Electromagnetism and general relativity can both be thought of as gauge theories, in which case there is a natural analogy between them:

enter image description here

(Strictly speaking, the gauge symmetry of diffeomorphism invariance of the metric $g_{\mu \nu}$ reduces to the global symmetry $\phi(x) \to \phi(x) + \text{const.}$ in the Newtonian limit.)

The theories can also both be thought of as connections on fiber bundles, in which case there is a different natural analogy between them:

enter image description here

Here (deep breath) $A_\mu = A$ is the electromagnetic vector potential, $g_{\mu \nu}$ is the metric, $\phi$ is the Newtonian gravitational potential field, $F_{\mu \nu} = F$ is the electromagnetic field strength tensor, $\Gamma^\mu_{\nu \rho}$ is the Christoffel connection, ${\bf g}$ is the Newtonian gravitational field, $J^\mu$ is the electric four-current, $G_{\mu \nu}$ is the Einstein tensor, $T_{\mu \nu}$ is the stress-energy tensor, $\rho$ is the spatial mass density, $L$ stands for "length", and $R_{\mu \nu \rho \sigma}$ is the Riemann curvature tensor (whew!).

Unfortunately, as noted in Dimensional analysis of metric tensor, these two analogies don't match up! The electromagnetic four-potential $A_\mu$ and field-strength tensor $F_{\mu \nu}$ correspond to the metric $g_{\mu\nu}$ and Christoffel connection $\Gamma^\mu_{\nu \rho}$ respectively in the first analogy, and to the Christoffel connection and Riemann curvature tensor $R_{\mu \nu \rho \sigma}$ respectively in the second. This doesn't seem right - surely there should be a single unified formulation of each theory in which the gauge-theory and fiber-bundle stories are naturally compatible. So what the heck is going on?

  • $\begingroup$ Maybe we can even make it more complicated if we consider the geometry of quantum computation as formulated by M. Nielsen in his paper arxiv.org/abs/quant-ph/0603161. Here he formulated the quantum computation complexity in an exact way of the GR geometry and optimal quantum algorithm is an analogy of geodesic in GR. Then GR can be compared with quantum mechanics or quantum computation. In fact this is my original motivation to ask the question in my post Dimensional analysis of metric tensor. $\endgroup$
    – XXDD
    Jun 20, 2017 at 7:34

3 Answers 3


There is an additional structure at play here. Depending on which level you're looking at, this is either the fact that the tangent connection $\nabla$ admits the exchange of arguments (if $\nabla_XY$ is well defined, then so is $\nabla_YX$), or that there is a soldering form $\theta$ on the princpal bundle of frames for GR, while there is no such soldering form for the connection in EM.

At any rate, both of these mean that there is a difference between the gauge symmetries.

  • For local Lorentz symmetries, the difference is that the local Lorentz symmetry is an "external" symmetry, while the $U(1)$ symmetry is an internal one. We may convert the "external" symmetry into an internal one precisely by the use of vielbeins: $A^a=A^\mu\theta^a_\mu$. But then we need to find dynamics for the vielbein $\theta^a_\mu$ as well. This vielbein is precisely the local appearance of the above mentioned soldering form.

  • For diffeomorphism symmetries, this is totally different. The local Lorentz symmetry and the $U(1)$ symmetry are both localizable. You can make a LL transformation or an $U(1)$ transformation at a point. You cannot make a diffeo at a point. I am not an expert in this bit, frankly, I have always disliked when people say that GR is a gauge theory with $\text{Diff}(M)$ being the gauge group. Sure, you can look at it like that, but that's gonna create an irreconcilable difference with the $U(1)/SU(2)/SU(3)$ gauge theories of the standard model.

The only somewhat unified language is when you consider the gauge group to be the Lorentz group. Moreover, the existence of spinor fields also seem to prefer the Lorentz group interpretation.

If you want some deep answer, I cannot give you one. There is a violation of this gauge analogy precisely because the Lorentz symmetry is an external spacetime symmetry.

An internal symmetry with gauge group $G$ will give you a connection $D$ that acts on sections of the associated vector bundles (associated to the principal $G$-bundle where the connection lives).

The fact that this is a $G$-connection rather than a $GL(k,\mathbb{C})$-connection is because the form of the Lagrangian for the matter field will will usually involve a fiber metric (for scalar QED, this is the "inner product" $q(\phi,\phi)=\phi^\dagger\phi$), usually a Hermitian metric, which, because we want the connection to be metric compatible, result in $GL(k,\mathbb{C})$ being reducible to some unitary group.

