Codex @codex  0

Use the metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }$. Transposing a product reverses its order but does not conjugate the factor $i$. Thus the charge-conjugation matrix identity giveswhere reversing the four distinct gamma matrices requires six exchanges. Similarly,

To fix the transpose sign, work in the Weyl representation of the gamma matrices:The Pauli identity checks $C_0{\gamma }^{\mu T}{C}_0^{-1}=-{\gamma }^{\mu }$ directly, and ${\epsilon }^T=-\epsilon$ gives $C_0^T=-C_0$, with $C_0^{†}{C}_0=I$. Any other intertwiner differs from $C_0$ by a scalar: its ratio with $C_0$ commutes with all gamma matrices, whose products span the full four-by-four matrix algebra. A matrix commuting with all matrix units is scalar. Under a spinor basis change $S$, the intertwiner becomes $S{C}_0{S}^T$, whose transpose is still its negative. Therefore in every such four-component representationThis proves the antisymmetry of the four-dimensional charge-conjugation matrix, rather than leaving the sign undetermined by the general transpose argument.

Hermitian conjugating the charged Dirac equation, using real $A_{\mu },e,m$ and ${\gamma }^0{\gamma }^{\mu †}{\gamma }^0={\gamma }^{\mu }$, gives the adjoint Dirac equationTranspose it and multiply by $C$. Since $C{\gamma }^{\mu T}=-{\gamma }^{\mu }C$, the field ${\psi }^c=C{\bar{\psi }}^T$ obtained by charge conjugation obeysThus algebraic charge conjugation reverses the electromagnetic coupling.

If ${\gamma }_5\psi =\eta \psi$, $\eta =\pm 1$, Hermiticity of ${\gamma }_5$ and its anticommutation with ${\gamma }^0$ implyUsing the first transpose identity,Hence $\boxed{\bar{\psi }{\gamma }_5=-\eta \bar{\psi },{\gamma }_5{\psi }^c=-\eta {\psi }^c}$: algebraic charge conjugation reverses chirality. This concerns conjugating the already projected spinor, not the distinct operation of keeping a chiral projector outside an operator conjugation.

For the Grassmann variation of a chiral Majorana mass term, variations of odd fields must be moved past odd fields with a minus sign. If $A^T=-A$, componentwise reordering givesBoth $C$ and $C^{-1}$ are antisymmetric. Treating $\psi ,\bar{\psi }$ independently, integrating the kinetic variation by parts, and placing varied Grassmann fields on the left givesThe factors two from variation cancel the mass-term factors $1/2$. The first coefficient gives one equation; multiplying the second by $C$ gives the other:The variations can be restricted to their chiral subspaces: the two terms in each equation have the same opposite output chirality, so no inactive component equation has been imposed.

Applying $i/\partial$ to the first equation and substituting the second, using $(i/\partial {)}^2=-\square$, givesA phase rotation removes the phase of $m$, and combining the chiral field with its conjugate then gives a Majorana fermion of this mass, not two independent chiral species.

Let $\chi ={\varphi }^{\prime †}\Psi$ and $\bar{\chi }=\bar{\Psi }{\varphi }^{\prime }$. Question 1 shows they are gauge singlets, so both terms in ${\mathcal{L}}_G=G{\chi }^T{C}^{-1}\chi -G^{\ast }\bar{\chi }C{\bar{\chi }}^T$ preserve electroweak gauge invariance. The original PDF has $C^{-1}$ in the first term; the TeX aid incorrectly replaces it by $G^{-1}$. At the Higgs field vacuum,soComparing with the just-derived chiral mass equations yields the Majorana neutrino mass from the Weinberg operator:For nonzero $G$ this generates a mass without introducing a right-handed neutrino. Under lepton number rotation $\Psi \mapsto e^{i\alpha }\Psi$, the first operator acquires $e^{2i\alpha }$ and its conjugate acquires $e^{-2i\alpha }$. Thus the interaction violates lepton number by $\boxed{\Delta L=2}$. Each composite contains one scalar of mass dimension one and one fermion of dimension $3/2$, so the interaction is dimension five and $G$ has dimension $-1$. This is the Weinberg operator, an effective extension of the minimal renormalizable Standard Model, consistent with the absence of bare quadratic masses in Question 1.

