In a gauge theory like QED a gauge transformation transforms one mathematical representation of a physical system to another mathematical representation of the same system, where the two mathematical representations don't differ at all with respect to observables. Gauge transformations are therefore manifestations of a true redundandcy of the mathematical descriptions.

In GR the role of diffeomorphisms is different, a diffeomorphism represents a change of the reference frame. Different observers in different reference frames will of course have different results when they measure the same observable/event. In this sense I agree with Raymond Streater (see Diff(M)) that it is misleading to say that Diff(M) is a gauge group (if you disagree with me please explain).

In AQFT one associates observables (selfadjoint operators) with bounded open subsets of a spacetime, these observables represent what is observable in the given domain of space and time. A detector that is operated for two hours in a laboratory would be, for example, represented by such an observable (this is only approximately true, because the Reeh-Schlieder theorem says that it is not possible to use a truly localized observable, but one has to use an "approximately local observable" instead).

I think that this line of reasoning will stay true even if one day there is a theory of quantum gravitation. But from time to time I read statements like "local observables are not gauge invariant in a theory with (quantum) gravity and therefore cannot exist/ are not valid observables". (If my phrasing of the statement is wrong, please explain and correct it.)

I have never read about an explanation of this statement and would like to hear about one. Isn't a detector, for example, a (approximatly) local observable and won't detectors exist within a theoretical framework of (quantum) gravity?

Edit: A little of explanation of "observables" in GR: I am aware that in GR only "events" make sense as an observable, but of course not, for example, the spacetime coordinates of a point of spacetime, see the discussion of Einstein's hole argument on the nLab:

  • spacetime, see paragraph "Einstein's hole argument".

When a detector detects a particle, I assume that this is an event that is observable because it is defined by the proximity of a localized field excitation and the detector, and the fact that the detecter makes "ping" is a fact that all observers in all reference frames agree upon.

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    $\begingroup$ Excellent question +1. I think the resolution lies in seeing that in gravity physical observables are relational observables and these should indeed be invariant under diffeos. See Carlo Rovelli's work on partial or Dirac observables for more details. $\endgroup$
    – user346
    Feb 1, 2011 at 9:19
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    $\begingroup$ Also Raymond Streater's list of "lost causes" that you link to appears to be more a list of topics that Streater is personally peeved at than failed ideas of any sort. I humbly disagree with Streater's classification of many of these topics as "lost causes". $\endgroup$
    – user346
    Feb 1, 2011 at 9:28
  • $\begingroup$ Sure, why not, I linked to the "Diff(M)" part because I agree with Streater on this one and did not want to repeat what he wrote. (I also happen to agree with him on the other points, but that's not important for the question at hand :-) $\endgroup$ Feb 1, 2011 at 9:40
  • $\begingroup$ I think the key point is the "(approximately)" in your "(approximately) local observable"; most accounts of this will probably invoke some thought experiment showing that if you try to define precisely local operators you begin to create black holes, so things that are local in QFT are necessarily "fuzzy" at distances of order the Planck length in the presence of gravity. But, obviously, GR describes what happens if an apple falls to the ground, and to any reasonable standard of precision I could care about this is a local question. $\endgroup$
    – Matt Reece
    Feb 2, 2011 at 7:25
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    $\begingroup$ I think what confuses me about this business is the whole idea that diffeomorphisms are the same as gauge-degrees of freedom - classical GR is 'covariant' under diffeomorphisms, meaning that representations of physical objects (observables) will transform appropiately - On the other hand, gauge-invariance is just flat out unmeasurable by definition; all the quantities (like the A vector potential in EM) that are 'covariant' under U(1) are actually not observable! ... Because of this i have a bad time believing this conclusion that gravity theories cannot have local observables $\endgroup$
    – lurscher
    Feb 7, 2011 at 21:48

1 Answer 1


At the level of representation theory, the diffeomorphism group in general relativity and its extensions plays the very same role as Yang-Mills symmetries in gauge theories. In both cases, one may define the action of the symmetry on the operators.

In both cases, only the gauge-invariant states - singlets - are allowed in the physical spectrum. That's why the representation theory of the gauge group plays no role at the level of the Hilbert space - only the singlets are relevant. That's why the "gauge symmetries" always just reduce the number of independent degrees of freedom and many people prefer to call them "gauge redundancies" rather than "gauge symmetries".

In both cases, the latter condition of gauge invariance ("physical states are singlets") must totally hold for transformations that converge to the identity at infinity. In both cases, there are subtleties for transformations that change the asymptotic region (fields at spatial infinity). In both cases, the gauge invariance of the physical states arises as the quantum version of the Gauss's law - the subset of Maxwell's or Einstein's equations that don't contain any time derivatives and may therefore be considered constraints on the initial state rather than evolution equations: it's the $\mbox{div} D=\rho$ equation in the case of electromagnetism. The corresponding operator $(\mbox{div} D-\rho)$ has to annihilate the physical states $|\psi \rangle$ in the quantum theory (which is non-trivial if the quantum theory is formulated in terms of the redundant fields, the gauge potentials). A total analogy exists for Einstein's equations (extrinsic curvature on the slice enters the constraint).

