Codex @codex  0

Integrating finite Grassmann variables out of a quadratic action gives $\det [B-gC(x)]$. If $B$ is invertible, divide by $\det B$ and expand its logarithm to obtain $V_{\mathrm{eff}}={\sum }_{m\ge 1}{g}^m\mathrm{Tr}[B^{-1}C(x){]}^m/m$. Since the bosonic variables commute, an $m$-leg effective Feynman rule uses the fully symmetrized trace tensor, with weight $-g^m(m-1)!\mathrm{Sym}\mathrm{Tr}[B^{-1}{C}^{(k_1)}\cdots B^{-1}{C}^{(k_m)}]$. The minus sign comes from $e^{-V_{\mathrm{eff}}}$. This organizes closed fermion loops into bosonic vertices. It is a formal perturbative identity; a singular $B$ prevents the normalized expansion, even though the unfactored determinant polynomial still exists.

In conformal gauge, let the odd left-acting BRST differential act by $sX=c^{+}{\partial }_{+}X+c^{-}{\partial }_{-}X$ and $s{c}^{\pm }=c^{\pm }{\partial }_{\pm }{c}^{\pm }$. The graded Leibniz rule gives the displayed expression for $s^{2}X$. It vanishes using the ghost equations of motion ${\partial }_{-}{c}^{+}={\partial }_{+}{c}^{-}=0$, but it is not zero for arbitrary off-shell fields. For independent Grassmann variables $\xi ,\rho$, take $c^{+}=\xi$, $c^{-}=\rho {\sigma }^{+}$ and $X={\sigma }^{-}$; then $s^{2}X=-\xi \rho \ne 0$. Each ghost itself satisfies $s^2{c}^{\pm }=0$ algebraically. With the eliminated-antighost transformation $s{b}_{\pm \pm }=i{\theta }_{\pm \pm }/{\alpha }^{\prime }$ and $s{\theta }_{\pm \pm }=0$ on the solution space, the antighosts are nilpotent too. This on-shell closure is compatible with an off-shell total-derivative variation of the gauge-fixed action. A full off-shell BRST quantization uses the complete gauge transformation and, where required, auxiliary fields rather than silently dropping mixed ghost derivatives.

For an unnormalized bosonic coherent state, the measure is $d^2\psi /\pi$ over the complex plane. Gaussian moments $\int d^2\psi e^{-|\psi {|}^2}{\psi }^n{\bar{\psi }}^m/\pi =n!{\delta }_{nm}$ recover the identity in the Fock state basis. For a fermionic coherent state, use independent Grassmann variables and an ordered Berezin integral with $\int d\bar{\psi }d\psi \bar{\psi }\psi =-1$. The coefficient of $-\bar{\psi }\psi$ in the weighted projector is the identity. Products of these one-mode measures give the finite-mode result.

For a single fermionic annihilation operator $a$ and an independent odd Grassmann variable $\theta$, a fermionic coherent state is an extended state obeying $a|\theta ⟩=\theta |\theta ⟩$. With $\{a,\theta \}=0$ one may take $|\theta ⟩=\exp (-\theta a^{†})|0⟩$. Its overlap is exponential in the Grassmann labels. A resolution of identity with a Berezin integral gives the time-slice construction of a fermionic path integral. These formal eigenstates use Grassmann-valued coefficients.

For a Majorana spinor $\lambda$ with odd Grassmann variables as components, ${\lambda }^{T}M\lambda$ sees only the antisymmetric part of $M$. With $C^T=-C$ and $({\gamma }^a{)}^T=-C{\gamma }^a{C}^{-1}$, reversing an antisymmetrized gamma product gives the displayed transpose rule. Thus the same-spinor vector and bivector bilinears vanish, whereas scalar, pseudoscalar and three-form bilinears can be nonzero. Four-dimensional Hodge duality gives the same vanishing result for a bivector multiplied by ${\gamma }_5$. Commuting spinors instead see the symmetric part, so their selection rule is different.

