Solution
= Solution
Take $\partial_\pm=(\partial_\tau\pm\partial_\sigma)/2$ on the strip $0\leq\sigma\leq\pi$. Varying the Grassmann-odd <worldsheet Majorana fermions> gives $\partial_-\psi_1=0$, $\partial_+\psi_2=0$ and, up to the orientation of the strip boundary, the surface term
$$
\delta I_{\rm bdry}=\frac{i}{2\pi}\int d\tau\,
\left(\psi_2\cdot\delta\psi_2-\psi_1\cdot\delta\psi_1\right)\Big|_0^\pi.
$$
At each endpoint it vanishes if $\psi_1=\eta\psi_2$ with $\eta=\pm1$, with the same relation on variations. A common sign can be absorbed by redefining one chiral field, so set $\eta_0=+1$. Since $\psi_1=f(\tau+\sigma)$ and $\psi_2=f(\tau-\sigma)$, the second endpoint imposes $f(u+2\pi)=\eta_\pi f(u)$. Equal endpoint signs give the periodic <Ramond sector>, and opposite endpoint signs give the antiperiodic <Neveu–Schwarz sector>:
$$
\boxed{\psi_1(0)=\psi_2(0),\quad
\psi_1(\pi)=+\psi_2(\pi)\ (R),\quad
\psi_1(\pi)=-\psi_2(\pi)\ (NS).}
$$
These gluing choices preserve the Neumann-endpoint <worldsheet supersymmetry> conditions and give integer or half-integer <worldsheet Majorana fermion> modes, respectively.
First fix a normalization issue in the original PDF. With its stated canonical <string oscillator> algebra, the standard <superconformal current> mode is
$$
G_r=\sum_{m\in\mathbb Z}:\alpha_m\cdot\psi_{r-m}:,
$$
without the printed factor $1/2$. To check the <Ramond supercurrent zero-mode square> directly, pair positive and negative integer modes:
$$
G_0=\alpha_0\cdot\psi_0+\sum_{m>0}Q_m,
\qquad Q_m=\alpha_{-m}\cdot\psi_m+\alpha_m\cdot\psi_{-m}.
$$
The zero-mode Clifford relations give $(\alpha_0\cdot\psi_0)^2=\alpha_0^2/2$. Different $Q_m$ anticommute, and so do $Q_m$ with the zero-mode term. The two same-sign terms in $Q_m^2$ vanish by antisymmetry of the <fermions> and symmetry of the bosonic product. The cross terms give
$$
Q_m^2=\alpha_{-m}\cdot\alpha_m+m\psi_{-m}\cdot\psi_m.
$$
The second term comes from commuting $\alpha_m$ through $\alpha_{-m}$, so the frequency factor $m$ is essential. Therefore
$$
G_0^2=\frac12\alpha_0^2+
\sum_{m>0}\left(\alpha_{-m}\cdot\alpha_m+m\psi_{-m}\cdot\psi_m\right)
=L_0^R,
\qquad\boxed{\{G_0,G_0\}=2L_0^R.}
$$
Here $L_0^R$ uses vacuum-annihilating Ramond <normal ordering>; its <string intercept> is zero. A plane-to-cylinder central shift must be included consistently if a different zero-mode convention is used.
\b[For the mode $F_r=G_r/2$ actually printed, the algebra instead gives $\{F_0,F_0\}=L_0^R/2$.] Already the oscillator-free term gives $\{F_0,F_0\}=\alpha_0^2/4$, whereas $2L_0^R=\alpha_0^2$. Thus the requested identity is not valid with both that prefactor and the supplied <string oscillator> algebra. Removing the prefactor gives the identity just derived. Renaming $L_0^R/4$ as $L_0$ would fix this isolated equality but would spoil the usual Virasoro normalization and the subsequent mass formula. We use the standard normalized modes $G_r$ in the remaining physical constraints; their vanishing is unaffected by a common nonzero rescaling.
