I allow myself to repost some questions from the General Relativity section, as they may fit in better here:(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Bivectors, Cartan Geometry and Curvature

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