argument about fallacy of diff(M) being a gauge group for general relativity

I want to outline a solid argument (or bulletpoints) to show how weak is the idea of diff(M) being the gauge group of general relativity.

basically i have these points that in my view are very solid but i want to understand if there are misconceptions on my part that i'm simply not getting and if its so, i ask for help to make the case more solid, or understanding why it doesn't apply (to gravity):

• gauge groups are not the same as a symmetry group (thanks to Raymond Streater for making that point completely clear)

• gauge invariance in electrodynamics is an observation that physical observables are unchanged after a gauge transformation without changing coordinate frame ( we are ask to believe that in gravity someone did the same? that is, someone made the observation that physical observables are unchanged after a diffeomorphism-gauge-transformation, only to later argue that because of this, that there are no physical observables to begin with, that doesn't make a lot of sense, to not say that its just plain stupid circular argument )

• classic electrodynamics is also invariant (as in symmetry invariant, not as gauge-invariant) under Diff(M). The invariance is of course broken when the theory is quantized and $\hbar$ makes an appearance, because it assumes a preferred scale for certain energies. The key point here is: classical gravity is not special regarding having diff(M) as a symmetry group

• from bulletpoints 2 and 3, if i cannot infer that Diff(M) is a gauge-invariance of electrodynamics, the same should apply to gravity

For this question, i would say that a valid answer would either disprove any of the arguments as fallacies themselves (hence showing a solid argument why gravity is special and diff(M) is without a doubt its gauge group), or improve the argument for making it bullet-proof (sorry for the pun)

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yes, at that time i didn't understand the point (of gauge invariance) very well, and the point i didn't understand at that time is that the statement (that A physical observable should be invariant under any gauge transformation) followed from the definition of gauge symmetry; but this question is actually about something i thought i understood (that Diff(M) is a gauge symmetry) but i definitely didn't (and now i perceive as wrong). In short, that previous question manifested my confusion on the subject, but i believe i now have a better idea from where that confusion stems –  lurscher Jul 19 '11 at 15:44
If you could define invariance in such a way, any differential equation would become invariant, because it is known, how to rewrite differential equations for different coordinate systems. So Maxwell equations should not be considered as diff(M) invariant only because you may write that in curved coordinates. –  Alex 'qubeat' Jul 19 '11 at 16:31
@Alex, exactly - so you are now feeling my confusion as well - so why expect gravity observables have to be invariant under diff(M)? –  lurscher Jul 19 '11 at 16:34

I will mostly talk about the classical physics as this is complicated enough already (I might mention something about quantum stuff at the end). So, let us first get all the relevant terms straight, so that we avoid any further confusion. In particular, we need to be precise about what we mean by invariance because already two different notions have been thrown into one bag.

• A symmetry group of a physical system is a group of transformations that leave the system invariant. E.g. the electric field of a point charge is invariant under rotations w.r.t. to that point. In other words, we want the group to act trivially. But that means that this immediatelly rules out any equations that carries tensorial indices (i.e. transforms in a non-trivial representation of the rotation group). For these equations, if you perform a rotation, the equation will change. Of course, it will change in an easily describable manner and a different observer will agree. But the difference is crucial. E.g. in classsical quantum mechanics we require the equations to be scalar always (which is reflected in the fact that Hamiltonian transforms trivially under the symmetry group).

• Carrying a group action is a broader term that includes tensorial equations we have left out in the previous bullet point. We only require that equations or states are being acted upon by a group. Note that the group action need not have any relation to the symmetry. E.g. take the point charge and translate it. This will certainly produce a different system (at least if there is some background so that we can actually distinguish points).

• A gauge group of a system is a set of transformations that leave the states invariant. What this means is that the actual states of the system are equivalence classes of orbits of the gauge group. Explicitly, consider the equation ${{\rm d} \over {\rm d} x} f(x) = 0$. This has a solution $g(x) \equiv C$ for any $C$. But if we posit that the gauge group of the equation consists of the transformations $f(x) \mapsto f(x) + a$ then we identify all the constant solutions and are left with a single equivalence class of them -- this will be the physical state. This is what gauge groups do in general: they allow us to treat equivalence classes in terms of their constituents. Obviously, gauge groups are completely redundant. The reason people work with gauge groups in the first place is that the description of the system may simplify after introduction of these additional parameters that "see" into the equivalence class. Of course historical process went backwards: since gauge-theoretical formulation is simpler, this is whan people discovered first and they only noticed the presence of the gauge groups afterwards.

