The curved vector multiplet

On a round sphere, a constant vector scalar with zero field strength need not be a supersymmetric configuration. Curvature changes the gaugino variation, and a compensating auxiliary field is required. We will calculate that correction and check how the curved transformations close into an isometry and a gauge transformation.

The prerequisites are the flat vector multiplet and the Killing spinor equations. We use their three-dimensional $\mathcal N=2$ theory, with four real Lorentzian supercharges before continuation, on an oriented Riemannian three-manifold. The gauge group here is Abelian $U(1)$; the formulae below contain no non-Abelian commutators.

Keep two kinds of data distinct. The rigid supergravity sources $(g_{\mu\nu},A_\mu,V_\mu,H,C_\mu)$ are fixed. The vector multiplet $(a_\mu,\sigma,\lambda,\widetilde\lambda,D)$ is initially a set of dynamical fields. In Euclidean calculations the gauginos are independent, and the bosonic fields may be complexified. The transformation algebra does not itself choose an integration contour.

The background and the component rules

The commuting spinor labels $\zeta,\widetilde\zeta$ carry R-charges $+1,-1$ and solve

$$\begin{aligned} \nabla_\mu\zeta &=i(A_\mu-V_\mu)\zeta-\frac H2\gamma_\mu\zeta -\frac12\epsilon_{\mu\nu\rho}V^\nu\gamma^\rho\zeta,\\ \nabla_\mu\widetilde\zeta &=-i(A_\mu-V_\mu)\widetilde\zeta-\frac H2\gamma_\mu\widetilde\zeta +\frac12\epsilon_{\mu\nu\rho}V^\nu\gamma^\rho\widetilde\zeta. \end{aligned}$$

These are P’s equations (1.4)–(1.5). The labels generate odd operations $\delta_\zeta,\delta_{\widetilde\zeta}$; their mixed algebra is an anticommutator. Their products with the Grassmann-odd gauginos obey the graded product rule. We hold labels and background sources fixed when varying the dynamical fields.

Use $\gamma_a\gamma_b=\delta_{ab}+i\epsilon_{abc}\gamma_c$, with $\epsilon_{123}=+1$. For lower-column spinors the contraction is $uv=u^{\mathsf T}Cv$, with $C=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$ and $\gamma_a^{\mathsf T}C=-C\gamma_a$. The epsilon with coordinate indices is the metric volume tensor.

The component gauge leaves the familiar redundancy $a_\mu\mapsto a_\mu+\partial_\mu\Lambda$; $\sigma,\lambda,\widetilde\lambda,D$ are gauge neutral. It is often called Wess–Zumino gauge. No superspace construction is needed to use it: it specifies the displayed component field list and its remaining gauge freedom. Supersymmetry includes the compensation needed to remain in this gauge.

Define

$$f_{\mu\nu}=\partial_\mu a_\nu-\partial_\nu a_\mu,\qquad b^\mu=\frac12\epsilon^{\mu\nu\rho}f_{\nu\rho}.$$

The complete Abelian rules, with $\delta=\delta_\zeta+\delta_{\widetilde\zeta}$, are

$$\begin{aligned} \delta a_\mu &=-i(\zeta\gamma_\mu\widetilde\lambda +\widetilde\zeta\gamma_\mu\lambda),\\ \delta\sigma&=-\zeta\widetilde\lambda+\widetilde\zeta\lambda,\\ \delta\lambda &=i\zeta(D+\sigma H)-i\gamma^\mu\zeta\,b_\mu -\gamma^\mu\zeta(i\partial_\mu\sigma-V_\mu\sigma),\\ \delta\widetilde\lambda &=-i\widetilde\zeta(D+\sigma H)-i\gamma^\mu\widetilde\zeta\,b_\mu +\gamma^\mu\widetilde\zeta(i\partial_\mu\sigma+V_\mu\sigma),\\ \delta D &=\nabla_\mu(\zeta\gamma^\mu\widetilde\lambda -\widetilde\zeta\gamma^\mu\lambda)\\ &\quad-iV_\mu(\zeta\gamma^\mu\widetilde\lambda +\widetilde\zeta\gamma^\mu\lambda) -H(\zeta\widetilde\lambda-\widetilde\zeta\lambda). \end{aligned}$$

These are P, (6.18)–(6.20). In particular, the derivative in $\delta D$ acts on the whole bilinear, including the label. Each bilinear is R-neutral, so its derivative needs no additional R-connection. The ordinary spin and vector connections are already contained in $\nabla$.