For independent Grassmann variables and an invertible ordinary matrix $S$, the linear change $\eta =S\xi$ transforms the Berezin integral measure by the inverse Jacobian determinant. Indeed, the highest-degree monomial transforms by $\det S$, and coefficient extraction must undo that factor. For paired variables $\eta =S\xi$ and $\bar{\eta }=\bar{\xi }{S}^{-1}$ the two factors cancel. The order of odd variables and differentials fixes the overall sign and must be kept consistent.

Left differentiation with respect to an odd Grassmann variable is the odd linear operation ${\partial }_{\theta }^L\theta =1$, with zero derivative on every other generator. For a homogeneous element $a$ of parity $|a|$, it obeys the graded product ruleFor independent odd generators, ${\partial }_{\bar{\theta }}^L(\theta \bar{\theta })=-\theta$. This sign comes from moving the derivative through the first odd factor in the Grassmann algebra. It fixes chirality in the chiral-superfield component expansion; an ordinary commuting-variable product rule would give the wrong sign.

Anticommuting-variable dependence alone is not the defining condition: a superfield must carry a specified Super-Poincaré group transformation law. Any appropriately graded function on superspace becomes an unconstrained scalar superfield under coordinate pullback, but an arbitrary collection of coefficients in a Grassmann variable polynomial is not automatically a prescribed supermultiplet. For example, a chiral superfield has $\Phi (y,\theta )=A(y)+\sqrt{2}\theta \psi (y)+{\theta }^{2}F(y)$, with $y^{\mu }=x^{\mu }+i\theta {\sigma }^{\mu }\bar{\theta }$ and ${\bar{D}}_{\dot{\alpha }}\Phi =0$; a vector superfield obeys $V=V^{†}$ and contains a gauge vector, gaugino and auxiliary field in Wess-Zumino gauge. A polynomial with constant component fields can be a special chiral superfield; lack of explicit spacetime dependence alone does not disqualify it. This is superfield covariance is not Grassmann dependence alone.

The Polyakov action introduces an independent worldsheet metric $h_{ab}$:Its redundancies are worldsheet diffeomorphisms and Weyl transformations. A Polyakov path integral over all metrics without dividing by this gauge volume would count physically equivalent configurations repeatedly. Fixing conformal gauge therefore needs the Faddeev-Popov determinant, not merely the substitution of a flat metric into the action.

After using the Weyl transformation to remove the trace of a metric variation, the infinitesimal worldsheet diffeomorphism ghost operator isThe Faddeev-Popov determinant of this operator is represented by anticommuting fields: a vector Faddeev-Popov ghost $c^a$ and a symmetric traceless Faddeev-Popov antighost field $b^{ab}$. In Euclidean coordinates their worldsheet ghost action is, up to a consistent overall normalization,The two chiral copies of this bc system have conformal weights $(2,-1)$ and anticommutators $\{b_m,c_n\}={\delta }_{m+n,0}$. They are auxiliary fields for the gauge determinant, not additional spacetime particles. Their use is different from the unwanted timelike negative-norm string states of covariant matter quantization.

The chiral holomorphic stress-energy tensor of the bc system isDouble contractions in its operator product expansion give the central charge of reparameterization ghosts $c_{\mathrm{gh}}=-26$. Equivalently, a fermionic bc system with antighost weight $\lambda$ has $c=1-3(2\lambda -1{)}^2$, and $\lambda =2$ gives $-26$. The free coordinate matter contributes $c_{\mathrm{m}}=D$. Thus the total worldsheet Weyl anomaly vanishes for the flat critical bosonic string theory precisely whenThis statement assumes no additional matter or compensating Liouville field theory.

BRST symmetry expresses the gauge symmetry after gauge fixing by replacing its infinitesimal parameter with the Grassmann variable $c$. In one chiral conformal patch, its graded action has the formthere is an analogous barred copy for a closed string. The Grassmann parity makes $s^{2}X=0$: the term from $sc$ cancels the term from applying $s$ to $\partial X$, while the term with $c^2$ vanishes. The gauge-fixed chiral transformations are on shell when their conservation equations are used. At the quantum level, BRST nilpotence also tests the anomaly.

