Build invariant F-terms and D-terms

The curved chiral multiplet gives transformations that close on a supersymmetric background. We now need an action. Closure alone does not say which component can be integrated, or which background terms must accompany it.

The organizing principle is simple: construct a multiplet whose appropriate highest component varies into a divergence. Its integral is invariant when the resulting boundary flux vanishes. We will do this first for a product of two chiral multiplets and then for a neutral general multiplet.

This lesson uses the three-dimensional Euclidean conventions of superspace, with a flat $\mathcal N=2$ parent containing four real Lorentzian supercharges. On a curved manifold, the available Killing spinors determine which operations survive. There is no dynamical gauge sector in the examples. The metric and the background fields $A_\mu,V_\mu,H,C_\mu$ are fixed; no gravitational auxiliary equation is imposed.

Multiply two chiral multiplets

Let $\Phi_i=(\phi_i,\psi_i,F_i)$, for $i=1,2$, have bottom-component R-charges $r_i$ and central-charge eigenvalues $z_i$. Matter spinors are odd. The Killing spinor labels $\zeta,\widetilde\zeta$ commute, so their supersymmetry operations are odd and obey

$$\delta(XY)=(\delta X)Y+(-1)^{|X|}X\,\delta Y.$$

Spinor contractions remain $uv=u^TCv$ with $C=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$ and $\gamma_\mu^TC=-C\gamma_\mu$. In particular, the scalar contraction of two odd spinors is symmetric: $\psi_1\psi_2=\psi_2\psi_1$. This does not make their individual components commuting.

The product starts with $\phi_{12}=\phi_1\phi_2$. Varying it fixes its fermion:

$$\delta\phi_{12} =\sqrt2\,\zeta(\phi_1\psi_2+\phi_2\psi_1).$$

Thus define $\psi_{12}=\phi_1\psi_2+\phi_2\psi_1$. The $\zeta$ part of its variation contains

$$\sqrt2\left[ (\zeta\psi_1)\psi_2+(\zeta\psi_2)\psi_1 +\zeta(\phi_1F_2+\phi_2F_1)\right].$$

The two-component identity

$$(\zeta\psi_1)\psi_2+(\zeta\psi_2)\psi_1 =-\zeta(\psi_1\psi_2)$$

follows by expanding each lower spinor component and moving the odd coefficients to the same order. Therefore the complete product multiplet is

$$\boxed{ \begin{aligned} \phi_{12}&=\phi_1\phi_2,\\ \psi_{12}&=\phi_1\psi_2+\phi_2\psi_1,\\ F_{12}&=\phi_1F_2+\phi_2F_1-\psi_1\psi_2. \end{aligned}}$$

Its bottom charges are

$$r_{12}=r_1+r_2,\qquad z_{12}=z_1+z_2.$$

The fermion and auxiliary have R-charges $r_{12}-1$ and $r_{12}-2$; all three components have central charge $z_{12}$. These formulas agree with the chiral embedding P v3, (6.7) specialized in the product rule (6.16). We have derived the needed chiral product rather than requiring the full general multiplication rule.

The auxiliary sign can also be checked directly from flat superspace. The highest coefficient in $(\phi_1+\sqrt2\theta\psi_1+\theta^2F_1) (\phi_2+\sqrt2\theta\psi_2+\theta^2F_2)$ is exactly $F_{12}$ because $2(\theta\psi_1)(\theta\psi_2)=-\theta^2\psi_1\psi_2$. The curved construction keeps this algebraic product. Curvature changes its transformation law.

Check the highest-component variation

For a field of charges $(q,z_X)$, the matter derivative is

$$D_\mu^{(q,z_X)} =\nabla_\mu-iq\left(A_\mu-\frac12V_\mu\right)-iz_XC_\mu.$$

For brevity put $M_i=z_i-r_iH$. The two relevant chiral rules from P v3, (6.8) are

$$\begin{aligned} \delta\psi_i &=\sqrt2\,\zeta F_i -\sqrt2 iM_i\widetilde\zeta\phi_i -\sqrt2 i\gamma^\mu\widetilde\zeta D_\mu\phi_i,\\ \delta F_i &=\sqrt2 i(M_i+2H)\widetilde\zeta\psi_i -\sqrt2 iD_\mu(\widetilde\zeta\gamma^\mu\psi_i). \end{aligned}$$

The last derivative acts on the whole bilinear, including the label. It has the charges of $F_i$. The labels solve the background Killing spinor equations; we do not replace their covariant derivatives by zero.

