Spinors as bundle sections

In the radial frame of Covariant derivatives of spinors, a parallel spinor acquired a minus sign when its angular coordinate increased by $2\pi$. Was the spinor ill-defined? The answer requires comparing components using the transition between the spin frames at the two ends.

A bundle section is a geometric field described by compatible local representatives. This lesson turns that definition into an overlap calculation. We use neutral test spinors on a fixed Euclidean metric in dimension $d=3$ or $4$, without a gauge field or a supersymmetry assumption. The transition matrices are even, so the same geometry also applies to Grassmann-odd fields.

A field can have more than one local description

Cover a manifold $M$ by open patches $U_i$. On each patch choose an oriented orthonormal coframe $e_i^a$. At a point belonging to both $U_i$ and $U_j$, the two coframes satisfy

$$e_j^a=R_{ji}{}^a{}_b e_i^b,\qquad R_{ji}(x)\in SO(d).$$

The matrix $R_{ji}$ rotates orthonormal frames; it is not the coordinate Jacobian. If the coordinate charts also differ, all expressions are first evaluated at the same geometric point.

Let $\gamma^a$ obey $\{\gamma^a,\gamma^b\}=2\delta^{ab}\mathbf1$. The local spinor columns must be related by a spin lift:

$$\psi_j=S_{ji}\psi_i,\qquad S_{ji}^{-1}\gamma^aS_{ji}=R_{ji}{}^a{}_b\gamma^b.$$

This is exactly the lift convention of the preceding lesson. The group $\operatorname{Spin}(d)$ implements these rotations on spinors and maps two-to-one onto $SO(d)$: both $S_{ji}$ and $-S_{ji}$ lift the same rotation.

The signs cannot be selected independently without a consistency check. On a triple overlap, passing from $i$ to $j$ to $k$ and back must give

$$R_{ik}R_{kj}R_{ji}=\mathbf1,\qquad S_{ik}S_{kj}S_{ji}=\mathbf1.$$

The second condition is extra information: a product of candidate lifts can be $-\mathbf1$ even when the product of the frame rotations is $+\mathbf1$. A consistent choice of lifts, up to changes of local spin frames, is a spin structure.

Once that choice is made, the columns $\psi_i$ with these transition laws define a section of the spinor bundle $\mathscr S$. Its fiber at a point is the spinor representation space: complex dimension two for a three-dimensional spinor, or four for a four-dimensional Dirac spinor. A four-dimensional Weyl representation has a two-dimensional fiber. These fiber dimensions are not a count of globally preserved supersymmetries.

This description assumes the required spin structure exists; an arbitrary oriented four-manifold need not admit one. Our explicit example below has two choices, which we will construct directly.

The differential equation must respect the overlaps

Write the local spin connection as the matrix-valued one-form

$$\mathcal A_i=\frac14\Omega_{i\,ab}\gamma^a\gamma^b,\qquad \nabla_i\psi_i=d\psi_i+\mathcal A_i\psi_i.$$

Here $\Omega$ is the coframe connection defined by $de^a+\Omega^a{}_b\wedge e^b=0$; $\mathcal A$ is not an additional gauge source. Differentiating the overlap relation gives

$$\begin{aligned} \nabla_j\psi_j &=d(S_{ji}\psi_i)+\mathcal A_jS_{ji}\psi_i\\ &=S_{ji}\nabla_i\psi_i \end{aligned} \quad\Longleftrightarrow\quad \mathcal A_j =S_{ji}\mathcal A_iS_{ji}^{-1} -(dS_{ji})S_{ji}^{-1}.$$

Thus $\nabla_i\psi_i=0$ implies $\nabla_j\psi_j=0$ when the representatives are correctly joined. A parallel spinor solves this equation in every direction.

This covariance does not guarantee that independently chosen local solutions agree. Suppose $\psi_i$ and $\psi_j$ solve their local equations. Their difference on the overlap is meaningful only after transporting their components to the same frame: $\psi_j-S_{ji}\psi_i$. It must vanish for them to describe one global solution.

