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In the Weyl representation of the gamma matrices, $C=i{\gamma }^2{\gamma }^0=\mathrm{diag}(\epsilon ,-\epsilon )$ with $\epsilon =i{\tau }_2$. The antisymmetry of $\epsilon$ gives $C^T=-C$, and direct multiplication gives $C{\gamma }^{\mu T}{C}^{-1}=-{\gamma }^{\mu }$ and $C^{†}C=I$. Any two such intertwining matrices differ by a scalar: their ratio commutes with all gamma matrices, whose products span the full four-by-four matrix algebra. Changing spinor basis sends $C$ to $SC{S}^T$, so transpose antisymmetry is basis-independent. Its sign is therefore fixed, not merely one of the two alternatives supplied by the general transpose symmetry of a charge-conjugation matrix argument.

After complexifying a nondegenerate metric in $D=2k$ dimensions, pair gamma matrices into $k$ fermionic creation and annihilation operators. Their binary occupation states span $2^k$ dimensions, and the gamma matrices generate all endomorphisms of this space. The Dirac module is therefore irreducible as a Clifford module. The even algebra preserves the chirality grading, splitting its Lorentz-spinor representation into two irreducible Weyl modules of dimension $2^{k-1}$. In $D=2k+1$, adjoining the old chirality matrix as the extra gamma gives an irreducible $2^k$-dimensional Spin group representation with no Weyl splitting. The two possible signs of that extra gamma distinguish Clifford modules but give equivalent Spin group modules.

An orthonormal coframe converts fixed flat gamma matrices to coordinate-index Clifford multiplication. Its metric identity gives the curved anticommutation relation. The spin connection is chosen so that these matrices are covariantly constant under the combined spin and tensor connection, which is the tetrad postulate in Clifford form.

For $g=\mathrm{diag}(1,-1,-1,-1)$ and the block-diagonal standard representation of the gamma matrices, write a positive-energy Dirac spinor as $(\xi ,\eta {)}^T$. The Dirac equation gives $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\xi /(E+m)$. Choosing two orthonormal two-spinors ${\chi }_s$ yields the displayed positive-frequency solutions. Negative-frequency solutions are $v_s=\sqrt{E+m}((\mathbf{\sigma }\cdot \mathbf{p}){\eta }_s/(E+m),{\eta }_s{)}^T$ multiplying $e^{ip\cdot x}$. They obey $(/p+m)v_s=0$. Both have norm $2E$; their Dirac adjoints give ${\overline{u}}_r{u}_s=2m{\delta }_{rs}$ and ${\overline{v}}_r{v}_s=-2m{\delta }_{rs}$. Spatial rotations act on each two-spinor with generators ${\sigma }_j/2$, explaining spin one-half.

With signature $(+,-,-,-)$, one standard representation is ${\tilde{\gamma }}^0=\mathrm{diag}(I_2,-I_2)$ and ${\tilde{\gamma }}^i=\left(\begin{array}{cc}0& {\sigma }^i\\ -{\sigma }^i& 0\end{array}\right)$. Block multiplication and the Pauli matrix multiplication law give time-time anticommutator $2I_4$, mixed anticommutators zero, and spatial anticommutators $-2{\delta }^{ij}{I}_4$. Thus these gamma matrices represent the spacetime Clifford algebra.

The arrow on a charged-fermion line in a Feynman diagram tracks fermion-number flow and fixes the order of the gamma matrices between external Dirac spinors. For an external particle it agrees with the particle direction; for an external antiparticle it is opposite to the antiparticle direction. It therefore need not point along the assigned four-momentum of every external leg.

Feynman slash notation contracts a four-vector with the gamma matrices. With mostly-minus signature, the Clifford algebra gives $/p^2=p^{2}I$, so $(/p-m{)}^{-1}=(/p+m)/(p^2-m^2)$ away from the mass shell.

For gamma matrices with $\{{\gamma }^a,{\gamma }^b\}=2{\eta }^{ab}$ and ${\gamma }^{ab}=[{\gamma }^a,{\gamma }^b]/2$, contracted Clifford algebra gives ${\gamma }_a{\gamma }^m{\gamma }^a=(2-D){\gamma }^m$. Expanding the two antisymmetrized factors gives ${\gamma }_a{\gamma }_b{\gamma }^m{\gamma }^a{\gamma }^b-D{\gamma }^m$. The five-gamma product equals $[2D-(D-2{)}^2]{\gamma }^m$, by moving the middle contracted gamma through and using the three-gamma contraction twice. Subtraction yields the displayed identity. A convention without the factor one-half in each bivector multiplies the result by four.

