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The Pauli matrix multiplication law is ${\sigma }_i{\sigma }_j={\delta }_{ij}{I}_2+i{ϵ}_{ijk}{\sigma }_k$, so $\{{\sigma }_i,{\sigma }_j\}=2{\delta }_{ij}{I}_2$. Block multiplication of the chiral gamma-matrix representation givesandThus the mixed anticommutators vanish and the spatial ones are $-2{\delta }_{ij}{I}_4$. With the Minkowski metric convention $\eta =\mathrm{diag}(1,-1,-1,-1)$,These matrices therefore represent the required Clifford algebra.

Take ${\partial }_{\pm }=({\partial }_{\tau }\pm {\partial }_{\sigma })/2$ on the strip $0\le \sigma \le \pi$. Varying the Grassmann-odd worldsheet Majorana fermions gives ${\partial }_{-}{\psi }_1=0$, ${\partial }_{+}{\psi }_2=0$ and, up to the orientation of the strip boundary, the surface termAt each endpoint it vanishes if ${\psi }_1=\eta {\psi }_2$ with $\eta =\pm 1$, with the same relation on variations. A common sign can be absorbed by redefining one chiral field, so set ${\eta }_0=+1$. Since ${\psi }_1=f(\tau +\sigma )$ and ${\psi }_2=f(\tau -\sigma )$, the second endpoint imposes $f(u+2\pi )={\eta }_{\pi }f(u)$. Equal endpoint signs give the periodic Ramond sector, and opposite endpoint signs give the antiperiodic Neveu–Schwarz sector:These gluing choices preserve the Neumann-endpoint worldsheet supersymmetry conditions and give integer or half-integer worldsheet Majorana fermion modes, respectively.

First fix a normalization issue in the original PDF. With its stated canonical string oscillator algebra, the standard superconformal current mode iswithout the printed factor $1/2$. To check the Ramond supercurrent zero-mode square directly, pair positive and negative integer modes:The zero-mode Clifford relations give $({\alpha }_0\cdot {\psi }_0{)}^2={\alpha }_0^2/2$. Different $Q_m$ anticommute, and so do $Q_m$ with the zero-mode term. The two same-sign terms in $Q_m^2$ vanish by antisymmetry of the fermions and symmetry of the bosonic product. The cross terms giveThe second term comes from commuting ${\alpha }_m$ through ${\alpha }_{-m}$, so the frequency factor $m$ is essential. ThereforeHere $L_0^R$ uses vacuum-annihilating Ramond normal ordering; its string intercept is zero. A plane-to-cylinder central shift must be included consistently if a different zero-mode convention is used.

For the mode $F_r=G_r/2$ actually printed, the algebra instead gives $\{F_0,F_0\}=L_0^R/2$. Already the oscillator-free term gives $\{F_0,F_0\}={\alpha }_0^2/4$, whereas $2L_0^R={\alpha }_0^2$. Thus the requested identity is not valid with both that prefactor and the supplied string oscillator algebra. Removing the prefactor gives the identity just derived. Renaming $L_0^R/4$ as $L_0$ would fix this isolated equality but would spoil the usual Virasoro normalization and the subsequent mass formula. We use the standard normalized modes $G_r$ in the remaining physical constraints; their vanishing is unaffected by a common nonzero rescaling.

The RNS string physical-state conditions in the two sectors areThe last condition is the fermionic zero-mode constraint; its square implies the Ramond mass-shell condition. Physical null string states are quotiented. Matter central charge is $3d/2$, while the reparameterization worldsheet ghost fields and superconformal ghosts contribute $-26+11=-15$. Thus $d=10$ is the critical dimension of the RNS superstring, with $a_{NS}=1/2$ and $a_R=0$.

To solve the constraints, choose $X^{\pm }=(X^0\pm X^9)/\sqrt{2}$, so ${\eta }_{+-}=-1$, and work on a patch with ${\alpha }_0^{+}=\sqrt{2{\alpha }^{\prime }}{p}^{+}\ne 0$. In light-cone gauge in string theory fix all nonzero ${\alpha }_n^{+}$ and all ${\psi }_r^{+}$ to zero. For the Ramond sector, the printed restriction to nonzero fermion modes is not by itself complete: one must also reduce the longitudinal fermionic zero modes using the zero-mode supersymmetry constraint. The usual reduced light-cone description sets the remaining ${\psi }_0^{+}$ to zero as well and retains the transverse Clifford algebra. Covariantly this reduction is the Dirac equation on the ground-state particle polarization; it is not an extra unconstrained canonical string oscillator.

