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With Hermitian $b_0,c_0$ and $\{b_0,c_0\}=1$, a ket annihilated by $b_0$ has zero bare full-ghost norm. Nonzero amplitudes need FP ghost zero-mode saturation. Low-level Virasoro descendant calculations can instead normalize the highest-weight coefficient and compute $L_n{L}_{-n}|\Omega ⟩$ algebraically. Such normalized contravariant coefficients are not positive-definite Hilbert norms. The pairing convention must be stated before using FP ghost descendants in a norm argument.

Use $[L_m,L_n]=(m-n)L_{m+n}+\frac{c}{12}(m^3-m){\delta }_{m+n,0}$ and $L_n^{†}=L_{-n}$. A normalized highest-weight vector obeys $L_0|h⟩=h|h⟩$ and $L_n|h⟩=0$ for $n>0$. Its Virasoro descendants are products of negative modes. Their inner products are determined entirely by $c,h$ and the Virasoro algebra; for example $\Vert L_{-1}|h⟩{\Vert }^2=2h$, so unitarity requires $h\ge 0$. An irreducible unitary highest-weight Virasoro module is the Virasoro Verma module quotient by its null submodules, and its surviving inner product must be positive definite.

The classification for $c<1$ is the Unitary Virasoro discrete series:where $m$ is an integer at least $2$, $1\le r\le m-1$, and $1\le s\le m$. The identification $(r,s)\sim (m-r,m+1-s)$ leaves $m(m-1)/2$ distinct modules. The two generating singular-vector levels are $rs$ and $(m-r)(m+1-s)$; all their Virasoro descendants are removed in the irreducible quotient. For $m=2$ the only module is the trivial $c=h=0$ module; for $m\ge 3$ these are the nontrivial unitary discrete-series modules. There are no other irreducible unitary highest-weight representations with $c<1$. This is the classification theorem, rather than a condition inferred only from finitely many descendant norms.

At $c=1/2$, $m=3$ and the possible weights areConstruct them with a single free chiral Majorana fermion:The free chiral Majorana fermion conformal field theory has $T=-:\psi \partial \psi :/2$. In the Neveu–Schwarz sector, $r\in \mathbb{Z}+1/2$, while in the Ramond sector, $r\in \mathbb{Z}$. With positive modes moved to the right and the zero-mode product antisymmetrized, the corresponding generators areThe oscillator relation gives $[L_n,b_r]=-(r+n/2)b_{n+r}$. The stress-tensor operator product expansion, or the reordering of the quadratic generators, then gives the Virasoro algebra with $c=1/2$. One can check both constants directly in the ground states. In the NS sector,In the R sector, $b_0^2=1/2$ and $L_{-1}|R⟩=\frac{1}{2}{b}_{-1}{b}_0|R⟩$, whose norm squared is $1/8$. The relation $[L_1,L_{-1}]=2L_0$ requires $2h_R=1/8$, so $h_R=a=1/16$. Thus the Ramond sector zero-mode shift is essential.

The three constructions in Ising Virasoro modules from a Majorana fermion are explicit. The NS vacuum is a highest-weight vector of weight $0$. The state $b_{-1/2}|0⟩$ is a highest-weight vector of weight $1/2$: applying any positive $L_n$ gives $-(n/2-1/2)b_{n-1/2}|0⟩=0$, including the zero coefficient for $n=1$. Since $L_n$ is quadratic in fermions, the two cyclic modules lie in the even and odd fermion parity sectors, respectively. For R, choose a normalized ground state annihilated by all $b_n$ with $n>0$ and with $b_0|R⟩=\pm |R⟩/\sqrt{2}$. All positive $L_n$ annihilate it and its weight is $1/16$. The two zero-mode sign choices give isomorphic Virasoro modules; one choice suffices to construct the third module.

