Frames and spin connections

In general relativity, a metric tells us how to measure lengths, while a connection tells us how to compare vectors at neighboring points. A local orthonormal frame puts those measurements into a familiar Euclidean or Lorentzian basis. It also introduces a choice: that basis can rotate from point to point. The connection must account for its rotation.

We will calculate this effect in ordinary flat space written in cylindrical coordinates. The answer will be nonzero in a radial frame and zero in a Cartesian frame. Seeing both answers describe the same geometry is the useful checkpoint. The calculation uses basic GR and a short introduction to differential forms; it requires no earlier supersymmetry lesson.

A coframe is a local set of measuring rods

Begin with a fixed, positive-definite metric in three dimensions. Coordinate indices are $\mu,\nu=1,2,3$, and orthonormal-frame indices are $a,b=1,2,3$. An orthonormal coframe is a set of one-forms

$$e^a=e^a{}_\mu\,dx^\mu$$

whose components form an invertible matrix and reconstruct the metric:

$$g=\delta_{ab}\,e^a\otimes e^b, \qquad g_{\mu\nu}=\delta_{ab}e^a{}_\mu e^b{}_\nu.$$

A one-form maps a vector to a number. Thus $e^a(V)$ gives the $a$th orthonormal component of a vector $V$. The tensor product in the metric means that $g(V,W)=\delta_{ab}e^a(V)e^b(W)$. A coframe is often called a vielbein; “frame” is also frequently used for its components. We will distinguish it from its dual vector basis,

$$E_a=e_a{}^\mu\partial_\mu, \qquad e^a(E_b)=\delta^a{}_b, \qquad e^a{}_\mu e_b{}^\mu=\delta^a{}_b.$$

The inverse relation is $e_a{}^\mu e^a{}_\nu=\delta^\mu{}_\nu$. Coordinate and frame indices describe the same tangent space in different bases; they are not interchangeable labels. The metric reconstruction agrees with the three-manifold paper, equation (A.7).

Our example is the patch $\rho>0$ with

$$ds^2=d\rho^2+\rho^2d\theta^2+dz^2, \qquad e^1=d\rho,\quad e^2=\rho\,d\theta,\quad e^3=dz.$$

Here $\rho,z$ have dimensions of length, and $\theta$ is dimensionless. Squaring the coframe immediately gives the metric above. Its dual is

$$E_1=\partial_\rho,\qquad E_2=\frac1\rho\partial_\theta,\qquad E_3=\partial_z.$$

The factor $1/\rho$ matters: $\partial_\theta$ has squared length $\rho^2$, whereas $E_2$ has unit length. We work locally in the angular coordinate. The radial frame does not extend through the axis $\rho=0$, where the polar coordinate system fails.

What the connection measures

Let $\nabla_\mu$ be the Levi-Civita connection of the metric. Define connection one-forms $\Omega^a{}_b=\Omega_\mu{}^a{}_b\,dx^\mu$ by

$$\nabla_\mu E_b=\Omega_\mu{}^a{}_b E_a, \qquad \Omega_\mu{}^a{}_b =e^a{}_\nu \left(\partial_\mu e_b{}^\nu+ \Gamma^\nu{}_{\mu\lambda}e_b{}^\lambda\right).$$

The last expression connects the new notation to the Christoffel symbols you already know. The first frame index labels the output direction; the second labels the vector being differentiated.

Write a vector as $V=V^aE_a$. The product rule gives

$$\nabla_\mu V= \left(\partial_\mu V^a+\Omega_\mu{}^a{}_bV^b\right)E_a.$$

Thus the ordinary variation of its components and the variation of its basis occur together. The matrix-valued one-form $\Omega$ is the orthonormal-frame connection. It is commonly called the spin connection because the same local rotation connection can act on spinor components through the spin representation. We will first determine it using vectors.

Metric compatibility says that orthonormal vectors remain orthonormal under parallel comparison. Differentiating $g(E_a,E_b)=\delta_{ab}$ gives

$$\Omega_{ab}=-\Omega_{ba}, \qquad \Omega_{ab}:=\delta_{ac}\Omega^c{}_b.$$

Antisymmetry is a constraint on the two frame indices, with the one-form index left untouched.

