# Are Metric Balls on a Riemannian Manifold Embedded Submanifolds?

Let $$(M,g)$$ be a connected $$n$$-dimensional Riemannian manifold. We can define a topological metric on $$M$$ by $$d_g(p,q) = \inf\{\text{Length}(\gamma) \mid \gamma \ \text{is a piecewise smooth curve from p to q}\}.$$

Consider the closed ball of radius $$R$$ centred at $$p \in M$$, denoted $$B_R(p)$$. Is this an $$n$$-dimensional embedded submanifold with boundary of $$M$$?

My thoughts:

If $$R$$ is less than the injectivity radius at $$p$$, the answer is yes because the metric ball will coincide with the geodesic ball, which is diffeomorphic to the standard closed ball in $$\mathbb{R}^n$$.

For $$R$$ larger or equal to the injectivity radius at $$p$$ I am not so sure anymore.

I would be grateful for any help!

• You should replace "submanifold" with "submanifold with boundary." In any case, the answer is negative already in dimension 2. I suggest you think about the case when $(M,g)$ is a cylinder. Commented Apr 1 at 20:31
• @MoisheKohan if I understand correctly, in the case of the cylinder (say of radius 1), in Euclidean space it could happen (e.g. if $R=\pi$) that a „rolled up“ disk where two antipodal points are touching is the resulting metric ball? This is clearly not a manifold at all. Commented Apr 1 at 20:51
• Right. You can now write your own answer and accept it. Commented Apr 1 at 20:52

In the case of the cylinder (say of radius 1) in Euclidean space it could happen (e.g. if $$R=\pi$$) that a „rolled up“ disk where two antipodal points are touching is the resulting metric ball. This is clearly not a manifold at all.