Supergravity as a system of background sources

In a curved space, the metric alone usually leaves uncancelled terms in a supersymmetry variation. Extra background fields can supply the missing terms. But which extra fields belong together, and what does it mean to keep them fixed while supersymmetry acts?

The preparation is Why minimal coupling is not enough and Sources, currents, and generating functionals. We will use the source calculus to construct a rigid theory from a system whose sources themselves transform. The decisive calculation is a variation of the action with the sources first allowed to vary and then frozen.

Our setting is three-dimensional Euclidean space or a Riemannian three-manifold. The matter theory has a continuous R-symmetry and flat-space $\mathcal N=2$ supersymmetry, meaning four real Poincaré supercharges before Euclidean continuation. The free, gauge-neutral chiral multiplet is the simplest example; there is no dynamical gauge group in that example. No extra supersymmetry is being assumed. The background may preserve only a subset, which we must determine.

A source for the supersymmetry current

A current measures the response to a source. The stress tensor couples to a change of metric; an internal current couples to a connection. A supercurrent is the fermionic Noether current associated with supersymmetry. Its source is consequently fermionic.

In the three-manifold paper, there are two independent Euclidean supercurrents. Write them as $\mathcal S^\mu$ and $\widetilde{\mathcal S}^{\mu}$. Their fermionic sources are vector-spinors $\Psi_\mu$ and $\widetilde\Psi_\mu$, called gravitini. A vector-spinor has both a spacetime index and a spinor index. We capitalize the gravitini here to distinguish them from a matter fermion $\psi$.

Near flat space, the part of the source coupling that we need is

$$\mathcal L_{\rm source} =-\frac12\mathcal S^\mu\Psi_\mu +\frac12\widetilde{\mathcal S}^{\mu}\widetilde\Psi_\mu.$$

Spinor contractions use the same order as displayed. These coefficients and relative signs are those of the paper, equation (2.10); the R-multiplet calculation derives them by multiplying the source and current superfields. The gravitini are external probes in this construction; introducing their sources does not put propagating spin-$3/2$ particles into the matter theory.

There is a useful parity convention for the next calculation. Use Grassmann-odd spinor parameters $\varepsilon,\widetilde\varepsilon$ and denote the resulting even infinitesimal variation by $\Delta$. An even variation obeys the ordinary Leibniz rule. A commuting spinor label $\zeta$ used in earlier lessons is recovered by writing $\varepsilon=\eta\zeta$ with an external Grassmann-odd constant $\eta$. The same construction applies to the independent tilded sector. Factoring out that constant will leave the ordinary spinor equation for $\zeta$.

When moving $\eta$ to the far left, remember that the current is odd: $\mathcal S^\mu\partial_\mu(\eta\zeta)=-\eta\mathcal S^\mu\partial_\mu\zeta$. Keep the displayed contraction order when comparing this even variation with an odd, fixed-label variation.

Promote the parameters of a flat-space supersymmetry transformation to functions of position. Define the supercurrent normalization so that, up to a total derivative,

$$\Delta\mathcal L_0 =\mathcal S^\mu\partial_\mu\varepsilon -\widetilde{\mathcal S}^{\mu}\partial_\mu\widetilde\varepsilon.$$

This is the local-parameter Noether identity in equation (2.12). The uncancelled derivatives of the parameters are the same kind of obstruction found by varying the minimally coupled chiral action.

Now set the sources to zero after varying them, and choose

$$\Delta\Psi_\mu=2\partial_\mu\varepsilon,\qquad \Delta\widetilde\Psi_\mu=2\partial_\mu\widetilde\varepsilon.$$

At zero sources, the terms containing $\Delta\mathcal S^\mu$ multiply zero, while

$$\left.\Delta\mathcal L_{\rm source}\right|_{\Psi=\widetilde\Psi=0} =-\mathcal S^\mu\partial_\mu\varepsilon +\widetilde{\mathcal S}^{\mu}\partial_\mu\widetilde\varepsilon =-\Delta\mathcal L_0.$$

The factor two is fixed by the one-half in the coupling. The source values can vanish while their variations do essential work.

This is a first-order calculation around flat space, not the full nonlinear supergravity action. Its purpose is to show why the source for a supercurrent has a transformation involving a derivative of the supersymmetry parameter. The nonlinear construction replaces this flat-space expression with a covariant one containing the other background fields.

