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The Polyakov action is invariant under worldsheet diffeomorphisms and Weyl transformations, ${\gamma }_{ab}\mapsto e^{2\omega }{\gamma }_{ab}$. A two-dimensional metric has three local components: two coordinate transformations and one Weyl function remove these locally. In conformal gauge the factor $\sqrt{-\gamma }{\gamma }^{ab}$ is independent of the conformal scale, and on a flat coordinate patch the matter action isThus it is the action of $d$ free scalar fields, with their target-space contractions using the Minkowski metric. On a Euclidean sphere there are no unpunctured conformal moduli: a metric can be put into the conformal class of the round metric, and in stereographic patches the scalar action has this free-field form. This does not mean that the sphere has a globally flat metric. A spherical worldsheet here is understood after Wick rotation; it has no nonsingular global Lorentzian metric. The residual Möbius transformations and their zero modes still have to be divided out or fixed by vertex positions.

To derive the worldsheet ghost action, consider an infinitesimal diffeomorphism generated by $v^a$ and a Weyl variation $\omega$. The metric variation separates asIntegration over the Weyl parameter removes the trace gauge condition. The trace-free part produces the Faddeev-Popov determinant of an operator mapping vectors to symmetric traceless tensors. This is trace elimination in string Faddeev-Popov gauge fixing. The stated Grassmann Gaussian integral exponentiates this determinant with an odd vector $c^a$ and an odd symmetric trace-free antighost $b_{ab}$.

The normalization of the operator and antighost can be chosen to match the printed convention. For example, in a flat patch take $P=-P_1/2$. It differs from the orbit operator only by a field-independent determinant factor. Tracelessness and symmetry give $b_{ab}(Pc{)}^{ab}=-b_{ab}{\partial }^a{c}^b$. Integration by parts, followed by exchanging two odd factors, gives $\int b_{ab}{\partial }^a{c}^b=\int c^a{\partial }^b{b}_{ab}$. Consequently the determinant exponent is $+(2\pi {)}^{-1}\int c^a{\partial }^b{b}_{ab}=i{I}_{\mathrm{gh}}$ withOverall constant normalization does not affect the determinant's dependence on the fields. The Faddeev-Popov ghost fields are vectors, while the antighost fields are symmetric trace-free tensors; they are not extra target-space string excitations.

Now use ${\sigma }^{\pm }=\tau \pm \sigma$, ${\partial }_{\pm }=({\partial }_{\tau }\pm {\partial }_{\sigma })/2$ and $c^{\pm }=c^0\pm c^1$. In these coordinates ${\eta }_{+-}={\eta }_{-+}=-1/2$ and ${\partial }^{\pm }=-2{\partial }_{\mp }$. Tracelessness implies $b_{+-}=0$, so directlyThereforeFor a Cartesian check, $b_{11}=b_{00}$ and $b_{++}=(b_{00}+b_{01})/2$, $b_{--}=(b_{00}-b_{01})/2$ reproduce the same result.

The original PDF repeats ${\delta }_{\eta }{c}^{+}$ for its final ghost transformation. The second occurrence must be ${\delta }_{\eta }{c}^{-}$; otherwise the displayed rules give two incompatible transformations for one ghost and none for the other. Write the corrected transformation as ${\delta }_{\eta }\Phi =\eta s\Phi$, with $\eta$ on the left. Then the odd left-acting BRST differential satisfies the graded Leibniz rule $s(AB)=(sA)B+(-1{)}^{|A|}A(sB)$. Its local rules areThe last two expressions are $s{b}_{\pm \pm }=i{\theta }_{\pm \pm }/{\alpha }^{\prime }$ with the matter-plus-ghost worldsheet stress tensor in the question.

Here is an off-shell check of the action variation, including the Grassmann signs. Temporarily set $c=c^{+}$, $f=c^{-}$, $b=b_{++}$, $h=b_{--}$, $u={\partial }_{+}X$ and $v={\partial }_{-}X$. Apart from the common factor $1/\pi$, the total Lagrangian is $\mathcal{L}=u\cdot v/{\alpha }^{\prime }+i(c{\partial }_{-}b+f{\partial }_{+}h)$. Its matter variation isThe matter terms in $sb$ and $sh$ give the ghost-action contribution ${\alpha }^{\prime -1}[c{\partial }_{-}(u^2)+f{\partial }_{+}(v^2)]$. Combining these terms with the last two matter terms produces ${\alpha }^{\prime -1}[{\partial }_{-}(c{u}^2)+{\partial }_{+}(f{v}^2)]$.

