# Which global symmetry of Minkowski space (if any) gets gauged to the diffeomorphism invariance of general relativity?

Minkowski space has both translational and Lorentz symmetry, which together give Poincare symmetry. (It also has some discrete symmetries like parity and time-reversal that I won't be concerned with.) In some senses, it seems natural to think of the diffeomorphism invariance/general covariance of general relativity as the "gauged" version of some of these symmetries. But which ones?

1) The equivalence principle is often stated as "spacetime always looks locally like Minkowski space," or "the value of a scalar contracted from Lorentz-covariant tensors at the same point in spacetime is coordinate-invariant," or something along those lines. It seems to me that if you only look at an infinitesimal patch of spacetime, then you can't really talk about translational invariance (which would move you outside the patch), so the symmetry group of that tiny region of spacetime should be thought of as the Lorentz group rather than the Poincare group. If Lorentz symmetry now holds locally at each point in spacetime, then you can say that we have "gauged" the Lorentz group.

2) On the other hand, the "conserved current" in GR is the local conservation of the stress-energy tensor $\nabla_\mu T^{\mu \nu} = 0$.

(I know this will unleash a torrent of commentary about whether GR is a gauge theory, and Noether's first vs. second theorem, and conservation laws that are mathematical identities vs. those only hold under the equations of motion, and so on. Those question have all been beaten to death on this site already, and I don't want to open up that Pandora's box. Let's just say that there is a formal similarity between $J^\mu$ in E&M and $T_{\mu \nu}$ in GR, in that their conservation is trivially true under the equations of motion $\partial_\nu F^{\mu \nu} = J^\mu$ and $G_{\mu \nu} \propto T_{\mu \nu}$ and leave it at that.)

But this is just the diffeomorphism-covariant version of the result $\partial_\mu T^{\mu \nu} = 0$ in Minkowski-space field theory, which is the Noether current corresponding to translational symmetry (As opposed to generalized angular momentum, which corresponds to Lorentz symmetry.) This seems to imply that the natural interpretation of diffeomorphism invariance is as the gauged version of translational symmetry.

Is it more natural to think of diffeomorphism invariance (or the general covariance of GR, which depending on your definitions may or may not be the same thing) as the gauged version of (a) Lorentz symmetry or (b) translational symmetry? Or (c) both (i.e. Poincare symmetry)? Or (d) neither, and these are just vague analogies that can't be made rigorous? If (a) or (b), then why does only a proper subgroup of the Poincare group get gauged? And if (c), then why does only the translational part of the gauge group seem to correspond to a conserved current?

(BTW, I'm looking for a high-level, conceptual answer, rather that one with a lot of math jargon.)

First, general relativity is not a gauge theory in the narrow sense (of having a gauge field) if you consider the second-order formalism in which only the metric is dynamical. The Einstein-Hilbert action conceived of as an action where the only dynamical field is $g$ still has spacetime dependent symmetries ($\mathrm{GL}(n)$-valued transformations acting like the Jacobians of diffeomorphisms on all fields), so it has gauge symmetries and consequently gauge freedom (e.g. the one used below in the spin connection formalism to "diagonalize the metric" at every point), but it does not have a dynamical gauge field. However, there are (at least) two ways to formulate the theory of the Einstein-Hilbert action in terms of a gauge field:

General relativity is a gauge theory with either the general linear group $\mathrm{GL}(n)$ or the Lorentz group $\mathrm{SO}(n-1,1)$ playing the role of the gauge group, depending on your formulation, if you're willing to relax the usually strict requirement that only gauge-invariant quantities are physically meaningful - while Lorentz invariant quantities are more useful in generic computations than, say, vectors, no one claims you can't measure a vector in a given frame. Additionally, GR is not a "free" gauge theory (in the sense of Yang-Mills or Chern-Simons) coupled to something, the gauge field is never the sole dynamical variable, but always coupled to either the metric or the vielbein, so there's another sense in which it doesn't conform to our usual notion of gauge theory.

The two formulations are as follows:

1. Classical (Palatini) formalism: In the first-order formulation (Palatini formalism, so also this question) of GR, the dynamical fields are the metric and the Christoffel symbols. Examining the transformation behaviour of the Christoffels (as I do in this answer), it is straightforward to see that they transform precisely like a $\mathrm{GL}(n)$-gauge field. It is rather crucial to note that diffeomorphism invariance is not the same as gauged $\mathrm{GL}(n)$-invariance - the former is a basic aspect of all "coordinate-invariant physics", while the latter essentially arises because the Ricci scalar in the Einstein-Hilbert action is analogous to the gauge-invariant $\mathrm{Tr}(F)$ terms in ordinary gauge theories. Yes, this is often claimed otherwise, and yes, I am sure that diffeomorphisms are not gauged versions of anything. However, diffeomorphisms induce $\mathrm{GL}(n)$ gauge transformations through their Jacobians, see again the answer about the transformation behaviour of the Christoffels I linked above.

