Why minimal coupling is not enough

Replacing ordinary derivatives by covariant derivatives makes an action geometrically meaningful. Does it also preserve its supersymmetry? We can answer by repeating the free chiral variation and keeping the terms that flat space allowed us to discard.

The preparation is R-symmetry and flavor symmetry and Covariant derivatives of spinors. We use the action and parity conventions of Build the free chiral theory. No sphere geometry is needed: our diagnostic metric will be derived directly.

Our starting theory is the massless, gauge-neutral three-dimensional chiral multiplet, whose flat-space supersymmetry is $\mathcal N=2$, with four real Lorentzian supercharges. We continue its fields to independent Euclidean variables. There is no gauge sector or superpotential; the scalar R-charge $r$ is arbitrary. On a curved background, the number of preserved symmetries is a result to determine.

Specify the proposed covariantization

Keep a fixed positive-definite metric and set every additional background field to zero:

$$A_\mu=V_\mu=C_\mu=H=0,\qquad z=0.$$

The minimally coupled candidate action is

$$S_{\min}=\int_M d^3x\,\sqrt g\,\mathcal L_{\min},\qquad \mathcal L_{\min} =\nabla_\mu\widetilde\phi\,\nabla^\mu\phi -i\widetilde\psi\gamma^\mu\nabla_\mu\psi -\widetilde F_\Phi F_\Phi.$$

For a scalar, $\nabla_\mu\phi=\partial_\mu\phi$. For a spinor, $\nabla_\mu=\partial_\mu+\mathcal A_\mu$, with $\mathcal A_\mu=\Omega_{\mu ab}\gamma^a\gamma^b/4$ in the frame convention of the preceding lesson. Coordinate gamma matrices are $\gamma^\mu=e_a{}^\mu\gamma^a$, and their full covariant derivative vanishes.

Let $\zeta(x)$ and $\widetilde\zeta(x)$ be commuting spinor labels. Define the proposed odd operation $\delta=\delta_\zeta+\delta_{\widetilde\zeta}$ by

$$\begin{aligned} \delta\phi&=\sqrt2\zeta\psi,\\ \delta\psi&=\sqrt2\zeta F_\Phi -\sqrt2i\gamma^\mu\widetilde\zeta\,\nabla_\mu\phi,\\ \delta F_\Phi&=-\sqrt2i\widetilde\zeta\gamma^\mu\nabla_\mu\psi,\\ \delta\widetilde\phi&=-\sqrt2\widetilde\zeta\widetilde\psi,\\ \delta\widetilde\psi&=\sqrt2\widetilde\zeta\widetilde F_\Phi +\sqrt2i\gamma^\mu\zeta\,\nabla_\mu\widetilde\phi,\\ \delta\widetilde F_\Phi&=-\sqrt2i\zeta\gamma^\mu\nabla_\mu\widetilde\psi. \end{aligned}$$

We obtained this ansatz by taking the flat rules with derivatives acting on matter fields and replacing those derivatives by $\nabla$. The metric, frame, and labels are fixed during the variation.

This specification matters. For constant labels in flat space, $\widetilde\zeta\gamma^\mu\partial_\mu\psi$ and $\partial_\mu(\widetilde\zeta\gamma^\mu\psi)$ coincide. For varying labels they differ. The paper’s full curved rules in equations (6.8) and (6.10) differentiate complete bilinears and include further background terms. We are testing the stated naive ansatz, not those completed rules.

The scalar and auxiliary fields are even; the matter spinors are odd. Hence

$$\delta(\widetilde\psi\gamma^\mu\nabla_\mu\psi) =(\delta\widetilde\psi)\gamma^\mu\nabla_\mu\psi -\widetilde\psi\gamma^\mu\nabla_\mu(\delta\psi).$$

The labels are commuting but the operation is odd. An ordinary even variation is $\Delta=\rho\delta$, with an independent odd constant $\rho$ placed on the left. The conclusions below are multiplied by $\rho$ without changing.

