Covariant derivatives of spinors

A spinor’s components depend on a choice of local orthonormal frame. If that frame rotates with position, differentiating the components also differentiates the choice of basis. The spin connection removes that extra change. Our task is to derive its matrix coefficient and check that it gives the same answer in two frames.

The preparation is Frames and spin connections. The general derivation and exercises below work in three or four dimensions and require no three-dimensional spinor conventions. Only the final explicit three-dimensional example uses Spinors in three dimensions. The field here is a neutral spinor on a fixed metric; its geometric transformation law applies both to ordinary spinor wavefunctions and to Grassmann-odd field components.

Derive the connection from frame rotations

Initially use Euclidean signature in dimension $d=3$ or $d=4$. Recall the coframe $e^a=e^a{}_\mu dx^\mu$ and its connection

$$de^a+\Omega^a{}_b\wedge e^b=0, \qquad \Omega_{ab}=-\Omega_{ba}.$$

An orientation-preserving orthogonal frame change $e'=Re$, with $R\in SO(d)$, sends vector components to $V'=RV$ and the connection to

$$\Omega'=R\Omega R^{-1}-dR\,R^{-1}.$$

Spinor components transform by a different matrix, $\psi'=S\psi$. We must specify the relationship between $S$ and $R$ before determining a sign.

Take constant matrices $\gamma^a$ satisfying the Clifford algebra

$$\gamma^a\gamma^b+\gamma^b\gamma^a=2\delta^{ab}\mathbf1.$$

They can be two-by-two matrices in three dimensions or four-by-four Dirac matrices in four dimensions. The derivation uses the algebra, not a particular realization. A spinor representation of the frame rotation satisfies

$$S^{-1}\gamma^a S=R^a{}_b\gamma^b.$$

This convention fits vector bilinears: if a row spinor transforms as $\chi'=\chi S^{-1}$, then $\chi'\gamma^a\psi'=R^a{}_b\chi\gamma^b\psi$. No Hermitian-conjugation relation between $\chi$ and $\psi$ is needed for that statement.

Consider an infinitesimal rotation $R=1+\lambda$, with $\lambda_{ab}=-\lambda_{ba}$, and write $S=1+\sigma(\lambda)$. The defining relation requires

$$[\sigma(\lambda),\gamma^c]=-\lambda^c{}_b\gamma^b.$$

To find $\sigma$, calculate one commutator directly:

$$\begin{aligned} [\gamma^a\gamma^b,\gamma^c] &=\gamma^a(2\delta^{bc}-\gamma^c\gamma^b) -\gamma^c\gamma^a\gamma^b\\ &=2\delta^{bc}\gamma^a -(\gamma^a\gamma^c+\gamma^c\gamma^a)\gamma^b\\ &=2(\delta^{bc}\gamma^a-\delta^{ac}\gamma^b). \end{aligned}$$

Contracting this expression with $\lambda_{ab}$ yields $-4\lambda^c{}_b\gamma^b$. Therefore

$$\sigma(\lambda)=\frac14\lambda_{ab}\gamma^a\gamma^b =\frac18\lambda_{ab}[\gamma^a,\gamma^b].$$

Every ordered pair $a,b$ is summed here. Equivalently, the first expression is $\frac12\sum_{a<b}\lambda_{ab}\gamma^a\gamma^b$. This double counting is the source of a common factor-of-two mistake.

The antisymmetric gamma products are the spin rotation generators. For comparison, Bilal’s equation (2.17) writes Lorentz generators as $\Sigma^{ab}=\frac{i}{4}[\gamma^a,\gamma^b]$ in signature $(+---)$. With that generator normalization, $\sigma(\lambda)=-\frac{i}{2}\lambda_{ab}\Sigma^{ab}$ is the whole infinitesimal transformation matrix, with the rotation parameters already included. The Clifford algebra and signature must accompany either convention.

Replace the infinitesimal rotation parameter by the frame connection. Its action on spinors is the matrix-valued one-form

$$\mathcal A=\frac14\Omega_{ab}\gamma^a\gamma^b, \qquad \nabla_\mu\psi= \left(\partial_\mu+\mathcal A_\mu\right)\psi.$$

The symbol $\mathcal A_\mu$ denotes this spin connection matrix, not an independent Abelian background field. Gauge or R-symmetry connections would be additional terms when those couplings are specified.