Afterwards, we specify a Lagrangian for the connection too, because it needs to be a dynamical field (for unitary connections, this is the Yang-Mills Lagrangian), and then we have our classical gauge theory.

For gravity, we first need a vielbein, to make the external symmetry internal. The vielbein needs to have dynamics, since otherwise how would you specify it? But let's ignore this for now. Now we have an internal Lorentz-symmetry, with fiber metric $\eta$, so we have an $\eta$-compatible connection.

We need to specify the dynamics of both $\theta^a_\mu$ and $\omega^{\ a}_{\mu\ \ b}$. Metric compatibility and torsionlessness enforces that $\omega$ must be nondynamical, so we need to cook up a Lagrangian for $\theta$. But Ostrogradsky instability forces us to look for second-order field equations, so the curvature (second order expression) cannot be a dynamical variable$^*$ (unlike Yang-Mills, where the Ostrogradsky instability won't prevent this, since there the curvature is first-order).

Even if you don't try to enforce compatibility and torsionlessness, it won't change anything. If you enforce torionlessness, but not compatibility, then you get Palatini-formalism, which is equivalent. If you don't enforce torsionlessness, you get differences (Einstein-Cartan theory) that only spinors feel. Even then, you cannot get rid of $\theta$ as the "potential" for the theory.

The conclusion is, the vielbein is a necessary dynamical object to reach gravity, and it ruins the analogy.

*: By saying that curvature cannot be dynamical, I mean that it cannot appear in the field equations in differentiated form.

  • $\begingroup$ One quick question, for general metric tensor if we examine the space of vierbein, it should have an invariant subspace w.r.t. Lorentz group which keep the metric tensor invariant, right? $\endgroup$
    – XXDD
    Jun 20, 2017 at 7:50
  • $\begingroup$ @X.Dong I'm not sure what you mean, but an orthonormal veilbein specifies the metric uniquely. And orthonormality can be kinda-sorta defined without making reference to a metric in the sense that your veilbein is orthonormal if $\theta^{a'}=\Lambda^{a'}_{\ a}\theta^a$ where $\Lambda$ is a local Lorentz transform, and then you can define $g_{\mu\nu}=\eta_{ab}\theta^a_{\mu}\theta^b_\nu$. $\endgroup$ Jun 20, 2017 at 8:05
  • $\begingroup$ Yes, that's what I mean. So the space of vielbein has a kind of fibre bundle structure with the Lorentz group as the fibre. For me GR has a very complex hierarchical symmetry structure from the Gaussian SU(2) symmetry, Lorentz symmetry , Diff symmetry. I am wondering if the Lorentz group is generated by SU(2)/SL(2) as local operation on a single spinor or a single qubit, then the GL(4) can be regarded as operation on 2 qubits. Then is there an essential difference between them? $\endgroup$
    – XXDD
    Jun 20, 2017 at 8:24
  • $\begingroup$ I mean if the space time geometry is characterized by $GL(4)$ transformations (as a generalized operation on qubits) at each point, can the diff symmetry corresponds to an invariant subgroup of $GL(4)$ at each point? $\endgroup$
    – XXDD
    Jun 20, 2017 at 8:31
  • $\begingroup$ @X.Dong I have no idea what you mean on qubits here, but the fiber bundle structure is given by $P$ being a principal Lorentz bundle over $M$ with a soldeing form $\theta\in\Gamma(\bigwedge^1P,\mathbb{R}^n)$. The soldering form has the interpretation that for arbitrary $X\in\Gamma(TP)$ vector field, $\theta_u(X)$ are the components of $\pi_*X$ in the orthonormal frame $u$. This defines $u$'s existence as a frame, so $\theta$ is the object that identifies $P$ as $O(M)$ the orthonormal frame bundle. $\endgroup$ Jun 20, 2017 at 8:35

