The curved supersymmetry algebra

In flat space, two supersymmetry operations produce a translation, possibly accompanied by a central-charge transformation. On a curved background there may be no translation symmetry. The replacement is an isometry together with internal rotations whose coefficients depend on the fixed sources. We will compute those coefficients, rather than guess them from the metric.

The required preparation is the Killing spinor equation and superspace as a component package. Our calculation uses a general multiplet, so it does not assume chirality or a matter equation of motion.

Choose the operations and the component package

The parent theory is three-dimensional $\mathcal N=2$, with four real Lorentzian Poincaré supercharges. We work with its Euclidean complexification on a real Riemannian metric. The fixed sources $A,V,H$ may be complex, with $\nabla\cdot V=0$ and $V=-i*dC$, $H=i*dB$ locally. No dynamical gauge group is needed.

Choose commuting solutions $\zeta,\widetilde\zeta$ of the two Killing spinor equations, with R-charges $+1,-1$ and zero central charge. The associated operations $\delta_\zeta,\delta_{\widetilde\zeta}$ are odd. Their appropriate bracket is an anticommutator. The sources and labels do not transform under these matter operations; derivatives of a label need not vanish.

Write a general multiplet as

$$\mathcal S=(X,\chi,\widetilde\chi,M,\widetilde M, a_\mu,\sigma,\lambda,\widetilde\lambda,D).$$

We call its bottom component $X$; the paper calls it $C$, which would conflict here with the background one-form $C$. The multiplet is the curved counterpart of an unconstrained complex superfield. Its flat polynomial begins

$$\mathcal S=X+i\theta\chi+i\widetilde\theta\widetilde\chi +\frac i2\theta^2M+\frac i2\widetilde\theta^2\widetilde M +(\theta\gamma^\mu\widetilde\theta)a_\mu -i\theta\widetilde\theta\,\sigma+\cdots .$$

The omitted higher coefficients complete the package. In particular, $a_\mu$ is an independent vector component, not automatically a gauge connection.

Let $X$ have R-charge $q$ and central-charge eigenvalue $z$. All components have the same $z$; their R-charges are

Components	R-charge
$X,a_\mu,\sigma,D$	$q$
$\chi,\widetilde\lambda$	$q-1$
$\widetilde\chi,\lambda$	$q+1$
$M,\widetilde M$	$q-2,\ q+2$

Here $q$ is dimensionless and $z$ has mass dimension one. If $X$ has engineering dimension $\Delta$, $\chi,\widetilde\chi$ have dimension $\Delta+1/2$, and $M,\widetilde M,a_\mu,\sigma$ have dimension $\Delta+1$. These are assignments in an off-shell package, not a count of propagating particles.

For a field of charges $(q,z)$, use

$$D_\mu=\nabla_\mu-iq\mathcal A_\mu-izC_\mu, \qquad \mathcal A_\mu=A_\mu-\frac12V_\mu . \tag{R07.1}$$

The symbol $\mathcal A$ is only a convenient name for the matter R connection. It does not replace $A$ in the Killing equation or $A-3V/2$ in the current-source coupling.

Compose the general scalar transformations

Only the first three rules of P (6.5) are needed:

$$\begin{aligned} \delta X&=i\zeta\chi+i\widetilde\zeta\widetilde\chi,\\ \delta\chi&=\zeta M-\widetilde\zeta\bigl[\sigma+(z-qH)X\bigr] -\gamma^\mu\widetilde\zeta(D_\mu X+ia_\mu),\\ \delta\widetilde\chi&=\widetilde\zeta\widetilde M -\zeta\bigl[\sigma-(z-qH)X\bigr] -\gamma^\mu\zeta(D_\mu X-ia_\mu). \end{aligned} \tag{R07.2}$$

Setting one label to zero selects one operation. These are supplied curved transformation rules; composing them checks their algebra, not their entire supergravity derivation.

