Reconstruct the background fields

The transverse integrability condition tells us which geometric data can support a preserved supercharge. It does not choose the background sources. How do we actually find them? In an adapted frame the answer becomes a sequence of ordinary calculations: solve Cartan’s equation, use the lower spinor components to choose $V,H$, and use the upper components to determine $A$.

Prepare with adapted coordinates and the background field dictionary. We work locally on a contractible patch. Global spinor signs, R-bundle transitions and flux conditions require the subsequent patching analysis.

Choose the metric and the spinor

The physical parent is three-dimensional $\mathcal N=2$, with four real Poincaré supercharges. We construct one commuting charge-$+1$ Killing-spinor solution. This neither assumes a second independent solution nor chooses a dynamical gauge group.

Use the preceding lesson’s coordinates $z=x-iy$ and its real positive metric,

$$\eta=d\tau+h\,dz+\bar h\,d\bar z,\qquad g=\eta\otimes\eta+c^2\,dz\,d\bar z,\qquad c>0. \tag{M05.1}$$

Here $dz\,d\bar z$ means $(dz\otimes d\bar z+d\bar z\otimes dz)/2$. Thus it is $dx^2+dy^2$, not twice that tensor. Set $p=2\operatorname{Re}h$, $q=2\operatorname{Im}h$, so $\eta=d\tau+p\,dx+q\,dy$. For coordinates with length units, $p,q,c$ are dimensionless.

Choose the oriented orthonormal coframe and its dual:

$$\begin{aligned} e^1&=\eta,&e^2&=c\,dx,&e^3&=c\,dy,\\ E_1&=\partial_\tau,& E_2&=c^{-1}(\partial_x-p\partial_\tau),& E_3&=c^{-1}(\partial_y-q\partial_\tau). \end{aligned} \tag{M05.2}$$

The distinguished unit vector is $\xi=E_1$. Choose a smooth nonzero complex function $\chi$ and seek $\zeta=\chi(1,0)^T$ in this spin frame. Its normalized Hermitian direction is $\eta$, independently of $\chi$. An overall constant can carry the label’s engineering dimension; $d\log\chi$ always means $d\chi/\chi$.

The charge-one equation to be solved is

$$(\nabla_a-iA_a)\zeta =-\frac H2\gamma_a\zeta-iV_a\zeta -\frac12\epsilon_{abc}V_b\gamma_c\zeta . \tag{M05.3}$$

We retain $\epsilon_{123}=1$ and $\gamma=(\sigma_3,-\sigma_1,-\sigma_2)$. The backgrounds may be complex. The real metric is fixed; no auxiliary field is eliminated by an equation of motion.

Compute the frame derivatives once

It is useful to name six real derivative coefficients. They all have dimension inverse length:

$$\begin{aligned} a&=\frac{\partial_\tau p}{c},& b&=\frac{\partial_\tau q}{c},& \sigma&=\partial_\tau\log c,\\ u&=E_2\log c,& v&=E_3\log c,& t&=\frac{\partial_xq-\partial_yp +q\,\partial_\tau p-p\,\partial_\tau q}{c^2}. \end{aligned} \tag{M05.4}$$

The lowercase $a,b$ in this calculation are frame coefficients, not the R-connection or a squashing parameter. Exterior differentiation gives

$$\begin{aligned} de^1&=a\,e^1\wedge e^2+b\,e^1\wedge e^3+t\,e^2\wedge e^3,\\ de^2&=\sigma\,e^1\wedge e^2-v\,e^2\wedge e^3,\\ de^3&=\sigma\,e^1\wedge e^3+u\,e^2\wedge e^3. \end{aligned} \tag{M05.5}$$

For example, the $q\,\partial_\tau p-p\,\partial_\tau q$ term in $t$ comes from replacing $d\tau=e^1-(p/c)e^2-(q/c)e^3$ inside $d\eta$. Also $\nabla_\mu\xi^\mu=2\sigma$ since the real-coordinate volume density is $c^2$, and $*(\eta\wedge d\eta)=t$. Thus $\sigma$ measures transverse expansion and $t$ measures contact twist; neither is assumed constant.