First, $\delta_\zeta F_{12}=0$. The variations of $\phi_1F_2+\phi_2F_1$ cancel those of the fermion product, using

$$\delta_\zeta(\psi_1\psi_2) =(\delta_\zeta\psi_1)\psi_2-\psi_1\delta_\zeta\psi_2.$$

For the other family, $\delta_{\widetilde\zeta}\phi_i=0$. The algebraic terms in $\delta_{\widetilde\zeta}F_{12}$ combine as

$$\begin{aligned} \delta_{\widetilde\zeta}F_{12}\big|_{\rm algebraic} ={}&\sqrt2 i(M_2+2H+M_1)\phi_1\widetilde\zeta\psi_2\\ &+\sqrt2 i(M_1+2H+M_2)\phi_2\widetilde\zeta\psi_1\\ ={}&\sqrt2 i\left[z_{12}-(r_{12}-2)H\right] \widetilde\zeta\psi_{12}. \end{aligned}$$

The extra $M_1$ and $M_2$ terms came from varying $-\psi_1\psi_2$. Omitting that term would already spoil the curved result.

The derivative terms are

$$-\sqrt2 i\left[ \phi_1D_\mu(\widetilde\zeta\gamma^\mu\psi_2) +\phi_2D_\mu(\widetilde\zeta\gamma^\mu\psi_1) +(D_\mu\phi_1)\widetilde\zeta\gamma^\mu\psi_2 +(D_\mu\phi_2)\widetilde\zeta\gamma^\mu\psi_1 \right].$$

Their connections add by the product rule. Combining both parts gives

$$\boxed{ \delta F_{12} =\sqrt2 i\left[z_{12}-(r_{12}-2)H\right] \widetilde\zeta\psi_{12} -\sqrt2 iD_\mu^{(r_{12}-2,z_{12})} (\widetilde\zeta\gamma^\mu\psi_{12}).}$$

This is the auxiliary rule for a chiral multiplet with the summed charges. No matter equation of motion was used.

From a component to an invariant integral

For the standard curved F-term prescription, require

$$r_{12}=2,\qquad z_{12}=0.$$

Then $F_{12}$ is a neutral scalar, and the algebraic term in its variation vanishes. The bilinear $\widetilde\zeta\gamma^\mu\psi_{12}$ is also neutral: its R-charge is $-1+(2-1)=0$. Its derivative is the ordinary covariant divergence. Hence

$$\delta\int_M d^3x\,\sqrt g\,F_{12} =-\sqrt2 i\int_{\partial M}d^2x\,\sqrt h\, n_\mu\widetilde\zeta\gamma^\mu\psi_{12}.$$

Here $n_\mu$ is the outward unit normal and $h$ the induced boundary metric. On a closed manifold this vanishes. On a noncompact space or a space with boundary, the corresponding falloff or boundary-flux condition is part of the action’s invariance.

The charge conditions do more than remove a coefficient. They make the integrand and current globally neutral, so their local expressions patch as a scalar and a vector. We assume the original fields and Killing spinors are globally defined sections of their appropriate bundles. A cancellation such as $z_{12}=(r_{12}-2)H$ on one special background is not a general substitute for neutrality.

For the independent antichiral partners,

$$\begin{aligned} \widetilde\phi_{12}&=\widetilde\phi_1\widetilde\phi_2,\\ \widetilde\psi_{12} &=\widetilde\phi_1\widetilde\psi_2 +\widetilde\phi_2\widetilde\psi_1,\\ \widetilde F_{12} &=\widetilde\phi_1\widetilde F_2 +\widetilde\phi_2\widetilde F_1 +\widetilde\psi_1\widetilde\psi_2. \end{aligned}$$

Their bottom charges are $(-2,0)$ when the chiral charges sum to $(2,0)$. The plus sign in the auxiliary product is the convention already established in the superspace lesson. Its variation is $\delta\widetilde F_{12} =-\sqrt2 i\nabla_\mu(\zeta\gamma^\mu\widetilde\psi_{12})$. Consequently,

$$\mathcal L_F=c F_{12}+\widetilde c\,\widetilde F_{12}$$

is invariant after integration under the stated conditions, for constant neutral couplings $c,\widetilde c$. They are independent Euclidean parameters unless an additional reality prescription is imposed. This is P v3, (6.14).

