I have some questions related to bivector space, the curvature tensor and Cartan geometry.(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Bivectors, Cartan Geometry and Curvature

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