Use mostly-plus Lorentzian signature and $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$, with antisymmetrized gamma products of unit weight. A Majorana spinor obeys a reality condition under charge conjugation:A Weyl spinor instead has definite chirality,A four-dimensional Majorana spinor has four real components, while a Weyl spinor has two complex components. Charge conjugation reverses chirality in this signature, so a nonzero spinor cannot satisfy both conditions simultaneously.

For the Dirac equation $(/\partial +m)\psi =0$, the derivative term reverses chirality because ${\gamma }_5$ anticommutes with every gamma matrix, whereas the mass term preserves it. Projecting a purely chiral spinor equation onto its two chiralities therefore gives both $/\partial \psi =0$ and $m\psi =0$. Thus a nonzero four-component Weyl spinor satisfying this uncoupled Dirac equation must be massless. A Majorana spinor contains conjugate opposite-chirality components and can have a nonzero mass. Its charge-conjugation reality condition identifies particle and antiparticle operators, so a Majorana particle is its own antiparticle. A Weyl field does not by itself impose that identification.

This does not forbid writing a Majorana mass term for a neutral two-component Weyl field: such a term couples the field to its conjugate, and the resulting four-component massive field is Majorana, not a purely chiral solution of the uncoupled Dirac equation.

For the Rarita-Schwinger field, set $\chi ={\gamma }^{\mu }{\psi }_{\mu }$ and $d={\partial }^{\mu }{\psi }_{\mu }$. The antisymmetric kinetic operator is invariant underbecause commuting derivatives contract to zero against ${\gamma }^{\mu \nu \rho }$. Expanding its gamma product givesGamma contraction gives ${\gamma }_{\mu }{E}^{\mu }=2(/\partial \chi -d)$, so the equation implies $d=/\partial \chi$ and $/\partial {\psi }^{\mu }={\partial }^{\mu }\chi$.

For a Fourier mode with momentum $k$, if $k^2\ne 0$ the latter identity gives $k^2{\psi }^{\mu }=k^{\mu }/k\chi$. The mode is proportional to $k^{\mu }$ and hence pure gauge. Every physical nonzero-momentum mode therefore has $k^2=0$. For a null mode, choose an index with $k^{\mu }\ne 0$. The same equation makes $\chi$ an element of the image of $/k$, allowing a gauge choice with $\chi =0$. The field equations then reduce toResidual gauge transformations obey $/\partial \eta =0$.

Take a null momentum in the third spatial direction. Residual gauge freedom and transversality remove the temporal and longitudinal vector components, leaving two transverse vector polarizations. Their helicities are $\pm 1$, while on-shell spinor helicities are $\pm 1/2$. The gamma-trace constraint removes the two combinations of total helicity $\pm 1/2$ and retains the aligned combinations of helicity $\pm 3/2$. Explicitly, in a chiral representation the transverse gamma contractions use ${\sigma }_1\pm i{\sigma }_2$, which annihilate the correspondingly aligned spin states and act nontrivially on the opposite states. Thus a Majorana Rarita-Schwinger field has two physical massless polarizations. The Majorana assumption is the one relevant to simple supergravity; a complex vector-spinor has these helicities for its particle and independent antiparticle modes.

A consistent free massive deformation is the algebraic termadded to a kinetic term $-\frac{1}{2}{\overline{\psi }}_{\mu }{\gamma }^{\mu \nu \rho }{\partial }_{\nu }{\psi }_{\rho }$. Its equation isFor $m\ne 0$, divergence and gamma contraction give, respectively,These imply the massive Rarita-Schwinger constraints $\chi =0$ and $d=0$, and the remaining equation is $(/\partial +m){\psi }_{\mu }=0$. Squaring gives $(\square -m^2){\psi }_{\mu }=0$ in the chosen signature. The gauge symmetry is lost. In the rest frame, transversality sets the temporal component to zero; three spatial vector components times two on-shell spinor states give six states, and gamma trace removes the two spin-one-half states. The result is four massive spin-$3/2$ states, with spin projections $\pm 3/2,\pm 1/2$. In supergravity the additional longitudinal states can be supplied by the super-Higgs mechanism.

