Einstein Gravity in Tetrad Variables

A tetrad replaces the metric by an orthonormal coframe and exposes a local Lorentz gauge symmetry. This is more than a change of notation: spinors couple naturally to the coframe and spin connection, while orientation, invertibility, and boundary data become independent assumptions that must be stated before the tetrad and metric descriptions can be called equivalent.

Required background. Frames, Coframes, and Anholonomy supplies coframes and local Lorentz freedom, The Gibbons-Hawking-York Boundary Term supplies the metric variational problem, and Presymplectic Potentials and Currents supplies the action-to-phase-space boundary decomposition.

Helpful background. The Einstein Equation supplies the metric field equation, Hypersurface Normals and Tangential Projectors supplies boundary and slicing geometry, and Formulations of General Relativity places tetrads among equivalent classical formulations.

Oriented coframes and the metric

Let $e^I=e^I{}_{\mu}dx^\mu$ be a smooth, invertible coframe on an oriented, time-oriented four-dimensional spacetime. Internal indices are raised with

$$\eta_{IJ}=\operatorname{diag}(-1,1,1,1),$$

and the spacetime metric is

$$g_{\mu\nu}=\eta_{IJ}e^I{}_{\mu}e^J{}_{\nu}.$$

Fix the internal symbol by $\epsilon_{0123}=+1$ and select coframes whose orientation agrees with the spacetime orientation. The volume form is then

$$\boldsymbol\epsilon =\frac1{4!}\epsilon_{IJKL} e^I\wedge e^J\wedge e^K\wedge e^L.$$

A proper, orthochronous local Lorentz transformation acts as

$$e^I\longmapsto e'^I=\Lambda^I{}_J(x)e^J, \qquad \Lambda^T\eta\Lambda=\eta,$$

and leaves both $g_{\mu\nu}$ and the chosen orientation unchanged. Thus the six Lorentz components of the sixteen coframe components are gauge. Improper or time-reversing transformations are not silently included in this gauge group, because they can change the selected orientation or time orientation.

Curvature in an orthonormal frame

A Lorentz connection satisfies $\omega^{IJ}=-\omega^{JI}$ and has curvature

$$F^{IJ}=d\omega^{IJ}+\omega^I{}_K\wedge\omega^{KJ}.$$

On this page $\omega=\omega(e)$ is the unique torsion-free connection compatible with an invertible coframe. Its curvature encodes the Riemann tensor through

$$F^{IJ}=\frac12R^{IJ}{}_{KL}\,e^K\wedge e^L.$$

The local Lorentz transformation of the connection is inhomogeneous,

$$\omega' =\Lambda\omega\Lambda^{-1}-d\Lambda\,\Lambda^{-1},$$

so that $F'=\Lambda F\Lambda^{-1}$. A position-dependent rotation of a coframe therefore cannot be made while leaving the spin connection fixed.

The tetrad Einstein action

Write $\kappa=8\pi G$. With the stated orientation and curvature convention, the second-order tetrad action is

$$S[e] =\frac1{4\kappa}\int_{\mathcal M} \epsilon_{IJKL} e^I\wedge e^J\wedge \left(F^{KL}[\omega(e)] -\frac{\Lambda}{6}e^K\wedge e^L\right) +S_{\partial\mathcal M}.$$

For an invertible coframe, its bulk integrand equals $\sqrt{-g}({}^{(4)}R-2\Lambda)d^4x/(2\kappa)$. The surface functional $S_{\partial\mathcal M}$ is fixed by the boundary data; replacing the metric by a tetrad does not remove the need for a well-posed variational principle.

Varying the coframe gives the Einstein three-form equation

$$\epsilon_{IJKL}e^J\wedge \left(F^{KL}-\frac{\Lambda}{3}e^K\wedge e^L\right)=0.$$

Multiplication by the inverse tetrad converts this equation into $G_{\mu\nu}+\Lambda g_{\mu\nu}=0$. Six combinations are Lorentz Noether identities, leaving the ten independent metric equations. The action and its first-order extension are developed with explicit boundary variations in Corichi, Rubalcava-García, and Vukašinac 2016, §IV.A, pp. 22–24, PDF.

The equivalence just used has a domain: $e^I{}_{\mu}$ must be invertible and orientation-compatible, the connection must be the torsion-free compatible one, and the boundary and matter couplings must be translated consistently. Degenerate coframes do not define an inverse metric and can introduce additional solution branches in a first-order theory.

Flat-coframe check

For Cartesian Minkowski space,

$$e^0=dt, \qquad e^1=dx, \qquad e^2=dy, \qquad e^3=dz,$$

the compatible connection and curvature vanish. With $\Lambda=0$, the tetrad field equation is therefore satisfied. A local Lorentz transformation can make $de'^I$ nonzero without changing the metric; the transformed connection then cancels that anholonomy and keeps the curvature zero.

Common pitfalls

Treating the tetrad as unique. Metrics determine Lorentz-gauge orbits of coframes, not a preferred coframe.

Ignoring orientation and degeneracy. The epsilon action distinguishes an orientation, and its metric equivalence uses the inverse tetrad at a decisive step.

Exercise: a local boost of the Minkowski coframe

Let $\Lambda(x)$ be a boost of rapidity $\chi(x)$ in the internal $0$–$1$ plane. Starting from the Cartesian coframe and $\omega=0$, find the connection after the boost and verify that torsion and curvature remain zero.

Solution

The transformed fields are

$$e'=\Lambda e, \qquad \omega'=-d\Lambda\,\Lambda^{-1}.$$

Using $d e=0$,

$$\begin{aligned} T'&=de'+\omega'\wedge e'\\ &=d\Lambda\wedge e -d\Lambda\,\Lambda^{-1}\wedge\Lambda e=0. \end{aligned}$$

The connection is pure gauge, so the Maurer–Cartan identity gives

$$F'=d\omega'+\omega'\wedge\omega'=0.$$

For the boost generator $J_{01}$ one may write $\omega'=-(d\chi)J_{01}$. The exercise shows why setting the connection to zero in every tetrad gauge would incorrectly create torsion.

References

• Corichi, Alejandro, Irais Rubalcava-García, and Tatjana Vukašinac. “Actions, Topological Terms and Boundaries in First Order Gravity: A Review.” International Journal of Modern Physics D 25, no. 4 (2016): 1630011. DOI. Open PDF.