The curved chiral multiplet

How does a chiral multiplet respond to a curved supersymmetric background? Replacing partial derivatives by covariant derivatives is only part of the answer. The background scalar $H$ also enters the transformations, and the spinor labels themselves vary under covariant differentiation. These two effects must cooperate if two supersymmetry operations are to give a spacetime symmetry.

Start with R-symmetry and flavor charges and the Killing spinor equations. We will extend the flat chiral multiplet and its off-shell closure. The first task is to use the general transformation rules correctly. The main calculation then proves closure for the constant-$H$ backgrounds needed by the round-sphere action.

Fields, labels, and the connection they use

The parent theory is three-dimensional $\mathcal N=2$, with four real Poincaré supercharges in Lorentzian signature. We use its Euclidean complexification. There is no dynamical gauge multiplet in this lesson. The background may preserve fewer supercharges; its Killing spinor solutions determine which transformations exist.

A chiral multiplet consists of $\Phi=(\phi,\psi,F)$. Its anti-chiral partner $\widetilde\Phi=(\widetilde\phi,\widetilde\psi,\widetilde F)$ is independent in Euclidean signature. Tildes do not impose a complex-conjugation condition. The bosons are even and the matter spinors are odd.

Component	Spin	Engineering dimension	R-charge	Central-charge eigenvalue
$\phi$	$0$	$1/2$	$r$	$z$
$\psi$	$1/2$	$1$	$r-1$	$z$
$F$	$0$	$3/2$	$r-2$	$z$
$\widetilde\phi$	$0$	$1/2$	$-r$	$-z$
$\widetilde\psi$	$1/2$	$1$	$1-r$	$-z$
$\widetilde F$	$0$	$3/2$	$2-r$	$-z$

These are free-field engineering dimensions, independent of the choice of $r$. They do not identify an interacting scaling dimension. The eigenvalue $z$ has mass dimension one; the R-charge is dimensionless.

We keep the convention of commuting spinor labels $\zeta,\widetilde\zeta$ and odd operations $\delta_\zeta,\delta_{\widetilde\zeta}$. Each label has dimension $-1/2$; their R-charges are $+1,-1$, respectively. They have zero central charge. The background and the labels are fixed under these operations. Their spacetime derivatives, however, need not vanish. To obtain an ordinary even variation, multiply an odd operation by an independent odd parameter on the left.

For a component of charges $(q,z_X)$, write the matter derivative explicitly as

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

Here $\nabla$ includes the spin connection when $X$ is a spinor. This is the matter connection from the background-field dictionary. It differs from both the R-current source $A-\tfrac32V$ and the connection $\nabla-iA$ appearing on the left of the $\zeta$ Killing spinor equation. The superscript will be suppressed only when the field or complete product makes its charges unambiguous.

For local algebra use lower-index spinor columns, with

$$uv=u^{\mathsf T}Cv,\qquad C=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad \gamma^\mu\gamma^\nu=g^{\mu\nu} +i\varepsilon^{\mu\nu\rho}\gamma_\rho,\qquad (\gamma^\mu)^{\mathsf T}C=-C\gamma^\mu.$$

Thus commuting labels obey $\widetilde\zeta\zeta=-\zeta\widetilde\zeta$, while $\widetilde\zeta\gamma^\mu\zeta=\zeta\gamma^\mu\widetilde\zeta$. The same exchange signs apply between a commuting label and an odd matter spinor; exchanging them produces no additional Grassmann sign.

