= Solution
\b[Solving the massive subsidiary conditions.] In <light-cone coordinates>, use $ds^2=-2dX^+dX^-+\delta_{IJ}dX^IdX^J$ and write $d=D-2$. The divergence condition reads
$$
-\partial_-h_{+n}-\partial_+h_{-n}+\partial_Ih_{In}=0.
$$
Thus, when $\partial_-$ is invertible,
$$
h_{+n}=\partial_-^{-1}\!\left(-\partial_+h_{-n}+\partial_Ih_{In}\right).
$$
First apply this with $n=-$, then with $n=I$, and finally with $n=+$; symmetry supplies the mixed components already determined. The <trace> condition becomes
$$
\delta^{IJ}h_{IJ}=2h_{+-}
=2\partial_-^{-1}\!\left(-\partial_+h_{--}+\partial_Ih_{I-}\right).
$$
The independent components are $h_{--}$, $h_{-I}$ and the <trace>-free part of $h_{IJ}$. Under transverse rotations they form a <scalar representation>, a <vector representation>, and a <symmetric traceless rank-two tensor>. Therefore
$$
\boxed{1+d+\left[\frac{d(d+1)}2-1\right]
=\frac{d(d+3)}2=\frac{(D-2)(D+1)}2.}
$$
This is the <light-cone decomposition of a massive spin-two field> for $D\geq3$. In $D=2$, the <trace> and divergence give $h_{+-}=h_{++}=0$ and $\partial_+h_{--}=0$. The massive wave equation then forces $h_{--}=0$, so there are no polarizations, consistent with the zero value of the printed count. The <massive particle little group> is $SO(D-1)=SO(d+1)$, and its <symmetric traceless square> branches as
$$
\operatorname{Sym}_0^2(\mathbb R^{d+1})\downarrow SO(d)
=\operatorname{Sym}_0^2(\mathbb R^d)\oplus\mathbb R^d\oplus\mathbb R.
$$
The mixed components with the extra direction give the vector; one independent <trace> combination gives the scalar. These are exactly the polarizations of a <massive spin-two field>. The remaining independent components retain the massive <Klein-Gordon equation>.
\b[Transverse bosonic modes and mass levels.] The variables $\alpha_k^I$ are Fourier amplitudes of the physical transverse <open-string mode expansion>. Classically, reality requires $\alpha_{-k}^I=(\alpha_k^I)^*$. The symplectic term in the action fixes their quantum <commutators>:
$$
[\alpha_k^I,\alpha_l^J]=k\,\delta^{IJ}\delta_{k+l,0},\qquad
(\alpha_k^I)^\dagger=\alpha_{-k}^I.
$$
With <string tension> convention $\alpha'=(2\pi T)^{-1}$, the zero mode of the constraint gives
$$
\mathcal M_{\mathrm{cl}}^2=\frac1{\alpha'}\sum_{k>0}\alpha_{-k}\cdot\alpha_k.
$$
The longitudinal nonzero modes have already been removed in <light-cone gauge in string theory>. Quantum <normal ordering> introduces the <string intercept> $a$, giving the <open bosonic string mass spectrum>
$$
\boxed{M^2=\frac{N-a}{\alpha'},\qquad
N=\sum_{k>0}\alpha_{-k}\cdot\alpha_k
=\sum_{k>0}k\,a_k^\dagger\cdot a_k,\quad
a_k^I=\alpha_k^I/\sqrt k.}
$$
Each <bosonic occupation number> is a nonnegative integer, so $N$ is a nonnegative integer weighted by oscillator frequency.
Suppressing the common momentum label, the <lowest light-cone levels of an open bosonic string> are
$$
\begin{aligned}
N=0:&\quad|0;p\rangle,\\
N=1:&\quad\alpha_{-1}^I|0;p\rangle,\\
N=2:&\quad\alpha_{-2}^I|0;p\rangle,\qquad
\alpha_{-1}^I\alpha_{-1}^J|0;p\rangle.
\end{aligned}
$$
The <oscillator vacuum> is annihilated by every positive $\alpha_k^I$. At level one there are only $d$ vector polarizations. For a Lorentz-consistent vector, these are the transverse polarizations of a massless particle, transforming under the rotation part of its <massless particle little group>. A <massive vector> would need $D-1$ polarizations, including a scalar under $SO(d)$ that is absent here. Thus \b[the first bosonic vector level must be massless], fixing $a=1$.
At level two the commuting <creation operators> give a <symmetric square>. Its scalar <trace> and <symmetric traceless square>, together with the mode-two vector, are the massive-spin-two decomposition above. In the consistent bosonic theory they form one <massive spin-two field> with $m^2=1/\alpha'$. The covariant equations describe its propagation while eliminating the redundant components. For $D=26$ the transverse counts are $300+24=324$, the <symmetric traceless rank-two tensor> dimension of $SO(25)$.
\b[Half-integer fermionic modes.] The <Neveu–Schwarz sector> has antiperiodic <worldsheet Majorana fermions>. Its <Neveu–Schwarz fermionic oscillators> obey
$$
\{b_r^I,b_s^J\}=\delta^{IJ}\delta_{r+s,0},\qquad
(b_r^I)^\dagger=b_{-r}^I,\qquad r\in\mathbb Z+\tfrac12.
$$
The <oscillator vacuum> satisfies $\alpha_k^I|0;p\rangle=0$ for $k>0$ and $b_r^I|0;p\rangle=0$ for $r>0$. A negative fermion mode is a <fermionic creation operator> for a transverse worldsheet excitation. The <Neveu–Schwarz level operator> and mass condition are
$$
N=\sum_{k>0}\alpha_{-k}\cdot\alpha_k+\sum_{r>0}r\,b_{-r}\cdot b_r,\qquad
M^2=\frac{N-a_{\mathrm{NS}}}{\alpha'}.
$$
The smallest positive frequency is $1/2$, so the only first-excited states are
$$
\boxed{b_{-1/2}^I|0;p\rangle,\qquad N=\tfrac12.}
$$
The same vector-polarization argument requires them to be massless in Lorentz-consistent quantization, fixing $a_{\mathrm{NS}}=1/2$.
At $N=1$, $\alpha_{-1}^I|0;p\rangle$ is a vector, while $b_{-1/2}^Ib_{-1/2}^J|0;p\rangle$ is the <exterior square>: interchanging the indices changes the sign and equal indices give zero. Together they branch from an antisymmetric tensor:
$$
\boxed{\mathbb R^d\oplus\Lambda^2(\mathbb R^d)
=\Lambda^2(\mathbb R^{d+1})\downarrow SO(d).}
$$
Thus the <Neveu–Schwarz level-one massive tensor> has $(D-1)(D-2)/2$ polarizations and mass squared $1/(2\alpha')$. It differs from spin two because the two-fermion tensor is antisymmetric and has neither the symmetric <trace>-free representation nor its scalar <trace>. At $D=10$, the count is $8+28=36$, compared with $44$ for a massive spin-two field. The specified states are \b[before the GSO projection]; the usual tachyon-removing <GSO projection> also removes this integer level.
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