# convexity and center of mass in riemannian manifolds

every riemannian manifold M is locally convex. Let $U$ be an open convex subset of $M$. Let $x_0, \ldots, x_n$ be points in $U$. Consider the map $\sigma \: \Delta \to U$ (where $\Delta$ is the standard $n$-simplex, i.e. the convex hull of the vectors $0 = e_0, e_1, \ldots, e_n$ in $\mathbb{R}^n$) such that $\sigma(e_i) = x_i$ for $0 \le i \le n$, extended over the whole $\Delta$ using "barycentric coordinates" (see "center of mass" on the reference below):

$$\sigma(\lambda_1 e_1 + \cdots + \lambda_n e_n) = \sum_{i=0}^n \lambda_i x_i$$ (where $\sum_i \lambda_i \le 1$, $0 \le \lambda_i$).

Let $\tau \: I \to \Delta$ a constant speed segment. Is $\sigma \circ \tau$ a (constant speed) geodesic? In other words: is the simplex $\sigma$ geodesic?

See more at page 245 of berger: a panoramic view on riemannian geometry.

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 I believe (hope) that this is true because it seems intuitively true, it would be nice, and there is a passage on Berger's book where, after he defines (what I call) the barycentric coordinates, saying: " the resulting points for all possible weights fill out the \emph{convex closure} of the set {x_i}"

This is already false for surfaces. As an example, take any connected surface $(M,g)$ on nonconstant curvature. If your conjecture were true then every point $p\in M$ would admit a neighborhood $U$ and a geodesic mapping $f: U\to R^2$, i.e., a diffeomorphism to its image sending geodesics to geodesics. It was proven in Beltrami in 19th century that existence of such mapping forces $g$ to have constant curvature on $U$. See Theorem 95.1 in "Differential Geometry", by Erwin Kreyszig.

• interesting! A little bit disappointing, too!! May 1, 2014 at 20:49
• For surfaces we do at least have that the image is convex; it is the triangle bounded by the minimal geodesics between the vertices. In higher dimensions, the image-simplices are not even convex in general, but the "edges" are still minimal geodesics. Nov 30, 2017 at 8:46

This should be a comment, but I do not have the reputation to do so.

I do not know about the construction of "barycentric coordinates" you are talking about, but I guess that, whatever the construction, the answer should be no. Indeed, correct me if I am wrong, but I think that if every geodesic in $\Delta$ mapped to a geodesic in $M$, then the two spaces would be locally isometric; this would mean that every Riemannian manifold is locally isometric to the euclidean space, i.e., flat.

I think that the most you can do is to take a normal neighborhood of a point, and use geodesic coordinates. In that case you have something more like a starred set (say, a ball) than a simplex $\Delta$, and you have that every segment passing through the origin maps to a geodesic (i.e. you can force geodesic along at most one direction at a time, to respect the curvature condition).

• mmmhhh.. I think you haven't read the question carefully enough! :) Apr 30, 2014 at 15:22
• That is probably true, but I don't see why you say so :-) if $\sigma$ sends geodesics to geodesics, for every $a,b \in \Delta$, $d(a,b) = d(\sigma(a),\sigma(b))$, that is, $\sigma$ is an isometric embedding of $\Delta$ (with its flat metric) into $M$, isn't it?
– Spi
Apr 30, 2014 at 15:31
• I don't think that σ is an isometry! Even if $d(x_i, x_j) = d(e_i, e_j)$ for every $i$, $j$. Note that a (constant speed) geodesic may not be a unit speed geodesic. Apr 30, 2014 at 15:58
• Ok I'm sorry, you're definitely right, my mind had substituted the "constant speed" with "unit speed" without me noticing it ;-) sorry for the waste of time!
– Spi
Apr 30, 2014 at 17:58