Patch the spinor and the R-symmetry bundle

A local solution of the Killing-spinor equation can fail to return to itself after transport around a closed curve. Conversely, components that acquire a phase on an overlap can describe a perfectly regular global spinor. The question is whether the spin rotation and the R-symmetry transformation fit together.

Prepare with local background reconstruction, spinors as bundle sections and connections, flux and holonomy. We will join two explicit local solutions, extract the general bundle condition, and then test information that curvature alone misses.

Cancel a spin connection on two patches

The parent is three-dimensional $\mathcal N=2$, with four real Poincaré supercharges. We construct a commuting charge-$+1$ Euclidean Killing spinor $\zeta$. The construction establishes this preserved solution; it does not infer a complete supersymmetry count from the name of the manifold.

Take a circle of circumference $L_\tau$ times a sphere of radius $\ell$:

$$g=d\tau^2+\ell^2(d\theta^2+\sin^2\theta\,d\varphi^2), \qquad \tau\sim\tau+L_\tau,\qquad H=V=0. \tag{M06.1}$$

Choose the periodic spin structure on the circle for now. The R symmetry is a compact $U(1)$ with $\zeta$ of unit charge; $A$ is its background connection, not a dynamical gauge field. Its orthonormal components have mass dimension one, while its line integral and flux are dimensionless.

Away from the poles use the oriented coframe $e^1=d\tau$, $e^2=\ell\,d\theta$, $e^3=\ell\sin\theta\,d\varphi$. We retain the matrices $(\gamma^1,\gamma^2,\gamma^3)=(\sigma_3,-\sigma_1,-\sigma_2)$. Thus $\gamma^2\gamma^3=i\gamma^1$. Cartan’s equation gives $\Omega_{23}=-\cos\theta\,d\varphi$. With $\nabla=d+\Gamma$ and $\Gamma=\Omega_{ab}\gamma^a\gamma^b/4$, the spin connection is

$$\Gamma_{\rm p}=-\frac{i}{2}\cos\theta\,\gamma^1d\varphi. \tag{M06.2}$$

Repeated frame indices in this definition are summed in both orders. The factor $1/4$ consequently becomes $1/2$ for the single independent pair $23$. In this polar frame, $A_{\rm p}=-\cos\theta\,d\varphi/2$ cancels $\Gamma_{\rm p}$ on the upper column $u_+=(1,0)^T$. But neither this coframe nor this gauge is regular at a pole.

Write $R(\beta)$ for the transverse coframe rotation $(e^2,e^3)^T\mapsto \left(\begin{smallmatrix}\cos\beta&-\sin\beta\\ \sin\beta&\cos\beta\end{smallmatrix}\right) (e^2,e^3)^T$. Its spin lift in our convention is $S(\beta)=\exp(-i\beta\gamma^1/2)$. Choose the north frame by $R(\varphi)$ and the south frame by $R(-\varphi)$, leaving $e^1$ unchanged. These frames are regular on their respective hemispheres. For example, near the north pole, with $X=\ell\theta\cos\varphi$, $Y=\ell\theta\sin\varphi$, the north coframe approaches $(dX,dY)$. Near the south pole the analogous south coframe approaches $(-dX,dY)$, with $X=\ell(\pi-\theta)\cos\varphi$, $Y=\ell(\pi-\theta)\sin\varphi$.

The transformed connection is $\Gamma'=S\Gamma S^{-1}-dS\,S^{-1}$. Therefore choose

$$\begin{aligned} A_N&=\frac{1-\cos\theta}{2}\,d\varphi, &\Gamma_N&=i A_N\gamma^1,\\ A_S&=-\frac{1+\cos\theta}{2}\,d\varphi, &\Gamma_S&=i A_S\gamma^1,\\ \zeta_N&=\zeta_S=u_+ . \end{aligned} \tag{M06.3}$$

The apparent angular singularities disappear: near a pole the corresponding potential is proportional to $\rho^2d\varphi=X\,dY-Y\,dX$. On both patches $(d+\Gamma-iA)\zeta=0$, the Killing equation for $H=V=0$.