Keep the derivatives of the first spinor

Set $\widetilde\zeta=0$. Write the action’s three terms as $\mathcal L_s,\mathcal L_\psi,\mathcal L_F$. Their variations are

$$\begin{aligned} \delta_\zeta\mathcal L_s &=\sqrt2(\nabla^\mu\widetilde\phi) \left[(\nabla_\mu\zeta)\psi+\zeta\nabla_\mu\psi\right],\\ \delta_\zeta\mathcal L_\psi &=\sqrt2(\nabla_\nu\widetilde\phi) (\gamma^\nu\zeta)\gamma^\mu\nabla_\mu\psi\\ &\quad+\sqrt2i\widetilde\psi\gamma^\mu \left[(\nabla_\mu\zeta)F_\Phi+\zeta\nabla_\mu F_\Phi\right],\\ \delta_\zeta\mathcal L_F &=\sqrt2iF_\Phi\zeta\gamma^\mu\nabla_\mu\widetilde\psi. \end{aligned}$$

The new pieces are the two occurrences of $\nabla_\mu\zeta$. We must keep them when rearranging the familiar flat-space cancellations.

Recall $uv=u^TCv$, with $C=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$, and the local orthonormal matrices $\gamma^a=(\sigma_3,-\sigma_1,-\sigma_2)$. They obey

$$(\gamma^a)^TC=-C\gamma^a,\qquad \gamma^\nu\gamma^\mu =g^{\nu\mu}\mathbf1+i\epsilon^{\nu\mu\rho}\gamma_\rho.$$

Here $\epsilon^{\nu\mu\rho}$ is the covariantly constant oriented volume tensor, with $\epsilon^{123}=+1$ in an oriented orthonormal frame. The transpose identity gives $(\gamma^\nu\zeta)\gamma^\mu\nabla_\mu\psi =-\zeta\gamma^\nu\gamma^\mu\nabla_\mu\psi$. Its metric part cancels the term containing $\zeta\nabla_\mu\psi$ in the scalar variation.

For the antisymmetric part, integration by parts differentiates the label too:

$$\begin{aligned} &-\sqrt2i\epsilon^{\nu\mu\rho} (\nabla_\nu\widetilde\phi)\zeta\gamma_\rho\nabla_\mu\psi\\ &=\nabla_\mu\!\left[ -\sqrt2i\epsilon^{\nu\mu\rho} (\nabla_\nu\widetilde\phi)\zeta\gamma_\rho\psi\right] +\sqrt2i\epsilon^{\nu\mu\rho} (\nabla_\nu\widetilde\phi)(\nabla_\mu\zeta)\gamma_\rho\psi. \end{aligned}$$

The omitted scalar-Hessian contribution is zero: $\epsilon^{\nu\mu\rho}\nabla_\mu\nabla_\nu\widetilde\phi=0$ for a torsion-free connection. No commutator of derivatives on a spinor has been used.

The auxiliary pieces still form a complete product derivative. Because $C\gamma^\mu$ is symmetric and the label is even, $\widetilde\psi\gamma^\mu\nabla_\mu\zeta =(\nabla_\mu\zeta)\gamma^\mu\widetilde\psi$. Thus their sum is $\nabla_\mu(\sqrt2iF_\Phi\zeta\gamma^\mu\widetilde\psi)$, including the label derivative.

Combining the remaining label term from $\delta\mathcal L_s$ with the antisymmetric term gives the result

$$\delta_\zeta\mathcal L_{\min} =\nabla_\mu B_\zeta^\mu+\mathcal E_\zeta, \qquad \mathcal E_\zeta =\sqrt2(\nabla_\nu\widetilde\phi) (\nabla_\mu\zeta)\gamma^\nu\gamma^\mu\psi,$$ $$B_\zeta^\mu =-\sqrt2i\epsilon^{\nu\mu\rho} (\nabla_\nu\widetilde\phi)\zeta\gamma_\rho\psi +\sqrt2iF_\Phi\zeta\gamma^\mu\widetilde\psi.$$

The remainder is precisely what assuming a constant spinor too early would hide.

Check the second spinor independently

Set $\zeta=0$. The auxiliary contributions $-\sqrt2i\widetilde F_\Phi\widetilde\zeta\gamma^\mu\nabla_\mu\psi$ and its opposite cancel directly. The other terms give

$$\begin{aligned} \delta_{\widetilde\zeta}\mathcal L_{\min} &=-\sqrt2\left[ (\nabla_\mu\widetilde\zeta)\widetilde\psi +\widetilde\zeta\nabla_\mu\widetilde\psi\right]\nabla^\mu\phi\\ &\quad+\sqrt2\widetilde\psi\gamma^\mu\gamma^\nu (\nabla_\mu\widetilde\zeta)\nabla_\nu\phi\\ &\quad+\sqrt2\widetilde\psi\gamma^\mu\gamma^\nu \widetilde\zeta\,\nabla_\mu\nabla_\nu\phi. \end{aligned}$$