Two angular patches and one parallel spinor

Consider punctured Euclidean space

$$M=(\mathbb R^2\setminus\{0\})\times\mathbb R^{d-2},\qquad ds^2=d\rho^2+\rho^2d\theta^2+\sum_{a=3}^{d}(dz^a)^2.$$

The two angular charts have branches

$$U:\ -\pi<\theta_U<\pi,\qquad V:\ 0<\theta_V<2\pi.$$

Their overlap has two disconnected parts. On the upper half-plane $\theta_V=\theta_U$; on the lower half-plane $\theta_V=\theta_U+2\pi$.

The Cartesian coframe is smooth across the omitted axis if that axis is restored. Start with the spin structure inherited from this unpunctured flat space. Its Cartesian spin-frame transitions are the identity on both overlap components, so a constant column $c$ represents a global parallel spinor.

Now use the radial coframe on each angular patch:

$$e^1=d\rho,\qquad e^2=\rho\,d\theta,\qquad e^a=dz^a\quad(a\ge3).$$

Set $G=\gamma^1\gamma^2$. The Clifford algebra gives $G^2=-\mathbf1$. The radial-to-Cartesian coframe rotation and its lift are

$$R(\theta)= \begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix}, \qquad S(\theta)=e^{-\theta G/2},$$

with the rotation acting trivially on spectator directions. Since $\psi_{\rm Cart}=S(\theta)\psi_{\rm rad}$, the same spinor has local representatives

$$\psi_U=e^{\theta_U G/2}c,\qquad \psi_V=e^{\theta_V G/2}c.$$

The spin connection is $\mathcal A_\theta=-G/2$. Direct differentiation gives $(\partial_\theta-G/2)e^{\theta G/2}c=0$ on either patch.

The overlap calculation now determines the spin transitions:

Overlap component	Angular branches	Radial spinor relation
Upper half-plane	$\theta_V=\theta_U$	$\psi_V=\psi_U$
Lower half-plane	$\theta_V=\theta_U+2\pi$	$\psi_V=-\psi_U$

For the second row,

$$e^{(\theta_U+2\pi)G/2} =e^{\theta_U G/2} \left(\cos\pi\,\mathbf1+\sin\pi\,G\right) =-e^{\theta_U G/2}.$$

The radial coframes themselves agree on both overlaps. Their spin frames agree up to the indicated signs. That is allowed because $+\mathbf1$ and $-\mathbf1$ both lift the identity frame rotation.

For a concrete two-component realization take $G=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$ and $c=(1,0)^T$. Then

$$\psi_U= \begin{pmatrix}\cos(\theta_U/2)\\ \sin(\theta_U/2)\end{pmatrix}, \qquad \psi_V= \begin{pmatrix}\cos(\theta_V/2)\\ \sin(\theta_V/2)\end{pmatrix}.$$

Their norm is one; the minus sign on the lower overlap is exactly the required transition. These are compatible descriptions of a nonzero global spinor, not a discontinuous physical field.

Change the spin structure, not the local equation

Because the axis is missing, there is another spin structure. To see the choice without using a classification theorem, keep the Cartesian coframes but take their spin transition to be $+1$ on the upper overlap and $\eta$ on the lower one, where $\eta=\pm1$.

Changing the sign of either local spin frame multiplies both transitions by the same sign. Their relative sign $\eta$ cannot change this way. The two values therefore give inequivalent lifts of the frame bundle. The value $\eta=+1$ is the structure inherited from unpunctured flat space; $\eta=-1$ does not extend across the restored axis.