The difference between the two gauge groups is just in the "details" - how they act on the spacetime.

The Yang-Mills transformations are local, so $\phi(x,y,z,t)$ only changes by other fields at the same point $(x,y,z,t)$: imagine a phase transformation of a charged field $\phi$, for example. In the case of diffeomorphism, the change is nonlocal: the field at one point depends on the fields at another point $(x',y',z',t')$ before the transformation.

This technical difference changes the character of gauge-invariant observables - and only gauge-invariant observables may correspond to numbers that make a physical sense and may be measured. Because the gauge transformations in the gauge theory are local, one may construct gauge-invariant local operators such as $\mbox{Tr}(F_{\mu\nu} (x,y,z,t)F^{\mu\nu}(x,y,z,t))$, to choose a random example.

In the case of general relativity, such quantities are not gauge-invariant because e.g. the Ricci scalar $R(x,y,z,t)$ is transformed to the Ricci scalar at another point $R(x',y',z',t')$, so even the Ricci scalar at a given point is not gauge-invariant. To construct gauge-invariant observables in general relativity, one has to be sensitive to - e.g. integrate over - the whole space (or spacetime). For example, the ADM energy is gauge-invariant in asymptotically flat backgrounds.

A physical apparatus that exists in a gravitational theory is not represented by any local observable - in the technical sense of "local" - because its location is not a gauge-invariant quantity. If you have a small device in the vicinity of $(x,y,z,t)=(0,0,0,0)$, its measured results may be expressed as a functional of the fields in the vicinity of $(0,0,0,0)$. However, that's only true in one coordinate system. In other words, it is only true before you make a general gauge transformation. After the gauge transformation, the form of the observable corresponding to the quantity measured by the apparatus is expressed by a different formula involving the physical fields - the new formula depends on fields at different values of $(x,y,z,t)$.

You may "know" that $(0,0,0,0)$ is "physically" the same point as $(7,2,-3,5)$ in some new coordinates but the mathematics doesn't know it: the form of the expression is changed so the quantity is not gauge-invariant. In the same way, you could claim that you know that a "red quark field" before the $SU(3)$ transformation is "physically" the same thing as a "green quark field" after the transformation - because you also know the transformation. But exactly because the form of the field that corresponds to the "same physical thing" depends on the transformation, we say that colorful fields in QCD - and local operators in GR - are not gauge-invariant.

You may pinpoint the location of the gadget by defining its proper distances from $d-1=3$ points A,B,C at "infinity" or "far enough" where you already require the legitimate gauge transformations (coordinate redefinitions) to be trivial. But the calculation of the point - and the fields at this point - will depend on the metric tensor between the apparatus and the points A,B,C. So the definition of the observables that represent the values measured by the apparatus is manifestly and inevitably non-local.

Again, there are no gauge-invariant observables in a theory with a coordinate reparametrization symmetry. I claim that the text above makes it totally clear but if it is not clear, please write down something that you think is a gauge-invariant local quantity - as a function of the basic degrees of freedom - in a theory with the general covariance and I will show you why it is not a gauge-invariant local quantity. There aren't any.

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    $\begingroup$ Right, of course. But I wrote other things, too. Another basic conclusion is that there are no local gauge-invariant observables in GR while there are many of them in Yang-Mills theories. $\endgroup$ Feb 1, 2011 at 9:36
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    $\begingroup$ What about the event that a detector makes "ping" because it detects a particle (i.e. a sharply localized excitation of the appropriate quantum field)? Isn't this a local event that all observers will agree upon? $\endgroup$ Feb 1, 2011 at 9:46
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    $\begingroup$ In GR, there are no "God-given coordinates" that could be used across space or spacetime - the very point of the gauge group of GR is that these coordinates may be changed arbitrarily. So you can't describe the beep in your lab by $(3,6,-2,3)$. Instead, you have to say that the beep A or B occurred 35 miles from the Empire State building, 250 miles from the Hancock tower, at the distance of 6,378,012 meters from the center of Earth, 13,700,002,011.063257 years after the Big Bang. To measure all those distances, you need to map the metric tensor in a big part of space(time). $\endgroup$ Feb 1, 2011 at 11:07
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    $\begingroup$ I have problems with the coordinate description, so please let me try to reformulate. Consider we have a spacetime manifold $M$. First we are in a 'world' without $Diff(M)$-gauge. Then we can say an event happens at a point $m\in M$. This point can be described by different charts, hence different coordinates, say $(0,0,0,0)$ as well as $(7,2,-3,5)$. But both 4-tupels represent the same point in the manifold. So this is not a gauge transformation, but just a different way of expressing the same point of $M$. .. $\endgroup$
    – Nevermind
    Sep 2, 2014 at 23:42
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    $\begingroup$ @LubošMotl I think you didn't distinguish passive transformation from active transformation. When you said "The Yang-Mills transformations are local", you actually considered a passive transformation, but when you talked about the transformation of GR, you thought of active transformation. So really, you could not conclude that GR is not a "gauge theory". $\endgroup$ Oct 10, 2014 at 3:24

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