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Does anyone know what is the relation between universal covering space and curvature of base manifolds?

Edit: For example If universal covering of a complete $3$-manifold $(M,g)$ isometric to a Riemann product $N^2\times \Bbb R$ where $N^2$ is a complete $2$-manifold with non-negative sectional curvature then what we can say about $(M,g)$?

Thanks

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    $\begingroup$ This is a rather broad and vague question. There is no single answer. Can you make your question more specific? $\endgroup$ – Lee Mosher Oct 4 '17 at 13:22
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    $\begingroup$ In the 3-manifold case, we know what the possible geometries are (on the universal cover). See homepages.warwick.ac.uk/~masgar/Teach/2012_MA4J2/geometry.pdf $\endgroup$ – Steve D Oct 4 '17 at 17:13
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The universal covering space is a topological construction that makes no reference to smooth or Riemannian structures. If the base is a smooth manifold, then the universal covering space has a natural smooth structure defined by pulling back the smooth structure on the base by the universal covering map, and by construction the universal covering map becomes a local diffeomorphism.

If the base is a Riemannian manifold, then the pullback of the metric by the universal covering map is again a Riemannian metric, making the universal covering a simply connected Riemannian manifold, the universal covering map a local isometry, and the deck group acts by isometries.

If the base has a Riemannian metric with constant sectional curvature, then the universal covering space is isometric to one of the constant-curvature space forms and the base can be realized as the quotient of a space form by a representation of its fundamental group in the isometry group of the space form. This is very useful in low dimensions for translating topological questions into a form amenable to analysis by geometric tools.

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