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I am trying to understand how one would use differential forms to calculate the components of the connection and the curvature tensor given a metric. Can anyone point me to relevant resources for the same? I have a basic understanding of differential geometry.

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marked as duplicate by sammy gerbil, Kyle Kanos, John Rennie, Danu, JamalS Oct 20 '17 at 12:44

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There is an example of deriving the Schwarzschild metric using tetrad formalism in Wald's General Relativity. Note that he uses abstract index notation instead of differential forms notation, however this is purely a notational difference on his part, the procedure is completely analogous.


The general methodology is however as follows:

1) Cartan's first equation of structure is $$ T^a=d\theta^a+\omega^a_{\ b}\wedge\theta ^b. $$ For a torsionless connection, this is $$ 0=d\theta^a+\omega^a_{\ b}\wedge\theta^b, $$ where $\theta^a$ is the covielbein and $\omega^a_{\ b}$ is the connection form.

If the connection is also metric compatible, then we have $$ (d^\nabla g)_{ab}=dg_{ab}-\omega^c_{\ b}g_{ac}-\omega^c_{\ a}g_{cb}=0, $$ which imply that for an orthonormal/rigid frame ($g_{ab}=\eta_{ab}$) we have $$ \omega_{ab}=-\omega_{ba}, $$ so in this case the connection form is $\mathfrak{o}(1,3)$-valued (skew-symmetric when both indices are lowered or raised).

We know from pseudo-Riemannian geometry, that the torsionlessness and metric compatibility condition determine the connection uniquely, so together with the antisymmetry condition, Cartan's first structure equation can be used to calculate the connection.

2) The equation $$ 0=d\theta_a +\omega_{ab}\wedge\theta^b $$ consists of 4 2-form equations. By writing out the basis expansion of $\omega_{ab}$ explicitly, you can obtain all the connection 1-forms (I have purposefully lowered everything, as the antisymmetry condition can be used more effectively this case).

3) To calculate the curvature, the second structure equation of Cartan is directly utilized: $$ \Omega_{ab}=d\omega_{ab}+\omega_{ac}\wedge\omega^c_{\ b}. $$

Basically, everything's known here, so you can just plug in everything and calculate separately $\Omega_{01},\Omega_{02},\Omega_{03},\Omega_{12},\Omega_{13}$ and $\Omega_{23}$. You're done.

Notes:

  • It is preferable to express the differential form using the coordinate basis, so what you'll get for the connection form for example is $\omega_{\mu\ b}^{\ a}\mathrm{d}x^\mu$.

  • Calculating the connection form is the "hard" part here - it is only slightly easier/faster than using the coordinate basis formula for the Christoffel symbols - however calculating the curvature is MUCH easier. In the coordinate basis approach you need to manually ennumerate all independent components and calculate them, which is a hassle, while here you only need to calculate six 2-forms, you are essentially bundling together multiple calculations efficiently into strings of components.

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  • $\begingroup$ Thank you uldreth for your answer. I will read the section from Wald. I think the differential forms method is good in the way that components that are zero are automatically eliminated unlike the coordinate approach where one needs a bit more work. $\endgroup$ – Abhishek Pal Oct 20 '17 at 13:11

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