The general transformation rules

Let $\delta=\delta_\zeta+\delta_{\widetilde\zeta}$. Each separate operation is obtained by setting the other label to zero. For a background satisfying the Killing spinor equations, the chiral transformations are

$$\begin{aligned} \delta\phi&=\sqrt2\,\zeta\psi,\\ \delta\psi&=\sqrt2\,\zeta F -\sqrt2i(z-rH)\widetilde\zeta\phi -\sqrt2i\gamma^\mu\widetilde\zeta D_\mu^{(r,z)}\phi,\\ \delta F&=\sqrt2i\bigl[z-(r-2)H\bigr]\widetilde\zeta\psi -\sqrt2iD_\mu^{(r-2,z)} (\widetilde\zeta\gamma^\mu\psi). \end{aligned}$$

The anti-chiral transformations, using the same $r,z$ to label the chiral partner, are

$$\begin{aligned} \delta\widetilde\phi&=-\sqrt2\,\widetilde\zeta\widetilde\psi,\\ \delta\widetilde\psi&=\sqrt2\,\widetilde\zeta\widetilde F +\sqrt2i(z-rH)\zeta\widetilde\phi +\sqrt2i\gamma^\mu\zeta D_\mu^{(-r,-z)}\widetilde\phi,\\ \delta\widetilde F&=\sqrt2i\bigl[z-(r-2)H\bigr]\zeta\widetilde\psi -\sqrt2iD_\mu^{(2-r,-z)} (\zeta\gamma^\mu\widetilde\psi). \end{aligned}$$

These are the component rules of Closset–Dumitrescu–Festuccia–Komargodski, version 3, equations (6.6), (6.8), and (6.10). In particular, the same displayed coefficient $z-(r-2)H$ occurs in both auxiliary rules with this labeling. Primary source, §6.2

The equations pass useful checks before any composition. The term $H\widetilde\zeta\phi$ has dimension $1-\tfrac12+\tfrac12=1$ and R-charge $-1+r=r-1$, as required for $\delta\psi$. The product $\widetilde\zeta\gamma^\mu\psi$ has R-charge $r-2$ and dimension $1/2$, so its derivative has the dimension and charge of $\delta F$. Its central charge remains $z$.

Most importantly, that derivative acts on the whole bilinear:

$$\begin{aligned} D_\mu^{(r-2,z)}(\widetilde\zeta\gamma^\mu\psi) ={}&(D_\mu^{(-1,0)}\widetilde\zeta)\gamma^\mu\psi\\ &+\widetilde\zeta\gamma^\mu D_\mu^{(r-1,z)}\psi. \end{aligned}$$

The gamma matrix is covariantly constant. The two connections add because the two R-charges add. On a background with $V\ne0$, even the derivative of the label in this identity must use the displayed matter connection; it cannot simply be read off as the left side of the Killing spinor equation.

With $H=A=V=C=z=0$ and constant labels, all six rules reduce to the flat transformations. Their curvature corrections cannot be discarded when the labels are nonparallel.

A first closure check on any allowed background

Define the scalar and vector bilinears

$$s=\zeta\widetilde\zeta,\qquad K^\mu=\zeta\gamma^\mu\widetilde\zeta.$$

An anticommutator is the appropriate composition of the two odd operations. Since $\delta_{\widetilde\zeta}\phi=0$, applying the second operation to $\delta_\zeta\phi$ gives

$$\begin{aligned} \{\delta_\zeta,\delta_{\widetilde\zeta}\}\phi &=\sqrt2\,\zeta\,\delta_{\widetilde\zeta}\psi\\ &=-2i\left[K^\mu D_\mu^{(r,z)}\phi +s(z-rH)\phi\right]. \end{aligned}$$

There is already more than covariant translation: an algebraic background-dependent rotation acts on the charged scalar. The anti-chiral calculation gives

$$\{\delta_\zeta,\delta_{\widetilde\zeta}\}\widetilde\phi =-2i\left[K^\mu D_\mu^{(-r,-z)}\widetilde\phi -s(z-rH)\widetilde\phi\right].$$

These follow directly from the rules, without a field equation. Checking spinors requires accounting for the rotation of their local frame as well.

A background family with a complete local check

For the rest of the closure proof, impose

$$A=V=C=0,\qquad H,z\ \text{constant},\qquad \nabla_\mu\zeta=-\frac H2\gamma_\mu\zeta,\qquad \nabla_\mu\widetilde\zeta=-\frac H2\gamma_\mu\widetilde\zeta.$$

We assume the stated spinor equations have solutions. This is a restriction on the supplied geometry, not a claim that arbitrary metrics admit them. It includes the round-sphere background used below. The proof in this section concerns this family; the general transformations above remain valid on the more general allowed backgrounds.

