# Convex hulls of compact sets in a 2-manifold

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 $$K$$.

• I think $conv(K)$ need not be compact in general, for example let $g_0$ be the usual Euclidean metric and let $g(x,y)=f(\sqrt{x^2+y^2})*g_0(x,y)$, where $f$ is some smooth decreasing function which decreases very fast in $[1,2]$ and is constantly $1$ in $[3,\infty]$. Then it seems to me that the convex closure of the compact ball $\overline{B(0,1)}$ should be an open ball $B(0,r)$ for some $r$ between $2$ and $3$. But it is not obvious and I can't think about the details until the weekend Jan 26, 2023 at 4:00

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.

• Thanks a lot! Regarding the boundedness, how do you know that 𝑐𝑜𝑛𝑣(𝐾) is contained in the set you defined? The set of all minimizing geodesics between points in 𝐾 is not necessarily 𝑐𝑜𝑛𝑣(𝐾), which might be larger...
– C M
Jan 26, 2023 at 11:28
• @CM You're right - I was a bit careless there. I'll think about it some more. Jan 26, 2023 at 11:29
• @CM I've now added an example to show that the convex hull need not be bounded either. Jan 26, 2023 at 13:59
• Thanks! Excellent examples :) I am not sure how to prove large balls are convex in the non-radially-symmetric case (when the manifold is eventually flat), but I think I can prove boundedness in this case another way.
– C M
Jan 26, 2023 at 18:27
• @CM I meant metrics that are Euclidean outside of a ball, but the wording is a clumsy. I'll make sure to fix that. I'm not sure whether it's true if $M$ is just flat; I'd have to think about it some more. Jan 26, 2023 at 18:53