Critical Dimension Flat Bosonic String

By [A1], [F2], and [F3], cancellation of the total central charge requires $0=c_m+c_{gh}=D-26$, hence $D=26$. This is the critical dimension of this free-boson model; a different matter CFT would replace $D$ by its central charge.

For $ϵ>0$, the geometric series gives ${\sum }_{n\ge 1}{e}^{-ϵn}=e^{-ϵ}\mathrm{/}(1-e^{-ϵ})$; differentiating in $ϵ$ gives ${\sum }_{n\ge 1}n{e}^{-ϵn}=e^{-ϵ}\mathrm{/}(1-e^{-ϵ}{)}^2=1\mathrm{/}(4{\sinh }^2(ϵ\mathrm{/}2))$. Taylor expansion $\sinh x=x+x^3\mathrm{/}6+O(x^5)$ yields $1\mathrm{/}(4{\sinh }^2(ϵ\mathrm{/}2))={ϵ}^{-2}-1\mathrm{/}12+O({ϵ}^2)$. Thus the declared finite part is $-1\mathrm{/}12$. The $D-2$ transverse oscillator families have vacuum energy $E_0=\frac{D-2}{2}(-1\mathrm{/}12)=-(D-2)\mathrm{/}24$, so the intercept that subtracts this zero-point contribution is $a=-E_0=(D-2)\mathrm{/}24$.

For nonzero $n$, first at transverse momentum zero, put ${\mathrm{\ell }}_n=\frac{1}{2}{\sum }_{r\in \mathbb{Z}}:{\alpha }_{n-r}^a{\alpha }_r^a:,{\alpha }_0^a=0,N={\sum }_{n>0,a}{\alpha }_{-n}^a{\alpha }_n^a,$ $E_i={\sum }_{n>0}\frac{{\alpha }_{-n}^i{\mathrm{\ell }}_n-{\mathrm{\ell }}_{-n}{\alpha }_n^i}{n},B_{m,ij}={\alpha }_{-m}^i{\alpha }_m^j-{\alpha }_{-m}^j{\alpha }_m^i,A_{ij}={\sum }_{m>0}\frac{B_{m,ij}}{m}\mathrm{.}$ All sums act locally finitely and preserve level. The transverse oscillator computation in the existing Virasoro supplier applies with Euclidean metric and $d$ components: $[{\mathrm{\ell }}_n,{\alpha }_m^a]=-m{\alpha }_{n+m}^a$ and $[{\mathrm{\ell }}_m,{\mathrm{\ell }}_n]=(m-n){\mathrm{\ell }}_{m+n}+d(m^3-m){\delta }_{m+n,0}\mathrm{/}12$. Also $E_i^{†}=-E_i$, $E_i\mathrm{\mid }0⟩=0$ and $[N,E_i]=0$. We prove the complete operator identity (1): $[E_i,E_j]=-2N{A}_{ij}+{\sum }_{m>0}[2m-\frac{d}{12}(m-\frac{1}{m})]B_{m,ij}\mathrm{.}$

