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 $K$.

  • $\begingroup$ 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 $\endgroup$
    – Saúl RM
    Jan 26, 2023 at 4:00

1 Answer 1


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.

  • $\begingroup$ 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... $\endgroup$
    – C M
    Jan 26, 2023 at 11:28
  • $\begingroup$ @CM You're right - I was a bit careless there. I'll think about it some more. $\endgroup$
    – Leo Moos
    Jan 26, 2023 at 11:29
  • $\begingroup$ @CM I've now added an example to show that the convex hull need not be bounded either. $\endgroup$
    – Leo Moos
    Jan 26, 2023 at 13:59
  • $\begingroup$ 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. $\endgroup$
    – C M
    Jan 26, 2023 at 18:27
  • $\begingroup$ @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. $\endgroup$
    – Leo Moos
    Jan 26, 2023 at 18:53

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