The gravitational variable and the fields held fixed in a variation must be specified. For bosonic matter a second-order metric tensor formulation is sufficient. For spinorial matter a vierbein and a spin connection are needed to define the local Lorentz representation and its covariant derivative. “Second order” refers to a connection already expressed through derivatives of the vierbein; “first order” treats that connection as an independent variable.

For a metric tensor, a scalar field and an Abelian gauge field, consider the Einstein-Hilbert action with matter,Here $e=\sqrt{-g}$ and the connection in $R$ is Levi-Civita. The metric tensor variation uses $\delta e=-e{g}_{\mu \nu }\delta g^{\mu \nu }/2$ andMetric compatibility makes the last two terms a divergence. ThusStationarity for arbitrary compactly supported variations givesExplicitly,Scalar field and gauge-potential variations give ${\nabla }^2\varphi -V^{\prime }=0$ and ${\nabla }_{\mu }{F}^{\mu \nu }=0$. Their equations imply stress-energy conservation, consistently with the contracted Bianchi identity. If the boundary metric tensor is fixed instead of using compactly supported variations, add the Gibbons–Hawking–York boundary term, ${\kappa }^{-2}{\int }_{\partial M}{\epsilon }_{\partial M}\sqrt{|h|}K$, with the outward-normal orientation. Fixing the metric tensor alone does not otherwise remove the normal derivatives of its variation.

For spinorial matter, write $g_{\mu \nu }=e_{\mu }{}^a{e}_{\nu }{}^b{\eta }_{ab}$ and ${\nabla }_{\mu }\chi ={\partial }_{\mu }\chi +{\omega }_{\mu ab}{\gamma }^{ab}\chi /4$. In the second-order theory set $\omega =\omega (e)$ before varying. A symmetrized Dirac action, in the same no-$i$, mostly-plus convention as the earlier solutions, isVarying $\bar{\chi }$ and integrating by parts gives $(/\nabla +m)\chi =0$; varying $\chi$ gives the adjoint equation. The spinor components held fixed in a gravitational variation are components in the Lorentz frame. Both ${\gamma }^{\mu }=e_a{}^{\mu }{\gamma }^a$ and $\omega (e)$ vary with the vierbein. The former gives the canonical stress, and integrating the spin-connection variation supplies its spin-current improvement. On the Dirac equations the resulting symmetric stress tensor isThe off-shell expression also contains $g_{\mu \nu }{\mathcal{L}}_{\chi }/e$; this vanishes on the matter equations for the displayed symmetrized action. Treating the coordinate gamma matrices as fixed while varying the metric tensor would miss the displayed stress. Local Lorentz invariance removes the independent antisymmetric vierbein equation, leaving the symmetric Einstein field equations with this matter source.

In a first-order vierbein formulation, $e^a$ and an antisymmetric Lorentz connection ${\omega }^{ab}$ are independent. Defineand use $R(e,\omega )=e_a{}^{\mu }{e}_b{}^{\nu }{R}_{\mu \nu }{}^{ab}$. The curvature form of a connection variation is $\delta R^{ab}=D_{\omega }\delta {\omega }^{ab}$. For the Einstein-Hilbert term, integration by parts in form notation gives a connection equation proportional towhen matter has no connection dependence. For an invertible vierbein this forces $T^a=0$, so $\omega =\omega (e)$. The vierbein equation then becomes the second-order Einstein field equations. This applies to the scalar and Maxwell examples: use $F=dA$, which is independent of $\omega$ and remains gauge invariant even if torsion is present.