Introduce two mass-dimension-one coefficients,

$$M=z-rH,\qquad N=z-\left(r-\frac12\right)H.$$

Why does the second coefficient contain a half? Transpose the gamma matrix before contracting the derivative of the label:

$$\begin{aligned} (\nabla_\mu\widetilde\zeta)\gamma^\mu\psi &=-\frac H2(\gamma_\mu\widetilde\zeta)\gamma^\mu\psi\\ &=\frac H2\widetilde\zeta\gamma_\mu\gamma^\mu\psi =\frac{3H}{2}\widetilde\zeta\psi. \end{aligned}$$

Consequently the expanded auxiliary rules are

$$\begin{aligned} \delta F&=\sqrt2iN\,\widetilde\zeta\psi -\sqrt2i\,\widetilde\zeta\gamma^\mu\nabla_\mu\psi,\\ \delta\widetilde F&=\sqrt2iN\,\zeta\widetilde\psi -\sqrt2i\,\zeta\gamma^\mu\nabla_\mu\widetilde\psi. \end{aligned}$$

The shift from $z-(r-2)H$ to $N$ has absorbed the label derivative. It is not a new choice of transformation.

Differentiate the label bilinears using the same transpose identity:

$$\begin{aligned} \nabla_\mu s&=0,\\ \nabla_\mu K_\nu &=\frac H2\zeta[\gamma_\mu,\gamma_\nu]\widetilde\zeta =iH\varepsilon_{\mu\nu\rho}K^\rho. \end{aligned}$$

Thus $\nabla_{(\mu}K_{\nu)}=0$: $K$ generates an isometry, possibly complex. On a scalar its Lie derivative is $K^\mu\partial_\mu$. On a spinor one must also rotate the orthonormal frame. The antisymmetric frame rotation contributes $\tfrac18(\nabla_\mu K_\nu)[\gamma^\mu,\gamma^\nu]$, giving

$$\mathcal L_K\psi =K^\mu\nabla_\mu\psi +\frac{i}{4}(\nabla_\mu K_\nu) \varepsilon^{\mu\nu\rho}\gamma_\rho\psi =K^\mu\nabla_\mu\psi-\frac H2K^\mu\gamma_\mu\psi.$$

The last equality uses $\varepsilon_{\mu\nu\rho}\varepsilon^{\mu\nu\sigma}=2\delta_\rho{}^\sigma$. This defines the spinorial Lie derivative needed here and fixes its sign in the chosen convention. Primary source, equation (A.13)

Fermion closure and the frame rotation

Two small matrix identities organize the calculation. Define the outer products

$$a=\widetilde\zeta\zeta^{\mathsf T}C,\qquad b=\zeta\widetilde\zeta^{\mathsf T}C,\qquad K\!\!\!/\,=K^\mu\gamma_\mu.$$

Every two-by-two matrix expands in $I,\gamma^\mu$. Taking its trace and gamma traces gives

$$a=\frac12(K\!\!\!/\,+sI),\qquad b=\frac12(K\!\!\!/\,-sI).$$

For example, $\operatorname{tr}a=s$ and $\operatorname{tr}(\gamma^\mu a)=K^\mu$ determine the first expansion. The Clifford algebra then gives, with repeated orthonormal indices summed,

$$\gamma^\mu a+b\gamma^\mu=K^\mu I,\qquad \gamma_\mu a\gamma^\mu=2sI-a.$$

Now compose the fermion transformations. Keeping the derivative of $\zeta$ visible, the result is

$$\begin{aligned} \{\delta_\zeta,\delta_{\widetilde\zeta}\}\psi=-2i\bigl[ &(\gamma^\mu\widetilde\zeta)(\zeta\nabla_\mu\psi) +\zeta(\widetilde\zeta\gamma^\mu\nabla_\mu\psi)\\ &+M\widetilde\zeta(\zeta\psi)-N\zeta(\widetilde\zeta\psi) +(\gamma^\mu\widetilde\zeta)((\nabla_\mu\zeta)\psi) \bigr]. \end{aligned}$$

