Given that spacetime is not affine Minkowskispace, it does of course not possess Poincare symmetry. It is still sensible to speak of rotations and translations (parallel transport), but instead of

$$[P_\mu, P_\nu] = 0$$

translations along a small parallelogram will differ by the curvature. I have not thought carefully about rotations and translations, but basically you could look at the induced connection on the frame bundle, to figure out what happens.

This is all to say that spacetime has obviously not exact Poincare symmetry, although the corrections are ordinarily very small. Most QFT textbooks seem to ignore this. Of course it is possible to formulate lagrangians of the standard theories in curved space and develop perturbation theory, too. But since there is no translation invariance, one can not invoke fourier transform.

My questions are:

  • Why is it save to ignore that there is no exact poincare symmetry? Especially the rampant use of fourier transforms bothers me, since they do require exact translation invariance.
  • How does one treat energy momentum conservation? Presumably one has to (at least) demonstrate that the covariant derivative of the energy momentum tensor is zero.

Any references that discuss those issues in more detail are of course appreciated.

  • $\begingroup$ It is not clear for me, whether what you asking is just general relativity in QFT or this is an extension of GR in QFT, but obviously your question has something to do with this topic. In both cases, merging GR with QFT is non-trivial task. Some people even say that they are incompatible. Modern textbooks usually have few paragraphs about the problems. $\endgroup$ – Misha Apr 22 '12 at 5:08
  • $\begingroup$ My question is about neither, I want to understand QFT in a curved background. I will try to edit my question to clarify that. $\endgroup$ – orbifold Apr 22 '12 at 11:42
  • $\begingroup$ The field theories are defined in a continuum limit, so you can treat the curvature as an external perturbation when you are done--- this is the interaction of gravity waves with matter. You can also formulate deSitter special relativity, to make the theory deSitter space from the beginning, but this is not a fundamental thing. $\endgroup$ – Ron Maimon May 5 '12 at 2:02

When done properly, none of the problems exists and some of your assumptions are invalid. First, concerning the two questions,

  • in topologically trivial but arbitrarily curved spacetimes, the Poincaré symmetry holds in the sense that it is a small subgroup of the infinite-dimensional group of all diffeomorphisms; general relativity and all of its extensions are invariant under arbitrary coordinate transformations which – for spacetimes of normal topology – include the Lorentz transformations as well.

  • on the contrary, the energy-momentum conservation law doesn't hold locally in general relativity. There's no coordinate-system-independent, nonzero definition of the stress-energy tensor in general relativity whose divergence using partial derivatives would vanish; this is related to the breaking of the translational symmetries by general backgrounds because these translational symmetries are needed, by Noether's theorem, for the conservation laws to exist. In GR in asymptotically flat or otherwise translationally symmetric backgrounds, one may still define a conserved total ADM mass and other things.

I would also like to correct some other statements:

  • the existence of a Fourier transform doesn't depend on the Lorentz symmetry. In $\exp(ip\cdot x)$, the inner product is really an action of a linear form on a vector (they live in mutually dual spaces) so one doesn't need an inner product on either of the two spaces.

  • general relativity, especially in the presence of fermions etc., is often formulated using tetrads/vierbeins/vielbeins (they're really needed for the fermions). Then the total gauge symmetry group (under which the physical states must be invariant) includes not only the diffeomorphisms but also local Lorentz transformations.

  • quantum gravity expanded around the Minkowski space does exactly preserve the global Lorentz symmetry as well, despite the fact that coherent states of gravitons are able to add curvature into the spacetime and turn it into a Lorentz-symmetry-violating geometry. That's because these gravitons may still be treated as spin-2 fields that exist in the pre-existing background.

  • the gauge symmetries have to annihilate physical states; there may be exceptions for symmetry generators that move the asymptotic region of the spacetime (at infinity); the physical states may carry charges under these generators but these generators must still be isometries of the background (and the corresponding superselection sector of the Hilbert space).

What's true about Nestoklon's comment is that one faces problems when he tries to quantize Einstein's equations including all loop corrections; the theory is non-renormalizable. But these problems don't get imprinted to none of the hypothetical problems you have sketched as long as one is satisfied with the one-loop approximation, for example.

  • $\begingroup$ Thank you for your answer. I do not understand your answer to my first question however. Do you mean by topological trivial, that it can be covered by a single coordinate chart? In that case you can certainly pull back the action of the poincare group, but this does not seem to yield a very geometric action. I thought that the right generalization was to consider the levi-civita connection, which at least gives you infitesimal translations. In any case we cannot apriori assume that spacetime is topological trivial. $\endgroup$ – orbifold Apr 22 '12 at 21:46
  • $\begingroup$ Lubos, there is no covariant coordinate conserved stress energy, but you can do Noether's procedure in the dumb way on GR action to get the pseudo-stress tensor which is coordinate conserved, and is useful. This is Einstein's way. $\endgroup$ – Ron Maimon May 5 '12 at 2:00

It is safe to ignore curvature at the length scales of particle physics, as in the relevant region of space-time one can approximate the manifold very well by its tangent space, which is a flat Minkowski space with Poincare symmetry.

For the same reason, engineers do not use general relativity but work with special relativirty (or even Newton's laws). Carrying the extra conceptual burden would only complicate things without any benefit.

But if curvature is important, one can use a curved version of the Fourier transform ((see, e.g., work by Fonarev, gr-qc/9309005).

  • $\begingroup$ What I am really trying to understand is how the usual constructions can be sensible, although they do not exist in curved space (for example the S-Matrix). What is more, as I understood Poincare symmetry so far, it is a symmetry of an affine Space. I would expect translations to correspond to parallel transport along geodesics. To identify that with translations in the tangent space seems artificial. Some problems with QFT in curved space are outlined in: Christian Bär, Klaus Fredenhagen (Editors) Quantum Field Theory on Curved Spacetimes: Concepts and Mathematical Foundations $\endgroup$ – orbifold Apr 29 '12 at 22:42
  • $\begingroup$ To clarify: I am perfectly happy to accept that since space is almost flat, we can approximately neglect curvature. Similiarly how one can demonstrate that GR gives back Newtonian gravity in suitable limit. But in all other cases I know how to argue that these effects can be neglected, whereas in QFT some of the standard constructions seem to depend on the global structure of spacetime. This belief might be completely misguided, but I do want to understand why. $\endgroup$ – orbifold Apr 29 '12 at 22:47
  • $\begingroup$ For a well-defined $S$-matrix, you need an unbounded space and an unbounded time (both in the past and in the future), not a Poincare group. So yes, it depends on the global structure. But as a scattering process happens essentially locally, one can approximate it by an S-matrix on the tangent space as one is not interested in the behavior at huge times (where the global S-matrix which might not exist applies) but at the behavior at times large enough to cross the experimental setting, where the local $S$-matrix is a much better choice. $\endgroup$ – Arnold Neumaier Apr 30 '12 at 11:29

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