A Hermitizing matrix for a set of gamma matrices is an invertible Hermitian matrix $H$ satisfying ${\gamma }^{a†}=H{\gamma }^a{H}^{-1}$. Equivalently, each $H{\gamma }^a$ is Hermitian. It defines a covariant adjoint $\bar{\psi }={\psi }^{†}H$. If ${\gamma }^{\prime a}=M{\gamma }^a{M}^{-1}$, then $H^{\prime }=M^{-†}H{M}^{-1}$ has the same property. In a standard Lorentzian Dirac basis, $H={\gamma }^0$ is a conventional choice.

For action $-\int \bar{\psi }(\gamma \cdot \partial +m)\psi$, choose gamma matrices with $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$ and signature $(-,+,\dots ,+)$. The Fourier kernel of $D=\gamma \cdot \partial +m$ is $i\gamma \cdot p+m$. Its algebraic inverse is $(-i\gamma \cdot p+m)/(p^2+m^2)$. The Grassmann Gaussian integral for $e^{-i\bar{\psi }D\psi }$ has covariance $-i{D}_F^{-1}$, giving the displayed Dirac propagator. Its denominator has positive-energy poles below and negative-energy poles above the real energy axis by the Feynman i-epsilon prescription. Both the metric and the factor of $i$ in the derivative kernel must be specified when comparing propagator formulas.

Use natural units and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. Multiplication of the block gamma matrices givesThe Pauli matrices obey ${\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j=2{\delta }_{jk}{I}_2$, so the spatial anticommutators are $-2{\delta }_{jk}{I}_4$, the mixed ones are zero, and the temporal one is $2I_4$. ThusBecause partial derivatives commute, only the symmetric part of ${\gamma }^{\mu }{\gamma }^{\nu }$ survives in their contraction. Multiplying the Dirac equation on the left by its opposite-mass factor therefore gives the flat-space Dirac factorizationHence every component satisfies $\boxed{(\square +m^2)\psi =0}$, the Klein-Gordon equation. This implication does not remove the first-order spinor constraints imposed by the Dirac equation.

For Lorentz covariance, transform coordinates by $x^{\prime }=Lx$ and the Dirac spinor by ${\psi }^{\prime }(x^{\prime })=S(L)\psi (x)$. The chain rule gives ${\partial }_{\mu }^{\prime }=(L^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. ConsequentlyThe cancellation uses $L^{\mu }{}_{\rho }(L^{-1}{)}^{\nu }{}_{\mu }={\delta }^{\nu }{}_{\rho }$. Thus the Lorentz transformation maps every solution to another, establishing the Lorentz covariance of the Dirac operator.

The infinitesimal identity follows directly from the Clifford algebra. Moving ${\gamma }^{\rho }$ past the two factors givesSubtracting the same identity with $\mu ,\nu$ interchanged proves the spinor Lorentz generator commutatorContracting with the antisymmetric tensor ${\omega }_{\mu \nu }$ yieldsIt also verifies $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }+O({\omega }^2)$ for the infinitesimal Spinor representation of the Lorentz group.

For a positive-energy four-momentum, put $\psi =u(p)e^{-ip\cdot x}$, $E=\sqrt{{\mathbf{p}}^2+m^2}>0$. The Dirac equation becomes $(/p-m)u=0$, where the Feynman slash notation uses $/p={\gamma }^{\mu }{p}_{\mu }=E{\gamma }^0-\mathbf{\gamma }\cdot \mathbf{p}$. Splitting $u=(\xi ,\eta {)}^T$ givesThe second equation sets $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\xi /(E+m)$, and the first is then the mass shell, since $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2={\mathbf{p}}^2{I}_2$. Choosing two orthonormal two-spinors ${\chi }_s$ gives the Dirac plane waves in the standard representation:The negative-frequency solutions are independentlywith orthonormal ${\eta }_s$. They are the antiparticle modes after field quantization. The normalization gives $u_r^{†}{u}_s=v_r^{†}{v}_s=2E{\delta }_{rs}$ and ${\overline{u}}_r{u}_s=2m{\delta }_{rs}$, ${\overline{v}}_r{v}_s=-2m{\delta }_{rs}$, using the Dirac adjoint $\overline{u}=u^{†}{\gamma }^0$.