Now perform the overlap calculation. Going from south components to north components requires

$$A_N=A_S+d\varphi,\qquad S_{NS}=e^{-i\varphi\gamma^1},\qquad \zeta_N=e^{i\varphi}S_{NS}\zeta_S=u_+ . \tag{M06.4}$$

Both actual overlap functions are single-valued under $\varphi\mapsto\varphi+2\pi$. The half-angle lifts from the singular polar frame were only local aids; they are not separate global transitions. More explicitly, $e^{i\varphi}S_{NS}=\operatorname{diag}(1,e^{2i\varphi})$. The selected component glues with no phase, even though neither the spin line nor the R line is trivial on the sphere.

The same calculation fixes the flux sign:

$$F_A=dA_N=dA_S=\frac12\sin\theta\,d\theta\wedge d\varphi, \qquad \frac1{2\pi}\int_{S^2}F_A=1. \tag{M06.5}$$

The sphere is oriented by $\sin\theta\,d\theta\wedge d\varphi$. There cannot be a single smooth potential $A$ on the whole sphere, since its exterior derivative would integrate to zero. The globally defined object is the connection with the transitions in (M06.4).

This also recovers the previous lesson’s local construction. In north stereographic coordinates $z=x-iy=e^{-i\varphi}\tan(\theta/2)$, $c=2\ell/(1+x^2+y^2)$ and $\eta=d\tau$. Its formula $A=(\partial_y\log c\,dx-\partial_x\log c\,dy)/2$ becomes $(x\,dy-y\,dx)/(1+x^2+y^2)=A_N$.

Patch the entire covariant equation

Let $S_{ji}$ lift the change of orthonormal frame from patch $i$ to patch $j$, and let $e^{i\lambda_{ji}}$ be the R transition. At the same geometric point, $\zeta_j=e^{i\lambda_{ji}}S_{ji}\zeta_i$ and $A_j=A_i+d\lambda_{ji}$. Ordinary R gauge transformations have real $\lambda_{ji}$: if $A$ is complex, only its real part transforms as a connection, while $\operatorname{Im}A$ is a global one-form. This is the global convention in the source’s section 3.

To check the full Killing equation, define its algebraic one-form

$$\mathcal T= \frac12H e^a\gamma^a+iV\,\mathbf1+ \frac12\epsilon_{abc}e^a V^b\gamma^c, \qquad \mathcal D=d+\Gamma-iA\,\mathbf1+\mathcal T . \tag{M06.6}$$

Here $V=V_a e^a$ is the global one-form dual to $V^\mu$, $H$ is a scalar, and $\epsilon_{123}=1$. The equation is $\mathcal D\zeta=0$. Clifford multiplication transforms by conjugation, so $\mathcal T_j=S_{ji}\mathcal T_iS_{ji}^{-1}$. Using the spin-connection transformation above,

$$\begin{aligned} (d+\Gamma_j-iA_j)(e^{i\lambda_{ji}}S_{ji}\zeta_i) &=e^{i\lambda_{ji}}S_{ji}(d+\Gamma_i-iA_i)\zeta_i,\\ \mathcal D_j\zeta_j &=e^{i\lambda_{ji}}S_{ji}\mathcal D_i\zeta_i . \end{aligned} \tag{M06.7}$$

The $i\,d\lambda_{ji}$ term from differentiating the phase cancels the gauge shift; $dS_{ji}$ cancels the inhomogeneous spin-connection term. Thus local zeroes of the equation agree on overlaps. This argument requires the sources themselves to patch, as tensors or as a connection, as well as compatible local spinor values.

On a triple overlap the product of transitions must be the identity. After fixing a spin structure, the spin lifts obey their cocycle condition, and the R phases separately define a line bundle $L$. The spinor is a section of $L\otimes S$, where $S$ is the rank-two spin bundle. One may instead formulate the construction with only the combined spin and R transitions, a $\operatorname{Spin}^c$ structure. We use a fixed spin structure here so that the separate flux statement (M06.5) is unambiguous.

Read the bundle constraint from the bilinear

The nowhere-vanishing spinor determines a transverse complex line of anti-holomorphic covectors, denoted $\bar K$. In an adapted chart this line is spanned by $d\bar z$. It is a line inside the complexified cotangent bundle, not an additional background gauge group.

