# To which extent is general relativity a gauge theory?

In quantum mechanics, we know that a change of frame -- a gauge transform -- leaves the probability of an outcome measurement invariant (well, the square modulus of the wave-function, i.e. the probability), because it is just a multiplication by a phase term.

I was wondering about general relativity. Is there something left invariant by a change of frame? (of course, energy, momentum, ... are invariants of Lorentz transform, but these are special relativity examples. I guess there is something else, more intrinsic and related to the mathematical structure of the theory, like a space-time interval or something).

I've tried looking at the Landau book on field theory, but it is too dense for me to have a quick answer to this question. I have bad understanding about GR -- I apologize for that. I'm trying to understand to which respect one calls the GR theory a gauge theory: for me a gauge transform leaves something invariant.

Best regards.

EDIT: Thanks to the first answers, I think I should refine my question and first ask this: To which extent is general relativity a gauge theory? If you have good references to this topic, that would be great(The Wikipedia pages are obscure for me for the moment). Thanks in advance. Best regards.

• Parallel transport involves moving around in a manifold, while a change of frame is, well just that. The title of the question is a bit misleading. – user10851 Dec 9 '12 at 2:13
• Thanks Chris White, that's partly my problem: I see the parallel transport as a change of frame with one extra condition (the moving frame remains as much as possible parallel to the original one). My second problem was that in QM, the invariant quantity is an extra structure (the wave function modulus, whereas the wave function is displaced / covariant). This you partly answered with Namehere: only scalar are invariant, tensor are covariant. I was asking if it exists an extra structure kept invariant when changing frame of reference. Thanks to both of you. – FraSchelle Dec 9 '12 at 9:40
• @Oaoa Tensors are not covariant, they are invariant. It is their components that are covariant. And by the way there are many scalars that can be constructed from tensors, such as norms of vectors, traces of matrices and others. Oh, and was there anything else you think I missed or could improve in my answer(I didn't edit my question to include Chris White's comment since I felt it would be like stealing)? – resgh Dec 9 '12 at 9:59
• I faintly remember that there is a nice way to think about GR as a gauge theory (or gauge theories as geometry), and it had to do with viewing the Levi-Civita connection as a gauge field. Unfortunately I don't know enough GR to write down the argument. – jdm Dec 9 '12 at 12:10
• Related discussion here physics.stackexchange.com/q/12461 – twistor59 Dec 9 '12 at 15:30

Consider a gauge theory with gauge group $GL(n,R)$.

First of all, let me remind you the basics of gauge transformations:

Let $G$ be a gauge group, $g(x)∈ G$ be an element of $G$. Then:

$\psi (x)→g(x)\psi (x)$

$A_\alpha→g(x)A_\alpha g^{-1}(x)-\frac{∂g(x)}{∂x^{\alpha}}g^{-1}(x)$

is a gauge transformation, and covariant derivative defined as

${\nabla}_{\alpha}\psi = \frac{∂\psi}{∂x^{\alpha}}+A_\alpha \psi$

Now consider coordinates $(x^1, ..., x^n)$ in region $U$. They define basis of vectors space $\frac{∂}{∂x^1},...,\frac{∂}{∂x^n}$. So, tangent vector fields in region $U$ can be considered as vector-valued functions: $\xi = (\xi^1,...,\xi^n)$. Change of coordinates in $U$: $x^{\nu}→x^{{\nu^{\prime}}}=x^{{\nu^{\prime}}}(x)$ defines local transformation:

$\xi^\nu→\xi^{\nu^\prime} = \frac{∂x^{\nu^\prime}}{∂x^\nu}\xi^\nu = g(x)\xi$.

Here matrix $g(x) = (\frac{∂x^{\nu^\prime}}{∂x^\nu})$ belongs to $GL(n,R)$, and inverse matrix has the form $g^{-1}(x)=(\frac{∂x^\nu}{∂x^{\nu^\prime}})$.

Lie algebra of $GL(n,R)$ is formed by all matrices of degree $n$, so the "gauge field" $A_\mu (x)$ is also matrix of degree $n$. Let us denote it's elements as follows:

$(A_\mu)^{\nu}_{\lambda}=\Gamma^{\nu}_{\lambda \mu}$.

Covariant derivative of the vector $\xi$ reads as follows:

$(\nabla_{\mu}\xi)^\nu=\frac{∂\xi^\nu}{∂x^\mu}+\Gamma^{\nu}_{\lambda \mu}\xi^\lambda ↔ \nabla_\mu \xi=\frac{∂\xi}{∂x^\mu}+A_\mu \xi$ (right side is in matrix form!)