The <RNS string> physical-state conditions in the two sectors are
$$
\begin{array}{ll}
NS:&(L_0-\tfrac12)|\Phi\rangle=0,\quad
L_{n>0}|\Phi\rangle=0,\quad G_{r>0}|\Phi\rangle=0,
\quad r\in\mathbb Z+\tfrac12,\\
R:&L_0^R|\Phi\rangle=0,\quad
L_{n>0}|\Phi\rangle=0,\quad G_{n>0}|\Phi\rangle=0,
\quad G_0|\Phi\rangle=0.
\end{array}
$$
The last condition is the fermionic zero-mode constraint; its square implies the Ramond <mass-shell condition>. Physical <null string states> are quotiented. Matter <central charge> is $3d/2$, while the reparameterization <worldsheet ghost fields> and <superconformal ghosts> contribute $-26+11=-15$. Thus $d=10$ is the <critical dimension of the RNS superstring>, with $a_{NS}=1/2$ and $a_R=0$.
To solve the constraints, choose $X^\pm=(X^0\pm X^9)/\sqrt2$, so $\eta_{+-}=-1$, and work on a patch with $\alpha_0^+=\sqrt{2\alpha'}p^+\ne0$. In <light-cone gauge in string theory> fix all nonzero $\alpha_n^+$ and all $\psi_r^+$ to zero. For the <Ramond sector>, the printed restriction to nonzero <fermion> modes is not by itself complete: one must also reduce the longitudinal fermionic zero modes using the zero-mode supersymmetry constraint. The usual reduced light-cone description sets the remaining $\psi_0^+$ to zero as well and retains the transverse <Clifford algebra>. Covariantly this reduction is the <Dirac equation> on the ground-state <particle polarization>; it is not an extra unconstrained canonical <string oscillator>.
With this reduction the <string oscillator> generators have the form
$$
\begin{aligned}
L_n&=-\alpha_0^+\alpha_n^-+L_n^\perp,\\
L_n^\perp&=\frac12\sum_{m\in\mathbb Z}:\alpha_{n-m}^i\alpha_m^i:
+\frac12\sum_r\left(r-\frac n2\right):\psi_{n-r}^i\psi_r^i:,\\
G_r&=-\alpha_0^+\psi_r^-+G_r^\perp,
\qquad G_r^\perp=\sum_m\alpha_m^i\psi_{r-m}^i.
\end{aligned}
$$
Repeated $i$ runs over the eight transverse directions. The coefficient in $L_n^\perp$ comes from the <fermion> <worldsheet stress tensor> $-\psi\partial\psi/2$; for $n=0$, positive and negative $r$ combine to give the occupation energy $r\psi_{-r}\psi_r$. Thus the <light-cone super-Virasoro constraints> are explicitly solved by
$$
\boxed{\alpha_n^-=\frac1{\alpha_0^+}\left[
\frac12\sum_m:\alpha_{n-m}^i\alpha_m^i:
+\frac12\sum_r\left(r-\frac n2\right):\psi_{n-r}^i\psi_r^i:
-a\delta_{n0}\right],\qquad
\psi_r^-=\frac1{\alpha_0^+}\sum_m\alpha_m^i\psi_{r-m}^i.}
$$
Here $a$ is $1/2$ in the <NS sector> and zero in the <R sector>. The zero-mode equation says $2p^+p^-=p_i^2+(N-a)/\alpha'$. All longitudinal excitations are consequently determined by transverse bilinears. The critical dimension and <string intercepts> are necessary for the resulting quantum theory to preserve the target <Lorentz algebra>.
The frequency-weighted <Neveu–Schwarz level operator> is
$$
N_{NS}=\sum_{n=1}^\infty\alpha_{-n}^i\alpha_n^i
+\sum_{r=1/2,3/2,\ldots}r\psi_{-r}^i\psi_r^i.
$$
\b[The PDF's displayed number operator omits the <fermion> frequency weights and the $r=1/2$ term.] With the stated anticommutators that cannot be a level operator: it would assign no level to the first fermionic excitation. The bosonic term already includes its frequency through $[\alpha_n,\alpha_{-n}]=n$, whereas the fermionic term must include its frequency explicitly.