For example, take $r_1=r_2=1$, $z_1=\nu$, and $z_2=-\nu$. The individual chiral auxiliaries are not neutral; the product auxiliary is. A bilinear superpotential $W=c\,\Phi_1\Phi_2$ gives

$$\mathcal L_F =c(\phi_1F_2+\phi_2F_1-\psi_1\psi_2) +\widetilde c\left( \widetilde\phi_1\widetilde F_2+ \widetilde\phi_2\widetilde F_1+ \widetilde\psi_1\widetilde\psi_2\right).$$

The background-dependent algebraic coefficient cancels because $M_1+M_2+2H=(\nu-\nu)-(1+1-2)H=0$. With the free-field engineering dimensions of the constituent fields, $[\phi_{12}]=1$, $[\psi_{12}]=3/2$, and $[F_{12}]=2$. Thus $[c]=[\widetilde c]=1$ makes this a dimension-three Lagrangian. The R-charge two of the bottom component is a different statement from its mass dimension.

In the flat superspace convention of the preceding lesson, the same density is

$$c\int d^2\theta\,\Phi_1\Phi_2 -\widetilde c\int d^2\widetilde\theta\, \widetilde\Phi_1\widetilde\Phi_2.$$

The minus before the antichiral integral translates its highest polynomial coefficient into $+\widetilde F_{12}$. On a curved background we use the component prescription just proved; we have not asserted that the flat Berezin measure itself is a curved superspace measure.

Why the curved D-term needs a completion

Now take a neutral general multiplet $\mathcal S$, whose bottom charges are $r_{\mathcal S}=z_{\mathcal S}=0$. We need only a few of its components: the scalar $\sigma_{\mathcal S}$, vector $a_{\mathcal S\mu}$, highest scalar $D_{\mathcal S}$, and odd spinors $\chi_{\mathcal S},\widetilde\chi_{\mathcal S},\lambda_{\mathcal S},\widetilde\lambda_{\mathcal S}$. These are the component names of the general package in the superspace lesson. In particular, $a_{\mathcal S\mu}$ is not automatically a gauge connection, and $D_{\mathcal S}$ need not be an independent elementary auxiliary.

Define the following neutral bilinears, suppressing the subscript $\mathcal S$ on the right:

$$\begin{aligned} X&=\zeta\chi-\widetilde\zeta\widetilde\chi, &Y&=\zeta\widetilde\lambda-\widetilde\zeta\lambda,\\ B^\mu&=\zeta\gamma^\mu\widetilde\lambda -\widetilde\zeta\gamma^\mu\lambda, &P^\mu&=\zeta\gamma^\mu\widetilde\lambda +\widetilde\zeta\gamma^\mu\lambda. \end{aligned}$$

Here $P^\mu$ is only a temporary vector bilinear, not a momentum generator. For a neutral general multiplet, the needed transformations in P v3, (6.5) reduce to

$$\begin{aligned} \delta a_{\mathcal S\mu}&=-iP_\mu+\nabla_\mu X,\\ \delta\sigma_{\mathcal S}&=-Y,\\ \delta D_{\mathcal S}&=\nabla_\mu B^\mu-iV_\mu P^\mu-HY. \end{aligned}$$

The last line displays the problem: $D_{\mathcal S}$ alone does not vary into a divergence. Since the backgrounds are fixed, varying $D_{\mathcal S}-a_{\mathcal S\mu}V^\mu-\sigma_{\mathcal S}H$ gives

$$\begin{aligned} \delta(D_{\mathcal S}-a_{\mathcal S\mu}V^\mu-\sigma_{\mathcal S}H) ={}&\nabla_\mu B^\mu-iV_\mu P^\mu-HY\\ &-V^\mu(-iP_\mu+\nabla_\mu X)-H(-Y)\\ ={}&\nabla_\mu B^\mu-V^\mu\nabla_\mu X. \end{aligned}$$

