# Why is there no gauge-invariant local operator in GR?

I have a hard time understanding why the bulk locality is a question. I know some operator which depends on a particular coordinate $$x$$, $$O(x)$$, and its correlation function like $$\langle O(x)O(y) \rangle$$ for example can’t be diffeomorphism invariant as the operator itself becomes different in some other coordinate patch, while nonlocal operators (i.e. operators integrated over whole spacetime region) is gauge invariant.

Then, why don’t we think an operator $$O(p)$$ where $$p \in M$$ on a spacetime manifold $$M$$? At the manifold level, we are yet to introduce a coordinate system and thus this kind of operator should be naturally gauge invariant.

For gauge theories, observables (schematically) are functions of the form $$$$\mbox{observable}\colon \frac{\mbox{phase space}}{\mbox{gauge symmetries}}\to\mathbb{R},$$$$ which maps distinct physical states ["gauge orbits"] to "some number" (see arXiv:hep-th/0203027 for a review of the geometry of gauge systems, or the first 5 chapters of Henneaux and Teitelboim's Quantization of Gauge Systems). This condition is satisfied if the observable commutes with the Poisson bracket for each generator of the gauge symmetry, i.e., every first-class constraint. General relativity has the diffeomorphism group for its gauge symmetry, which causes complications: the algebra of constraints is infinite-dimensional.

Torre (arXiv:gr-qc/9306030) provides a "no-go" theorem denying the existence of any nontrivial local and diffeomorphism-invariant observables. This is on "closed" universes, i.e., the spatial hypersurfaces decomposition of the spacetime into $$\mathbb{R}\times\Sigma$$ has the space-like hypersurface $$\Sigma$$ be compact without boundary.

We can try to construct a class of observables by considering integrals over spacetime of scalar functions, but this family of observables do not posses a local interpretation. The scalar must commute with the generators of the diffeomorphism constraints. For a given scalar field $$\phi(x)$$, the diffeomorphism group acts as $$\partial_{\mu}\phi(x)$$ which vanishes if and only if $$\phi(x)=\phi_{0}$$ is a constant. Torre has shown general relativistic observables must include an infinite number of derivatives and hence are very nonlocal.

The proposed routes around this problem attacks the notion of "locality".

One possible way to resolve this is with complete observables. The idea is to consider gauge-invariant relations between gauge-dependent fields. The approach is motivated by the idea that all that matters in general relativity are relations between dynamical quantities. For a review of this, see Tamborino (arXiv:1109.0740) and for applications to general relativity (arXiv:gr-qc/0610060 and arXiv:gr-qc/0702093).

Another approach uses holonomies for observables (see J.N. Goldberg, J. Lewandowski and C. Stornaiolo, "Degeneracy in loop variables". Commun.Math.Phys. 148, no.377 (1992) doi:10.1007/BF02100867, Project Euclid's Eprint.). This uses a change of variables from the metric (and its conjugate momenta) to spin-connections (and its conjugate momenta) popularly called "Ashtekar variables", swapping the (local) Lorentz group $$O(3,1)$$ for the simpler $$SL_{2}(\mathbb{C})$$ group. These $$SL_{2}(\mathbb{C})$$-connections's holonomies are clearly nonlocal quantities, thus circumventing Torre's "no-go" theorem.

One last approach worth mentioning [arXiv:gr-qc/9708041 and arXiv:gr-qc/9906044] is to use the eigenvalues for the Dirac operator defined on a given spacetime. Although not all spacetimes admit a Dirac operator, the general idea may be generalized to the Klein-Gordon operator.

• Thanks for the answer. I have 3 questions. First, spacetime with a boundary (e.g. AdS) is not the case and there is some nontrivial local and diffeomorphism-invariant operator allowed? Second, I understood a uniform field is only gauge-invariant scalar. But then why this concludes an infinite number of derivatives should be included in observables? The previous observation seems only the 0th order term i.e. a constant (uniform) scalar is the only choice. Lastly, can’t we discuss local operators without coordinates from the beginning? e.g. by defining such operators on a manifold as in my post. – Amplituhedron Sep 5 '19 at 18:17
• (1) For gravity on a "spacetime with a boundary", anything remotely resembling a local observable would live on the boundary, yes that is correct. But you run into problems when you try leaving it. (2) Because of diffeomorphism invariance. If a scalar field (the simplest candidate for an observable) is a local observable, it must be invariant under diffeomorphisms...which requires its derivatives to vanish...which implies it's a constant. The "other way out" is to have the scalar field be nonlocal. (3) Coordinates simplify the argument. Local coordinates do help address local concerns. – Alex Nelson Sep 5 '19 at 23:35
• (1) What do you mean by “resembling a local observable?” Do you mean it is not local in an exact way? And if I leave it, how can I be in trouble? Also “Bulk locality and the quantum error correction” paper by Harlow et al. addresses the question of bulk locality and they say it is resolved by restricting the boundary Hilbert space within the code subspace. Is this another way to tackle the problem other than you mentioned? – Amplituhedron Sep 9 '19 at 6:10