2. Spin connection formalism: Instead of conceiving of the tangent bundle as associated to a $\mathrm{GL}(n)$-frame bundle, a manifold of signature $p,q$ has naturally a reduction of the frame bundle to a $\mathrm{SO}(p,q)$ frame bundle, which you may think of as just the bundle of all orthonormal bases relative to the given metric of signature $p,q$, whereas the $\mathrm{GL}(n)$ bundle is the bundle of all bases. The physicist knows this reduction as the tetrad or vielbein formalism, and it allows us to reduce the $\mathfrak{gl}(n)$-valued gauge field $\Gamma$ that is the Christoffels to a $\mathfrak{so}(p,q)$-valued gauge field that is the spin connection $\omega$ essentially by a smooth choice of orthonormal (non-coordinate) basis all over spacetime, which I explain in a bit more detail in this answer. The dynamical fields in the spin connection formalism are the spin connection and the vielbein.

As supplementary evidence that the slogan that "diffeomorphism invariance is a gauge invariance" is false, I urge you to consider that ordinary Yang-Mills theory is also perfectly "diffeomorphism invariant": The Yang-Mills action $$\int_M \mathrm{tr}(F\wedge{\star}F)$$ has no dependence on coordinates whatsoever either, it is not more or less "diffeomorphism invariant" than the Einstein-Hilbert action is. The significance of "diffeomorphism invariance" in GR is really much more that, as I said above, the Jacobians of diffeomorphisms are the natural source for the gauge transformations of the Christoffels, and that the theory would also be separately invariant just under the $\mathrm{GL}(n)$ transformations without considering an underlying diffeomorphism.

• Then how does one explain the formal similarity between the conservation of $T_{\mu \nu}$ in GR and in translationally-invariant theories on flat spacetime? – tparker Jul 21 '17 at 21:02
• @tparker I don't think it is a "formal similarity" so much as that the latter is simply a special case of the former: In a flat spacetime, we simply have $\nabla_\mu = \partial_\mu$ (in some coordinates). – ACuriousMind Jul 21 '17 at 22:16
• I disagree that it's a special case. In GR, the conservation of $T_{\mu\nu}$ follows from the Einstein equations. In the case of flat spacetime, Einstein's equations reduce to the completely trivial theory $T_{\mu\nu} \equiv 0$. On the other hand, the conservation of $T_{\mu\nu}$ holds (on-shell) for any flat-spacetime theory whose action is translationally invariant. Neither result's hypothesis is contained within the other. – tparker Jul 21 '17 at 23:20
• @tparker Oh, I misunderstood what you meant, then. In that case, I'm not sure what you think there is to explain, though, or rather why you think the gauge theory has something to do with that - the conservations laws of electromagnetism don't contain any gauge covariant derivatives, nor do the tautological (and gauge-variant) analogues in Yang-Mills theories, so why would you connect the appearance of the covariant derivative in that law in GR to the gaugey aspect? – ACuriousMind Jul 22 '17 at 0:03
• @tparker I see no reason at all why Noether's theorem would yield a covariant derivative acting on the conserved current, since the conserved currents follow from Noether's first theorem applied to the global version of the symmetry and are unrelated to the gauge theory (Noether's second theorem for gauge symmetries yields off-shell identities unrelated to conservation), and the sentence in Srednicki mystifies me. Do you have any reference that derives (instead of just claims, as Srednicki does) the appearance of this covariant derivative? – ACuriousMind Jul 22 '17 at 15:48

As a supplement to ACuriousMind's answer, I'd point out that there is literally zero content in diffeomorphism-invariance or "local $U(1)$ invariance" or anything of the sort.

The real devil behind the scene is background independence, when suitably interpreted.

For any system that consists of local fields on a manifold, there is diffeomorphism invariance. Classical mechanics is diffeomorphism-invariant. Classical electrodynamics is diffeomorphism-invariant. You can make pretty much anything that uses a smooth manifold as a background arena diffeomorphism-invariant.

For theories that involve globally linear objects (momentum vectors, etc.), this is more difficult, because diffeomorphisms can kill the linear structure of a vector space, but then you can use density fields instead of "integral quantities", which are now local fields and the problem is partially solved.

Likewise, using scalar QED (before quantization, so really scalar CED but with matter fields) as an example, the same is true for local gauge-invariance.