Check covariance rather than guessing a coefficient

The required transformation law follows by differentiating $\psi'=S\psi$:

$$\begin{aligned} \nabla'_\mu\psi' &=\partial_\mu(S\psi)+\mathcal A'_\mu S\psi\\ &=S\partial_\mu\psi+ \left(\partial_\mu S+\mathcal A'_\mu S\right)\psi. \end{aligned}$$

This equals $S\nabla_\mu\psi$ precisely when

$$\mathcal A'_\mu =S\mathcal A_\mu S^{-1} -(\partial_\mu S)S^{-1}.$$

The connection built above has this law. To verify it locally, expand the frame transformation to first order:

$$\Omega'=\Omega+[\lambda,\Omega]-d\lambda.$$

The map $\lambda\mapsto\sigma(\lambda)$ preserves commutators. This follows by using the gamma commutator just derived on each factor in $\gamma^a\gamma^b$. Consequently its image is

$$\mathcal A'=\mathcal A+[\sigma(\lambda),\mathcal A] -d\sigma(\lambda),$$

which is exactly the first-order expansion of the spinor transformation law. Composing smooth local rotations gives the finite law, on a patch where a continuous choice of $S$ has been made.

There is a second compatibility check. Define coordinate-index gamma matrices by

$$\gamma^\mu(x)=e_a{}^\mu(x)\gamma^a, \qquad \{\gamma^\mu(x),\gamma^\nu(x)\}=2g^{\mu\nu}(x)\mathbf1.$$

The tetrad postulate and the identity $[\mathcal A_\mu,\gamma^a]=-\Omega_\mu{}^a{}_b\gamma^b$ give

$$\partial_\mu\gamma^\nu +\Gamma^\nu{}_{\mu\lambda}\gamma^\lambda +[\mathcal A_\mu,\gamma^\nu]=0.$$

Both the coordinate-vector index and the spinor matrix action are accounted for. This is why contracting the derivative into the Dirac operator, $\not\nabla=\gamma^\mu\nabla_\mu$, is geometrically consistent.

A worked calculation in the radial frame

Use the cylindrical coframe from the preceding lesson:

$$e^1=d\rho,\quad e^2=\rho\,d\theta,\quad e^3=dz, \qquad \Omega^1{}_2=-d\theta,\quad\Omega^2{}_1=d\theta.$$

The patch has $\rho>0$. Both ordered frame-index pairs contribute:

$$\mathcal A_\theta =\frac14(-\gamma^1\gamma^2+\gamma^2\gamma^1) =-\frac12\gamma^1\gamma^2, \qquad \mathcal A_\rho=\mathcal A_z=0.$$

Thus

$$\nabla_\rho=\partial_\rho,\qquad \nabla_\theta=\partial_\theta-\frac12\gamma^1\gamma^2, \qquad \nabla_z=\partial_z.$$

The dimensionless angular derivative has a dimensionless connection coefficient. Its contribution to a derivative per unit physical length includes the inverse-frame factor $1/\rho$.

The coframe rotation into $(dx,dy,dz)$ is

$$R(\theta)= \begin{pmatrix} \cos\theta&-\sin\theta&0\\ \sin\theta&\cos\theta&0\\ 0&0&1 \end{pmatrix}.$$

Set $G=\gamma^1\gamma^2$. The Clifford algebra gives $G^2=-1$. The spin rotation lifting this coframe rotation is

$$S(\theta)=e^{-\theta G/2} =\cos\frac\theta2\,\mathbf1-\sin\frac\theta2\,G.$$

One can multiply these matrices to check, for example, $S^{-1}\gamma^1S=\cos\theta\,\gamma^1-\sin\theta\,\gamma^2$. Since $(\partial_\theta S)S^{-1}=-G/2=\mathcal A_\theta$, the transformed connection is zero.

A spinor with constant components $\psi_0$ in the Cartesian spin frame therefore has radial-frame components

$$\psi_{\rm rad}(\theta)=S^{-1}(\theta)\psi_0 =e^{\theta G/2}\psi_0.$$

Direct differentiation gives $(\partial_\theta-G/2)\psi_{\rm rad}=0$. The components vary, but the spinor is parallel. This is the spinor counterpart of the constant Cartesian vector checked in the frame lesson.

Notice that $S(\theta+2\pi)=-S(\theta)$. The rotating spin frame is defined here on an angular patch; identifying its ends requires a spin-frame transition. The minus sign of these components does not mean that flat space has acquired curvature. Global choices of spin structure require the additional patching discussion developed later.