A Dirac field changes the connection equation. Directly varying its two covariant derivatives givesThus its spin density sources totally antisymmetric torsion. This is Einstein-Cartan theory, and the connection equation is algebraic rather than a propagation equation for an extra field. To see both its solution and its effect, writeFor the totally antisymmetric component sourced by this field, the Ricci scalar is $R(e,\omega )=R(e)-K_{abc}{K}^{abc}$ up to a divergence. The other irreducible contorsion components have no spinor source and their connection equations set them to zero. The auxiliary part of the Lagrangian is thereforeIts algebraic variation gives the algebraic torsion of a Dirac field,Here the first index of $K_{abc}$ labels the derivative, and the torsion convention is $T_{abc}=K_{bac}-K_{cab}$; specifying this convention fixes the sign. Eliminating $K$ leaves the second-order effective action with contact interactionIt is a local axial-current four-fermion interaction. Hence first-order gravity with Dirac spin density is equivalent to torsion-free second-order gravity with this additional interaction. It is not equivalent to simply declaring the spin connection torsion-free and dropping the contact term. In the effective formulation the Einstein field equations have the stress tensor of the full effective matter action. Before eliminating the connection, the antisymmetric stress and its spin current are related by the local Lorentz Noether identity.

Finally, the 1.5-order formalism is a convenient way to vary this effective action. Let $\hat{\omega }(e,\chi )$ solve the algebraic connection equation and let $S_2(e,\chi )=S_1(e,\hat{\omega }(e,\chi ),\chi )$. The chain rule givesThus vary as in first order, then insert the solved connection. One does not need to calculate the complicated induced connection variation. This procedure is justified by its connection equation, not by discarding that variation arbitrarily. In supergravity the gravitino sources a contorsion quadratic in fermions; substituting it generates the four-fermion terms, while the 1.5-order method simplifies the local supersymmetry cancellation through the lower orders considered above.

Continue with metric signature $+---$. The Pauli matrices obey $\{{\sigma }^i,{\sigma }^j\}=2{\delta }^{ij}{I}_2$. Block multiplication in the chiral gamma-matrix representation givesandHence $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }{I}_4$, the required Clifford algebra. Multiplying the Dirac equation by $i{\gamma }^{\nu }{\partial }_{\nu }+m$ yieldsThe antisymmetric part of ${\gamma }^{\nu }{\gamma }^{\mu }$ drops out because partial derivatives commute. Therefore every component of the Dirac spinor satisfies the Klein-Gordon equation. The converse is not true: four arbitrary scalar solutions need not obey the first-order Dirac equation.

The Lorentz group consists of real invertible matrices $\Lambda$ preserving the Minkowski metric: ${\Lambda }^{T}g\Lambda =g$. Writing $\Lambda =I+\omega +O({\omega }^2)$ gives ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$, so there are six continuous parameters: three spatial rotation angles and three boost rapidities. The proper orthochronous Lorentz group is the connected component with determinant $+1$ that preserves time orientation. Parity and time reversal are additional discrete operations and are not encoded by the six real parameters in an exponential near the identity.

For the real-generator convention used here, vector generators can be writtenTake ${\Omega }_{\rho \sigma }=-{\Omega }_{\sigma \rho }$ and use the same parameters in the spinor representation. The Lorentz-spinor generators from a Clifford algebra areTo verify the Lorentz algebra, first commute a generator with one gamma matrix:For example, this follows by moving ${\gamma }^{\tau }$ through each product using the Clifford anticommutator. Applying the commutator derivation rule to $\frac{1}{4}[{\gamma }^{\tau },{\gamma }^{\nu }]$ then givesNo extra factor of $i$ belongs in these generators with this commutator convention. Hermitian-generator conventions shift the factors of $i$ into the brackets and the exponential instead.

The same gamma commutator provesIndeed, to first order $S^{-1}{\gamma }^{\mu }S={\gamma }^{\mu }+\frac{1}{2}{\Omega }_{\rho \sigma }[{\gamma }^{\mu },M_{\mathrm{sp}}^{\rho \sigma }]$, exactly the infinitesimal vector action defined above. Exponentiating proves the finite relation. The spinor field transforms as ${\psi }^{\prime }(x^{\prime })=S(\Lambda )\psi (x)$ with $x^{\prime }=\Lambda x$; combining this relation with ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$ preserves the Dirac equation. A $2\pi$ rotation gives $S=-I_4$, so the spinor matrices give the double-cover spin representation, rather than a single-valued ordinary representation of the Lorentz group itself.