Now, having said this, let's look at the electromagnetism (first in the flat space). What symmetries are the Maxwell equations invariant under? One would like to say under Lorentz group but this actually not the case. Let's look at this more closely. As alluded to previously the equation $$\partial_{\mu} F^{\mu \nu} = J^{\nu}$$ can't really be invariant since it carries a vector index. It transforms in the four-vector representation of the Lorentz group, yes -- but it is certainly not invariant. Contrast this with the Minkowski space-time itself which is left invariant by the Lorentz group.

We also have ${\rm d} F = 0$ and therefore (in a contractible space-time) also $F = {\rm d} A$ which is obviously invariant w.r.t. to $A \mapsto A + {\rm d} \chi$. In terms of the above discussion the equivalence class $A + {\rm d} \Omega^0({\mathbb R}^{1,3})$ is the physical state and the gauge transformation lets us distinguish between its constituents

Let's move to a curved spacetime now. Then we have $$\nabla_{\mu} F^{\mu \nu} = J^{\nu}$$ Again, this is not invariant under ${\rm Diff}(M)$. But it transforms under an action of ${\rm Diff}(M)$. The only thing in sight that is invariant under ${\rm Diff}(M)$ is $M$ itself (by definition).

In the very same way, GR is not invariant under ${\rm Diff}(M)$ but only transforms under a certain action of it (different than EM though, since GR equations carry two indices). Also, ${\rm Diff}(M)$ can't possibly be a gauge group of any of these systems since it would imply that almost all possible field configurations get cramped to a single equivalence class possibly indexed by some topological invariant which can't be changed by a diffeomorphism. In other words, theory with ${\rm Diff}(M)$ as a gauge group would need to be purely topological with no local degrees of freedom.

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i understand that tensor covariance is not exactly the same as invariance, but rather well-geometrically-defined-non-invariance (in this case, of tensors with 1 and 2 indices). I'm not sure i understand the idea that "Hamiltonian is a (geometrical) scalar always"; i definitely can take eigenstates of an atom and apply a Lorentz boost, measure them, and i will see a $\gamma$ factor shifting all the eigenvalues, isn't that how a 4-vector should behave? - I agree with the cramping of the fields and no local degrees of freedom, but isn't that a hint that this approach is wrong? –  lurscher Jul 19 '11 at 19:47
@lurscher: ah, good point. I was only talking about classical QM; I will make that explicit. As for the last sentence, what do you mean by wrong? In topological theories of gravity (such as GR in three dimensions) this is precisely what happens. But if you are referring to quantum gravity, I am not sure. I have no idea what kind of observables people use in different kind of theories. It might be perfectly possible that the only reasonable observables are of the topological type. –  Marek Jul 19 '11 at 20:04
wrong in the following sense: all quantum theories i know of produce classical observables from aggregation or reduction of quantum observables (averages, projectors, traces, etc.) but you cannot take a number of topological invariants and even in principle have them say anything about particular configurations of curvature, black holes, or Newtonian limits. So it would effectively partition physics in observables that are not measurable and measurables that are not observables - a sort of blind alley –  lurscher Jul 19 '11 at 20:29
@lurscher: that is actually not the case if the observable is understood in the sense of a Hermitian operator. Consider QFT: there is no notion of position operator there since there is no way to consistently talk about position of a particle. Yet this information somehow magically appears when one restricts to low energies (implying both non-relativistic limit and restriction to some $n$-particle portion of the Fock space). Tell me why you think something similar can't happen in quantum gravity. –  Marek Jul 19 '11 at 20:30
@lurscher, okay, your edited comment makes more sense. Now, tell me why you think that topological information on quantum scale can't determine macroscopic geometry? Here's one (arguable very naive) way of how to do this: suppose there were holes in the space-time roughly Planck length appart. There would be such a huge number of them that surely one could encode properties of the macroscopic system in a purely topological way. Once again, it's not obvious to me that something like this can't be done. –  Marek Jul 19 '11 at 20:43

Correct me if I am wrong. Gauge transformations are for when you have a bundle over a manifold. GR does not yet have a non-trivial bundle over a manifold, so it is not yet a gauge theory. If in the future someone alters it, they may supply gauge transformations of the enlarged theory

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The fiber would be the space of pseudo-Riemannian metrics, the underlying base manifold would be spacetime, and you got yourself a nifty bundle. –  Alex Nelson May 30 at 2:50
Is that really quite right? a metric tensor is a section of the bundle of tensors: at each point of M, the fibre of this bundle are tensors on the tangent space at the bundle. So the fibre is not the space of metrics. Still, your point is that this is indeed a bundle. –  joseph f. johnson May 30 at 21:59