The component R-charges are $(0,0,+1,-1,0)$. With the gauge coupling absorbed into $a_\mu$, engineering dimensions are

Fields or labels	Dimension
$a_\mu,\sigma,A_\mu,V_\mu,H$	$1$
$\lambda,\widetilde\lambda$	$3/2$
$D,b_\mu,\partial_\mu\sigma$	$2$
$\zeta,\widetilde\zeta$	$-1/2$

For example, $\sigma H$ has dimension two and the same charges as $D$; multiplying by a label gives a gaugino variation of dimension $3/2$. Every line reduces to the earlier flat rules when the sources vanish and the labels are constant. That limiting check would not, by itself, fix the new coefficients.

Read the gaugino variations as matrices

Introduce two useful combinations,

$$K_D=D+H\sigma,\qquad \mathcal B_\mu=b_\mu+i\sigma V_\mu.$$

$K_D$ is a shifted auxiliary variable, not an extra field. $\mathcal B$ is a shifted field strength, generally complex. Regrouping the two gaugino rules gives

$$\begin{aligned} \delta_\zeta\lambda &=i\bigl[K_D I-\gamma^\mu(\mathcal B_\mu+\partial_\mu\sigma)\bigr]\zeta,\\ \delta_{\widetilde\zeta}\widetilde\lambda &=-i\bigl[K_D I+\gamma^\mu(\mathcal B_\mu-\partial_\mu\sigma)\bigr]\widetilde\zeta. \end{aligned}$$

For instance, $-i\gamma^\mu(i\sigma V_\mu)\zeta=+\sigma V_\mu\gamma^\mu\zeta$, reproducing the sign in the original variation. The opposite family does not vary each of these gauginos.

The same auxiliary combination appears in P’s gauge-invariant component collection (6.21): its scalar, vector, and auxiliary entries are $\sigma,-ib_\mu,K_D$. The vector entry is conserved by the Abelian Bianchi identity, $\nabla_\mu b^\mu=0$. This supplies a second reason to keep $K_D$ together; it does not require introducing the full linear-superfield formalism here.

Differentiate the label bilinears before closing the algebra

Set

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

Because the labels commute, $\widetilde\zeta\zeta=-s$ but $\widetilde\zeta\gamma^\mu\zeta=K^\mu$. Neither $s$ nor $K$ is assumed real.

Apply the product rule to $s$. The $A-V$ terms cancel between the opposite charges. Transposing a gamma matrix through $C$ reverses its sign, so the two $H$ terms also cancel. The two epsilon terms add:

$$\begin{aligned} \partial_\mu s &=(\nabla_\mu\zeta)^{\mathsf T}C\widetilde\zeta +\zeta^{\mathsf T}C\nabla_\mu\widetilde\zeta\\ &=\epsilon_{\mu\nu\rho}V^\nu K^\rho. \end{aligned}$$

Thus a scalar bilinear that was constant in flat space need not remain constant. Similarly, differentiating $K_\nu$ gives

$$\begin{aligned} \nabla_\mu K_\nu &=\frac H2\zeta[\gamma_\mu,\gamma_\nu]\widetilde\zeta +\frac12\epsilon_{\mu\alpha\rho}V^\alpha \zeta\{\gamma^\rho,\gamma_\nu\}\widetilde\zeta\\ &=\epsilon_{\mu\nu\rho}(iH K^\rho-sV^\rho). \end{aligned}$$

The right side is antisymmetric in $\mu,\nu$, proving that $K$ is a Killing vector, possibly complex. This verifies the geometry needed for the closure, rather than assuming that the flat translation vector remains constant.