The analogies presented in the question are, I think, flawed. I agree with Weinberg, as mentioned in @tparker's answer, that the analogy is between the Christoffel symbols ($\Gamma^\nu_{\mu\sigma}$) and the and the Gauge potential ($A_\mu$). This parallel becomes more clear if we compare gravity with non-abelian gauge theories and don't suppress the group indices. In particular, compare the covariant derivatives (applied to "rank 1" tensors, for concreteness): \begin{align} [D_\mu]^i_{\hphantom{i}j} &= \delta^i_{\hphantom{i}j} \partial_\mu - i g [A_\mu]^i_{\hphantom{i}j} \\ [\nabla_\mu]^\nu_{\hphantom{\nu}\sigma} & = \delta^\nu_{\hphantom{\nu}\sigma} \partial_\mu + [\Gamma_\mu]^\nu_{\hphantom{\nu}\sigma}. \end{align} Note that I've modified standard notation for both items. I modified the Yang-Mills notation to include a distinction between whether indices are up or down. I changed the GR covariant derivative to separate the index that defines the direction in which parallel transport is happening ($\mu$) from the ones that describe the transformation of the vector along the path ($\nu$, $\sigma$).

Put another way, the basis vectors of the vector spaces at a point in space-time, $x^\mu$, is related to the basis vectors at a point $x^\mu + \operatorname{d}x^\mu$ to first order by: \begin{align} \delta^i_{\hphantom{i}j} & + ig [A_\mu(x)]^i_{\hphantom{i}j} \operatorname{d}x^\mu \\ \delta^\nu_{\hphantom{\nu}\sigma} & - [\Gamma_\mu]^\nu_{\hphantom{\nu}\sigma} \operatorname{d}x^\mu. \end{align} Note how the bases transform in the opposite sense from the vectors. Importantly, this implies that each second term above is an infinitesimal generator of transformations in the relevant symmetry space, meaning that they are, overall, anti-Hermitian in their open indices. The covariant derivatives above are, obviously, the generalization for translating an entire vector-valued function, as opposed to a single vector. Translating by a finite distance requires building up a structure that is called a Wilson Line (see Peskin & Schoeder page 495 and surrounding) in non-abelian gauge theories.

All of that said, here are the non-notational differences between ordinary gauge theories and gravity:

  1. the metric in Yang-Mills theories is constant,
  2. the group space in gravity is tangent to the space-time manifold,
  3. the degrees of freedom of gravity include diffeomorphisms of the space-time manifold, and
  4. the Lagrangian in gravity is linear in the curvature instead of quadratic.

The metric in Yang-Mills theories are typically not even discussed because it is usually isomorphic to a constant Euclidean metric. Even the abelian Yang-Mills theory, electromagnetism, has a metric. The defining representation of the $U(1)$ group, for example, is the transformations that preserve the norm of complex numbers, $\rho^2 = z^\star z$. Writing this norm in terms of a metric gives a vector $z = x + iy$, a $2$-dimensional vector ($[z]^i$), and the metric maps that to the covector $z^\star = x - iy$ ($[z]_i$). Given the mutliplication rules of complex numbers, this maps the metric to a $2$-$d$ identity matrix, $g_{ij} = \delta_{ij}$.

Point being: it is not studied, as far as I know, for a Yang-Mills theory to have a metric that is a function of space-time.

Gravity is also special for, in some sense, the group representating a transformation tangent to the manifold. In other words, it is significant that the group indices (indices outside of square brackets above) are in the same space as the symmetries on the manifold, making them tangent to the space in a way that even a group that is just isomorphic to those symmetries is not. This somewhat ties into the idea that gravity is tied into diffeomorphisms of the space-time manifold, itself, and is more than just a gauging the $\operatorname{SL}(1,3)$ group.

It is for this reason that we require that the connection be "metric compatible" in gravity. Adding the torsion free condition, this moves all of the degrees of freedom from the connection into the metric. This is why the Newtonian gravitational potential appears there.

As a consequence of all of the above, the curvature becomes quadratic in the metric, instead of linear in the connection. Of particular import, it has two derivatives instead of one. This is one of the big reasons for using a Lagrangian that is linear in the curvature - there are unresolved theoretical difficulties with theories that have more than two derivatives in the Lagrangian; this review by Woodard (2007) has a good overview of the difficulties of this related to $f(R)$ gravity theories.

I wish I could make this more clear with examples, but my experience with Yang-Mills theories is deeper, and I haven't completely crystallized my thinking on this topic, and what the implications are.

