Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 57 (Supergravity)

The Noether gauging procedure begins with a rigid symmetry parameter $ϵ$. Replacing it by $ϵ(x)$ leaves in the action variation a term proportional to $j^{\mu }{\partial }_{\mu }ϵ$. Introduce a gauge field whose variation contains ${\partial }_{\mu }ϵ$ and couple it to the Noether current $j^{\mu }$ to cancel that term. The new coupling produces further variations, so both the action and transformation laws are corrected successively in the coupling constant until local invariance is obtained. Covariant derivatives organize this iterative completion.

For the Wess–Zumino model, the parameter is a spinor and the current is the supercurrent. Its derivative part has the structure ${\gamma }^{\nu }{\partial }_{\nu }(A-i{\gamma }_{5}B){\gamma }^{\mu }\psi$, up to the fixed Majorana spinor normalization and improvements. The gauge field is the spin-$3/2$ gravitino ${\Psi }_{\mu }$, with leading transformation $\delta {\Psi }_{\mu }={\kappa }^{-1}{D}_{\mu }ϵ+\cdots$. A coupling of ${\bar{\Psi }}_{\mu }$ to that current cancels the derivative of the local parameter. Since the commutator of two local supersymmetries includes a spacetime transformation, the vierbein and gravity must participate as well. This gives the Wess-Zumino chiral multiplet coupled to supergravity.

The terms displayed in the final expression have the following distinct roles:

The last two interactions are part of the higher-order local supersymmetry completion. They can be organized through the composite Kähler connection in the fermion derivatives, or derived by eliminating the corresponding supergravity auxiliary connection. Matter fermions and gravitini carry the appropriate chiral Kähler weights, fixing the relative current couplings. They are not new adjustable Yukawa interactions. In reading this organization, one must not include the composite connection in $D_{\mu }$ and then count its separately displayed current terms a second time.

Finally the spin connection itself acquires gravitino-induced torsion and matter spin-current contributions. Eliminating this algebraic connection, or solving for auxiliary fields, can make further fermion contact terms implicit in the chosen covariant derivatives and component convention. The key mechanism is current coupling at first order, followed by the covariant and higher-order terms required by local supersymmetry, rather than merely changing ordinary derivatives in the original matter Lagrangian.

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 $S={\Phi }_0$, $u={\Phi }_{+}$, $v={\Phi }_{-}$, $M=M_p$, $P=uv-\zeta$, $x=|S{|}^2/M^2$ and $q=|u{|}^2+|v{|}^2$. Assume $g\ne 0$ and, initially, $\zeta \ne 0$. The supergravity F-term potential isThe canonical Kähler potential gives the identity Kähler metric. The three Kähler covariant derivatives of a superpotential areSubstituting, including the negative term rather than treating the potential as just a sum of squares, gives the exact answerHere $(1+x{)}^2-3x=1-x+x^2$ explains the first coefficient. It is strictly positive for real $x$, since $1-x+x^2=(x-1/2{)}^2+3/4$. Thus this model's scalar potential is nonnegative everywhere.

In the stated limit $q/M^2\ll 1$, the exponential can be replaced by $e^x[1+O(q/M^2)]$. The other displayed terms retain the effects of the Kähler covariant derivatives of a superpotential. Small waterfall fields alone do not justify discarding $\zeta /M^2$ in those derivatives, or expanding in $x$. If also $|\zeta |/M^2\ll 1$ and $x\ll 1$, the leading result reduces to the global F-term scalar potentialNo gauge sector or gauge coupling is supplied, so there is no specified D-term contribution to add.

The supersymmetric vacuum manifold of a bilinear hybrid superpotential isIndeed $W=0$ and all three Kähler covariant derivatives of a superpotential vanish there, so all supergravity auxiliary fields vanish and supersymmetry is unbroken. The nonnegative scalar potential makes this a global minimum. Conversely zero potential requires $uv=\zeta$, and then $|S{|}^{2}q=0$; for $\zeta \ne 0$ this forces $S=0$. With just the supplied fields this is a complex vacuum manifold before imposing any gauge quotient: the product is fixed but the ratio of $u$ and $v$ is not. More precisely, it is an F-flat vacuum manifold whether or not a gauge symmetry is specified. If one additionally gauges opposite charges, D-flatness typically imposes $|u|=|v|$, giving $|u|=|v|=\sqrt{|\zeta |}$ up to phases and gauge equivalence; that is an extra assumption, not a consequence of $K$ and $W$ alone.

The other familiar branch of F-term hybrid inflation has $u=v=0$. Its nonzero derivative $D_{S}W=-g\zeta (1+x)$ establishes supersymmetry breaking, and its exact scalar potential isThe canonical-Kähler correction to an F-term hybrid valley begins asIn particular the inflaton quadratic mass cancels at this order; the first lift of the global flat valley is quartic in $|S|$.

To identify when this branch is a transverse minimum, rephase the waterfall fields so that $\zeta$ is real positive. To quadratic order in $u,v$, the global scalar potential isThe normalized combinations $(u\pm \bar{v})/\sqrt{2}$ diagonalize this quadratic form, withThus $|S{|}^2>|\zeta |$ gives a valley of transverse minima, while below the threshold the negative eigenvalue produces the waterfall instability in F-term hybrid inflation.