Why the sources must form a multiplet

Under supersymmetry, a supercurrent is related to other operators, including the stress tensor. Transforming its source coupling therefore produces terms involving their sources. The metric, gravitini, and additional bosonic sources must transform together. An off-shell supergravity multiplet provides such transformation laws, including auxiliary fields, without using gravitational field equations to close the symmetry transformations.

“Off shell” and “external” answer different questions. Off shell concerns whether equations of motion were needed in the symmetry algebra. External means a field is prescribed rather than integrated over. A field can be both.

Let $\Phi$ denote all matter variables, including matter auxiliary fields when an off-shell matter description is used. Collect bosonic gravitational sources into $\mathcal B$ and fermionic gravitational sources into $\Psi_{\rm grav}=(\Psi_\mu,\widetilde\Psi_\mu)$. A supersymmetric coupling has the form

$$S[\Phi;\mathcal B,\Psi_{\rm grav}].$$

The semicolon separates matter variables from sources. Before freezing the sources, its symmetry identity is

$$0=\Delta_{\rm full}S =\Delta_{\rm matter}S+\Delta_{\rm sources}S,$$

with boundary terms either absent or treated by suitable boundary conditions. Here $\Delta_{\rm sources}S$ means the functional chain-rule variation caused by changing the sources while holding the matter fields fixed. It includes fermionic source terms with their displayed order.

This identity holds for arbitrary matter configurations in the off-shell description. It does not instruct us to solve for a classical matter solution.

Freeze the background and find the surviving transformations

Choose a bosonic source profile $\mathcal B_0(x)$ and set both gravitini to zero:

$$S_{\mathcal B_0}[\Phi] \equiv S[\Phi;\mathcal B_0,0].$$

The new theory integrates over $\Phi$ and holds $\mathcal B_0$ fixed. Its candidate supersymmetry transformations are the matter transformations inherited from the coupled system, evaluated at those background values.

To be a symmetry of this one theory, a transformation must preserve the selected background. Moving to a different source profile relates two members of the family of theories; it is not a symmetry of the member we chose.

On a bosonic background, supersymmetry variations of the bosonic supergravity fields contain fermionic supergravity fields. They therefore vanish when the gravitini vanish. The remaining conditions are

$$\left.\Delta\Psi_\mu\right|_{\mathcal B_0,0}=0,\qquad \left.\Delta\widetilde\Psi_\mu\right|_{\mathcal B_0,0}=0.$$

An independent parameter can be set to zero: preserving a transformation in the $\zeta$ sector does not require a nonzero solution in the $\widetilde\zeta$ sector. We require at least one nonzero parameter to obtain a supersymmetry.

Factor out the external odd constants. The conditions become linear differential equations for commuting spinors, schematically

$$\mathcal D_\mu[\mathcal B_0]\zeta=0,\qquad \widetilde{\mathcal D}_\mu[\mathcal B_0]\widetilde\zeta=0.$$

The operators include the spin connection and algebraic terms made from the extra bosonic backgrounds. These are the Killing spinor equations of the background construction. Their name does not mean that $\zeta$ must be parallel under the Levi-Civita spin connection alone.

For a solution, $\Delta_{\rm sources}S=0$. The full symmetry identity then immediately gives

$$\Delta S_{\mathcal B_0} =\left.\Delta_{\rm matter}S\right|_{\mathcal B_0,0}=0.$$

This proves the mechanism, assuming the supersymmetric source coupling has been supplied. It does not derive every coefficient of that coupling or solve the background equations. The three-dimensional gravitino transformations required for that next step are given in equation (2.15).

At the quantum level the same argument gives a symmetry of the regulated matter functional integral when its measure, counterterms, and boundary conditions preserve the transformation. The classical source identity by itself does not prove those additional properties.

The flat background as a complete first test

Take $M=\mathbb R^3$ with its Cartesian frame, metric $g_{\mu\nu}=\delta_{\mu\nu}$, and zero additional bosonic sources. In particular, the paper’s fields $A_\mu,V_\mu,H$ vanish, as do the underlying gauge potentials in this example. The matter theory is the massless free chiral theory, with its chosen continuous R-symmetry.

In this background the full gravitino variations reduce to the terms calculated above. Thus

$$\partial_\mu\zeta=0,\qquad \partial_\mu\widetilde\zeta=0,$$

and the solutions are constant spinors. They are globally defined on $\mathbb R^3$. All the supersymmetries of the flat matter theory survive.

Each independent Euclidean spinor has two complex components. The two complex solution directions in each sector describe the complexification of the original four-real-supercharge algebra; they do not redefine three-dimensional $\mathcal N=2$ as eight real physical supercharges.

