# 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, 2012 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. Dec 9, 2012 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)? Dec 9, 2012 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, 2012 at 12:10
• Related discussion here physics.stackexchange.com/q/12461 Dec 9, 2012 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. Dec 9, 2012 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. Dec 9, 2012 at 21:28
• @AlexNelson Thank you for this comment. Indeed, that's part of the problem when trying to use gauge theory. Dec 10, 2012 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. Dec 10, 2012 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) Dec 11, 2012 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. Jun 7, 2017 at 15:22
• @jinawee what's the most important between the two ? May 1 at 21:09

The precise sense in which general relativity is a gauge theory has been known (but apparently largely overlooked) for decades. The original sources for what I'm summarizing in the following are a series of papers by Andrzej Trautman, in particular "Fibre bundles associated with space-time", "On Gauge Transformations and Symmetries" and "Fiber Bundles, Gauge Fields and Gravitation". This answer will be a bit lengthy since it rehashes the essential bits of the theory of bundles necessary to get at least an idea of what we're doing. The upshot is this:

Both Yang-Mills theories and general relativity are "gauge theories" if we pick an idea of a gauge theory where some bundle - a structure "over spacetime" - has transformations, and some of these transformations can preserve the physics we care about. Yang-Mills theories are cases where these transformations split neatly into finite-dimensional "spacetime" and infinite-dimensional "internal" symmetries, while general relativity produces infinite-dimensional spacetime and no internal symmetries. This lack of internal symmetries - usually seen as the hallmark of "gauge theories" - explains why there are many claims that general relativity is "not a gauge theory": This claim is correct if one takes the definition in terms of internal symmetries.

In any case, our generic notion of a gauge theory which will encompass both Yang-Mills type theories and GR is that of a $$G$$-principal bundle $$\pi : P\to M$$ over our spacetime $$M$$ on which we have a connection 1-form $$\omega$$. $$\omega$$ descends for charts $$\phi_i : U_i \to M$$ over which $$P$$ is trivial to $$\mathfrak{g}$$-valued 1-forms $$\omega_i$$ on $$M$$, these $$\omega_i$$ are what we usually call "the gauge field". We have $$\omega_i = (\omega_i)_\mu^a T^a\mathrm{d}x^\mu$$ for $$T^a$$ a set of generators for the group, the $$(\omega_i)_\mu^a$$ are the connection coefficients. In the case of a Yang-Mills theory, we usually write these as $$A_\mu^a$$ and call them the gauge field or the gauge potential.

The group of gauge transformations $$\mathscr{G}$$ is the group of all bundle automorphisms $$P\to P$$ (this full group we will just write as $$\mathrm{Aut}(P)$$) that additionally preserve the "fundamental structures" of our theory. A diffeomorphism $$f : P\to P$$ is a bundle automorphism if there is a diffeomorphism of the base $$\bar{f} : M \to M$$ such that $$\pi\circ f = \bar{f}\circ \pi$$. A vertical or pure gauge transformation $$g : P\to P$$ is one that preserves the fibers $$\pi\circ g = \pi$$, i.e. acts only in the "internal space" of the gauge theory. We call the group of pure gauge transformations $$\mathscr{G}_0$$. It is usually the case that $$\mathscr{G_0}$$ connects "physically equivalent" configurations, while $$\mathscr{G}/\mathscr{G}_0$$ is some sort of spacetime symmetry. Note that by the definition of a bundle automorphism, we have a map $$\bar{} : \mathscr{G}\to \mathrm{Diff}(M)$$, and two gauge transformations that differ only by a pure gauge transformation have the same image under this map, so we can think of $$\mathscr{G}/\mathscr{G}_0\subset \mathrm{Diff}(M)$$.

For instance, for a Yang-Mills type theory, the essential structure that needs to be preserved is the Yang-Mills Lagrangian $$\mathrm{tr}(F\wedge{\star}F)$$, where $$F = \mathrm{d}_\omega \omega$$ is the curvature associated with a connection $$\omega$$. All vertical transformations preserve this since the trace is invariant under the action of the gauge group, so $$\mathscr{G}_0$$ is just all smooth functions $$g : P\to G$$, which are in bijection with local gauge transformations $$g_i : U_i\to G$$ for trivializing patches $$U_i$$. The Hodge star $$\star$$ involves the metric of the underlying space, so $$\mathscr{G}/\mathscr{G_0}$$ is just the isometries of $$M$$ connected to the identity.1 For the case of $$M=\mathbb{R}^{3,1}$$ and a trivial bundle $$M\times G$$, we have $$\mathscr{G} = \mathscr{G}_0 \ltimes \mathrm{P}_0(3,1)$$, where $$\mathrm{P}_0$$ is the identity component of the Poincaré group.

