The four-dimensional new minimal rigid background

Replacing ordinary derivatives by covariant derivatives does not usually preserve supersymmetry. The metric rotates spinors, and a flat-space constant supersymmetry parameter need not remain parallel. Can extra fixed sources compensate this rotation?

We will obtain the precise compensation equation and solve it on $S^3\times S^1$. The preparation is four-dimensional supersymmetric actions, spinor covariant derivatives, sources and currents, and stress-tensor improvements. No three-dimensional spinor calculation is required.

Our matter theory has four-dimensional Lorentzian $\mathcal N=1$ supersymmetry: four real Poincaré generators, with that entire algebra manifest. It must possess a conserved continuous R-symmetry admitting an R-current multiplet. We continue the background problem to Euclidean signature. The matter gauge group is arbitrary and is not the background R-symmetry group; no gauge-rank assumption enters the calculation.

Which sources must remain fixed?

A current multiplet packages the stress tensor, supersymmetry currents, and their supersymmetric partners. Choosing the R-current multiplet makes the conserved R-current one of these operators. Its background partner is called new minimal supergravity. We use this name for its source transformation rules, without making gravity dynamical.

Its bosonic sources are a real positive-definite metric $G_{\mu\nu}$, an R connection $A_\mu$, and a conserved vector $V^\mu$. The fermionic sources are two gravitini $\psi_\mu,\widetilde\psi_\mu$. Locally a two-form potential $B_{\mu\nu}$ describes $V$:

$$V^\mu=\frac12\varepsilon^{\mu\nu\rho\sigma}\partial_\nu B_{\rho\sigma}, \qquad \nabla_\mu V^\mu=0. \tag{B02.1}$$

The conservation law follows because antisymmetrized second derivatives vanish; equivalently the associated three-form field strength is closed. In a physical orthonormal frame, $[A_a]=[V_a]=1$ and $[B_{ab}]=0$ in mass units. Coordinate components also depend on whether the coordinates carry length units. We allow $A,V,B$ to be complex after Euclidean continuation.

The rigid limit freezes these sources and sets the gravitini to zero. It does not extremize the action with respect to $A,V$ or impose Einstein’s equation. Their prescribed values remain in the matter action. This differs from eliminating the matter auxiliaries $F,D$ when finding a vacuum. Festuccia–Seiberg, section 2 and section 6

An R-current improvement changes the current multiplet and its source couplings, just as a stress-tensor improvement changes the curved extension. We must select that R-symmetry before choosing the background. A formal charge assigned to a fixed coupling does not by itself produce a conserved current.

A dictionary before a Killing equation

Let $\tau_i$ be the Pauli matrices, and take Euclidean orientation $\varepsilon_{1234}=+1$. The continuation established in the Weyl-spinor lesson uses

$$s^\mu=(i\tau_1,i\tau_2,i\tau_3,\mathbf1), \qquad \widetilde s^\mu=(-i\tau_1,-i\tau_2,-i\tau_3,\mathbf1).$$

Their mixed anticommutator is $+2\delta^{\mu\nu}$. Define the two Weyl rotation matrices

$$\Sigma^{\mu\nu}=\frac14(s^\mu\widetilde s^\nu-s^\nu\widetilde s^\mu), \qquad \widetilde\Sigma^{\mu\nu}=\frac14(\widetilde s^\mu s^\nu-\widetilde s^\nu s^\mu).$$

The three-manifold paper’s Appendix C instead inherits the matrices $\sigma_P^\mu=(\tau_i,-i\mathbf1)$ and $\widetilde\sigma_P^\mu=(-\tau_i,-i\mathbf1)$. Thus

$$\sigma_P=-is,\qquad \widetilde\sigma_P=-i\widetilde s, \qquad \sigma_P^{\mu\nu}=-\Sigma^{\mu\nu},\qquad \widetilde\sigma_P^{\mu\nu}=-\widetilde\Sigma^{\mu\nu}. \tag{B02.2}$$

Both one-index matrices acquire $-i$, so their products acquire a minus sign. This is a matrix convention, not a rephasing of mutually adjoint Lorentzian supercharges. In Euclidean signature $\zeta$ and $\widetilde\zeta$ are independent two-component columns of opposite chirality. A tilde does not mean complex conjugation. Dumitrescu–Festuccia–Seiberg, Appendix A.1