With this reduction the string oscillator generators have the formRepeated $i$ runs over the eight transverse directions. The coefficient in $L_n^{\perp }$ comes from the fermion worldsheet stress tensor $-\psi \partial \psi /2$; for $n=0$, positive and negative $r$ combine to give the occupation energy $r{\psi }_{-r}{\psi }_r$. Thus the light-cone super-Virasoro constraints are explicitly solved byHere $a$ is $1/2$ in the NS sector and zero in the R sector. The zero-mode equation says $2p^{+}{p}^{-}=p_i^2+(N-a)/{\alpha }^{\prime }$. All longitudinal excitations are consequently determined by transverse bilinears. The critical dimension and string intercepts are necessary for the resulting quantum theory to preserve the target Lorentz algebra.

The frequency-weighted Neveu–Schwarz level operator isThe PDF's displayed number operator omits the fermion frequency weights and the $r=1/2$ term. With the stated anticommutators that cannot be a level operator: it would assign no level to the first fermionic excitation. The bosonic term already includes its frequency through $[{\alpha }_n,{\alpha }_{-n}]=n$, whereas the fermionic term must include its frequency explicitly.

One independent way to obtain the string intercept is the transverse zero-point energy. Each periodic boson contributes $\frac{1}{2}{\sum }_{n\ge 1}n=-1/24$. Each antiperiodic real fermion contributes $-\frac{1}{2}{\sum }_{r\ge 1/2}r=-1/48$, with the minus sign due to fermionic normal ordering. For example a common exponential cutoff givesSubtracting the local divergent terms leaves $8(-1/24-1/48)=-1/2$, so the NS string intercept is $a=1/2$. In the Ramond sector the integer fermions instead contribute $+1/24$ per component and cancel the bosonic contribution, giving $a_R=0$.

The unprojected NS sector oscillator vacuum has $N=0$ andIt is a spacetime scalar tachyon. The first excitation ${\psi }_{-1/2}^i|0;p⟩$ has $N=1/2$ and is massless, with eight vector particle polarizations. Covariantly its particle polarization obeys $p\cdot ϵ=0$ modulo the longitudinal null particle polarization, leaving the same eight degrees of freedom.

The R sector oscillator vacuum is also massless, but it is degenerate because $\{{\psi }_0^{\mu },{\psi }_0^{\nu }\}={\eta }^{\mu \nu }$. The operators ${\Gamma }^{\mu }=\sqrt{2}{\psi }_0^{\mu }$ give the target Clifford algebra, and $G_0|u;p⟩=0$ becomes $p_{\mu }{\Gamma }^{\mu }u=0$. At nonzero null momentum this Dirac equation leaves sixteen unprojected on-shell spinor states. Equivalently the eight transverse zero modes give the sixteen-dimensional spin representation $8_s\oplus 8_c$ of $SO(8)$. Thus the Ramond ground state describes spacetime fermions, not a single scalar oscillator vacuum.

To define the GSO projection explicitly, let $f_{NS}$ count excited NS fermions and give its vacuum even parity. ThenIt retains odd fermion number, removes the tachyon, and retains the eight massless vector particle polarizations. In the Ramond sector let ${\Gamma }_{\ast }$ be transverse chirality and $f_R$ count nonzero-mode fermions. Choose one consistent chirality sign $s=\pm 1$ and useEach operator squares to itself. At the ground level $P_R$ retains one eight-dimensional spinor chirality, so the retained spectrum has eight massless bosonic and eight massless fermionic states. The parity convention can instead assign the NS vacuum odd parity and write a plus projector; the retained states are identical.

Equality continues at every mass level. Let $q$ count ${\alpha }^{\prime }{M}^2$, suppressing common momentum and endpoint-label factors, and set $D(q)={\prod }_{n\ge 1}(1-q^n{)}^8$. The projected string oscillator generating functions areThe Ramond factor eight follows because either chirality projector selects eight ground states for every nonzero-oscillator parity. The Jacobi abstruse identity makes the bracket equal to $16q^{1/2}{\prod }_{n\ge 1}(1+q^n{)}^8$, soThis proves level-by-level equality of bosonic and fermionic multiplicities. The appropriate spacetime supercharge from an RNS spin field relates the two sectors; the projection ensures that it acts within the retained state space. These sector and chirality conventions agree with the primary superstring lecture notes. Equal total counts alone are not the definition of the projection; they follow from the stated parity selection and the string oscillator identity.