Let us take a complex Klein-Gordon field $$\mathcal{L}=\partial^\mu\phi^\dagger\partial_\mu\phi+\phi^\dagger\phi.$$ (I am not bothering with constants.)

You can acutely observe that this field theory is invariant under global $U(1)$ transformations but not local ones, so we do the usual procedure to introduce a gauge connection $$D_\mu=\partial_\mu+iA_\mu.$$

The Lagrangian $$\mathcal L=(D^\mu\phi)^\dagger D_\mu\phi+\phi^\dagger\phi$$ is now local $U(1)$-invariant.

Do you have electrodynamics now? Nope. For example, instead of specifying a Lagrangian for $A_\mu$, you can proclaim that

• there exists an internal reference frame (eg. a gauge), in which $A_\mu=0$.

Now, you have a family of internal reference frames, in which $D_\mu=\partial_\mu$, and members of this family are related by gauge transforms $e^{i\chi}$, where $\chi$ is a constant, since we know that $A$ transforms as $A'=A+\mathrm d\chi$.

But as long as we know that there exists one such internal reference frame, we absolutely don't have to stick with these frames. We can instead perform a gauge transform from such a frame by $e^{i\chi(x)}$, then in the new frame, $A'=\mathrm d\chi$, and we need to use the covariant derivative $D_\mu=\partial_\mu+i\partial_\mu\chi$ to get sensible results.

However, we then calculate the curvatue form and get $$F=\mathrm dA=\mathrm d\mathrm d\chi=0,$$ which shows that we don't have an electromagnetic field at all! Yet our theory is still locally $U(1)$-invariant.

The real difference happens when instead of specifying $A$ to be either zero or pure guage, we make $A$ dynamical and impose field equations on it!

Then we can say that our theory of electromagnetism is background independent, because, although there is a principal $U(1)$-bundle over spacetime with an Ehresmann-connection $A$, the geometry of this principal bundle is not a-priori defined, because our connection is not fixed, but dynamical. So the theory is background independent.

The same is true for GR regardless of considering diffeomorphism-invariance or non-absolute-parallel moving frames. What gives us gravity isn't diffeomorphism invariance or local Lorentz-invariance, we can make diffeomorphisms of Minkowski spacetime or take moving orthonormal frames on Minkowski spacetime, and then Minkowski spacetime will still be flat and absent of gravity. What gives us gravity is specifying this background geometry as unknown and providing field equations that govern the dynamics of it.

• While I think this answer is "morally" correct, the mental gymnastics required to claim that "$D_\mu \phi^\dagger D_\mu \phi + \phi^\dagger \phi$ is gauge invariant" and have $A$ not be dynamical are simply wrong from a formal standpoint: A field transformation - such as a gauge transformation - acts by definition on the coordinates and the dynamical variables. If you say that $A_\mu$ transforms under a gauge transformation, then either $A_\mu$ is dynamical or you are using a non-standard notion of what a transformation/symmetry is. – ACuriousMind Jul 22 '17 at 14:22
• @ACuriousMind If $A_\mu$ is a non-dynamical background field, then we can think of it as simply being a certain function of the coordinates. Then isn't a gauge transformation that transforms $A$ simply a coordinate transformation? On the other hand, I suppose that specifying how some function of the coordinates transforms still leaves some ambiguity in how the coordinates themselves transform. – tparker Jul 22 '17 at 15:06
• @ACuriousMind Why would it be wrong? $A$ is a local expression for a section of an affine bundle over $M$, and the "gauge transformation" is nothing more than the effect of the change of reference frame on the local representative. $A$ is dynamical by definition if it is determined by a differential equation. The two concepts are entirely unrelated. – Bence Racskó Jul 22 '17 at 15:33
• @ACuriousMind Of and btw, the gauge transformation does act on a dynamical variable in my example, which is $\phi$. The point is, $U(1)$ can act locally even if we keep the connection fixed and flat, in the abstract sense, so local $U(1)$ invariance does not immediately imply that $A$ is dynamical. It allows $A$ to be. – Bence Racskó Jul 22 '17 at 15:36
• What I'm saying is that when physicists talk of a "transformation" of an action, or its invariance under such a transformation, said transformation acts solely on the dynamical variables. For instance, if you examine the formal statements of Noether's theorem, you'll note that it considers purely variations of the coordinates and the dynamical fields (if you vary something that's not dynamical, its corresponding Lagrange expression has no reason to vanish and Noether's (first) theorem has no reason to hold). Having $A$ transform but not be dynamical is inconsistent with this usage. – ACuriousMind Jul 22 '17 at 15:51