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In the absence of a metric, it is not clear to me to what extent does knowing the curvature tensor determine its associated connection? I would be satisfied knowing this for zero torsion. I'd like to know the answer for manifolds in 4 or more dimensions.

I've seen several related questions but have not found a clear answer to this.

MidwestGeek
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  • It could only possibly determine the connection up to an automorphism of the tangent bundle, since changing by such an automorphism preserves the curvature. Even then, I doubt you can get anything, even locally. –  Feb 12 '19 at 15:39
  • Isn't an automorphism equivalent to a change of basis or, if holonomic, a change of coordinate frame? Of course, I meant given the curvature tensor modulo changes such as these, which I would regard as trivial invariances of the tangent bundle. – MidwestGeek Feb 12 '19 at 17:23
  • Yes, these are essentially uninteresting. But it wasn't clear from your post that you knew about them. :) –  Feb 12 '19 at 17:27
  • Thanks for responding. So you think the connection is essentially uniquely determined by the curvature? Isn't that rather different from other gauge symmetries, such as for Lie groups, where the curvature F^a{\mu\nu}_ does not determine the gauge potential A^a\mu_? Maybe I am confused about fiber bundles? – MidwestGeek Feb 12 '19 at 23:43
  • That is not what I said. Here are some more details. I do not have time to write a good answer now, but have bookmarked your question and hope to later (though this shouldn't stop anybody else from doing so!) 1) This is precisely a special case of the notion of connection on a principal bundle; the story here will be no different from the general case. 2) The largest problem with curvature not determining the potential (aka, connection) is the gauge group, in both cases. 3) Globally, this is not the only obstruction: if $A$ is a connection on a principal bundle over a closed manifold $M$... –  Feb 13 '19 at 00:00
  • then the expected dimension of the space of connections $A'$ with $F_{A'} = F_A$, after quotienting by gauge transformations, is $\dim \mathcal H^1_A(M)$ - the space of harmonic ($\mathfrak g$-valued) 1-forms on $M$, with respect to the connection $A$. Certainly if $M$ has interesting topology this may be nontrivial; flat connections are an easy example, and the space of flat connections modulo gauge is an interesting object. However, I am less knowledgeable about the local story, and it could be (though I still feel skpetical) that... –  Feb 13 '19 at 00:01
  • if $A$ is close to $A'$ and $F_A = F_{A'}$, then $A$ is locally gauge equivalent to $A'$. This is true for flat connections, but I make nooooo promises about the general case. –  Feb 13 '19 at 00:01
  • Thanks for taking time to respond. I roughly understand what you are driving at. The dependence on global topology is interesting, but at the moment, I am focused on local requirements. My question is closely related to a previous one by @TheQuantumMan (https://math.stackexchange.com/questions/2500089/since-the-curvature-tensor-depends-on-a-connectionnot-metric-is-it-the-releva) Since the definition of the covariant derivative may be shifted by an arbitrary tensor, $\tilde{\nabla}-\nabla=\tilde{\Gamma}-\Gamma=C,$ one might think that the curvature would inherit such an invariance as well. – MidwestGeek Feb 13 '19 at 21:49
  • Evidently, it does not. Since the difference depends on $C,$ $\tilde{R}-R=\nabla C+C\cdot C,$ for a fixed section (i.e., choice of frame.) If I require $\tilde{R}=R,$ is there any residual freedom in the choice of $C?$ – MidwestGeek Feb 13 '19 at 21:50

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