There is only one thing left to check, namely the form of gauge field transformation.

Using general rule of transforming gauge field we obtain:

$\Gamma^{\nu}_{\lambda \mu}→\Gamma^{\nu^\prime}_{\lambda^\prime \mu}=\frac{∂x^{\nu^\prime}}{∂x^\nu}\Gamma^{\nu}_{\lambda \mu}\frac{∂x^\lambda}{∂x^{\lambda^\prime}}+\frac{∂x^{\nu^\prime}}{∂x^\nu}\frac{∂}{∂x^\mu}(\frac{∂x^\nu}{∂x^{\lambda^\prime}})$.

Since $A_\mu$ is a covariant vector, then $A_{\mu^\prime}=\frac{∂x^\mu}{∂x^{\mu^\prime}}A_\mu$. Hence we obtain:

$\Gamma^{\nu^\prime}_{\lambda^\prime \mu^\prime}=\frac{∂x^\mu}{∂x^{\mu^\prime}}\frac{∂x^{\nu^\prime}}{∂x^\nu}\Gamma^{\nu}_{\lambda \mu}\frac{∂x^\lambda}{∂x^{\lambda^\prime}}+\frac{∂x^{\nu^\prime}}{∂x^\nu}\frac{∂^2 x^\nu}{∂x^{\lambda^\prime}∂x^{\mu^\prime}}$.

Q.E.D.

And final remark: the commutator of two covariant derivatives leads to expression of the Riemann tensor:

$(F_{\mu\nu})^\rho_\lambda = R^\rho_{\lambda ,\mu\nu}$

EDIT:

Dear Oaoa,

I’m not a GR specialist, so what I have written below might be wrong.

My first advise is as follows: do not read Landau who is mixing up two fundamental concepts: connection and metric.

In order to answer your question let us first separate the roles of connection and metrics.

1. Connection is used for parallel transport and enables to compare two vectors in different points. Important consequence is that using connection one can introduce the curvature tensor (that can further be contracted to curvature scalar). Curvature appears when you transport vector along the closed curve and then compare with the initial vector. Curvature scalar is then used to construct “field action” just like in all gauge theories.

As shown in Schrödinger’s book, connection can also be used to measure distance along geodesic line (it worth noting that the expression for such “measure” is so much similar to the expression of Feynman’s path integral action!). But in general, connection cannot be used for measuring distances between arbitrary points.

1. Metric is introduced for measuring distances between arbitrary points and defining vector products.

2. Connection and Metric are independent concepts. Only additional condition of their consistency (i.e. when you require that vector product is invariant when both vectors are parallel transported) allows to express connection via metric tensor.

Let’s get back to your question now. All that is written about $GL(n,R)$ above is related to connection only. In the first place, it allows expressing “field action” in terms of a scalar curvature. But what you are most interested in is probably not this, but conservation laws related to matter fields. In the theory with point particles functions $\xi$ (or $\psi$) can be associated with vectors $\frac{dx^\nu}{ds}$. I’m not sure, but consequent conservation law is probably energy-momentum conservation. I think it is the same in Special relativity where space is flat and all connections are zero, but indirectly the conservation of energy-momentum in SR might be a consequence of “conservation” of null curvature by Lorentz transformations (please note that homogeneity of space-time means zero curvature). I know you expect to see some other conserved quantities similar to “electric charge” conservation in Dirac theory of electron. But please note that in Dirac theory the “global” conservation of “charge” is practically indistinguishable from conservation of energy-momentum. As for local theories – I do not know, concrete model need to be considered.

• thanks. So you've shown that the Christofell symbols behave mathematically like a connection under a general GL gauge group transform. What is the field $\psi(x)$ in your notations ? Is it by some extent conserved ? Thanks in advance. – FraSchelle Dec 9 '12 at 17:21
• But $GL(n)$ doesn't admit finite dimensional unitary representations, which cause problems when we want to write down components of fields...because you end up with an infinite number of components...which is why people usually use $SO(4)$ (or $SO(3,1)$ depending on your religion) instead. – Alex Nelson Dec 9 '12 at 21:28
• @AlexNelson Thank you for this comment. Indeed, that's part of the problem when trying to use gauge theory. – FraSchelle Dec 10 '12 at 9:31
• @MurodAbdukhakimov Thank again and so much for your new update. Thanks also for the two references (I actually tried to read Utiyama paper last year, without big success). I will try to read the book by Schrödinger for the holidays. Thanks a lot again. – FraSchelle Dec 10 '12 at 9:33
• I think one has to be a little bit careful here. The above construction essentially shows that the Christoffel symbols can be thought of as components of a connection on the tangent bundle of the space-time manifold (as you mention one has additional conditions such that this is compatible with the metric, leading to the Levi-Civita metric). But GR is NOT like a Yang-Mills theory since its action cannot be written in a form which resembles such types of gauge theories (of spin-one fields). (continued) – Heidar Dec 11 '12 at 0:35

GR has some formal resemblance to Yang-Mills gauge theory. But it isn't quite the same thing.

