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Let $(M, g_{ab})$ be a 4 dimensional spacetime with metric tensor $g_{ab}$, where $a,b$ are the abstract indices. Choose a tetrad basis $\{(e_\mu)^a\}$, so that $\eta_{\mu\nu}=g_{ab}(e_\mu)^a(e_\nu)^b$. The connection 1-form is given by

$$\omega_{a\mu\nu}=(e_\mu)^b\nabla_a(e_\nu)_b$$

Here, $(e_\nu)_b=g_{bc}(e_\nu)^c$. If we choose a different basis, $\{(e'_\mu)^a\}$, we get a different connection 1-form $\omega'_{a\mu\nu}$. Suppose $(e'_\mu)^a=(e_\mu)^a+\delta(e_\mu)^a$ with $\delta(e_\mu)^a$ representing an infinitesimal change, which induces an infinitesimal change in the connection 1-form, too,

(*) $$\delta\omega_{a\mu\nu}=\delta(e_\mu)^b\nabla_a(e_\nu)_b+(e_\mu)^b\nabla_a\delta(e_\nu)_b$$

But in this paper, the variation of the connection 1-form is given by Equation (58) in page 12, which is rather different from the one I put here. (Of course, the authors used a different convention. Instead of the abstract index notation, they used "conceret index notation", i.e., their latin indices correspond to my greek indices. And they took the dual basis $\theta^a$)

$$\omega^{ab}=\nabla^a\delta\theta^b-\nabla^b\delta\theta^a,$$

translated in the abstract index notation,

$$\delta\omega_a^{\phantom a\mu\nu}=\nabla^\mu\delta(e^\nu)_a-\nabla^\nu\delta(e^\mu)_a$$

where greek indices are raised by $\eta^{\mu\nu}$. This expression is very different from the equation (*) above. I believe these two equations are equivalent, but I am never able to prove it. Should you please take a look at it? Thank you very much!

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