The scalar Hessian is symmetric, so its gamma product reduces to $g^{\mu\nu}$. Also $\widetilde\psi\widetilde\zeta=-\widetilde\zeta\widetilde\psi$: this minus sign is from the antisymmetric contraction matrix, with one even and one odd spinor. The first and last lines therefore form a product derivative:

$$\delta_{\widetilde\zeta}\mathcal L_{\min} =\nabla_\mu B_{\widetilde\zeta}^\mu+\mathcal E_{\widetilde\zeta},$$ $$B_{\widetilde\zeta}^\mu =-\sqrt2(\widetilde\zeta\widetilde\psi)\nabla^\mu\phi, \qquad \mathcal E_{\widetilde\zeta} =\sqrt2(\nabla_\nu\phi)\widetilde\psi\gamma^\mu\gamma^\nu \nabla_\mu\widetilde\zeta.$$

Both remainders have dimension three and R-charge zero. For example, $[\nabla\widetilde\phi]=3/2$, $[\nabla\zeta]=1/2$, and $[\psi]=1$, while their R-charges add as $-r+1+(r-1)=0$.

The action variation is consequently

$$\delta S_{\min} =\int_{\partial M}d\Sigma_\mu (B_\zeta^\mu+B_{\widetilde\zeta}^\mu) +\int_M d^3x\,\sqrt g\, (\mathcal E_\zeta+\mathcal E_{\widetilde\zeta}).$$

The $B^\mu$ are the boundary currents from this variation, not definitions of conserved Noether supercurrents. Compact support, adequate falloff, or compatible periodic conditions remove the first integral. They do not remove the bulk remainder. No matter field equation has been imposed.

Where curvature enters

Parallel spinors, satisfying $\nabla_\mu\zeta=0$ and similarly for the second label, make the remainders vanish. In a rotating frame they need not have constant components. The condition is covariant, unlike $\partial_\mu\zeta=0$.

For this particular off-shell ansatz, parallelism is also necessary if the action variation is to vanish for every compactly supported matter configuration. The two remainders contain independent matter fields, so they cannot cancel each other. In the second one, choose the scalar gradient and the undifferentiated spinor freely at a point. With $u_\mu=\nabla_\mu\widetilde\zeta$, its coefficient must obey

$$\sum_\mu\gamma^\mu\gamma^\nu u_\mu=0 \quad\text{for each }\nu.$$

In an orthonormal frame define $w=\sum_\mu\gamma^\mu u_\mu$. The Clifford relation turns this equation into $2u_\nu-\gamma^\nu w=0$. Multiply by $\gamma^\nu$ and sum: $2w-3w=0$, so $w=0$ and every $u_\nu=0$. Transposing the first remainder with $C$ gives the same condition for $\zeta$.

Can a proposed metric admit such a spinor? Commuting its derivatives supplies a separate test:

$$[\nabla_\mu,\nabla_\nu]\zeta =\mathcal R_{\mu\nu}\zeta,\qquad \mathcal R_{\mu\nu} =\partial_\mu\mathcal A_\nu-\partial_\nu\mathcal A_\mu +[\mathcal A_\mu,\mathcal A_\nu].$$

For a parallel spinor, $\mathcal R_{\mu\nu}\zeta=0$. This is the curvature obstruction. It is distinct from the first-derivative remainder derived above; we have not inserted an unexplained scalar-curvature term into that variation. The matrix connection agrees with the paper’s equation (A.11) after the frame-index translation taught in the spinor-derivative lesson.

A curved example with a nonzero action variation

Take $M=\mathbb R_x\times T^2_{y,z}$, with periodic spin structure in the two periodic directions, and

$$ds^2=dx^2+e^{2\kappa x}dy^2+dz^2,\qquad \kappa>0.$$

Set $f=e^{\kappa x}$ and choose $e^1=dx$, $e^2=f\,dy$, $e^3=dz$. Since $de^2=\kappa e^1\wedge e^2$, Cartan’s equation gives $\Omega^2{}_1=\kappa e^2$ and $\Omega^1{}_2=-\kappa e^2$. Both ordered frame pairs contribute to the spin connection:

$$\mathcal A_y=-\frac{\kappa f}{2}\gamma^1\gamma^2,\qquad \mathcal A_x=\mathcal A_z=0.$$

Therefore

$$\mathcal R_{xy} =-\frac{\kappa^2f}{2}\gamma^1\gamma^2,\qquad (\gamma^1\gamma^2)^2=-\mathbf1.$$

This curvature matrix is invertible. It annihilates no nonzero spinor, so this metric admits no parallel label for our ansatz, even locally.