Use four-dimensional Minkowski spacetime with $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$ and ${\gamma }^{\mu \nu \rho }={\gamma }^{[\mu }{\gamma }^{\nu }{\gamma }^{\rho ]}$. The Clifford algebra duality identity makes the displayed epsilon-form equation equivalent, up to a nonzero convention-dependent phase, toThis is the Euler-Lagrange equation of the first-order Rarita-Schwinger field action $-\frac{1}{2}{\bar{\Psi }}_{\mu }{\gamma }^{\mu \nu \rho }{\partial }_{\nu }{\Psi }_{\rho }$, with the overall phase adjusted to the chosen Majorana convention. Varying the Majorana spinor action accounts for both occurrences of the field and cancels the one-half normalization.

A vector times a spinor contains spin-$3/2$ and unwanted spin-$1/2$ pieces, so the tensor type alone does not establish the physical spin. The field equation has gauge invariancesince the two derivatives contract an antisymmetric pair of gamma indices. Put $\chi ={\gamma }^{\mu }{\Psi }_{\mu }$ and $d={\partial }^{\mu }{\Psi }_{\mu }$. Gamma matrix multiplication givesHence $d=/\partial \chi$. Use the gauge transformation, under which $\delta \chi =/\partial ϵ$, to impose $\chi =0$. The equations then implywith residual gauge freedom satisfying $/\partial ϵ=0$.

For a plane wave with nonzero momentum satisfying the null condition along the third axis, this residual freedom sets the temporal component to zero. Transversality then removes the longitudinal component, leaving the tensor product of two transverse vector polarizations with the two massless spinor helicities. The possible total helicities are $\pm 3/2$ and $\pm 1/2$. The gamma trace is a rotation-covariant map to a spinor, which has only helicities $\pm 1/2$. It therefore vanishes on the aligned $\pm 3/2$ combinations and removes the two mixed combinations. Equivalently, the transverse gamma trace condition has rank two on the four remaining complex positive-frequency coefficients. The massless Rarita-Schwinger polarization count is thusThe Majorana spinor reality condition relates negative-frequency coefficients to their conjugates; it does not introduce a separate antiparticle multiplet. A complex vector-spinor would also have independent antiparticle states.

A nonzero standard Rarita-Schwinger field mass term destroys the massless gauge invariance and adds the two longitudinal helicities. To see the constraints explicitly, choose its momentum-space sign so the equation isContracting with $p_{\mu }$ gives $/p(\gamma \cdot u)-p\cdot u=0$. Its gamma trace then gives $3m\gamma \cdot u=0$. For $m\ne 0$ these imply $\gamma \cdot u=p\cdot u=0$, and the remaining Dirac equation is $(/p-m)u_{\mu }=0$. In the rest frame $u_0=0$. Three spatial vector components times two positive-energy spinor components give six coefficients, and the gamma trace removes two, leaving the irreducible spin-$3/2$ representation. ThereforeIn supergravity a consistent spontaneous mass generation is the super-Higgs mechanism, which supplies those extra states from the goldstino.

The massless graviton also has two physical helicities, so the graviton and gravitino already match on shell. The reason for auxiliary fields is the off-shell component count of minimal supergravity mismatch, not a mismatch of these physical states. After removing gauge functions but before using field equations, a metric has $10-4=6$ independent bosonic components, while a Majorana vector-spinor has $16-4=12$ fermionic components. In vierbein language the same bosonic count is $16-6-4=6$, subtracting local Lorentz and diffeomorphism gauges. Old-minimal supergravity adds a complex scalar and a real vector, supplying six real nonpropagating bosonic components. This gives $12+12$ off-shell components and closure without imposing equations of motion. Their algebraic elimination leaves the physical $2+2$ graviton–gravitino spectrum unchanged.

Write $I=I_2$. The Pauli matrix multiplication law gives ${\sigma }^i{\sigma }^j+{\sigma }^j{\sigma }^i=2{\delta }^{ij}I$. Block multiplication of the Weyl representation of the gamma matrices givesandThus the mixed time-space anticommutators vanish, the time-time anticommutator is $2I_4$, and the spatial anticommutators are $-2{\delta }^{ij}{I}_4$. Together these prove the Clifford algebra identityThe negative spatial sign comes from the lower-left block of each spatial gamma matrix.