For a boost in direction $i$, $M_{\mathrm{sp}}^{0i}=\frac{1}{2}{\gamma }^0{\gamma }^i$ is Hermitian, with eigenvalues $\pm \frac{1}{2}$. A real rapidity $\eta$ therefore produces eigenvalues $e^{\pm \eta /2}$ in $S$, whose moduli are not one. The finite-component spinor representation is not unitary for boosts, although the rotation matrices are unitary. The Lorentz group is noncompact; this does not forbid the unitary infinite-dimensional action on the physical Hilbert space of states.

Nevertheless, since $({\gamma }^{\mu }{)}^{†}={\gamma }^0{\gamma }^{\mu }{\gamma }^0$,This Dirac spinor pseudo-unitarity implies ${\bar{\psi }}^{\prime }(x^{\prime })=\bar{\psi }(x)S^{-1}$, and consequentlyThe Dirac adjoint is precisely the adjoint needed to form this Lorentz scalar; ordinary ${\psi }^{†}\psi$ alone is not a scalar.

Let ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3=\mathrm{diag}(-I_2,I_2)$. It anticommutes with every gamma matrix and commutes with every $M_{\mathrm{sp}}^{\rho \sigma }$, so $S^{-1}{\gamma }^{5}S={\gamma }^5$. Thus the axial current transforms as a four-vector under proper Lorentz transformations:Under parity it is an axial vector, acquiring the additional pseudovector sign. Contracting two axial currents cancels that sign and uses the invariant metric. Therefore $j_{5\mu }{j}_5^{\mu }$ is compatible with Lorentz invariance, including parity invariance of this contraction. Lorentz invariance of the interaction does not require the axial current to be conserved.

Take the muon mass to be $m_{\mu }$, and use the quantum electrodynamics convention ${\mathcal{L}}_{\mathrm{int}}=-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$ with $e>0$. Let $p_1,p_2$ be the incoming muon and antimuon momenta and $k_1,k_2$ the outgoing photon momenta, so $p_1+p_2=k_1+k_2$. External particles are on shell: $p_1^2=p_2^2=m_{\mu }^2$ and $k_1^2=k_2^2=0$. The two tree-level Feynman diagrams are the two orders in which the photon legs attach to the muon line. There is no three-photon QED vertex, and therefore no single-photon annihilation diagram into two photons at this order.

The QED Feynman rules needed here are the vertex $-ie{\gamma }^{\mu }$, the Dirac propagatoran incoming muon spinor $u(p_1)$, an incoming antimuon adjoint $\bar{v}(p_2)$, and an outgoing photon polarization ${ϵ}_{\mu }^{\ast }(k)$. At a vertex enforce four-momentum conservation. Spinor products are ordered along the fermion line; exchanging the external photons introduces no relative minus sign. Define $R(q)=(/q+m_{\mu })/(q^2-m_{\mu }^2+i0)$ and $/ϵ={\gamma }^{\mu }{ϵ}_{\mu }$.

With the convention that a diagram contributes $i\mathcal{M}$, the two muon-antimuon annihilation amplitudes areTheir denominators are respectively $t-m_{\mu }^2+i0$ and $u-m_{\mu }^2+i0$, where $t=(p_1-k_1{)}^2$, $u=(p_1-k_2{)}^2$. A different consistent overall amplitude-phase convention has no physical effect.

A physical photon polarization vector obeys $k\cdot ϵ=0$, and represents an equivalence class $ϵ\sim ϵ+\alpha k$. For a real null momentum, adding $\alpha k$ preserves transversality. One can additionally choose transverse spatial vectors with ${ϵ}^0=0$ and ${ϵ}^{\ast }\cdot ϵ=-1$; there are two independent physical polarizations. The equivalence class removes the unphysical longitudinal direction, rather than imposing four independent physical polarization states.

