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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.

Kevin Ye
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  • 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. – user110373 Aug 03 '17 at 19:43
  • I've love to see a full answer, but I think I can see how this is $\sigma_1^2+\sigma_2^2+\sigma_3^2$, where those are the left-invariant forms on the 3-sphere. – levitopher May 22 '19 at 20:45

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