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.

  • $\begingroup$ 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. $\endgroup$ – user98602 Feb 12 at 15:39
  • $\begingroup$ 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. $\endgroup$ – MidwestGeek Feb 12 at 17:23
  • $\begingroup$ Yes, these are essentially uninteresting. But it wasn't clear from your post that you knew about them. :) $\endgroup$ – user98602 Feb 12 at 17:27
  • $\begingroup$ 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? $\endgroup$ – MidwestGeek Feb 12 at 23:43
  • $\begingroup$ 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 $M$... $\endgroup$ – user98602 Feb 13 at 0:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.