The exact supergravity correction to this stability test is also available without taking $x$ small. Put $b=|\zeta |/M^2$. Expansion of the exact supergravity F-term potential about $u=v=0$ gives equal diagonal coefficients $|g{|}^2{M}^2{e}^x[x+b^2(1+x^2)]$ and off-diagonal magnitude $|g{|}^2{M}^2{e}^{x}b(1+x+x^2)$. HenceFor the usual sub-Planckian symmetry-breaking scale $0<b\ll 1$, the lower eigenvalue is $|g{|}^2{M}^2{e}^x(1-b)[x-b(1+x^2)]$. It is positive for the familiar sub-Planckian valley $b\ll x\ll 1$, and changes sign near $x=b$.

The printed assertion of two full minima requires qualification. The broken branch is a transverse inflationary valley, not generally a second stationary minimum of the complete supergravity potential. In factIts only stationary point in $S$ is the origin. For $0<|\zeta |<M^2$ that point has a negative waterfall eigenvalue $|g{|}^2(|\zeta {|}^2/M^2-|\zeta |)$. A direct counterexample is the path $S=0$, $u=v=a$ with real small $a$, after rephasing $\zeta >0$:The negative quadratic coefficient proves it is not a minimum. No loop corrections or additional stabilizing interactions were specified that could create the requested second stationary vacuum. The mathematically supported interpretation is a SUSY-breaking transverse valley and a zero-energy supersymmetric vacuum minimum manifold. The limit on $u,v$ alone does not fix the parameter range $|\zeta |/M^2$; the familiar waterfall conclusion uses the sub-Planckian range just stated.

In the corresponding global theory the exponential prefactor, Kähler corrections to $D_{i}W$ and universal $-3|W{|}^2/M^2$ term are absent. Its $u=v=0$ valley has exactly constant tree-level energy $|g\zeta {|}^2$ for all $S$, whereas canonical supergravity lifts it by the quartic correction above. The supersymmetric vacuum $S=0$, $uv=\zeta$ and its zero energy persist in both descriptions. If $\zeta =0$, the origin and the $S$ axis instead have vanishing F-terms; the broken-valley interpretation does not apply.

Work in the supplied Planck units and write $t=T+\bar{T}>0$. The nonzero components of the Kähler metric and its inverse areThe Kähler covariant derivatives of a superpotential areIn the supergravity F-term potential, $K^{T\bar{T}}|D_{T}W{|}^2=3|W{|}^2$ cancels its negative universal term. This is the no-scale supergravity identity. The exact answer is thereforeIn particular $C=0$ is a global minimum for every $T$ in the domain, with zero vacuum energy.

Use the standard convention for a supergravity auxiliary field, $F^i=-e^{K/2}{K}^{i\bar{j}}\overline{D_{j}W}$. In this conventionAt the $C=0$ minimum these becomeThus supersymmetry breaking occurs if $B\ne 0$, despite the vanishing vacuum energy. Indeed $K_{T\bar{T}}|F^T{|}^2=3|B{|}^2/t^3$ is precisely canceled by the negative gravitino contribution. For $B=0$ the asserted breaking does not occur: both supergravity auxiliary fields vanish at $C=0$, so this vacuum is supersymmetric.

For completeness, the full set of finite zero-energy minima includes additional branches. For $C=r{e}^{i\theta }\ne 0$, the condition $D_{C}W=0$ reduces, after factoring $r{e}^{-i\theta }$, toFor every $B\ne 0$ there is a unique positive solution of $r(3+r^2)=|B|$, since the left side increases strictly from zero to infinity. There are three associated phases,These also give $V=0$, and on each branch $W=3B/(3+r^2)\ne 0$. Consequently $F^C=0$ but $F^T\ne 0$, so these are also supersymmetry breaking no-scale vacua with a cubic matter superpotential. A supersymmetric finite vacuum would require $D_{T}W=0$, hence $W=0$, and then $D_{C}W=3C^2=0$; this is possible only at $C=0$ with $B=0$.

The scalar potential is independent of $\mathrm{Im}T$ everywhere. On every zero-energy vacuum branch it is also independent of $\mathrm{Re}T$, so both real components of $T$ are flat directions of a scalar potential at the vacuum. Away from $D_{C}W=0$, the real direction is not flat: ${\partial }_{t}V=-3V/t$, giving a runaway towards $t\to \infty$ rather than a finite positive-energy stationary point. The zero-energy branches have no continuous flat direction of a scalar potential in $C$; their allowed $C$ values are isolated. This is why the vacuum does not select a unique numerical gravitino mass: the modulus $t$ is unfixed.

The gravitino mass from a superpotential is $m_{3/2}=e^{K/2}|W|$. On the conventional $C=0$ branch,On a nonzero-$C$ branch it is insteadWhen $B=0$, the only finite zero-energy matter solution is $C=0$ and $m_{3/2}=0$, with unbroken supersymmetry. These results exhibit the distinction between vanishing vacuum energy and vanishing supersymmetry breaking auxiliary fields in a no-scale supergravity model.