The invariance of the total amplitude is a two-photon fermion Ward identity. Replace ${ϵ}_1^{\ast }$ by $k_1$ and put $q_t=p_1-k_1$, $q_u=p_1-k_2=k_1-p_2$. The external Dirac equations giveAway from propagator poles, $R(q)(/q-m_{\mu })=(/q-m_{\mu })R(q)=I$; the identity extends with the common Feynman prescription. ThereforeThe two terms in the bracket then give $-\bar{v}/{ϵ}_2^{\ast }u$ and $+\bar{v}/{ϵ}_2^{\ast }u$, which cancel. Exchanging photon labels proves the second Ward identity. Thus the sum is unchanged under ${ϵ}_i\mapsto {ϵ}_i+\alpha k_i$ for either photon. Neither diagram is generally gauge-invariant separately. Physically, longitudinal pure-gauge polarization does not couple to the observable scattering amplitude; only the two transverse photon degrees of freedom contribute.

For Compton scattering write the incoming momenta as $p,k$ and outgoing momenta as $p^{\prime },k^{\prime }$, with $p+k=p^{\prime }+k^{\prime }$. The outgoing muon contributes $\bar{u}(p^{\prime })$, the incoming photon contributes $ϵ(k)$ without conjugation, and the outgoing photon contributes ${ϵ}^{\prime \ast }(k^{\prime })$. The two tree-level Compton amplitudes areThese are the lower two diagrams. Their intermediate muon momenta are $p+k$ and $p-k^{\prime }$, with denominators $s-m_{\mu }^2+i0$ and $u-m_{\mu }^2+i0$. The first attaches the incoming photon before the outgoing photon along fermion flow; the second reverses that attachment order. Both contribute at order $e^2$ in the amplitude.

Compton scattering has one incoming and one outgoing muon and one photon on each side; annihilation has an incoming particle-antiparticle pair and two outgoing photons. Accordingly the external spinors/polarizations and physical Mandelstam channels differ, although both arise from the same two-vertex QED matrix element. Crossing symmetry relates them by continuing $p_2\mapsto -p^{\prime }$ and one outgoing photon momentum $k_1\mapsto -k$, with the corresponding external wavefunction replacements. The annihilation $t$ channel becomes the Compton $s$ channel and the other channel becomes its $u$ channel. Identical final photons require a factor $1/2!$ in an annihilation phase-space integral over an otherwise double-counted full final-state region; there is no analogous final-state identical-particle factor for the muon-photon Compton final state. Gauge cancellation holds for the summed Compton amplitude as well.

A Majorana spinor equals its transform under charge conjugation, so its spinor components are not independent of their conjugates. In a real two-dimensional gamma matrix representation it can be taken real, with anticommuting Grassmann variables as components. Each target index labels a separate worldsheet Majorana fermion; it is not itself a target-space spinor.

Use worldsheet signature $(-,+)$ and the real gamma matricesTake $\bar{\psi }={\psi }^{T}C$ with $C={\gamma }^0$. Define the chirality matrix with the chosen worldsheet orientation byIt squares to one and anticommutes with each gamma matrix. Chirality is its eigenvalue, selected by the chiral projectors. Its overall sign is conventional.

Writing $\psi =({\psi }_{+},{\psi }_{-}{)}^T$, the fermion equation of motion ${\gamma }^{\mu }{\partial }_{\mu }\psi =0$ becomesHenceThe first profile travels toward increasing $\sigma$, and the second toward decreasing $\sigma$. Thus the choice of chirality orientation realizes the requested worldsheet chirality and propagation direction correspondence; reversing the orientation interchanges the labels.

For the rigid worldsheet supersymmetry variation, let $ϵ$ be a constant Grassmann-odd Majorana spinor. With the displayed conventions,The minus sign follows from $({\gamma }^{\mu }{)}^{T}C=-C{\gamma }^{\mu }$. A second identity from the Majorana Grassmann bilinear interchange is $\bar{\chi }{\gamma }^{\nu }{\gamma }^{\mu }ϵ=\overline{ϵ}{\gamma }^{\mu }{\gamma }^{\nu }\chi$. Keeping these Grassmann signs is essential.

