Here's a question I was wondering about this week. Not sure how interesting it is, but I thought it was kind of curious.

**Question:** Given $k$, is there a number $N=N(k)$ such that if a closed orientable hyperbolic surface X is the union of at most $k$ embedded metric balls, then the genus of $X$ is at most $N$?

(Here, an embedded metric ball is an pathwise isometric embedding of a ball in $\mathbb H^2$.)

**Some half baked comments:**

It's well known that any $X$, of any genus, can be covered by $3$ *topological* balls. You can convince yourself of this pretty easily: draw a couple embedded arcs that cut $X$ into disks, thicken them to get the first two balls, and then make the third ball by taking all the complementary disks and connecting them together in the pattern of some sort of dual spanning tree. So, the metric content above is essential.

If you make a graph G by taking a vertex for every ball and an edge for every connected component of intersection of two balls, then $\pi_1 G $ will surject on $\pi_1 X$. So, if you could bound the number of connected components of the intersection of each pair of balls, you can bound the genus. However, I think two balls in a hyperbolic surface of genus g can have something on the order of g connected components, which isn't helpful. For an example, take a right angled regular polygon and double it along every other side. This gives a surface with boundary, and then you can glue on whatever to make it closed. If you take the largest embedded balls around the two copies of the center of the polygon, they'll have a component of intersection for every edge of the polygon.

Another reason you're never going to get a universal bound on the number of connected components of intersections is that this would give you that $N(k)$ is linear in $k$, which I think is impossible. Namely, there are some examples out there of genus g surfaces where the injectivity radius is everywhere at least $a \log g$, for some fixed constant a>0.

(See e.g. https://arxiv.org/pdf/math/0505007.pdf, where they give examples with $a=2/3$, although their examples may have cusps, I haven't checked. There are other closed examples too though.)

You can then construct a maximal set $S $ of points in $X$ such that the distance between any two points is at least $a \log g$. The balls of radius $a/2 \log g$ around the points of S are disjoint and have volume at least $g^{a/2}/100$ or something, so there are at most $1000 g^{1-a/2}$ points in $S$, by Gauss Bonnet. By maximality of S the radius $a \log g$ balls around the points of S cover X. They are embedded, but there are only around $g^{1-a/2}$ of them. That is, you can cover a genus $g$ surface with a number of balls that's sublinear in $g$.