Derive and interpret the Killing spinor equations

A frozen background need not admit a parallel spinor. Its extra sources can compensate for the spin connection, so that a position-dependent supersymmetry label still generates a symmetry. The equation describing that compensation is called a Killing spinor equation. It is an equation for a symmetry parameter, not an equation of motion for a propagating fermion.

The required preparation is the background-field dictionary and the covariant spinor derivative. We will obtain the equation from a stated gravitino transformation, verify its gauge covariance, and solve a flat example. The reconstruction of the entire supercurrent multiplet is not needed for this calculation.

We work on a smooth, oriented Riemannian three-manifold with real positive metric. The parent theory has three-dimensional $\mathcal N=2$ supersymmetry: four real Poincaré supercharges before Euclidean continuation. Its full algebra is manifest in the two label families used below; a particular background may preserve fewer solutions. No gauge rank enters this background problem.

The transformation supplied by supergravity

Write the background gravitini as $\Psi_\mu,\widetilde\Psi_\mu$, to distinguish them from the matter spinor $\psi$ of a chiral multiplet. Set both background gravitini to zero. Let $\zeta$ and $\widetilde\zeta$ be commuting two-component spinor labels of R-charge $+1$ and $-1$ respectively. They are independent complex fields in Euclidean signature.

The needed input is the rigid-limit gravitino transformation, equation (2.15) of the three-manifold paper, arXiv v3:

$$\begin{aligned} \delta\Psi_\mu &=2(\nabla_\mu-iA_\mu)\zeta +H\gamma_\mu\zeta+2iV_\mu\zeta +\epsilon_{\mu\nu\rho}V^\nu\gamma^\rho\zeta,\\ \delta\widetilde\Psi_\mu &=2(\nabla_\mu+iA_\mu)\widetilde\zeta +H\gamma_\mu\widetilde\zeta-2iV_\mu\widetilde\zeta -\epsilon_{\mu\nu\rho}V^\nu\gamma^\rho\widetilde\zeta . \end{aligned}$$

Terms proportional to a background gravitino have already vanished. The coefficients in this transformation are supplied physical input; setting the variation to zero does not itself derive the off-shell supergravity transformation law.

Here $\nabla_\mu$ contains the Levi-Civita spin connection, $\gamma_\mu=e_\mu{}^a\gamma_a$, and $\gamma_a\gamma_b=\delta_{ab}+i\epsilon_{abc}\gamma_c$ with $\epsilon_{123}=+1$ in an oriented orthonormal frame. The coordinate epsilon is the metric volume tensor. As in the preceding lesson, $A,V,H$ may be complex, $\nabla_\mu V^\mu=0$, and R-gauge transformations have real parameters.

All terms have the same dimension. With length coordinates, $[\nabla]=[A]=[V]=[H]=1$ and $[\zeta]=[\widetilde\zeta]=-1/2$, so each displayed gravitino variation has dimension $1/2$. The labels carry no central charge; $C_\mu$ enters their equation through $V$, not through an additional $-izC_\mu$ connection.

The vanishing of both variations is the requirement that the chosen bosonic background stay fixed. To preserve only one label family, the other label can be set to zero. Bosonic background variations are then zero because they contain background fermions. The matter transformations and action must still be the corresponding completed curved ones.

Divide by two and keep every sign

Move the three nonderivative terms to the right and divide by two:

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

These are the paper’s equations (1.4)–(1.5), repeated as (3.1). The two factors of one-half come from the coefficient two of the derivative in the input variation. The $H$ term has the same sign in both equations.

For calculation it helps to combine the two equations without losing their charges. Put $\varepsilon_+=\zeta$, $\varepsilon_-=\widetilde\zeta$, and let $s=+1,-1$ label these choices. Define

$$\mathscr K_\mu^{(s)} =\nabla_\mu-isA_\mu+\frac H2\gamma_\mu +isV_\mu+\frac s2\epsilon_{\mu\nu\rho}V^\nu\gamma^\rho.$$

Then the complete condition is $\mathscr K_\mu^{(s)}\varepsilon_s=0$ for every $\mu$. A label which solves only one component is not yet a preserved symmetry.

