Ben Niehoff said:
Sorry, I still have no idea what you're trying to get at with your use of "path dependence". I would recommend you stick to standard terminology.
Gauge connections parallel-propagate vectors in an "internal" vector space in exactly the same way that the Levi-Civita connection parallel-propagates vectors in the tangent space. There is no fundamental difference between the two concepts.
I don't believe we are on the same page at all, because you're continuing to say things that are obviously false:
Fair enough, I actually agree with what you explain here so I'm obviously doing a bad job trying to get across my point. Let me try and clarify once more. Also please be a bit indulgent with me here:I have received certain physics instruction long ago but I'm basically trying to express complex things from a layman point of view, and it is not easy at all. I'm here to correct my misunderstandings relying on people that is professionally dedicated to this like may be your case so the last thing I'd want is to appear as argumentative.
Gauge bundles do not have to be trivial, and gauge invariance does not require that the base manifold be flat Minkowski!
You are of course right, actually, for instance, global(spacetime) anomalies in QFT are examples of nontrivializations. But let me give you the context in which I refer to trivial bundles in physics(which I also tried earlier with the reference to the stack.exchange page) and please let me know if I'm misunderstanding it. I'm referring only to QFT, and the standard model of particle physics, and I understand this gauge QFT does use flat minkowski space as base manifold. If this were so(please correct it if this is not so), there is a theorem that says: "Given a fibre bundle with base space X and structure group G, if either X or G is contractible then the bundle is trivial." Since Minkowski space is contractible I understand that the bundle is trivial, but I'm open to be corrected if I'm wrong following this reasoning.
Also, ALL metric-compatible connections (of which the Levi-Civita connection is one) preserve both lengths AND angles. That's what metric-compatible means.
EDITED: I originally wrote something confusing and that deviates from the main discussion.
Yes, the inner product is preserved by the connectio along the specific curve, and depending on the specific path. But only line lengths are preserved independent of curve, position or direction(path).
The distinction I was doing was because in flat space the inner product is preserved independently of path.I was referring to what can be preserved independent of path.
I think an important distinction here between the Riemannian covariant derivative and the gauge covariant derivative is that the connection in the first case is not a tensorial object on the manifold, the Christoffel symbols are not tensors, the curvature is tensorial; while the pricipal connection used in gauge theory is an Ehresmann connection and is constructed as a tensorial object in the bundle(in the total space instead of in the manifold), and in this sense it is frame independent already from the parallel transport moment, this was the difference I was trying to stress when referring to path dependence/independence, is this any clearer?
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Of course a U(1) gauge bundle can have curvature! That's what electromagnetism is!
Yes, what I meant was that in the U(1) case being abelian there is no contribution to the curvature from non-commutativity, as it happens for Yang-Mills fileds. So the curvature is computed only with the term corresponding to the exterior derivative of the connection ##dA##.
Also gauge covariance has absolutely nothing to do with flatness of the base or triviality of the gauge bundle.
Right. I was again referring to the special case that doesn't enter into nontrivialities of the bundle.
But I'd rather wait till I get some feed-back before I attempt to explain again the link I see and how it relates to path dependence/independence in order to avoid saying wrong things if it turns out I'm actually misunderstanding the basics.
Thanks for the assistance.