Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 60 (Supergravity)

Use signature $(-,+,+,+)$, $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$, and unit-weight antisymmetrized gamma products. The massless Rarita-Schwinger field has the gauge freedomsince the antisymmetric ${\gamma }^{\mu \nu \rho }$ annihilates ${\partial }_{\nu }{\partial }_{\rho }ϵ$. This gauge freedom is essential to remove the lower-spin components of the vector-spinor.

Put $\chi ={\gamma }^{\mu }{\psi }_{\mu }$. Contracting the field equation with ${\gamma }_{\mu }$ and using ${\gamma }_{\mu }{\gamma }^{\mu \nu \rho }=2{\gamma }^{\nu \rho }$ givesExpanding the original three-gamma product then reduces its equation toA gauge transformation changes $\chi$ by $/\partial ϵ$. Locally solve $/\partial ϵ=-\chi$ to impose gamma-trace gauge. The two equations becomeApplying $/\partial$ again gives $\square {\psi }_{\mu }=0$. Nonzero propagating Fourier modes therefore have $p^2=0$: the physical modes are massless.

For the massless Rarita-Schwinger polarization count, choose a nonzero future null momentum $p^{\mu }=(\omega ,0,0,\omega )$. The remaining gauge parameters satisfy $/pϵ=0$. Transversality gives ${\psi }_0+{\psi }_3=0$, and this residual gauge removes ${\psi }_0$, hence also ${\psi }_3$, because ${\psi }_0$ itself satisfies $/p{\psi }_0=0$. Only ${\psi }_1,{\psi }_2$ remain, withThe kernel of $/p$ on four-component spinors has complex dimension two, so there are exactly two independent polarization amplitudes.

To identify their helicities, the spinor rotation generator about the third axis in this convention is $S_3=-i{\gamma }^1{\gamma }^2/2$. Its two eigenvalues on that kernel are $h_s=\pm 1/2$. If ${\psi }_1$ is such an eigenvector, then ${\psi }_2=2i{h}_s{\psi }_1$. Its transverse vector polarization is consequently $(1,+i)$ for $h_s=+1/2$ or $(1,-i)$ for $h_s=-1/2$. These carry vector helicity $+1$ or $-1$, respectively. Adding the vector and spinor angular momenta leavesThe unwanted opposite-sign vector-spinor combinations, which would have helicity $\pm 1/2$, violate the gamma-trace constraint. For the gravitino of simple supergravity, its Majorana spinor reality condition identifies the negative-frequency antiparticle modes, giving two propagating degrees of freedom. If one instead chooses a complex Dirac vector-spinor, its independent antiparticle sector doubles the particle-plus-antiparticle count; the equation alone does not impose a Majorana condition.

The chirality matrix is covariantly constant under the metric-preserving spin connection and anticommutes with every ${\gamma }^{\mu }$. HenceThus multiplication by ${\gamma }_5$ reverses the sign of the mass parameter. At $m=0$ it maps solutions into solutions of the same Dirac equation.

Use the spinor curvature identityHere the rough Laplacian includes the affine connection on the index of the first derivative. Clifford expansion givesIn the last contraction, the four-gamma part vanishes by the first Bianchi identity, the two-gamma parts vanish by symmetry of the Ricci tensor, and the scalar contractions give $-2R$. Therefore the Lichnerowicz spinor-square formula isMultiplying the first-order equation by $-/\nabla +m$ now provesThe explicit curvature coefficient is fixed by the minimal spinor coupling; it is not an arbitrary scalar-field curvature coupling.

For the general metric-preserving connection, define $T^{\rho }{}_{\mu \nu }={\Gamma }_{\mu \nu }^{\rho }-{\Gamma }_{\nu \mu }^{\rho }$ and let $R^T$ be its curvature. Distinguish the spin connection acting on the spinor from the full second derivative acting also on its derivative index. Their antisymmetric relation isThus the torsion correction to the Dirac square gives the exact squared field equationGeneral torsion removes the curvature symmetries used in the preceding simplification. The curvature contraction may now contain nonscalar Clifford terms, in addition to the displayed first-order torsion term; simply replacing $R$ by a torsionful scalar in the torsion-free answer is not valid.

Equivalently, write the connection as Levi-Civita plus contorsion tensor $C_{\mu ab}$. Then $/{\nabla }_T=/{\nabla }_0+Q$, where $Q={\gamma }^{\mu }{C}_{\mu ab}{\gamma }^{ab}/4$, and the square contains $\{/{\nabla }_0,Q\}+Q^2$. This exhibits the extra couplings through derivatives of contorsion, first-order spinor derivatives and quadratic contorsion. Metric compatibility still preserves ${\gamma }_5$, so the mass-sign result remains valid. Spin therefore couples directly to curvature through its connection and the universal $R/4$ term, and to torsion through additional spin-dependent terms. The scalar reduction in the torsion-free square does not mean the full spinor transport is insensitive to the remaining Riemann curvature.

