Check integrability and verify a proposed background

A proposed metric and a list of sources are not yet a supersymmetric background. The Killing equation must have a nonzero solution, and that solution must return consistently around every closed curve. Curvature provides a fast way to reject many candidates before solving any differential equation. It also has a limitation: an algebraic kernel at each point need not fit together into a parallel spinor.

Prepare with the Killing spinor equations and spinor rearrangements. We will derive the obstruction, repair an explicit failed background, and then test global data that local curvature cannot see.

Treat the equation as a connection

The parent theory is three-dimensional $\mathcal N=2$, with four real Poincaré supercharges. We seek a commuting charge-$+1$ Euclidean label $\zeta$ on a fixed oriented Riemannian manifold. Its opposite-charge counterpart obeys a separate equation. The sources $A,V,H$ may be complex; $V$ must satisfy $\nabla_\mu V^\mu=0$.

Use $\epsilon_{123}=1$ and $\gamma=(\sigma_3,-\sigma_1,-\sigma_2)$, so $\gamma_a\gamma_b=\delta_{ab}+i\epsilon_{abc}\gamma_c$. Bring the charge-one equation to the left:

$$\mathscr D_\mu\zeta=0,\qquad \mathscr D_\mu=\nabla_\mu+N_\mu,\qquad N_\mu=-i(A_\mu-V_\mu)\mathbf1 +\frac H2\gamma_\mu +\frac12\epsilon_{\mu\nu\rho}V^\nu\gamma^\rho . \tag{R09.1}$$

Here $\nabla$ is the spin connection of the real metric. The extra matrix-valued one-form $N$ has dimension inverse length. Thus $\mathscr D$ is a connection on the charged spinor bundle, although it need not be unitary. “Parallel” below means parallel for this complete connection, not necessarily for $\nabla$ alone.

Along a path $x^\mu(t)$, the equation is the linear ordinary differential equation $d\zeta/dt=-\dot x^\mu(\Gamma_\mu+N_\mu)\zeta$. Its transport matrix is invertible. Consequently, a solution which vanishes at one point vanishes throughout its connected domain. A preserved nonzero solution has no isolated zero where an incompatible equation can be ignored.

Commute the derivatives

Applying two derivatives in opposite orders gives the necessary condition

$$\mathscr F_{\mu\nu}\zeta=0,\qquad \mathscr F_{\mu\nu} =[\nabla_\mu,\nabla_\nu] +\nabla_\mu N_\nu-\nabla_\nu N_\mu+[N_\mu,N_\nu]. \tag{R09.2}$$

The first term is an operator on spinors; the derivative of $N$ includes both its one-form and matrix indices. This formula follows by applying each operator to a test spinor: the terms containing one derivative of that spinor cancel in the commutator.

In three dimensions it is convenient to dualize the two-form: $\mathscr I_a=\epsilon_{abc}\mathscr F_{bc}$. There is no factor $1/2$ in this definition. All three conditions $\mathscr I_a\zeta=0$ are equivalent to (R09.2).

Work at a point in a normal oriented frame, where the connection vanishes but its curvature does not. The Clifford algebra gives

$$\begin{aligned} \epsilon_{abc}(\nabla_bN_c-\nabla_cN_b) ={}&-2i\epsilon_{abc}\nabla_b(A_c-V_c)\mathbf1\\ &+\bigl(\nabla_dV_a-\delta_{ad}\nabla_bV_b -\epsilon_{adc}\nabla_cH\bigr)\gamma_d,\\ \epsilon_{abc}[N_b,N_c] ={}&i\bigl(H^2\delta_{ad}+V_aV_d -H\epsilon_{adc}V_c\bigr)\gamma_d . \end{aligned} \tag{R09.3}$$

For example, the $H^2$ term is $\epsilon_{abc}H^2[\gamma_b,\gamma_c]/4=iH^2\gamma_a$. The identity part of $N$ commutes with every matrix, so it contributes only through its exterior derivative.

