Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Currently, (classical) gravity (General Relativity) is NOT a gauge theory (at least in the sense of a Yang-Mills theory).

Why should "classical" gravity be some (non-trivial or "special" or extended) gauge theory? Should quantum gravity be a gauge theory?

Remark: There are some contradictory claims in the literature to this issue. To what extent gravity is "a gauge" theory?Obviously, GR is not a YM theory. So, why do some people say that gravity "IS" a gauge theory? I found this question relevant, e.g., then we handle GR in the Einstein-Cartan theory or any other theory beyond GR, like teleparallel theories or higher-derivative gravitational theories. So I think it could be useful to discuss the "gauge flavor" of gravity here.

share|improve this question
Why do you say GR is not a gauge theory? I've seen claims both that it is and that it isn't, but the distinction seems somewhat technical to a non-specialist like me. –  John Rennie Jul 18 '13 at 10:37
It depends how rigidly you define gauge theory. Typically gauge theories have compact Lie groups. Gravitational gauge theory, such as it is, is based on the non-compact diffeomorphism group (note I'm really talking about the individual fibres of whatever bundle it is). There is also no analogy to the metric/tetrad in standard gauge theories. Those are the basic differences. It all gets very complicated from there. –  Michael Brown Jul 18 '13 at 10:43
It probably is the KISS principle followed by unification theorists. If unification means all forces become one at very high energies it seems simpler to assume that gravity will wear the same costume as the other three forces. –  anna v Jul 18 '13 at 10:57
As Michael Brown said, the definition of a gauge theory is a key point. I think "gauge theory = a Yang Mills theory" is too restrictive. People also define gauge theories in Hamiltonian terms - phase space/primary constraints/gauge orbits. Not necessary to have the Yang Mills apparatus there. I think some of the arguments in the literature/blogs could be avoided if people gave up front the definition of gauge theory which they were using. –  twistor59 Jul 18 '13 at 12:37
show 4 more comments

3 Answers 3

Gravity isn't Yang-Mills theory in the narrow sense – well, except for equivalences such as AdS/CFT or Matrix theory that imply that a quantum gravitational theory is fully equivalent to a gauge theory living in a different space (e.g. in AdS/CFT, on the boundary of the AdS space).

However, gravity is a gauge theory in the broader sense because it's conveniently formulated using the diffeomorphism symmetry group. The diffeomorphisms identify physical configurations that are physically equivalent, just like in the Yang-Mills case, so although they are not of the Yang-Mills type, they have to be treated just like Yang-Mills symmetries in Yang-Mills theories.

share|improve this answer
Hi, Lubos. Yes, I do knot those known facts. Moreover, I also know that, in the case of gravity, a Lie-group like formalism can be built, but it seems "a formidable mathematical" challenge, since, instead of Lie "structure constants", we are left with Lie "structure functions". I have read and studied this issue, after all, many people have argued and speculated that the issue to "quantize" gravity could be related to the way in which gravity is an extended "gauge gravity". You have also recalled that link, due to the gauge/gravity duality! –  riemannium Jul 18 '13 at 16:02
Indeed, I posted a related question to F.Wilczek twitter account and he (gently) answered me that he believed that GR could be some "kind" of non-linear sigma model. Of course, I have never read a paper about it, but it seems "plausible" (specially due to the string theory connection). Lubos, do you think that Quantum Gravity or gravity itself can be some "deformed" type of gauge theory? –  riemannium Jul 18 '13 at 16:05
Dear Riemannium, the diffeomorphism group is an infinite-dimensional Lie group with its infinite-dimensional Lie algebra that isn't quite the Yang-Mills gauge group but isn't too different. There are various relationships between nonlinear sigma models and gravity but I don't know in what sense they are "the same thing". Not in any sense I am aware of. Gravity is surely a "deformed gauge theory" to the extent that the character of the deformation remains sufficiently vague. ;-) –  Luboš Motl Jul 20 '13 at 15:19
add comment

A theory is usually denominated a 'gauge theory' if all the interactions in that theory are introduced by promoting global symmetries to gauge symmetries. Note that a gauge theory is a gauge invariant theory, but a gauge invariant theory doesn't has to be a gauge theory (for example, the Standard Model is gauge invariant, but it's not a gauge theory since the scalar self-interaction doesn't enlarge the gauge symmetry of the model). Yang-Mills theory is an example of gauge theory, but not all gauge theories are Yang-Mills'. General Relativity is a gauge theory in three different senses, namely:

  1. Invariance under diffeomorphims. Diffemorphism may be seen as a local (gauged) version of translations $\delta x^{\mu}\rightarrow a^{\mu}(x)$. In order for the theory to be diff. invariant, a covariant derivative $\nabla$ must replace partial derivatives $\partial$ (a general, dynamic metric $g$ tensor must replace Minkowski metric $\eta$ as well). Here, the most similar field to Yang-Mills' is the Levi-Civita connection $\Gamma$ (note that in Palatini's formulation this field is independent of the metric), which transforms as a tensor plus a term that involves the derivative of $a(x)$, similar to the transformation of a non-abelian field.

  2. Invariance under infinitesimal diff. One can split $g$ in a fixed background and a dynamical perturbation $h$, and the action of an infinitesimal diff. on the perturbation turns out to be $\delta h_{\mu\nu}=\partial_{\mu}a_{\nu}+\partial_{\nu}a_{\mu}$, which is also a gauge symmetry. This is the gauge symmetry connected with the masslessnes of gravitons (much like $SU_c(3)$ is related to the masslessnes of gluons and $U_{em}(1)$ to the masslenes of photons). Here, the most similar field to Yang-Mills' is $h$, which transforms similarly to the electromagnetic potential, even though $h$ is not a connection in any sense I am aware of.

  3. Invariance under local Lorentz transformations. It turns out that in order for couple spinors to be coupled to the gravitational field, it is convenient to introduced the tetrad formulation. In this approach, there is a gauge symmetry related to the freedom that one has to choose different basis in different space-time points. One has to introduce a covariant derivative (different from the first one in this answer) that allows us to change basis. This formulation is the closest to Yang-Mills theory (well, Ashtekar variables are probably closer). The main difference is that in GR, besides a dynamical connection (equivalent to the gauge field in Yang-Mills), there is a tetrad field (due to the fact that the metric is a dynamical field in gravity) that doesn't have a counterpart in Yang-Mills. Here, the closest field to Yang-Mills' is the before-mentioned spin connection, which transforms as a tensor plus a term that involves the derivative of the local Lorentz transformation, very similarly to a Yang-Mills field.

share|improve this answer
Wait, with the term "diffeomorphism invariance", doesn't it refer to the symmetry $g_{\mu\nu}\to g_{\mu\nu}-\nabla_{\mu}\xi_{\nu}-\nabla_{\nu}\xi_{\mu}$, i.e., the Lie Derivative along $\xi$ of the metric vanishes? (See section 3.3 of these MIT notes). –  Alex Nelson Jul 19 '13 at 0:37
Hello @AlexNelson . Yes, in the first case, the metric tensor transform as you write. In the second — linearized diff — $h$ transforms with the covariant derivatives replaced by partial derivatives. The metric $g$ is invariant only in the case of a isometry. –  drake Jul 19 '13 at 2:34
Drake, the tetrad field can be viewed as a part of the Yang-Mills connection for Poincare or (anti)-de Sitter group. In this case, tetrad and spin-connection are different parts of a single connection. Then gravity looks almost like YM-type gauge theory. There are important differences however. –  John Jul 19 '13 at 19:59
Hello @John Your comment seems very interesting to me. Can you expand it? Some link or reference? Name of the formulation or key word to search? –  drake Jul 19 '13 at 20:06
You can look in Blagojevic, "gravitation and gauge symmetries", which is freely downloadable. The matter is quite old. Take connection of the Poincare group $A=\frac12\omega^{a,b}L_{ab}+e^a P_a$, $L_{ab}$ and $P_a$ are Lorentz and translation generators. Then spin-connection and tetrad are the gauge fields corresponding to these generators. Compute the Yang-Mills field strength $F=dA+A^2=R^{a,b}L_{ab}+T^a P_a$. Then $T^a$ is a torsion two-form and $R^{a,b}$ is a Riemann two-form. Einstein-Hilbert action reads $\int R^{a,b} e^c...e^u \epsilon_{abc...u}$ –  John Jul 19 '13 at 21:33
show 1 more comment

Gravity can be seen as a gauge theory of the Lorentz group (which acts on the tangent space). These was pointed out by Kibble and Sciama during the 50s and 60s.

As John said before, it's better seen in terms of differential forms.

Another reference you might find interesting is the Lecture notes on Chern-Simons gravity by Jorge Zanelli (available in arXiv).

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.