# Differential forms manipulation in Cartan's formalism

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.

• If I understand your question correctly, I think this is explained in Zee's 'Einstein's Gravity' in the Differential Forms chapter. – bolbteppa Nov 16 '18 at 23:21
• 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. – E. Marc. Nov 17 '18 at 7:55
• What do you mean by to explicit $\omega^a_b$? – MBN Nov 17 '18 at 8:47
• I mean to reduce the expression above to something similar to $\omega^a_b=... \wedge ...$ – E. Marc. Nov 17 '18 at 9:04

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 $$\omega^a{}_b$$.
Assuming $$=\sum_{i}e^{a}_{i}e^{i}_{b}=\delta^{a}_{b}$$, 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.
$$\omega^{a}_{\;b}\wedge e^b=e^{a}_{jb}e^{b}_{i}dx^{j}dx^{i}$$
• 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. – Cinaed Simson Aug 19 '19 at 0:04