For a spatial rotation, ${\sigma }^{ij}={ϵ}_{ijk}\mathrm{diag}({\sigma }_k,{\sigma }_k)$. Hence the intrinsic rotation generators areFor a massive particle in its rest frame, the two independent spinors therefore carry the two-dimensional spin one-half representation, with $J_3$ eigenvalues $\pm 1/2$ and Casimir $s(s+1)=3/4$. A $2\pi$ rotation acts as $-I$, the spinor sign under a full spatial rotation. Thus the particle has spin $\boxed{s=1/2}$; four spinor components encode particle and antiparticle sectors, not four spin states of one particle. In the massless limit the corresponding physical labels are helicities $\pm 1/2$ rather than a rest-frame spin basis.

The mass statement needs a nonzero parallel spinor and an isolated asymptotically flat spacetime; the zero spinor solves the equation on every geometry and cannot imply anything about mass. Assume the usual complete regular spin initial data and dominant energy condition used by the positive energy theorem, with the spinor tending to a nonzero constant ${ϵ}_{\infty }$ at infinity. Let $V^a=\overline{ϵ}{\gamma }^aϵ$ be its future causal Dirac current.

The Nester two-form can be written, up to normalization, asIt is bilinear in $ϵ$ and $\nabla ϵ$. For a spacetime-parallel spinor it vanishes identically, and hence so does its asymptotic flux. The ADM boundary term of the Nester two-form identifies that flux, up to a fixed positive normalization, asThe positive energy theorem gives $E\ge |P|$. Since $V_{\infty }$ is nonzero future causal, a future timelike ADM momentum would have strictly positive contraction with it. Consequently the ADM momentum is zero or null:In a rest frame, when one exists, this immediately says $E=0$. The usual regular asymptotically flat rigidity conclusion also excludes a nontrivial null-momentum configuration and gives flat initial data. This is the spinorial boundary-charge proof, not an inference that every Lorentzian manifold with a parallel spinor is flat. Without the asymptotic and global hypotheses the local spinor equation alone does not define, let alone determine, an ADM mass.

For the modified connection, it is important to fix the Clifford algebra normalization. WriteThe compatible spinor curvature identity isIndeed, rescaling conventional gamma matrices by $\sqrt{s}$ rescales ${\gamma }_{ab}$ by $s$, while leaving the geometric spin connection unchanged. Since the connection is torsion-free and $\nabla \gamma =0$, the cross terms cancel in the modified commutator:Thus the Killing-spinor integrability with rescaled gamma matrices equation isA factor of two in the convention for antisymmetrization multiplies the whole zero equation and cannot change this relative coefficient.

The two printed coefficients are consistent with $s=2$, namely $\{{\gamma }_a,{\gamma }_b\}=4g_{ab}I$. In that convention the displayed integrability equation becomes $R_{abcd}{\gamma }^{cd}+16c^2{\gamma }_{ab}=0$ on each solution. With the customary convention $s=1$, the coefficient is instead $8c^2$, and the final Ricci coefficient below is $-12c^2$. The PDF does not state its Clifford normalization, so these alternatives must be distinguished rather than mixing them.

In four spacetime dimensions the complex Dirac spinor fibre has dimension four. Four independent solutions of $Dϵ=0$ span that fibre at every point: a solution vanishing at one point vanishes everywhere by parallel transport. Therefore the integrability matrix annihilates every spinor and is the zero matrix. The six bivector matrices ${\gamma }^{cd}$ are linearly independent, as is seen by taking traces against them; their trace pairing is a nondegenerate multiple of the metric on two-forms. Expressing ${\gamma }_{ab}=g_{ac}{g}_{bd}{\gamma }^{cd}$ therefore givesThis proves that maximal Killing spinors force constant negative curvature, not merely an Einstein Ricci tensor. Contracting in four dimensions givesFor $s=2$ this is the requested Einstein equation. Since $c>0$, its Ricci tensor automatically has rank four; under the maximal-spinor hypothesis the additional rank assumption is redundant. If the stated four-dimensional solution space is interpreted directly as the pointwise kernel of the algebraic integrability equation, the same spanning and trace argument applies.