We formulate YM in terms of gauge fields, AKA, connections on G-bundles on our manifold. We also make use of a connection when we formulate GR, the Levi-Civita connection on the tangent bundle of our spacetime, which is determined by the metric and some assumptions (metric is covariantly constant, no torsion). But the metric is the more fundamental degree of freedom, and there's nothing like this in YM theory. (You can do functional integration on the space of YM fields with enough mathematical rigor to satisfy most physicists, but in 4d, it's not possible to do this with metrics.)

Another similarity: The observables in YM theory are invariants of the group of gauge transformations. Similarly, in GR, the true observables are generally supposed to be invariant under the group of spacetime diffeomorphisms (n.b., not the same thing as the group of gauge transformations for the tangent bundle). These observables aren't generally local observables, like the curvature at a point, but instead more complicated expression constructed from local observables, like the average of the curvature over the spacetime. This is also in contrast to Yang-Mills theory, where there are plenty of local observables, like the energy density of the YM field at a point.

The common theme is that we have to introduce weird auxilliary unphysical variables into both theories to make locality and Lorentz invariance manifest; physical observables are then obtained by forgetting the redundant information.

• Your answer seems to indicate that "gauge theory = Yang-Mills", which might confuse people. – jinawee Jun 7 '17 at 15:22

In general relativity, tensor fields(e.g. the metric tensor, the Riemann curvature tensor, the energy momentum tensor) are left unchanged by changes of frames. Tensors are naturally unchanged under changes of frames. They form the basis of general relativity and is the key to how general relativity is generally covariant.

Update:

A note on the invariance of tensors: while tensors are invariant their components certainly do vary under different frames. There are also scalars which are 'single numbers' that don't change at all.

I do not think general relativity can be considered a gauge theory at all. In general, there are no local or global symmetries in the Langrangian of general relativity. Any spacetime can occur in general relativity; just define the energy momentum tensor according to Einstein's field equations.

• Also, while tensors as a whole are "invariant" in their own geometric sense, their components certainly change. If you want a single number to be unchanged, then you have to restrict yourself to scalars. – user10851 Dec 9 '12 at 2:16
• @ChrisWhite Yes, I should have included that too. Thank you for reminding me. – resgh Dec 9 '12 at 2:54
• @Oaoa No problem :) – resgh Dec 9 '12 at 13:25
• @namehere Feel free to incorporate comments (mine at least) into your answers at your discretion. Single coherent answers are usually better than scattered comments, and besides the comment record is preserved :) – user10851 Dec 9 '12 at 20:50
• But the metric tensor (the field for GR), in different coordinates, should give the same physical result. This is a gauge invariance! – Alex Nelson Dec 10 '12 at 1:09

In arXiv:physics/9801019, gravity is classified as a parameterized classical field theory with dynamic metric.

The following quote is the best one I could find on short notice:

Consider a classical field theory with gauge group $\mathcal G$. Suppose that $\mathcal G ⊂ \mathrm{Aut}(Y)$, the automorphism group of the covariant configuration bundle $Y$. We may distinguish two basic types of field theories based upon the relationship between the gauge group $\mathcal G$ and the (spacetime) diffeomorphism group $\mathrm{Diff}(X)$.

The first consists of those which are parametrized in the sense that the natural homomorphism $\mathrm{Aut}(Y) \to \mathrm{Diff}(X)$ given by $η_Y \mapsto η_X$ maps $\mathcal G$ onto $\mathrm{Diff}(X)$ (or at least onto a “sufficiently large” subgroup thereof, such as the compactly supported diffeomorphisms). This terminology reflects the fact that such a theory is invariant under (essentially) arbitrary relabeling of the points of the parameter “spacetime” $X$. (In relativity theory, cf. Anderson , one would say that $X$ is a relative object in the theory.) The parametrized theory par excellence is of course general relativity, in which case $\mathcal G$ equals the spacetime diffeomorphism group.