For the scalar, direct composition gives

$$\begin{aligned} \{\delta_\zeta,\delta_{\widetilde\zeta}\}\sigma &=-\zeta\,\delta_{\widetilde\zeta}\widetilde\lambda +\widetilde\zeta\,\delta_\zeta\lambda\\ &=i(s-s)K_D+i(K^\mu-K^\mu)\mathcal B_\mu -2iK^\mu\partial_\mu\sigma\\ &=-2i\mathcal L_K\sigma. \end{aligned}$$

Here $\mathcal L_K\sigma=K^\mu\partial_\mu\sigma$. No field equation was used.

For the gauge potential, substitute into

$$\{\delta_\zeta,\delta_{\widetilde\zeta}\}a_\mu =-i\zeta\gamma_\mu\delta_{\widetilde\zeta}\widetilde\lambda -i\widetilde\zeta\gamma_\mu\delta_\zeta\lambda.$$

The Clifford identity gives the two combinations

$$\begin{aligned} \zeta\gamma_\mu\gamma_\nu\widetilde\zeta +\widetilde\zeta\gamma_\mu\gamma_\nu\zeta &=2i\epsilon_{\mu\nu\rho}K^\rho,\\ \zeta\gamma_\mu\gamma_\nu\widetilde\zeta -\widetilde\zeta\gamma_\mu\gamma_\nu\zeta &=2s g_{\mu\nu}. \end{aligned}$$

The auxiliary terms cancel. Expanding $\mathcal B=b+i\sigma V$ in the remaining result yields

$$\begin{aligned} \{\delta_\zeta,\delta_{\widetilde\zeta}\}a_\mu &=-2iK^\nu f_{\nu\mu} +2s\partial_\mu\sigma +2\sigma\epsilon_{\mu\nu\rho}V^\nu K^\rho\\ &=-2iK^\nu f_{\nu\mu}+\partial_\mu(2s\sigma). \end{aligned}$$

The last equality is exactly where the derivative of the labels matters. Dropping it would lose the $V$ term and give an incorrect gauge parameter.

For a one-form, $\mathcal L_K a_\mu=K^\nu\partial_\nu a_\mu+a_\nu\partial_\mu K^\nu =K^\nu f_{\nu\mu}+\partial_\mu(K^\nu a_\nu)$. Consequently,

$$\boxed{ \{\delta_\zeta,\delta_{\widetilde\zeta}\}a_\mu =-2i\mathcal L_Ka_\mu+\partial_\mu\Lambda_{\rm cl}, \qquad \Lambda_{\rm cl}=2iK^\nu a_\nu+2s\sigma.}$$

The scalar sees an isometry; the potential also sees the allowed gauge redundancy. Taking an exterior derivative removes $d\Lambda_{\rm cl}$, giving $\{\delta_\zeta,\delta_{\widetilde\zeta}\}f=-2i\mathcal L_K f$. These are off-shell checks of the bosonic algebra, not an action-invariance proof.

Freeze a vector multiplet without imposing its equations of motion

A dynamical field need not be invariant under supersymmetry: the symmetry normally moves it to another configuration. By contrast, a vector multiplet used as a fixed flavor source must itself be left unchanged. Set $\lambda=\widetilde\lambda=0$ and require both gaugino variations to vanish for the supercharges to be retained.

Suppose the rigid background admits two independent solutions in each label family. Their values span the two-component spinor fiber. Preserving all four therefore requires both matrices in the gaugino equations to vanish. The matrices $I,\gamma^1,\gamma^2,\gamma^3$ are linearly independent: their coefficients can be extracted by traces. Thus

$$K_D=0,\qquad \mathcal B_\mu+\partial_\mu\sigma=0,\qquad \mathcal B_\mu-\partial_\mu\sigma=0.$$

Adding and subtracting gives

$$\partial_\mu\sigma=0,\qquad D=-H\sigma,\qquad b_\mu=-i\sigma V_\mu.$$

The background definition $V=-i*dC$ implies $*dC=iV$. For constant $\sigma$, the last condition is therefore

$$d(a+\sigma C)=0,\qquad a=-\sigma C+a_{\rm flat}.$$

This reproduces P, (6.26). Locally $a_{\rm flat}$ is pure gauge. Globally it can retain holonomy, and the combined potentials must obey the chosen bundle and flux conditions. The local equation does not guarantee that arbitrary global source choices are allowed.