Cartan’s equation turns the calculation into exterior algebra

The wedge product is the antisymmetric product of forms:

$$dx^\mu\wedge dx^\nu=-dx^\nu\wedge dx^\mu.$$

In particular, a one-form wedged with itself vanishes. The exterior derivative of $A=A_\mu dx^\mu$ is

$$dA=\partial_\mu A_\nu\,dx^\mu\wedge dx^\nu =\frac12(\partial_\mu A_\nu-\partial_\nu A_\mu) dx^\mu\wedge dx^\nu.$$

This formula is enough to do the example; no integration theorem is needed.

The relation between the coordinate and frame connections can also be written as the tetrad postulate:

$$\partial_\mu e^a{}_\nu -\Gamma^\lambda{}_{\mu\nu}e^a{}_\lambda +\Omega_\mu{}^a{}_b e^b{}_\nu=0.$$

“Tetrad” is the four-dimensional name, but this relation holds in any dimension. To see why it follows, covariantly differentiate the identity $e^a(E_b)=\delta^a{}_b$ and use the definition of $\Omega$. Antisymmetrizing in $\mu,\nu$ removes the Christoffel term because the Levi-Civita connection has zero torsion. The result is Cartan’s torsion-free equation:

$$de^a+\Omega^a{}_b\wedge e^b=0.$$

Use this equation together with $\Omega_{ab}=-\Omega_{ba}$. Torsion freedom alone does not impose metric compatibility. Together the two conditions determine the unique Levi-Civita connection for an invertible coframe.

For the cylindrical coframe,

$$de^1=0,\qquad de^2=d\rho\wedge d\theta=\frac1\rho e^1\wedge e^2, \qquad de^3=0.$$

The second equation is canceled by $\Omega^2{}_1=d\theta$. Antisymmetry then requires $\Omega^1{}_2=-d\theta$. Set the components involving the third direction to zero and substitute into all three Cartan equations:

$$\begin{aligned} de^1+\Omega^1{}_2\wedge e^2 &=-d\theta\wedge\rho\,d\theta=0,\\ de^2+\Omega^2{}_1\wedge e^1 &=d\rho\wedge d\theta+d\theta\wedge d\rho=0,\\ de^3&=0. \end{aligned}$$

We have exhibited a metric-compatible, torsion-free solution, so uniqueness identifies it as the required connection:

$$\Omega= \begin{pmatrix} 0&-d\theta&0\\ d\theta&0&0\\ 0&0&0 \end{pmatrix}.$$

The same entry is $\Omega^2{}_1=\rho^{-1}e^2$. Its coefficient is dimensionless in the $d\theta$ basis and has inverse-length dimension in the orthonormal basis. Dimensional checks must say which basis is being used.

Check the answer in coordinates

An independent route begins with the coordinate metric, without using Cartan’s equation. The usual formula

$$\Gamma^\lambda{}_{\mu\nu} =\frac12g^{\lambda\sigma} \left(\partial_\mu g_{\sigma\nu} +\partial_\nu g_{\sigma\mu} -\partial_\sigma g_{\mu\nu}\right)$$

gives the only nonzero coefficients

$$\Gamma^\rho{}_{\theta\theta}=-\rho, \qquad \Gamma^\theta{}_{\rho\theta} =\Gamma^\theta{}_{\theta\rho}=\frac1\rho.$$

Apply them to the dual frame:

$$\begin{aligned} \nabla_\theta E_1 &=\frac1\rho\partial_\theta=E_2,\\ \nabla_\theta E_2 &=\frac1\rho(-\rho)\partial_\rho=-E_1,\\ \nabla_\rho E_2 &=-\frac1{\rho^2}\partial_\theta +\frac1\rho\frac1\rho\partial_\theta=0. \end{aligned}$$

All other derivatives of these frame vectors vanish. Therefore $\Omega_\theta{}^2{}_1=1$ and $\Omega_\theta{}^1{}_2=-1$, exactly as above. The cancellation in $\nabla_\rho E_2$ shows why differentiating the inverse vielbein itself cannot be omitted.