There is also a connection dictionary. Use the ordinary Cartan convention $de^a+\Omega^a{}_b\wedge e^b=0$. Then

$$\nabla_\mu\zeta =\partial_\mu\zeta+\frac12\Omega_{\mu ab}\Sigma^{ab}\zeta, \qquad \nabla_\mu\widetilde\zeta =\partial_\mu\widetilde\zeta +\frac12\Omega_{\mu ab}\widetilde\Sigma^{ab}\widetilde\zeta. \tag{B02.3}$$

Every ordered pair $a,b$ is summed. The paper uses the input-first connection $\omega_{\mu a}{}^b=e^b{}_\nu\nabla_\mu e_a{}^\nu$: with both frame indices lowered, $\omega_{\mu ab}=-\Omega_{\mu ab}$. Its expression $\omega_{ab}\sigma_P^{ab}/2$ therefore equals ours: the two minus signs cancel. Dumitrescu–Festuccia–Seiberg, (A.7)–(A.9)

Freeze the background and set its fermionic variation to zero

We take the following supergravity transformation as input, with terms containing gravitini already set to zero:

$$\begin{aligned} \delta\psi_\mu &=-2(\nabla_\mu-iA_\mu)\zeta -2iV_\mu\zeta-2iV^\nu\sigma_{P\mu\nu}\zeta,\\ \delta\widetilde\psi_\mu &=-2(\nabla_\mu+iA_\mu)\widetilde\zeta +2iV_\mu\widetilde\zeta +2iV^\nu\widetilde\sigma_{P\mu\nu}\widetilde\zeta . \end{aligned} \tag{B02.4}$$

This is the Euclidean column-spinor form of the new minimal variation. Festuccia–Seiberg give the Lorentzian, index-position-dependent form in (6.8); the explicit Euclidean form is Dumitrescu–Festuccia–Seiberg (1.10). The latter’s definition (B02.1) uses twice the coefficient of $B$ printed in Festuccia–Seiberg section 6: equal $V$ corresponds to $B_{\rm FS}=2B$. Our equations use $V$ itself.

The variations of bosonic supergravity sources vanish when their fermionic partners vanish. For the whole fixed background to be invariant, (B02.4) must vanish too. Divide each line by $-2$, and use (B02.2). The result is

$$\boxed{ \begin{aligned} (\nabla_\mu-iA_\mu)\zeta &=-iV_\mu\zeta+iV^\nu\Sigma_{\mu\nu}\zeta,\\ (\nabla_\mu+iA_\mu)\widetilde\zeta &=+iV_\mu\widetilde\zeta -iV^\nu\widetilde\Sigma_{\mu\nu}\widetilde\zeta . \end{aligned}} \tag{B02.5}$$

Replacing $\Sigma,\widetilde\Sigma$ by minus the paper’s matrices gives exactly Appendix C, (C.1). We have derived the condition from the supplied transformation, not derived the supergravity transformation from an action.

The spinor labels have R-charges $+1,-1$. Consequently $A\mapsto A+d\alpha$, $\zeta\mapsto e^{i\alpha}\zeta$, $\widetilde\zeta\mapsto e^{-i\alpha}\widetilde\zeta$ preserves both equations. With $A=V=0$, they reduce to ordinary parallel-spinor equations. Each chirality has at most two independent complex solutions: a parallel-transport equation fixes a solution everywhere on a connected region from its two initial components. Curvature and global holonomy can reduce that number.

Solve a curved example completely

Take a sphere of radius $\ell$ and a circle of length $L$:

$$ds^2=\ell^2\,d\Omega_3^2+d\tau^2,\qquad \tau\sim\tau+L,\qquad A=V=-\frac{i}{\ell}\,d\tau . \tag{B02.6}$$

These imaginary Euclidean sources are precisely the choice in Festuccia–Seiberg (7.2)–(7.3). They are not a requirement that real Lorentzian sources be imaginary. No background flavor or dynamical gauge field is turned on.