Use the antisymmetric contraction matrix $C=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$. For the local solution $\zeta=\chi u_+$ and $e^2-ie^3=c\,dz$, direct matrix multiplication gives

$$P=\zeta^TC\gamma_a\zeta\,e^a =\chi^2(e^2+ie^3)=c\chi^2d\bar z . \tag{M06.8}$$

It has R-charge two. Thus $P$ is a nonzero section of $L^2\otimes\bar K$, not an ordinary uncharged one-form. A nowhere-zero section trivializes a complex line bundle, giving $L^2\otimes\bar K\simeq\mathbf1$, or $2c_1(L)+c_1(\bar K)=0$ in integral cohomology. This is the bundle statement immediately before equation (4.15). Integral cohomology matters: flux integrals alone do not detect torsion classes.

One can see the compensating phase without invoking bundle notation. Under $z'=f(z)$ write $f'=|f'|e^{i\alpha}$. The adapted metric has $c'=c/|f'|$. If $P=p\,d\bar z$, then before an R transformation $p'=p/\overline{f'}$. Hence

$$s=\frac{p}{c}=\chi^2,\qquad s'=e^{i\alpha}s,\qquad \chi'=e^{i\alpha/2}\chi . \tag{M06.9}$$

The last equation uses a chosen spin lift on the overlap. An R phase $e^{-i\alpha/2}$ cancels it. This is how a spinor can give a scalar supercharge under the combined transformations. The factors of $c$ are essential: $p$ itself is a covector component and does not transform by a pure phase.

The relation involving $L^2$ does not choose the square root, its overlap signs or its connection holonomies. The bilinear loses $\zeta\mapsto-\zeta$. Consequently, a correct bilinear and correct fluxes are necessary checks, not a substitute for gluing the original spinor. In the sphere example the selected spin line has connection $\Gamma_+=iA$; written as $d-ia_{\rm spin}$ it has $a_{\rm spin}=-A$ and Chern number $-1$. Tensoring with the R line of Chern number $+1$ cancels it.

Check a closed curve as well as a flux

For a contractible sphere loop bounding oriented solid angle $\mathcal A$, the R parallel-transport phase is $e^{i\mathcal A/2}$. The selected spin line has the inverse phase. Their product is one. For the equator, $\mathcal A=2\pi$: both factors are $-1$. Ignoring spin lifts because the corresponding vector rotation is a full turn would miss this cancellation.

There is also the noncontractible circle. Add the globally defined flat term $\nu\,d\tau$ to $A$ without changing (M06.5). The local equation becomes $\partial_\tau\zeta=i\nu\zeta$. With periodic spin structure, its nonzero solution is global only if $e^{i\nu L_\tau}=1$. Flux is unchanged for every $\nu$, but most choices lose this solution. The exercise below checks both spin structures and also a complex $\nu$.

Finally, a background for the supercharge must support the field theory’s other charged fields. With the compact R group and normalization fixed above, scalar representations have integer R charge. A fractional trial charge used in a local calculation does not automatically define a power of this nontrivial line bundle. A different global symmetry group, or an explicitly specified combined gauge/flavor bundle, requires its own transition check. The geometry does not silently supply that extra data.

Exercises

1. Change the R flux

Replace the sphere potentials by $A_N^{(n)}=nA_N$ and $A_S^{(n)}=nA_S$, with integer $n$, keeping $H=V=0$ and the same metric. For which $n$ can the charge-$+1$ equation have a nonzero sphere solution? Find the selected spinor component in each case. Explain why allowing nonconstant local components cannot rescue another $n$.

Hint 1

Apply the curvature of $d+\Gamma-iA^{(n)}$ to a putative solution before trying to solve the differential equations.

Hint 2

The spin curvature is $iF_A\gamma^1$. Its two eigenvalues test the two components separately.

Solution

The curvature condition is $iF_A(\gamma^1-n\mathbf1)\zeta=0$. Since $F_A$ is nonzero everywhere as an area two-form, a nonzero component requires $n=+1$ or $n=-1$. For $n=+1$ the solution is $u_+$, as above. For $n=-1$ it is $u_-=(0,1)^T$. Its overlap is $e^{-i\varphi}e^{-i\varphi\gamma^1}u_-=u_-$. Each local covariant derivative vanishes directly.