The expression for $E_i$ is the normally ordered cubic $E_i=-\frac{1}{2}{\sum }_{\genfrac{}{}{0px}{}{r+s+t=0}{r,s,t\mathrm{\ne }0}}\frac{1}{r}:{\alpha }_r^i{\alpha }_s^a{\alpha }_t^a:\mathrm{.}$ There is no internal contraction: a paired opposite pair would force the third index to zero. A commutator of two such cubics has only quartic, quadratic and scalar normal-ordered terms. Its quartic part is obtained by one oscillator contraction, equivalently by the commutative Poisson bracket with $\{x_r^a,x_s^b\}=r{\delta }^{ab}{\delta }_{r+s,0}$. Here is the complete cancellation of that quartic bracket. Write $l_n=\frac{1}{2}{\sum }_{r+s=n}{x}_r^a{x}_s^a$, with $x_0=0$, and $e_i=-\frac{1}{2}{\sum }_{r+s+t=0}{x}_r^i{x}_s^a{x}_t^a\mathrm{/}r$. Its derivative is $\frac{\mathrm{\partial }{e}_i}{\mathrm{\partial }{x}_r^a}=-\frac{{\delta }_{ai}}{r}{l}_{-r}-{\sum }_{u+v=-r}\frac{x_u^i{x}_v^a}{u}\mathrm{.}$ The two mixed products in ${\sum }_{r,a}r(\mathrm{\partial }{e}_i\mathrm{/}\mathrm{\partial }{x}_r^a)(\mathrm{\partial }{e}_j\mathrm{/}\mathrm{\partial }{x}_{-r}^a)$ combine to ${\sum }_{\genfrac{}{}{0px}{}{u,v\mathrm{\ne }0}{u+v\mathrm{\ne }0}}(\frac{1}{v}-\frac{1}{u})x_u^i{x}_v^j{l}_{-u-v}\mathrm{.}$ In the product of the two sums, symmetrize the two equally colored contracted variables. Their coefficient becomes $(v-u)\mathrm{/}(2uv)$, giving ${\sum }_{u,v\mathrm{\ne }0}(\frac{1}{u}-\frac{1}{v})x_u^i{x}_v^j{l}_{-u-v}\mathrm{.}$ These cancel for $u+v\mathrm{\ne }0$. At $v=-u$ the remaining sum is ${\sum }_{u\mathrm{\ne }0}(2\mathrm{/}u)x_u^i{x}_{-u}^j{l}_0=-2N{A}_{ij}$. Thus the quartic normal-ordered part of $[E_i,E_j]$ is exactly $-2:N{A}_{ij}:$. This coefficient calculation is legitimate term by term: each fixed quartic monomial has finitely many contractions, and all resulting operators act locally finitely on each finite level. The scalar part vanishes since both $E_i$ annihilate the vacuum. Every remaining quadratic term has total mode index zero. Inspecting its Kronecker contractions leaves only $B_{m,ij}$: terms proportional to ${\delta }_{ij}$ cancel by antisymmetry, while the two free indices are $i,j$. Hence it remains to compute one coefficient for each $m$, with no uncomputed higher-degree operator remainder.

On each level in [L1], $N$ is the real scalar $N$, $A_{ij}$ and $E_i(0)$ from step 1.3 are finite matrices with $A_{ij}^{†}=-A_{ij}$ and $E_i(0{)}^{†}=-E_i(0)$. Put $E_i(k)=E_i(0)+k_a{A}_{ia}$ and $P^{-}=(k_a{k}_a\mathrm{/}2+N-a)\mathrm{/}s$. They act by smooth matrix multiplication on $\mathrm{\Omega }$. Multiplication by $s^{-1}$, momentum polynomials, $N$, and these matrices, and differentiation in $s,k_i$, preserve $\mathcal{D}$. All compositions and commutators below therefore share that invariant domain. Finite-level truncation followed by compact cutoff and scalar mollification in each coefficient approximates any vector in $\mathcal{H}$, so $\mathcal{D}$ is dense. With Lebesgue measure, integration by parts without boundary terms gives $V_i^{†}=-V_i-\mathrm{div}{V}_i$ on test functions for $V_i=k_i{\mathrm{\partial }}_s+P^{-}{\mathrm{\partial }}_{k_i}$, and $\mathrm{div}{V}_i={\mathrm{\partial }}_s{k}_i+{\mathrm{\partial }}_{k_i}{P}^{-}=k_i\mathrm{/}s$. Consequently $i(V_i+k_i\mathrm{/}(2s)-E_i(k)\mathrm{/}s)$ is symmetric on $\mathcal{D}$: both $V_i+\mathrm{div}{V}_i\mathrm{/}2$ and $-E_i(k)\mathrm{/}s$ are skew-symmetric. Here symmetric means equality of the two integrated pairings on $\mathcal{D}$, not an assertion of a self-adjoint extension or global unitary Lorentz representation. In the closed sector the two commuting spin matrices are added on the level-matched tensor factor and the orbital derivative occurs once; the same domain and symmetry argument applies.

