In the context of the 3 + 1 decomposition of spacetime needed for a Hamiltionian formulation of general relativity, quantities with so called internal indices are introduced (in the book I am reading on p.43). For such quatities $G^i$ , some kind of a "covariant derivative" is defined:

$D_aG^i = \partial_a G^i + \Gamma _{aj}^iG^i$

Using this derivative, a corresponding "curvature tensor" $\Omega_{ab}^{ji}$ is then calculated by

$D_aD_b - D_bD_a = \Omega_{ab}^{ji}G^i$

My quastions about this are:

1) Why is $\Gamma _{aj}^i$ called spin connection; it has to do with the spin of what ...?

2) How is the so called curvature of connection $\Omega_{ab}^{ji}$ related to the "conventional" curvature tensor ?

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    $\begingroup$ Which book are you using? (and by any chance does this have to do with the spin networks of loop quantum gravity?) $\endgroup$ – David Z Mar 11 '12 at 6:03
  • $\begingroup$ Ha ha @DavidZaslavsky, Im reading this [childrens book](backreaction.blogspot.com/2012/02/…) :-P and they introduce the notation and concepts I`m asking about around p.43. $\endgroup$ – Dilaton Mar 11 '12 at 9:44
  • $\begingroup$ ... the link in my above comment does not work as I expected it; You have to scroll down to the bottom of the article to see why it is a children`s book :-) $\endgroup$ – Dilaton Mar 11 '12 at 9:59

1) The spin connection allows you to define covariant derivatives of spinors in curved spacetime. For example, to do this, you want a set of gamma matrices which are covariantly constant, so you use the combinations

$\Gamma^i=\gamma^a E_a^i$

where $\gamma^a$ are the usual flat space gamma matrices and $E_a^i$ are the tetrad components, i.e.


where $E_a$ is the tetrad basis for the tangent space.

2) The differential geometric relations between the vielbein formalism and the "standard" one is described in detail here. Section IV B describes the curvature relationship I think you're looking for.

  • $\begingroup$ Thanks @twistor59 this is very helpful; and the whole paper looks interesting. Although I`ll probably have to brush up my knowledge about the tetrade formalism (from my relativity demystified book) a bit first to fully understand it ... :-) $\endgroup$ – Dilaton Mar 11 '12 at 9:35

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