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Bivectors, Cartan Geometry and Curvature

  1. Nov 9, 2009 #1
    I allow myself to repost some questions from the General Relativity section, as they may fit in better here:

    I have 3 questions related to bivector space, the curvature tensor and Cartan geometry.

    1) Because of its antisymmetric properties

    [tex] R_{\mu\nu\alpha\beta}=-R_{\nu\mu\alpha\beta}[/tex] , [tex] R_{\mu\nu\alpha\beta}=-R_{\mu\nu\beta\alpha}[/tex] ,

    the Riemann curvature tensor can be regarded as a second-rank bivector [tex] R_{AB} [/tex] in six-dimensional space (in case of spacetime dimension four). Due to the symmetry

    [tex] R_{\mu\nu\alpha\beta}=R_{\alpha\beta\mu\nu} [/tex] ,

    one can also conclude that [tex] R_{AB}=R_{BA}[/tex] . My question now is, which of the symmetry properties remain when extending Riemannian geometry to Cartan geometry with a non-symmetric Ricci-Tensor? Is it correct that one can still obtain a bitensor [tex] R_{AB}[/tex] , which then however is non-symmetric?

    2) The six-dimensional space is of signature (+++---). Is there any analogue to Lorentz transformations in this space?

    3) The metric [tex] g_{AB}[/tex] in bivector space can be constructed by

    [tex] g_{\mu\nu\rho\sigma} = g_{\mu\rho}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho} [/tex] .

    I guess from that one can derive a curvature tensor [tex] R_{ABCD} [/tex] for the six-dimensional space. Is that correct? And is there any interpretation for the bitensor representation [tex] R_{AB} [/tex] of [tex] R_{\mu\nu\alpha\beta}[/tex] ?

    Any answers highly appreciated!

    Cheers
     
  2. jcsd
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