Construct the Lorentz-spinor generators from a Clifford algebra asThey are antisymmetric in $\rho ,\sigma$. Reordering a third gamma matrix with the Clifford algebra yieldsUse the commutator derivation identity on the two factors in $M^{\tau \nu }=\frac{1}{4}[{\gamma }^{\tau },{\gamma }^{\nu }]$:This derives every Lorentz algebra commutation relation, so the matrices $M^{\rho \sigma }$ form its spinor representation. Their finite exponentials describe the connected Spinor representation of the Lorentz group, more precisely the lift to its double cover.

For $x^{\prime }=\Lambda x$, choose the corresponding spin liftThe spinor field transforms asThe infinitesimal Clifford algebra identity from part (b) exponentiates to $S^{-1}{\gamma }^{\mu }S={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$, establishing Lorentz covariance of the Dirac operator. At fixed coordinates the infinitesimal variation also includes the orbital term $-{\Omega }^{\mu }{}_{\nu }{x}^{\nu }{\partial }_{\mu }\psi$.

A spatial rotation generator $M^{ij}$ is anti-Hermitian, but a Lorentz boost generator is Hermitian. For example,Its eigenvalues are $\pm \frac{1}{2}$. A nonzero rapidity $\zeta$ gives boost eigenvalues $e^{\pm \zeta /2}$, which are not all on the unit circle. This proves that the finite-dimensional component Spinor representation of the Lorentz group cannot be unitary for a positive-definite component inner product, even after a change of basis. It does not contradict a unitary action on the physical space of quantum states.

Consequently ${\psi }^{\prime †}(x^{\prime }){\psi }^{\prime }(x^{\prime })={\psi }^{†}{S}^{†}S\psi$ need not equal ${\psi }^{†}\psi$: it is the time component of the Dirac current, not a Lorentz scalar. The appropriate invariant form is indefinite. From $({\gamma }^{\mu }{)}^{†}={\gamma }^0{\gamma }^{\mu }{\gamma }^0$ one obtainsThis Dirac spinor pseudo-unitarity gives ${\bar{\psi }}^{\prime }(x^{\prime })=\bar{\psi }(x)S^{-1}$ and hence the Dirac scalar bilinear

Use the momentum-space Feynman rules with all momenta conserved at each vertex. The free scalar propagator, free Dirac propagator and pseudoscalar Yukawa interaction vertex are respectivelyThe Dirac propagator follows by multiplying the inverse kinetic matrix $/p-m$ by $/p+m$, using the Clifford algebra to obtain $(/p-m)(/p+m)=(p^2-m^2)I_4$. The vertex follows by expanding $e^{i{S}_{\mathrm{int}}}$ once: there is no factorial because it has one scalar and two distinct spinor field factors. An internal scalar has no spinor indices; an internal fermion line carries the displayed matrix with its momentum directed along fermion flow.

These rules use the coefficient exactly as printed. With the standard Hermitian ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$, the Hermiticity of a pseudoscalar Dirac bilinear gives $(\bar{\psi }{\gamma }^5\psi {)}^{†}=-\bar{\psi }{\gamma }^5\psi$. A Hermitian physical interaction therefore requires ${\lambda }^{\ast }=-\lambda$; writing $\lambda =ig$ with real $g$ changes the vertex to $g{\gamma }^5$. This convention issue does not alter the formal rules above, and no unstated real-coupling assumption is needed.

Take ${\eta }^{\alpha \beta }=\mathrm{diag}(-1,1)$ with coordinate zero equal to $\tau$ and coordinate one to $\sigma$. In the supplied representation setThey satisfy the Clifford algebra $\{{\rho }^{\alpha },{\rho }^{\beta }\}=2{\eta }^{\alpha \beta }1$. For anticommuting spinor components and an ordinary matrix $B$, relabelling the two summed indices givesNow $C^T=-C$, whereas $C{\rho }^0=-1$ and $C{\rho }^1=\mathrm{diag}(-1,1)$ are symmetric. Therefore the Majorana Grassmann bilinear interchange yieldsThese signs require fermionic, Grassmann-odd fields; commuting spinors would have the opposite interchange signs.