Use the Minkowski metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and Feynman slash notation $/p={\gamma }^{\mu }{p}_{\mu }$. The Pauli matrices satisfy ${\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j=2{\delta }_{jk}{I}_2$. In the Dirac representation of the gamma matrices, direct block multiplication givesandThus the spatial anticommutators are $-2{\delta }_{jk}{I}_4$, proving the Clifford relationsSince partial derivatives commute, their symmetric product selects the symmetric part of the gamma matrices. Multiplying the Dirac equation by $i{\gamma }^{\mu }{\partial }_{\mu }+m$ givesHence every spinor component satisfies the Klein-Gordon equation with the same mass.

For parity, let $x_P=(t,-\mathbf{x})$ and ${\psi }_P(x)={\gamma }^0\psi (x_P)$. The chain rule gives ${\partial }_0\psi (x_P)=({\partial }_0\psi )(x_P)$ and ${\partial }_j\psi (x_P)=-({\partial }_j\psi )(x_P)$. Combining these signs with ${\gamma }^j{\gamma }^0=-{\gamma }^0{\gamma }^j$ provesThis establishes parity symmetry in quantum field theory for the free Dirac equation, rather than merely asserting its covariance.

A positive-energy plane wave $\psi =u(p)e^{-ip\cdot x}$ solves $(/p-m)u=0$. Write $u=(\chi ,\eta {)}^T$. The two block equations areThe second gives $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\chi /(E_p+m)$. In the first, $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2={\mathbf{p}}^2{I}_2$ and $E_p^2-{\mathbf{p}}^2=m^2$ make the remaining coefficient zero. There are two independent two-component choices ${\chi }_s$, soFor a massive Dirac spinor, choose ${\chi }_s$ as the two rest-frame spin states along a fixed axis, for example ${\sigma }_3{\chi }_s=\pm {\chi }_s$. The label $s$ distinguishes these canonical spin states after boosting; it is not automatically a helicity label.

Now $p\cdot x_P=p_P\cdot x$ and ${\gamma }^0{u}_s(p)=u_s(p_P)$ for the same fixed ${\chi }_s$. ThereforeThe parity action on canonical massive spin states leaves the canonical spin label unchanged while reversing the momentum. Spin is an axial vector under parity; its orientation is unchanged. Consequently helicity, the projection of spin along momentum, reverses. If $s$ had instead been chosen to denote helicity, the transformed label would have the opposite sign.

The Dirac adjoint is $\bar{\psi }={\psi }^{†}{\gamma }^0$, with ${\gamma }^{0†}={\gamma }^0$. ThusThe chirality matrix has ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3=\left(\begin{array}{cc}0& I_2\\ I_2& 0\end{array}\right)$ in this representation, so $({\gamma }^5{)}^2=I_4$ and it anticommutes with every ${\gamma }^{\mu }$. In particular ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$. The parity transformation of a Dirac bilinear therefore yieldsThe first bilinear is a scalar under parity, whereas the second is a pseudoscalar.

For the requested gamma matrix trace identities, use the cyclic property of the trace. Conjugation of a product of $n$ gamma matrices by ${\gamma }^5$ multiplies it by $(-1{)}^n$, but leaves its trace unchanged. All odd products therefore have zero trace. Taking the trace of the Clifford algebra relation and using $\mathrm{Tr}{I}_4=4$ gives $\mathrm{Tr}({\gamma }^{\alpha }{\gamma }^{\beta })=4g^{\alpha \beta }$. For four factors, move the first gamma matrix successively past the next three using the anticommutators. If $A=\mathrm{Tr}({\gamma }^{\alpha }{\gamma }^{\beta }{\gamma }^{\rho }{\gamma }^{\delta })$, this givesThus $A=4(g^{\alpha \beta }{g}^{\rho \delta }-g^{\alpha \rho }{g}^{\beta \delta }+g^{\alpha \delta }{g}^{\beta \rho })$. Contracting with the relevant four-vectors proves all four answers:

The original PDF contains three algebraic misprints in this question. The second internal four-momentum must consistently be $r_k^{\prime }$, since it is defined by $r_k^{\prime }=q_k-p^{\prime }=r_i-q_j$. Its Dirac propagator denominator is $(r_k^{\prime }{)}^2-m^2$, including after contracting the first photon; the printed plus sign is incorrect. Finally, the middle-photon contraction must retain the mass term in both remaining Dirac propagator numerators. The following calculation keeps the electron mass and proves the corrected identities.

Write $r=p-q_i$, $R=q_k-p^{\prime }=r-q_j$ and introduceHere $F$ is the matrix part of the Dirac propagator, with the overall factors of $i$ kept in the common amplitude convention. The Clifford algebra proves $D(l)F(l)=F(l)D(l)=I$. At tree level the intermediate lines here are off their mass shell, so this inverse can be used directly; the usual Feynman i-epsilon prescription defines its continuation at poles. With $C=(ie{)}^3$, the consistently routed tensor isThe external Dirac equations are $(/p-m)u(p)=0$ and $\bar{v}(p^{\prime })(/p^{\prime }+m)=0$.

For the photon at the electron end, $/q_i=/p-/r$ and hence $/q_{i}u=-D(r)u$. Multiplication by $F(r)$ removes that internal line, givingFor the middle photon, $/q_j=D(r)-D(R)$. Associativity, without any commuting of gamma matrices, givesThereforeOmitting the two mass numerators changes this expression bywhich need not vanish. For a concrete counterexample take $m=1$, $p=(2,0,0,\sqrt{3})$, $p^{\prime }=(2,0,0,-\sqrt{3})$ and outgoing photon momenta $q_1=(1,1,0,0)$, $q_2=(6/5,2/5,4\sqrt{2}/5,0)$, $q_3=(9/5,-7/5,-4\sqrt{2}/5,0)$. All external momenta are on shell and conserve four-momentum. Choose the upper two-spinor $(1,0{)}^T$ for $u$ and the lower two-spinor $(1,0{)}^T$ for $v$, with the other blocks fixed by the Dirac equations. For $(i,j,k)=(1,2,3)$ and $(\mu ,\sigma )=(0,3)$, the correct middle contraction divided by $C$ is $1/3$, whereas the expression with the mass numerators omitted is $5/27$. Thus the printed middle identity is not valid for a massive Electron.

At the positron end, $q_k=R+p^{\prime }$. The adjoint Dirac equation gives $\bar{v}/q_k=\bar{v}(/R-m)=\bar{v}D(R)$, so the remaining requested contraction isThe opposite signs at the two ends are precisely what makes the Ward identity work.

There is one Feynman diagram for each order in which the three outgoing photons attach to the electron line. Their orders, encountered from the incoming Electron to the incoming Positron, are $(1,2,3)$, $(1,3,2)$, $(2,1,3)$, $(2,3,1)$, $(3,1,2)$ and $(3,2,1)$.

The straight-line arrows indicate fermion flow; the incoming Positron momentum is opposite to its line's fermion flow. Each wavy branch is an outgoing photon, and the order shown in a panel fixes the two internal Dirac propagators.

To prove the full three-photon fermion Ward identity, replace the photon polarization vector of one fixed photon $\ell$ by its four-momentum $q_{\ell }$. Fix the order $(a,b)$ of the other two photons and write ${\Gamma }_a=/{ϵ}_a$, ${\Gamma }_b=/{ϵ}_b$. There are exactly three positions for photon $\ell$ relative to that order. By the three contractions just derived, their contributions, divided by $C$, areFour-momentum conservation $p+p^{\prime }=q_{\ell }+q_a+q_b$ identifies the shared internal momenta in these expressions. The three terms cancel exactly. Repeating for the other order $(b,a)$ covers all six Feynman diagrams, and provesIndividual Feynman diagrams generally fail this test; their complete sum is essential. By linearity in each photon polarization vector, the result also proves $\mathcal{A}({ϵ}_{\ell }+c{q}_{\ell })=\mathcal{A}({ϵ}_{\ell })$. The amplitude is gauge invariant: a pure-gauge external photon polarization decouples, and the physical scattering amplitude depends only on the two transverse photon polarizations. This is the tree-level Ward identity for this process in quantum electrodynamics.

The RNS string adds worldsheet Majorana fermions ${\psi }^{\mu }$ to the coordinate fields $X^{\mu }$. The new fields are spinors on the worldsheet but vectors under the target Lorentz group; this distinction matters when interpreting their oscillator states. In a flat target, a convenient Lorentzian convention for the free gauge-fixed action iswhere ${\rho }^a$ are two-dimensional gamma matrices. The bosonic and fermionic kinetic terms are related by worldsheet supersymmetry. Before fixing local worldsheet supersymmetry, one introduces a worldsheet gravitino along with the worldsheet metric. After superconformal gauge fixing, the residual matter constraints are the Virasoro constraints and the constraints generated by the superconformal current $G\propto \psi \cdot \partial X$.

Each real worldsheet Majorana fermion contributes $1/2$ to the chiral central charge, while each coordinate contributes one. The reparameterization bc system contributes $-26$, and fixing local worldsheet supersymmetry also produces commuting superconformal ghosts of weights $(3/2,-1/2)$, with central charge $11$. ThusIn light-cone gauge in string theory there are eight physical transverse coordinate fields and eight transverse worldsheet Majorana fermions.

On the doubled spatial interval of an open string, the two allowed fermionic periodicities areThey arise from the relative signs of the left-right fermion gluing conditions at the two ends. Thus the Ramond sector has integer-mode oscillators $d_n$, including $d_0$, and the Neveu–Schwarz sector has half-integer oscillators $b_r$, $r\in \mathbb{Z}+1/2$, without zero modes. In either sector,A closed string has independent left- and right-moving periodicities, giving four sector choices rather than two.

The Ramond zero-mode Clifford algebra isHence its ground states carry a Spinor representation of the Lorentz group. The zero-mode supercurrent constraint reduces on a ground state to the Dirac equation $p_{\mu }{\Gamma }^{\mu }u=0$. In the Neveu–Schwarz sector the oscillator vacuum instead carries a scalar representation. Consequently the open string NS sector describes spacetime bosons, while the R sector describes spacetime fermions. Exciting a target-vector oscillator does not change this statistics assignment.

The string intercepts follow from the transverse zero-point energy. Per boson the integer-mode contribution is $\frac{1}{2}\zeta (-1)=-1/24$. Per fermion it is the negative of the corresponding half-sum. Since $\zeta (-1,1/2)=1/24$, the eight transverse pairs giveThus $a_{\mathrm{NS}}=1/2$, $a_{\mathrm{R}}=0$, and the open string mass formula is ${\alpha }^{\prime }{M}^2=N-a$. This distinguishes the lowest states explicitly:

The supersymmetric GSO projection keeps odd oscillator fermion number in the NS sector. Defining $F_{\mathrm{osc}}$ so that the NS vacuum has $F_{\mathrm{osc}}=0$, its projector isIt removes the scalar tachyon and keeps the massless vector. In the R sector, fermion parity contains the zero-mode chirality operator, so one may writeHere $F_{\mathrm{osc}}$ counts nonzero-mode fermionic excitations. On the ground states, this keeps one Majorana-Weyl spinor chirality: 16 real covariant components, or eight on-shell polarizations after the Dirac equation. The projected massless open-string states are an eight-polarization gauge vector and an eight-polarization gaugino. The compatible projected tower exhibits spacetime supersymmetry; unprojected worldsheet supersymmetry by itself did not eliminate the tachyon or equalize these massless counts.

For a closed string, both chiral zero-mode constraints must hold:This shifted closed-string level matching is particularly important in the mixed sectors. The lowest states and their spacetime nature are as follows.

For type IIA superstring theory, the two R sector ground-state chiralities are opposite. The Ramond–Ramond potentials can be represented by $C_1,C_3$, with transverse counts $8+56=64$. The two mixed-sector gravitinos have opposite chiralities, and the ten-dimensional theory is nonchiral. For type IIB superstring theory, the two R sector chiralities agree; the Ramond–Ramond potentials are $C_0,C_2,C_4$, with self-dual five-form field strength, giving $1+28+35=64$. Its two gravitinos have the same chirality. Thus both projected type II theories have 128 massless bosonic and 128 massless fermionic polarizations and no tachyon, while their chiralities and form spectra differ.