For $i\mathrm{\ne }j$, let $\mathrm{\mid }{i}_m⟩={\alpha }_{-m}^i\mathrm{\mid }0⟩$, whose norm is $m$. Directly from $[{\mathrm{\ell }}_n,{\alpha }_{-m}^i]=m{\alpha }_{n-m}^i$, $E_i\mathrm{\mid }{j}_m⟩=m{\sum }_{n=1}^{m-1}\frac{{\alpha }_{-n}^i{\alpha }_{-(m-n)}^j}{n}\mathrm{\mid }0⟩-{\delta }_{ij}{\mathrm{\ell }}_{-m}\mathrm{\mid }0⟩\mathrm{.}$ Put $F_i=m{\sum }_{n=1}^{m-1}{n}^{-1}{\alpha }_{-n}^i{\alpha }_{-(m-n)}^i\mathrm{\mid }0⟩$ and $L={\mathrm{\ell }}_{-m}\mathrm{\mid }0⟩$. The exact Wick contractions give $⟨F_i,F_j⟩=0(i\mathrm{\ne }j),⟨F_i,L⟩=m{\sum }_{n=1}^{m-1}(m-n)=\frac{m^2(m-1)}{2},$ $⟨L,L⟩=\frac{d}{2}{\sum }_{n=1}^{m-1}n(m-n)=\frac{d}{12}(m^3-m),$ $⟨E_j{i}_m,E_i{j}_m⟩=m^2{\sum }_{n=1}^{m-1}1=m^2(m-1)\mathrm{.}$ Using $E_i^{†}=-E_i$ therefore gives $⟨i_m,[E_i,E_j]j_m⟩=2m^2(m-1)-\frac{d}{12}(m^3-m)\mathrm{.}$ The normal-ordered quartic term kills every one-oscillator state, and $⟨i_m,B_{m,ij}{j}_m⟩=m^2$. Thus the remaining quadratic coefficient is exactly $2(m-1)-d(m-1\mathrm{/}m)\mathrm{/}12$. Finally normal ordering the product $N{A}_{ij}$ gives $N{A}_{ij}=:N{A}_{ij}:+{\sum }_{m>0}{B}_{m,ij}\mathrm{.}$ Substitution proves (1) for every mode, including $m=1$ with its empty finite sums. For $i=j$ both sides vanish. When $d=1$ there is no pair $i\mathrm{\ne }j$, and this particular Lorentz commutator is identically zero; no incorrect low-dimensional uniqueness claim is needed below.

