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What is the gauge group of general relativity?

  1. Sep 10, 2007 #1
    I tried to convince myself it is Diff(M), but I failed.

    Most books say Bianchi Identities reduce the independent equations in Einstein's equations by 4, therefore there are some redundancies in the metric variables. As a result, there could be many solutions that correspond to one physical state. There are two arguments which show the gauge group is Diff(M).

    Hole argument: a diffeomorphism that preserves a Cauchy surface and moves other points will cause the Cauchy data to not have a unique evolution, therefore two metrics related by a diffeomorphism must correspond to one physical state. Objection: the set of all diffeomorphisms that preserve a Cauchy surface is not Diff(M), there exists diffeomorphisms that move all points around.

    Dirac constraint analysis: GR has two first class constraints, one corresponds to 3D diffeomorphism of the hypersurface (Diffeomorphism constraint), one corresponds to diffeomorphism that moves forward in time (Hamiltonian constraint). Objection: the two constraints generate diffeomorphisms on a hypersurface (at an instant of time), not the whole of M.

    So I believe the gauge group of GR is not Diff(M), it should be a subset of it.
    Last edited: Sep 10, 2007
  2. jcsd
  3. Sep 16, 2007 #2


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    Gauge transformations are defined at a fixed spacetime (or base space) point, but the diffeomorphism group maps points in spacetime to other points in spacetime. It seams to me that if you want to preserve the geometric formulation of gauge theories in general relativity the diffeomorphism group cannot be by any means a gauge group. Surely others can expand or explain on this point better than me.
    Last edited: Sep 16, 2007
  4. Sep 16, 2007 #3
    Sl(2, C)
  5. Sep 16, 2007 #4


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  6. Sep 17, 2007 #5
    Sorry for the terse answer -- was in a hurry. What I should have said, is that it is *possible* to create a theory of gravity, which is identical to Einstein's GR in all current predictions, such that it is a gauge theory, and its gauge group is SL(2,C), which is the double cover of SO(3,1) (and they share the same Lie algebra -- the crucial bit of information). The gauge theory is (of course) not Yang-Mills theory. Furthermore, it's possible to come up with non-gauge theories, or even perhaps (but I don't know of any) as gauge theories with a different gauge (actually, don't the teleparallel formulation use translations, and the Cartan one use Poincare+translations?)
  7. Sep 17, 2007 #6

    George Jones

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    This thread, particularly posts #2 and #3, is relevant, although it might not answer kakarukeys' questions.
  8. Sep 30, 2007 #7
    I believe you are giving me an internal gauge group of GR, made manifest because of a reformulation. E.g. for spinor formulation, we have SL(2,C), LQG, we have either SO(3), or SU(2).

    I just want to know the status of Diff(M).

    I saw the thread before....can't find my answer

    Here I talk about gauge theory in broadest sense not in the Yang Mill's sense, gauge freedom = redundancy in the description of the system.
  9. Sep 30, 2007 #8


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  10. Oct 2, 2007 #9
    yes I read that one too...
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