Let $M$ be a smooth Riemannian manifold with riemannian measure $\mu$. I don't suppose that $M$ is complete. Can we find a finite or countable disjoint collection of open (or closed) and relatively compact geodesic balls$(B_n)_{n\in\mathbb{N}}$ such that :

$$\mu\left(M\setminus\bigsqcup_{n\in\mathbb{N}}B_n\right) = 0$$ ? We might assume that $M$ has a bounded curvature. The $M$ I'm interested in is the intersection of a (complete) submanifold of $\mathbb{R}^d$ with a ball. The compactness restriction ensures that each ball is a "real" ball, and hasn't been cropped by the "edge" of $M$.

  • $\begingroup$ It seems to me we should be able to actually cover $M$ by a countable collection of closed (compact) balls. $\endgroup$ – Kevin Casto Apr 4 at 21:14
  • $\begingroup$ I forgot the disjoint hypothesis in my question.. $\endgroup$ – Pii_jhi Apr 5 at 8:16
  • 1
    $\begingroup$ In the case of $E^n$'s, such packings do exist. For instance, Apollonian packing does the job for the Euclidean plane. Not sure what happens if one requires zero Hausdorff dimension of the residual set. $\endgroup$ – Moishe Kohan Apr 5 at 8:49
  • $\begingroup$ The residual set will contain the boundary of the ball (if we consider open balls), so it will be of Hausdorff dimension $d-1$ at least and I think it has to be strictly greater than $d-1$. In the case of Apollonian packing, which feels close to an"optimal" packing, has Hausdorff dimension $\simeq 1.3$.. $\endgroup$ – Pii_jhi Apr 5 at 10:26
  • $\begingroup$ @Pii_jhi: Of course, I meant closed balls, as in your question. In all packing constructions I know, the residual set has positive H.D. $\endgroup$ – Moishe Kohan Apr 7 at 4:45

I believe that this is true. If you look at Lemma 1.10. Of Introduction to Smooth Manifolds By John Lee,

Lemma 1.10. Every topological manifold has a countable basis of precompact coordinate balls.

If you follow the proof using an open cover of $M$ by Riemannian normal coordinate charts, you should end up with a countable basis of precompact geodesic balls of $M$ so that the union of all such balls is equal to $M$. Bounded curvature and completeness are not necessary for the proof.

I'm not sure what you mean by the last two sentences, but the fact that $M$ is topologically embedded into $\mathbb{R}^d$ should imply that the intersection of $M$ with a ball in $\mathbb{R}^d$ is an open set in $M$.

| cite | improve this answer | |
  • $\begingroup$ Thank you for your answer. In fact I forgot the disjointness hypothesis in my question, which makes it less trivial. I believe it is true on R^n and belong to the family of Vitali covering theorem. $\endgroup$ – Pii_jhi Apr 5 at 8:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.