The oscillator inner products are positive, and each cyclic highest-weight representation just constructed is irreducible. Indeed any nonzero proper invariant graded submodule would have a lowest-energy vector $w$ of weight greater than the cyclic ground weight. Every positive $L_n$ annihilates $w$; thus $w$ is orthogonal to every negative-mode descendant of the original ground vector, by moving the modes across the inner product. It is also orthogonal to that ground vector by its different $L_0$ eigenvalue. Since those Virasoro descendants span the cyclic module, $w$ would have zero norm, a contradiction. This proves irreducibility directly inside the positive fermionic Fock space.

In fact the cyclic modules fill the entire even NS, odd NS and chosen R Fock spaces. Otherwise an orthogonal remaining lowest vector would be the ground vector of a further unitary highest-weight Virasoro module; its weight would lie in ${\mathbb{Z}}_{>0}$, $1/2+{\mathbb{Z}}_{>0}$ or $1/16+{\mathbb{Z}}_{>0}$, respectively. None belongs to the classified list $\{0,1/2,1/16\}$. As a check, their Virasoro characters, defined by ${\chi }_h(q)=\mathrm{Tr}{q}^{L_0-c/24}$, areThe NS signs project onto even and odd occupation number; the R product uses the single chosen ground state. Null vectors of the abstract Virasoro Verma modules are automatically zero in this construction. In particular $L_{-1}|0⟩=0$, and the weight-$1/2$ and weight-$1/16$ ground states obey, respectively, $(L_{-2}-\frac{3}{4}{L}_{-1}^2)|1/2⟩=0$ and $(L_{-2}-\frac{4}{3}{L}_{-1}^2)|1/16⟩=0$, as is also verified by their level-two Gram matrices.

In conformal gauge, a free closed bosonic string has $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^{\mu }=0$, periodicity in $\sigma$, and the two Virasoro constraints $(\dot{X}\pm X^{\prime }{)}^2=0$. Its left- and right-moving string oscillator families have independent Virasoro generators $L_n$ and ${\tilde{L}}_n$. In old covariant string quantization, a physical string state obeyswith $a=1$ in the critical bosonic string theory. The two zero-mode conditions also impose closed-string level matching. For an open string, keep just one string oscillator family. Negative-mode constraints are adjoints on bras, not additional annihilation conditions on physical string states.

A spurious string state is a Virasoro descendant orthogonal to every physical string state: if $|s⟩={\sum }_{n>0}{L}_{-n}|{\chi }_n⟩$, then $⟨\Phi |s⟩=0$ follows from $L_n|\Phi ⟩=0$ and $L_n^{†}=L_{-n}$. A null string state is a spurious state that is itself physical. It consequently has zero norm and is removed in the null-state quotient of a string. The calculations below concern one string oscillator family and use

For $|n_1⟩=L_{-1}|{\chi }_1⟩$ with $L_0|{\chi }_1⟩=0$ and all positive modes zero, $L_0|n_1⟩=|n_1⟩$. AlsoIt is therefore physical for string intercept one and spurious by construction. In particular $\Vert n_1{\Vert }^2=⟨{\chi }_1|2L_0|{\chi }_1⟩=0$. This proves the requested level-one descendant is null.

For the scalar descendant, write $|v_b⟩=(L_{-2}+b{L}_{-1}^2)|{\chi }_2⟩$ with $L_0|{\chi }_2⟩=-|{\chi }_2⟩$ and positive modes zero. Direct commutation givesFor instance $[L_1,L_{-1}^2]|{\chi }_2⟩=-2L_{-1}|{\chi }_2⟩$ and $L_2{L}_{-1}^2|{\chi }_2⟩=-6|{\chi }_2⟩$. Modes $m\ge 3$ annihilate $|v_b⟩$ because every remaining positive mode annihilates its parent. Hence the physical scalar descendant requires $b=3/2$ and $d=26$:It is spurious, so physicality also proves orthogonality and zero norm without an extra positivity assumption.