In flat conformal gauge, suppress the common factor $(4\pi {\alpha }^{\prime }{)}^{-1}$ and write the combined density as $\mathcal{L}=-{\partial }_{\alpha }X\cdot {\partial }^{\alpha }X+\bar{\psi }\cdot {\rho }^{\alpha }{\partial }_{\alpha }\psi$. The supersymmetry variation including its odd parameter is an even derivation. From $({\rho }^{\alpha }{)}^{T}C=-C{\rho }^{\alpha }$ we obtainAlso $(C{\rho }^{\alpha }{\rho }^{\beta }{)}^T=-C{\rho }^{\beta }{\rho }^{\alpha }$, so interchanging odd spinors gives $\bar{\chi }{\rho }^{\alpha }{\rho }^{\beta }ϵ=\overline{ϵ}{\rho }^{\beta }{\rho }^{\alpha }\chi$. The two density variations, with target contractions understood, are thereforeIn the second term the symmetric second derivative annihilates the antisymmetric part of ${\rho }^{\beta }{\rho }^{\alpha }$. Using $2{\eta }^{\alpha \beta }-{\rho }^{\beta }{\rho }^{\alpha }={\rho }^{\alpha }{\rho }^{\beta }$ now gives the exact off-shell identityThus the action is invariant when this boundary flux vanishes and the transformation preserves the field boundary conditions, for example with suitable falloff or compatible supersymmetric endpoint conditions. On a spatial circle, periodic fields admit the constant parameter; antiperiodic fermions with periodic bosons require the local superconformal treatment, since a nonzero constant parameter would make the boson variation antiperiodic. The constant parameter and ordinary derivatives describe rigid supersymmetry in flat gauge; a locally supersymmetric curved-worldsheet formulation includes the worldsheet gravitino. The result is the rigid worldsheet supersymmetry boundary term without using field equations.

For the commutator on $X$, keep the order of the odd parameters:For the fermion, the order is equally important. Moving ${ϵ}_2$ through both odd factors in $({\overline{ϵ}}_1{\partial }_{\alpha }\psi ){ϵ}_2$ incurs two minus signs. ConsequentlyTo check the spinor matrix explicitly, set ${ϵ}_1=(a,b{)}^T$, ${ϵ}_2=(c,d{)}^T$, with all four components anticommuting and monomials ordered $a,b,c,d$. ThenDirect multiplication gives ${\rho }^{\alpha }M+M{\rho }^{\alpha }=2V^{\alpha }1$. HenceVariation of the fermion action gives the Dirac equation ${\rho }^{\alpha }{\partial }_{\alpha }\psi =0$. On this equation the second term vanishes, provingThis is on-shell closure of rigid worldsheet supersymmetry; the action's invariance was off shell, but its fermionic transformation algebra closes in this field content only on shell.

The associated superconformal current is proportional to $\psi \cdot \partial X$, of conformal weight $3/2$. Its fermionic modes $G_r$ extend the stress-tensor generators to the N=1 super-Virasoro algebra:The modes are half-integral in the Neveu–Schwarz sector and integral in the Ramond sector. Their anticommutator giving a Virasoro generator is the conformal-mode counterpart of the translation commutator just derived. Ten coordinate bosons and ten Majorana fermions give matter central charge $c=10+10/2=15$, canceled by the usual reparametrization and superconformal ghosts in the critical RNS string.

The Noether gauging procedure starts by allowing a constant symmetry parameter to depend on position. For an internal symmetry written as $\delta {\varphi }^i={ϵ}^A{T}_A{\varphi }^i$, integration by parts identifies the Noether current:up to a boundary term. Introduce a gauge field $A_{\mu }^A$ and add $-g{A}_{\mu }^A{j}_A^{\mu }$, with the inhomogeneous gauge transformation $\delta A_{\mu }^A=g^{-1}{\partial }_{\mu }{ϵ}^A+\cdots$. This cancels the displayed variation. The current coupling has its own variation, so one iterates: add the needed higher-order interactions and corrections to the gauge transformations, and require closure of the local symmetry algebra. For an ordinary gauge theory, this reorganizes derivatives into gauge covariant derivatives and fixes the nonlinear gauge field strength. Gauging is therefore more than replacing a constant parameter by a function.