For every other integer the curvature matrix is invertible, forcing $\zeta=0$ pointwise. Changing its local dependence cannot undo this algebraic integrability obstruction. This calculation concerns the charge-$+1$ equation with these fixed $H,V$; it does not classify other backgrounds on the same manifold.

2. Keep the flux but change the circle data

On the $n=1$ background add $A_\tau=\nu$, where $\nu$ may be complex. Let the spin boundary sign be $\varepsilon=+1$ or $-1$. Derive the condition for the selected solution, identify the allowed real $\nu$ modulo single-valued R gauge transformations, and compare it with the condition obtained by squaring the spinor.

Hint 1

Use $\zeta(\tau)=e^{i\nu\tau}u_+$ and impose the boundary condition on the original spinor, not only on $P$.

Hint 2

Separate the modulus and phase of $e^{i\nu L_\tau}=\varepsilon$. A gauge function $e^{2\pi i m\tau/L_\tau}$ shifts $\nu$ by $2\pi m/L_\tau$.

Solution

Gluing the endpoints requires $e^{i\nu L_\tau}=\varepsilon$. Its modulus sets $\operatorname{Im}\nu=0$. The remaining conditions are $\nu L_\tau=2\pi m$ for the periodic spin structure and $\nu L_\tau=(2m+1)\pi$ for the antiperiodic one. Modulo the stated gauge transformations, the allowed holonomies are respectively $+1$ and $-1$.

The bilinear instead tests $e^{2i\nu L_\tau}=1$. It permits both answers without knowing which spin structure was chosen, so it cannot complete the global check. Equivalently, for real $\nu$ the eigenvalues of the circle operator $-i(\partial_\tau-i\nu)$ are $(2\pi m+\delta)/L_\tau-\nu$, with $\delta=0$ or $\pi$. A zero eigenvalue occurs exactly under the original spinor condition. All these connections have the same sphere flux.

3. Test a charged scalar

Keep the compact R group, unit supercharge and $n=1$ flux. A scalar of R charge $q$ has overlap $\phi_N=e^{iq\varphi}\phi_S$. Check $q=2$ and $q=1/2$. For the allowed case, can a smooth scalar section be nowhere zero on the sphere? Contrast your answer with the nowhere-zero charged spinor constructed above.

Hint 1

First test whether the overlap itself is single-valued around the equator.

Hint 2

A nowhere-zero section of a complex line bundle gives a global basis for that line. Compute the line’s Chern number before assuming such a section exists.

Solution

For $q=2$ the transition is single-valued and defines $L^2$, whose Chern number is two. Sections exist, but none can be everywhere nonzero: such a section would trivialize $L^2$, contradicting its nonzero Chern class. This is a statement about sections, not a claimed scalar equation of motion.

For $q=1/2$, increasing $\varphi$ by $2\pi$ changes the proposed transition’s sign. It is not a representation of the specified compact R group, and $L$ with Chern number one has no square-root line bundle. The proposed scalar is therefore not globally defined with these data. Additional global-group or bundle choices would be a different problem.

The unit-charge spinor has a spin transition as well as an R transition. On its selected component the two cancel, as (M06.4) demonstrates. Treating it as an ordinary charge-one scalar would discard the mechanism that made the global solution possible.

The practical exit test is now concrete: give the local spinor, its spin and R overlap matrices, and its return phase around every noncontractible cycle. The source’s construction of the global scalar supercharge uses precisely the compensation between the adapted spin frame and the R bundle. Finding an additional solution imposes further conditions on the same sources. Two supercharges and their Killing vector derives the isometry selected by an opposite-charge pair. Squashed spheres provides a complete family whose global solution spaces can be written explicitly.

The geometric-dependence dossier applies global spin and source data to a finite curved-torus family and a controlled change of its foliation-compatible metric.

Paper guides

Connect this lesson to the original papers and their notation.

• Paper guide: Section 4, transverse holomorphic geometry