Teleparallel Equivalent of General Relativity

The teleparallel equivalent of general relativity (TEGR) describes the same classical metric dynamics as general relativity using a flat, metric-compatible connection whose torsion is nonzero. The equivalence rests on an exact identity: the Levi-Civita Ricci scalar equals minus a particular torsion scalar plus a total divergence. The boundary term, inertial spin connection, and variational data are therefore part of the theorem, not optional details.

Required background. Frames, Tetrads, and Spinors supplies tetrads and Lorentz connections; Differential Forms and Integration supplies curvature and torsion forms; and Gravitational Actions and Boundary Terms supplies the boundary-data comparison required for action equivalence.

Helpful background. Einstein Equations gives the target field equations, while Formulations of General Relativity clarifies what classical equivalence does and does not assert.

Teleparallel data and theorem

Assume on an oriented four-dimensional spacetime region that $e^a{}_{\mu}$ is an invertible $C^2$ tetrad with $e=\det(e^a{}_{\mu})\gt0$ and $\omega^a{}_{b\mu}$ is a $C^1$ flat Lorentz spin connection. They define

$$g_{\mu\nu}=\eta_{ab}e^a{}_{\mu}e^b{}_{\nu}, \qquad \Gamma^\rho{}_{\mu\nu} =e_a{}^\rho\left( \partial_\mu e^a{}_{\nu} +\omega^a{}_{b\mu}e^b{}_{\nu} \right).$$

The connection is metric compatible, and flatness means $R^\rho{}_{\sigma\mu\nu}(\Gamma)=0$. Its torsion and trace are

$$T^\rho{}_{\mu\nu}=2\Gamma^\rho{}_{[\mu\nu]}, \qquad T_\mu=T^\nu{}_{\nu\mu}.$$

For the site signature $(-,+,+,+)$, define the torsion scalar as the complete package

$$\mathcal T =\frac14T^\rho{}_{\mu\nu}T_\rho{}^{\mu\nu} +\frac12T^\rho{}_{\mu\nu}T^{\nu\mu}{}_{\rho} -T_\rho T^\rho.$$

Then the exact curvature–torsion identity is

$$\boxed{ \mathring R=-\mathcal T+2\mathring\nabla_\mu T^\mu } =-\mathcal T+\frac{2}{e}\partial_\mu(eT^\mu), \qquad e=\det(e^a{}_{\mu})=\sqrt{-g}.$$

No field equation is assumed. The identity and the TEGR equations are derived in the opposite-signature source Bahamonde et al. 2023, §4.3.1, pp. 67–70, Eqs. (4.147)–(4.168), PDF; the torsion scalar, trace, spin connection, and action above have been translated together to the site convention.

Proof from flatness

Metric compatibility gives the unique decomposition

$$\Gamma^\rho{}_{\mu\nu} =\mathring\Gamma^\rho{}_{\mu\nu}+K^\rho{}_{\mu\nu},$$

with contorsion determined algebraically by $T^\rho{}_{\mu\nu}$. Insert this into the difference formula for two curvatures:

$$R^\rho{}_{\sigma\mu\nu}(\Gamma) =\mathring R^\rho{}_{\sigma\mu\nu} +2\mathring\nabla_{[\mu}K^\rho{}_{\nu]\sigma} +2K^\rho{}_{[\mu|\lambda|}K^\lambda{}_{\nu]\sigma}.$$

The left side vanishes by the teleparallel hypothesis. Contract $\rho$ with $\mu$, then contract the remaining indices with $g^{\sigma\nu}$. The two derivative terms combine into $-2\mathring\nabla_\mu T^\mu$. Substituting

$$K^\rho{}_{\mu\nu} =\frac12\left( T^\rho{}_{\mu\nu} +T_\mu{}^\rho{}_{\nu} +T_\nu{}^\rho{}_{\mu} \right)$$

in the quadratic terms gives precisely $\mathcal T$. Therefore

$$0=R(\Gamma)=\mathring R+\mathcal T-2\mathring\nabla_\mu T^\mu,$$

which proves the stated identity. Finally, $\mathring\nabla_\mu T^\mu=e^{-1}\partial_\mu(eT^\mu)$ turns the last term into a coordinate density divergence. This proof is local; a global teleparallel description additionally requires a global tetrad, or compatible patching of a flat Lorentz connection, so topology cannot be ignored.

Actions, boundaries, and the exact scope

With $\kappa=8\pi G$ and zero cosmological constant for brevity,

$$S_{\mathrm{EH}} =\frac{1}{2\kappa}\int_{\mathcal M}d^4x\,e\,\mathring R, \qquad S_{\mathrm{TEGR}} =-\frac{1}{2\kappa}\int_{\mathcal M}d^4x\,e\,\mathcal T.$$

The theorem gives

$$S_{\mathrm{EH}}-S_{\mathrm{TEGR}} =\frac{1}{\kappa} \int_{\mathcal M}d^4x\,\partial_\mu(eT^\mu).$$

This equality also gives a short field-equation transfer argument. Vary both sides within the stated class of tetrads and flat Lorentz connections. For compactly supported variations, the variation of the last integral vanishes, so the bulk tetrad equation from $S_{\mathrm{TEGR}}$ is the Einstein equation written in tetrad variables. The spin-connection variation carries no independent bulk dynamics in TEGR: subject to flatness, it changes the action through the same boundary structure. This conclusion would not follow merely from evaluating the two actions on one solution; it follows because their difference is a divergence throughout the allowed configuration space.

