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Can the convex hull of $3$ points in a Cartan-Hadamard manifold be smooth?

A Cartan-Hadamard manifold $M$ is a complete simply connected manifold with nonpositive curvature (so it includes the Euclidean and hyperbolic spaces as special examples). A subset of $M$ is convex if it contains the geodesic connecting every pair of its points. The convex hull of a set is the intersection of all convex sets which contain it. A convex set is smooth if its boundary is a differentiable hypersurface, i.e., the tangent cones of the boundary are all flat.

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    $\begingroup$ Presumably it’s not smooth at the three points. Any minimal radius ball containing the three points will contain one of the points in the boundary. By varying the center of the ball, one ought to be able to show that the tangent cone at the point is not flat. $\endgroup$
    – Ian Agol
    Commented Nov 24, 2023 at 4:19
  • $\begingroup$ @IanAgol: Can you elaborate more? I am not sure I understand the setup even before you "varying the center of the ball". $\endgroup$ Commented Nov 24, 2023 at 6:01
  • $\begingroup$ @IanAgol I am trying to remember whether in a Hadamard manifold all metric balls are convex. If so, it would be enough to show that there are several metric balls that contain all three points in their boundaries, such that the tangent cone at a boundary point of their intersection does not contain a halfspace. This might settle the question in $\dim\ge 3$, and for $\dim=2$ it seems to be clear. $\endgroup$ Commented Nov 24, 2023 at 8:36

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I believe that the idea described by Ian Agol works, and can be elaborated on as follows. The general fact we want to establish is that the convex hull of a finite collection $X$ of points in $M$ is not smooth. Actually, the so called horro-hull, that is the intersection of all horro-balls which contain $X$ is not smooth. It follows that the convex hull is not smooth, because it lies in the horro-hull.

First let us recall that balls in $M$ are convex, because the distance function from a point in $M$ is convex. Given a ball $B$ in $M$ and a point $x$ on the boundary $\partial B$ of $B$, we can always find a larger ball $B'$ which contains $B$ and intersects $\partial B$ only at $x$. Letting the radius of $B'$ go to infinity yields a horro-ball which contains $X$ and whose boundary passes through $x$.

Let $B$ be the ball of smallest radius containing $X$. Then some point $x$ of $X$ lies on $\partial B$. Let $B'$ be a ball which contains $B$ and intersects $\partial B$ only at $x$. Then all points of $X$ other than $x$ lie in the interior of $B'$. It follows that every point in an open neighborhood of the center of $B'$ is the center of a ball which contains $X$ and whose boundary passes through $x$. Hence the normal cone of the horro-hull of $X$ at $x$ must have interior points, which yields that the tangent cone of the boundary of the horro-hull at $x$ cannot be flat.

Addendum (12/8/2023): More generally, if we have a finite collection of points in $M$ which are convexly independent, i.e., no point is contained in the convex hull of others, then every point is a singularity of the convex hull. Establishing this fact requires a bit more work, because in general there may not exist a sphere which passes through each of the points and contains the other points in the interior of the ball that it bounds.

To prove this, given a point $p$ from the collection $X$ of points, consider the convex hull $C$ of $X-\{p\}$. By assumption $p$ has distance $d>0$ from $C$. Let $C_d$ be the outer parallel body of $C$ at distance $d$. Then the boundary of $C_d$ forms a smooth ($C^{1,1}$) convex surface passing through $p$, while $X-\{p\}$ lies in the interior of $C_d$ (convexity of $C_d$ is due to the fact that the distance function from a convex set in a Cartan-Hadamard manifold is convex, and smoothness follows from the fact that a ball of radius $d$ rolls freely inside $C_d$).

Now note that by perturbing the elements of $X-\{p\}$ we may construct many different smooth convex surfaces which pass through $p$ and contain $X-\{p\}$. All these surfaces contains the convex hull of $X$. Hence as discussed earlier, the tangent cone of the boundary of the convex hull of $X$ at $p$ cannot be flat. Indeed the normal cone of the convex hull at $p$ (which is dual to the tangent cone of the convex hull) has full dimension.

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  • $\begingroup$ It all makes sense until the last sentence which looks unjustified. BTW, where did the proof use finiteness of $X$ (you only used that it is compact)? $\endgroup$ Commented Nov 25, 2023 at 0:09
  • $\begingroup$ @Moishe Kohan: The normal cone has interior because it includes the normals to the spheres passing through $x$. This in turn yields that the tangent cone has nonempty interior because the tangent and normal cones are dual to each other. Finiteness of the set $X$ ensures that the points in the interior of $B'$ have finite distance from the boundary. Hence the center of $B'$ can be perturbed. $\endgroup$ Commented Nov 25, 2023 at 1:37
  • $\begingroup$ I see now, thank you for the clarification. $\endgroup$ Commented Nov 25, 2023 at 2:06

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