Let $\Omega_{ab}=-\Omega_{ba}$ be the Cartan connection, $de^a+\Omega^a{}_b\wedge e^b=0$. Solving (M05.5) yields

$$\begin{aligned} \Omega_{12}&=-a\,e^1-\sigma\,e^2+\frac t2e^3,\\ \Omega_{13}&=-b\,e^1-\frac t2e^2-\sigma\,e^3,\\ \Omega_{23}&=-\frac t2e^1+v\,e^2-u\,e^3. \end{aligned} \tag{M05.6}$$

Substitution into Cartan’s equation checks all three signs. This is the standard Cartan index order: the paper’s input-first, all-lowered connection is $\omega_{ab}=-\Omega_{ab}$. The spin connection is consequently $\Gamma=\frac14\Omega_{ab}\gamma^a\gamma^b =\frac i2(\Omega_{23}\gamma_1+\Omega_{31}\gamma_2+\Omega_{12}\gamma_3)$, where both ordered index pairs are included in the first expression.

Solve the lower and upper components

On the column $(1,0)^T$, the lower components of $\Gamma_a$ are

$$(\Gamma_1)_{21}=-\frac{a+ib}{2},\qquad (\Gamma_2)_{21}=-\frac{\sigma}{2}-\frac{it}{4},\qquad (\Gamma_3)_{21}=\frac t4-\frac{i\sigma}{2}. \tag{M05.7}$$

The derivative of $\chi$ and the connection $A$ multiply the upper component only. The lower components of the right side of (M05.3), divided by $\chi$, are $(iV_2-V_3)/2$, $(H-iV_1)/2$, and $(iH+V_1)/2$. A compatible choice is therefore

$$\boxed{ V=t\,e^1-b\,e^2+a\,e^3=*d\eta,\qquad H=-\sigma+\frac{it}{2}.} \tag{M05.8}$$

Here and below $V$ denotes the one-form; $V^\sharp$ is its metric-dual vector. This is the representative in P (4.5).

The upper spin-connection components are $-it/4$, $iv/2$, and $-iu/2$. The upper $a=1$ equation, for instance, reads

$$E_1\log\chi-\frac{it}{4} =i(A_1-t)-\frac H2.$$

Solving it and the other two upper equations gives

$$\boxed{ A=\left(t+\frac{i\sigma}{2}\right)e^1 +\left(\frac v2-b-\frac{ia}{2}\right)e^2 +\left(-\frac u2+a-\frac{ib}{2}\right)e^3 -i\,d\log\chi .} \tag{M05.9}$$

Equations (M05.7)–(M05.9) verify all six component equations, not only a contracted condition. They apply to arbitrary smooth real $p,q$, positive $c$ in the adapted metric, and any nonzero local $\chi$.

For comparison with P (4.14) and (4.19)–(4.21), its spinor multiplier is $\sqrt{s_{\!P}}=\chi$. Its auxiliary connection and one-form in this frame are $\widehat A=(v e^2-u e^3)/2-i\,d\log\chi$ and $W=-t\eta/4$. Substituting them into the paper’s reconstruction reproduces (M05.9). The symbol $W$ here is a connection-related one-form, not a superpotential.

Check that the sources are field strengths

The Killing equation is not the only constraint in the background dictionary: $V=-i*dC$ requires $\nabla_\mu V^\mu=0$. Our representative satisfies it identically because $*V=d\eta$ and $d^2\eta=0$. A convenient potential is

$$C=i\eta,\qquad -i*dC=*d\eta=V. \tag{M05.10}$$

Similarly $H=i*dB$ requires $dB=-iH\,\mathrm{vol}_g$. On a contractible three-dimensional patch this three-form is closed and has a local primitive. These local statements do not assert a single global $B$ potential on a compact manifold with nonzero integrated flux.

A phase change $\chi\mapsto e^{i\beta}\chi$ changes $A\mapsto A+d\beta$, the ordinary unitary R-gauge transformation. A positive change $\chi\mapsto e^f\chi$ instead changes $A\mapsto A-i\,df$. It changes a gauge-invariant imaginary source, not merely a unitary gauge choice.

A background with expansion and twist

Take real constants $\lambda,s$ of inverse-length dimension and choose

$$\eta=d\tau+\lambda x\,dy,\qquad c=e^{s\tau},\qquad \chi=e^{s\tau/2}. \tag{M05.11}$$

The coefficients are $a=b=u=0$, $\sigma=s$, $v=-s\lambda x e^{-s\tau}$ and $t=\lambda e^{-2s\tau}$. The reconstructed sources are

$$\begin{aligned} V&=\lambda e^{-2s\tau}\eta,\\ H&=-s+\frac{i\lambda}{2}e^{-2s\tau},\\ A&=V-\frac{\lambda s x}{2}\,dx +\frac{i\lambda s x}{2}\,dy. \end{aligned} \tag{M05.12}$$

The imaginary vertical term in (M05.9) cancels the vertical part of $-i\,d\log\chi$; its remaining horizontal part is the last term in $A$. This is why simply setting $A=V$ would fail when both parameters are nonzero.