The source paper’s curvature convention has negative Ricci scalar on a round sphere. With that convention its spin-curvature identity is $\epsilon_{abc}[\nabla_b,\nabla_c] =\frac i2(2R_{ad}-R\delta_{ad})\gamma_d$. Combining this with (R09.3) and the required $\nabla\cdot V=0$ yields

$$\begin{aligned} \mathscr I_a={}& \frac i2(2R_{ad}-R\delta_{ad})\gamma_d -2i\epsilon_{abc}\nabla_b(A_c-V_c)\mathbf1\\ &+\bigl[ \nabla_dV_a-\epsilon_{adc}\nabla_cH +iH^2\delta_{ad}+iV_aV_d -iH\epsilon_{adc}V_c \bigr]\gamma_d . \end{aligned} \tag{R09.4}$$

This is P (5.8), written as an operator whose common kernel must contain $\zeta$. The spin-curvature sign follows P (A.8)–(A.12). Substituting a positive-sphere Ricci tensor into (R09.4) without translating the convention would give the wrong source condition.

What a curvature test does and does not prove

At each point, calculate $\ker\mathscr I_1\cap\ker\mathscr I_2\cap\ker\mathscr I_3$. An invertible matrix among the three immediately excludes a solution. But checking that each determinant vanishes is too weak: the matrices $\operatorname{diag}(0,1)$ and $\operatorname{diag}(1,0)$ are individually singular and have no common nonzero kernel.

Even a common line at every point is only a candidate. It must be carried into itself by $\mathscr D$. Differentiating $\mathscr F_{\mu\nu}\zeta=0$ along an actual parallel solution gives another necessary condition, $(\mathscr D_\rho\mathscr F_{\mu\nu})\zeta=0$, where the derivative acts on the curvature as an endomorphism-valued tensor. Exercise 2 constructs a simple connection that fails precisely this test.

If the whole connection is flat, $\mathscr F=0$, it does admit a parallel basis on a contractible patch: path-independent transport constructs the solutions from arbitrary initial columns. That is a sufficient local condition, but it is stronger than preserving one column. Conversely, two linearly independent solutions of the same charge equation force its curvature matrix to vanish, because that matrix annihilates a basis. Neither statement settles global holonomy.

The practical sequence is therefore: check the source constraints, find the common curvature kernel, solve the first-order equation on that candidate, and finally check its transitions and closed-path transport.

Repair a curved candidate

Consider coordinates $(r,y,z)$ with length units and

$$g=dr^2+e^{2\kappa r}dy^2+dz^2,\qquad H=V=0,\qquad \kappa>0. \tag{R09.5}$$

The orthonormal coframe is $e^1=dr$, $e^2=e^{\kappa r}dy$, $e^3=dz$. Set $f=e^{\kappa r}$ for the next calculation. Cartan’s equation gives $\Omega_{12}=-f'\,dy$, with other independent components zero. With the standard Cartan index order, $\Gamma=\Omega_{ab}\gamma^a\gamma^b/4$, including both ordered pairs, so

$$\Gamma_y=-\frac{i f'}2\gamma_3,\qquad [\nabla_r,\nabla_y]=-\frac{i f''}2\gamma_3 . \tag{R09.6}$$

This also follows directly from $\Gamma^r{}_{yy}=-ff'$ and $\Gamma^y{}_{ry}=f'/f$ by the vielbein postulate. The source convention gives $R_{ab}=\operatorname{diag}(\kappa^2,\kappa^2,0)$, $R=2\kappa^2$.

With $A=0$, the matrix in (R09.6) is invertible everywhere. No nonzero spinor solves the proposed equation, even locally.