In the target null basis of the Statement, the gauge in [A3] gives ${\alpha }_n^{+}=0$ for $n\mathrm{\ne }0$ and $k^{+}={\alpha }_0^{+}\mathrm{\ne }0$. The mode expansions and constraints in [F5]--[F6] then give, in each chiral sector, $0=L_n^m=-k^{+}{\alpha }_n^{-}+L_n^{\perp }$ for $n\mathrm{\ne }0$, where $L_n^{\perp }=\frac{1}{2}{\sum }_{r\in \mathbb{Z}}{\sum }_{i=1}^d:{\alpha }_{n-r}^i{\alpha }_r^i:$. Thus ${\alpha }_n^{-}=L_n^{\perp }\mathrm{/}{k}^{+}$. The zero-mode constraint is $0=L_0^m-a=-k^{+}{k}^{-}+k_i{k}_i\mathrm{/}2+N-a$, so $k^{-}=(k_i{k}_i\mathrm{/}2+N-a)\mathrm{/}{k}^{+}$. In a closed string the same formulas hold separately for the tilded chiral sector. This derives the longitudinal constraint solution on the stated chart. For general transverse momentum write $L_n^{\perp }={\mathrm{\ell }}_n+k_a{\alpha }_n^a$ and $E_i(k)=E_i(0)+k_a{A}_{ia}$. The elementary rotation commutators, obtained from the same oscillator CAR, give identity (2): $[E_i(k),E_j(k)]=[E_i(0),E_j(0)]-k_i{E}_j(k)+k_j{E}_i(k)-k_a{k}_a{A}_{ij}\mathrm{.}$ For example $[A_{ia},E_j(0)]={\delta }_{aj}{E}_i(0)-{\delta }_{ij}{E}_a(0)$, and the quadratic $k$ term follows from $[A_{ia},A_{jb}]={\delta }_{aj}{A}_{ib}-{\delta }_{ij}{A}_{ab}+{\delta }_{ib}{A}_{aj}-{\delta }_{ab}{A}_{ij}$. Thus (2) includes, rather than discards, all transverse momentum terms. Put $P^{-}=(k_a{k}_a\mathrm{/}2+N-a)\mathrm{/}{k}^{+}$ and use the symmetric dynamical generator on $\mathcal{D}$ established in step 2.2, $J^{-i}=i[k_i{\mathrm{\partial }}_{k^{+}}+P^{-}{\mathrm{\partial }}_{k_i}+\frac{k_i}{2k^{+}}-\frac{E_i(k)}{k^{+}}]\mathrm{.}$ The first three terms are the symmetrically ordered orbital generator; the last is the light-cone oscillator spin with ${\alpha }_n^{-}=L_n^{\perp }\mathrm{/}{k}^{+}$. The sign convention is $J^{\mu \nu }=k^{\mu }{q}^{\nu }-k^{\nu }{q}^{\mu }$ as in the source. The vector fields $V_i=k_i{\mathrm{\partial }}_{k^{+}}+P^{-}{\mathrm{\partial }}_{k_i}$ commute, since $V_i(P^{-})=0$ and $V_i(k_j)=P^{-}{\delta }_{ij}$. Their half-divergence terms have symmetric derivatives and contribute no antisymmetric commutator. Also ${\mathrm{\partial }}_{k_i}{E}_j-{\mathrm{\partial }}_{k_j}{E}_i=-2A_{ij}$. Direct differential-operator multiplication consequently gives $[J^{-i},J^{-j}]=-(k^{+}{)}^{-2}([E_i(k),E_j(k)]+k_i{E}_j(k)-k_j{E}_i(k)+2k^{+}{P}^{-}{A}_{ij})\mathrm{.}$ Substitute (1) and (2). All level operators, quartic pieces and transverse momentum pieces cancel, leaving the complete anomaly (3): $[J^{-i},J^{-j}]=\frac{1}{(k^{+}{)}^2}{\sum }_{m>0}[(\frac{d}{12}-2)m+\frac{2a-d\mathrm{/}12}{m}]B_{m,ij}\mathrm{.}$ The other generators are the orbital momentum derivatives and the exact transverse rotations $A_{ij}$. Their commutators with $E_i$ are the rotation relation above, and with $P^{-}$ are its displayed first derivatives; hence all the other Lorentz relations have their classical values. No further multiple contraction can occur in those rotation or first-order momentum actions. Equation (3) is therefore the only possible light-cone Lorentz anomaly on this common invariant test domain $\mathcal{D}$.

At $d=24,a=1$ every coefficient in (3) vanishes. If $d\ge 2$, independence of the $B_{m,ij}$ (detected on $\mathrm{\mid }{j}_m⟩$) shows that Lorentz closure requires each coefficient to vanish. Modes $m=1$ and $m=2$ give respectively $2(a-1)=0$ and $d=24$. Thus $D=26,a=1$. For the usual closed string, two independent chiral spin sums give the two copies of (3); on the level-matched subspace the same cancellation applies, and the two independent oscillator families cannot cancel a nonzero anomaly. No all-level condition is inferred only from finitely many low-level states: the finite-state tests determine coefficients only after the complete operator identity has been proved. The original Statement requires both this Lorentz closure and covariant conformal-anomaly cancellation. The already proved total central charge is $D-26$, so their joint conditions force $D=26$ even in the low-dimensional cases where the particular $[J^{-i},J^{-j}]$ test is vacuous. The declared finite-part zero-point prescription gives $a=(D-2)\mathrm{/}24=1$, and the corrected-consistent BRST nilpotence theorem gives the same intercept. This proves the unchanged standard-flat-model criticality claim, not a claim about arbitrary internal matter CFTs or Liouville strings.