Try instead $\zeta(x)=f(x)\zeta_0$ with $\zeta_0\ne0$ constant. The source variation is proportional to $2(\partial_\mu f)\zeta_0$. It vanishes only when $f$ is constant. The full source system allowed local supersymmetry parameters because the sources transformed too. The frozen theory retains only the parameters that preserve its background. On a curved space such a permitted parameter can depend on position, but that dependence is constrained.

This check also identifies a common mistake. Declaring $\Psi_\mu=0$ is a choice of background value. It does not imply $\Delta\Psi_\mu=0$. The latter is a condition on a transformation, and must be tested.

A fixed auxiliary field is not eliminated

The rigid limit removes gravitational fluctuations while retaining prescribed bosonic sources. It does not impose Einstein equations or auxiliary equations from a dynamical gravity action on those sources. The background still has to satisfy the preservation equations for the desired supersymmetry.

This distinction is central to Festuccia and Seiberg’s construction, introduction and section 2: the metric can remain nontrivial while gravitational fluctuations decouple, and auxiliary background values remain available. We use that construction here, not its four-dimensional spinor coefficients.

An algebraic example makes the difference operational. This example is a finite-dimensional Gaussian, with dimensionless real variables, and is not itself a supersymmetric field theory. Let $x$ be a matter variable and $u$ an auxiliary variable, with

$$S(x,u)=\frac a2x^2+\frac12u^2+\lambda ux,\qquad a>\lambda^2,\quad a,\lambda\in\mathbb R.$$

If $u$ is part of the dynamical system, its algebraic equation is

$$\frac{\partial S}{\partial u}=u+\lambda x=0.$$

Eliminating it gives

$$u=-\lambda x,\qquad S_{\rm elim}(x)=\frac12(a-\lambda^2)x^2.$$

The same result follows from integrating over $u$, apart from an $x$-independent normalization, because

$$S(x,u)=\frac12(u+\lambda x)^2 +\frac12(a-\lambda^2)x^2.$$

If instead $u=u_0$ is a prescribed source, the matter action is

$$S_{\rm fixed}(x;u_0) =\frac a2x^2+\lambda u_0x+\frac12u_0^2.$$

There is no variation with respect to $u$ in the matter equations. In particular, the relation $u_0=-\lambda x$ is not imposed on every configuration in the matter integral. Doing so would let the alleged source follow the fluctuating matter field.

For $a=2$, $\lambda=1$, and $u_0=1$, the fixed-source matter equation is $2x+1=0$, giving $x=-1/2$. Eliminating $u$ first instead gives $x=0$. Both operations are well defined, but they produce different matter theories. The exercise below compares their fluctuations too.

The source-independent-of-$x$ term $u_0^2/2$ does not change normalized matter expectation values at fixed $u_0$, but it does change derivatives of the free energy with respect to $u_0$. This is the same reason that source-only terms must be tracked when calculating contact responses.

Background auxiliaries in supergravity are more structured than this one-variable model. The example establishes the logical distinction between fixing and eliminating; it is not a proposed action or reality condition for the paper’s scalar $H$.

Practice

1. Determine the gravitino coefficient

Keep the source-current coupling and the local-parameter matter variation used above, but suppose

$$\Delta\Psi_\mu=c\,\partial_\mu\varepsilon,\qquad \Delta\widetilde\Psi_\mu=c\,\partial_\mu\widetilde\varepsilon.$$

Determine $c$. Explain why setting both gravitini to zero before applying the variation would miss the result.

First hint

At zero source values, the variation of the current itself multiplies zero. The variation of the source does not.

Second hint

Keep the coefficient of each independent current times parameter derivative. Both coefficients must vanish.

Complete solution

The total variation at zero sources is

$$\Delta(\mathcal L_0+\mathcal L_{\rm source}) =\left(1-\frac c2\right) \mathcal S^\mu\partial_\mu\varepsilon +\left(-1+\frac c2\right) \widetilde{\mathcal S}^{\mu}\partial_\mu\widetilde\varepsilon.$$

It vanishes for arbitrary currents and parameter derivatives precisely when $c=2$. For example, $c=-2$ would double both original residual terms instead of cancelling them in these conventions.

This sign belongs to the three-dimensional paper’s current/source normalization. A differently normalized supergravity formalism can use another sign, but its coupling and transformation must be translated together.