The same calculation also determines the Dirac operator:

$$\begin{aligned} \not\nabla &=\gamma^1\partial_\rho+ \frac{\gamma^2}{\rho} \left(\partial_\theta-\frac12\gamma^1\gamma^2\right) +\gamma^3\partial_z\\ &=\gamma^1\left(\partial_\rho+\frac1{2\rho}\right) +\frac{\gamma^2}{\rho}\partial_\theta+\gamma^3\partial_z. \end{aligned}$$

In the last step, $\gamma^2\gamma^1\gamma^2=-\gamma^1$. Replacing Cartesian derivatives by inverse-frame derivatives alone would miss the $1/(2\rho)$ term. Exercise 2 checks its coefficient through integration by parts.

What changes in four dimensions

In four Euclidean dimensions add $e^4=dw$ and use four gamma matrices. The cylindrical rotation still acts only in the $1$–$2$ plane, and the Dirac operator gains $\gamma^4\partial_w$. The derivation of the connection and the parallel spinor uses exactly the same Clifford relations.

For Lorentzian signature choose, for example, $\eta_{ab}=\operatorname{diag}(-1,1,1,1)$, as in the frame lesson, and matrices $\Gamma^a$ satisfying $\{\Gamma^a,\Gamma^b\}=2\eta^{ab}$. Lower the first connection index with $\eta$:

$$\Omega_{\mu ab}=\eta_{ac}\Omega_\mu{}^c{}_b, \qquad \nabla_\mu=\partial_\mu+\frac14\Omega_{\mu ab}\Gamma^a\Gamma^b.$$

The commutator calculation now uses $\eta$ instead of $\delta$. Spatial rotations and boosts are both included in this formula, with local frame transformations restricted to the connected proper, time-oriented Lorentz group. The spinor matrix for a boost need not be unitary. The inverse $S^{-1}$, rather than an assumed Hermitian adjoint, is what appears in the covariance proof.

This is a signature declaration for the Clifford matrices, not a prescription to reuse unchanged Euclidean matrices. It is also different from Bilal’s $(+---)$ choice; the abstract connection formula works with either signature when the Clifford algebra and index lowering are changed consistently.

Exercises using the general construction

1. Detect a missing factor of two

Suppose a calculation gives $\mathcal A_\theta=cG$ in the cylindrical coframe, where $G=\gamma^1\gamma^2$. Determine $c$ by requiring that the radial representation of a constant Cartesian spinor be parallel. Test the incorrect value $c=-1/4$.

First hint

The required spinor is $e^{\theta G/2}\psi_0$. Differentiate it before adding the connection.

Second hint

The same coefficient must work for every $\psi_0$. Compare your result with the sum over both ordered pairs $(1,2)$ and $(2,1)$.

Solution

Its derivative is $G\psi_{\rm rad}/2$, so

$$\nabla_\theta\psi_{\rm rad} =\left(\frac12+c\right)G\psi_{\rm rad}.$$

For this to vanish for every constant $\psi_0$, we need $c=-1/2$. With $c=-1/4$, the residual is $G\psi_{\rm rad}/4$, which is nonzero for a nonzero spinor because $G^2=-1$. The incorrect coefficient arises if one keeps only $\Omega_{12}\gamma^1\gamma^2/4$ and forgets the equal contribution from the ordered pair $(2,1)$. Covariance fixes the normalization without a field equation.

2. Change the circumference and check the measure

For $e^1=d\rho$, $e^2=f(\rho)d\theta$, $e^3=dz$, with $f>0$, the frame lesson gives $\Omega^2{}_1=f'd\theta$. Find the Dirac operator. Then check its radial term by integration by parts in the Euclidean measure $f\,d\rho\,d\theta\,dz$, using Hermitian gamma matrices and smooth test spinors supported away from the radial boundary.

First hint

First calculate $\mathcal A_\theta$, and then contract with the inverse coframe. The angular direction contributes a factor $1/f$.

Second hint

Integration by parts differentiates both the spinor and the factor $f$ in the measure. Thus the formal adjoint of $\partial_\rho$ is $-\partial_\rho-f'/f$. Determine the multiplication term that makes the full radial derivative anti-Hermitian.

Solution

Both antisymmetric entries give $\mathcal A_\theta=-f'\gamma^1\gamma^2/2$. Therefore

$$\not\nabla =\gamma^1\left(\partial_\rho+\frac{f'}{2f}\right) +\frac{\gamma^2}{f}\partial_\theta+\gamma^3\partial_z.$$

To derive the stated adjoint, integrate $\int f\,u^\dagger\partial_\rho v$ by parts. The boundary term vanishes by the support assumption, and differentiation of $f u^\dagger$ gives both $f\partial_\rho u^\dagger$ and $f'u^\dagger$. For a real multiplication term $h(\rho)$,

$$(\partial_\rho+h)^\dagger =-\partial_\rho-\frac{f'}f+h.$$

It equals $-(\partial_\rho+h)$ precisely when $h=f'/(2f)$, matching the connection calculation. This checks formal anti-Hermiticity of the Euclidean kinetic derivative. With periodic angular dependence and suitable behavior in $z$, the other two terms are also anti-Hermitian. Taking $f=\rho$ recovers the cylindrical result, while constant $f$ removes the radial correction.

