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Bivectors, Cartan Geometry and Curvature

  1. Nov 7, 2009 #1
    I have some 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
  3. Nov 8, 2009 #2
  4. Nov 9, 2009 #3
    Sorry, but the link doesn't work. If further elaboration on the questions is of help, or if my understanding is flawed, please tell. You're also welcome to point me to resources dealing with these topics. I couldn't find anything on these questions though.
     
  5. Nov 9, 2009 #4
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