For rigid supersymmetry, the corresponding Noether current is the spinor-valued supercurrent $S^{\mu }$. A local Majorana spinor parameter gives a variation proportional to $({\partial }_{\mu }\overline{ϵ})S^{\mu }$. A gravitino ${\Psi }_{\mu }$, with an inhomogeneous transformation $\delta {\Psi }_{\mu }\sim {\kappa }^{-1}{\partial }_{\mu }ϵ$, cancels it through a coupling proportional to $\kappa {\bar{\Psi }}_{\mu }{S}^{\mu }$. The commutator of two rigid supersymmetry transformations is a translation. With local parameters, the translation parameter becomes position dependent, so the completion must also gauge translations: it introduces the vierbein, diffeomorphism invariance of general relativity, and local Lorentz transformations. The gravitino and graviton thus belong to the same supergravity multiplet.

For pure minimal four-dimensional supergravity, start with the free massless Fierz-Pauli action and the free Rarita-Schwinger field kinetic term. The Noether gauging procedure replaces the former by the Einstein-Hilbert action and couples the latter to the vierbein and spin connection. In a canonical normalization the leading transformations can be writtenAn overall sign or phase changes with the curvature and Majorana spinor conventions, but the relative normalization must be used consistently. To see why these terms fit together, first work to quadratic order in fermions, so the spin connection is torsion free. Varying the Rarita-Schwinger field kinetic term under $\delta {\Psi }_{\mu }={\nabla }_{\mu }ϵ/\kappa$ produces an antisymmetric double spinor covariant derivative. The spinor curvature identity and the Einstein tensor contraction of spin curvature give, in the matching curvature convention,The resulting Einstein tensor term cancels the variation of the Einstein-Hilbert action under $\delta e_{\mu }{}^a$. This is the essential first nontrivial cancellation in the construction.

At higher fermion order, the variations of the vierbein and spin connection produce further terms. Treating the spin connection as independent makes its Euler-Lagrange field equation algebraic. Its solution is a gravitino-induced torsion connection. With ${\Psi }_a=e_a{}^{\mu }{\Psi }_{\mu }$ and ${\gamma }^{ab}=[{\gamma }^a,{\gamma }^b]/2$, a conventional canonical normalization givesEquivalently, with $T^a=d{e}^a+{\tilde{\omega }}^a{}_b\wedge e^b$, the torsion tensor obeysThe Majorana Grassmann bilinear interchange makes the right-hand side antisymmetric in $\mu ,\nu$. Eliminating the spin connection generates the four-fermion completion. In the 1.5-order formalism one substitutes its solution but varies the other fields holding the connection fixed: the omitted chain-rule term is $(\delta S/\delta \omega )\delta \tilde{\omega }=0$. The remaining cubic fermion variations cancel by the Clifford algebra and Fierz rearrangements. This produces the displayed supergravity action in its compact connection-dependent form. Calling this formulation on shell means that no independent auxiliary fields have been retained and that closure of the supersymmetry algebra uses the propagating Euler-Lagrange field equations; it does not mean that action invariance is obtained only by setting every variation to zero on a solution.

In the antisymmetrized kinetic term, ${\Psi }_{\sigma }$ is a spinor-valued one-form and the derivative is conventionally writtenThis is the Lorentz-covariant exterior-derivative convention for that term. If one instead uses the full affine covariant derivative of a vector-spinor, a coordinate-index connection is also present and the antisymmetrization must include its torsion tensor correction. The crucial physical distinction from pure metric general relativity is the spin connection contains gravitino-induced contorsion, rather than being the torsion-free Levi-Civita connection. It reduces to the ordinary spinor covariant derivative when the gravitino vanishes.

Use the Clifford algebra $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$ and unit-weight antisymmetrization. The duality identity between ${\gamma }_5{\gamma }_{\nu }$ and three antisymmetrized gamma matrices turns the equation intoThe overall nonzero factor in this identity depends on the orientation and ${\gamma }_5$ convention and does not change its solutions. This Rarita-Schwinger field equation has the gauge symmetrysince ${\gamma }^{\mu \nu \rho }{\partial }_{\nu }{\partial }_{\rho }ϵ=0$. It is the free-field limit of local supersymmetry, so it is suitable for the gravitino.

To check which spin it propagates, put $A={\gamma }^{\mu }{\Psi }_{\mu }$ and $B={\partial }^{\mu }{\Psi }_{\mu }$. Expanding the three gamma matrices using the Clifford algebra givesin four dimensions. Thus the Euler-Lagrange field equation requires $B=/\partial A$. Under the gauge transformation, $A\mapsto A+/\partial ϵ$; locally one may choose gamma-trace gauge fixing, $A=0$, by solving $/\partial ϵ=-A$. The Euler-Lagrange field equation then yieldsThese are the massless Dirac equation, transversality and gamma-trace constraints that remove the lower-spin part of a vector-spinor.