3. Differentiate a bilinear

Let a row spinor transform as $\chi'=\chi S^{-1}$ and a column as $\psi'=S\psi$. Find the derivative of $\chi$ that makes $\partial_\mu(\chi\psi)=(\nabla_\mu\chi)\psi+\chi\nabla_\mu\psi$. Then prove that $V^a=\chi\gamma^a\psi$ has the vector covariant derivative from the frame lesson.

First hint

Choose the sign of the row connection so the connection terms cancel in the scalar product.

Second hint

For the vector product, the uncanceled matrix is $\gamma^a\mathcal A_\mu-\mathcal A_\mu\gamma^a=-[\mathcal A_\mu,\gamma^a]$. Use the gamma commutator without exchanging the two spinor fields.

Solution

The row derivative is

$$\nabla_\mu\chi=\partial_\mu\chi-\chi\mathcal A_\mu.$$

The terms $-\chi\mathcal A_\mu\psi$ and $+\chi\mathcal A_\mu\psi$ cancel in the scalar product. For the vector product,

$$\begin{aligned} (\nabla_\mu\chi)\gamma^a\psi +\chi\gamma^a\nabla_\mu\psi &=\partial_\mu V^a-\chi[\mathcal A_\mu,\gamma^a]\psi\\ &=\partial_\mu V^a+\Omega_\mu{}^a{}_b V^b. \end{aligned}$$

This is the required vector derivative. Spacetime differentiation and the connection are even operations, so this calculation also holds for odd spinor fields with the written row-column order. It neither exchanges those fields nor identifies one with a complex conjugate.

Specialize to the three-dimensional matrices

This section uses the conventions from Spinors in three dimensions:

$$\gamma^a=(\sigma_3,-\sigma_1,-\sigma_2),\qquad \gamma^a\gamma^b=\delta^{ab}\mathbf1+ i\epsilon^{abc}\gamma_c,\qquad \epsilon^{123}=1.$$

The symmetric term drops out of the antisymmetric connection sum. Thus

$$\mathcal A_\mu=\frac{i}{4}\Omega_{\mu ab}\epsilon^{abc}\gamma_c.$$

The target paper’s input-first connection is $\omega_{\mu a}{}^b=\Omega_\mu{}^b{}_a$. With Euclidean frame indices lowered this becomes $\omega_{\mu ab}=-\Omega_{\mu ab}$. Substitution gives

$$\nabla_\mu\psi= \left(\partial_\mu-\frac{i}{4} \omega_{\mu ab}\epsilon^{abc}\gamma_c\right)\psi,$$

in exact agreement with equation (A.11), using (A.8), of the paper’s version 3. Its gamma product is equation (A.4). The minus sign in the paper’s expression and the plus sign in the generic expression refer to opposite orderings of the frame indices.

For the cylindrical example, $\gamma^1\gamma^2=i\gamma^3$, so

$$\mathcal A_\theta=-\frac{i}{2}\gamma^3 =\frac12\begin{pmatrix}0&1\\-1&0\end{pmatrix}.$$

Choose the Cartesian spinor $\psi_0=(1,0)^T$. The radial spinor and its ordinary derivative are

$$\psi_{\rm rad}= \begin{pmatrix}\cos(\theta/2)\\\sin(\theta/2)\end{pmatrix}, \qquad \partial_\theta\psi_{\rm rad} =\frac12\begin{pmatrix}-\sin(\theta/2)\\\cos(\theta/2)\end{pmatrix}.$$

Multiplication by $\mathcal A_\theta$ gives the negative of the displayed derivative. Its norm is one, and its covariant derivative vanishes. This explicit two-component check verifies both the frame-index translation and the factor of one-half.

The same method now applies to a genuinely curved coframe. Compute the geometry of the round three-sphere to construct a global frame, solve its torsion-free connection, and convert that connection to spin matrices. Determining which curved backgrounds admit the spinors needed for supersymmetry is the next physical question.

For the global meaning of the radial spinor’s minus sign, continue with Spinors as bundle sections. It joins two angular patches and separates a frame-dependent component sign from the return map on the spinor fiber.

Paper guides

Connect this lesson to the original papers and their notation.

• Paper guide: Appendices A and B, conventions and geometry