Can you cover a genus a billion hyperbolic surface with 15 balls?

Question

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 $g$.

Answer 1

Your conjecture is false. Every nonorientable closed connected surface of negative Euler characteristic, admits a hyperbolic metric such that the surface is covered by 3 embedded disks. Hence, for each $p\ge 2$, there is a closed connected orientable hyperbolic genus $p$ surface covered by 6 embedded disks. (One can likely reduce this number to 3 as well.) Here is a construction. Start with a regular hyperbolic $2n$-gon $P$ with $n\ge 4$ and the vertex angle $2\pi/n$. (I assume that $P$ is closed and 2-dimensional.) Let $R$ denote the radius of the inscribed circle in $P$ and $a$ the common length of the sides of $P$.

Denote its vertices $z_kw_k$, indexed cyclically by $k\in {\mathbb Z}_n$. Then there is a collection of hyperbolic isometries $g_k$ sending (the oriented segment) $z_kw_k$ to $z_{k+1}w_{k+1}$ and, respectively, $h_k$ sending $w_kz_{k+1}$ to $w_{k+1}z_{k+2}$ and mapping the interior of $P$ to the exterior of $P$. Note that under the identification via these isometries, there are two equivalence classes of vertices of $P$. I assume, you are familiar with Poincare's Fundamental Domain Theorem: It shows that the isometries $g_1,...,g_n, h_1,...,h_n$ generate a discrete torsion-free isometry group of the hyperbolic plane with fundamental domain $P$ and quotient surface $X$ homeomorphic to $N_p$ (the nonorientable surface of genus $p$), $p=n-1$. Just in case, my favorite reference is

Beardon, Alan F., The geometry of discrete groups, Graduate Texts in Mathematics, 91. New York - Heidelberg - Berlin: Springer-Verlag. XII, 337 p. DM 108.00; $ 44.60 (1983). ZBL0528.30001.

Next, take the (open) inscribed hyperbolic disk $B$ in $P$ (of the radius $R$), centered at the center $o$ of $P$. Furthermore, take closed hyperbolic disks $A_k, C_k, k=1,...,n$ centered at, respectively, the vertices $w_k, z_k$ of $P$ and whose radii equal $a/2$. All these $2k+1$ disks completely cover $P$ and the $A$-disks are pairwise disjoint and $C$-disks are pairwise disjoint. Intersect these disks with $P$. (I will retain the names $A_k, C_k$ for the intersections. A picture would be great here but would take too much effort to draw.)

The unions $A_1\cup...\cup A_n$, $C_1\cup...\cup C_n$ project to two isometrically embedded closed disks $A, C$ in $X$; I will denote the projection of $B$ again by $B$; it is again isometrically embedded since I chose $B$ to be open. Then $A\cup B\cup C=X$. Thus, we got two closed and one open disk in $X$ covering $X$. In order to get three open disks with the same properties, expand slightly $A$ and $C$ and take the interiors of the expanded disks.

Question: Is there a function $V(n,m)$ such that for every compact hyperbolic $n$-manifold $X$, $n\ge 3$, covered by $m$ embedded hyperbolic metric disks, the volume of $M$ is $\le V(n,m)$?

Answer 2

Consider a regular hyperbolic $4n+2$-gon with vertex angle of genus $3$. I claim we can glue $3$ copies of this polygon together to form an oriented hyperbolic surface of genus $n$, without gluing any polygon to itself.

This will form a decomomposition of the surface with $4n+2$ vertices, $6n+3$ edges, and $3$ faces, for an Euler characteristic of $4n+2 + 3 - (6n + 3 ) = 2-2n$.

I will describe it by first describing the vertices, then the edges, then the cyclic ordering of the edges at each vertex, and finally the faces.

The vertices are indexed by the numbers $0$ to $4n+1$, mod $4n+2$. There are edges from vertex $i$ to vertices $i+1, i-1, i+2n+1$. Thus each vertex has three edges.

The three edges leaving $i$ towards vertices $i+1, i-1, i+2n+1$ are in clockwise order.

One face has boundary vertices $0,1,2,3,\dots, 4n+1$ in order (clockwise around the face. Another has boundary the vertices $$ 2n+1 , 2n , 4n+1, 4n, 2n-1, 2n-2, 4n-1, 4n-2, 2n-3,2n-4, \dots, 1,0$$ in order (clockwise around the face).

The last one has boundary the vertices $$ 4n+1, 2n, 2n-1, 4n, 4n-1 , 2n-2, 2n-3, \dots, 2n+1,0$$ in order (clockwise around the face).

Each edge meets two of the three faces and each vertex meets all three faces. So indeed the polygons glue to a smooth hyperbolic surface.

Now take a disc whose center is the center of each polygon, whose radius is at least the distance from the center to a vertex but less than the distance from a center to a vertex plus half the length of an edge). The disc will contain the polygon, hence three will cover the surface, but each disc will only extend a short distance from each polygon into the next and thus will not intersect itself.