One independent way to obtain the <string intercept> is the transverse zero-point energy. Each periodic <boson> contributes $\tfrac12\sum_{n\ge1}n=-1/24$. Each antiperiodic real <fermion> contributes $-\tfrac12\sum_{r\ge1/2}r=-1/48$, with the minus sign due to fermionic <normal ordering>. For example a common exponential cutoff gives
$$
\sum_{n\ge1}n e^{-\varepsilon n}=\frac1{\varepsilon^2}-\frac1{12}+O(\varepsilon^2),
\qquad
\sum_{r\ge1/2}r e^{-\varepsilon r}=\frac1{\varepsilon^2}+\frac1{24}+O(\varepsilon^2).
$$
Subtracting the local divergent terms leaves $8(-1/24-1/48)=-1/2$, so \b[the NS <string intercept> is $a=1/2$]. In the <Ramond sector> the integer <fermions> instead contribute $+1/24$ per component and cancel the bosonic contribution, giving $a_R=0$.
The unprojected <NS sector> <oscillator vacuum> has $N=0$ and
$$
\boxed{M_{NS,0}^2=-\frac1{2\alpha'}=-1\quad(\alpha'=1/2).}
$$
It is a spacetime scalar <tachyon>. The first excitation $\psi_{-1/2}^i|0;p\rangle$ has $N=1/2$ and is massless, with eight vector <particle polarizations>. Covariantly its <particle polarization> obeys $p\cdot\epsilon=0$ modulo the longitudinal null <particle polarization>, leaving the same eight degrees of freedom.
The <R sector> <oscillator vacuum> is also massless, but it is degenerate because $\{\psi_0^\mu,\psi_0^\nu\}=\eta^{\mu\nu}$. The operators $\Gamma^\mu=\sqrt2\psi_0^\mu$ give the target <Clifford algebra>, and $G_0|u;p\rangle=0$ becomes $p_\mu\Gamma^\mu u=0$. At nonzero null momentum this <Dirac equation> leaves sixteen unprojected on-shell spinor states. Equivalently the eight transverse zero modes give the sixteen-dimensional spin representation $8_s\oplus8_c$ of $SO(8)$. Thus the Ramond ground state describes spacetime <fermions>, not a single scalar <oscillator vacuum>.
To define the <GSO projection> explicitly, let $f_{NS}$ count excited NS <fermions> and give its vacuum even parity. Then
$$
P_{NS}=\frac12\left(1-(-1)^{f_{NS}}\right).
$$
It retains odd <fermion> number, removes the <tachyon>, and retains the eight massless vector <particle polarizations>. In the <Ramond sector> let $\Gamma_*$ be transverse <chirality> and $f_R$ count nonzero-mode <fermions>. Choose one consistent <chirality> sign $s=\pm1$ and use
$$
P_R=\frac12\left(1+s\Gamma_* (-1)^{f_R}\right).
$$
Each operator squares to itself. At the ground level $P_R$ retains one eight-dimensional spinor <chirality>, so the retained spectrum has eight massless bosonic and eight massless fermionic states. The parity convention can instead assign the NS vacuum odd parity and write a plus projector; the retained states are identical.
Equality continues at every mass level. Let $q$ count $\alpha'M^2$, suppressing common momentum and endpoint-label factors, and set $D(q)=\prod_{n\ge1}(1-q^n)^8$. The projected <string oscillator> <generating functions> are
$$
Z_{NS}(q)=\frac{q^{-1/2}}{2D(q)}
\left[\prod_{r\ge1/2}(1+q^r)^8-\prod_{r\ge1/2}(1-q^r)^8\right],
\qquad
Z_R(q)=\frac8{D(q)}\prod_{n\ge1}(1+q^n)^8.
$$
The Ramond factor eight follows because either <chirality> projector selects eight ground states for every nonzero-oscillator parity. The <Jacobi abstruse identity> makes the bracket equal to $16q^{1/2}\prod_{n\ge1}(1+q^n)^8$, so
$$
\boxed{Z_{NS}(q)=Z_R(q).}
$$
This proves level-by-level equality of bosonic and fermionic multiplicities. The appropriate <spacetime supercharge from an RNS spin field> relates the two sectors; the projection ensures that it acts within the retained state space. These sector and <chirality> conventions agree with https://arxiv.org/pdf/hep-th/9709062[the primary superstring lecture notes]. Equal total counts alone are not the definition of the projection; they follow from the stated parity selection and the <string oscillator> identity.