)\!:.
$$
The exponentials have <conformal weight> $\alpha'p_L^2/4=1$. Free-boson contractions give the <operator product expansions>
$$
J^3(z)J^3(u)\sim\frac1{2(z-u)^2},\qquad
J^3(z)J^\pm(u)\sim\frac{\pm J^\pm(u)}{z-u},
$$
$$
J^+(z)J^-(u)\sim\frac1{(z-u)^2}+\frac{2J^3(u)}{z-u}.
$$
They 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. Consequently
$$
\boxed{U(1)_L\times U(1)_R\longrightarrow SU(2)_L\times SU(2)_R}.
$$
Changing $R$ 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 $(\pm2,0),(0,\pm2)$. 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 $M^2=-4/\alpha'$. The enhancement establishes the additional massless vectors, not stability of the entire bosonic vacuum.
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