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Is there a way to find a function $F:\mathbb S^2 \rightarrow \mathbb R^3$ of class $C^1$, minimizing $$\int_{\mathbb S^2\times\mathbb S^2}(d(F(x),F(y))−\delta(x,y))^2 dx dy$$ , where $d$ stands for the euclidean distance in $\mathbb R^3$ and $\delta$ the geodesic distance on the sphere $\mathbb S^2$?
Or $d$ could stand for the squared euclidean distance, and $\delta$ the square geodesic distance, if this makes the problem simpler. The goal is thus to approximate geodesic distances by euclidean distances of transformed points.

I tried to perform a Multi-Dimensional Scaling to get this least square solution for a finite set of point, but it seems that the solution was just the identity (or a uniform scaling)... is that right?

Thanks!

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Also asked on MO: "Isometric embedding of a sphere," mathoverflow.net/questions/67139 . –  Joseph O'Rourke Jun 8 '11 at 13:39
    
yep indeed. I just realized stackexchange was more active that mathoverflow which is why I wanted to give a try here. –  WhitAngl Jun 8 '11 at 13:47
    
I posted a suggestion on MO, not a true answer, but ... –  Joseph O'Rourke Jun 8 '11 at 22:44
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The overwhelming symmetry of the problem certainly suggests that the solution ought to be a sphere. It would be slightly a enlarged one, because the Euclidean distance underestimates geodesic distances between almost all pairs of points. –  Rahul Aug 7 '11 at 19:36
    
The radius of the sphere mentioned in Rahul Narain's comment would be $(6\pi-8)/9\doteq 1.20551$. The square root of the mean quadratic deviation would be $0.168261$, resulting mainly from pairs $(x,y)$ with large $\delta(x,y)$. –  Christian Blatter Jul 15 '12 at 9:42
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1 Answer

Hints in order for you to solve this problem:

(1) Let $x, y \in S^2$. The segment from $x$ to $y$ that minimizes the distance on $S^2$ must be contained on a great circle.

(2) Look at the plane that contains $x, y$ and the origin of $\mathbb{R}^3$. There the problem is reduced in dimension (as Anton Petrunin pointed out in MO).

(3) Draw the circle and mark $x$ and $y$. The euclidean distance is the length of the straight segment connecting then. Use the law of cosines to relate this distance to the length of the circle arc connecting $x$ to $y$.

I realize this doesn't give the function you're asking for, but this gives you a formula for $d(x,y)$ in terms of $\delta(x,y)$. Is that the point? If it is, you should also check this. Otherwise, I have to think a bit more about it.

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I think I don't understand : why would the euclidean distance between 2 points be a multiple of the geodesic distance ? This would mean that $\|u-v\|=\alpha.acos(u.v)$... which is false. –  WhitAngl Jun 8 '11 at 15:58
    
I caught a naive mistake. I'm editing it now. –  user8126 Jun 8 '11 at 16:34
    
It's not really what I am looking for : I have some euclidean geometry pipeline set up, and now I'd like to work on the sphere with minimal changes in the pipeline... that why I was looking for an (almost-)isometry, to transform my points on the sphere to points in the euclidean space. –  WhitAngl Jun 8 '11 at 17:53
    
I think I'm starting to get it now (despite never having heard the term "geometry pipeline" before). It seems to me that the answer would be $F(x,y,z) = (x,y, 0)$, but I don't know how to prove it right now. –  user8126 Jun 8 '11 at 19:49
    
by pipeline, I mean some geometry algorithms/implementations. I don't necessarily need the points in 2D as in your suggestion : it could be a transformation in 3D (if it is better like that). I guess increasing the dimension also increases the precision of the least square estimation. –  WhitAngl Jun 8 '11 at 20:02
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