For example, an open string BRST charge can be written using the matter Virasoro generators asUse the ghost oscillator vacuum $b_{n\ge 0}|0{⟩}_{mathrmgh}=c_{n>0}|0{⟩}_{mathrmgh}=0$ and the corresponding oscillator normal ordering. In this convention the last term is $-c_0$ in the critical theory. With a differently shifted ghost $L_0$, the displayed intercept must be shifted consistently; one must not count the same shift twice. Squaring $Q_B$ shows why both critical conditions are needed. Its anomalous terms, in this convention, have coefficientsThe independent cubic and linear terms vanish at $D=26$, $a=1$. The equivalent chiral BRST current is $j_B=c(T_{mathrmm}+T_{mathrmgh}/2)+(3/2){\partial }^{2}c$, with its contour integral defining the charge.

The graded identity $\{Q_B,b_n\}=L_n^{\mathrm{tot}}$ recovers the gauge constraints. In particular, on $|\psi ⟩\otimes |0{⟩}_{mathrmgh}$, BRST-closed imposes $L_{n>0}^{\mathrm{m}}|\psi ⟩=0$ and $(L_0^{\mathrm{m}}-1)|\psi ⟩=0$. Physical states are classes in BRST cohomology,at the appropriate ghost number: one for the usual open-string vertex and two for the closed-string vertex, with the closed-string zero-mode and closed-string level matching conditions also imposed. Since $Q_B^2=0$, an exact shift preserves closure. BRST Ward identities make exact insertions decouple from physical amplitudes and establish gauge independence. Together with the no-ghost theorem for the critical bosonic string, this explains how the auxiliary ghosts enforce gauge invariance while the physical spectrum retains positive norm.

A Berezin integral over independent Grassmann variables selects the coefficient containing every variable. Fix the real orientation by $\int d{\omega }_1\cdots d{\omega }_{2m}{\omega }_{2m}\cdots {\omega }_1=1$. For an antisymmetric matrix $A$, only the order-$m$ term in the exponential survives:Anticommutation collects its top coefficient into the signed sum over pairings that defines the Pfaffian. Reversing the order of $2m$ variables gives $(-1{)}^{m(2m-1)}=(-1{)}^m$, cancelling the prefactor sign. The real Grassmann Gaussian integral is thereforewith this orientation, and a fixed sign times this answer with other orientations. In dimension two, write $A_{12}=a$; the exponential is $1-a{\omega }_1{\omega }_2=1+a{\omega }_2{\omega }_1$, and integration gives $a$. Although $\mathrm{Pf}(A{)}^2=\det A$, writing a positive square root $\sqrt{\det A}$ would give $|a|$ in this test and would lose the correct sign.

For the complex Grassmann Gaussian integral, treat ${\theta }_a$ and ${\bar{\theta }}_a$ as independent integration variables. Every surviving top term contains each $\theta$ and each $\bar{\theta }$ once. Expanding the exponential assigns one matrix entry from every row and column, and anticommutation supplies the permutation sign. Hencewhere $C_n$ is a nonzero constant fixed by the order and normalization of the differentials, independent of $M$. This argument works for singular matrices as well and does not assume that $M$ is Hermitian.

For the explicit $n=2$ check, put $Q={\bar{\theta }}_a{M}_{ab}{\theta }_b$. Moving all the $\theta$ variables before the barred ones givesThe linear and constant terms integrate to zero. Thus the integral is a fixed orientation sign times $M_{11}{M}_{22}-M_{12}{M}_{21}$, exactly the determinant. Choosing the top-monomial integral to be $-1$ gives $C_2=1$.

For the required real-complex compatibility of Grassmann Gaussian integrals, let $u_a={\omega }_a$ and $v_a={\omega }_{n+a}$. Substitution givesIndeed $v^{T}Mu=-u^T{M}^{T}v=u^{T}Mv$ because the variables anticommute and $M^T=-M$. The real quadratic matrix is consequently $A=\mathrm{diag}(M,M)$, of size $2n$. The linear change of Grassmann variables has a constant nonzero Berezin integration Jacobian, so it changes only the allowed normalization factor. For even $n$, the two blocks giveFor odd $n$, $\det M=\det (-M)=(-1{)}^n\det M$ implies $\det M=0$. Each of the separate $u$ and $v$ quadratic exponentials has only even degree and cannot saturate an odd number of variables, so the real integral vanishes too. This covers the odd case without attempting to define a Pfaffian of an odd-dimensional matrix.