Here is a global frame that makes the solution visible. Embed the unit sphere as $y_0^2+y_1^2+y_2^2+y_3^2=1$ and form the unitary matrix $g=y_0\mathbf1+i y_a\tau_a$. Define

$$g^{-1}dg=\frac{i}{\ell}e^a\tau_a,\qquad e^a=\ell(y_0dy_a-y_a dy_0+\varepsilon_{abc}y_bdy_c), \qquad e^4=d\tau .$$

Pauli multiplication shows $\sum_a(e^a)^2=\ell^2\sum_{A=0}^3dy_A^2$ on tangent vectors. Differentiating $g^{-1}dg$ gives $d(g^{-1}dg)+(g^{-1}dg)\wedge(g^{-1}dg)=0$. Thus

$$de^a=\frac1\ell\varepsilon_{abc}e^b\wedge e^c, \qquad \Omega_{ab}=\frac1\ell\varepsilon_{abc}e^c, \qquad \Omega_{a4}=0.$$

Both ordered terms occur in the first epsilon sum. The Weyl matrices obey $\Sigma^{ab}=\widetilde\Sigma^{ab}=i\varepsilon_{abc}\tau_c/2$ for $a,b\leq3$, whereas $\Sigma^{a4}=i\tau_a/2$ and $\widetilde\Sigma^{a4}=-i\tau_a/2$. Equation (B02.3) therefore gives, for either chirality,

$$\nabla_a=E_a+\frac{i}{2\ell}\tau_a,\qquad \nabla_4=\partial_\tau ,$$

where $E_a$ is the frame dual to $e^a$. Substituting (B02.6) into (B02.5) now yields

$$E_a\zeta=E_a\widetilde\zeta=0,\qquad \partial_\tau\zeta=\partial_\tau\widetilde\zeta=0. \tag{B02.7}$$

The connection and auxiliary terms cancel with their displayed coefficients. For the periodic spin structure on the circle, all constant columns $c,\widetilde c\in\mathbb C^2$ give global solutions. The simply connected sphere has a unique spin structure, and its global oriented frame lifts to a spin frame. Hence the solution dimensions are $(2,2)$, not merely a pointwise count. This is the full complexification of the four-real-generator $\mathcal N=1$ algebra, not four-dimensional $\mathcal N=2$.

An independent coordinate check is useful. In a stereographic chart with real coordinates $u_i$, put $D=1+u_iu_i$ and $e_{\rm st}^i=2\ell\,du_i/D$. Cartan’s equation gives $\Omega_{ij}=2(u_i du_j-u_jdu_i)/D$. The matrix

$$U(u)=\frac{\mathbf1+i u_i\tau_i}{\sqrt D}$$

satisfies

$$\left(\partial_{u_i} -\frac{i}{D}\varepsilon_{ijk}u_j\tau_k\right)U =\frac{i}{D}\tau_i U . \tag{B02.8}$$

Thus $\zeta_{\rm st}=Uc$ and $\widetilde\zeta_{\rm st}=U\widetilde c$ solve the same equations in that frame. Their nonconstant components reflect a frame change; the apparent stereographic infinity is not a singularity of the global solutions.

The auxiliary field also has global content. With our orientation, (B02.1) and $V_4=-i/\ell$ require $dB=i\,\mathrm{vol}_{S^3}/\ell$. Its integral is $2\pi^2i\ell^2$, so $B$ cannot be one smooth global two-form. For polar angle $0\leq\chi\leq\pi$ and unit-sphere area form $\omega_2$, local potentials are

$$B_N=i\ell^2\left(\frac\chi2-\frac{\sin2\chi}{4}\right)\omega_2, \qquad B_S=B_N-\frac{i\pi\ell^2}{2}\omega_2 .$$

They are regular at their respective poles: near the north pole the coefficient is $i\ell^2\chi^3/3+O(\chi^5)$, and $\omega_2=\varepsilon_{abc}X_a\,dX_b\wedge dX_c/(2|X|^3)$ in radial Cartesian coordinates. Their difference is locally exact on angular overlap patches. These potentials describe the smooth closed three-form background. They do not assign an unmentioned compact two-form gauge group or quantize its periods; such extra global data must be checked if a particular theory requires them.

Practice

1. Change the circle holonomy

Keep $V$ as in (B02.6), but take $A=V+a\,d\tau$ with constant complex $a$. Let the spin boundary condition be $\zeta(\tau+L)=\epsilon\zeta(\tau)$ and likewise for $\widetilde\zeta$, where $\epsilon=\pm1$. Find all global solutions. For real $a$, identify the large R-gauge equivalence if the R group has period $2\pi$ and all charges are integers.

Hint 1

The sphere equations are unchanged. Solve the two circle equations before imposing their boundary conditions.

Hint 2

A nonzero initial column returns multiplied by $e^{iaL}$ or its inverse. Both must be compared with the chosen spin sign, not automatically with one.