Work through terms quadratic in the fermions, with Grassmann-odd Majorana spinors ${\psi }_{\mu }$ and $ϵ$. At this order the spin connection can be taken torsion-free: eliminating its gravitino-induced torsion changes the action only at four-fermion order. The Christoffel term needed to turn the printed $D_{\nu }{\psi }_{\sigma }$ into a full vector-spinor covariant derivative drops out after contraction with ${\gamma }^{\mu \nu \sigma }$, since the connection is symmetric in $\nu ,\sigma$.

First establish the Einstein tensor contraction of spin curvature. Expanding the Clifford product and using the first Bianchi identity, its five-gamma and antisymmetric three-gamma curvature contributions vanish, leavingSince $[{\nabla }_{\nu },{\nabla }_{\rho }]=R_{\nu \rho ab}{\gamma }^{ab}/4$, it follows thatThe factor $1/2$ arises from antisymmetrizing the two derivatives, not from a choice of Einstein-Hilbert normalization.

For the Einstein-Hilbert action, the metric variation isVariation with respect to the covariant metric, including the determinant variation and discarding the usual boundary divergence, givesNo gravitational field equation has been assumed here.

For the Rarita-Schwinger field action, the determinant, gamma-matrix and connection variations each multiply a term already quadratic in $\psi$. Since $\delta e$ is bilinear in $\psi ,ϵ$, these contributions have four fermionic factors and are outside the requested order. Retain only the variations of the two explicit gravitino factors. The four-dimensional Majorana bilinear interchange gives $\bar{\chi }{\gamma }^{\mu \nu \rho }\eta =\bar{\eta }{\gamma }^{\mu \nu \rho }\chi$ for odd spinors. After integration by parts, the first gravitino variation iswhere the last step swaps $\mu ,\rho$ and reverses the three-gamma sign. It is exactly equal to the variation of the second gravitino factor. ThereforeThe curvature contribution cancels $\delta S_{\mathrm{EH}}$ pointwise. Thusup to boundary terms and the omitted four-fermion-order variations. This directly proves the stated local supersymmetry invariance with its printed coefficients. At higher order, the torsion and quartic-fermion completion of minimal four-dimensional supergravity is needed; it cannot be inferred by dropping the terms indefinitely.

On a bosonic background, a Killing spinor is a nonzero spinor parameter that makes all fermionic supersymmetry variations vanish. It identifies an unbroken supersymmetry of the background; the number of independent solutions counts preserved supercharges. In simple ungauged minimal four-dimensional supergravity, with the gravitino set to zero, its equation is ${\nabla }_{\mu }ϵ=0$ and the bosonic field equations are $R_{\mu \nu }=0$. In theories with additional fields or a cosmological deformation, the Killing-spinor equation instead uses the corresponding modified connection.

Use commuting spinors to form ordinary geometric bilinears, and orient the Dirac current to the future with $\overline{ϵ}=-{ϵ}^{†}{\gamma }^{\hat{0}}$ in mostly-plus signature. Since the spinor and gamma matrices are parallel,Thus $K$ is actually parallel, and in particular is a Killing vector field. In a local orthonormal frame,For every spatial unit vector $u$, $u_i{\alpha }^i$ is Hermitian and squares to one. Its eigenvalues are $\pm 1$, so $|u_i{K}^i|\le K^0$. Taking $u$ along the spatial current givesA nonzero parallel spinor cannot vanish at an isolated point, so this causal Dirac current is nonzero everywhere on a connected solution. Its squared norm is constant: it cannot alternate between timelike and null regions.

If $K$ is timelike, normalize its constant norm. Its parallel dual one-form is closed, and its orthogonal distribution is integrable. Adapted local coordinates put the metric in the product formThe time independence follows because $K$ is Killing, and the unit lapse and zero cross terms follow from parallelism. The vacuum equations give $R_{ij}(h)=0$. In three dimensions the Riemann tensor is completely determined by the Ricci tensor and its scalar contraction, so $h$ is flat. ThereforeConstant spinors in its flat frame provide all local Killing spinors. Global identifications are possible only when they preserve a compatible spin structure and the required spinors.