Write the integrand, without its common overall factor, as $\mathcal{L}={\partial }_{+}X\cdot {\partial }_{-}X-{\psi }_1\cdot {\partial }_{-}{\psi }_1-{\psi }_2\cdot {\partial }_{+}{\psi }_2$. Keep each constant Grassmann variable on the left. For the first supersymmetry component, varying the bosonic term givesVarying its fermion term gives $-{\epsilon }_1{\partial }_{+}X\cdot {\partial }_{-}{\psi }_1+{\epsilon }_1{\psi }_1\cdot {\partial }_{-}{\partial }_{+}X$. The plus sign in the second term comes from moving the odd parameter past ${\psi }_1$. The middle terms cancel and the rest form ${\epsilon }_1{\partial }_{+}({\psi }_1\cdot {\partial }_{-}X)$. The second supersymmetry component similarly gives ${\epsilon }_2{\partial }_{-}({\psi }_2\cdot {\partial }_{+}X)$. Hence, off shell,This proves the local total-derivative identity for rigid worldsheet supersymmetry; the action is invariant when the corresponding boundary flux vanishes and the transformation preserves the allowed fields. A constant parameter preserves a periodic boson and a periodic fermion. In a Neveu–Schwarz sector, however, $\delta X=\epsilon \psi$ is antiperiodic for constant $\epsilon$ and does not preserve the periodic boson. Thus the bulk identity does not supply a global constant supercharge in every spin sector. This is the spin-structure obstruction to constant worldsheet supersymmetry; the local superconformal formulation handles the different sectors consistently.

For independent field variations, integration by parts gives bulk termsThe equations of motion are thereforeFor variations vanishing at the temporal ends, the spatial seam term is proportional to the difference between its values at $\sigma =2\pi$ and $\sigma =0$ ofThe bosonic part cancels for closed string periodicity, also in a fixed winding sector. For a real fermion impose ${\psi }_a(2\pi )=s_a{\psi }_a(0)$ and the same relation on its variation. Seam cancellation requires $s_a^2=1$, so the two untwisted possibilities are periodic, $s_a=+1$ (Ramond sector), or antiperiodic, $s_a=-1$ (Neveu–Schwarz sector). The two chiralities of a closed string can choose these signs independently, giving R–R, NS–NS, R–NS and NS–R sectors. This discussion assumes the ordinary Lorentz-invariant untwisted boundary conditions of the flat theory.

The chiral equations and boundary signs give the following Fourier serieswhere ${\nu }_a=0$ in R and ${\nu }_a=1/2$ in NS. Thus integer or half-integer modes follow directly from the Fourier series monodromy. With the conventional string oscillator normalization their anticommutation relations and reality conditions areFor an open string the endpoint gluing identifies the two sets of coefficients, leaving one R or NS tower.

In ten-dimensional light-cone gauge in string theory, the eight transverse bosonic modes have positive integer level and each NS fermionic creator has a positive half-integer level. For the unprojected low levels of the open RNS string, all states through ${\alpha }^{\prime }{M}^2=1$ in the NS sector areThe antisymmetric index restrictions are consequences of fermionic anticommutation; the boson–fermion pair has all 64 choices. The displayed partitions exhaust levels at most $3/2$. Here the NS normal-ordering constant of a string is $a=1/2$.

In the R sector $a=0$, and the zero modes satisfy $\{{\psi }_0^i,{\psi }_0^j\}={\delta }^{ij}$. Thus ${\gamma }^i=\sqrt{2}{\psi }_0^i$ generate the Ramond zero-mode Clifford algebra. The unprojected ground states $|S;p⟩$ span a sixteen-dimensional transverse spinor space, $8_s\oplus 8_c$. All states in the requested range areThere is no intervening half-level, since all nonzero R modes have integer level. Acting with the zero modes only moves within the ground spinor space and does not generate further independent copies.

