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this may be a silly question but I'm not seeing any easy answer.

Right now I'm dealing with the condition of zero-torsion of the spin connection in General Relativity: $$ 0=de^a +\omega{^a_b}\wedge e^b $$ and I wonder if there is a way to explicit $\omega^a_b$, without exploiting the tensor formalism, but only the differential forms formalism.

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  • $\begingroup$ If I understand your question correctly, I think this is explained in Zee's 'Einstein's Gravity' in the Differential Forms chapter. $\endgroup$
    – bolbteppa
    Nov 16, 2018 at 23:21
  • $\begingroup$ Thanks, I checked that out, but it actually seems to give some hints on how to do calculations with the vierbeins. Although the identity given by me is present, the book does not give an explicit formula for omega in the forms formalism. $\endgroup$
    – E. Marc.
    Nov 17, 2018 at 7:55
  • $\begingroup$ What do you mean by to explicit $\omega^a_b$? $\endgroup$
    – MBN
    Nov 17, 2018 at 8:47
  • $\begingroup$ I mean to reduce the expression above to something similar to $\omega^a_b=... \wedge ...$ $\endgroup$
    – E. Marc.
    Nov 17, 2018 at 9:04

2 Answers 2

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If the connection is metric compatible, then $\omega_{ab}=-\omega_{ba}$. Thus, $$ e_b\rfloor d\vartheta_a = \omega_{cab}\vartheta^c+\omega_{ab}\\ e_a\rfloor d\vartheta_b = \omega_{cba}\vartheta^c+\omega_{ba}\\ e_a\rfloor e_b\rfloor d\vartheta_c\wedge\vartheta^c=(\omega_{cba}-\omega_{cab})\vartheta^c $$ where $\rfloor$ denotes the interior product. Then, you can easily see that $$ \omega_{ab}=\tfrac{1}{2}(e_b\rfloor d\vartheta_a-e_a\rfloor d\vartheta_b+e_a\rfloor e_b\rfloor d\vartheta_c\wedge\vartheta^c) $$

Notation $\vartheta^a$ are the form components of the coframe $\vartheta=\vartheta^a\otimes e_a$. $e_a$ are the components of the frame $e$, such that $e_a\rfloor \vartheta^b=\delta^b_a$ (duality). $\omega^a{}_{bc}$ are the components of $\omega^a{}_b$.

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Assuming $<e^a,e_{b}>=\sum_{i}e^{a}_{i}e^{i}_{b}=\delta^{a}_{b}$, then I presume you may be looking for this form $$\omega^{ab}_{i}=\frac{1}{2}[e^{ak}(\partial_{i} e^{b}_{k}-\partial_{k} e^{b}_{i})-e^{bk}(\partial_{i} e^{a}_{k}-\partial_{k} e^{a}_{i})-e^{aj}e^{bl}(\partial_{j} e_{cl}-\partial_{i} e_{cj})e^{c}_{i}]$$ where $a,b,$ and $c$ are the frame indices, and $i,j$ and $k$ are the tangent indices.

And that you're not looking for this

$$\omega^{a}_{\;b}\wedge e^b=e^{a}_{jb}e^{b}_{i}dx^{j}dx^{i}$$

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  • $\begingroup$ If you need spin the connection in the form $\omega^{a}_{\;b}$ instead of $\omega^{ab}$ or $\omega_{ab}$, use the frame metric $\eta$ to lower it or raise it. $\endgroup$ Aug 19, 2019 at 0:04
  • $\begingroup$ There are two kinds of connections on a Riemannian manifold - spin and Christoffel. The tangent space indices are needed so the spin connection is compatible with the Riemannian connection - in addition to being metric compatiable and torsion free. The spin connection is calculated in terms of frame fields and their derivatives - in a manner similar to the way the Christoffel connection is calculated in terms of the metric tensor and it's derviatives. $\endgroup$ Aug 19, 2019 at 0:08

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