The first matrix identity combines the first line into $K^\mu\nabla_\mu\psi$. In the second line, the label derivative becomes $\tfrac H2\gamma_\mu a\gamma^\mu\psi$. Its entire coefficient matrix is therefore

$$\begin{aligned} Ma-Nb+\frac H2(2sI-a) &=-\frac H2K\!\!\!/\, +s\bigl[z-(r-1)H\bigr]I. \end{aligned}$$

The first term is exactly the frame rotation in $\mathcal L_K$. The second knows the fermion’s R-charge $r-1$. We have obtained

$$\boxed{\{\delta_\zeta,\delta_{\widetilde\zeta}\}\psi =-2i\left[\mathcal L_K\psi +s\bigl[z-(r-1)H\bigr]\psi\right].}$$

The auxiliary field supplied the term containing $N$. No Dirac equation was imposed, so this is off-shell closure.

Auxiliary closure and the complete multiplet

The scalar auxiliary calculation is shorter and checks the coefficient independently:

$$\begin{aligned} \{\delta_\zeta,\delta_{\widetilde\zeta}\}F ={}&2iN(\widetilde\zeta\zeta)F -2i\,\widetilde\zeta\gamma^\mu \nabla_\mu(\zeta F)\\ ={}&-2i\left[K^\mu\partial_\mu F +s\left(N+\frac{3H}{2}\right)F\right]\\ ={}&-2i\left[\mathcal L_KF +s\bigl[z-(r-2)H\bigr]F\right]. \end{aligned}$$

Here $\widetilde\zeta\gamma^\mu\nabla_\mu\zeta=3Hs/2$, and $\mathcal L_KF=K^\mu\partial_\mu F$. Together with the scalar and fermion results, the answer has a single charge-based form:

$$\boxed{ \{\delta_\zeta,\delta_{\widetilde\zeta}\}X =-2i\left[\mathcal L_KX+s(z_X-q_XH)X\right]. }$$

Use the table’s $(q_X,z_X)$ for each of the six fields. In particular, the independent tilded sector obeys

$$\begin{aligned} \{\delta_\zeta,\delta_{\widetilde\zeta}\}\widetilde\psi &=-2i\left[\mathcal L_K\widetilde\psi -s\bigl[z-(r-1)H\bigr]\widetilde\psi\right],\\ \{\delta_\zeta,\delta_{\widetilde\zeta}\}\widetilde F &=-2i\left[\mathcal L_K\widetilde F -s\bigl[z-(r-2)H\bigr]\widetilde F\right]. \end{aligned}$$

One obtains the first by the same two matrix expansions with the roles of the labels exchanged; the second is derived in Exercise 2.

Each single-label operation is nilpotent. On $\phi$, for example, $\delta_\zeta^2\phi=2(\zeta\zeta)F=0$. The longest same-label check is on the opposite auxiliary field. Its scalar Hessian contracts with an antisymmetric gamma product, its scalar terms contain $\zeta\zeta=0$, and its remaining scalar-gradient coefficient is proportional to

$$M-N+\frac H2=0.$$

These exhaust that variation; the tilded calculation has the same cancellation. Since the statements hold for arbitrary commuting solutions, replacing a label by a sum of two solutions also gives the vanishing same-family anticommutators. The resulting algebra is the specialization of the paper’s equations (6.1)–(6.2) to the background family just checked. Primary source, §6.1

Worked example: supplied round-sphere data

Take a round three-sphere of radius $\ell>0$, with

$$A=V=C=0,\qquad H=-\frac i\ell,\qquad z=0.$$

Use an oriented global orthonormal frame in which the spin connection acts on a constant-component spinor as $i\gamma_a/(2\ell)$. Such a spinor satisfies the stated equation $\nabla_a\zeta=i\gamma_a\zeta/(2\ell)$, and likewise for $\widetilde\zeta$. The geometric construction of that frame is a separate background calculation; here these are supplied data whose substitution can be checked directly.