If $K$ is null, its parallelism first gives a Brinkmann coordinate system with $K={\partial }_v$. The parallel-spinor condition further restricts the local spin holonomy to the stabilizer of that spinor. In four dimensions, for a nonzero Weyl component represented by $(1,0{)}^T$, its stabilizer in $SL(2,\mathbb{C})$ is the upper-triangular group $\left(\begin{array}{cc}1& b\\ 0& 1\end{array}\right)$, $b\in \mathbb{C}$: the two real null rotations, with no transverse rotation. Consequently the transverse frame is flat and the allowed curvature is of the wave type $R_{uiuj}$. Using the coordinate freedom to remove transverse cross terms gives the local plane-fronted gravitational wave formHere $R_{uiuj}=-H_{ij}/2$ and the only potentially nonzero Ricci component is $R_{uu}=-(H_{xx}+H_{yy})/2$, so the last equation is exactly the vacuum equation. Conversely, in the coframe $e^{+}=du$, $e^{-}=dv-Hdu/2$, $e^i=d{x}^i$, the nonzero spin-connection matrices are proportional to $H_i{\gamma }^{+i}du$. Constant spinors with ${\gamma }^{+}ϵ=0$ are therefore parallel. Every harmonic transverse profile, with arbitrary $u$ dependence, gives such a supersymmetric vacuum wave; the flat profile is included. This verifies both the form and sufficiency in the parallel-spinor classification of vacuum four-geometries.

There is a reality-condition subtlety in interpreting the branches. A single commuting four-dimensional Majorana spinor has a null current of a commuting Majorana spinor: its two-component representation gives $|{\xi }^{†}\mathbf{\sigma }\xi {|}^2=({\xi }^{†}\xi {)}^2$. A timelike current can instead come from a Dirac Killing spinor or a sum of independent real-spinor currents. The conditional timelike classification above remains valid, but does not assert a generic timelike current for a single real Majorana parameter.

For the final question use constant sectional curvature $-1/a^2$, so $R_{\mu \nu }=-3g_{\mu \nu }/a^2$. The spinor curvature identity givesDefine the Anti-de Sitter Killing-spinor connection ${\mathcal{D}}_{\mu }={\nabla }_{\mu }+{\gamma }_{\mu }/(2a)$. Since ${\nabla }_{\mu }{\gamma }_{\nu }=0$ and the connection is torsion-free,Thus the modified connection is flat. On a simply connected region, transport any initial spinor with this connection; flatness makes the result path independent and supplies actual nonzero solutions, not merely a necessary integrability condition. Hencehas a full local space of solutions on Anti-de Sitter spacetime, and globally on its universal cover. Global quotients require compatible spinor monodromy. These are Killing spinors of the cosmological deformation, not parallel-spinor vacua of the undeformed zero-cosmological-constant theory.

Take a complete smooth asymptotically flat spacelike initial-data hypersurface $\Sigma$, carrying a spin structure, without inner boundary. Let $n$ be its future unit normal, $h$ its induced metric, and use the extrinsic curvature convention $k_{ij}=-g({\nabla }_{i}n,e_j)$. Assume the Einstein constraints, the dominant energy condition, and the standard decay sufficient for finite ADM charges. In one asymptotically Cartesian end it is enough here to work with $h_{ij}={\delta }_{ij}+q_{ij}$, $q=O(r^{-1})$, $\partial q=O(r^{-2})$, $k=O(r^{-2})$, and the corresponding differentiated decay and integrability assumptions. The desired conclusion is positivity of the whole ADM four-momentum, not merely of the energy in one selected frame.

Introduce a commuting auxiliary spinor $ϵ$; this is not an anticommuting gravitino field. With $\overline{ϵ}=-{ϵ}^{†}{\gamma }^{\hat{0}}$ its causal Dirac spinor current is $K^{\mu }=\overline{ϵ}{\gamma }^{\mu }ϵ$, future causal. The Nester two-form isIt is real in these conventions and antisymmetric in $\mu ,\nu$. Differentiating, the two terms quadratic in first derivatives become equal after exchanging $\nu ,\rho$. The remaining terms are curvature commutators. The Einstein tensor contraction of spin curvature gives ${\gamma }^{\mu \nu \rho }{\nabla }_{\nu }{\nabla }_{\rho }=G^{\mu }{}_{\lambda }{\gamma }^{\lambda }/2$; the adjoint term supplies the other half. ThusThis is the central spinor-curvature identity, derived by differentiation rather than posited as a positivity formula.