The passage from the first equation to the second is

$$\zeta\longmapsto\widetilde\zeta,\qquad A\longmapsto-A,\qquad V\longmapsto-V,\qquad H\longmapsto H.$$

This is the opposite-charge substitution in the paper’s (3.2). In a complex Euclidean background it is not permission to set $\widetilde\zeta=\zeta^*$: conjugating a solution also conjugates the background and gamma matrices. Nor does a solution on $(A,V,H)$ automatically produce an opposite-charge solution on those same fields. Both equations must be checked on the chosen background.

Recall also the distinction between a label and its operation. The commuting label $\zeta$ has charge $+1$, while the fixed-label odd transformation $\delta_\zeta$ lowers a matter field’s R-charge by one. An ordinary even supersymmetry variation includes an external odd parameter. The Killing equation is the linear equation for its commuting coefficient.

Check local R-gauge covariance

Take a real function $\lambda(x)$ and change

$$A_\mu'=A_\mu+\partial_\mu\lambda,\qquad \varepsilon_s'=e^{is\lambda}\varepsilon_s, \qquad V'=V,\quad H'=H .$$

The derivative term transforms as

$$\begin{aligned} (\nabla_\mu-isA_\mu')\varepsilon_s' &=e^{is\lambda} \bigl[\nabla_\mu+is(\partial_\mu\lambda) -isA_\mu-is(\partial_\mu\lambda)\bigr]\varepsilon_s\\ &=e^{is\lambda}(\nabla_\mu-isA_\mu)\varepsilon_s. \end{aligned}$$

Every algebraic term acquires the same phase. Thus $\mathscr K_\mu^{(s)\prime}\varepsilon_s' =e^{is\lambda}\mathscr K_\mu^{(s)}\varepsilon_s$. The cancellation tests the relative sign of the label charge and $A_\mu$.

It does not fix the coefficients of $V$ or $H$: those fields are already gauge invariant. Their coefficients come from the supplied transformation law. In particular, replacing $A$ by the matter connection $A-V/2$ while leaving the rest of the Killing equation unchanged changes the equation. Exercise 3 gives the correct rearrangement.

A flat solution with nonconstant components

First take Cartesian $\mathbb R^3$, a constant frame, and $A=V=H=0$. Both equations reduce to $\partial_\mu\varepsilon_s=0$. Each family has the two independent constant complex columns $(1,0)^T$ and $(0,1)^T$.

Now let

$$\lambda=b\,x^1x^2,\qquad A=d\lambda=b\,x^2dx^1+b\,x^1dx^2,\qquad V=H=0,$$

where $b$ is real with dimension two. The field strength is zero: $F^A_{12}=\partial_1A_2-\partial_2A_1=b-b=0$. For an arbitrary constant spinor $\xi_s$,

$$\varepsilon_s(x)=e^{is b x^1x^2}\xi_s$$

solves all three equations. Indeed, $\partial_1\varepsilon_s=is b x^2\varepsilon_s$, $\partial_2\varepsilon_s=is b x^1\varepsilon_s$, and $\partial_3\varepsilon_s=0$. The first two terms cancel $-isA_\mu\varepsilon_s$ component by component. The spinors are not constant, but the background is related to the zero-source background by an allowed R-gauge transformation.

This also gives a useful failure test. Keep $H=V=0$ on flat space but choose an R-connection with nonzero field strength. Commuting the two equations yields

$$[\partial_\mu-isA_\mu,\partial_\nu-isA_\nu]\varepsilon_s =-isF^A_{\mu\nu}\varepsilon_s=0.$$

If any component of $F^A$ is nonzero at a point, this scalar multiple of the identity forces the spinor to vanish there. A solution cannot then become nonzero elsewhere on the connected background, as we show next. This obstruction applies to the stated $H=V=0$ example; additional sources change the test.

Local solutions, global solutions, and their number

In a chosen frame write $\nabla_\mu=\partial_\mu+\mathcal A_\mu$. Our equation is

$$\partial_\mu\varepsilon_s+\mathcal M_\mu^{(s)}\varepsilon_s=0,\qquad \mathcal M_\mu^{(s)} =\mathcal A_\mu-is(A_\mu-V_\mu)\mathbf1 +\frac H2\gamma_\mu +\frac s2\epsilon_{\mu\nu\rho}V^\nu\gamma^\rho .$$

Along a smooth curve $x(t)$ this becomes the ordinary matrix differential equation $d\varepsilon_s/dt=-\dot x^\mu\mathcal M_\mu^{(s)}\varepsilon_s$. The background is smooth, so an initial value has a unique continuation along that curve. Zero initial data continue to zero. Therefore a nonzero solution on a connected manifold is nowhere zero, and specifying its two complex components at one point determines it wherever it exists.

