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There is a family of comparison theorems in Riemannian geometry (Rauch, Günther-Bishop, Gromov, Toponogov-Cheng) that all rely on two hypotheses: some boundedness of the sectional or Ricci curvature, and completeness.

All the places that I have seen assume completeness, while only Chavel ("Riemannian Geometry: A Modern Introduction", 2nd ed., page 128) briefly says "for convenience, we assume that $M$ is complete". Since I am working with manifolds not necessarily complete, could anyone please firmly clarify whether completeness is necessary, or just convenient? I wouldn't want to base my work on flawed results. (I am interested mostly in the Bishop-Gromov volume comparison theorem.)

Thank you.

** Later edit **

After further reflection, I realize that I may have misunderstood Chavel's words: I have interpreted them as "completeness is convenient for the proof, but we can drop it and not lose too much", but now I realize that it might also mean "completeness is convenient for the proof, but we can replace it with other assumptions and still get the results, but through uglier proofs".

As such, I would like to ask whether requiring bounded Ricci curvature (both above and below) would still allow us to get the aforementioned theorems, in particular Bishop-Gromov.

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  • $\begingroup$ In my very rough "counter-example" below, Ricci is bounded from below and also from above. Time permitting, I will elaborate on it. $\endgroup$ – Raziel Sep 16 '15 at 19:56
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It is best to read the proofs and see where and how completeness is needed. Some local aspects remain true for noncomplete manifolds. Global ones often fail.

For example, Bishop-Gromov volume comparison immediately implies that the volume growth of any complete open manifold of nonnegative Ricci curvature has is at least linear. Clearly if you remove a large closed subset from the manifold the volume growth may change while the curvature bound will not.

It is instructive to see how the basic Sturm comparison theorem for ODE on the line (which underlines the Rauch comparison theorem) fails when the domain is disconnected. A local differential inequality propagates along a connected interval to a certain cumulative effect in the long run. There is no accumulation if instead you look at a sequence of intervals. They are independent.

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As far as I can understand, you need completeness to have existence of minimizers, and thus a 1:1 parametrization of balls through the exponential map (up to a set of zero measure).

I think that you can build examples of manifolds with $Ricci \geq K >0$, non-complete and with infinite diameter (this is a "counter-example" to the Bonnet-Myers theorem without completeness assumption).

Think at a seashell-like surface in $\mathbb{R}^3$. Something like this picture:

enter image description here

You must arrange the arms of the "spiral" to have roughly the same curvature.

Playing with the "rate of spiraling" should give you a surface with $Ricci \geq K >0$, infinite diameter and possibly infinite volume.

This also gives a counter-example to Bishop-Gromov without completeness assumption.

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