8
$\begingroup$

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 $K$.

$\endgroup$
1
  • $\begingroup$ 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

5
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.