On a hyperbolic surface, is a compact set $K$ necessarily contained in a larger compact set $K’$ so that any two points of $K$ can be joined by a minimising geodesic within $K’$?
The question came out when I was studying the Pick Theorem in the book Dynamics in One Complex Variable by John Milnor (Page22).
At the end of the proof, it seems that the answer of my question is obviously yes so that one can choose such a larger compact set $K’$ to conclude the proof. But I am just unsure about the existence of such a larger compact set $K’$.
Here’s my attempt. Let $G$ be the union of all geodesics of $K$, that is, $x$ is in $G$ if and only if $x$ is on a minimising geodesic joining some two points of $K$. If we can prove $G$ to be bounded(I still cannot work this out), then we can take the closed ball containing $G$ as the wanted $K’$.
Any help and discussion will be sincerely appreciated.
Here are two relevant notions of convexity for complete Riemannian manifolds $M$:
a. A subset $C\subset M$ is weakly convex if for all pairs of points $x, y\in C$ there exists a geodesic $c$ in $M$ connecting $x, y$, such that $c\subset C$.
b. A subset $C\subset M$ is strongly convex if for all pairs of points $x, y\in C$ there exists a minimizing geodesic $c$ in $M$ connecting $x, y$, such that $c\subset C$.
(The terminology "weakly" and "strongly" convex is mine, as there is no consistent terminology in the literature.)
Lemma 1. Suppose that $M$ is a complete (connected) hyperbolic surface. Then for every compact $K\subset M$ there exists a weakly convex compact subset $C\subset M$ containing $K$.
Proof. There exists a compact subset $\tilde{K}$ of the hyperbolic plane ${\mathbb H}^2$ (the universal covering space of $M$) which projects onto $K$ under the universal covering map $\pi: {\mathbb H}^2\to M$. Since $\tilde{K}$ is compact, there exists a closed metric ball $B\subset {\mathbb H}^2$ containing $\tilde{K}$. Then $C=\pi(B)$ contains $K$. Let us check that $C$ is weakly convex. Take any pair of points $x, y\in C$. Since $\pi(B)=C$, there are points $p, q\in B$ such that $\pi(p)=x, \pi(q)=y$. Since $B$ is convex, the hyperbolic geodesic $pq$ connecting $p, q$, is contained in $B$. Thus, $\pi(pq)$ is a geodesic in $M$ connecting $x, y$ and contained in $C$. (To be more precise, one should work with geodesic paths rather than images of geodesics.) Hence, $C$ is weakly convex. qed
Lemma 2. There are complete hyperbolic surfaces $M$ and compacts $K\subset M$ such that $K$ is not contained in any strongly convex compact subset of $M$.
Proof. Let $M$ be the quotient of the hyperbolic upper half-plane by the cyclic group $\Gamma$ of isometries generated by $z\mapsto z+1$. The group $\Gamma$ preserves the horocycle $H\subset {\mathbb H}^2$ given by the equation $Im(z)=1$. The projection of $H$ to the surface $M$ is a smooth circle $K\subset M$ (actually, also a horocycle in $M$). The subset $K$ is, of course, compact. Suppose that $C\subset M$ is the smallest closed strongly convex subset containing $K$. Then the preimage $\tilde{C}$ of $C$ in ${\mathbb H}^2$ is also convex. One verifies that $\tilde{C}$ has to be invariant under the group $G$ of all translations $z\mapsto z+t$, $t\in {\mathbb R}$ (since $H$ is $G$-invariant). Thus, $\tilde{C}$ has to be of the form $U_{b}$, $$ \{z: 1\le Im(z)\le b\} $$ or $$ U=\{z: 1\le Im(z)\}. $$ In the former case, $U_{b}$ is not convex. In the latter case, $\pi(U)\subset M$ is noncompact. qed