In radial frames the transitions become $+1$ and $-\eta$. Local parallel solutions are still

$$\psi_U=e^{\theta_U G/2}c_U,\qquad \psi_V=e^{\theta_V G/2}c_V.$$

The upper overlap requires $c_V=c_U$. On the lower overlap, the same formulas give $\psi_V=-\psi_U$, whereas the prescribed transition requires $\psi_V=-\eta\psi_U$. Hence

$$(\eta-1)c_U=0.$$

For $\eta=+1$, every constant column joins to a global parallel spinor. For $\eta=-1$, only the zero parallel solution survives. The local metric, curvature, and differential equation are identical in these two cases.

This failure concerns parallel sections, not the existence of sections. For $\eta=-1$, for example, a nonzero constant radial column joins globally because both radial transitions are $+1$. It simply has $\nabla_\theta\psi=-G\psi/2\ne0$.

Compare the endpoints in the same fiber

Parallel transport solves an ordinary differential equation along a path. Its holonomy around a loop is the resulting linear map on the starting spinor fiber after all end-frame identifications have been included.

Unwrap the radial angle from $0$ to $2\pi$. Solving the local parallel equation gives the transport matrix

$$U_{\rm rad}=e^{\pi G}=-\mathbf1.$$

But the endpoint components are identified with the starting ones by the radial seam transition $t_{\rm rad}=-\eta\mathbf1$. The return map in one fiber is consequently

$$\mathcal H=t_{\rm rad}^{-1}U_{\rm rad}=\eta\mathbf1.$$

In Cartesian frames the local connection vanishes, so $U_{\rm Cart}=\mathbf1$, while the seam transition is $t_{\rm Cart}=\eta\mathbf1$. This gives the same $\mathcal H$. For the spin structure extending through the axis, $\mathcal H=+\mathbf1$, despite the radial components’ minus sign. For the other structure, $\mathcal H=-\mathbf1$, despite zero local curvature.

A global parallel spinor must be fixed by the return map of every loop. Flatness removes local curvature obstructions; it does not remove the noncontractible loop around a missing axis.

What changes for charged spinors?

An additional phase transition is different from choosing a spin lift. If a field also has unit $U(1)$ charge, its overlap law can be

$$\psi_j=e^{i\lambda_{ji}}S_{ji}\psi_i.$$

The phases describe a complex line bundle $L$, whose fiber is one complex dimension. The combined spinor is a section of $L\otimes\mathscr S$; charge $-1$ uses the inverse phases and $L^{-1}\otimes\mathscr S$. This is the bundle assignment for the two supersymmetry parameters in the three-manifold paper, section 3, printed pages 10–11.

Our neutral calculation fixed a spin structure separately. The paper also allows a combined structure called $\operatorname{Spin}^c$, where the product transitions are meaningful without making that separation. The separate factors must not be assumed globally defined in that more general setting. Connections, flux, and holonomy computes the additional gauge phases and their quantization; no such phase was added to our neutral example.

Local uniqueness is not global existence

The spinor equations relevant to the paper determine all directional derivatives: $\nabla_\mu\psi=M_\mu(x)\psi$, with smooth matrix coefficients $M_\mu$. Along any smooth path $x(s)$, such an equation becomes a homogeneous matrix ODE,

$$\frac{d}{ds}\psi(x(s))=B(s)\psi(x(s)).$$

Uniqueness of that ODE means that zero initial data stays zero. On a connected manifold, a global solution that vanishes at one point must vanish everywhere. A nonzero solution is therefore nowhere zero. This is the argument in section 3, footnote 5.

It does not prove that arbitrary nonzero initial data extends globally. Compatibility between directions, and the loop return maps, still restrict it. Nor does the argument apply to every first-order field equation: the contracted Dirac equation does not specify every directional derivative. Exercise 3 makes that distinction explicit.

Exercises

1. Repair three individually valid lifts

On a triple overlap, suppose the frame transitions $R_{21},R_{32},R_{13}$ are each rotations through $2\pi/3$ in the same plane. Their product is the identity. Choose initially $S_{21}=S_{32}=S_{13}=e^{-\pi G/3}$, with $G^2=-\mathbf1$. Check the spin cocycle and repair it without changing any frame rotation.