We can also see the failure in the action itself. Let $u(x)$ be a real, rapidly decaying profile and take

$$\zeta=t\begin{pmatrix}1\\0\end{pmatrix},\qquad \widetilde\phi=u(x),\qquad \psi=\vartheta u'(x)\begin{pmatrix}0\\1\end{pmatrix},$$

with all other matter fields and $\widetilde\zeta$ zero. Here $t$ commutes and $\vartheta$ is an odd amplitude. This is an off-shell test in the independent Euclidean variables, before choosing an integration contour. With $[u]=1/2$, both $[t]$ and $[\vartheta]$ are $-1/2$.

For fields depending only on $x$, the Dirac operator is

$$\not\nabla=\gamma^1\left(\partial_x+\frac\kappa2\right).$$

The angular connection term produces $\kappa/2$, despite the absence of $y$ dependence. Directly varying the two kinetic terms now gives

$$\delta_\zeta\mathcal L_s=-\sqrt2t\vartheta\,u'u'', \qquad \delta_\zeta\mathcal L_\psi =\sqrt2t\vartheta\,u'\left(u''+\frac\kappa2u'\right).$$

Thus

$$\delta_\zeta\mathcal L_{\min} =\frac{\kappa t\vartheta}{\sqrt2}(u')^2.$$

For these spinor components $B_\zeta^\mu=0$ identically. The surviving variation is a bulk term. If the coordinate periods are $L_y,L_z$ and $u=v e^{-x^2/\ell^2}$, with $\ell>0$, its invariant-volume integral is

$$\delta_\zeta S_{\min} =\frac{\kappa t\vartheta L_yL_z v^2\sqrt\pi}{2\ell} e^{\kappa^2\ell^2/8} \left(1+\frac{\kappa^2\ell^2}{4}\right).$$

To check the integral, complete the square in $-2x^2/\ell^2+\kappa x$ and differentiate the Gaussian normalization twice with respect to $\kappa$. The coefficient of $t\vartheta$ is nonzero for $v\ne0$. The Gaussian falloff controls both ends of the noncompact direction. Taking $\kappa\to0$ removes both the connection correction and the variation, as the flat-space proof requires.

What must change

The calculation explains the qualification in the three-manifold paper’s introduction: metric coupling generally requires additional terms and deformed supercharges. The corresponding discussion in Festuccia–Seiberg, introduction and equation (1.1) motivates adding controlled background-dependent couplings.

The next task is to organize those couplings and transformation changes together. A prescribed connection or auxiliary background can alter the equation for the supersymmetry label, while additional matter couplings can cancel the new variation. Merely guessing a curvature mass does not establish that cancellation.

This is not a universal prohibition on minimal coupling: if suitable global parallel spinors exist and boundary conditions are compatible, the action and ansatz tested here do work. Nor does failure of this ansatz prove that the same metric cannot support a completed rigid supersymmetric theory.

Exercises

1. A varying label can fail even in flat space

On a flat torus, let $x=x^1$ have period $2\pi/k$, with $k>0$. Choose

$$\phi=v\sin(kx),\qquad \widetilde\psi=\vartheta\begin{pmatrix}0\\1\end{pmatrix}, \qquad \widetilde\zeta=t\sin(kx)\begin{pmatrix}1\\0\end{pmatrix},$$

and set the remaining fields and $\zeta$ to zero. Compute the action variation per unit transverse area. Is curvature responsible?

First hint

Use $\gamma^1\gamma^1=\mathbf1$ and $\left(\begin{smallmatrix}0&1\end{smallmatrix}\right) C\left(\begin{smallmatrix}1\\0\end{smallmatrix}\right)=1$ in $\mathcal E_{\widetilde\zeta}$.

Second hint

All profiles are periodic, so the integral of the boundary-current derivative vanishes. The period integral of $\cos^2(kx)$ is $\pi/k$.

