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This is a math question; however given the subject are physicists are more likely to be familiar with the math.

Let's consider the standard, local, conformal transformation of a metric:

$$g_{\mu\nu}=\Omega^{2}(x^{i})\tilde{g}_{\mu\nu}$$

More specifically let's consider a metric $g_{\mu\nu}$ that is conformally flat. We can then always find \Omega such that:

$$g_{\mu\nu}=\Omega^{2}(x^{i})\eta_{\mu\nu}$$

Where $\eta$ as usual denotes the Minkowski metric (in some arbitrary coordinate system). In this frame the metric appears flat in a local area. This is entirely similar to a tetrad or frame field in which:

$$g_{\mu\nu}=e_{\mu}^{a}e_{\nu}^{b}\eta_{ab}$$

In the latter two equations the distinction between lorentzian indices and coordinate ones (Latin and Greek respectively as is standard) seems to be null as far as the Two Minkowski metrics are concerned. (They would be numerically identical for any given coordinate system). This seems to imply that:

$$e_{\mu}^{a}e_{\nu}^{b}=\Omega^{2}(x^{i})$$

However; that doesn't seem right since we need a scalar unless we write it as:

$$e_{\mu}^{a}e_{a}^{\mu}=e_{\mu}^{a}(e_{\mu}^{a})^{T}=\Omega^{2}$$

where we've disregarded indices and there seems to be a natural requirement that all other terms would be vanish.

I'm not sure where to go from here but there seems to be the implication that the conformal factor can be represented by tetrad fields for any conformally flat space. Any input on whether this is correct?

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  • $\begingroup$ There could be topological obstructions. $\endgroup$ – Qmechanic Mar 3 '18 at 10:04
  • $\begingroup$ @Qmechanic Example or elaboration please? Much appreciated, This pertains to a "project" I'm working on. $\endgroup$ – R. Rankin Mar 6 '18 at 19:59
  • $\begingroup$ I don't think the relation in the end is correct. Basically, the only thing we know from sure from this is that the "Minkowski matrix" $\eta_{ab}$ is an "eigenvector" of the "operator" $e^a_\mu e^b_\nu$ with eigenvalue $\Omega^2$. I don't think we can infer anything else from that. $\endgroup$ – Bence Racskó Mar 7 '18 at 10:16
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There could be topological obstructions. In order for a globally defined vielbein to exists the manifold should be a spin manifold, which means that the first & second Stiefel-Whitney classes should vanish.

Counterexamples:

  1. A sphere with the stereographic metric, which is conformally flat.

  2. The Moebius strip with a flat metric, which is non-orientable.

References:

  1. M. Nakahara, Geometry, Topology and Physics, 2003.
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  • $\begingroup$ After some thought I feel like I should point out that neither of these counter examples is conformal to minkowski space. While I didn't explicitely say conformal to Minkowski space, it was implied in the question/equations (through the use of $\eta$ ). Thank you again though, I've to think some more on it.. $\endgroup$ – R. Rankin Mar 11 '18 at 8:28

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