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Discussion indicates the convex hull of a finite set can fail to be closed in a complete Hadamard space. Hence the following question should have a negative answer:

Suppose K is a compact convex set in the complete Hadamard (or $\operatorname{CAT}(0)$) space $X$, and suppose $K$ is the closed convex hull of the finite point set $\{x_1,x_2,\dots,x_N\}$. Suppose $x$ is in $X$ and $C$ is the cone over $K$ with respect to $x$, i.e. $C$ is the union of all the geodesics connecting $x$ to $K$. Must $C$ be convex?

If not, which value of $N$ yields a minimal counterexample? Is it $N=3$? Can someone describe a simple counterexample?

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I learned the following example from Stéphane Félix:

Already $N=2$ is enough (i.e., the cone over a geodesic need not be convex).

Remove one quadrant from $\mathbb{R}^3$, say the part where $x$- and $y$-coordinates are both negative, and equip the resulting space $X$ with the induced length metric. You can check that this is a $\operatorname{CAT}(0)$–space, for example by Reshetnyak's theorem on gluing along complete convex subsets, Theorem 11.1 on page 347 of Bridson–Haefliger: $X$ can also be described by gluing the half-spaces $\{x \geq 0\}$ and $\{y\geq 0\}$ in $\mathbb{R}^3$ along the quadrant $\{x,y \geq 0\}$).

Draw a geodesic from $x_1 = (-1,0,0)$ to $x_2 = (0,-1,1)$ which will consist of two line segments connecting $x_1$ and $x_2$ to the point $p = (0,0,1/2)$ on the $z$-axis. Now let $x$ be a point in the quadrant $\{x,y \gt 0\}$ from which the geodesic $[x_1,x_2]$ appears broken, so $x$ should be a point outside the Euclidean plane spanned by $x_1,x_2$ and $p$. Since the triangles $\Delta(x_1,p,x)$ and $\Delta(x_2,p,x)$ lie entirely in the isometrically embedded half-spaces $\{y \geq 0\}$ and $\{x \geq 0\}$, they are the cones of $[x_i,p]$ with respect to $x$ and their union is the cone $C$ of $[x_1,x_2]$ over $x$. Since the triangles do not lie in the same plane, the cone $C$ is not convex.

Stéphane Félix's counterexample

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Fantastic! Thanks t.b! – Paul Fabel Jun 25 '12 at 23:20
Glad you liked it. Note that this also yields an example of three points whose closed convex hull is $3$-dimensional. Similar things happen for example in complex hyperbolic space. The argument I have in mind involves non-constant sectional curvature and is a bit more cumbersome to flesh out in detail. – t.b. Jun 25 '12 at 23:32
Thanks! Indeed it is disconcerting that `convex hull' can raise dimension in this manner. The mathoverflow discussion suggests it is still a mystery to understand in general the convex hull of a compact set in a complete Hadamard space. – Paul Fabel Jun 25 '12 at 23:50

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