For a more explicit massless Rarita-Schwinger polarization count, take a positive-frequency plane wave with null momentum along the third axis, in signature $(-,+,+,+)$. Transversality relates its temporal and longitudinal components. Residual gauge transformations with $/pϵ=0$ remove both. Write the surviving components as $u_1,u_2$; their gamma-trace condition reads ${\gamma }^1{u}_1+{\gamma }^2{u}_2=0$, henceThe massless Dirac equation leaves two complex positive-frequency amplitudes in $u_1$. If $i{\gamma }^1{\gamma }^2{u}_1=s{u}_1$, $s=\pm 1$, then $(u_1,u_2)$ is proportional to $(1,-is)$ in the transverse vector index. With spatial rotation generators $J_3^{\mathrm{vec}}{}_{12}=-i$ and $J_3^{\mathrm{spin}}=-i{\gamma }^1{\gamma }^2/2$, the vector has helicity $-s$ and the Majorana spinor has helicity $-s/2$, giving total helicity $-3s/2$. The Majorana spinor condition fixes the negative-frequency coefficients from these amplitudes. Therefore a massless gravitino has spin $3/2$, with the two physical helicities $\pm 3/2$. A complex vector-spinor would also have independent antiparticle modes.

The requested count is instead the off-shell component count of minimal supergravity, before imposing any Euler-Lagrange field equations. A four-dimensional Majorana spinor has four real components. Its vector index supplies four copies, and local supersymmetry supplies four real gauge functions:One must not additionally subtract transversality and the massless Dirac equation in this count: those are on shell conditions. The gamma-trace gauge fixing already uses the same four gauge functions subtracted above. For the graviton, a symmetric metric tensor has ten real components, with four diffeomorphism gauge functions. Equivalently, a vierbein has sixteen real components, minus six local Lorentz transformation and four diffeomorphism functions:The gravitino and graviton each have two propagating states on shell, but their off-shell counts disagree by six.

An auxiliary field has an algebraic Euler-Lagrange field equation and no propagating kinetic term. In old-minimal supergravity, a complex scalar $M$ contributes two real bosonic components and a real vector $b_{\mu }$ contributes four, so the bosonic count becomes $6+2+4=12$, matching the gravitino. Their supersymmetry transformations also absorb the terms that would otherwise require the propagating Euler-Lagrange field equations to close the local algebra. Thus auxiliary fields allow off-shell supersymmetry closure and matching component counts without adding physical particles. Eliminating them recovers an on shell formulation.

The supplied matrices are the chiral gamma-matrix representation. The Pauli matrix multiplication law implies $\{{\sigma }^i,{\sigma }^j\}=2{\delta }^{ij}{I}_2$. Block multiplication gives $({\gamma }^0{)}^2=I_4$, $\{{\gamma }^0,{\gamma }^i\}=0$, and $\{{\gamma }^i,{\gamma }^j\}=-2{\delta }^{ij}{I}_4$. Hence their Clifford algebra relation isSince partial derivatives commute, multiplying the Dirac equation by $i{\gamma }^{\nu }{\partial }_{\nu }+m$ givesThus every component of the Dirac spinor satisfies $\boxed{(\square +m^2)\psi =0}$, the Klein-Gordon equation with the same mass.

For a Lorentz transformation $x^{\prime }=\Lambda x$, set ${\psi }^{\prime }(x^{\prime })=S(\Lambda )\psi (x)$. The derivative transforms as ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. With the given spinor-matrix identity,This proves the Lorentz covariance of the Dirac operator explicitly.

For parity, let $x_P=(x^0,-\mathbf{x})$ and ${\psi }_P(x)={\gamma }^0\psi (x_P)$. Spatial differentiation brings a minus sign, which is canceled by ${\gamma }^i{\gamma }^0=-{\gamma }^0{\gamma }^i$. ThereforeSince ${\gamma }^0$ is Hermitian and squares to one, the Dirac adjoint transforms asThe chirality matrix ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$ anticommutes with ${\gamma }^0$. The parity transformation of a Dirac bilinear consequently givesThe first bilinear is a Lorentz scalar even under parity, and the second is a pseudoscalar.