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A unit n-sphere is defined as $$\mathcal{S}^n = \{\mathbf{p} \in \mathbb{R}^{n+1}: \|\mathbf{p}\| = 1\}$$

The distance between two points $\mathbf{p}$, $\mathbf{q}$ on $\mathcal{S}^n$ is the great-circle distance: $$\rho(\mathbf{p},\mathbf{q}) = \arccos(\mathbf{p}^T \mathbf{q})$$ where $\arccos(\cdot): [-1,1] \to [0,\pi]$ is the inverse cosine function.

Given a set of points on $\mathcal{S}^n$, I would like to compute the mean (centroid) of these points. Since the surface of a unit n-sphere is not a Euclidean space, I guess we cannot use arithmetic mean in this case.

Question: how do we compute this mean point $\boldsymbol{\mu} \in \mathcal{S}^n$, taking into account the spherical geometry?

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  • $\begingroup$ What is the centroid in your setting? $\endgroup$
    – Igor Rivin
    Commented Feb 18, 2016 at 13:13
  • $\begingroup$ The mean/average of a given set of points on the sphere $\endgroup$
    – nino
    Commented Feb 18, 2016 at 13:26
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    $\begingroup$ In $R^n$ the mean is the value of $\mu$ that minimizes $\sum (x_i-\mu)^2.$ For your case the minimizer computed similarly from $\rho$ won't necessarily be unique, nor on the sphere, I think. $\endgroup$
    – kodlu
    Commented Feb 18, 2016 at 13:55
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    $\begingroup$ @kodlu I don't understand the "nor": if you allow a minimiser outside the sphere, then you can pick a unique one. $\endgroup$ Commented Feb 18, 2016 at 16:39
  • $\begingroup$ you're right, of course... $\endgroup$
    – kodlu
    Commented Feb 19, 2016 at 7:20

3 Answers 3

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The mean on a Riemannian manfold is called Karcher-mean (or Frechet mean on metric spaces). It minimizes the sum of the squares of geodesic distances to the data.

It is no longer unique, nor does it depend continuously on the data: it may jump. But if the data points are near to each other, then it is unique. There is also the notion of a sticky mean which stays constant under each small deformation of the data (but not on the sphere).

  • H. Karcher, Riemannian center of mass and mollifier smoothing, Communications on Pure and Applied Mathematics, vol xxx (1977), 509-541

  • Means in complete manifolds: uniqueness and approximation. Marc Arnaudon, Laurent Miclo. arXiv

  • Medians and means in Riemannian geometry: existence, uniqueness and computation. Marc Arnaudon, Frédéric Barbaresco, Le Yang. arXiv

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    $\begingroup$ From the description you give, the Karcher mean is continuous, at least in a weak sense: if a sequence of sets of $k$ elements converges to another set of $k$ elements, and if there exists a sequence of Karcher means that converges to a limit point, then that limit point is also a Karcher mean of the limit set. $\endgroup$ Commented Feb 18, 2016 at 16:38
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There cannot be a meaningful definition of a unique centroid for all sets of points on $S^n\subset\mathbb R^{n+1}$ that is invariant under isometries. To see this, consider the corners $p_0, \dots, p_{n+1}\in S^n$ of a regular $(n+2)$-simplex. There are isometries of $S^n$ that permute the corners in any possible way, and these form a representation of the symmetric group $\Sigma_{n+2}$. If $p\in S^n$ was the centroid and $\gamma\in\Sigma_{n+2}$ one of the isometries that fix $\{p_0,\dots,p_{n+1}\}$ as a set, then $\gamma(p)$ would also be a legitimate centroid. But the action of $S_{n+2}$ has no common fixpoint.

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    $\begingroup$ I got slightly confused by $S^n$ denoting the unit sphere by $S_n$ denoting the symmetric group… Oh, by the way: The $\gamma(x)$ should be $\gamma(p)$, I guess? In view of Peter Michor's answer you may rephrase "There cannot be a meaningful unique centroid…". $\endgroup$
    – Dirk
    Commented Feb 18, 2016 at 14:57
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    $\begingroup$ Wouldn't it be easier to just point out the example of a point and its antipode? $\endgroup$ Commented Feb 23, 2016 at 0:09
  • $\begingroup$ @StevenStadnicki Of course. Any example with a symmetry without fixpoints would do the job. $\endgroup$ Commented Feb 23, 2016 at 10:49
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I think what you're looking for is the field of statistics known as directional statistics. Even for the circle $S^1$ it's not obvious how things should be defined, but it is possible, depending on context.

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