Bosonic BRST Nilpotence

Write $Q_B=C+G-a{c}_0$, where $C={\sum }_r{c}_{-r}{L}_r^m$ and $G=-\frac{1}{2}{\sum }_{u,w}(u-w):c_{-u}{c}_{-w}{b}_{u+w}:$ uses the relative-vacuum ordering of [F1]. Put $v=\mathrm{\mid }\downarrow ⟩$ for the ghost vacuum. Anticommuting the ghosts gives $C^2=\frac{1}{2}{\sum }_{r,s}{c}_{-r}{c}_{-s}[L_r^m,L_s^m]$. Its noncentral part is ${\sum }_{r,s}r{c}_{-r}{c}_{-s}{L}_{r+s}^m$. Also $\{G,c_{-r}\}=-\frac{1}{2}{\sum }_{u+w=r}(u-w)c_{-u}{c}_{-w}$, so $\{C,G\}$ cancels that noncentral part. The remaining matter contribution, by [F2], is $\frac{D}{12}{\sum }_{n>0}(n^3-n)c_{-n}{c}_n$.

In the normally ordered expansion of $G^2=\{G,G\}\mathrm{/}2$, terms with no contractions cancel because each cubic monomial is odd. Single contractions leave three $c$ modes and one $b$ mode; after antisymmetrizing their coefficient is the Witt Jacobi sum $(r-s)(r+s-t)+(s-t)(s+t-r)+(t-r)(t+r-s)=0$. There are at most two contractions, since the two original monomials contain just two $b$ modes and only $b,c$ pairs contract. Double contractions therefore leave two $c$ modes. Each contraction preserves total mode index, so $G^2={\sum }_{n>0}{g}_n{c}_{-n}{c}_n$ for scalars $g_n$; the putative zero-index term is $c_0^2=0$. Normal-ordering contractions use just the CAR in [F1], with no inference from central charge alone.

The relative-vacuum ordering gives $Gv=0$: a nonzero cubic monomial must contain an annihilator, since two creation $c$ modes require nonpositive indices and their paired $b$ then has nonnegative index. For $n>0$, anticommuting $b_{-n}$ through $G$ and retaining the terms which survive on $v$ gives $G{b}_{-n}v=n{c}_0{b}_{-n}v+{\sum }_{r=1}^{n-1}(n+r)c_{-r}{b}_{-(n-r)}v$. Explicitly, $\{G,b_{-n}\}={\sum }_r(n+r):c_{-r}{b}_{r-n}:$; only $r=0,1,\dots ,n-1$ survive, since $b_{0}v=0$ and all positive modes annihilate $v$.

Extract the coefficient of the Fock basis vector $c_{-n}v$ after applying $G$ again to step 1.3. The identity in step 1.1 gives $\{G,c_0\}=-2{\sum }_{j>0}j{c}_{-j}{c}_j$, whose action on $b_{-n}v$ contributes $-2n{c}_{-n}v$. For $1\le r<n$, put $s=n-r$; the only terms in $\{G,c_{-r}\}{b}_{-s}v$ contributing $c_{-n}v$ have $(u,w)=(n,-s),(-s,n)$ in that identity and sum to $-(n+s)c_{-n}v=-(2n-r)c_{-n}v$. In $G(c_{-r}{b}_{-s}v)=\{G,c_{-r}\}{b}_{-s}v-c_{-r}G{b}_{-s}v$ (also for $r=0,s=n$), the last term has three creation modes by step 1.3 and contributes no one-creator vector. Consequently $g_n=-2n^2-{\sum }_{r=1}^{n-1}(n+r)(2n-r)=-(13n^3-n)\mathrm{/}6=-\frac{26}{12}(n^3-n)-2n$. Here the finite sum uses ${\sum }_{r=1}^{n-1}r=n(n-1)\mathrm{/}2$ and ${\sum }_{r=1}^{n-1}{r}^2=n(n-1)(2n-1)\mathrm{/}6$, including the empty sum at $n=1$. Step 1.2 justifies extracting this coefficient, since $c_{-j}{c}_j{b}_{-n}v={\delta }_{jn}{c}_{-n}v$.

Since $\{C,c_0\}=c_0^2=0$, the intercept contributes $-a\{G,c_0\}=2a{\sum }_{n>0}n{c}_{-n}{c}_n$. Combining steps 1.1, 1.2 and 2.1 yields the exact identity $Q_B^2={\sum }_{n>0}{c}_{-n}{c}_n(\frac{D-26}{12}(n^3-n)+2(a-1)n)$. This locally computed ghost coefficient agrees with [F3]; its linear part has now been obtained in the actual relative ordering. All expansions are locally finite on the stated algebraic domain: annihilator indices are bounded on each input, and after their contractions each fixed remaining level admits only finitely many creators. No analytic limit or regularization is invoked.

If $Q_B^2=0$, apply step 3.1 to a nonzero matter vacuum tensored with $b_{-n}v$. All terms except the $n$-th vanish, and $c_{-n}v\mathrm{\ne }0$. Taking $n=1$ therefore forces $2(a-1)=0$ and $a=1$.

With $a=1$, the same test at $n=2$ forces $(D-26)(8-2)\mathrm{/}12=(D-26)\mathrm{/}2=0$, hence $D=26$. Conversely $D=26,a=1$ makes every coefficient in step 3.1 vanish, giving $Q_B^2=0$ on every finite-particle vector.