For general relativity, the metric on the base space is dynamic and hence nothing we need to preserve. This makes the theory different from the outset, since nothing really seems to restrict $$\mathscr{G}/\mathscr{G}_0$$. Furthermore, the principal bundle $$P$$ in general relativity is the frame bundle $$FM$$, which is the bundle of all possible choices of bases for the tangent spaces of $$M$$. That is, a point in $$FM$$ is given by $$(x,v)$$, where $$x\in M$$ and $$v$$ is a basis of $$T_x M$$. We can think about the data necessary for a basis in terms of a linear isomorphism $$v : \mathbb{R}^{\mathrm{dim}(M)}\to T_x M$$, or as the images of the standard basis $$e_i$$ of $$\mathbb{R}^n$$, $$v_i := v(e_i)$$. In any case, the frame bundle is a $$\mathrm{GL}(n)$$-principal bundle over $$M$$ that inherits its bundle structure from the tangent bundle: If $$\phi_i : \pi_T^{-1}(U_i) \to U_i\times\mathbb{R}^n$$ are a trivialization of the tangent bundle, then $$F(\phi_i) : \pi_F^{-1}(U_i) \to U_i\times\mathrm{GL}(n), (x,v)\mapsto (x,\phi_{i,x}\circ v)$$ is a trivialization of the frame bundle. There are natural coordinates for the frame bundle induced by a coordinate chart of $$M$$, which means that we often don't really appreciate when an object lives in the frame bundle rather than on $$M$$ itself. 2

Since we have defined $$FM$$ via the tangent bundle $$TM$$, it is "attached" to $$TM$$ and therefore $$M$$ itself in a way that a generic principal bundle is not. We say that it is soldered to $$M$$, and this soldering is expressed by the solder form $$\theta_{(x,v)}(X) = v^{-1}\mathrm{d}\pi_F(X)$$, which is a $$\mathbb{R}^n$$-valued 1-form on $$FM$$. This form is horizontal and equivariant, and by general arguments it corresponds therefore to a $$TM$$-valued form on $$M$$. The form it corresponds to is just the identity $$TM\to TM$$, read as a form $$\mathrm{id}_x(X) = X$$ (or, in coordinates, $$\mathrm{id} = \delta^\mu_\nu \partial_\mu\otimes \mathrm{d}x^\nu$$).

This form is the crucial object that distinguishes general relativity from Yang-Mills type theories. It allows us to define a lot of the oft-studied objects such as the Ricci curvature and Ricci scalar: We can think of the solder form as a collection of $$\mathbb{R}$$-valued 1-forms $$\theta^\mu$$. These form a basis of horizontal equivariant 1-forms, i.e. we have $$G = G_{\mu_1\dots \mu_k} \theta^{\mu_1}\wedge\dots\wedge \theta^{\mu_k}$$ for any equivariant horizontal $$k$$-form $$G$$ and in particular $$x^\ast \theta^\mu = \mathrm{d}x^\mu$$ for any coordinate system $$x$$. This allows us to write things like $$F_\omega = {{F_\omega}^\mu}_\nu {T^\nu}_\mu = {R^\mu}_{\nu\sigma\rho}{T^\nu}_\mu (\theta^\sigma\wedge \theta^\rho)$$ and define the Riemann tensor with its usual four indices without reference to any particular coordinate system - its contractions then yields the usual quantities of interest. In order for these constructions to be preserved, gauge transformations need to preserve the solder form, i.e. $$g\in\mathscr{G}$$ must fulfill $$g^\ast\theta = \theta$$.

It turns out that this means $$\mathscr{G}\cong \mathrm{Diff}_0(M)$$ and $$\mathscr{G}_0\cong \{1\}$$ (see this answer of mine for an explanation of Trautman's argument for this), i.e. the group of gauge transformations is isomorphic to the group of small (= generated as the flow of vector fields) diffeomorphisms and contains no non-trivial pure gauge transformations. This explains a) how general relativity is a "gauge theory" in a broad sense and b) why we often find the puzzling claim that the crux of general relativity is "diffeomorphism invariance". What it is invariant under is not actually simple "diffeomorphisms", but transformations of the frame bundle that are in bijection to these diffeomorphisms. Infinitesimally and for a connection without torsion, the generators of these induced transformations of the frame bundle are given by a vector field $$X^\mu$$ and its covariant derivatives $$\nabla_\mu X^\nu$$, interpreted as a $$\mathfrak{gl}(n)$$ matrix.

Footnotes

1. Diffeomorphisms not connected to the identity cannot in general lift to bundle automorphisms, see this MO answer.

2. A coordinate chart $$x:\mathbb{R}^n\to U\subset M$$ defines a frame $$F(x)$$ on $$U$$ by $$e_i\mapsto \partial_i$$ and a frame $$v$$ in turn defines coordinates on $$FU$$ by expanding any $$(y,w)\in F_y M$$ into $$w_i = w^j_i v_j$$ and the $$w^j_i$$ are $$\mathbb{R}^{n^2}$$-valued coordinates for $$FU$$, so using this for the frame $$F(x)$$ gives coordinates $$(x^\mu, w^\nu_\sigma)$$. In these coordinates, we have a basis $$T^i_j$$ of the vertical part $$V(FM)\subset T(FM)$$ given by $$T^i_j = \partial_{w^i_j}$$, and expanding $$\omega$$ in this basis gives the Christoffel symbols: $$\omega = \Gamma_\mu \mathrm{d}x^\mu = {{\Gamma_\mu}^\nu}_\sigma \partial_{w^\nu_\sigma} \otimes \mathrm{d}x^\mu$$.

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, 2012 at 2:16
• @ChrisWhite Yes, I should have included that too. Thank you for reminding me. Dec 9, 2012 at 2:54
• @Oaoa No problem :) Dec 9, 2012 at 13:25
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.