“Four” here describes the full continued algebra, with two solutions in each of the two families; it does not double the original count of real Lorentzian supercharges. With fewer retained labels, a nonzero matrix may annihilate just those labels. The stronger conditions above are then not necessary.

Worked sphere calculation

Take the supplied round-sphere background of radius $\ell$,

$$A=V=C=0,\qquad H=-\frac{i}{\ell}, \qquad \nabla_\mu\zeta=\frac{i}{2\ell}\gamma_\mu\zeta,\qquad \nabla_\mu\widetilde\zeta=\frac{i}{2\ell}\gamma_\mu\widetilde\zeta.$$

The sphere has two independent global solutions in each family. These equations specify all spinor data needed for this calculation; no coordinate expression for the solutions is required. They are the negative-imaginary-$H$ choice of the sphere case in P, section 5.2.

For constant $\sigma=\sigma_0$ and $a=0$, setting $D=0$ gives

$$\delta_\zeta\lambda=\frac{\sigma_0}{\ell}\zeta, \qquad \delta_{\widetilde\zeta}\widetilde\lambda =-\frac{\sigma_0}{\ell}\widetilde\zeta.$$

Unless $\sigma_0=0$, this is not an invariant background. The corrected choice is

$$D=\frac{i\sigma_0}{\ell},$$

which sets $K_D=0$ and kills both variations. For real $\sigma_0$, this auxiliary value is imaginary in the Euclidean convention. It is a condition on a frozen supersymmetric source, not an instruction to integrate a dynamical auxiliary field over that value or to eliminate it through an action.

There is a second curvature check in $\delta D$. Expanding its divergence gives a term from each label derivative. For example,

$$\sum_\mu(\nabla_\mu\zeta)^{\mathsf T}C\gamma^\mu\widetilde\lambda =-\frac{i}{2\ell}\sum_\mu\zeta^{\mathsf T}C\gamma_\mu\gamma^\mu\widetilde\lambda =-\frac{3i}{2\ell}\zeta\widetilde\lambda.$$

Combining both such terms with the explicit $-H$ term gives

$$\boxed{ \delta D =\zeta\gamma^\mu\nabla_\mu\widetilde\lambda -\widetilde\zeta\gamma^\mu\nabla_\mu\lambda -\frac{i}{2\ell} (\zeta\widetilde\lambda-\widetilde\zeta\lambda).}$$

The coefficient is $-i/(2\ell)$, not the $+i/\ell$ obtained by dropping the label derivatives. This is a direct diagnostic for an incomplete covariantization of the flat rule. In the large-radius limit, both the gaugino shift and this correction vanish.

Exercises

1. Isolate the derivative of the label

At one point on the sphere choose $\zeta=(1,0)^T$, set the other label to zero, and take $\widetilde\lambda=(0,\eta)^T$, where $\eta$ is Grassmann odd. Set the covariant first derivatives of both gauginos to zero at that point. Calculate $\delta D$, $\delta\sigma$, and $\delta K_D$. Which part of $\delta D$ would be missed by differentiating only the gaugino?

Hint 1

Use the specified $C$ matrix to calculate $\zeta\widetilde\lambda$. Setting a field’s covariant first derivative to zero at one point does not set the label derivative to zero.

Hint 2

Keep the divergence contribution and the explicit $-H$ contribution separate. Since the background is fixed, $\delta K_D=\delta D+H\delta\sigma$.