The remaining contribution from the $c,b$ sector iswhere subscripts denote derivatives. In the last equality the terms $c_{-}{c}_{+}b$ and $c_{+}{c}_{-}b$ cancel by anticommutation. The $f,h$ sector has the same identity with $+\leftrightarrow -$. ThuswhereHence the total action is BRST invariant up to a boundary term, which vanishes on a closed worldsheet or under the usual compatible boundary and time-end conditions. This calculation requires no equations of motion.

Nilpotence needs a qualification for these particular chiral rules. For either ghost, $s(c\partial c)=(c\partial c)\partial c-c\partial (c\partial c)=0$, because $c^2=(\partial c{)}^2=0$. Thus $s^2{c}^{\pm }=0$ algebraically. For the embedding, however, the graded calculation givesMixed second derivatives of $X$ cancel because $c^{+}{c}^{-}=-c^{-}{c}^{+}$. The remaining terms vanish on the ghost equations of motion, ${\partial }_{-}{c}^{+}={\partial }_{+}{c}^{-}=0$, which follow by varying the antighosts in the chiral ghost action. They need not vanish off shell. An explicit counterexample uses independent Grassmann variables $\xi ,\rho$: take $c^{+}=\xi$, $c^{-}=\rho {\sigma }^{+}$ and one coordinate $X={\sigma }^{-}$. Then $s^{2}X=-\xi \rho \ne 0$.

For the antighosts, use the permitted stress-tensor identity $s{\theta }_{ab}=0$ on the classical solution space. Then $s^2{b}_{\pm \pm }=(i/{\alpha }^{\prime })s{\theta }_{\pm \pm }=0$. Indeed direct variation of the displayed stress tensors giveswhich vanish using the ghost equations and the free matter wave equation. Therefore two successive BRST transformations annihilate every field on shell. This is chiral conformal-gauge BRST nilpotence is on shell, not an unrestricted off-shell identity for arbitrary ghosts. Quantum BRST nilpotence additionally requires cancellation of the worldsheet anomaly, giving $d=26$ in the critical bosonic theory.

Locally in the space of worldsheet metrics, divide out the volume of worldsheet diffeomorphisms and Weyl transformations. With stabilizer zero modes omitted, the inverse Faddeev-Popov determinant at a metric $g$ isThis is the linearized orbit integral: the delta functional is the product of the metric-component delta functions, and its Jacobian is the inverse determinant of the gauge-orbit map. The omitted zero modes and any metric moduli must be handled separately; this local formula is not an integration over the physical moduli themselves.

The trace elimination in string Faddeev-Popov gauge fixing is essential. In two dimensions, split the variation asThe second term is trace-free. Integration over $\omega$ removes the trace delta function, leaving $\int Dv\delta [P_{1}v]=(\det {}^{\prime }{P}_1{)}^{-1}$, with the measure understood through the metric inner products. Equivalently, a Fourier representation of the metric delta functional introduces a commuting symmetric tensor ${\beta }^{ab}$. Integration over $\omega$ enforces $g_{ab}{\beta }^{ab}=0$, and the remaining exponent is proportional to $\int \sqrt{g}{\beta }^{ab}(P_{1}v{)}_{ab}$. This gives the same inverse Faddeev-Popov determinant and explains why its tensor variable must be trace-free.

For the Polyakov path integral, insert the gauge-fixing identitywhere $\hat{g}(m)$ is a representative depending on the worldsheet moduli $m$. Gauge invariance of the action, measure, and Faddeev-Popov determinant lets us change variables along the gauge orbit. Its volume cancels the original gauge-group denominator, leaving an integral over $X$, the moduli, and the determinant evaluated at $\hat{g}$. This is how the determinant enters the numerator of the gauge-fixed integral, despite the inverse determinant occurring in the orbit integral above.