The printed coefficient $3/4$ does not give this null string state. It instead gives $L_1|v_{3/4}⟩=\frac{3}{2}{L}_{-1}|{\chi }_2⟩$, generally nonzero. For a normalized highest-weight parent with weight $-1$ at $d=26$, the descendant Gram matrix isThus the printed vector has norm $9/4$, whereas the corrected vector has norm zero. This is a genuine PDF coefficient error. Away from 26 dimensions even the corrected coefficient does not give the claimed physical null string state; the remaining condition is $L_2|v_{3/2}⟩=(d-26)|{\chi }_2⟩/2$.

The level-two open string state also needs a source correction: its vector string oscillator must be ${\alpha }_{-2}^{\mu }$, not the printed ${\alpha }_{-1}^{\mu }$. A single ${\alpha }_{-1}$ has level one and cannot share the stated mass-shell condition with the two-${\alpha }_{-1}$ term. Put $q^{\mu }={\alpha }_0^{\mu }=\sqrt{2{\alpha }^{\prime }}{p}^{\mu }$, and write the genuine level-two state asThe mass-shell condition $L_0=1$ gives ${\alpha }^{\prime }{p}^2+2=1$, so $q^2=-2$. The commutators above giveFor the second formula, the contraction from the two ${\alpha }_{-1}$ operators is exactly $a^{\mu }{}_{\mu }$, not twice that value. Modes $m>2$ vanish automatically. We have derivedThe coefficient formulas in the question are consequently consistent after the string oscillator correction and with $d=26$.

To prove the full level-two open-string polarization decomposition, not merely check one candidate, defineThe constraints give $q\cdot b=-t/2$ and therefore $q\cdot A=0$. Contracting $c$ with $q$ givesIts trace is $t-t(26+3q^2)/20-2q\cdot A=0$. Thus $c$ is symmetric, traceless and transverse. Conversely, arbitrary $t$, a vector $A$ with $q\cdot A=0$, and such a tensor $c$ giveand direct contraction verifies both constraints. The number $20$ uses $d=26$; it would not reproduce the same unrestricted trace in arbitrary dimension.

Finally the two unphysical pieces are precisely the descendants already proved null. The vector part iswhose parent has $L_0=0$ and all positive modes zero because $q\cdot A=0$. For the scalar part useThenhas exactly the scalar tensor $t(3q_{\mu }{q}_{\nu }+{\eta }_{\mu \nu })/20$ and vector $t{q}_{\mu }/4$. The momentum vacuum has $L_0=-1$ and positive modes zero, as required. HenceThe physical quotient contains only the transverse traceless tensor. Its massive little-group space has dimension $25\cdot 26/2-1=324$, agreeing with the light-cone gauge in string theory count $24\cdot 25/2+24=324$ at level two.

The old covariant quantization of a free-ended bosonic string retains oscillators in all target directions. In mostly-plus Minkowski spacetime, set ${\alpha }_0^m=\sqrt{2{\alpha }^{\prime }}{p}^m$ andTime-coordinate excitations have negative norms in this covariant Fock space. The string ghost states in this discussion are unwanted negative-norm physical states, distinct from the anticommuting Faddeev-Popov ghost fields in the path integral. Constraints and the null-state quotient of a string remove unphysical polarizations while maintaining target-space Lorentz covariance.

Why only positive-mode constraints annihilate states. The quantum Virasoro algebra isThe physical conditions arewhere $a$ is the string intercept. If both positive and negative modes annihilated states, $[L_1,L_{-1}]=2L_0$ would force $a=0$. Then $[L_2,L_{-2}]=4L_0+D/2$ would force $D=0$. Thus the Virasoro central extension and shifted zero mode prevent imposing every classical constraint strongly in a nontrivial string. As in Gupta-Bleuler quantization, negative-mode conditions act on physical bras, rather than also annihilating physical kets. Physical null string states are quotiented because their inner products with all physical states vanish.

The intercept bound at level one. For $|ϵ;p⟩={ϵ}_m{\alpha }_{-1}^m|p⟩$, the constraints and norm areIf $a>1$, momentum is spacelike and its orthogonal complement contains a timelike negative-norm polarization. The level-one intercept bound in covariant string quantization is thereforeFor $a<1$, momentum is timelike and the $D-1$ orthogonal polarizations are positive. This avoids level-one ghosts but gives a massive vector with one more polarization than the transverse light-cone gauge in string theory spectrum. Ghost absence alone is weaker than equivalence.

