Diff(M) as a gauge group and local observables in theories with gravity

Question

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.

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.