The scalar rules are unchanged. The remaining transformations become

$$\begin{aligned} \delta\psi&=\sqrt2\zeta F +\frac{\sqrt2r}{\ell}\widetilde\zeta\phi -\sqrt2i\gamma^\mu\widetilde\zeta\nabla_\mu\phi,\\ \delta\widetilde\psi&=\sqrt2\widetilde\zeta\widetilde F -\frac{\sqrt2r}{\ell}\zeta\widetilde\phi +\sqrt2i\gamma^\mu\zeta\nabla_\mu\widetilde\phi,\\ \delta F&=-\frac{\sqrt2(r-\frac12)}{\ell}\widetilde\zeta\psi -\sqrt2i\,\widetilde\zeta\gamma^\mu\nabla_\mu\psi,\\ \delta\widetilde F&=-\frac{\sqrt2(r-\frac12)}{\ell}\zeta\widetilde\psi -\sqrt2i\,\zeta\gamma^\mu\nabla_\mu\widetilde\psi. \end{aligned}$$

All algebraic corrections vanish as $\ell\to\infty$. Notice the different coefficients $r$ and $r-\tfrac12$: the label derivative explains their difference.

For an explicit field configuration choose $\phi=v$ constant, $\psi=F=0$, and commuting frame components

$$\zeta=t\begin{pmatrix}1\\0\end{pmatrix},\qquad \widetilde\zeta=\widetilde t\begin{pmatrix}0\\1\end{pmatrix},\qquad \gamma_a=(\sigma_3,-\sigma_1,-\sigma_2).$$

The amplitudes $t,\widetilde t$ have dimension $-1/2$, while $v$ has dimension $1/2$. Direct multiplication gives $s=-t\widetilde t$ and $K^a=(t\widetilde t,0,0)$. Although $\phi$ is constant, its configuration is not generally fixed by supersymmetry:

$$\delta_{\widetilde\zeta}\psi =\frac{\sqrt2r}{\ell}\widetilde\zeta v,\qquad \{\delta_\zeta,\delta_{\widetilde\zeta}\}\phi =-\frac{2t\widetilde t\,r}{\ell}v.$$

The second result agrees with the charge rotation in the closure formula, since $K^\mu\partial_\mu v=0$. This is a test of a field configuration, not a claim that it solves an action’s equations or defines a vacuum. At $r=0$ this particular obstruction disappears. At finite radius and generic $r$, “constant scalar” is insufficient.

Practice

1. Give the entire product its charge

Write $a_\mu=A_\mu-\tfrac12V_\mu$ and $B^\mu=\widetilde\zeta\gamma^\mu\psi$. A calculation differentiates $B^\mu$ with $D_\mu^{(r,z)}$ instead of $D_\mu^{(r-2,z)}$. Find the error in $\delta F$. Explain why choosing a gauge with $a_\mu=0$ at one point does not repair the transformation law.

Hint 1

The R-charge of a product is the sum of the charges of its factors.

Hint 2

Subtract the two derivatives before multiplying by the auxiliary transformation’s coefficient $-\sqrt2i$.

Complete solution

The product has R-charge $-1+(r-1)=r-2$ and central charge $z$. The wrong derivative differs by

$$(D_\mu^{(r,z)}-D_\mu^{(r-2,z)})B^\mu =-2ia_\mu B^\mu.$$

Consequently

$$\delta F_{\mathrm{wrong}}-\delta F =-2\sqrt2\,a_\mu\widetilde\zeta\gamma^\mu\psi.$$

The sign follows from $(-i)(-2i)=-2$. Its dimension is $1-\tfrac12+1=3/2$, so dimensional analysis alone misses the error.

