Open Oldcovariant BRST Equivalence

In the target $M={\mathbb{R}}^{1,25}$, define $X^{\pm }=(X^0\pm X^{25})\mathrm{/}\sqrt{2}$, $p^{\pm }=(p^0\pm p^{25})\mathrm{/}\sqrt{2}$, and ${\alpha }_n^{\pm }=({\alpha }_n^0\pm {\alpha }_n^{25})\mathrm{/}\sqrt{2}$, with transverse indices $i=1,\dots ,24$; then ${\eta }_{+-}=-1$ and $[{\alpha }_m^{+},{\alpha }_n^{-}]=-m{\delta }_{m+n,0}$. Set $k^{\mu }=\sqrt{2{\alpha }^{\mathrm{\prime }}}{p}^{\mu }$ and work in the target momentum chart $p^{+}\mathrm{\ne }0$ (equivalently $k^{+}\mathrm{\ne }0$); no world-sheet gauge condition is imposed in this BRST comparison. Work at $D=26,a=1$, real fixed $k$ with $k^{+}\mathrm{\ne }0$. Let ${\mathcal{C}}_N(k)$ be the relative algebraic subspace generated over $\mathrm{\mid }k⟩\mathrm{\mid }\downarrow ⟩$ by negative-index matter modes and $b_{-n},c_{-n}$, $n>0$, of total oscillator level $N$. Its zero-mode equation is $L_0^{tot}=k^2\mathrm{/}2+N-1=0$. This space is finite dimensional, in all ghost numbers, and the relative differential $d$ is $Q_B$ with its $c_0,b_0$ terms removed. Nilpotence gives $d^2=0$ on this on-shell subspace. Assign degrees $\mathrm{deg}{\alpha }_n^{+}=\mathrm{deg}{c}_n=1$, $\mathrm{deg}{\alpha }_n^{-}=\mathrm{deg}{b}_n=-1$, $\mathrm{deg}{\alpha }_n^i=0$; the relative vacuum has degree zero. At fixed level this degree is bounded above and below. Expanding the quadratic matter Virasoro modes and the cubic ghost term gives $d=d_0+\delta ,d_0=-k^{+}{\sum }_{n\mathrm{\ne }0}{c}_{-n}{\alpha }_n^{-},$ where every term of $\delta$ raises degree by one or two. Explicitly the degree-one terms are $c$ times the transverse quadratic and ${\alpha }^{+}{\alpha }^{-}$ quadratic terms, the transverse momentum term, and $ccb$; the degree-two term is $-k^{-}\sum c_{-n}{\alpha }_n^{+}$. All preserve total level. This includes nonzero transverse momentum; it need not be set to zero. Every sum is locally finite.

The dagger in the existing definition is the antilinear Krein adjoint. The real-momentum charge is Hermitian: the matter term pairs index $n$ with $-n$, the cubic term does so after the CAR reversal, and its zero-mode coefficient is real. In the decomposition $Q_B=c_0{L}_0^{tot}-b_{0}M+d$ all three displayed terms are separately Hermitian; $d^{†}=d$ and $\{d,c_0\}=0$. Therefore for relative $\chi$ and closed $\psi$, $⟨d\chi ,c_0\psi {⟩}_K=⟨\chi ,d{c}_0\psi {⟩}_K=-⟨\chi ,c_{0}d\psi {⟩}_K=0.$ Thus exact vectors are orthogonal to all closed vectors for $h_{rel}$, and the form descends. No BPZ form is substituted here.

Define $K=\frac{1}{k^{+}}{\sum }_{n\mathrm{\ne }0}\frac{{\alpha }_{-n}^{+}{b}_n}{\mathrm{\mid }n\mathrm{\mid }}\mathrm{.}$ The CAR and oscillator relations give $\{d_0,K\}=\mathcal{N}={\sum }_{n>0}[-\frac{{\alpha }_{-n}^{+}{\alpha }_n^{-}+{\alpha }_{-n}^{-}{\alpha }_n^{+}}{n}+c_{-n}{b}_n+b_{-n}{c}_n]\mathrm{.}$ For clarity the two un-normal-ordered contributions are ${\sum }_{n\mathrm{\ne }0}\mathrm{sgn}(n)c_{-n}{b}_n$ and $-{\sum }_{n\mathrm{\ne }0}\mathrm{\mid }n{\mathrm{\mid }}^{-1}{\alpha }_{-n}^{+}{\alpha }_n^{-}$; their constants cancel when the negative indices are moved to down-vacuum normal order. Commuting $\mathcal{N}$ with each of ${\alpha }_{-n}^{+},{\alpha }_{-n}^{-},c_{-n},b_{-n}$ gives that same creator. Its value on the vacuum is zero. Thus its eigenvalue is exactly the total number of these four kinds of creators, a nonnegative integer, and its kernel is precisely the transverse oscillator space ${\mathcal{T}}_N(k)\mathrm{\mid }\downarrow ⟩$. Its ghost number is exactly one. Also $d_0^2=K^2=0$ and $[\mathcal{N},d_0]=[\mathcal{N},K]=0$. Let $\iota :{\mathcal{T}}_N\to {\mathcal{C}}_N$ be inclusion and $\pi$ the projection onto $\mathcal{N}=0$. Set $h=K\mathrm{/}\mathcal{N}$ on positive eigenspaces and $h=0$ on the kernel. Then $d_{0}h+h{d}_0=1-\iota \pi ,h^2=0,\pi h=h\iota =0.$ In particular $H(d_0)={\mathcal{T}}_N$, in ghost number one and in no other ghost number. This is a complete finite-level contraction; $1\mathrm{/}n$ rather than $1\mathrm{/}\mathrm{\mid }n\mathrm{\mid }$ would not give a nonnegative count and is not used.

