# Packing a Riemannian manifold with disjoints balls

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. I'm interested in the case when $$M$$ 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 $$M$$.

• It seems to me we should be able to actually cover $M$ by a countable collection of closed (compact) balls. Apr 4, 2020 at 21:14
• I forgot the disjoint hypothesis in my question.. Apr 5, 2020 at 8:16
• 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. Apr 5, 2020 at 8:49
• 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 $\simeq 1.3$.. Apr 5, 2020 at 10:26
• @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. Apr 7, 2020 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 $$M$$.

• 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. Apr 5, 2020 at 8:18

For any smooth Riemannian manifold $$M$$ there is a countable disjoint union of balls with complement of measure $$0$$.

First of all, for each $$p\in M$$ let $$B_p$$ be a precompact normal ball centered at $$p$$ so small that:

1. $$\mu(\partial B_p)=0$$ (we can do this because $$\{r>0;\mu(\partial B(p,r))>0\}$$ is countable).
2. The injectivity radius of points in $$B_p$$ is bounded below by some constant $$\varepsilon>0$$. We can prove that a small enough ball achieves this by changing the metric of $$M$$ far from $$p$$ so that $$M$$ becomes complete, and then using that in a complete Riemannian manifold the injectivity radius is continuous.
3. Sectional curvatures of points of $$B_p$$ are in some compact interval $$[a,b]$$.

Claim 1: If $$\delta>0$$ is small enough, then for any $$q\in B_p$$ we have $$\frac{\mu(B(q,\delta))}{\mu(B(q,2\delta))}>2^{-n-1}$$, where $$n$$ is the dimension of $$M$$.

Proof: This constant $$\delta$$ will be $$<\frac{\varepsilon}{2}$$, and to prove it exists, first note that due to Theorem 3.23 in [1] we have $$\frac{\mu(B(q,\delta))}{\mu(B(q,2\delta))}\geq\frac{V_b(\delta)}{V_a(2\delta)}$$, where $$V_\kappa(r)$$ is the volume of the ball of radius $$r$$ in the $$n$$-dimensional space of constant sectional curvature $$\kappa$$. Note that for any $$x>0$$ we have $$V_\kappa(xr)=x^nV_{\kappa x}(r)$$, because the space of constant curvature $$r$$ is obtained from multiplying the metric of the space of constant curvature $$xr$$ by $$x^2$$. So when $$\delta\to0$$, we have $$\frac{V_b(\delta)}{V_a(2\delta)}=\frac{V_{\delta b}(1)}{V_{\delta a}(2)}\to\frac{V_0(1)}{V_0(2)}=2^{-n}$$, so indeed for small $$\delta$$ we have $$\frac{\mu(B(q,\delta))}{\mu(B(q,2\delta))}\geq\frac{V_b(\delta)}{V_a(2\delta)}>2^{-n-1}$$. $$\square$$

Claim 2: For any open subset $$A$$ of $$B_p$$ there is a finite set of disjoint balls $$B_1,\dots,B_m$$ contained in $$A$$ such that $$\mu(\cup_{i=1}^m B_i)>\frac{1}{2^{n+2}}\mu(A)$$.

Proof: Let $$\delta$$ be so small that it satisfies the previous claim and such that, if $$B:=\{x\in A;d(x,M\setminus A)>\delta\}$$, then $$\mu(B)>\frac{1}{2}\mu(A)$$. Now consider a maximal $$2\delta$$-separated set $$\{x_1,\dots,x_m\}$$ in $$B$$, and let $$B_i:=B(x_i,\delta)$$. These balls are disjoint, and the balls $$B(x_i,2\delta)$$ cover $$B$$, so $$\sum_i\mu(B_i)>\frac{1}{2^{n+1}}\sum_i\mu(B(x_i,2\delta))\geq\frac{1}{2^{n+1}}\mu(B)\geq\frac{1}{2^{n+2}}\mu(A)$$. $$\square$$

We can also ensure that the boundaries of the balls $$B_i$$ of claim $$2$$ have measure $$0$$: if not, note that for each $$q\in M$$, the set $$\{r>0;\mu(\partial B(q,r))>0\}$$ is countable, so we can reduce the radii of the balls just a little bit so that the sum of their volumes is still $$>\frac{1}{2^{n+2}}\mu(A)$$.

Claim 3: We can cover any open set $$X\subseteq B_p$$ up to measure $$0$$ by a disjoint collection of balls contained in $$X$$.

Proof: Take $$A=X$$ in the previous claim, and find balls $$B_{0,1},\dots,B_{0,n_0}$$ with boundary of measure $$0$$ such that $$\mu(\cup_{i=1}^m B_i)>\frac{1}{2^{n+2}}\mu(X)$$. Now let $$X_1=X\setminus\cup_i\overline{B_{0,i}}$$, so that $$\mu(X_1)\leq(1-\frac{1}{2^{n+2}})\mu(X)$$. Applying the same to $$X_1$$ we can remove from it finitely many balls $$B_{1,1},\dots,B_{1,n_1}$$ to obtain some open $$X_2$$ with $$\mu(X_2)\leq(1-\frac{1}{2^{n+2}})\mu(X_1)$$. Repeating this step to obtain spaces $$X_n$$ for each $$n$$, we get that the balls $$\{B_{i,j}\}_{i\in\mathbb{N};j=1,\dots,n_i}$$ are pairwise disjoint, and $$\mu(X\setminus\bigcup_{i,j}B_{i,j})=\lim_{m\to\infty}\mu(X_m)\leq \lim_{m\to\infty}(1-\frac{1}{2^{n+2}})^m\mu(X)=0$$. $$\square$$

Claim 4: We can cover $$M$$ up to measure $$0$$ using a countable collection of disjoint compact balls.

Proof: Consider the collection of balls $$\mathcal{B}:=\{B_p;p\in M\}$$. As $$M$$ is second countable, we can find a countable subcover of $$\mathcal{B}$$, $$(B_n)_{n\in\mathbb{N}}$$. Moreover, for each $$n$$, we can cover $$B_n\setminus\bigcup_{i=1}^{n-1}\overline{B_i}$$ up to measure $$0$$ with a countable collection of disjoint compact balls. The union of these countable collections of balls covers $$B_n$$ up to measure $$0$$ for all $$n$$, thus it covers all $$X$$ up to measure $$0$$. $$\square$$

[1] Cornelia Druţu, Michael Kapovich, $$\textit{Geometric Group Theory}$$, Colloquium Publications. Volume: 63; 2018.