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ShayanJ

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- #1

ShayanJ

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- #2

Ben Niehoff

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The Riemann tensor can be thought of as the field strength of a GL(n) connection. When the connection is metric-compatible, this is an O(n) connection; and when the manifold is furthermore orientable, it is an SO(n) connection. This SO(n) connection is the (matrix-valued) connection 1-form in the orthonormal basis.

I don't see where the non-Abelian Stokes theorem is applicable when varying the EH action, though. I'd have to see the context.

- #3

ShayanJ

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Well...if for non-Abelian gauge fields and their strengths, its only correct to use non-Abelian Stokes theorem, and if the Riemann tensor can be thought of as the strength of the connection coefficients(as gauge fields), and if this will make GR a non-Abelian gauge theory, then the conclusion is that for Riemann and Ricci tensors, non-Abelian Stokes theorem should be used. And because there is a step in the variation of the EH action that uses Stokes theorem, then the conclusion should be that non-Abelian Stokes theorem is applicable to the variation of the EH action. So I'm confused that you don't think its applicable!I don't see where the non-Abelian Stokes theorem is applicable when varying the EH action, though. I'd have to see the context.

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Ben Niehoff

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ShayanJ

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It was in a talk and the one giving the talk, was just criticizing different things in physics and mathematics and this was one of the things he said, that its actually correct to use non-Abelian Stokes theorem in the variation of EH action instead of the Abelian case. He said its because the group of transformations(which he meant SO(3)) is non-Abelian, nothing more.

I should add that from the other things he said, I'm sure he was actually a crackpot. But I didn't know enough to think about this one claim on myself so I asked it here.

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