Use lower-index columns with contraction $uv=u^TC_{\rm spin}v$, where $C_{\rm spin}=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$. The identity $\gamma_\mu^TC_{\rm spin}=-C_{\rm spin}\gamma_\mu$ gives $\widetilde\zeta\zeta=-\zeta\widetilde\zeta$ and $\widetilde\zeta\gamma^\mu\zeta=\zeta\gamma^\mu\widetilde\zeta$. Define

$$s=\zeta\widetilde\zeta,\qquad K^\mu=\zeta\gamma^\mu\widetilde\zeta,\qquad \mu_X=z-qH .$$

The two orders of composition are

$$\begin{aligned} \delta_\zeta\delta_{\widetilde\zeta}X &=+is(\sigma-\mu_X X)-iK^\mu D_\mu X-K^\mu a_\mu,\\ \delta_{\widetilde\zeta}\delta_\zeta X &=-is(\sigma+\mu_X X)-iK^\mu D_\mu X+K^\mu a_\mu . \end{aligned} \tag{R07.3}$$

Thus the independent $\sigma$ and $a_\mu$ terms cancel:

$$\boxed{\{\delta_\zeta,\delta_{\widetilde\zeta}\}X =-2i\bigl[K^\mu D_\mu X+s(z-qH)X\bigr].} \tag{R07.4}$$

No constraint on those components and no field equation was used. For two labels of the same R-charge, the scalar calculation instead gives $i(\zeta\eta+\eta\zeta)M=0$; the opposite sector similarly gives zero with $\widetilde M$. Each definite-charge odd operation is therefore nilpotent on $X$.

For comparison with ordinary even variations, put independent odd parameters on the left: $\Delta_1=\epsilon_1\delta_1$, $\Delta_2=\epsilon_2\delta_2$. Moving an odd operation through the other parameter gives $[\Delta_1,\Delta_2]=-\epsilon_1\epsilon_2\{\delta_1,\delta_2\}$. This conversion prevents an extra sign from being mistaken for a change of the physics.

Interpret the geometric and internal transformations

For a scalar, its Lie derivative is $\mathcal L_KX=K^\mu\partial_\mu X$. For a covector it also transports the index:

$$(\mathcal L_Kb)_\mu =K^\nu\nabla_\nu b_\mu+(\nabla_\mu K^\nu)b_\nu.$$

For a spinor it includes the local frame rotation,

$$\mathcal L_K\psi =K^\mu\nabla_\mu\psi +\frac i4(\nabla_\mu K_\nu) \epsilon^{\mu\nu\rho}\gamma_\rho\psi . \tag{R07.5}$$

This is P (A.13), used here for a Killing vector and extended complex-linearly when necessary. It is not generally just $K^\mu\nabla_\mu$.

Why is our $K$ Killing? Substitute both Killing spinor equations into the derivative of its bilinear and use the Clifford product. The $A$ terms cancel between opposite charges; the remaining terms combine into

$$\nabla_\mu K_\nu =\epsilon_{\mu\nu\rho}(iH K^\rho-sV^\rho). \tag{R07.6}$$

Its symmetric part vanishes. This short identity supplies the isometry needed here; neither $K$ nor $s$ has been required to be real or constant.

For any spin and that component’s charges, define the covariant Lie derivative

$$\mathcal L'_K =\mathcal L_K-iqK^\mu\mathcal A_\mu-izK^\mu C_\mu .$$

The general algebra realized by the complete multiplet rules is

$$\{\delta_\zeta,\delta_{\widetilde\zeta}\} =-2i\mathscr B,\qquad \mathscr B=\mathcal L'_K+s(z-qH),\qquad \{\delta_\zeta,\delta_\eta\} =\{\delta_{\widetilde\zeta},\delta_{\widetilde\eta}\}=0. \tag{R07.7}$$

This is the sourced result P (6.1)–(6.2). Equation (R07.4) proves it on a general bottom scalar. Checking all higher components uses their additional transformation rules and the Killing equations; a bottom-scalar check alone does not establish that full result. The curved chiral lesson gives a complete component proof for its stated constant-$H$ family.

To read the internal rotations explicitly, define their action by $\delta_R(\Lambda)X=iq\Lambda X$ and $\delta_Z(\Omega)X=iz\Omega X$. Then

$$\begin{aligned} \mathscr B&=\mathcal L_K+\delta_R(\Lambda_R)+\delta_Z(\Lambda_Z),\\ \Lambda_R&=-K^\mu\mathcal A_\mu+isH,\qquad \Lambda_Z=-K^\mu C_\mu-is . \end{aligned} \tag{R07.8}$$

These are generally complex, position-dependent coefficients in the complexified algebra. The eigenvalue $z$ remains constant: a varying coefficient of its transformation is not a new position-dependent central charge. With labels of dimension $-1/2$, $K,s$ have dimension $-1$, $\Lambda_R$ is dimensionless, and $\Lambda_Z$ has length dimension, as required.

