# Unitary representations of the diffeomorphism group in curved spacetime

In (special) relativistic quantum mechanics there is a standard argument that says that the (rigged) Hilbert space of states $H$ should be equipped with a projective unitary representation $U$ of the Poincaré group $P$ that goes something like this:

Suppose we have two observers $O_1$ and $O_2$ and $x\in P$ is the Poincaré transformation that maps $O_1$s coordinate system into $O_2$s coordinate system. Then, if $O_1$ and $O_2$ attempt to measure the same thing, they will actual find different states, $\left| \psi _1\right>$ and $\left| \psi _2\right>$ respectively, in $H$ (for example, if you first find the vector $(1,0,0)$, rotate your axes by $\pi/2$ and measure the same vector, the numbers you now record are going to be $(0,1,0)$). In fact, this gives us a map of states: $\left| \psi _1\right> \mapsto \left| \psi _2\right>$ (it's only going to be well-defined up to phase). Let us call this map $U(x)$, so that $\left| \psi _2\right> =U(x)\left| \psi _1\right>$.

However, if we accept the principle of relativity that physics should be the same in two reference frames related by a Poincaré transformation, then we better have that $$\left| \left< \phi _1|\psi _1\right> \right| =\left| \left< \phi _2|\psi _2\right> \right| =\left| \left< U(x)\phi _1|U(x)\psi _1\right> \right|$$ for all (normalized) $\left| \psi _1\right>$ and $\left| \phi _1\right>$ because this represents a probability (of course, $\left| \psi _2\right> :=U(x)\left| \psi _1\right>$ and $\left| \phi _2\right> =U(x)\left| \phi _1\right>$).

Wigner's Theorem then tells us that this gives a projective unitary representation of $P$ on $H$. (Note: I am allowing some elements of $P$ to be represented by antiunitaries.)

Now, if you try to do the same argument in curved spacetime, you run into the obvious problem that you will not in general have an analogue of the Poincaré group (I believe there exist spacetimes which possess no Killing fields); however, it naively seems as if the principle of general covariance suggests that we should 'upgrade' the isometry group of spacetime to the entire diffeomorphism group of the spacetime in the above argument. (In particular, I don't see how this argument makes crucial use of the fact that both observes coordinate bases are orthonormal.) That then, would imply, that the Hilbert space of states in a quantum theory of curved spacetime should possess a projective unitary representation of the diffeomorphism group of that spacetime. This, however, at first glance, seems like it would be false.

So, where does the argument break down if you replace the Poincaré group in special relativity with the diffeomorphism group in general relativity? Or, is it in fact the case that we should obtain a unitary representation of the entire diffeomorphism group.

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A problem with your approach (but see also Dan's answer below), is that, with a Hilbert space formulation in curved ST, one immediately encounters problems with the appearance of non unitary equivalent representations of the observables, as soon as he switches on the curvature. The algebraic approach is much more useful. Nevertheless the group of diffeomorphisms does not admit a representation in terms of automorphisms of the algebra of observables in a given spacetime, since diffeomorphisms change the metric of that spacetime and the algebra "sees" the metric (causality relations). – Valter Moretti Feb 26 '14 at 8:20

No, you do not want representations of the diffeomorphism group for the same reason that you do not want representations of the gauged Lie group in Yang-Mills. The diffeomorphisms are a gauge symmetry, not a real symmetry of the theory. Gauge transformations act trivially on physical states, they map one redundant description of a state onto another. They are redundant descriptions of the physical degrees of freedom, while a real symmetry of the theory maps physical states to other physical states.

In the non-Abelian Yang-Mills case, you look for states in representations of the global SU(N) group. Such transformations are not gauge transformations because the gauge parameter does not tend to zero at infinity as gauge transformations must. These global transformations map one physical state onto a different physical state. The story is similar in most of the understood gravitational cases. The diffeomorphisms are gauge transformations, but there is an asymptotic symmetry group (essentially large gauge transformations) that are the real symmetry of the theory.

For example, in the case of asymptotically flat spacetimes, you put some boundary conditions on the metric at infinity. You then consider diffeomorphisms that leave these boundary conditions fixed. You then basically mod out by trivial transformations that do not touch the metric at infinity, and obtain the BMS group. The BMS group is essentially a semidirect product of $SL(2,\mathbb{C})$ with an infinite dimensional group of "supertranslations", 4 of which can be identified with the 4 global translations. These transformations leave the asymptotics of the metric fixed, but act non-trivially on the boundary data and thus on the states of the system. So you see that we actually encounter an even larger symmetry group in a non-flat spacetime.

Similar procedures can be applied to other spacetimes. You do not need the spacetime to have killing vectors (a generic spacetime has none), you just need the spacetime to have some specified asymptotic form and then you may find a group whose representations control the Hilbert space.

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Nice answer. It is true what you said about BMS group and QFT. Within the algebraic approach, it is possible to built up QF theories admitting that infinite dimensional group as invariance group and preserving several important properties, as short distance behaviour of two-point functions assuring feasibility of renormalization procedures. I wrote, together with some coworkers, several papers on these issues some years ago. – Valter Moretti Feb 26 '14 at 8:14
If this is the case, then I feel we should apply the Faddeev-Popov trick to the Einstein Hilbert action so as to find the corresponding BRST charge associated with this diffeomorphism symmetry, and perform the usual cohomological construction to find the space of states. Has this not been done before? What is the result? – Jonathan Gleason Feb 26 '14 at 19:01
This can certainly be done for the linearized theory. I haven't seen it done for the full non-linear theory, and I suspect it is more complicated. – Dan Feb 28 '14 at 0:22

The difference is that Poincare invariance is a global symmetry, so it acts nontrivially on the physical states. This has real physical consequences; for instance, if you act with a translation operator on the state of a particle localized at the origin, you get the state of a particle localized at some position other than the origin. Poincare invariance tells you that these two particles have the same energy.

Diffeomorphism invariance, on the other hand, is a local gauge symmetry in general relativity. This means that it acts trivially on physical gauge invariant states, and cannot have real consequences. It is a redundancy in the description. Of course, we can get a different description of the same physics by acting with a gauge transformation (i.e. going to a different observers frame), but this is not as powerful as a global symmetry, which organizes the Hilbert space into irreps.

As you alluded to above, this does not mean that there is not an analog of Poincare symmetry for curved backgrounds. For example, the global isometry group of AdS is the group of global conformal transformations in one less dimension. This organizes the physics of AdS into irreps of the conformal group.

Incidentally, you might wonder what happens if you try to come up with a theory with local Lorentz invariance. The answer is that you get general relativity.

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