First hint

The three matrices commute because they are exponentials of the same generator. Add their exponents.

Second hint

Multiplying one lift by $-1$ leaves its action on the gamma matrices unchanged.

Solution

The spin product is

$$S_{13}S_{32}S_{21}=e^{-\pi G}=-\mathbf1.$$

It fails the triple-overlap consistency condition. Individual validity as a lift does not suffice. Replace $S_{13}$ by $-e^{-\pi G/3}=e^{2\pi G/3}$; the product is then $+\mathbf1$. This is the lift of the reverse rotation by $-4\pi/3$, which projects to the same $SO(d)$ rotation as $2\pi/3$.

The check is also visible on a putative representative: the original transitions would demand $\psi_1=-\psi_1$ after a round trip through descriptions of the same point. They therefore cannot define the intended spin bundle. This is a transition inconsistency, not a dynamical obstruction to solving an equation.

2. A global section that is not a parallel section

Use $\eta=-1$ on the punctured space. Let the radial spinor components equal the same nonzero constant column $c$ on both angular patches. Show that they define a global section, compute its covariant derivative, and reconcile its existence with $\mathcal H=-\mathbf1$.

First hint

Check the prescribed radial transitions, which are $+1$ and $-\eta$. Do not replace “constant components” with “parallel.”

Second hint

The connection is still $\mathcal A_\theta=-G/2$. Since $G^2=-\mathbf1$, its kernel is zero.

Solution

Both radial transitions are $+\mathbf1$, so the constant columns agree correctly. Their covariant derivative is

$$\nabla_\theta\psi=-\frac12Gc,\qquad \nabla_\rho\psi=\nabla_{z^a}\psi=0.$$

The angular derivative is nonzero because $Gc=0$ would imply $c=0$ after multiplication by $G$. The holonomy test rules out a nonzero parallel section, which would have to satisfy $-c=c$. It does not rule out ordinary sections with nonzero covariant derivative.

In local Cartesian spin frames the same section has components $e^{-\theta G/2}c$. Those acquire the required Cartesian transition $-1$ around the omitted axis. A change of frame has not converted the section into a parallel one.

3. Does a Dirac solution have to be nowhere zero?

On unpunctured flat Euclidean space with Cartesian coordinates $x,y$ and any spectator coordinates, take a nonzero constant spinor $c$ and

$$\psi(x,y)=(x\gamma^1-y\gamma^2)c.$$

Show that $\gamma^\mu\partial_\mu\psi=0$ and that the solution vanishes at $x=y=0$. Explain why this does not contradict the preceding uniqueness argument.

First hint

Only the $x$ and $y$ derivatives contribute. Each squared gamma matrix is the identity.

Second hint

Compare a contracted equation with the separate equations for $\partial_x\psi$ and $\partial_y\psi$. The zero field and the proposed field share their value at the origin but not their derivatives.

Solution

Direct differentiation gives

$$\gamma^\mu\partial_\mu\psi =\left[(\gamma^1)^2-(\gamma^2)^2\right]c=0.$$

The field vanishes at $x=y=0$ but is not identically zero: for example, $\partial_x\psi=\gamma^1c\ne0$, since $(\gamma^1)^2=\mathbf1$.

The equation $\not\partial\psi=0$ only constrains a gamma-contracted combination of derivatives. It does not supply a homogeneous ODE for the spinor along every arbitrary curve. In contrast, $\nabla_\mu\psi=M_\mu\psi$ specifies every directional derivative; at a zero of $\psi$, all those derivatives must also vanish. The hypothesis needed for that uniqueness argument is absent here.

A local spinor formula becomes a global solution only after its transition laws, differential equation, and loop identifications agree. That is the check to perform before treating a locally constructed supersymmetry parameter as a symmetry of the full background.

Paper guides

Connect this lesson to the original papers and their notation.

• Paper guide: Section 3, geometry from one spinor