There can consequently be at most two complex linearly independent solutions in each family. There may be fewer: transport along different paths must be compatible, and after going around a closed cycle the spinor must satisfy the chosen boundary condition. An arbitrary initial column does not necessarily extend to a solution of all the partial differential equations.

For flat $\mathbb R^3$ both families attain the maximum. The four complex parameters describe the complexification of the parent algebra with four real Lorentzian supercharges. Euclidean independence does not turn the parent theory into an eight-real-supercharge theory. When discussing a curved background, state the dimensions of the two solution spaces rather than inferring them solely from the parent $\mathcal N$.

The circle exercise below separates local solvability from global boundary conditions. The round sphere will supply the complementary example: a nonzero spin connection compensated by a nonzero $H$.

Exercises

1. Can a constant auxiliary source preserve flat-space supersymmetry?

On Cartesian $\mathbb R^3$, set $A=V=0$ and take a constant complex $H=h\ne0$. Show that neither label family has a nonzero local solution. Why is checking only the first coordinate equation insufficient?

First hint

The equation is $(\partial_a+h\gamma_a/2)\varepsilon=0$. Commute the first and second operators.

Second hint

Use $[\gamma_1,\gamma_2]=2i\gamma_3$ and $\gamma_3^2=\mathbf1$.

Complete solution

The matrices and $h$ are constant, so

$$[\partial_1+h\gamma_1/2,\partial_2+h\gamma_2/2]\varepsilon =\frac{ih^2}{2}\gamma_3\varepsilon=0.$$

Since $h\ne0$ and $\gamma_3$ is invertible, $\varepsilon=0$. This applies to both charges because their equations coincide in this background.

The first equation alone has the solutions $\varepsilon=e^{-hx^1\gamma_1/2}\xi(x^2,x^3)$, but the other two equations are incompatible with a nonzero solution. A constant auxiliary field is not an arbitrary extra mass one can add while retaining every flat-space supercharge. On a curved manifold the spin-connection curvature contributes to the same commutator and can change the result.

2. A connection can compensate an antiperiodic boundary condition

Consider flat $S^1_L\times\mathbb R^2$, with length coordinate $x\sim x+L$, $A=a\,dx$, and $V=H=0$. Take $a$ real. In a constant frame, the chosen spin structure requires $\varepsilon_s(x+L)=\eta\varepsilon_s(x)$, where $\eta=+1$ is periodic and $\eta=-1$ is antiperiodic.

For this exercise assume a compact $U(1)_R$ whose $2\pi$ transformation acts trivially on every field: all matter charges are integers in the normalization where the labels have charges $\pm1$. This declares the global gauge period; the local Killing equation alone does not fix it.

Solve both equations and find the allowed values of $aL$. Can an allowed R-gauge transformation always remove $A$ without changing the spin structure?

First hint

The local solutions are $\varepsilon_s=e^{isa x}\xi_s$, with constant $\xi_s$. Apply the boundary condition to a nonzero column.

Second hint

An R-gauge transformation on the circle must have a single-valued phase $e^{i\lambda(x)}$. Test the candidate $\lambda=-ax$.

Complete solution

For either family, a nonzero global solution requires $e^{isaL}=\eta$. Since $\eta=\eta^{-1}$, the two charges give the same condition:

$$\begin{array}{ll} \eta=+1:&aL=2\pi n,\\ \eta=-1:&aL=(2n+1)\pi,\qquad n\in\mathbb Z. \end{array}$$

When it holds, each family has two independent complex solutions. Otherwise neither family has a nonzero global solution in this example. The curvature $dA$ vanishes in every case; the distinction is the phase acquired around the circle.

The transformation $\lambda=-ax$ removes $A$ locally. Its phase is single-valued only for $aL=2\pi n$. In the antiperiodic case it is not an allowed gauge transformation on the same spin bundle. A connection with phase $-1$ around the circle compensates that boundary condition, but it cannot be gauged away while retaining it. Large single-valued transformations $\lambda=2\pi n x/L$ shift $a$ by $2\pi n/L$ and leave these conditions unchanged.