As a Bianchi check, $V^\sharp=\lambda e^{-2s\tau}\partial_\tau$ and $\sqrt g=e^{2s\tau}$, hence $\partial_\tau(\sqrt g\,V^\tau)=0$. Explicit potentials on this patch are

$$C=i\eta,\qquad B=\left[\frac i2(e^{2s\tau}-1)+\frac{\lambda\tau}{2}\right]dx\wedge dy. \tag{M05.13}$$

Differentiation gives $dB=(is e^{2s\tau}+\lambda/2)d\tau\wedge dx\wedge dy =-iH\,\mathrm{vol}_g$.

Two limits test different terms. At $\lambda=0$, $A=V=0$, $H=-s$, and the nonconstant spinor solves the equation on $d\tau^2+e^{2s\tau}(dx^2+dy^2)$. At $s=0$, the spinor has constant components but the spin connection is nonzero: $A=V=\lambda\eta$ and $H=i\lambda/2$ cancel it. Only when both constants vanish do we recover the ordinary Cartesian zero-source equation.

Identify the remaining freedom

The sources above are a representative, not the unique background preserving this spinor. Let $U$ be a complex holomorphic vector, $\Phi U=iU$, and let $\kappa$ be a complex function. Then the shifts

$$\delta V^\sharp=U+\kappa\xi,\qquad \delta H=i\kappa,\qquad \delta A=\frac32(U^\flat+\kappa\eta) \tag{M05.14}$$

leave the same spinor equation invariant, provided the field-strength constraint also holds. “Holomorphic vector” here means the eigenspace condition; it does not require its coefficient functions to be holomorphic functions of $z$.

To check the cancellation, bring the equation to the left. Its change acting on $\zeta$ is

$$\delta\mathscr K_a\zeta= \frac12\left[ -i(U_a+\kappa\eta_a)+i\kappa\gamma_a +\epsilon_{abc}(U_b+\kappa\xi_b)\gamma_c \right]\zeta . \tag{M05.15}$$

The projector identities give $U^a\gamma_a\zeta=0$ and $\eta^a\gamma_a\zeta=\zeta$. Multiplying these by $\gamma_a$ with the Clifford algebra yields

$$\epsilon_{abc}U_b\gamma_c\zeta=iU_a\zeta,\qquad \epsilon_{abc}\xi_b\gamma_c\zeta =i(\eta_a-\gamma_a)\zeta .$$

Each term in (M05.15) now cancels. But $V$ must still be divergence free:

$$\boxed{\nabla_\mu(U^\mu+\kappa\xi^\mu)=0.} \tag{M05.16}$$

For example, writing $U=w(\partial_z-h\partial_\tau)$, this is $\partial_\tau[c^2(\kappa-hw)]+\partial_z(c^2w)=0$ in an adapted chart. This is the actual differential restriction in P (4.2)–(4.3). The shifts generally change source field strengths and couplings, so they are not ordinary gauge transformations.

The local reconstruction establishes existence when the sources can be chosen this way. It does not solve the equation for arbitrary preassigned sources, prove that a second charge survives, or supply global bundle transitions. The exit task is to reconstruct a source set, verify all components and its Bianchi identity, and identify the information still needed to extend it across patches.

Exercises

1. Cancel a curved spin connection

Take $\eta=d\tau$, $c=e^{f(x,y)}$ with real smooth $f$, and $\chi=1$. Find $A,V,H$ and the R-field strength. Explain how a spinor with constant frame components can solve the equation on a nonflat metric. For $f=\alpha x^2$, with $\alpha$ of dimension inverse length squared, determine whether $A$ is locally pure gauge.

Hint 1

Here $a=b=\sigma=t=0$. Only the transverse derivatives $u,v$ remain.

Hint 2

Write $\Omega_{23}$ and act with $\gamma_2\gamma_3=i\gamma_1$ on $(1,0)^T$. Compare the resulting upper-component connection with $iA$.