The Euler–Lagrange equations agree when variations have compact support, or when the two actions are supplemented and varied with boundary terms and boundary data that make this divergence harmless. On a manifold with boundary, simply deleting it without stating the variational problem is not a proof.

The inertial spin connection is also essential. Locally, flatness permits

$$\omega'=\Lambda\omega\Lambda^{-1}-(d\Lambda)\Lambda^{-1}, \qquad e'^a=\Lambda^a{}_b e^b.$$

The combined transformation leaves $\Gamma$ and $\mathcal T$ covariant. A gauge with $\omega=0$ exists locally in a proper frame, but imposing $\omega=0$ in every Lorentz-related tetrad mistakes a gauge choice for a tensor equation. The covariant construction and explicit Cartesian-versus-spherical tetrad check appear in Krššák and Saridakis 2016, §§II–IV, pp. 2–9, PDF.

Classical equivalence further assumes the same physical tetrad metric and compatible matter coupling. A direct nonminimal torsion coupling changes the theory. The identity alone establishes neither quantum equivalence nor equality of boundary charges under mismatched boundary prescriptions.

Flat-FLRW sign check

Take Cartesian flat FLRW data

$$e^a{}_{\mu}=\operatorname{diag}(1,a,a,a), \qquad \omega^a{}_{b\mu}=0, \qquad H=\frac{\dot a}{a}.$$

The only independent nonzero torsion components are $T^i{}_{0j}=H\delta^i_j$, and $T_0=-3H$. Direct contraction in the displayed definition gives

$$\mathcal T=6H^2.$$

Since $T^0=3H$ and $e=a^3$,

$$2\mathring\nabla_\mu T^\mu =\frac{2}{a^3}\frac{d}{dt}(3a^3H) =6\dot H+18H^2.$$

The identity therefore yields

$$-\mathcal T+2\mathring\nabla_\mu T^\mu =6(\dot H+2H^2)=\mathring R,$$

which independently fixes both the boundary sign and the site result $\mathcal T=6H^2$.

Why modified teleparallel gravity is different

Modified teleparallel gravity does not generally preserve the TEGR equivalence. TEGR is linear in $\mathcal T$. If $\mathcal T$ changes by a divergence under a pure-tetrad frame change, the linear action changes only at the boundary. For $f(\mathcal T)=\mathcal T+\alpha\mathcal T^2$, the extra term transforms as

$$(\mathcal T+\delta B)^2-\mathcal T^2 =2\mathcal T\,\delta B+(\delta B)^2,$$

which is not generally a total divergence. Keeping the inertial spin connection restores transparent local Lorentz covariance of the formulation, but it does not turn nonlinear $f(\mathcal T)$ into general relativity. This is the required counterexample to extending the theorem beyond TEGR.

Nor does the proof decide the number or stability of extra modes in a modified model. Those questions depend on constraints and background rank. Empirical bounds on model parameters belong to Tests of Gravity, and ultraviolet completion claims belong elsewhere; this page proves only the classical TEGR identity under the stated geometric and boundary hypotheses.

The converse also needs care. Equality of Einstein and TEGR bulk equations does not identify their canonical boundary generators term by term, nor does it say that every choice of tetrad and vanishing spin connection represents the same admissible configuration. Those questions require the matched boundary prescription and the flat inertial connection already included in the hypotheses.

Common pitfalls

A total divergence is not zero. It can affect boundary data, charges, and on-shell actions even when it leaves bulk field equations unchanged.

Covariance is not dynamical equivalence. A covariant tetrad–spin-connection formulation of $f(\mathcal T)$ is still a modified theory.

Exercises

Suppose $f(\mathcal T)=\mathcal T+\alpha\mathcal T^2$. Use the curvature–torsion identity to explain why no function $F(\mathring R)$ plus a single boundary divergence reproduces this action for arbitrary tetrads. Identify the exceptional value of $\alpha$.

Solution

Write $\mathcal T=-\mathring R+B$, where $B=2\mathring\nabla_\mu T^\mu$. The linear term is $-\mathring R+B$, so $B$ is removable under the stated boundary conditions. The nonlinear term is $\alpha(\mathring R^2-2\mathring R B+B^2)$. Its mixed and squared pieces depend on the tetrad-generated boundary scalar throughout the bulk and are not one overall divergence. Only $\alpha=0$ returns the TEGR action and the proved equivalence.

References

• Bahamonde, S., et al., “Teleparallel Gravity: From Theory to Cosmology,” Reports on Progress in Physics 86 (2023) 026901. DOI. Open PDF
• Krššák, M., and E. N. Saridakis, “The Covariant Formulation of $f(T)$ Gravity,” Classical and Quantum Gravity 33 (2016) 115009. DOI. Open PDF