Solution

Equation (B02.5) gives $\zeta=e^{ia\tau}c$ and $\widetilde\zeta=e^{-ia\tau}\widetilde c$. The condition is

$$e^{iaL}=\epsilon.$$

Since $\epsilon^{-1}=\epsilon$, both chiralities survive together. If it holds there are $(2,2)$ solutions; otherwise there are $(0,0)$. For complex $a$, its modulus first forces $\operatorname{Im}a=0$. The real values are $aL=2\pi n$ for periodic spinors and $aL=(2n+1)\pi$ for antiperiodic spinors.

The gauge transformation $e^{i\alpha}=e^{2\pi in\tau/L}$ shifts $a$ by $2\pi n/L$. This equivalence uses the stated charge lattice. It cannot identify the two spin structures. In particular $a=0$ with antiperiodic spinors has no global supersymmetry, although every small patch has the full local solution space. This conclusion concerns the combined spin and source holonomy; it does not equate an arbitrary supersymmetric twist with an ordinary thermal boundary condition.

2. Change the auxiliary magnitude

On the same sphere and circle, set $A=V=-i\kappa\,d\tau/\ell$ for constant complex $\kappa$ and use periodic spinors. Derive the curvature of the modified spinor connection and determine which $\kappa$ admit solutions. Find the solutions for the second allowed sign explicitly using $g$.

Hint 1

In the global frame each chirality obeys $\mathscr D_a\zeta=0$ with $\mathscr D_a=E_a+i(1-\kappa)\tau_a/(2\ell)$. This is a noncoordinate frame.

Hint 2

Cartan’s equation gives $[E_a,E_b]=-2\varepsilon_{abc}E_c/\ell$. Subtract this frame commutator when computing curvature. For $\kappa=-1$, compare the connection one-form with $g^{-1}dg$.

Solution

Writing $c=(1-\kappa)/(2\ell)$, the matrix part of the curvature is

$$\begin{aligned} \mathscr F_{ab} &=[ic\tau_a,ic\tau_b] +\frac2\ell\varepsilon_{abc}(ic\tau_c)\\ &=\frac{i(1-\kappa^2)}{2\ell^2}\varepsilon_{abc}\tau_c. \end{aligned}$$

If $\kappa^2\ne1$, already $\mathscr F_{12}$ is invertible, so no nonzero parallel spinor exists. For $\kappa=1$ we recover constant columns. For $\kappa=-1$ the connection is $g^{-1}dg$, and $d(g^{-1}c)+(g^{-1}dg)g^{-1}c=0$. Therefore $\zeta=g^{-1}c$ and $\widetilde\zeta=g^{-1}\widetilde c$ are smooth global solutions, independent of $\tau$. Both signs have $(2,2)$ solutions. Removing $V$ is $\kappa=0$, which fails even locally: the metric alone does not preserve this rigid algebra.

3. Does a flat-space model admit this source choice?

Take a canonical neutral chiral with fixed nonzero couplings $W=m\Phi^2/2+h\Phi^3/3$. Can one couple its continuous R-current to the background above by assigning one R-charge $r$ to $\Phi$? What changes if either coupling vanishes? Explain why treating both couplings as spurions does not solve the fixed-coupling question.

Hint 1

The superpotential must have R-charge two. Apply this separately to its two monomials.

Hint 2

A spurion transformation changes an external parameter. A symmetry of a fixed theory must leave its specified parameter values unchanged.

Solution

Nonzero fixed $m$ requires $2r=2$, whereas nonzero fixed $h$ requires $3r=2$. No $r$ satisfies both. Consequently this generic model does not have the continuous R-symmetry needed for this new minimal construction. This is not a claim that it has no supersymmetry in flat space, nor that every other curved construction is excluded.

With $h=0$ one can choose $r=1$; with $m=0$ one can choose $r=2/3$. These statements test the classical action; a quantum gauge theory must additionally have the appropriate conserved R-current rather than a gauge-anomalous candidate. When discussing large background gauge transformations, fractional written charges also require a compatible choice of the global R-group period.

Assigning $R(m)=2-2r$ and $R(h)=2-3r$ makes a formally covariant family of actions. A transformation then generally moves between different nonzero coupling values. It does not supply the conserved continuous current of the fixed generic model required here.

The central result is a reusable source equation with a global solution test. Appendix C of the three-manifold paper uses this four-dimensional input. Twisted reduction to three dimensions then derives the source and multiplet dictionary on a circle fibration, including the new charge and shifted auxiliary.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading Supersymmetric Field Theories on Three-Manifolds
• Paper guide: Appendices C and D, four-dimensional origin