# Non-Abelian Stokes theorem and variation of the EL action

• A
Gold Member
Today I heard the claim that its wrong to use Stokes(more specificly divergence/Gauss) theorem when trying to get the Einstein equations from the Einstein-Hilbert action and the correct way is using the non-Abelian stokes theorem. I can't give any reference because it was in a talk. It was the first time I heard about the non-Abelian Stokes theorem, so I checked and found this about it. But it seems this should be applied only when we're trying to integrate a e.g. matrix valued quantity or any other quantity that can be given a non-Abelian group structure. I understand this when we're dealing with non-Abelian gauge fields but does it apply to Ricci tensor or tensors on ## \mathbb R^n ## in general? Or...maybe the tensor structure is itself a non-Abelian group structure? or what?
Thanks

Ben Niehoff
Gold Member

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.

ShayanJ
Gold Member
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.
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!

Ben Niehoff