In reading Greg Kuperberg's partial answer to this question Convex hull in CAT(0) , I started wondering if the center of gravity is always contained in the closed convex hull.

More precisely, given $n+1$ points $x_0,\dots,x_n$ in a CAT(0) space $X$, there is a center-of-gravity map $c\colon\Delta_n \to X$ (where $\Delta_n$ is the standard $n$-simplex in $\mathbb{R}^{n+1}$), defined by $$ c(a_0,a_1,\dots,a_n) = \text{the point $x \in X$ minimizing }\sum_{i=0}^n a_i d(x,x_i)^2. $$ (In a CAT(0) space, the function being minimized is strictly convex, so the center of gravity is unique.)

Is $c(a_0,a_1,\dots,a_n)$ always contained in the closed convex hull of $x_0,\dots,x_n$?

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    $\begingroup$ Just an idea: Assume that $x$ is not in the convex hull and let $x'=Px$ be the projection of $x$ onto the convex hull. Then we can try to show that $d(x',x_i)<d(x,x_i)$ for all $i$ to get a contradiction. Looking at the geodesic triangle $x_i,x,x'$ we can maybe prove that $\angle x_ix'x>90^\circ>\angle x'xx_i$ and follow that $d(x',x_i)<d(x,x_i)$. However I am not that familiar with CAT(0) spaces and can not rigorously prove this. $\endgroup$
    – user35593
    May 7, 2015 at 12:36
  • $\begingroup$ This is exactly right, thank you! That is Proposition II.2.4 of Bridson and Häfliger. (The notion of angle has to be suitably defined, of course.) $\endgroup$ May 7, 2015 at 13:09
  • $\begingroup$ Now that I think a little more, are we guaranteed the existence of a center of gravity when $X$ is not locally compact? $\endgroup$ May 7, 2015 at 13:54
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    $\begingroup$ @DylanThurston: I think that compactness arguments can be replaced by completeness in this kind of situation. I would try to use the argument for uniqueness to show that a minimizing sequence of points is Cauchy, and then concludes under a mere completeness assumption. It would need a little bit of checking though. $\endgroup$ May 7, 2015 at 14:25
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    $\begingroup$ @DylanThurston: my suggestion does need some more argument, but it seems to me that the CAT(0) condition should precisely imply that the objective function is as least as convex as in Euclidean space. $\endgroup$ May 7, 2015 at 19:00

1 Answer 1


This is expanding on @user35593 's comment above.

Let $\bar{C}$ be the closure of the convex hull of $x_0,\dots,x_n$, and let $x'$ be the projection of $x$ onto $\bar{C}$. By Bridson and Haefliger, Proposition II.2.4 (which is easy), projection onto the convex hull is a retraction that does not increase distances. In particular, $d(x',x_i) \le d(x,x_i)$. Since $x$ was the unique minimizer of a positive combination of the distances, we must have $x=x'$.


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