# geodesic balls in Riemannian manifolds with bounded geometry

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!

• 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. – Yu Ding Mar 24 at 10:34
• Thank you --- I agree with this. Would there be any positive results at all? – Siran Victor Li Mar 25 at 3:04
• In this direction one might want to read Cheeger-Gromov's paper "Chopping Riemannian manifolds"... – Yu Ding Mar 25 at 3:10
• Thank you for the suggestion! – Siran Victor Li Mar 29 at 3:00