The remaining term is a divergence because $\nabla_\mu V^\mu=0$. This is the background field-strength identity in P v3, (1.1), not a matter equation. Therefore

$$\boxed{ \begin{aligned} \mathcal L_D[\mathcal S] &=-\frac12 \left(D_{\mathcal S} -a_{\mathcal S\mu}V^\mu-\sigma_{\mathcal S}H\right),\\ \delta\mathcal L_D &=-\frac12\nabla_\mu\left(B^\mu-V^\mu X\right). \end{aligned}}$$

This proves the curved D-term prescription P v3, (6.13) for every supplied background and surviving label for which the source transformations apply. It does not require $H$ or $V$ to be constant. As for the F-term, global fields and a vanishing integrated boundary flux are required.

If the bottom component has engineering dimension one, then $[a_{\mathcal S}]=[\sigma_{\mathcal S}]=2$ and $[D_{\mathcal S}]=3$. Both background completions have dimension three because $[V]=[H]=1$.

One neutral matter D-term

Apply the prescription to the particular product $\mathcal K=\widetilde\Phi\Phi$, where the chiral and antichiral bottom charges are $r,-r$ and the central charges vanish. This is a neutral general multiplet. We will use only its two components needed when

$$A_\mu=V_\mu=C_\mu=0,\qquad H=\text{constant}.$$

Assume the background admits the Killing spinors of the curved chiral lesson. These conditions restrict the geometry; they are not a claim that an arbitrary curved metric admits constant-$H$ supersymmetry.

For this calculation put $\rho=\widetilde\phi\phi$. The special chiral/antichiral embeddings in P v3, (6.7) and (6.9) give

$$\sigma_\Phi=-rH\phi,\qquad \sigma_{\widetilde\Phi}=-rH\widetilde\phi,\qquad D_\Phi=d_r\phi,\qquad D_{\widetilde\Phi}=d_r\widetilde\phi, \quad d_r=\frac r4R-\frac r2H^2.$$

These are components of each chiral multiplet embedded in a general multiplet. They are not additional independent fields in the chiral theory.

The specialized product rule then gives

$$\sigma_{\mathcal K}=-2rH\rho+i\widetilde\psi\psi.$$

To reconstruct $D_{\mathcal K}$, the contributions from P v3, (6.16) can be grouped by their origin:

Origin in the product	Contribution to $D_{\mathcal K}$
Highest components times bottom components	$2d_r\rho$
Scalar background components and their charge terms	$-2r^2H^2\rho$
Matter auxiliary product	$2\widetilde F F$
The two scalar derivative products	$-2\nabla_\mu\widetilde\phi\,\nabla^\mu\phi$
Fermion derivative terms	$i\widetilde\psi\gamma^\mu\nabla_\mu\psi-i(\nabla_\mu\widetilde\psi)\gamma^\mu\psi$
Fermion background terms	$2irH\widetilde\psi\psi$

Thus

$$\begin{aligned} D_{\mathcal K}={}& -2\nabla_\mu\widetilde\phi\,\nabla^\mu\phi+2\widetilde F F\\ &+i\widetilde\psi\gamma^\mu\nabla_\mu\psi -i(\nabla_\mu\widetilde\psi)\gamma^\mu\psi +2irH\widetilde\psi\psi\\ &+\left[\frac r2R-r(2r+1)H^2\right]\rho. \end{aligned}$$

The product calculation has supplied the curvature term before any action is guessed. Now use $\mathcal L_D=-D_{\mathcal K}/2+H\sigma_{\mathcal K}/2$. The two scalar coefficients and two fermion coefficients combine to give

$$\begin{aligned} \mathcal L_D[\mathcal K] ={}&\mathcal L_{\rm chiral} +\frac i2\nabla_\mu(\widetilde\psi\gamma^\mu\psi),\\ \mathcal L_{\rm chiral} ={}&\nabla_\mu\widetilde\phi\,\nabla^\mu\phi -i\widetilde\psi\gamma^\mu\nabla_\mu\psi-\widetilde F F\\ &+\left[-\frac r4R+r\left(r-\frac12\right)H^2\right]\rho -i\left(r-\frac12\right)H\widetilde\psi\psi. \end{aligned}$$

The explicit divergence accounts for placing the fermion derivative entirely on $\psi$. The displayed action agrees with the corresponding specialization of P v3, (6.32). Its invariance follows from the neutral-multiplet proof above. The complete term-by-term sphere variation has its separate teaching home in the round-sphere chiral action.