3. Translate between the three connections

For the charge-$s$, central-charge-zero label, define

$$D_\mu^{\mathrm{mat}}=\nabla_\mu-is(A_\mu-V_\mu/2),\qquad D_\mu^{R}=\nabla_\mu-is(A_\mu-3V_\mu/2).$$

Rewrite the Killing equation using each derivative. What error results from replacing only the derivative on its left-hand side?

First hint

Relative to $\nabla_\mu-isA_\mu$, the first derivative contains $+isV_\mu/2$ and the second contains $+3isV_\mu/2$.

Second hint

Add that difference to the right-hand side as well. Keep the epsilon term and $H$ term fixed.

Complete solution

The equivalent equations are

$$\begin{aligned} D_\mu^{\mathrm{mat}}\varepsilon_s &=-\frac H2\gamma_\mu\varepsilon_s -\frac{is}{2}V_\mu\varepsilon_s -\frac s2\epsilon_{\mu\nu\rho}V^\nu\gamma^\rho\varepsilon_s,\\ D_\mu^{R}\varepsilon_s &=-\frac H2\gamma_\mu\varepsilon_s +\frac{is}{2}V_\mu\varepsilon_s -\frac s2\epsilon_{\mu\nu\rho}V^\nu\gamma^\rho\varepsilon_s . \end{aligned}$$

Changing the derivative alone leaves a spurious residual $isV_\mu\varepsilon_s/2$ in the first case, or $3isV_\mu\varepsilon_s/2$ in the second. The same geometry has several useful ways of grouping its terms. Gauge covariance by itself does not make those groupings interchangeable.

Return from the linearized current construction

This optional return assumes the R-multiplet reconstruction. The main calculation above uses its resulting transformation as input. Here we identify the precise step which connects the linearized answer to that input, without repeating the current-multiplet derivation.

In symmetric frame gauge, write $g_{\mu\nu}=\delta_{\mu\nu}+2h_{\mu\nu}+O(h^2)$ and $e_\mu{}^a=\delta_\mu{}^a+h_\mu{}^a+O(h^2)$. The tetrad postulate and linearized Christoffel symbol give

$$\Omega_{\mu ab} =\partial_bh_{\mu a}-\partial_ah_{\mu b}+O(h^2), \qquad \mathcal A_\mu =-\frac i2\epsilon^{\nu\rho\lambda} \partial_\nu h_{\rho\mu}\gamma_\lambda+O(h^2).$$

For example, the second identity follows by inserting the first into $\mathcal A_\mu=\Omega_{\mu ab}\gamma^a\gamma^b/4$: the symmetric Clifford part vanishes and both antisymmetric ordered pairs contribute. This is precisely the connection in the parentheses of the paper’s linearized gravitino law (2.13). Adding the local-parameter term $2\partial_\mu\zeta$ gives $2\nabla_\mu\zeta$ to this order.

Covariantize that expression and the volume tensor. The dimension assignments of (2.14), together with the local R-charge and the off-shell formulation’s field content, constrain further regular local terms: each term in the variation is linear in a label and has one derivative or one dimension-one bosonic field. Those bosonic structures and their coefficients already occur in the linearized law. The remaining gravitino-dependent terms vanish on the bosonic background. This is the argument from (2.13)–(2.15), under its local supergravity assumptions; dimension counting alone would not fix the coefficients without the linearized result.

The outcome is the same input variation used at the start. Requiring its vanishing then gives the Killing equations. A complete solution still requires a metric, source fields, and global spinor boundary conditions. The round three-sphere calculation supplies all of them. Construct geometry from one spinor develops the tensors selected by a nonzero solution and distinguishes their pointwise identities from differential constraints. For a compact comparison of the connections used here and in the sequel, keep the background-field reference nearby.

To test a new proposal systematically, check the curvature and global transport of the complete spinor connection. That calculation explains both the power and the limits of integrability.

For a first reading assignment in the original source, the paper introduction guide compares the two charge sectors term by term and supplies a changed-source exercise.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading Supersymmetric Field Theories on Three-Manifolds
• Paper guide: Section 2, currents and rigid backgrounds
• Paper guide: Section 3, geometry from one spinor
• Paper guide: Section 4, transverse holomorphic geometry