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this is related to this question but is simpler, and hopefully is well-known. There are a number of references that say that the convex hull of a collection of points in a CAT(0) space need not be closed. I was wondering if anyone was aware of an explicit example ?

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There are such examples already in Riemannian world! In fact in any generic Riemannian manifold of dimension $\ge3$ convex hull of 3 points in general position is not closed. BUT it is hard to make explicit and generic at the same time :)

If it is closed then there are a lot of geodesics lying in its boundary --- that is rare! To see it do the following exercise first: Show that in generic 3-dimensional manifold, arbitrary smooth convex surface contains no geodesic. (Here geodesic = geodesic in ambient space.)

To make word "generic" more clear: show that any metric admits $C^\infty$-perturbation such that above property holds.

Semisolution: Assume that a geodesic $\gamma$ lies in the boundary of a convex set $K$ with smooth boundary. Let $N(t)$ be the outer normal vector to $K$ at $\gamma(t)$. Note that $N(t)$ is parallel. Further note that from convexity of $K$ we get that for any Jacoby field $J(t)$ such that $$\langle N(t_0),J(t_0)\rangle\le 0\ \text{and}\ \langle N(t_1),J(t_1)\rangle\le 0,$$ we have $$\langle N(t),J(t)\rangle\le 0\ \text{if}\ t_0<t<t_1.$$ Note that this condition does not hold if the curvature tensor on $\gamma$ is generic.

P.S. Roughly it means that convex hulls in Riemannian world are too complicated. But I know one example where it is used, see Kleiner's An isoperimetric comparison theorem. But he is only using that Gauss curvature of non-extremal points on the boundary of convex hulls is zero...

Appendix. (A construction of convex hull.) To construct convex hull you can do the following: start with some set $K_0$ and construct a sequence of sets $K_n$ so that $K_{n+1}$ is a union of all geodesics with ends in $K_n$. The union $W$ of all $K_n$ is convex hull. Now assume it coincides with its closure $\bar w$. In particular if $x\in\partial\bar W$ then $x\in K_n$ for some $n$. I.e. there is a geodesic in $\bar W$ passing through $x$ (if $x\not\in K_0$). From convexity, it is clear that such geodesic lies in $\partial \bar W$...

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this is very helpful. However, there are some things that I'm not entirely clear about. * I'm not sure why in your statement about generic 3D manifolds, the geodesic has to be in the ambient space ? * I don't see why this implies the result (I'm not doubting: merely not following) –  Suresh Venkat Dec 12 '09 at 6:34
    
@Anton, I don't get it. In any manifold, take a point $p$ and a nearby small ball $B$, and connect $p$ to all points of $B$ by geodesic segments. I believe the resulting set is convex if everything is small enough, yet the conical part of the boundary is filled by geodesics. –  Sergei Ivanov Dec 13 '13 at 17:49
    
@SergeiIvanov, I extended the hint; now it is called "semisolution". I hope it helps. –  Anton Petrunin Dec 14 '13 at 2:48
    
Thanks, now I understand. Indeed the Euclidean intuition fails. –  Sergei Ivanov Dec 14 '13 at 20:18

There is a natural example if you don't ask the collection of point to be finite (but where it is closed): take the subset $A_1$ of the real Hilbert space $X=L^2(\mathbb{R})$ consisting of function taking at most one value beside $0$. Then it is easily seen that the geodesic segments between pairs of points in $A_1$ cover the set $A_3$ of function taking at most $3$ values beside $0$. The convex hull of $A_1$ is then the set of functions taking a finite number of values, and is dense in $X$.

For the story, this is a variation on an example I came across in optimal transportation: $X$ was the Wasserstein space of the real line with quadratic cost, which is isometric to the subset of $L^2([0,1])$ consisting of non-decreasing functions.

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