Solution

The contraction is $\zeta\widetilde\lambda=-\eta$. Hence the divergence contributes $+3i\eta/(2\ell)$ and the explicit term contributes $-i\eta/\ell$. Their sum is

$$\delta D=\frac{i\eta}{2\ell},\qquad \delta\sigma=+\eta,\qquad \delta K_D=-\frac{i\eta}{2\ell}.$$

Differentiating only the gaugino would give zero for the divergence and incorrectly leave $\delta D=-i\eta/\ell$. These are permitted local first-derivative data for testing a transformation; we have not assumed that a nonzero gaugino is covariantly constant everywhere on the sphere. The values of the odd fluctuations test the transformation law, not a bosonic supersymmetric background.

2. Preserve one family without preserving both

On a Cartesian flat patch with $A=V=H=C=0$, take real nonzero $c$ and

$$a_2=c x^1,\qquad a_1=a_3=0,\qquad \sigma=c x^3,\qquad D=0,\qquad \lambda=\widetilde\lambda=0.$$

Determine which constant label families leave the configuration invariant. Why does this not contradict the constant-scalar condition derived for four supercharges?

Hint 1

Compute $b$ and $d\sigma$ separately before inserting them in the two gaugino matrices.

Hint 2

The two matrices contain $\mathcal B+d\sigma$ and $\mathcal B-d\sigma$. Use $\gamma_3^2=I$ to test whether a nonzero matrix has a kernel.

Solution

The only nonzero field-strength component is $f_{12}=c$, so $b=(0,0,c)=d\sigma$. Here $K_D=0$ and $\mathcal B=b$. Therefore

$$\delta_{\widetilde\zeta}\widetilde\lambda=0 \quad\text{for every constant }\widetilde\zeta,\qquad \delta_\zeta\lambda=-2ic\gamma_3\zeta.$$

Since $\gamma_3$ is invertible and $c\ne0$, the second expression vanishes only for $\zeta=0$. The whole opposite-charge family survives, while the $\zeta$ family does not. The bosonic variations and $\delta D$ vanish because the gauginos vanish.

The four-supercharge argument required both matrices to vanish on full spinor bases. Here only one matrix must do so, allowing $b=d\sigma\ne0$. This is a local background check; it does not claim finite action, a global compact solution, or a solution to a chosen dynamical field equation.

3. A flat connection need not be removable

Take a flat three-torus with periodic spin structure, vanishing rigid sources, and a coordinate $x^3\sim x^3+L$. Let $\sigma=\sigma_0$, $D=0$, the gauginos vanish, and $a=(\alpha/L)dx^3$ with real constant $\alpha$. Does this preserve the constant spinor labels? For a unit-charge probe, calculate the phase around this circle and determine when a single-valued gauge transformation removes $a$.

Hint 1

The gauginos of an Abelian vector multiplet have no charge under $a$. The local gaugino conditions involve its field strength.

Hint 2

Use the phase $\exp(i\oint a)$ and the candidate parameter $\Lambda=-\alpha x^3/L$. The gauge function $e^{i\Lambda}$, rather than the real-valued representative $\Lambda$, must be single-valued.

Solution

All rigid labels can be constant because the chosen spin structure is periodic and the R-connection vanishes. The vector configuration has $b=d\sigma=K_D=0$, so it preserves both complete families.

Its unit-charge holonomy is

$$\exp\left(i\int_0^L\frac{\alpha}{L}\,dx^3\right)=e^{i\alpha}.$$

The candidate removing transformation changes by $-\alpha$ around the circle. Its gauge function is single-valued precisely when $\alpha\in2\pi\mathbb Z$. More generally $\alpha$ and $\alpha+2\pi n$ describe the same holonomy. Supersymmetry therefore allows globally distinct flat vector sources. The local expression $a_{\rm flat}=d\Lambda$ on a contractible patch cannot discard those global data.

You can now check the shifted gaugino matrices, the curved gauge closure, and the difference between an arbitrary dynamical field and an invariant flavor source. Constructing gauge and matter actions is the next task: it requires invariant combinations of these multiplets, not merely a list of their transformations.

Paper guides

Connect this lesson to the original papers and their notation.

• Paper guide: Section 6, curved multiplets and actions