As is well known that the three sphere $S^3$ may be viewed as a $S^1$ bundle over base space $S^2$. And it is more interesting, but likewise confusing to me that the metric on $S^3$ may be written as:

\begin{equation} ds^2 = (d\psi - \cos \theta\ d\phi)^2 + d\theta^2 + \sin^2 \theta\ d\phi^2 \end{equation}

here $\cos \theta\ d\phi$ is exactly the Ricci form connection on the base space $S^2$. It seems here that the non-triviality of the fibre bundle can be seen from the metric, constrasting to the idea that the bundle is always locally trivial.

Thus my question is (1) given a specific bundle how to construct the metric on the total space, and how the connection of the base might come into play? (2) given a specific metric, how could we tell if the underlying manifold exhibits a bundle structure? It would be very illustrative if examples may be given in the meantime (for example the Hopf fibration above).

It would also be great if reference on the related topic can be provided. I have been searching online, but when come to bundles most reference tends to fall into topological discussions.

  • $\begingroup$ I actually doubt this can be done. If there is any method of seeing the non-triviality from the metric, one has to also see the triviality of $ds^2=(d\psi -\cosh\rho d\phi)^2+d\rho^2+\sinh^2\rho d\phi^2$, but this is far from being clear. $\endgroup$ – user110373 Aug 3 '17 at 19:43

No metric related the total space since we consider only configuration of points and line as property of projective geometry.

  • $\begingroup$ You've chosen to respond to an older Question (nearly three years old), so brevity and haste are of no merit. At best you've claimed that neither of the problems posed in the Question can be answered, but it is not a convincing demonstration for either part. $\endgroup$ – hardmath Aug 22 '17 at 21:51

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.