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Let $(\mathbb{R}^2,g)$ be a complete Riemannian manifold. Let $K\subset \mathbb{R}^2$ be a compact, connected set, and let $\text{conv}(K)$ be its convex hull, i.e., the intersection of all geodesically-convex sets containing $K$.
Is $\text{conv}(K)$ bounded? Compact?
If it helps, I am interested in cases where the curvature of $g$ has both signs, but is flat outside of $K$.
The convex hull of a compact set $K \subset M^2$ in a complete manifold need neither be bounded, nor closed.
Both counterexamples are rotationally symmetric, and the second has a Euclidean metric outside of a compact region.
To prove that a convex hull need not be bounded, consider a sphere with an infinitely long and thin spike attached at the north pole. (This is diffeomorphic to $\mathbf{R}^2$, and can be done with a complete metric.) For our compact set $K$ we take a loop $\gamma$ enclosing the spike, say at height $z = 10$. Any convex set containing $K$ also contains the strip $\{ 10 \leq z < 10 + \epsilon \}$. However, no region of the form $\{ 10 \leq z < h \}$ with $h > 10$ is convex. Therefore any convex set containing $K$ must also contain the whole spike, making it unbounded.
To find an example where $\operatorname{conv} K$ is not closed we construct a rotationally symmetric metric on $M = \mathbf{R}^2 \subset \mathbf{R}^3$ by gluing in a 'mushroom' with a thin waist $\gamma_1$ at height $z = 1$, of radius one for example, above which there lies a larger sphere, say of radius five. Outside of the unit disc $D \subset \mathbf{R}^2$, the Euclidean metric is unchanged.
Claim. The convex hull of $K = \{ z = 1/2 \}$ is $\{ 1/2 \leq z < 1 \}$.
Proof. Let $C$ be the convex hull of the set. This contains $\{ 1/2 \leq z < 1/2 + \epsilon \}$ for some small $\epsilon > 0$. None of the sets $\{ 1/2 \leq z < h \}$ with $h < 1$ is geodesically convex, so $\{ 1/2 \leq z < 1 \} \subset C$.
To show that $\{ 1/2 \leq z < 1 \}$ is geodesically convex, take two points $x,y$ lying in it, and let $\gamma$ be a minimizing geodesic connecting the two. This has length at most $4 \pi$, say. By preservation of angular momentum—Clairaut's relation—, if $\gamma$ crossed the waist, then it would cross into the large sphere, thus making it longer than allowed.
Still by Clairaut's relation, $\gamma$ has angular momentum strictly larger than that of the waist $\gamma_1$. (It cannot have angular momentum equal to it, because that would make it an unbounded geodesic converging to $\gamma_1$.) Therefore $\max z(\gamma) < 1$. Q.E.D.
