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In General Relativity, we generally assume that the derivative operator is torsion-free, i.e., second covariant derivatives commute on functions.

However, in Kerr black holes, spacetime is dragged (especially in the ergosphere), so it seems that there is torsion in this case.

Geometrically, why there isn't torsion in the Kerr spacetime? What is the geometric interpretation of torsion if Kerr spacetime doesn't have it?

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Torsion is not frame dragging. Torsion is having an anti-symmetric spacetime connection. As you do parallel transport in general relativity (GR) you drag frames the frames roll as they move. With torsion they would twist. The connection is GR is the Christopher symbols, symmetric in the two bottom indices. The torsion is an anti-symmetric tensor. It will give you an alternate theory to GR, not in any way part of GR. So Kerr and any other GR metric have no torsion. See the Wikipedia article on it at https://en.m.wikipedia.org/wiki/Torsion_tensor

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  • $\begingroup$ Also even in a case where torsion would be present, the Kerr solution is a vacuum solution, and vacuums do not carry any torsion. $\endgroup$ – Slereah May 20 '16 at 6:19
  • $\begingroup$ @ Bob Bee--it strikes me that you don't consider Einstein-Cartan theory to be in any way part of GR, because Poplawski, who (I think) has said that EC's theory of gravity is a "variant" or "version" of GR, uses CMB data to back up his cosmological theory, as in his "Non-parametric reconstruction of an inflaton potential". Wouldn't EC expressions describing many phenomena be identical to those in GR? $\endgroup$ – Edouard Feb 8 at 22:20

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