  • $\begingroup$ Interesting about the metric in YM theory. As far as I heard, the Fubini-Study/Bures metric is closely connected with the natural connection in YM theory (for example in a Hopf fibration). Can you help to clarify if the FS/Bures metric is compatible with the connection in YM theory, or what's their relationship? If we apply other ASD connections in YM theory, what will be the correspondent compatible metric? I am not familiar with these issues. Thanks. $\endgroup$
    – XXDD
    Jun 21, 2017 at 2:27
  • $\begingroup$ You mentioned that in GR we ask the connection should be metric compatible. Is there a fundamental reason for this? Usually we think the connection is more fundamental or physical. My questions: (1) Why we ask gravity diffeomorphic invarant? Just a reasonable assumption to make the theory work? (2) Does this assumption first determine the metric and then propagate to determine the connection (by compatibility and torsion free condition)? Or the diffeo-invariance first determine the connection then the metric? $\endgroup$
    – XXDD
    Jun 21, 2017 at 2:41
  • $\begingroup$ I wish I could answer your questions, @X.Dong, but this answer as is pretty much plumbs the depth of my current knowledge. $\endgroup$ Jun 21, 2017 at 3:32

Weinberg discusses these analogies a bit in Vol. II of his QFT series:

The analogy breaks down in one important respect: in general relativity the affine connection is itself constructed from first derivatives of the metric tensor, while in gauge theories the gauge fields are not expressed in terms of any more fundamental fields ... [pg. 7]

so he seems to be thinking of the Christoffel connection, not the metric, as the gauge field, as the second analogy. But then on pg. 13, he says that $J_\mu$ is analogous to $T_{\mu \nu}$, as in the first analogy.

It seems that, contrary to my first impression, both of the analogies above really are just analogies, and neither correspondence is particularly tight. Yang-Mills theory and GR really are qualitatively different kinds of ``gauge theories'', because in Yang-Mills theory the gauge-invariant field strength tensor $F_{\mu \nu}$ is formed from first derivatives of the fundamental fields, while in GR the gauge-invariant field strength tensor $R_{\mu \nu \rho \sigma}$ is formed from second derivatives of the fundamental fields. So both analogies above are useful in different contexts, but far from perfect.

However, we can make the second analogy tighter by considering the Palatini formalism of GR, where we treat the metric and the connection as independent fields and vary the action with respect to each separately. Under this formalism, the fiber bundle connection itself is a fundamental field, just as in Yang-Mills theory. (But there is still a major difference between the theories, which is that sections of the fiber bundle appear explicitly in GR as $g_{\mu \nu}$ but not in classical electromagnetism.)

Edit: The general consensus seems to be that this is the "best" analogy, where downward-pointing arrows denote derivatives and rightward-pointing arrows denote tensor contractions:

enter image description here

The strange thing about this analogy is that the equations relating the "field strength tensor" to the "matter source field" are very different in the two cases: the Yang-Mills equation of motion $d(*F) = J$ is a dynamical differential equation, while the analogous general relativity equation of motion $R^\rho_{\ \, \mu \rho \nu} - \frac{1}{2} g^{\sigma \lambda} R^\rho_{\ \, \sigma \rho \lambda} g_{\mu \nu} \propto T_{\mu \nu}$ is just an algebraic relation with no derivatives. I'm not sure what to make of that.

  • $\begingroup$ Do you think that $g_{uv}$ is a section of the fibre bundle, where the Riemannian connection is defined ? For me it's something like a section of an associated bundle. $\endgroup$
    – XXDD
    Jun 20, 2017 at 16:28
  • $\begingroup$ @X.Dong I don't know, my understanding of fiber bundles isn't strong enough. $\endgroup$
    – tparker
    Jun 20, 2017 at 17:20
  • $\begingroup$ @X.Dong The metric is a section of an associated bundle. It can be constructed explicitly. Let $G$ be the Lorentz-group and let $\rho:G\rightarrow GL(4,\mathbb{R})$. Let $\rho^*$ be the dual (contragredient) representation. We have $S^2T^*M=O(M)\times_{\rho^*\wedge\rho^*}(\mathbb{R}^{n*}\otimes\mathbb{R}^{n*})$, so it is an associated bundle. The metric tensor however appears in a completely different role than the associated bundles in say semiclassical EM. $\endgroup$ Jun 20, 2017 at 17:50
  • 1
    $\begingroup$ @tparker No, I am another X. Dong. :-) $\endgroup$
    – XXDD
    Jun 21, 2017 at 2:06

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