Project onto $\Sigma$ and put $b^i=n_{\mu }{B}^{\mu i}$. In an adapted orthonormal frame $n_{\mu }=(-1,0,0,0)$, Clifford algebra givesIndeed the expansion of $|{\gamma }^i{\nabla }_iϵ{|}^2$ is ${\sum }_i|{\nabla }_iϵ{|}^2+({\nabla }_iϵ{)}^{†}{\gamma }^{ij}{\nabla }_jϵ$. With a Gaussian normal extension of the slice, the symmetric second fundamental form has zero contraction with the antisymmetric $B$, so the projected divergence is the intrinsic spatial divergence. This proves the Witten-Nester identityHere $D_i$ on $b$ denotes the spatial metric derivative. On spinors, ${\nabla }_i$ is the Sen spinor connection, the restriction of the spacetime connection, not just the intrinsic three-dimensional one:The extrinsic-curvature term is essential when the data have momentum.

Choose $ϵ$ to solve the Witten spinor equation, $\mathcal{D}ϵ=0$, approaching a prescribed constant ${ϵ}_{\infty }$ in the chosen end and zero constants in any other ends. The Einstein constraints identify $G_{\mu \nu }{n}^{\mu }{K}^{\nu }=8\pi G{T}_{\mu \nu }{n}^{\mu }{K}^{\nu }$. The dominant energy condition makes this nonnegative because both $n$ and $K$ are future causal. Consequently the right-hand side of the identity becomes a sum of nonnegative terms.

Existence of that spinor is an analytic part of the proof, not something established by the positivity identity alone. Extend ${ϵ}_{\infty }$ smoothly using a cutoff and solve $\mathcal{D}u=-\mathcal{D}{ϵ}_0$ for a decaying correction. The spatial principal symbol ${\gamma }^i{\xi }_i$ squares to $|\xi {|}_h^2$, so the equation is elliptic. For a compactly supported spinor, the integrated identity givesTogether with the weighted asymptotic estimates this gives the coercivity needed for elliptic existence of a Witten spinor. A decaying homogeneous solution would be parallel by the same identity and hence zero, excluding the kernel; the appropriate weighted elliptic solvability gives the correction. Under the stated smooth decay assumptions it has $ϵ-{ϵ}_{\infty }=O(r^{-1})$ and first derivative $O(r^{-2})$. The rigorous analytic step is supplied by the spinor existence proof. This also explains why completeness, asymptotic data and the absence of uncontrolled inner boundary terms are hypotheses of the argument.

It remains to identify the boundary integral with the physical ADM charge. In the asymptotic symmetric coframe $e_i{}^a={\delta }_i{}^a+q_{ia}/2+O(r^{-2})$, the leading spatial spin connection isInsert this and the Sen correction into $b^i$. The real scalar part of ${\gamma }^{ij}{\gamma }^{kl}$ is ${\delta }^{il}{\delta }^{jk}-{\delta }^{ik}{\delta }^{jl}$; its two-gamma part has imaginary expectation and drops out. It follows that the intrinsic connection contributes $\frac{1}{2}({\partial }_j{q}_{ij}-{\partial }_i{q}_{jj})K_{\infty }^{\hat{0}}$. For the other term, symmetry of $k$ removes the three-spatial-gamma product, leaving $-(k_{ij}-k{h}_{ij})K_{\infty }^{\hat{j}}$. Thus the ADM boundary term of the Nester two-form has the expansionThe constant-spinor derivative contribution integrates to zero on a large coordinate sphere. Its coefficient is antisymmetric in $i,j$; the corresponding tangential vector on the sphere is divergence free, so integration by parts removes it. Equivalently, the flux of ${\gamma }^{ij}{\partial }_{j}u$ has identically zero divergence, with its asymptotic leading term having zero flux. The quadratic correction terms are $O(r^{-3})$ and their surface integrals vanish in the limit.

Define the ADM energy and ADM momentum with the same conventions:The boundary calculation then gives, including every normalization factor,This is the energy associated with the asymptotic spinor's causal translation. In the zero-momentum rest frame it is the ADM mass times $K_{\infty }^{\hat{0}}$; more generally it proves the full energy-momentum bound.

Finally apply the divergence theorem to the Witten identity. There is no inner boundary, and the spinors assigned zero at other ends give no charge there. ThereforeAsymptotic constant spinors can produce every future null direction. Normalize $K_{\infty }^{\hat{0}}=1$ and choose its spatial direction along $\mathbf{P}$; the inequality givesThis establishes the positive energy theorem and nonnegativity of the invariant mass when it is defined. For zero ADM energy, repeat the identity for a basis of asymptotic spinors: all become parallel and all matter contractions vanish. Their curvature integrability forces the spacetime curvature to vanish on the initial surface; the standard rigidity conclusion is that the data embed in Minkowski spacetime. Merely having a parallel null spinor in a local wave geometry would not by itself prove this global rigidity: completeness and asymptotic flatness remain essential.