1
$\begingroup$

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!

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$
    – Moishe Kohan
    Commented Apr 1 at 20:31
  • $\begingroup$ @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. $\endgroup$
    – Algebro1000
    Commented Apr 1 at 20:51
  • 1
    $\begingroup$ Right. You can now write your own answer and accept it. $\endgroup$
    – Moishe Kohan
    Commented Apr 1 at 20:52

1 Answer 1

Reset to default
2
$\begingroup$

The comment of Moishe Kohan was very enlightening in that he gave a hint towards a counter example:

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .