Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 56 (Supergravity)

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.

Choose the Riemann curvature tensor conventionwith positive scalar curvature for a round sphere. The connection preserves the gamma matrices. If $D={\gamma }^{\mu }{\nabla }_{\mu }$, its square isHere ${\nabla }^2$ is the rough Laplacian, including the connection on the derivative index. In the product of two antisymmetric gamma pairs, the fully antisymmetric four-gamma term vanishes by the first Bianchi identity. The two-gamma contractions vanish because the Ricci tensor is symmetric, while the scalar contractions give $-2R$. Thus the Lichnerowicz spinor-square formula isMultiplying $(D+m)\psi =0$ by $-D+m$, with constant $m$, provesIn Riemannian signature $D$ in this positive-Clifford convention is formally anti-self-adjoint; the more usual self-adjoint Dirac operator $iD$ has square $-{\nabla }^2+R/4$. Stating this convention avoids an apparent sign conflict between the two forms of the formula.

Now let $D\psi =0$ on the compact Riemannian spin manifold. Use the positive-definite spinor inner product and integrate the squared equation. Integration by parts has no boundary contribution, givingBoth terms are nonnegative. Smoothness then forces $\nabla \psi =0$ pointwise. Consequently every harmonic spinor is parallel, and $R|\psi {|}^2=0$. A nonzero parallel spinor has constant positive norm on its connected component, so its scalar curvature vanishes there. This is compact harmonic-spinor rigidity under nonnegative scalar curvature.

The full Ricci conclusion follows from integrability, not merely from vanishing scalar curvature. Since a parallel spinor has zero curvature for its spin connection, contract its Ricci identity with ${\gamma }^{\nu }$:The three-gamma term vanishes by the Bianchi identity; the remaining contractions give the displayed Ricci term. For each fixed $\mu$, put $v_{\nu }=R_{\mu \nu }$. Clifford multiplication twice givesPositive definiteness and $\psi \ne 0$ imply $v=0$. Thus the Riemannian parallel-spinor Ricci-flatness result isOn a connected manifold a nontrivial parallel solution is nowhere zero, so this holds everywhere. On a disconnected manifold it holds on every component supporting a nonzero spinor; the stated nowhere-vanishing hypothesis ensures all components are covered. The positive-definite metric tensor is essential to the last inference.

Use the mostly-plus Clifford algebra and Riemann curvature tensor conventions of the preceding solutions, with $e=\det (e_{\mu }{}^a)$, ${\kappa }^2=8\pi G$, and a Majorana gravitino. An action for minimal four-dimensional supergravity in the exact transformation normalization given isThe kinetic coefficient is tied to the normalization of the supersymmetry transformations: if ${\Psi }_{\mu }=2{\psi }_{\mu }$ and $\eta =2ϵ$, this becomes the canonical coefficient $-1/2$, with $\delta e_{\mu }{}^a=(\kappa /2)\bar{\eta }{\gamma }^a{\Psi }_{\mu }$ and $\delta {\Psi }_{\mu }={\nabla }_{\mu }\eta /\kappa$. Mixing those two normalizations would spoil the cancellation.

To the retained order, use the torsion-free connection. More generally the 1.5-order formalism allows its variation to be omitted after imposing its own algebraic equation; the resulting contorsion is quadratic in fermions and first affects the displayed action at quartic order. The metric tensor variation induced by the vierbein is $\delta g_{\mu \nu }=4\kappa \overline{ϵ}{\gamma }_{(\mu }{\psi }_{\nu )}$, so the Einstein-Hilbert action varies asFor the fermionic action, varying both Majorana factors gives the same contribution after integration by parts and the Grassmann bilinear interchange rule. ThusThe vector-index Riemann curvature tensor contribution vanishes by the first Bianchi identity. The spin-index contribution is determined byFor the last equality, expand the antisymmetric three-gamma/two-gamma product: the totally antisymmetric Riemann curvature tensor terms vanish by Bianchi, and the surviving terms are the Ricci contraction minus half its trace. Hencewhich cancels the gravitational variation. Variations of vierbeins, gamma matrices and the connection in the fermionic kinetic term carry three dynamical fermions. They lie beyond this linear-fermion cancellation and are cancelled by the omitted four-fermion completion. Thus the action is invariant up to boundary terms at the order specified, rather than claiming that a truncated action is exactly invariant at every fermion order.

For the cosmological extension, add a gravitino mass parameter $m$ and deform the derivative:Indeed the fermion terms combine as $-2{\bar{\psi }}_{\mu }{\gamma }^{\mu \nu \rho }{\mathcal{D}}_{\nu }{\psi }_{\rho }$ because ${\gamma }^{\mu \nu \rho }{\gamma }_{\nu }=-2{\gamma }^{\mu \rho }$. The modified spinor connection curvature isand contraction givesThe same action-variation cancellation therefore works with $G$ replaced by $G+\Lambda g$ precisely forThis is the Anti-de Sitter deformation of minimal supergravity. A cosmological term alone would leave its variation uncancelled; the mass term and the gamma shift of the gravitino transformation are both necessary. A positive cosmological constant cannot arise from this real unbroken minimal deformation.

For electromagnetism, retain $\mathcal{N}=1$ and introduce an Abelian vector multiplet $(A_{\mu },\lambda )$, with Majorana gaugino $\lambda$. A complete quadratic-fermion extension isThe accompanying leading transformations areThey preserve the rigid Maxwell-gaugino action for constant $ϵ$. For a local parameter, the variation containsTo obtain this, vary $F^2$, vary both gaugino factors, and use ${\gamma }^{ab}{\gamma }^{\mu }={\gamma }^{ab\mu }+g^{b\mu }{\gamma }^a-g^{a\mu }{\gamma }^b$. The terms with $\nabla \lambda$ cancel between the two actions; integrating the three-gamma term uses ${\nabla }_{[\mu }{F}_{\nu \rho ]}=0$, leaving exactly the displayed derivative of the parameter. The interaction is $-\kappa e{\bar{\psi }}_{\mu }{j}^{\mu }$, whose variation under $\delta {\bar{\psi }}_{\mu }={\nabla }_{\mu }\overline{ϵ}/\kappa$ cancels that term.

The other linear-fermion variation is also fixed. Put $\mathcal{F}=F_{ab}{\gamma }^{ab}$ and $T_{\mathrm{EM}}^{\mu \nu }=F^{\mu \rho }{F}^{\nu }{}_{\rho }-g^{\mu \nu }{F}^2/4$. Clifford multiplication givesVarying $\lambda$ in the gravitino-current interaction then gives $2\kappa e{T}_{\mathrm{EM}}^{\mu \nu }{\bar{\psi }}_{\mu }{\gamma }_{\nu }ϵ=-2\kappa e{T}_{\mathrm{EM}}^{\mu \nu }\overline{ϵ}{\gamma }_{\nu }{\psi }_{\mu }$, which cancels the Maxwell vierbein variation. This proves the required supergravity coupling of an Abelian vector multiplet through the retained order.

Beyond that order one uses a supercovariant gauge field strength, here ${\hat{F}}_{\mu \nu }=F_{\mu \nu }-2\kappa {\bar{\psi }}_{[\mu }{\gamma }_{\nu ]}\lambda$, and the associated higher-fermion transformation terms and four-fermion interactions. An auxiliary scalar may be retained for off-shell closure, or eliminated in this on-shell formulation. A Maxwell field without a fermionic partner cannot be coupled by merely adding $-F^2/4$ while leaving all transformations unchanged. An alternative is an $\mathcal{N}=2$ gravity multiplet, where the vector is a graviphoton and a second gravitino supplies its extra fermionic states; the explicit construction above is the minimal $\mathcal{N}=1$ matter extension.

A Killing spinor is a nonzero spinor parameter for which the fermionic supersymmetry variations vanish on a bosonic background. In minimal four-dimensional supergravity without a cosmological term this meansIt is an overdetermined equation and expresses an unbroken supersymmetry of the background. In the cosmological deformation the corresponding equation is ${\mathcal{D}}_{\mu }ϵ=0$; in theories with additional bosonic fields their contributions also enter the fermion variations.

For the undeformed theory the connection preserves both the spinor adjoint and gamma matrices. Consequently the Dirac currentobeys ${\nabla }_{\nu }{K}_{\mu }=0$. In particular ${\nabla }_{(\nu }{K}_{\mu )}=0$, so $K$ is a Killing vector, and in fact is parallel. For the real cosmological deformation, ${\nabla }_{\mu }ϵ=-(m/2){\gamma }_{\mu }ϵ$ and ${\nabla }_{\mu }\overline{ϵ}=(m/2)\overline{ϵ}{\gamma }_{\mu }$ instead give ${\nabla }_{\mu }{K}_{\nu }=(m/2)\overline{ϵ}[{\gamma }_{\mu },{\gamma }_{\nu }]ϵ$, whose symmetric part again vanishes.

The causal property uses an ordinary commuting spinor. Take an orthonormal frame with $({\gamma }^0{)}^{†}=-{\gamma }^0$, $({\gamma }^i{)}^{†}={\gamma }^i$, and $\overline{ϵ}={ϵ}^{†}{\gamma }^0$. Then $K^0=-{ϵ}^{†}ϵ$. Reverse its time orientation and write $V=-K$, soThe matrices ${\alpha }^i$ are Hermitian and obey $\{{\alpha }^i,{\alpha }^j\}=2{\delta }^{ij}$. For every spatial unit vector $u$, $(u_i{\alpha }^i{)}^2=1$, whence $|u_i{V}^i|\le V^0$. Choose $u$ along the spatial part of $V$ to obtainA nonzero Killing spinor cannot vanish at an isolated point: its defining connection equation transports its value along every curve. Thus on a connected background this is a nonzero, nowhere-spacelike causal Dirac spinor current. The estimate does not need the Majorana condition, but it does need commuting components; Grassmann bilinears have no such ordinary positivity order.

For the positive energy theorem, take a complete asymptotically flat spacelike hypersurface with a spin structure $\Sigma$, with suitable decay, Einstein constraint equations, and matter satisfying the dominant energy condition. Assume there are no inner boundaries, or that any inner-boundary contribution has the required nonnegative sign. The auxiliary commuting spinor approaches an arbitrary constant ${ϵ}_{\infty }$ at spatial infinity. Define the real Nester two-formAntisymmetry removes second derivatives except through the Riemann curvature tensor. Differentiation and the three-gamma spinor curvature identity used above giveThe first term comes from differentiating the two spinor factors. For the second term, replace ${\nabla }_{\nu }{\nabla }_{\rho }$ by half their commutator; its spin-curvature contraction is $G^{\mu \nu }{\gamma }_{\nu }/2$ from each conjugate term. This is the reason the Einstein tensor, and hence the matter energy density, enters the boundary-charge identity.

In an orthonormal frame adapted to the future normal of $\Sigma$, put ${\chi }_i={\nabla }_iϵ$. Since ${\gamma }^0{\gamma }^{0ij}=-{\gamma }^{ij}$,Here ${\nabla }_i$ is the spacetime spin connection pulled back to $\Sigma$, including its extrinsic curvature. Choose the auxiliary spinor to solve the elliptic Witten spinor equationThe standard asymptotic-flatness and completeness hypotheses permit this boundary-value problem; the positivity identity also excludes a decaying homogeneous zero mode. It is this hypersurface equation that removes the negative square. A spacetime Killing spinor is not assumed for a general initial-data set.

Stokes' theorem and $G_{\mu \nu }={\kappa }^2{T}_{\mu \nu }$ now giveThe dominant energy condition makes the second integrand nonnegative because $n$ and $V$ are future causal. Evaluating the two-form with the asymptotic spin connection identifies its surface integral with the ADM energy and momentum:For example its energy term uses $B^{0i}=\frac{1}{2}({\partial }_j{h}_{ij}-{\partial }_i{h}_{jj})|{ϵ}_{\infty }{|}^2$ to first order in the asymptotic spatial metric tensor perturbation. With ${\kappa }^2=8\pi G$ this is exactly the ADM normalization. Since $P_i{\alpha }^i$ has eigenvalues $\pm |\mathbf{P}|$, positivity for every boundary spinor provesKilling spinors describe saturation: their spatial derivative term vanishes, and the contracted matter term must also vanish for the corresponding charge to be zero. They characterize preserved supersymmetry rather than being prerequisites for positivity. Under the usual rigidity hypotheses, zero total four-momentum makes the nonnegative integral vanish for a full basis of asymptotic spinors; their parallel extensions force a vanishing Riemann curvature tensor and give the Minkowski vacuum. A single Killing spinor with null current alone does not justify asserting that every component of the total momentum vanishes.

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.