Now choose a real R connection

$$A=-\frac{f'}2\,dy+\nu\,dy, \qquad \nu\ \hbox{constant}. \tag{R09.7}$$

The constant $\nu$ has dimension inverse length. The complete operator and its only independent curvature component are

$$\begin{aligned} \mathscr D_r&=\partial_r,& \mathscr D_z&=\partial_z,\\ \mathscr D_y&=\partial_y -\frac{i f'}2(\gamma_3-\mathbf1)-i\nu\mathbf1,& \mathscr F_{ry}&=-\frac{i f''}2(\gamma_3-\mathbf1). \end{aligned} \tag{R09.8}$$

The curvature has a one-dimensional kernel, the $\gamma_3=+1$ eigenspace. It is preserved by every operator in (R09.8). With $u_+=(1,-i)^T/\sqrt2$, the solution is

$$\zeta(r,y,z)=c_0 e^{i\nu y}u_+ . \tag{R09.9}$$

Substitution checks all three first-order equations. Conversely, the curvature forces any solution into this line, after which the first-order equations determine exactly (R09.9). The complete connection is not flat: it acts nontrivially on the other eigenspace. One preserved solution did not require every curvature component to vanish as a matrix.

The repair has supplied an R-field strength $dA=-f''\,dr\wedge dy/2$ that cancels the spin curvature on the selected line. It has not flattened the metric or eliminated an auxiliary source dynamically.

Test the same solution around a circle

The metric and sources above descend to $y\sim y+L$. The displayed coframe is globally defined on $\mathbb R_r\times S^1_y\times\mathbb R_z$. Fix a spin structure with boundary sign $\varepsilon=\pm1$, and take a trivial R-bundle with the connection (R09.7). Equation (R09.9) is global precisely when

$$e^{i\nu L}=\varepsilon . \tag{R09.10}$$

For a periodic spin structure this means $\nu L=2\pi n$; for an antiperiodic one it means $\nu L=(2n+1)\pi$. If $\nu$ is allowed to be complex, (R09.10) also forces $\operatorname{Im}\nu=0$. Single-valued unit-charge R gauge transformations identify $\nu$ and $\nu+2\pi n/L$.

The local curvature is independent of $\nu$. Thus infinitely many locally equivalent candidates fail the chosen global condition. The relevant holonomy is that of the complete spinor connection: the $f'$ contribution to R transport cancels the selected spin transport. Checking the R phase alone would miss that cancellation.

This is the distinction between an integrability test and verification of a background. The former can rule out a candidate quickly; the latter supplies an actual solution and its global meaning. The opposite-charge equation is tested separately by $A\mapsto-A$, $V\mapsto-V$, with $H$ unchanged.

Exercises

1. Test the sphere source

On a locally round three-sphere of radius $\ell$, set $A=V=0$ and take constant $H$. In the source convention, $R_{ab}=-2\delta_{ab}/\ell^2$. Use (R09.4) to find the allowed values of $H$. Check the result directly from the spin curvature and $[H\gamma_a/2,H\gamma_b/2]$. What local conclusion follows, and what global conclusion has not yet been checked?

Hint 1

The Ricci scalar is $-6/\ell^2$. Each $\gamma_a$ is invertible.

Hint 2

For $a\ne b$, the round-sphere spin curvature is $[\nabla_a,\nabla_b]=\gamma_a\gamma_b/(2\ell^2)$. The source commutator has the same matrix structure.

Solution

Equation (R09.4) becomes $\mathscr I_a=i(\ell^{-2}+H^2)\gamma_a$. A nonzero common kernel therefore requires

$$H=\frac{i}{\ell}\quad\hbox{or}\quad H=-\frac{i}{\ell}.$$

Directly, $\mathscr F_{ab}=(\ell^{-2}+H^2)\gamma_a\gamma_b/2$ for $a\ne b$, because $[\gamma_a,\gamma_b]=2\gamma_a\gamma_b$. Both calculations agree. At either value the charge-one generalized connection is flat, so a parallel basis exists on a contractible patch. A real $H$ cannot pass the test on a sphere of finite radius.