As a coefficient check, the established round-sphere background has $H=-i/\ell$ and $R=-6/\ell^2$ in P’s curvature convention. Then

$$-\frac r4R+r\left(r-\frac12\right)H^2 =\frac{r(2-r)}{\ell^2}, \qquad -i\left(r-\frac12\right)H =-\frac{r-\frac12}{\ell}.$$

Both are the coefficients found by the direct sphere calculation. Sending $\ell$ to infinity at fixed $r$ removes them and recovers the flat free density. This is a consistency check on the multiplet construction, not a replacement for the sphere action’s detailed proof or contour analysis.

Compare with the flat superspace integrals

For a general flat superfield with bottom component $C_{\mathcal S}$, the normalized highest polynomial coefficient is

$$\int d^4\theta\,\mathcal S =-\frac12D_{\mathcal S}-\frac14\partial^2C_{\mathcal S}.$$

Therefore, when all background fields vanish,

$$\mathcal L_D[\mathcal S] =\int d^4\theta\,\mathcal S+\frac14\partial^2C_{\mathcal S}.$$

The two descriptions agree as actions under the specified boundary assumptions, although their local densities can differ. The $a_{\mathcal S}V$ and $\sigma_{\mathcal S}H$ terms are the curved completion. Merely replacing partial derivatives in the flat density by covariant derivatives would miss them and the curvature terms already contained in the embedded composite.

F-term and D-term invariance here are classical off-shell statements. They do not prove quantum nonrenormalization, choose a contour for complex Euclidean fields, or determine a partition function. They provide action densities to which those later questions can be applied.

Exercises

1. Test two superpotential monomials

On the R-symmetry background used here, consider one chiral multiplet $X$ of R-charge $r$ and zero central charge. Can $W=mX^2/2+gX^3/3$ with fixed, nonzero, neutral constants $m,g$ satisfy the standard curved F-term prescription? What changes if only one monomial is retained? Explain why assigning a formal charge to the symbol $m$ is not, by itself, a construction of an admissible background coupling.

Hint 1

The bottom component of each term in $W$ must have R-charge two. Equal mass dimension and equal R-charge are different requirements.

Hint 2

A spacetime-dependent or charged coupling must be treated as a background section or multiplet, with its own transformation and global conditions.

Solution

With neutral $m,g$, the two conditions are $2r=2$ and $3r=2$. They require $r=1$ and $r=2/3$, respectively, and cannot both hold. A purely quadratic term uses $r=1$; a purely cubic term uses $r=2/3$.

The engineering dimensions are separately $[m]=1$ and $[g]=1/2$ for $[X|]=1/2$. They make both terms dimension two, but do not cure the incompatible R-charges.

A spurion charge is useful bookkeeping for a transforming coupling. To turn it into the fixed charged background required by a local R-symmetry, one must supply a globally meaningful section or background multiplet and show that its chosen profile preserves the desired supersymmetry. A numerical constant with an assigned label does not perform those steps.

This conclusion concerns the stated R-symmetric curved prescription. In flat space, the generic quadratic-plus-cubic interaction was already shown to preserve supersymmetry even when it breaks a continuous R-symmetry. The curved construction imposes additional background-coupling conditions.

2. Omit the background scalar completion

On the round sphere, set $V=0$, $H=-i/\ell$. In a neutral general multiplet take $\chi=\widetilde\chi=\lambda=0$ and let $\widetilde\lambda=\eta$ have constant odd components in the global sphere spin frame. Let $\zeta$ have constant commuting components in that same frame and set $\widetilde\zeta=0$. Both satisfy $\nabla_\mu u=-H\gamma_\mu u/2$.