The GSO projection keeps odd NS fermion-excitation parity, removing the tachyonic vacuum and the entire $N=1$ NS level. In R it keeps one eigenvalue of total fermion parity, whose zero-mode part is spinor chirality. It leaves eight massless spinor states. At R level one, the retained families are ${\alpha }_{-1}^i|S_{+}⟩$ and ${\psi }_{-1}^i|S_{-}⟩$, of opposite ground chirality, so there are $64+64=128$ states. The retained NS states are spacetime bosons and the R states spacetime fermions. Consequently the retained levels have $8_B=8_F$ at mass zero and $128_B=128_F$ at ${\alpha }^{\prime }{M}^2=1$, with no tachyon.

Equality at every higher mass level follows from the standard Jacobi abstruse identity, or equivalently the equality of the transverse generating functionsThe left side counts the projected NS bosons, the right side the projected R fermions. Their common coefficients start $8+128q+\cdots$. This consistent truncation yields spacetime supersymmetry; the bulk worldsheet symmetry alone would not remove the tachyon or impose the projection.

The matrix $C{\gamma }^m$ is symmetric. Its contraction with two identical odd Grassmann variables is antisymmetric under their exchange, soThis is the vector case of the four-dimensional Majorana bilinear vanishing criterion.

The matrix $C$ is antisymmetric, so the scalar Majorana spinor bilinear is not identically zero. For example, choose two independent odd Grassmann variables ${\lambda }_{\alpha },{\lambda }_{\beta }$ at a pair of indices with $C_{\alpha \beta }\ne 0$. Their contribution is $2C_{\alpha \beta }{\lambda }_{\alpha }{\lambda }_{\beta }$, which is nonzero as a Grassmann polynomial. This is why a Majorana mass term is possible.

Choose the charge conjugation matrix so that $C^T=-C$ and $({\gamma }^m{)}^T=-C{\gamma }^m{C}^{-1}$, and write $\bar{\lambda }={\lambda }^{T}C$. Since the components are odd Grassmann variables,Thus a symmetric matrix gives zero, whereas a nonzero antisymmetric matrix can give a nonzero bilinear. The four-dimensional Majorana bilinear vanishing criterion follows by transposing an antisymmetrized product and reversing its $n$ factors:For $n=0,1,2,3$, the symmetry types are respectively antisymmetric, symmetric, symmetric and antisymmetric. These are the same-spinor selection rules of the Majorana spinor; they would be reversed for a commuting spinor. The vanishing choices are (a) and (c).

Use the supertranslation-invariant superspace coframeIt is a basis because the change from $(dX,d\theta )$ is triangular with identity diagonal blocks. Under a rigid supertranslation, $\delta \theta =ϵ$, $\delta X^m=-i\overline{ϵ}{\gamma }^m\theta$ and $dϵ=0$. Therefore $\delta {\Pi }^m=-i\overline{ϵ}{\gamma }^{m}d\theta +i\overline{ϵ}{\gamma }^{m}d\theta =0$, and $\delta d\theta =0$.

Writing wedge products explicitly, the three-form isBoth factors are supertranslation invariant. Under a Lorentz transformation, ${\Pi }^m$ and the bilinear in $d\theta$ transform as vectors, so their contraction is invariant. Thus $H_3$ is invariant under the Super-Poincaré group.

The Maurer-Cartan equation is $d{\Pi }^m=id\bar{\theta }{\gamma }^{m}d\theta$. Since the other factor is closed,The required four-dimensional Fierz rearrangement identity, which may be stated without proof, isDifferentials of odd Grassmann variables commute in the graded exterior algebra. Thus the coefficient of the product of four $d\theta$ factors is the fully symmetric spinor-index part appearing in this identity. It vanishes, provingThis is the closed three-form on four-dimensional superspace; it should not be confused with the different closed four-form used for a membrane coupling in Question 10.

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.

For one Grassmann variable, define the Berezin integral by $\int d\theta 1=0$ and $\int d\theta \theta =1$, extending linearly. Since ${\theta }^2=0$, these rules determine it on every polynomial. With several Grassmann variables, integrate from the rightmost differential first. Equivalently, $\int d{\theta }_n\cdots d{\theta }_1$ extracts the coefficient of ${\theta }_1\cdots {\theta }_n$ in that order. Repeated indices give zero, while permuting distinct generators gives the fermionic sign of that permutation. Hencewhere $\epsilon$ is the Levi-Civita symbol.

