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The Vielbein postulate

  1. Feb 16, 2010 #1

    haushofer

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    Hi,

    I have a question on the vielbein postulate. By this I mean

    [itex]
    \nabla_{\mu}e_{\nu}^a = \partial_{\mu}e_{\nu}^a - \Gamma_{\mu\nu}^{\rho}e_{\rho}^a + \omega_{\mu}^{\ a}_{\ b}e_{\nu}^b \equiv D_{\mu}e_{\nu}^a - \Gamma_{\mu\nu}^{\rho}e_{\rho}^a = 0
    [/itex]

    Someone like Carrol derives this from rewriting the covariant derivative of a vector field X in a coordinate basis and a general basis, so in that sense it's a statement that the index-free object [itex]\nabla X [/itex] doesn't care about being described by a coordinate basis or a general basis, right? He explicitly says,

    "Note that this is always true; we did not need to assume anything about the connection in order to derive it."

    So, covariance (you have the freedom to write any tensor in any basis you like) would then automatically imply the vielbein postulate. Somehow, I don't feel comfortable with this. In GR, saying that the metric is "covariantly constant", [itex]\nabla_{\rho}g_{\mu\nu}=0[/itex], enables us to express the Levi Civita connection in terms of the metric, which I'll call the metric postulate. We can do the same thing with the vielbeins by saying that the curvature of the vielbein disappears, [itex]R_{\mu\nu}(e_{\rho}^a)=0[/itex]. But doesn't the vielbein postulate already implies the metric postulate?

    So, I'm a little puzzled by the precise relation between the metric postulate and the vielbein postulate, and I'm wondering if the vielbein postulate follows from covariance. I ofcourse understand that in some sense the vielbein postulate is just a way of putting constraints on the vielbein and that antisymmetrizing this constraint gives you information about the torsion, but can someone shed a light on this?
     
  2. jcsd
  3. Feb 17, 2010 #2

    haushofer

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    No-one?
     
  4. Feb 17, 2010 #3

    Fredrik

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    It would help if you posted some definitions and part of the derivation that you don't like. (You don't have to define the connection, covariant derivative or the Christoffel symbol, but at least explain the e and the omega, and what you meant by rewrite in a coordinate basis and a general basis).
     
  5. Feb 17, 2010 #4

    atyy

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    According to to Wald (3.4.16) the antisymmetry of the Christoffel symbols implies torsion freeness, whereas the antisymmetry of the connection one-forms implies metric compatibility.
     
  6. Feb 19, 2010 #5

    haushofer

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    Ah, ok, sorry. The e is the vielbein [itex]e_{\mu}^a[/itex] with inverse [itex]e^{\mu}_a[/itex] satisfying

    [itex]
    g_{\mu\nu} = e_{\mu}^a e_{\nu}^b \eta_{ab}
    [/itex]

    and the omega is the spinconnection which can be defined by

    [itex]
    \nabla_{\mu}X^a = \partial_{\mu}X^a + \omega_{\mu}^a_{\ b}X^b
    [/itex]

    By a "general basis" I ment a "non-coordinate basis",

    [itex]
    \hat{e}_{a} = e_a^{\mu}\partial_{\mu}
    [/tex]

    I'll take a look at Wald, but I think I already start to see things here. :) The point is that a lot of people seem to "postulate" the vielbein"postulate" as a constraint, but as I now see it it's really a consequence of covariance.
     
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