Let $(M,g)$ be an open (:=complete, non-compact) Riemannian manifold with bounded geometry, in the sense that in some atlas of charts of radius $r_0>0$, the metric and all its derivatives are uniformly bounded (see, e.g. Cheeger--Gromov--Taylor). For simplicity, we only consider $\partial M=\emptyset$. My question is:

$\bullet$ For some $p\in M$ and a large number $R>0$, can we deduce that the geodesic ball $B(p,R)\subset M$ is a manifold-with-boundary of bounded geometry?

The boundedness of geometry for a manifold-with-boundary $\mathscr{N}$ is defined in the sense of T. Schick (https://arxiv.org/abs/math/0001108). It means that (1), the interior of $\mathscr{N}$ is of bounded geometry in the aforementioned sense; (2), $\partial \mathscr{N}$ can be flowed for a positive definite time along the inward unit normal; and (3), the second fundamental form of $\partial \mathscr{N}$ and all its derivatives are uniformly bounded, and the injectivity radius of $\partial \mathscr{N} \geq \iota_0 >0$.

I think this should be true, but I find it difficult to write down a prove. In particular, technicality arises if the geodesic sphere $\partial B(p,R)$ goes beyond the conjugate points of $p$.

A more general (but vague) question is:

$\bullet$ Let $p,R,M$ be as in the previous question. What can be said about the geometry of $\partial B(p,R)$?

Many Thanks!

  • $\begingroup$ At least one cannot expect $\partial B_R(p)$ be smooth. For example, let $M=R\times S^n$ be a cylinder, then for all big $R$, the sphere $ \partial B_R(p)$ has one nonsmooth point. $\endgroup$ – Yu Ding Mar 24 at 10:34
  • $\begingroup$ Thank you --- I agree with this. Would there be any positive results at all? $\endgroup$ – Siran Victor Li Mar 25 at 3:04
  • $\begingroup$ In this direction one might want to read Cheeger-Gromov's paper "Chopping Riemannian manifolds"... $\endgroup$ – Yu Ding Mar 25 at 3:10
  • $\begingroup$ Thank you for the suggestion! $\endgroup$ – Siran Victor Li Mar 29 at 3:00

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