It's always nice to point out the structural similarieties between (semi-)Riemannian geometry and gauge field theories alla Classical yang Mills theories. Nevertheless, I feel the relation between the gauge group and the gauge bosons "$A_{\mu}\times T^i_{\ j}$" are much simpler than the relation betreen the diffeomorphisms/isometries and the Christoffel symbols $\Gamma_{\mu\nu}^\rho$. At the very least, at first glance, I don't see the $SO(3)$ matrix lingering in $\Gamma$. I don't really know the technicalities of the Erlangen program and I specifically don't know how that "generating a manifold from a single point using the isometry group" (as I understand it) really works.

I'd like to know if it is really directly possible to view both, that is general relativity and the classical gauge theories, just as special cases of the general Cartan connection framework, or if there is a fundamental difficulty to (I guess) general relativty which makes this not work. Especially if the answer is no, then what is it that makes the metric theory so different? (At first I'd guess it's the fact that the GR manifold looks different in each patch due to $g$, but $A$ can also be difficult and both are related to what else is in space. So I suppose it's the interplay between the isometry group and the bigger diffeomorphism group.)

  • $\begingroup$ Related: physics.stackexchange.com/q/4359/2451 and physics.stackexchange.com/q/12461/2451 $\endgroup$ – Qmechanic Jul 9 '12 at 10:35
  • $\begingroup$ "generating a manifold from a single point using the isometry group": if the isometry group is transitive, you can move any point to an arbitrary point by applying a symmetry. E.g., in a Euclidean space you can move the origin anywhere by a translation. $\endgroup$ – Arnold Neumaier Jul 9 '12 at 13:13
  • $\begingroup$ @ArnoldNeumaier: Yeah, you can move one point to another if the space is already there, but the idea seems to be to generate the whole space from the point using the group. Seems strange to me. (Maybe it's by using the parametrization of the Lie-Group to map out the "space" and thereby giving meaning to it?) $\endgroup$ – Nikolaj-K Jul 9 '12 at 14:53
  • $\begingroup$ Apply the whole group to a single point, and you get the manifold. In modern terms, you can define any homogeneous space directly in terms of the group alone, by taking as points the coset of the point stabilizer. (What was revolutionary at Klein's time is today just a few lines of trivialites.) Indeed, this is the systematic way to generate the full classification of Riemannian symmetric spaces, say. $\endgroup$ – Arnold Neumaier Jul 9 '12 at 15:25

The major problem with this approach is that the gauge theory action is quadratic in the curvature, while the general relativity action is linear in the curvature. Further, the applications of gauge theory are all in the highly quantum limit, while GR is in the classical limit.

Still, you could cover gauge theory by starting with GR and doing a Kaluza Klein compactification on a symmetrical space. This gives rise to gauge theory in the low energy limit, but you need a stabilization mechanism for the internal dimensions. This was one of the paths by which gauge theory was understood to be natural, and I think it first appears (as an exercise) in unpublished lecture notes of deWitt in the 1960s.

The technical differences between an action quadratic in curvature and an action linear in curvature is big enough to make it natural to separate the two subjects in my opinion. It is also true that quantum gravity calculations are harder by orders of magnitude than gauge theory, although the two are unified somewhat in string theory.

  • $\begingroup$ What about the MacDowell–Mansouri action? This is quadratic in the curvature form, isn't it? $\endgroup$ – Marc Jun 21 '13 at 19:32

I too have worried about making that relationship more concrete. I wrote a paper 10 years ago that makes the parallel very explicit. Maybe it will help. See: http://uk.arxiv.org/abs/hep-th/0205250 published in JHEP.


You could certainly cover both GR and classical (Yang-Mills) gauge theory in a one semester course. Both theories are expressed in the language of modern differential geometry: manifolds, bundles, tensors & forms, metrics, connections, and curvature. Likewise, both theories have their dynamics specified by actions.

There are some crucial physical differences though. First, as Ron mentions, the Einstein-Hilbert action for GR is linear in the curvature. Second, there is not AFAIK, anything like the metric tensor (or the frame field, if you prefer) in Yang-Mills style gauge theory. In Yang-Mills, the connection is the basic variable. In GR, there's a connection but it's derived from a metric. Thirdly, gauge theories have local observables (e.g., the curvature), whereas in GR, treating diffeomorphisms as gauge symmetries basically forbids local observables.

The mathematical machinery is quite similar -- these are both classical field theories -- but the physical content is rather different.

  • $\begingroup$ You should compare Yang-Mills theory to Einstein-Cartan theory, where the basic object is the connection and not the metric. P.S.: What do you mean by GR has no local observables? Isn't something like the scalar curvature a nice local observable? $\endgroup$ – Marc Jun 21 '13 at 19:30
  • $\begingroup$ @Marc: I'm making the old 'diffeomorphisms are gauge transformations in GR, so there are no local observables' argument. The scalar curvature is certainly local, but it's not invariant under diffeomorphisms. $\endgroup$ – user1504 Jun 21 '13 at 20:02
  • $\begingroup$ I see... but when you count diffeomorphisms as gauge transformations in GR, why don't you count diffeomorphisms of the base or the whole principle bundle as gauge transformations in Yang-Mills theory? $\endgroup$ – Marc Jun 21 '13 at 20:08
  • $\begingroup$ @Marc In those cases, you've gauge-fixed by choosing a spacetime and a metric. This gets rids of the diffeomorphisms which approach the identity at infinity, and leaves behind (for example, on flat space) the Poincare group. $\endgroup$ – user1504 Jun 21 '13 at 21:05

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