Rotate the frame, keep the geometry

Set $x=\rho\cos\theta$ and $y=\rho\sin\theta$. The Cartesian coframe is

$$\begin{pmatrix}dx\\dy\\dz\end{pmatrix} =R(\theta)\begin{pmatrix}e^1\\e^2\\e^3\end{pmatrix}, \qquad R(\alpha)= \begin{pmatrix} \cos\alpha&-\sin\alpha&0\\ \sin\alpha&\cos\alpha&0\\ 0&0&1 \end{pmatrix}.$$

Since $R^TR=1$, the rotated coframe reconstructs the same metric. This is a change of orthonormal frame; we can still express every one-form in the original coordinates $(\rho,\theta,z)$.

For a general position-dependent rotation $e'=Re$, vector components obey $V'=RV$. Write $DV=dV+\Omega V$ for the column of covariant component differentials, and $D'V'=dV'+\Omega'V'$ in the new frame. Requiring $D'V'=R\,DV$ gives

$$d(RV)+\Omega'RV =R\,dV+(dR+\Omega'R)V.$$

Equating the remaining terms for every $V$ yields

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

The second term is essential. A connection does not transform as an ordinary tensor under a varying frame rotation.

Let

$$J= \begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix}.$$

Our cylindrical answer is $\Omega=J\,d\theta$. For $R=R(\theta)$, direct differentiation gives $dR\,R^{-1}=J\,d\theta$, while $RJR^{-1}=J$. Consequently $\Omega'=0$. The rotated coframe is $(dx,dy,dz)$, precisely where a zero connection is expected.

This is flat space with a rotating radial basis. Nonzero connection coefficients by themselves do not demonstrate curvature. Conversely, one cannot generally remove a connection throughout a curved neighborhood by a frame rotation. The example establishes flatness here through the explicit Cartesian coordinates and coframe.

Translate the paper’s index order before using it

The three-manifold paper defines its connection in Appendix A.2, equation (A.8) as

$$\omega_{\mu a}{}^b=e^b{}_\nu\nabla_\mu e_a{}^\nu.$$

The input frame index comes first in this notation. Comparing definitions gives the dictionary

$$\omega_{\mu a}{}^b=\Omega_\mu{}^b{}_a, \qquad \omega_{\mu ab}=\Omega_{\mu ba}=-\Omega_{\mu ab}.$$

The last equality uses the Euclidean frame metric and antisymmetry. Thus, in our example, the paper’s nonzero entries are $\omega_{\theta1}{}^2=1$ and $\omega_{\theta2}{}^1=-1$. There is no disagreement between the two answers: the displayed order of the frame indices changed. Keep this dictionary when inserting the connection into the paper’s spinor derivative.

The same construction in four dimensions

Nothing in the frame definition or Cartan equation required three-dimensional gamma matrices. For a four-dimensional Euclidean example, add $dw^2$ to the cylindrical metric and choose $e^4=dw$. The connection becomes the same three-by-three block with an additional zero row and column.

For Lorentzian geometry, use the constant orthonormal metric $\eta_{ab}$ in place of $\delta_{ab}$. For example, with signature $(-+++)$,

$$ds^2=-dt^2+d\rho^2+\rho^2d\theta^2+dz^2, \qquad e^0=dt,\ e^1=d\rho,\ e^2=\rho\,d\theta,\ e^3=dz.$$

The same spatial connection solves Cartan’s equation, and $\Omega^0{}_a=\Omega^a{}_0=0$. Metric compatibility means antisymmetry after lowering with $\eta$: $\Omega_{ab}=\eta_{ac}\Omega^c{}_b=-\Omega_{ba}$. Frame changes now preserve $\eta$ rather than a Euclidean identity matrix. This supplies the geometric entrance to four-dimensional spinor derivatives independently of the three-dimensional spinor lesson.

Practice

1. Change the circumference function

On an interval where a smooth function $f(\rho)$ is positive, take

$$ds^2=d\rho^2+f(\rho)^2d\theta^2+dz^2, \qquad e^1=d\rho,\quad e^2=f(\rho)d\theta,\quad e^3=dz.$$

Find the connection and its nonzero Christoffel symbols. Check the cases $f(\rho)=\rho$ and $f(\rho)=L$ with constant $L>0$.