There is a direct covariance test. Under local source transformations $A'=A+d\lambda$, $C'=C+d\omega$ and $X'=e^{i(q\lambda+z\omega)}X$, the two label phases cancel in $K,s$. The compensators change by $\Lambda_R'=\Lambda_R-K\lambda$ and $\Lambda_Z'=\Lambda_Z-K\omega$. Those terms cancel the derivative of the phase in $\mathcal L_KX'$, giving

$$\mathscr B'X'=e^{i(q\lambda+z\omega)}\mathscr BX . \tag{R07.9}$$

This is a local statement. Allowed transformations around noncontractible cycles also depend on the specified source groups and charge lattices.

A varying source whose R terms cancel

Consider flat coordinates $(\tau,x,y)$ with orientation $d\tau\wedge dx\wedge dy>0$. Take real $\kappa$ of mass dimension two and the smooth fixed background

$$\begin{aligned} H&=\kappa x,&V&=-iH\,d\tau,&A&=-\frac{3i}{2}H\,d\tau,\\ C&=\frac{\kappa x^2}{2}\,dy,& B&=\frac{i\kappa x^2}{2}\,d\tau\wedge dy . \end{aligned} \tag{R07.10}$$

Indeed $dC=\kappa x\,dx\wedge dy$ gives $V=-i*dC$, while $dB=-i\kappa x\,d\tau\wedge dx\wedge dy$ gives $H=i*dB$. Also $\nabla\cdot V=0$. The background is complex, as allowed in the Euclidean construction; no positivity claim about a matter integral is needed.

Use $\gamma=(\sigma_3,-\sigma_1,-\sigma_2)$ and constant labels $\zeta=u_+=(1,0)^T$, $\widetilde\zeta=u_-=(0,1)^T$, with a chosen common normalization. Write the full Killing operator as $\partial_\mu+\mathcal M_\mu^\pm$. Its matrices are

$$\begin{aligned} \mathcal M_\tau^\pm&=\frac H2(\gamma_1\mp\mathbf1),\\ \mathcal M_x^\pm&=\frac H2(\gamma_2\pm i\gamma_3),\\ \mathcal M_y^\pm&=\frac H2(\gamma_3\mp i\gamma_2). \end{aligned} \tag{R07.11}$$

All three plus matrices annihilate $u_+$; all three minus matrices annihilate $u_-$. Thus the labels solve the complete equations even though $dH\ne0$. Constant $H$ was never a general requirement for one solution of each charge.

Their bilinears are $K=\partial_\tau$ and $s=-1$. Since $\mathcal A=A-V/2=-iH\,d\tau$,

$$K^\mu D_\mu X=(\partial_\tau-qH)X,\qquad s(z-qH)X=(-z+qH)X.$$

Consequently

$$\boxed{\mathscr BX=(\partial_\tau-z)X.} \tag{R07.12}$$

The varying $qH$ terms cancel exactly. In (R07.8) this is $\Lambda_R=0$, $\Lambda_Z=i$. Replacing the matter connection by $A$ alone would leave a spurious $-qH/2$ in $\mathscr B$; using the current source $A-3V/2$ would instead leave $+qH$.

For $X=e^{ip\tau}f(x,y)$, the odd anticommutator is $2(p+iz)X$. It is an algebra action, not an asserted positive Hamiltonian: the two Euclidean operations have not been related by a Hilbert-space adjoint. The flat-source limit $\kappa=0$ retains the usual translation and central-charge term.

Exercises

1. Compose transformations on a product

Let $X,Y$ be even bottom components of general multiplets with charges $(q_X,z_X)$ and $(q_Y,z_Y)$. Show that the mixed anticommutator acts on $XY$ with the summed charges. Track the terms containing one odd variation of each field. What remains when both charge sums vanish?

Hint 1

An odd operation obeys $\delta(UV)=(\delta U)V+(-1)^{|U|}U\delta V$. After one operation, $\delta X$ and $\delta Y$ are odd.

Hint 2

Expand both orders of composition before adding them. The potentially troublesome cross terms have opposite signs.

