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

1. May 31, 2016

### ShayanJ

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

2. Jun 1, 2016

### Ben Niehoff

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. Jun 1, 2016

### ShayanJ

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!

4. Jun 2, 2016

### Ben Niehoff

I've never heard of any practical use for the non-Abelian Stokes theorem in physics at all. Perhaps you could share the context in which you saw it?

5. Jun 2, 2016

### ShayanJ

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.