Show that $\nabla_\mu(\zeta\gamma^\mu\eta)=0$. Compare the integrated variation of the correct D-term with that of the incomplete density $-D_{\mathcal S}/2$. Use $\operatorname{Vol}(S^3_\ell)=2\pi^2\ell^3$.

Hint 1

Transpose the gamma matrix when differentiating the first spinor in the bilinear. The two contributions to its divergence have opposite signs.

Hint 2

For this configuration $Y=\zeta\eta$, while $\delta D_{\mathcal S}=-HY$ and $\delta\sigma_{\mathcal S}=-Y$ once the divergence is evaluated.

Solution

Using $\gamma_\mu^TC=-C\gamma_\mu$ and $\gamma_\mu\gamma^\mu=3\mathbf1$ gives

$$\begin{aligned} \nabla_\mu(\zeta\gamma^\mu\eta) &=(\nabla_\mu\zeta)\gamma^\mu\eta +\zeta\gamma^\mu\nabla_\mu\eta\\ &=\frac{3H}{2}\zeta\eta-\frac{3H}{2}\zeta\eta=0. \end{aligned}$$

The correct density is $-(D_{\mathcal S}-H\sigma_{\mathcal S})/2$, so its variation vanishes even locally on this profile: $-(-HY-H(-Y))/2=0$. For the incomplete density, $\delta(-D_{\mathcal S}/2)=HY/2$. Its integral is

$$\delta S_{\rm incomplete} =\frac H2\,\zeta\eta\,2\pi^2\ell^3 =-i\pi^2\ell^2\zeta\eta.$$

It is nonzero for a spinor choice with $\zeta\eta\ne0$. The sphere has no boundary, so this discrepancy cannot be dismissed as a missing surface condition. It diagnoses the omitted background completion. The odd matter coefficient is a valid test of the polynomial variation; multiplying the operation by an external odd parameter gives the corresponding ordinary even variation.

3. Insert a curved linear multiplet

A neutral linear multiplet with bottom scalar $J$, vector $j_\mu$, and scalar $K$ is embedded in a general multiplet with

$$a_{\mathcal J\mu}=-j_\mu-V_\mu J,\qquad \sigma_{\mathcal J}=-K,\qquad D_{\mathcal J}=-V^\mu j_\mu-HK-\nabla^2J-V^\mu V_\mu J.$$

These are the relevant entries of P v3, (6.11). Evaluate $\mathcal L_D[\mathcal J]$. Is its integral a bulk kinetic action on a closed manifold? Reconcile its flat limit with the superspace result $\int d^4\theta\,\mathcal J=\partial^2J/4$.

Hint 1

Keep all three subtractions in $D_{\mathcal J}-a_{\mathcal J}\cdot V-\sigma_{\mathcal J}H$ before simplifying.

Hint 2

The component D-term and the flat highest polynomial coefficient differ by the bottom-component Laplacian.

Solution

Substitution gives

$$\begin{aligned} D_{\mathcal J}-a_{\mathcal J}\cdot V-\sigma_{\mathcal J}H ={}&-V\cdot j-HK-\nabla^2J-V^2J\\ &+V\cdot j+V^2J+HK\\ ={}&-\nabla^2J. \end{aligned}$$

Hence $\mathcal L_D[\mathcal J]=\nabla^2J/2$. For a globally defined smooth scalar $J$, its integral on a closed manifold vanishes. It supplies no bulk kinetic action.

In flat space the dictionary is $\mathcal L_D=\int d^4\theta\,\mathcal J+\partial^2J/4$. Together with the stated superspace coefficient, this gives $\mathcal L_D=\partial^2J/2$, in agreement. The different local coefficients multiply a boundary term; neither calculation contradicts the other or produces dynamics from a lone linear multiplet.

The useful distinction is now concrete. A chiral F-term requires the appropriate total charges and a neutral highest component. A general D-term requires a neutral multiplet and its background completion. Both proofs retain boundary currents. These prescriptions organize gauge and matter couplings, while the sphere action remains a fully expanded example against which to check them.

Paper guides

Connect this lesson to the original papers and their notation.

• Paper guide: Section 6, curved multiplets and actions