Solution

The sources reduce to

$$V=H=0,\qquad A=\frac12(f_y\,dx-f_x\,dy),\qquad dA=-\frac12(f_{xx}+f_{yy})\,dx\wedge dy.$$

The only nonzero connection one-form is $\Omega_{23}=f_y\,dx-f_x\,dy$. Thus $\Gamma\zeta=(i/2)\Omega_{23}\zeta=iA\zeta$, and $(\nabla-iA)\zeta=0$ despite $\partial_\mu\zeta=0$ in the selected frame. “Constant components” is not “covariantly constant under the Levi-Civita connection.”

For $f=\alpha x^2$, $A=-\alpha x\,dy$ and $dA=-\alpha\,dx\wedge dy$. It is not pure gauge when $\alpha\ne0$. The real metric is $d\tau^2+e^{2\alpha x^2}(dx^2+dy^2)$; no flatness or second-spinor assumption was used.

2. An algebraic shift that fails the Bianchi identity

Start with flat space, $\eta=d\tau$, $z=x-iy$, $\chi=1$ and zero sources. Let $L>0$ be a length. First try $U=0$ and $\kappa=-\tau/L^2$ in (M05.14). Why does this not define an admissible background, although it cancels in the Killing equation? Repair it using $U=(z/L^2)\partial_z$, then give the resulting $V,H,A$ and check the constraint.

Hint 1

In these coordinates $\partial_z=(\partial_x+i\partial_y)/2$. For $U=w\partial_z$, the divergence condition is $\partial_\tau\kappa+\partial_z w=0$.

Hint 2

Lowering the index of $\partial_z$ gives $(dx+i\,dy)/2$. The metric is real, but $U$ and its metric dual can be complex.

Solution

For the first attempt, $\nabla\cdot(\kappa\xi)=-1/L^2\ne0$. No smooth local $C$ can have $V=-i*dC$ with this divergence. The algebraic cancellation of $\delta\mathscr K\zeta$ alone missed that necessary background constraint.

For the repaired choice, $\partial_z(z/L^2)=1/L^2$ cancels the first term. The sources are

$$V=-\frac{\tau}{L^2}d\tau +\frac{z}{2L^2}(dx+i\,dy),\qquad H=-\frac{i\tau}{L^2},\qquad A=\frac32 V.$$

In real coordinates the divergence is $-1/L^2+1/(2L^2)+1/(2L^2)=0$. The holomorphic vector annihilates the reference spinor under Clifford multiplication, so the full Killing equation still vanishes. Since $d(*V)=0$, a local $C$ exists on this contractible patch. These nonconstant source changes are not the phase gauge transformation of the original zero-source background.

3. Add a reality requirement

Suppose, in addition to the assumptions of the lesson, one insists that $V$ be real and $H$ purely imaginary. Use the lower-component equations in a frame aligned with any nonzero $\zeta$ to show that its normalized direction must have zero expansion. Apply the result to (M05.11) with $s\ne0$. Could another choice within the source ambiguity restore both reality conditions while preserving the same spinor direction?

Hint 1

The normalized-bilinear calculation gives $\nabla_2\eta_2=\nabla_3\eta_3 =-\operatorname{Re}(H-iV_1)$ at the aligned point.

Hint 2

Take the trace, and remember that the derivative of a unit vector has $\nabla_a\eta_1=0$ in this frame. A condition derived from the equation itself applies to every representative.

Solution

The expansion is

$$\nabla_\mu\xi^\mu =-2\operatorname{Re}H-2\operatorname{Im}(\eta_\mu V^\mu).$$

If $H$ is imaginary and $V$ real, both terms vanish. The expanding example has $\nabla\cdot\xi=2s$, so for $s\ne0$ there is no such restricted source set preserving this direction. Changing $\chi$ does not change the normalized direction, and a source ambiguity cannot evade a necessary consequence of the same Killing equation.

The constructed background is consistent because it allowed complex sources without imposing this extra restriction: its $H$ has real part $-s$. Passing the transverse-integrability test guarantees the local construction with the allowed complex fields, not existence after arbitrary additional reality conditions.

Patch the spinor and the R-symmetry bundle next tests whether local solutions join on overlaps and return consistently around closed curves.

Paper guides

Connect this lesson to the original papers and their notation.

• Paper guide: Section 4, transverse holomorphic geometry
• Paper guide: Appendices A and B, conventions and geometry