A fermionic Gaussian integral exponentiates $\det {}^{\prime }{P}_1$ using an anticommuting vector $c^a$ and an anticommuting symmetric trace-free tensor $b^{ab}$:The constant $C$ can be absorbed into $b$. In local complex conformal gauge, trace-freeness sets $b_{z\bar{z}}=0$. The only surviving components of the worldsheet diffeomorphism ghost operator are proportional to $\bar{\partial }{c}^z$ and $\partial c^{\bar{z}}$. Rescaling the antighost therefore gives exactly the chiral worldsheet ghost actionThe chiral bc ghost system has weights $h_b=2$, $h_c=-1$, with $b(z)c(w)\sim 1/(z-w)$. Its worldsheet stress tensor is $T_{\mathrm{gh}}=-2:b\partial c:-:(\partial b)c:$. Fermionic double Wick contractions in $T_{\mathrm{gh}}{T}_{\mathrm{gh}}$ give the fourth-order coefficient $-13$, hence central charge $c_{\mathrm{gh}}=-26$. More generally this calculation gives $c=1-3(2h_b-1{)}^2$. Each embedding scalar contributes one, so the total chiral central charge is $D-26$. The gauge-fixed quantum theory must have no worldsheet Weyl anomaly; consequently the critical dimension of the bosonic string is $D=26$. Zero-mode insertions for Conformal Killing vector fields and moduli do not change this local anomaly calculation.

For the Veneziano amplitude, first examine an ordered beta-function term. A pole occurs when $-{\alpha }^{\prime }s-1=-n$, with $n=0,1,2,\dots$, so the exchanged masses follow the open bosonic string mass spectrumThere is a tachyon at level zero, a massless vector at level one, then an infinite tower with equally spaced squared masses. In oscillator quantization these levels are $N=n$; the poles alone do not determine all oscillator degeneracies or all states that might fail to couple to the chosen external particles.

The Veneziano amplitude pole residue gives considerably more information than a pole position. Near $s_n=(n-1)/{\alpha }^{\prime }$, use $\Gamma (-n+ϵ)\sim (-1{)}^n/(n!ϵ)$ and $ϵ=-{\alpha }^{\prime }(s-s_n)$. The finite Gamma function ratio is a product of $n$ linear factors. Thus, for generic $t$,The empty product at $n=0$ is one. This degree-$n$ angular polynomial decomposes into spin partial waves; it reveals the spins that couple, up to maximal spin $n$ in the ordered channel, and the corresponding products of three-point couplings. For example, the level-one linear residue encodes vector exchange. A single external-state amplitude need not separate all states with the same mass and spin.

There is a qualification specific to the sum printed in the question. For four identical on-shell tachyons, $s+t+u=-4/{\alpha }^{\prime }$. At ${\alpha }^{\prime }s=n-1$, put $P_n(t)={\prod }_{j=1}^n({\alpha }^{\prime }t+j+1)$. Then ${\alpha }^{\prime }u=-n-3-{\alpha }^{\prime }t$, and reversing the product gives $P_n(u)=(-1{)}^n{P}_n(t)$. The $t,u$ beta-function term is regular at a generic $s$-channel pole. The residue of the full symmetric amplitude is therefore proportional toThis is odd-level pole cancellation in the symmetric Veneziano amplitude. The fully symmetrized identical-tachyon amplitude has only even-level poles at generic angle, including the tachyon but not the massless vector. This is an external-state selection rule, not removal of odd-level states from the string theory. Ordered amplitudes, or suitable amplitudes with color factors and different external states, retain those exchanges. Thus interpreting every candidate Gamma function pole as a pole of the printed sum would give a false inference about that amplitude.

At high energy with $x=-t/s\in (0,1)$ fixed, the Gamma function reflection formula followed by the Stirling formula gives the fixed-angle softness of an open-string amplitude:up to oscillatory factors and possible additional cancellations. The bracket is positive at a genuinely nonforward fixed angle. This is the envelope away from resonance poles, or the corresponding analytically continued asymptotic regime; it is not a uniform bound through real-axis poles. String scattering is exponentially soft at fixed angle, unlike the typical power-law behavior of point-particle interactions. The finite string length distributes a hard collision over an extended object. This is evidence of improved ultraviolet behavior, not by itself a proof that every loop amplitude is finite. The fixed-angle limit also differs from the fixed-$t$ Regge limit.