Hint

Compute $de^2$ before guessing a connection. Its coefficient in the orthonormal basis is $f'/f$, where a prime means $d/d\rho$.

Solution

Metric reconstruction follows by squaring the stated coframe. Its derivative is

$$de^2=f'\,d\rho\wedge d\theta =\frac{f'}f e^1\wedge e^2.$$

The antisymmetric solution is

$$\Omega^2{}_1=f'\,d\theta=\frac{f'}f e^2, \qquad \Omega^1{}_2=-f'\,d\theta,$$

with every other entry zero. In the second Cartan equation, $f'd\rho\wedge d\theta$ cancels $f'd\theta\wedge d\rho$. In the first, the connection term wedges $d\theta$ with itself.

Computing independently from $g_{\theta\theta}=f^2$ gives

$$\Gamma^\rho{}_{\theta\theta}=-ff', \qquad \Gamma^\theta{}_{\rho\theta} =\Gamma^\theta{}_{\theta\rho}=\frac{f'}f.$$

For example, $\nabla_\theta E_1=(f'/f)\partial_\theta=f'E_2$, confirming the connection coefficient. With $f=\rho$ these are the cylindrical results. With $f=L$, both the Christoffel symbols and this frame connection vanish: locally $y=L\theta$ turns the metric into $d\rho^2+dy^2+dz^2$. If $\theta$ is periodic, the global identification of $y$ remains. A local calculation does not erase that identification.

2. Rotate twice as fast

Return to the cylindrical metric and replace the coframe by $e'=R(2\theta)e$. Find $\Omega'$ and explain why its sign can differ from $\Omega$ while the metric stays the same.

Hint

Use $R(\alpha)JR(\alpha)^{-1}=J$ and differentiate the angle $2\theta$. Do not drop the derivative term in the connection transformation.

Solution

The derivative is $dR(2\theta)R(2\theta)^{-1}=2Jd\theta$. Hence

$$\Omega'=Jd\theta-2Jd\theta=-Jd\theta.$$

The metric remains unchanged because the rotation is orthogonal. A direct check of torsion is possible without expanding every trigonometric coefficient:

$$\begin{aligned} de'+\Omega'\wedge e' &=dR\wedge e+R\,de +(R\Omega R^{-1}-dR R^{-1})\wedge Re\\ &=R(de+\Omega\wedge e)=0. \end{aligned}$$

This manipulation keeps the order of the matrices fixed. Antisymmetry also holds. The sign change describes how fast the chosen basis rotates; it does not change the metric or create curvature.

3. Constant components or a parallel vector?

The fixed Cartesian vector $V=\partial_x$ has cylindrical orthonormal components

$$V^a=(\cos\theta,-\sin\theta,0).$$

Check that $DV^a=dV^a+\Omega^a{}_bV^b=0$. Compare it with the vector $W=E_1$, whose orthonormal components are constant, and explain what “constant” fails to specify.

Hint

First multiply the two-component column by $J$. The differential of the components and the connection term must both be included.

Solution

For $V$,

$$dV= \begin{pmatrix}-\sin\theta\\-\cos\theta\\0\end{pmatrix}d\theta, \qquad \Omega V= \begin{pmatrix}\sin\theta\\\cos\theta\\0\end{pmatrix}d\theta.$$

They cancel. In the coordinate basis this vector is

$$V=\cos\theta\,\partial_\rho -\frac{\sin\theta}{\rho}\partial_\theta=\partial_x,$$

which independently explains why it is parallel in Euclidean space.

By contrast, $W^a=(1,0,0)$ has $dW^a=0$ but

$$DW^a=(0,d\theta,0).$$

The outward radial unit vector changes direction as $\theta$ changes. Constant components in a rotating frame need not describe a parallel vector; changing components can compensate exactly for the rotating basis. Both statements concern the patch $\rho>0$, where the chosen frame exists.

You can now reconstruct a metric, calculate its connection, and test the result in a second basis. Covariant derivatives of spinors asks how the same connection acts when a field’s components transform as a spinor rather than a vector. Before proceeding, check that you can recover the sign of the cylindrical connection from Cartan’s equation and translate it into $\omega_{\mu a}{}^b$ without memorizing either matrix.