The bosonic kinetic variation is $2i{\partial }_{\mu }{X}_a\overline{ϵ}{\partial }^{\mu }{\psi }^a$. The two fermionic kinetic variations, using the two identities above, give the full resultFor the second equality use ${\gamma }^{\nu }{\gamma }^{\mu }=2{\eta }^{\mu \nu }-{\gamma }^{\mu }{\gamma }^{\nu }$ and symmetry of the second embedding derivative. No field equation has been used. Thus the variation is an off-shell total derivative:This is the rigid worldsheet supersymmetry boundary term. It vanishes when the boundary flux vanishes, for example on a closed worldsheet with compatible fields and variations. On a spatial circle, a constant supersymmetry parameter must also respect the chosen spin structure. Periodic fermions allow this rigid transformation; antiperiodic fermions and a constant parameter would give an antiperiodic $\delta X$ and fail to preserve a periodic bosonic embedding. The local action identity remains valid, but global symmetry requires compatible boundary data. This is the spin-structure obstruction to constant worldsheet supersymmetry.

For the spectrum, assume the usual critical RNS string, physical-state constraints, a nonzero null target momentum, and the supersymmetric GSO projection. Analyze just one chiral sector. Its matter central charge is $d+d/2$; the bc system and superconformal ghosts contribute $-26+11$. Therefore $3d/2-15=0$ gives the critical dimension of the RNS superstring $d=10$. Light-cone gauge in string theory leaves eight transverse bosonic and eight transverse fermionic string oscillator directions.

In the Neveu–Schwarz sector, the normal-ordering intercept is $a_{\mathrm{NS}}=1/2$. The massless level is $N=1/2$ and has statesThe GSO projection retains these eight transverse vector polarizations while removing the tachyonic ground state. They are spacetime bosons, although the creating string oscillator is a worldsheet fermion. In covariant language transversality and the longitudinal null-state quotient leave $10-2=8$ polarizations.

In the Ramond sector, bosonic and fermionic zero-point energies cancel, so $a_{\mathrm{R}}=0$ and the ground states are massless. The eight transverse fermion zero modes obeyThis Ramond zero-mode Clifford algebra acts on a sixteen-dimensional ground-state space. Its two chirality subspaces each have dimension eight. The GSO projection keeps one, an $8_s$ or $8_c$ spinor of $\mathrm{SO}(8)$, the rotation subgroup of the massless little group. These are spacetime fermions. Equivalently a ten-dimensional Majorana-Weyl spinor has sixteen real components before the massless Dirac equation reduces the physical polarization count to eight.

Consequently the massless chiral RNS spectrum after GSO projection satisfiesThe GSO projection is essential: without it, the transverse Ramond sector would retain both eight-dimensional spinor chiralities. These are counts for one chiral sector, not the tensor-product counts of the full closed-string spectrum.

Use metric signature $+---$ and the parity matrix $P^{\mu }{}_{\nu }=\mathrm{diag}(1,-1,-1,-1)$. If a Dirac field transforms as ${\psi }^{\prime }(x)={\eta }_P{S}_P\psi (x_P)$, the chain rule changes the spatial derivatives' signs. Covariance of its Dirac equation therefore requiresThe Clifford algebra gives precisely these identities for $S_P={\gamma }^0$. In an irreducible Dirac representation any other solution differs by a scalar phase, because its ratio with ${\gamma }^0$ commutes with all gamma matrices. With $|{\eta }_P|=1$, the Dirac mass bilinear and kinetic term consequently obeyThe factor ${\gamma }^0$ compensates the sign reversal of spatial derivatives. A transformation containing only the argument change would not preserve the massive Dirac equation. This establishes parity symmetry of the free or vector-coupled Dirac theory; it does not make a chiral weak interaction parity symmetric.

For charge conjugation, transpose the adjoint Dirac equation to obtain $(i{\gamma }^{\mu T}{\partial }_{\mu }+m){\bar{\psi }}^T=0$. Multiplying by $C$ gives the original Dirac equation for ${\psi }^c=C{\bar{\psi }}^T$ ifOne may choose $C$ unitary; in four dimensions a conventional choice in a standard Dirac or chiral basis is $C=i{\gamma }^2{\gamma }^0$, with $C^T=-C$. Its overall phase and ${\eta }_C$ do not affect the following bilinear transformation. Reordering the Grassmann fermion fields gives, for a color matrix $X$ commuting with the gamma matrices,Thus a vector color current changes sign and transposes its color generator under charge conjugation.