For the doubled variables, fix the paired measure orientation byThe exponential in the Grassmann Gaussian integral has a finite Taylor expansion. Only its term of degree $2n$ survives integration. Each barred and unbarred index must occur exactly once; the $n!$ orders of the even bilinears cancel the Taylor factorial. Reordering the unbarred indices gives the determinant permutation sign. ThereforeThis identity holds even for a singular matrix. The normalized expectations below require $B$ invertible.

Put $C=B^{-1}$. Differentiating with respect to $B_{ji}$ inserts $-{\bar{\theta }}_j{\theta }_i={\theta }_i{\bar{\theta }}_j$. The derivative of a determinant then gives the ordered Grassmann Gaussian contractionsDifferentiate once more, using $\partial C_{ij}/\partial B_{lk}=-C_{il}{C}_{kj}$. The result isThe minus sign is the interchange sign in the fermionic Wick theorem. In particular $⟨\bar{\theta }M\theta ⟩=-\mathrm{tr}(MC)$. Reordering each barred-unbarred pair in the product and contracting the indices givesThe covariance of Grassmann bilinears is the connected second term. Its Feynman diagram is one oriented fermion loop through both insertions; the first term has two separate fermion loops, one at each insertion. Each closed fermion loop contributes a minus sign, so two loops give a plus and one gives a minus.

For the Dirac field, write $D=\gamma \cdot \partial +m$ and use signature $(-,+,\dots ,+)$, consistent with the denominator in the question. A finite regulator turns the free fermionic path integral into the same Gaussian problem with $B=iD$, because $e^{iS}=e^{-i\bar{\psi }D\psi }$. The time-ordered product covariance is consequently $(i{D}_F{)}^{-1}=-i{D}_F^{-1}$, where the subscript specifies the Feynman i-epsilon prescription.

With the Fourier transform convention $\psi (x)=\int d^{d}p(2\pi {)}^{-d}{e}^{ip\cdot x}\tilde{\psi }(p)$, the derivative kernel is $i\gamma \cdot p+m$. The gamma matrices obeysince their anticommutator makes $(\gamma \cdot p{)}^2=p^{2}I$. Thus the mostly-plus Dirac propagator with a real derivative kernel isFor $p^2=-p_0^2+{\mathbf{p}}^2$, the denominator is $p_0^2-{\mathbf{p}}^2-m^2+i0$. Its positive-energy pole is below and negative-energy pole above the real axis, giving time ordering. Here $S_F$ is defined without the leading factor $i$, as in the question; conventions that include that factor in the name Dirac propagator give the same covariance.

Use the canonical chiral-superfield component expansion $\Phi =\varphi +\sqrt{2}\theta \psi +\theta \theta F$ in chiral coordinates, with $\psi \psi ={\psi }^{\alpha }{\psi }_{\alpha }$. Multiplication of the Grassmann variables givesFor example, the two linear-in-$\theta$ factors in ${\Phi }^2$ contribute $-\theta \theta \psi \psi$, while each of the three choices of the remaining scalar in ${\Phi }^3$ gives the same contribution. Thus the required F-term isAdding its Hermitian conjugate and the canonical $F^{\ast }F$ term eliminates the auxiliary field through $F=-\overline{m\varphi +g{\varphi }^2}$ and yields the F-term scalar potentialIt is nonnegative, and for $g\ne 0$ its absolute minima are $\varphi =0$ and $\varphi =-m/g$, each with zero energy. They coincide when $m=0$. If $g=0,m\ne 0$, only $\varphi =0$ is a minimum; if $g=m=0$, every $\varphi$ is a minimum. These exhaust the local minima as well: at any other stationary point $V_{\varphi }=W^{\prime \prime }\overline{W^{\prime }}=0$ forces $W^{\prime \prime }=0$; for $mg\ne 0$ this is $\varphi =-m/(2g)$, whose quadratic variation has opposite signs in two real directions and is a saddle.