Setting $\Psi=\widetilde\Psi=0$ inside the source term before taking the full variation deletes the term that supplied the cancellation. After freezing, that deletion is legitimate only for transformations whose source variations already vanish. It is a restriction of the allowed transformations, not an identity following from the source values.

2. Measure the effect of eliminating an auxiliary variable

Add a real probe $-Jx$ to the Gaussian example. Calculate the mean and variance of $x$ when $u=u_0$ is fixed and when $u$ is integrated over. Use the weight $e^{-S+Jx}$, integration over the real line for each dynamical variable, and $a>\lambda^2$.

First hint

A normalized Gaussian proportional to $\exp[-b x^2/2+t x]$ has center $t/b$. Complete the square to find its variance.

Second hint

In the fixed theory the linear coefficient is $J-\lambda u_0$. In the integrated theory first shift $u$ to $u+\lambda x$.

Complete solution

For a fixed $u_0$, the partition function is

$$Z_{\rm fixed}(J;u_0) =\sqrt{\frac{2\pi}{a}}\, \exp\!\left[-\frac12u_0^2+ \frac{(J-\lambda u_0)^2}{2a}\right].$$

Differentiating $\log Z$ gives

$$\langle x\rangle_{\rm fixed}=\frac{J-\lambda u_0}{a}, \qquad \langle(x-\langle x\rangle)^2\rangle_{\rm fixed}=\frac1a.$$

Integrating $u$ first gives

$$Z_{\rm integrated}(J) =\frac{2\pi}{\sqrt{a-\lambda^2}}\, \exp\!\left[\frac{J^2}{2(a-\lambda^2)}\right],$$

so

$$\langle x\rangle_{\rm integrated}=\frac{J}{a-\lambda^2}, \qquad \langle(x-\langle x\rangle)^2\rangle_{\rm integrated} =\frac1{a-\lambda^2}.$$

At $a=2,\lambda=1,u_0=1,J=0$, the means are $-1/2$ and $0$, and the variances are $1/2$ and $1$. The quadratic Hessian of the joint system is

$$K=\begin{pmatrix}a&\lambda\\ \lambda&1\end{pmatrix}, \qquad K^{-1}=\frac1{a-\lambda^2} \begin{pmatrix}1&-\lambda\\-\lambda&a\end{pmatrix}.$$

Its $xx$ entry independently reproduces the integrated variance. The inequality $a>\lambda^2$ ensures convergence and positive variances. As $\lambda\to0$, both matter means and variances agree, as they must when $u$ decouples from $x$.

3. Change the global boundary condition

Replace $\mathbb R^3$ by $\mathbb R^2\times S^1$, with a circle coordinate $y\sim y+L$, $L>0$. Keep the metric flat and all additional background sources zero. A spin structure specifies whether spinors are periodic or antiperiodic around the circle:

$$\zeta(y+L)=s\,\zeta(y),\qquad s=+1\ \text{or}\ -1.$$

Which choices permit the constant spinor solutions? Why does this test add information beyond solving the local differential equation?

First hint

The local equation is still $\partial_\mu\zeta=0$. Evaluate the boundary condition on its constant solutions.

Second hint

For the antiperiodic choice the same constant vector must equal its own negative.

Complete solution

With $s=+1$, any constant two-component $\zeta$ is periodic, and the same holds for $\widetilde\zeta$. The flat supersymmetries survive when the matter fields and their boundary conditions transform consistently.

With $s=-1$, a constant solution must obey $\zeta_0=-\zeta_0$, hence $\zeta_0=0$. The tilded sector is obstructed in exactly the same way. There is no nonzero globally allowed parameter for this zero-source background.

A local solution becomes a symmetry only when it is a well-defined parameter on the whole space and preserves the chosen boundary conditions. The result assumes the additional background connections vanish; turning on their holonomies would change the combined transport condition and would be a different background.

Checkpoint

You should now be able to start with a source-dependent action, freeze a bosonic background, and explain why the surviving transformations are selected by vanishing gravitino variations. You should also be able to identify the extra, unjustified step in imposing a gravitational auxiliary equation on a prescribed source.

The remaining practical question is what each bosonic source contributes. The metric, R-symmetry connection, and auxiliary backgrounds enter the spinor equation in different ways; identifying those roles prepares us to write the explicit equation and solve it on a curved manifold.

To compare this construction with the original presentation, use the paper introduction guide, which maps the frozen-background argument to the two gravitino equations.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading Supersymmetric Field Theories on Three-Manifolds
• Paper guide: Section 2, currents and rigid backgrounds