Under an R-gauge change, $B$ transforms with charge $r-2$ and $a$ changes by the gradient of the gauge parameter. Only the derivative with that charge cancels the differentiated phase. A gauge in which the connection vanishes at one point conceals the mismatch there; it does not make the rule transform covariantly between gauges.

2. Close on the anti-chiral auxiliary

Use the constant-$H$ family to derive the mixed anticommutator on $\widetilde F$ directly. Keep $\widetilde F$ independent of $F$ and explain the sign of its charge rotation.

Hint 1

Only $\delta_{\widetilde\zeta}\delta_\zeta\widetilde F$ contributes. Substitute $\delta_{\widetilde\zeta}\widetilde\psi=\sqrt2\widetilde\zeta\widetilde F$.

Hint 2

In this order, $\zeta\widetilde\zeta=s$ and $\zeta\gamma^\mu\nabla_\mu\widetilde\zeta=-3Hs/2$.

Complete solution

The reduced auxiliary rule gives

$$\begin{aligned} \{\delta_\zeta,\delta_{\widetilde\zeta}\}\widetilde F &=2iN\,s\widetilde F -2i\zeta\gamma^\mu\nabla_\mu (\widetilde\zeta\widetilde F)\\ &=-2iK^\mu\partial_\mu\widetilde F +2i\left(N+\frac{3H}{2}\right)s\widetilde F\\ &=-2i\left[ \mathcal L_K\widetilde F -s\bigl[z-(r-2)H\bigr]\widetilde F\right]. \end{aligned}$$

The table gives $q_{\widetilde F}=2-r$ and $z_{\widetilde F}=-z$, so

$$z_{\widetilde F}-q_{\widetilde F}H =-\bigl[z-(r-2)H\bigr].$$

This reproduces the universal charge formula. The opposite sign follows from the partner’s charges, not from conjugating $H$ or the spinor equation. Setting $H=z=0$ leaves the ordinary flat translation.

3. Catch an omitted label derivative

Suppose someone replaces the whole derivative in $\delta F$ by a derivative of $\psi$ alone, while retaining the coefficient $z-(r-2)H$. On the constant-$H$ family, calculate the resulting error in the mixed closure on $\psi$. Give a choice of label and field components that detects it.

Hint 1

The incorrect reduced coefficient is $N_{\mathrm{wrong}}=N+\tfrac32H$.

Hint 2

The affected term in the mixed fermion closure comes from $\delta_{\widetilde\zeta}(\sqrt2\zeta F)$.

Complete solution

The auxiliary variation acquires

$$\delta_{\widetilde\zeta}F_{\mathrm{wrong}} -\delta_{\widetilde\zeta}F =\frac{3\sqrt2iH}{2}\widetilde\zeta\psi.$$

Hence the mixed fermion closure acquires

$$\Delta_{\mathrm{error}}\psi =3iH\,\zeta(\widetilde\zeta\psi)=3iH\,b\psi.$$

For $\zeta=t(1,0)^{\mathsf T}$ and $\widetilde\zeta=\widetilde t(0,1)^{\mathsf T}$, the matrix is $b=t\widetilde t\,\operatorname{diag}(1,0)$. Choose an odd spinor with nonzero first component. The error is then nonzero whenever $Ht\widetilde t\ne0$. On the sphere it is $3t\widetilde t\,\psi_1/\ell$ in that component.

The error vanishes at $H=0$, which explains why a flat-space check alone would not reveal it. At nonzero $H$, it changes the already fixed algebra; it is not removed by calling the spinor label “constant” in a frame. Constant frame components still have a spin connection derivative.

From transformations to an action

The supplied background determines the allowed labels; the component charges determine their connections; the complete auxiliary derivative makes the curved algebra work off shell. The complete round-sphere action uses these rules to determine and check the scalar and fermion terms. Closure is an essential algebraic check, but action invariance still requires its own variation. The systematic F- and D-term construction derives reusable invariant densities from multiplet products and their boundary currents.

Paper guides

Connect this lesson to the original papers and their notation.

• Paper guide: Section 6, curved multiplets and actions