If unoriented open strings are included in the complete theory, the supersymmetric example is type I string theory. Worldsheet orientation reversal projects the closed type IIB superstring theory spectrum and identifies the two mixed sectors; the surviving closed fields include the graviton, dilaton, a Ramond–Ramond potential $C_2$, and one gravitino and dilatino. The open states form vector and gaugino multiplets with Chan-Paton factors; tadpole and anomaly cancellation select $SO(32)$. This is an additional orientation projection, not part of the GSO projection itself. The heterotic string instead pairs one supersymmetric chiral side with a bosonic chiral side carrying an internal gauge sector; it does not have a Ramond choice on both sides.

Use signature $(-,+,+,+)$ and $c=1$. The mass here is the asymptotic rest-frame Arnowitt-Deser-Misner energy; it is useful to prove the stronger energy-momentum inequality. Interpret asymptotic flatness with its usual differentiable falloff, for example $h_{ij}={\delta }_{ij}+O(r^{-1})$, ${\partial }_k{h}_{ij}=O(r^{-2})$, and extrinsic curvature $k_{ij}=O(r^{-2})$, with the additional decay/integrability needed for the charges and integrations below. The complete nonsingular slice has no inner boundary. An oriented three-dimensional slice admits a spin structure. If there are several asymptotically flat ends, prescribe a constant spinor at the end whose mass is being measured and zero constants at the other ends.

Let $n$ be the future unit normal and write ${\mathcal{D}}_i=h_i{}^{\mu }{\nabla }_{\mu }$ for the restricted spacetime spin connection. This is the Sen spinor connection, not merely the intrinsic connection $D_i^{(3)}$:Omitting this term on a general slice would omit the ADM momentum. Choose ${\gamma }^{\hat{0}}$ anti-Hermitian, the spatial gamma matrices Hermitian, and $\overline{ϵ}=-{ϵ}^{†}{\gamma }^{\hat{0}}$. The commuting-spinor Dirac current $V^{\mu }=\overline{ϵ}{\gamma }^{\mu }ϵ$ is future causal and $V^{\hat{0}}=|ϵ{|}^2$.

The Nester two-form supplies the needed integration identity. DefineHere multiple gamma indices denote antisymmetrized products. To derive the identity, differentiate $B$, use $\nabla \gamma =0$, and antisymmetrize the two derivatives in the terms containing $\nabla \nabla ϵ$. Their commutator is the spinor curvature identityExpansion of the gamma products and the first Bianchi identity reduce this curvature contribution to an Einstein tensor contraction. It vanishes for the Vacuum Einstein equations. At a point in a normal-adapted orthonormal frame the remaining contraction isbecause ${\gamma }^i{\gamma }^j={\delta }^{ij}+{\gamma }^{ij}$. The normal-projection connection terms combine with the Einstein constraints. Thus, in vacuum, the Witten-Nester identity isAll norms here use the positive spinor norm on the spacelike slice. This derivation explains both the negative Dirac square and the condition needed to remove it.

Solve the Witten spinor equation $\mathcal{D}ϵ=0$ with $ϵ\to {ϵ}_{\infty }$. The existence step is part of the argument. Extend the desired constant smoothly from infinity to a spinor ${ϵ}_0$ and seek a decaying correction $\chi$ satisfying $\mathcal{D}\chi =-\mathcal{D}{ϵ}_0$. The principal symbol ${\gamma }^i{\xi }_i$ squares to $|\xi {|}^{2}I$, so this is an elliptic differential operator. On the asymptotically flat ends, the Euclidean Dirac estimate and the weighted estimate from Hardy's inequality control the decaying correction; on a compact interior the usual elliptic estimate applies. Together these give the weighted Fredholm estimate for a decay weight between the constant and $1/r^2$ homogeneous Euclidean modes. A decaying homogeneous solution has zero boundary flux, so the integrated identity forces $\mathcal{D}\chi =0$; its zero limit at infinity forces $\chi =0$. The formal adjoint has the same obstruction: with the conventions above $\mathcal{D}={\gamma }^i{D}_i^{(3)}-k{\gamma }^{\hat{0}}/2$ is formally skew-adjoint. Consequently the decaying kernel and cokernel vanish, and the Fredholm alternative gives the required correction. Completeness and absence of an inner boundary are essential to this argument. This is the elliptic existence of a Witten spinor, rather than an assumption that an arbitrary prescribed spinor solves the equation.