= Solution
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 is
$$
I_X=\frac1{4\pi\alpha'}\int d\sigma d\tau\,\big(\dot X^2-X'^2\big)
=\frac1{\pi\alpha'}\int d\sigma d\tau\,\partial_+X\cdot\partial_-X.
$$
Thus 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 as
$$
\delta\gamma_{ab}=(2\omega+\nabla\cdot v)\gamma_{ab}+(P_1v)_{ab},
\qquad (P_1v)_{ab}=\nabla_av_b+\nabla_bv_a-\gamma_{ab}\nabla_cv^c.
$$
Integration 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 $+(2\pi)^{-1}\int c^a\partial^b b_{ab}=iI_{\rm gh}$ with
$$
\boxed{I_{\rm gh}=-\frac{i}{2\pi}\int d\sigma d\tau\,c^a\partial^b b_{ab}.}
$$
Overall 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 directly
$$
c^a\partial^b b_{ab}=-2c^+\partial_-b_{++}-2c^-\partial_+b_{--}.
$$
Therefore
$$
\boxed{I_{\rm gh}=\frac{i}{\pi}\int d\sigma d\tau\,\big(c^+\partial_-b_{++}+c^-\partial_+b_{--}\big).}
$$
For 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 are
$$
\begin{aligned}
sX&=c^+\partial_+X+c^-\partial_-X,& sc^\pm&=c^\pm\partial_\pm c^\pm,\\
sb_{++}&=\frac{i}{\alpha'}(\partial_+X)^2+c^+\partial_+b_{++}+2(\partial_+c^+)b_{++},\\
sb_{--}&=\frac{i}{\alpha'}(\partial_-X)^2+c^-\partial_-b_{--}+2(\partial_-c^-)b_{--}.
\end{aligned}
$$
The last two expressions are $sb_{\pm\pm}=i\theta_{\pm\pm}/\alpha'$ 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 $\mathcal L=u\cdot v/\alpha'+i(c\partial_-b+f\partial_+h)$. Its matter variation is
$$
s\mathcal L_X=\frac1{\alpha'}\left[\partial_+(c\,u\cdot v)+\partial_-(f\,u\cdot v)+(\partial_-c)u^2+(\partial_+f)v^2\right].
$$
The matter terms in $sb$ and $sh$ give the ghost-action contribution $\alpha'^{-1}[c\partial_-(u^2)+f\partial_+(v^2)]$. Combining these terms with the last two matter terms produces $\alpha'^{-1}[\partial_-(cu^2)+\partial_+(fv^2)]$.
The remaining contribution from the $c,b$ sector is
$$
\begin{aligned}
i\left[(cc_+)b_- -c\partial_-(cb_++2c_+b)\right]
&=-i\left(cc_+b_-+cc_-b_++2cc_{+-}b\right)\\
&=-i\partial_-(cc_+b)-i\partial_+(cc_-b),
\end{aligned}
$$
where 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 $+\leftrightarrow-$. Thus
$$
\boxed{s\mathcal L=\partial_+K^++\partial_-K^-,}
$$
where
$$
\begin{aligned}
K^+&=\frac{c\,u\cdot v+fv^2}{\alpha'}-i\big(c\partial_-c\,b+f\partial_-f\,h\big),\\
K^-&=\frac{f\,u\cdot v+cu^2}{\alpha'}-i\big(c\partial_+c\,b+f\partial_+f\,h\big).
\end{aligned}
$$
Hence \b[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 $s^2c^\pm=0$ algebraically. For the embedding, however, the graded calculation gives
$$
\boxed{s^2X=-c^+\partial_+c^-\partial_-X-c^-\partial_-c^+\partial_+X.}
$$
Mixed 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 $s^2X=-\xi\rho\ne0$.
For the antighosts, use the permitted stress-tensor identity $s\theta_{ab}=0$ on the classical solution space. Then $s^2b_{\pm\pm}=(i/\alpha')s\theta_{\pm\pm}=0$. Indeed direct variation of the displayed stress tensors gives
$$
\begin{aligned}
s\theta_{++}&=2(\partial_+c^-)\partial_+X\cdot\partial_-X+2c^-\partial_+X\cdot\partial_+\partial_-X,\\
s\theta_{--}&=2(\partial_-c^+)\partial_+X\cdot\partial_-X+2c^+\partial_-X\cdot\partial_+\partial_-X,
\end{aligned}
$$
which vanish using the ghost equations and the free matter wave equation. Therefore \b[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.
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