Solution

Write the two operations as $\delta,\widetilde\delta$. In one order,

$$\delta\widetilde\delta(XY) =(\delta\widetilde\delta X)Y -(\widetilde\delta X)(\delta Y) +(\delta X)(\widetilde\delta Y) +X(\delta\widetilde\delta Y).$$

The other order has the negative of the two middle terms. Therefore $\{\delta,\widetilde\delta\}(XY)= (\{\delta,\widetilde\delta\}X)Y+ X(\{\delta,\widetilde\delta\}Y)$. Both the connection and the algebraic term add:

$$\mathscr B(XY)= K^\mu D_\mu^{(q_X+q_Y,z_X+z_Y)}(XY) +s[z_X+z_Y-(q_X+q_Y)H]XY .$$

For a neutral product this reduces to $\mathcal L_K(XY)$. The cancellation did not require $\chi$ or $\widetilde\chi$ to vanish. Using an ordinary ungraded product rule at the second step would leave false fermion bilinears.

2. Detect the missing spin rotation

As a kinematic check of (R07.5), take Cartesian flat space and the rotation $K=x\partial_y-y\partial_x$, where $x,y$ are the second and third coordinates. Evaluate $\mathcal L_K$ on a constant spinor and its finite rotation. Explain why replacing $\mathcal L_K$ by $K\cdot\nabla$ fails. Does this test by itself establish a new supersymmetry generator on the zero-source flat background?

Hint 1

The two nonzero entries are $\partial_xK_y=1$ and $\partial_yK_x=-1$. Both ordered index pairs contribute.

Hint 2

The constant Cartesian components have zero directional derivative. Compare a rotation through $2\pi$ with its spin lift.

Solution

The spin term is $i\gamma_1/2$, so

$$\mathcal L_K\psi=\frac i2\gamma_1\psi,\qquad e^{t\mathcal L_K}\psi=e^{it\gamma_1/2}\psi$$

for a constant spinor. At $t=2\pi$ the lift is $-\mathbf1$. The directional derivative alone would predict no change and miss this sign.

This is a test of geometric spin transport, not a new solution of the Killing spinor equations. Zero-source Cartesian Killing spinors are constant, and their mixed vectors are constant translations. A rotation field supplied independently does not establish that it occurs as such a bilinear on that background.

3. Check a central Wilson line on the circle

Make $\tau$ periodic with length $L$ in (R07.10), choose periodic spin structure, and add the flat term $c\,d\tau$ to $C$, with real $c$. The potentials and constant labels are global on $S^1_L\times\mathbb R^2$.

For this exercise specify a compact central-source group with a positive fundamental charge $z_0$, of mass dimension one. The normalized connection is $z_0C$ and all central charges are $z=N z_0$, $N\in\mathbb Z$. Find the algebra action on a periodic scalar mode $e^{2\pi in\tau/L}f(x,y)$. Check its invariance under the allowed large transformation $\omega=2\pi w\tau/(z_0L)$, $w\in\mathbb Z$.

Hint 1

The new term changes $K\cdot D$ by $-izc$. It does not change $V$ or the Killing spinor equations.

Hint 2

Transform the mode as well as the connection: $X'=e^{iz\omega}X$. Its Fourier integer changes.

Solution

The bosonic operator in the odd bracket is

$$\mathscr B=\partial_\tau-izc-z,\qquad \mathscr B X_n= \left[i\left(\frac{2\pi n}{L}-zc\right)-z\right]X_n.$$

Thus the odd anticommutator has eigenvalue $2(2\pi n/L-zc)+2iz$. The allowed transformation changes $c'=c+2\pi w/(z_0L)$ and $n'=n+Nw$. It follows that $2\pi n'/L-zc'=2\pi n/L-zc$. The full eigenvalue is invariant, although a fixed untransformed Fourier label would suggest otherwise.

The period and charge lattice were additional global data in this exercise. The local algebra did not derive them. All background field strengths and both Killing spinors are unchanged by this flat central connection because the labels have $z=0$.

The useful output is an operator with three identifiable parts: geometric transport, an R rotation, and a central-charge transformation. Keeping their source dependence and component charges explicit makes subsequent invariant-action calculations possible without confusing off-shell closure with a matter equation of motion.

Recover the linearized flat-space description next uses the current multiplet to construct a first-order background deformation and checks its component supersymmetry preimage.

Paper guides

Connect this lesson to the original papers and their notation.

• Paper guide: Section 6, curved multiplets and actions
• Paper guide: Appendices A and B, conventions and geometry