Complete solution

The only nonzero label derivative is $\partial_1\widetilde\zeta=tk\cos(kx)(1,0)^T$. Hence

$$\mathcal E_{\widetilde\zeta} =\sqrt2tv\vartheta k^2\cos^2(kx),\qquad B_{\widetilde\zeta}^1 =\sqrt2tv\vartheta k\sin(kx)\cos(kx).$$

The current is periodic. Integrating its derivative gives zero, while

$$\frac{\delta_{\widetilde\zeta}S_{\min}} {\text{transverse area}} =\sqrt2\pi tv\vartheta k.$$

The metric has zero curvature. The failure comes from choosing a nonparallel label for a rigid symmetry: the flat theory permits constant supersymmetry parameters, not arbitrary local functions with unchanged transformation rules. This does not contradict flat-space supersymmetry.

2. Does adding a curvature mass fix the ansatz?

Add $\Delta\mathcal L=cR\widetilde\phi\phi$, where $R$ is the fixed metric’s scalar curvature and $c$ is a constant. Keep precisely the same proposed transformations. Compute its $\zeta$ variation. Can this one added term cancel $\mathcal E_\zeta$ for arbitrary off-shell matter fields on a background with a nonparallel label?

First hint

The metric and $R$ are fixed, while $\delta_\zeta\widetilde\phi=0$.

Second hint

At a point, the value of a scalar can vanish while its first derivative is arbitrary. Compare which of those data occurs in the two terms.

Complete solution

The added variation is

$$\delta_\zeta\Delta\mathcal L =\sqrt2cR\widetilde\phi\,\zeta\psi.$$

Its dependence on the undifferentiated scalar differs from $\mathcal E_\zeta=\sqrt2(\nabla_\nu\widetilde\phi) (\nabla_\mu\zeta)\gamma^\nu\gamma^\mu\psi$. Choose $\widetilde\phi=0$ at a point and select its gradient and $\psi$ so the latter coefficient is nonzero. The mass variation vanishes there for every $c$, while the derivative remainder does not.

Because $\psi$ is undifferentiated in both bulk terms, invariance for arbitrary compactly supported $\psi$ would require the combined local coefficient to vanish. The independent scalar value and gradient show why it cannot. This rules out this isolated repair with unchanged rules; it does not rule out a coordinated modification of the action, transformations, and background equation.

3. Test a proposed nonparallel spinor

Suppose a nonzero label obeys, on a region,

$$\nabla_\mu\zeta=\lambda\gamma_\mu\zeta,$$

with a constant $\lambda$ of dimension one. This is an assumption to test, not a claim that an arbitrary metric admits such a spinor. Set $\widetilde\zeta=0$ and simplify $\mathcal E_\zeta$. Does satisfying this equation alone make the minimally coupled action invariant for arbitrary matter fields?

First hint

Move the gamma matrix off the first spinor using $(\gamma^\mu)^TC=-C\gamma^\mu$.

Second hint

In an orthonormal frame, use the Clifford relation to compute $\sum_\mu\gamma^\mu\gamma^\nu\gamma^\mu$. There are three values of $\mu$.

Complete solution

For each fixed $\nu$,

$$\sum_\mu\gamma^\mu\gamma^\nu\gamma^\mu =\sum_\mu(2\delta^{\mu\nu}-\gamma^\nu\gamma^\mu)\gamma^\mu =2\gamma^\nu-3\gamma^\nu=-\gamma^\nu.$$

Substituting the assumed spinor derivative and using its transpose sign gives

$$\begin{aligned} \mathcal E_\zeta &=\sqrt2\lambda(\nabla_\nu\widetilde\phi) (\gamma_\mu\zeta)\gamma^\nu\gamma^\mu\psi\\ &=-\sqrt2\lambda(\nabla_\nu\widetilde\phi) \zeta\gamma_\mu\gamma^\nu\gamma^\mu\psi\\ &=\sqrt2\lambda(\nabla_\nu\widetilde\phi)\zeta\gamma^\nu\psi. \end{aligned}$$

For $\lambda\ne0$ this is generally nonzero. A differential equation for the label is useful only together with an action and transformation rules whose variations cancel under that equation. Setting $\lambda=0$ recovers the parallel-label result. This calculation makes no assertion about the existence or global behavior of solutions to the assumed equation.

The question for a completed background

We have identified the missing terms rather than inferred failure from curvature alone. The next question is which additional fixed sources and spinor equation make the action variation vanish without requiring a parallel spinor. Supergravity as a system of background sources organizes the action, transformation rules, and background conditions as one construction.