For $a=1$, momentum is null. Its orthogonal complement contains $D-2$ positive directions and the null direction $p$. The state $p\cdot {\alpha }_{-1}|p⟩$, proportional to $L_{-1}|p⟩$, is physical, spurious and null. Removing it givesThe quotient has exactly the $D-2$ massless transverse vector polarizations. Thus equivalence at level one selects $a=1$ and the null-state quotient of a string, not just the inequality.

Level two and the dimension. Set $a=1$. At level two, $p^2=-1/{\alpha }^{\prime }$ and $k=\sqrt{2{\alpha }^{\prime }}p$ has $k^2=-2$. A general state isUsing $[L_m,{\alpha }_n^a]=-n{\alpha }_{m+n}^a$, the nontrivial positive-mode conditions areThere are $D-1$ vector null string states $L_{-1}(v\cdot {\alpha }_{-1}|p⟩)$ with $k\cdot v=0$. The parent has $L_0=0$, so the descendant norm is zero. Quotienting these leaves the massive symmetric traceless rank-two tensor plus one additional scalar.

The level-two scalar in covariant string quantization can be chosen, for every $D$, asOn its three displayed structures, $L_1$ gives respectively $2,-4,2$ times $k\cdot {\alpha }_{-1}|p⟩$, and $L_2$ gives $D,-2,-4$ times the vacuum. The coefficients make both combinations vanish. The first two structures have norms $2D,8$ and cross inner product $-4$; the mode-two structure has norm $-4$ and is orthogonal to them. For $B=(D+4)/10$, $C=(D-1)/5$, this yieldsAbove 26 this is a physical negative-norm string state. Below 26 it is an extra positive-norm scalar, which cannot be discarded just to force the ordinary light-cone state count. At 26 it becomes null and can be removed. Thus level-two ghost absence gives $D\le 26$, whereas equivalence to the ordinary transverse spectrum requires $D=26$.

The critical scalar is also a Virasoro descendant. For $L_0|p⟩=-|p⟩$, the level-two scalar Virasoro null state candidateobeysAt $D=26$ it is physical and null, with $|S_{26}⟩=2|Z⟩$. At other dimensions it is not physical, so its norm must not be treated as a physical ghost test; the already-physical $|S_D⟩$ gives the correct test.

Finally, the covariant level-two oscillator space has dimension $D(D+3)/2$. The $D$ conditions from $L_1$ and one from $L_2$ leave $(D-1)(D+2)/2$ physical components. Quotienting $D-1$ vector null states and the critical scalar null state leavesthe massive spin-two field count and the level-two light-cone gauge in string theory count. The vector null descendants change the components of $\zeta$ orthogonal to $k$, and the critical scalar null descendant changes its component along $k$. Thus $\zeta$ can be set to zero. A representative then has $\zeta =0$, $k^m{ϵ}_{mn}=0$ and ${ϵ}^m{}_m=0$. The two approaches consequently agree on the massless level-one vector and massive level-two spin-two tensor for $a=1,D=26$.

In old covariant quantization, an open-string physical string state obeyswith analogous left- and right-moving conditions for a closed string. A spurious string state is a Virasoro descendant orthogonal to every physical state. A null string state is both physical and spurious; it has zero norm and zero inner product with every physical state, so quotienting by null states removes gauge redundancy without removing observable states.

A unitary highest-weight Virasoro module is generated by a normalized vector satisfying the displayed relations, with a positive-definite inner product and $L_n^{†}=L_{-n}$. Its Virasoro descendants have $L_0$ eigenvalues $h+N$ for nonnegative integral levels $N$. The commutator gives $\Vert L_{-1}|h⟩{\Vert }^2=2h$, so unitarity requires $h\ge 0$. Such a module is a quotient of a Virasoro Verma module; its Unitary Virasoro discrete series describes the cases with central charge below one.