The operator $h\delta$ strictly raises degree, hence is nilpotent on this finite-level space. Define the finite sums $I=(1+h\delta {)}^{-1}\iota ={\sum }_{r\ge 0}(-h\delta {)}^r\iota ,U=I\pi +dh+h{d}_0\mathrm{.}$ One has $\pi \delta I=0$: $I$ has leading degree zero and strictly positive higher degrees, $\delta$ raises degree, whereas $\pi$ has image only in degree zero. From $I=\iota -h\delta I$, the contraction identity and $d_0\delta +\delta d_0+{\delta }^2=0$ give $(1+h\delta )dI=\iota \pi \delta I=0.$ Thus $dI=0$. Moreover $U=1+(I-\iota )\pi +\delta h$ differs from the identity by a strictly degree-raising operator. Its inverse is a finite geometric sum. Direct multiplication, using $d^2=d_0^2=0$, gives $dU=dh{d}_0=U{d}_0\mathrm{.}$ This is an actual invertible similarity $U^{-1}dU=d_0$ on every on-shell finite-level relative complex, with no term described merely as trivial on physical states. In particular $U\iota =I$ identifies the complete relative cohomology with ${\mathcal{T}}_N$ in ghost number one, all other relative ghost numbers being zero.

The vacuum value is $⟨\downarrow \mathrm{\mid }{c}_0\mathrm{\mid }\downarrow {⟩}_K=⟨0\mathrm{\mid }{c}_{-1}{c}_0{c}_1\mathrm{\mid }0{⟩}_K=1$. In the relative oscillator basis, a nonzero Krein contraction pairs ${\alpha }^{+}$ with ${\alpha }^{-}$ and $c$ with $b$. Dagger preserves the declared degree. Consequently two homogeneous vectors have zero $c_0$ pairing unless their degrees sum to zero. Every $It$ is $t\mathrm{\mid }\downarrow ⟩$ plus strictly positive-degree terms. It follows exactly that $h_{rel}([It],[Is])=⟨t,s{⟩}_{\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{s}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{s}\mathrm{e}}\mathrm{.}$ The transverse Heisenberg norm is positive definite: a monomial with occupation numbers $r_{n,i}$ has squared norm ${\prod }_{n,i}{n}^{r_{n,i}}{r}_{n,i}!$, and distinct normalized occupation monomials are orthogonal. This proves relative BRST positivity in all levels without claiming that $U$ is unitary on the ambient Krein space.

There is also a ghost-free choice of these representatives. Let ${\mathcal{F}}^{+}$ be the matter subspace generated only by transverse and ${\alpha }_{-n}^{+}$ creators. For $v\in {\mathcal{F}}^{+}$, the relative charge on $v\mathrm{\mid }\downarrow ⟩$ is ${\sum }_{n>0}{c}_{-n}{L}_n^{m}v\mathrm{\mid }\downarrow ⟩$. The positive-mode ghost cubic annihilates the down vacuum. Both $d_0$ and this expression have matter coefficients still in ${\mathcal{F}}^{+}$: a positive ${\alpha }^{-}$ can contract a plus creator, while a remaining positive ${\alpha }^{+}$ annihilates the whole subspace. Applying $h$ contracts each $c_{-n}$ with $b_n$ and inserts ${\alpha }_{-n}^{+}$; the other half of $K$ has a positive ${\alpha }^{+}$ and vanishes. Dividing by the diagonal count $\mathcal{N}$ changes neither matter nor ghost support. Hence $I$ maps every transverse $t$ to $v_t\mathrm{\mid }\downarrow ⟩$ with $v_t\in {\mathcal{F}}^{+}$, and its zero-plus-oscillator part is exactly $t$. Since $dI=0$, independence of the $c_{-n}$ ghost states gives $L_n^m{v}_t=0$ for every $n>0$. Its level and on-shell equation give $(L_0^m-1)v_t=0$. Thus $v_t$ is an old-covariant physical representative.

For every old-covariant physical matter vector $v\in P_N(p)$ define $j(v)=v\mathrm{\mid }\downarrow ⟩$. The same explicit charge action shows that this is relative closed; no extra ghost correction is required. Its pairing is exactly $h_{rel}([j(v)],[j(w)])=⟨v,w{⟩}_K\mathrm{.}$ Step4 shows that every BRST class is $[j(v_t)]$, so the map $j:P_N\to H_{rel}^1$ is surjective. If $j(v)$ is exact, the descent identity makes its pairing with every $j(w)$ zero, so $v\in \mathrm{Rad}(P_N)$. Conversely, if $v$ is in that radical, its class pairs to zero with every BRST class by surjectivity. Positive definiteness established in step4 implies $[j(v)]=0$. Thus $\ker j=\mathrm{Rad}(P_N),P_N\mathrm{/}\mathrm{Rad}(P_N)\stackrel{\sim }{\to }{H}_{rel}^1$ is the actual natural isometric bijection. It proves old-covariant positivity as well as BRST positivity; none is assumed to prove another. Each level includes $N=0$, and every nonzero momentum is covered by the existing spatial-rotation $p^{+}\mathrm{\ne }0$ chart supplier. Real spatial rotations preserve all commutators, adjoints, vacuum pairings and the relative map, transporting the result back. The excluded $p=0$ sector is not added; no closed-string claim or infinite-level convergence is used.