This local calculation does not specify a global quotient, its spin structure, or its connection holonomies. These must still be checked before retaining every locally available solution. It also illustrates why changing the Ricci convention alone would incorrectly replace imaginary $H$ by real $H$.

2. A common kernel that does not survive transport

The following is a mathematical rank-two connection, used to test the logic of the curvature criterion; it is not asserted to come from the supersymmetric source ansatz. On a plane with length coordinates $x,y$, let

$$D_x=\partial_x,\qquad D_y=\partial_y+\frac1L \begin{pmatrix}0&0\\1&x/L\end{pmatrix}, \qquad L>0.$$

Compute its curvature and common kernel. Then solve the first-order equations on that kernel. Finally evaluate $D_yF_{xy}$ as an endomorphism and explain what extra information it detects.

Hint 1

Since the $x$ connection vanishes, the curvature is just the $x$ derivative of the matrix in $D_y$.

Hint 2

If $F_{xy}\psi=0$, write $\psi=f(x,y)(1,0)^T$ and inspect both components of $D_y\psi$.

Solution

The curvature is $F_{xy}=L^{-2}\operatorname{diag}(0,1)$, with the nonzero common kernel spanned by $(1,0)^T$ at every point. But

$$D_y\begin{pmatrix}f\\0\end{pmatrix} =\begin{pmatrix}\partial_yf\\f/L\end{pmatrix}.$$

The lower component forces $f=0$. There is no nonzero parallel section. The derivative of the curvature, including its matrix commutator, is

$$D_yF_{xy}=\partial_yF_{xy}+[M_y,F_{xy}] =-\frac1{L^3}\begin{pmatrix}0&0\\1&0\end{pmatrix}.$$

It does not annihilate the candidate line. The obstruction is that parallel transport moves the line out of itself, not that the pointwise kernel was computed incorrectly.

3. A flat quotient with a nontrivial spin lift

Start with flat $\mathbb R^3$, $H=V=0$, and $A=\nu\,dz$. Form the smooth quotient by the freely acting screw motion

$$(x,y,z)\longmapsto(-x,-y,z+L).$$

Choose its spin lift $S=-i\gamma_3$ and no additional R transition. The metric rotation has angle $\pi$; its spin lift has half that angle. Find the real values of $\nu$ for which a nonzero charge-one solution descends, and identify the selected $\gamma_3$ eigenspace. Explain why $\nu=0$ fails although the generalized curvature is zero.

Hint 1

On the covering space, $\zeta=e^{i\nu z}\zeta_0$. The quotient condition is $\zeta(-x,-y,z+L)=S\zeta(x,y,z)$.

Hint 2

On a $\gamma_3$ eigenvector of eigenvalue $\lambda=\pm1$, $S$ has eigenvalue $-i\lambda$. Test the quotient generator itself, not only its square.

Solution

The condition is $e^{i\nu L}\zeta_0=-i\gamma_3\zeta_0$. The $\lambda=+1$ eigenspace survives when $\nu L=-\pi/2+2\pi n$; the $\lambda=-1$ eigenspace survives when $\nu L=\pi/2+2\pi n$. Neither survives at $\nu=0$, because the spin lift has no eigenvalue one.

The other possible lift, $-S$, exchanges the two assignments. Squaring the chosen lift gives $S^2=-\mathbf1$, so applying the screw motion twice requires $e^{2i\nu L}=-1$. This squared condition is necessary but does not select which component survives; the generator condition does. The quotient is free because of the nonzero translation along $z$, so there is no geometric singularity responsible for the obstruction.

All local curvature tests pass for every $\nu$. The failure at $\nu=0$ is global holonomy, and the allowed R connection compensates it on one spinor line.

Use the Section 5 paper guide to locate this calculation in the original argument and compare its local and global assumptions.

Paper guides

Connect this lesson to the original papers and their notation.

• Paper guide: Section 5, additional supercharges