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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.